mcquiston heating ventilating air conditioning 6th solutions

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SOLUTION MANUAL

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Chapter 1

1-1 (a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K)

(b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K

(c) 0.04 Ibm/(ft-hr)3600 sec/hr

x1.488 = 16.5 2Ns

mµ

(d) 1050 BtuIbm

x 41

9.48x10− JBtu

x 2.20462 Ibmkg

= 2.44 MJkg

(e) 12,000 BtuIbm

x 13.412

= 3.52 kW

(f) 14.7 2Ibfin

x 6894.76 = 101 kPa

1-2 (a) 120 kPa x 2lbf / in

6.89476kPa = 17.4 lbf/in2

(b) 100 Wm K−

x 0.5778 = 57.8 Btu/hr-ft-F

(c) 0.8 2W

m K− x 0.1761 = 0.14 Btu/hr-ft2-F

(d) 10-6 N-s/m2 x 11.488

= 6.7 x 10-7 lbmft sec−

(e) 1200 kW x 3412 = 4.1 x 10-6 Btu/hr

2

(f) 1000 kJkg

x 1 Btu1.055 kJ

x 1 kg2.2046 lbm

= 430 Btulbm

1-3 Hp = 50 (ft) x 0.3048 ( mft

) = 15.2 m

∆P = 15.2 m1000 Pa/kPa

x 9.8071

( Nkg

) x 1000 (kg/m3) = 149 kPa

1-4 P = ∆4

12 (ft) x 0.3048 ( m

ft) x 9.807

1 ( N

kg) x 1000 ( 3

kgm

)

∆P = 996 Pa 1.0 kPa ≈ 1-5 TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE

+ METER CHARGE

( ) ( ) ( ) ( )96,000 kw - hrs 0.045 $ /kw hr + 624 kw 11 50 $ /kw− − + $68 = $4,320 + $7,176 + $68 = $11,564

1-6 7 AM to 6 PM 11 hrs/day, 5 days/wk

hrs days(11) (22) 242 hrs /monthday months

=

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

3

( )

( )( )

624 kwratio = 1.57

96,000 kw hr242 hr

=⎛ ⎞−⎜ ⎟⎝ ⎠

1-7 This is a trial and error solution since eq. 1-1 cannot be solved

explicitly for i.

Answer converges at just over 4.2% using eq. 1-1

1-8 Determine present worth of savings using eq. 1-1

( )( )( )12 120.012$1000 1- 1+

12P =

0.01212

P $134,000

−⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠

=

1-9 (a) Q VA= = 2 x 3.08 x 10-3 = 6.16 x 10-3m3/s m 6.16 x 10Q ρ= = -3 x 998 = 6.15 kg/s

(b) A= 4π (0.3)2 = 7.07 x 10-2 m2

Q 7.07x10= -2 x 4 = 0.283 m3 / s; ρ = 1.255 kq/m3

m = 1.225 x 0.283 = 0.347 kg/s

1-10 V = 3x10x20 = 600m3


4

= 600 x iQ 14

x 13600

= 4.17 x 10-2 m3/s

1-11

p p

3

q = mc T c = 4.183 kJ/(kg-K)

= 983.2 kg/m ρ

∆

1-11 (cont’d)

( ) ( ) ( ) ( )

3 c3

m kg kJq = 1 983.2 4.183 5 20,564s kg Km

q = 20,564 kw

=−

kJs

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted

1-12 = watq airq−

11,200(1)(10) =

= 25000x60x14.7x144x0.24(t 50)(53.35x510)

−

11,200 = 5601.5 (t2-50); t2 = (11,200/5601.5) + 50 = 70 F 1-13 Diagram as in 1-12 above. q wat = -q air

1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000) 6279(90-t2) = 29,400

by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Excerpts from this work may be reproduced by instructors for distribution on a not- l purposes only to students enrolled in courses for which the textbook has been adopted. this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permissi

5

t2 = 90 - 29,4006279

= 85.3 C

1-14 q hA(t= s- ) t∞ A= π (1/12) x 10 = 2.618 ft2

st = tsur 212 F ≈ = 10x2.618x(212-50) = 4241 Btu/hr q

1-15 A= π x 0.25x4 = 3.14 16 m2

hA(tq = s- ) t∞

h= s

qA(t -t )∞

= 12503.1416(100 10)−

; h = 4.42 W/(m2 – C)

1-16 (tpq mc= 2-t1) ; m Q x ρ= ρ = P/RT = 14.7x144/53.35(76+460) ρ = 0.074 lbm/ft3

m = 5000x0.074x60 = 22,208 lbm/hr = 0.24 Btu/lbm-F pc q= 22,208x0.24(58-76) = -95,939 Btu/hr Negative sign indicates cooling

1-17 (t1 pm c 3-t1) +

for-profit basis for testing or instructionaAny other reproduction or translation of

on of the copyright owner is unlawful.

6

(t2 p2m c 3-t2) = 0 = p1c p2c

t3 = 1 1 2 2

1 2

(m t m t )(m m )

++

1 2m Q 1ρ= = 1000x 14.7x14453.35(460 50)+

= 73.5 lbm/min

1-17 (cont’d)

2 2m Q 2ρ= = 600x 14.7x14453.35(460 50)+

= 46.7 lbm/min

3(73.5x80) (46.7 x 50)t 68.3 F

(73.5 46.7)+

= =+


7


Chapter 2

2-1 through 2-20 Solutions are not furnished since many acceptable responses exist

for each problem. It is not expected that the beginning student can handle

these questions easily. The objective is to make the student think about

the complete design problem and the various functions of the system.

These problems are also intended for use in class discussions to enlarge

the text material.



Chapter 3 3-1 (a) Pv = =srPφ 0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psia Pa = 101 – 1.43 = 99.57 kPa or 14.696-0.196 = 14.5 psia

(b) =v

v

Pρ RvT or = = = 3v

v vv

P 1430ρ ; ρ 0.0104 kg/mR T 462.5(297)

or =0.196(144) 0.0006285.78(535)

lbv/ft3

(c) W = 0.6219 (1.43)

(99.57) = 0.00893 kgv/kga

or 0.6219(0.196) 0.00854 lbv/lba

14.5=

3-2 (a) English Units – t = 80F; P = 14.696 psia; Pv = 0.507 psia Table A-1a

W = 0.6219 a

vPP

= 0.6219 (0.507)(14.696 0.507)−

= 0.0222 lbv/lba

i = 0.24t + W(1062.2 + 0.444t) i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm

8

v = a

a

R T 53.35(460 80)P (14.696 0.507)144

+=

− = 13.61 ft3/lbm

(b) English Units – 32F, 14.696 psia Pv = 0.089 psia (Table A-1) 3-2 (cont’d)

W = 0.6219(0.089) lbmv 0.00379

(14.696 0.089) lbma=

−

i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbma

v = 53.35(492)

(14.696 0.089)144− = 12.48 ft3/lbma

3-2 (a) SI Units – 27C; 101.325 kPa Pv = 3.60 kPa, Table A-1b

W = 0.6219 v

a

P 0.6219(3.6) kgv0.0229P (101.325 3.6) kga

= =−

i = 1.0t + W(2501.3 + 1.86t) kJ/kga i = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kga

v = 3a

a

R T 0.287(300) = =0.88 m /kgaP (101.325 - 3.6)

(b) SI Units 0.0C; 101.325 kPa Pv = 0.61 kPa, Table A-1b

W = 0.6219(0.61) =0.00377 kgv/kga

(101.325 - 0.61)


9

i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga

v = 30.287(273) 0.778 m /kga(101.325 - 0.61)

=

3-3 (a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hg t = 80 F; Pv = 0.507 psia (Table A-1a)

W = 0.6219 v

a

P 0.6219(0.507) = P (12.24 - 0.507) = 0.0269 lbv/lba

i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma

v = a

a

R T 53.35(540) = P (12.24 - 0.507) 144

= 17.05 ft3 / lbma

(b) English Units – t = 32 F, Pv = 0.089 psia ( Table A-1a)

W = 0.6219(0.089)

(12.24 0.089) − = 0.00456 lbmv/lbma

i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma

v = 53.35(492)

(12.24 0.089)144− = 15.00 ft3/lbma

3-3 (a) SI Units -27 C, 1500 m elevation P = 99.436 + 1500(-0.01) = 84.436 kPa Pv = 3.60 kPa, Table A-1b


10

W = 0.6219x3.60 0.0277 kgv/kga(84.436 3.60)

=−

i = 27 + 0.0277 (2501.3 + 1.86 x 27) = 97.68 kJ/kga 3-3 (cont’d)

v = 30.287x300 1.065 m / kga(84.436 - 3.60)

=

(b) SI Units – 0.0C; 1500m or 84.436 kPa

Pv = 0.61 kPa; Table A-1b

W = 0.6219 x 0.61 0.00453 kgv / kga

(84.436 - 0.61)=

i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kga

v = 0.287 x 273

(84.436 - 0.61) = 0.935 m3 / kga

3-4 (a) English Units – 70F, Pv = 0.363 psia Pv = φ Pg = 0.75(0.363) = 0.272 psia

W = 0.6219 (0.272) 0.0117 lbmv / lbma

(14.696 - 0.272) =

i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)] 29.58 Btu / lbma= (b) Pv = 0.75 (0.363) = 0.272 psia; P = 12.24 psia


11

W = 0.6219 (0.272)(12.24 - 0.272)

= 0.0141 lbmv / lbma

i = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)] 32.20 Btu/ lbma= 3-4 SI Units – (a) 20C, 75% RH, Sea Level 3-4 (cont’d) Ps = 2.34 kPa; Pv = 0.75 x 2.34 = 1.755 kPa

0.6219 x 1.755W = =(101.325 - 1.755)

0.0110 kgv / kga

i = 1.0 t + W(2501.3 + 1.86t) i = 20 + 0.0110(2501.3 + 1.86 x 20) = 47.92 kJ / kga (b) 20C, 75% RH, 1525m P = 99.436 – 0.01 x 1525 = 84.186 kPa Ps = 2.34 KPa; Pv = 0.75 x 2.34 = 1.755 kPa

W = 0.6219 x 1.755(84.186 - 1.755)

= 0.0132 kgv / kga

i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga 3-5 English Units – t = 72 Fdb; psia 14.696 P %; 50 ==φ

svs

v P P or PP φφ == ; Pv = 0.5(0.3918) = 0. 196 psia


12

Air dewpoint = saturated temp. at 0.196 psia = 52.6 F Moisture will condense because the glass temp. 40 F is below the dew point temp. 3-5 SI Units – t = 22C ; 50% ; P = 100 kPa Pv = φ Ps ; Pv = 0.5(2.34) = 1.17 kPa 3-5 (cont’d)

Air dewpoint = sat.temp. at 1.17 kPa = 9.17 C Glass temp. of 4 C is below the dewpoint of 9.17 C, therefore, moisture will ccondense on the glass

3-6 English Units - (a) At 55F, 80% RH, va = 13.12 ft3 / lba and ρ a = 0.0752 lbma / ft3

= 22,860 lbma / hr am 5000 (0.0762) 381 lbma / min= = (b) Using PSYCH ρ a = 0.0610 lbma / ft3 or va = 16.4 ft3 / lba = 5000 (0.061) = 305 lbma / min am 18,300 lbma / hr= 3-6 SI Units – (a) t = 13 C and relative humidity 80% then va 0.820 m≈ 3 / kga; am 2.36 / 0.82 2.88 kga / s= = (b) Assuming same conditions ; 3

av 0.985 m / kga= am 2.36 / 0.985 2.40 kga / s= =


13

3-7 English Units – t = 80F, 60% RH (a) v sP P 0.6 (0.507) 0.304 psiaφ= = = = 64.5 F dp sat vt (t @ P= ) (b) Same as (a) above 3-7 SI Units – (a) 27 C, 60% RH, Sea Level Ps = 3.57 kPa; Pv = 0.6 x 3.57 = 2.14 kPa dp sat vt =(t at P ) 18.4 C≈

(b) Same as (a) above 3-8 dpt 9C (48F)≤

42%φ ≤ ; W 0.0071 kgv / kga (lbv / lba)≤

Chart 1a & 1b


14

10 15 20 25

30

35

40

45

50

55

55

60

60

ENTHALPY - BTU PER POUND OF DRY AIR

15

20

25

30

35

40

45

50ENTHALPY -

BTU PER POUND O

F DRY A

IR

SATURATION TEMPERATURE - °

F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028

10% RELATIVE HUMIDITY

20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80

85 WET BULB TEMPERATURE - °F

85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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S M

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DR

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dp Room

Problem 3-8

W=0.0071

72 (22)48 (9)

42 %

R R

ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE

BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL

0

1.0 1 .0

2.04.08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt

ENTHALPYHUMIDITY RATIO

hW

3-9 (a,b,d) Using the Properties option of PSYCH: Relative Humidity = 0.59 or 59%

Enthalpy = 30.4 Btu/lbma

Humidity Ratio = 0.0114 lbu/lba

(c) Again using the Properties option

At W=0.0114 lbv/lba; RH = 1.00 or 100%

The dew point = tdb or twb = 59.9 F Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

15

3-9 (cont’d)

(e) Using the Density of Dry Air option:

Mass Density = 0.070 lba/ft3 3-10 Using program PSYCH (a) tdb = 102.6; twb = 81.1F 75 Fdb; 65 fwb; 14.2 psia (b) m 58.7ν = lbm/hr

Q 2 = 1027 cfm 3-11 t1 = 80 / 67 F; t2 = 55 F and sat.; assume std. barometer (a) W1 – W2 = 0.0112 – 0.0092 = 0.002 lbv / lba (b) lq 31.5 - 29.3 2.2 Btu / lba= = (c) qs = 29.3 – 23.2 = 6.1 Btu / lba (d) q = l sq q 8.3 Btu / lba+ =

3-12 (a) *2

0.6219 (0.3095)W 0.0134 kgv / kga(14.696 0.3095)

= =−

10.24 (65 - 80) ( 0.0134 x 1056.5)W 0.00993 lbv / lba

(1096 - 33)+

= =

also W1 = 0.6219 Pv1 / (P – Pv1)


16

Pv1 = (0.00993 x 14.696) / ( 0.6219 + 0.00993) = 0.231 psia 3-12 (cont’d)

10.231 0.46 or 46%0.507

φ = =

(b) P = 29.42 – (0.0009 x 5000) = 24.92 in.Hg. or P = 12.24 psia

*2

0.6219 x (0.3095)W 0.01613(12.24 - 0.3095)

= = lbv/lba

W1 = 0.24(65 80) (0.01613 x 1056.5) 0.01265 lbv / lba

( 1096 - 33)− +

=

or kgv / kga Pv1 = 0.01265 x 12.24 / ( 0.6219 + 0.01265) = 0.244 psia

10.244 0.48 or 48%0.507

φ = =

3-13 (a) Sea Level

Dry Bulb, F

Wet Bulb, F

Dew point

F

Humid. Ratio, lba/lbv

EnthalpyBtu/lba

Rel. Humid., %

Mass Density lba/ft3

85 60 40.6 0.0053 26.6 21 0.072 75 59.6 49.2 0.0074 26.1 40 0.073

74.6 65.1 60.1 0.0111 30 60 0.073 88.6 70 60.9 0.01143 33.8 40 0.071 100 85.8 81.7 0.0235 50 56 0.068


17

(a) 5000 ft.

Dry Bulb, F

Wet Bulb, F

Dew point

F

Humid. Ratio, lba/lbv

EnthalpyBtu/lba

Rel. Humid., %

Mass Density lba/ft3

85 60 45.1 0.0076 28.7 25 0.060 75 58.6 49.2 0.0089 27.7 40 0.061

71.2 61.6 56.7 0.0118 30 60 0.061 102.7 70 55.8 0.01143 37.3 22 0.058 100 81.3 76.1 0.0235 50 47 0.057

(c) Note effect of barometric pressure. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

18

3-14

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50

ENTHALPY - BTU PER P

OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028

10% RELATIVE H UMIDITY

20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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dp Room

Problem 3-14

72 (22)52 (11)

Max RH=49.6 %W=0.0083

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0.5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-15 Use Chart 1b, SI (a) td = 10 C; SHF = 0.62

(b) 1 22.4q m (i i ) (57.1 - 34)

0.867= − = = 63.95 kJ / s = 63.95 k W

sq 63.95 (0.62) 39.65 kW= = 3-15 Use Chart 1a, IP (a) td = 52 F; SHF = 0.63 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

19

3-15 (cont’d)

(b) 5000(60)q = (32 - 22.6)= 203,317. Btu/hr

13.87

10 15 20 25

30

35

40

45

50

55

55

60

60

ENTHALPY - BT U PER POUND OF DRY AIR

15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

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1

ADP2

Problem 3-15

80 (27)55 (13)52 (10)

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0.5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-200 0

-1000

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt

ENTHALPYHU MIDITY RATIO

hW

sq 203,317 (0.63) 128,089. Btu/hr= =

3-16 (a) i1 = 30 Btu / lba; v1 = 13.78 ft3 / lba; W = 0.0103 lbalbv ; 50% 1 =φ

(b) i1 = 51.6 kJ / kga v1 = 0.86 m3 / kga


20

3-16 (cont’d)

W1 = kgakgv 0103.0

50% 1 =φ 3-17 Use the Heat Transfer option of program PSYCH:

q = 148,239 Btu/hr sq 102,235 Btu/hr=

SHF = 0.69

3-18 Use the Heat Transfer option of program PSYCH for sensible heat transfer only:

sq 178,911 Btu/hr= −

Negative sign indicates heating. 3-19 Use the program PSYC to compute the various properties at 85/68 F; sea level and 6000 ft elevation.

Elevation ft

Enthalpy Btu/lbm

Rel. Humpercent

Hum. Ratio lbv/lba

Density lba/ft3

0 32.2 42 0.0107 0.072 6000 36.3 45 0.0144 0.058

At sea level: am 5000 x 0.072 x 60 21,600 lba/hr= =


21

3-19 (cont’d) At 6000 feet:: am 5000 x 0.057 x 60 17,100 lba/hr= = Percent Decrease at 6000 ft:

(21,600 17,100)100PD 20.8%21,600−

= =

3-20 Use the program PSYC to compute the heat transfer rates at 1000 and 6000 feet elevation: (a) at 1000 ft, q 200,534 Btu/hr= (b) at 6000 ft, q 190,224 Btu/hr=

(c) PD = (200,534 190,224)100 5.1 %200,543−

=

3-21 (a) English Units – ; = 0 in.Hg. 29.92 PB = q

wi i 180.2 0.8 (970.2)

W∆

= = +∆

iw = 956.4 Btu / lbv From chart 1a; t2 = 91.5 F


22

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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DR

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1

2

Problem 3-21

98 (38)91.5 (32)

60 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-21 (a) SI Units – PB = 101.325 kPa

wi i 419.04 (0.8 x 2257)

W∆

= = +∆

iW = 2224.6 kJ / kg From chart 1b; t2 = 32 C (b) Use Humidification (adiabatic) option to obtain 91.5 F db. 3-22 PB = 29.92 in.Hg.; 0 q = (a) Using chart 1a Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

23

3-22 (cont’d)

wi i 1090 Btu / lbm

W∆

= =∆

From table A-1

f

fg

i-i 1090 - 196.1x = i 960.

=1

x = 0.931 or about 93 %

(b) x will be the same


10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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1

a

bProblem 3-22

80

60

1090

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-100 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

24

3-23 Assume PB = 101.325 kPa; 0 q =

w∆i 272.1 i kJ / kg∆W 1000

= =

iw = 0.272 (on scale) t2 = 22.6 C

10 20 30 40 50

60

70

80

90

100

110

110

120

120

ENTHALPY - KJ PER KILOGRAM OF DRY AIR

10

20

30

40

50

60

70

80

90

100

ENTHALPY - KJ P

ER KILOGRAM O

F DRY AIR

SATURATIO

N TEMPERATURE - °

C

5 10 15 20 25

30 35 40 45 50

DR

Y B

ULB

TE

MP

ER

ATU

RE

- °C

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

10% RELATIVE HUMID ITY

20%

30%

40%

50%

60%

70%

80%

90%

5

5

10

10

15

15

20

20

25

25

30 WET BULB TEMPERATURE - °C

30

0.78

0.80

0.82

0.84

0.86 VO

LUM

E - CU

BIC

ME

TER

PER

kg DR

Y A

IR

0.88

0.90

0.92

0.94

HU

MID

ITY

RA

TIO

- G

RAM

S M

OIS

TUR

E P

ER K

ILO

GR

AM D

RY

AIR

1

2

Problem 3-23

3822.6

20

80 %

0.272

R R


BAROMETRIC PRESSURE: 101.325 kPaCopyright 1992


SEA LEVEL

0

1.0 1 .0

1 .52.0

4 .0

-4.0-2 .0-1.0

-0 .5

-0.2

0 .1

0.2

0.3

0.4

0.5

0.60.7

0 .8-5.0

-2.0

0.0

1 .0

2.02.5

3.0

4.0

5.0

10.0

- SENSIBLE HEAT QsTOTAL HEAT Qt


hW

3-24 For adia. humidification

(a) w∆i = i 1131 Btu / lbw∆W

=


25

3-24 (cont’d) c a 2q = m (i - i )1 am 2000 x 60 / 13.14= am 9132 lba / hr= 1 2i 18.1 Btu / lba ; i 29.7 Btu / hr= = cq 9132 (29.7 - 18.1) 105,931 Btu / hr= = w a 3 2 3 2m m (W - W ) ; W = 0.0167; W 0.0032 lbv/lba= = wm 9132 (0.01 67 - 0.0032) 123.3 lbw / hr= =


26

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

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S M

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E P

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OU

ND

DR

Y A

IR

1

2

3

Problem 3-24

1131

30 %

110 (43)60 (16)

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-200 0

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

(b) Solution similar to (a) 3-25 English Units – See diagram for construction on chart 1a.

1

3

Q32 2000 2=3000 3Q12

= =

Layout 2L/3 on the chart and read: W3 = 0.007 lbv/lba Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

27

I3 = 22.2 Btu/lba

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50ENTHALPY -

BTU PER POUND O

F DRY A

IR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

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TUR

E P

ER P

OU

ND

DR

Y A

IR

1

2

3

Problem 3-25

40 (4) 100 (38)58.4 (15)

35

77

52

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-200 0

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-25 SI Units – Same procedure as above, read: 3i 34 kJ / kga= 3W 0.007 kgv / kga= 3-26 English Units – Layout the given data on Chart 1a as shown for problem 3-25. a1m 2000(60) 12.66 9,479lba hr= = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

28

3-26 (cont’d) a2m 1000(60) 14.44 4,155lba hr= =

a1

a1 a2

m32 9479= 0.695m +m 9479 415512

= =+

Layout distance 32 on line from 1 to 2 to locate point 3 for the mixture. Read: i3 = 21.5 Btu/lbm W3 = 0.0067 lbu/lba

For W, % Error = (0.007 0.0067)100 4.5

0.0067−

=

For I, % Error = (22.2 21.5)100 3.3

21.5−

=

3-27 250,000SHF 0.8200,000

= =

or SHF = 59 .8173

=


29

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

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S M

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DR

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1

2

Problem 3-27

75 (24)

50 %

53 (12)

0.8

21.5

28.2

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1000

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-28 Refer to diagram for 3-27 (a) a 1 2 1 2q = m (i - i ); i 28.2; i 21.5= = am 250,000 / (28.2 - 21.5) 37,313 lba / hr= = 3

a 2Q = m v 37,313 x 13.09 / 60 8,140 ft / min= = (b) similar procedure; 3Q 3.85 m / s= Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

3-29 (a) Use the AirQuantity option of program PSYCH, iterating on the relative humidity and setting the minimum outdoor Air Quantity to 0.01, NOT ZERO.

Use the properties option to find the entering wet bulb temperature of 62.6F. Then

φ = 0.852 (iterated) ts = 56F

= 9,360 cfm sQ (b) Proceed as above

φ = 0.882 ts = 56F

= 10,014 cfm sQ 3-30 Proceed as in 3-29 above.

φ = 0.92 ts = 56.1 56 F ≈

= 11,303 cfm sQ ≤

3-31 (a) 91.0000,550000,500SHF ==


31

3-31 (cont’d) (b) a 2 1q = m (i -i ) or a 2m = q/(i -i )1

a550,000m

(34.3 22.8)=

−

am =47,826lba hr

a 22

m v 47,826Q = = x 14.62=11,654 cfm60 60

or 5.5 m3/s


32

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

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DR

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IR

1

2

Problem 3-31

0.91

115 (46)72 (22)

30 %

22.8

34.3

R R




SEA LEVEL

0

1.0 1.0

2.04.08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-32 a 2 1q = m (i -i )

2a

qi = +im 1

a1400 x 60m 5,915.5

14.2=

2

-5 x 12,000i = +38.55,915.5

Btu/lba 2i 2 8 .3= 6 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

33

Then from Chart 1a, t2= 67F

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

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S M

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DR

Y A

IR

1

2ADP

Problem 3-32

90

75

67

28.4

55

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0.2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-33 Use Adiabatic Mixing option of PSYCH with the Properties option to enter requested data. Assume volume flow rates of 3 to 1 to obtain. Tmix,db = 84.2 F Tmix,wb = 71.3 F 3-34 Use Program PSYCH at Sea Level elevation

Iteration on the supply volume flow rate is required. This is the same as the

leaving air quantity for the coil. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

34

3-34 (cont’d)

(a) Supply air quantity is 9,384 cfm.

(b) The outdoor air quantity is 938 cfm.

(c) Air enters the coil at 74.6 F db, 60.5 F wb at a rate of 9,740 cfm

(d) The coil capacity is 248,256 Btu/hr.

The amount of air returned is: (9,740 – 939) = 8,802 cfm.

3-35 Use Program PSYCH at 5,000 ft elevation

Iteration on the supply volume flow rate is required. This is the same as the

leaving air quantity for the coil.

(a) Supply air quantity is 11,267 cfm.

(b) The outdoor air quantity is 1,127 cfm.

(c) Air enters the coil at 74.6 F db, 62.1 F wb at a rate of 11,697 cfm

(d) The coil capacity is 334,143 Btu/hr.

The amount of air returned is: (11,697 – 1,127) = 10,570 cfm. 3-36 cfm 1000Q0 = (a) From Chart 1a st =120 / 74 F

ss r

q 200,000m =(i -i ) (37.2 22.8)

=−


35

= 13,889 lb/hr = 1m 3

s s s sQ = m v = m (14.78)/60 = 3,421 ft /min (b) o o om = Q /v 1000 x 60 / 12.61 4758 lb/hr= =

r

1

m 13,889 4758 0.66;m 13,889

−= = 1From Chart 1a t 61/ 47 F=

3 1t - t (119 61)= − (c) w s s 2m = m (W -W ) 13,889 (0.0075 - 0.0036) = = 54.2 lbm/hr (d) f 1 3 1q = m (i -i ) =13,889 (32.8 18.6) 197,224 Btu/hr− =


36

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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RA

TIO

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DR

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IR

r

s

0

1

31 3

Problem 3-36

12072

30 %

40 61

47

0.8

1150

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-37 (a) st 120 / 71.4 F Use Chart 1Ha= s 1m 200,000 /(38.7 24.0) 13,605 lba/hr m= − = = sQ 13,605 x 17.85 / 60 4048 cfm= = (b) lba/hr 3947 60 x )2.15/1000(m0 ==

r1

1

m 13,605 3947 0.71; t 62.8 / 47 Fm 13,605

−= = =

3 1t -t (119.5 62.8)= − (c) w s s 1m =m (w -W ) 13,605 (0.0088 - 0.0046) 57.14 lbw/hr= =Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

37

(d) fq 13,605 (33.8 - 20.2) 185,028 Btu/hr= = 3-38 Assume fan power and heat gain are load on the space

s9384m x 60 = 42,915 lbm/hr; Prob 3-3413.12

=

fan duct s s cW q m (i i+ = − ) = (4 x 2545) + 1000 = 11,180 Btu / hr

c11,180i 20.8 20.54 Btu/lbm42,915

= − =

State c is required condition leaving coil Part a, b, and c are same as prob. 3-34; (d) coil 1 1 cq =m (i -i ) 42,915 (26.8 - 20.54) 268,648 Btu/hr= =


38

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

s

r

0

1

c

1

Problem 3-38

1007255

20.54

50 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0.5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-39 )ii(mW );ii(mq cssfansrsr −=−=

(a) c ri 28 Btu/lbm; i 33.7 Btu/lbm= = Using Chart 1Ha rq 1,320,000 Btu/hr= fanW 30 x 2545 76350 Btu/hr= = fan a s cW 30 x 2545 76,350 = m (i -i )= =


39

s aq = 1,320,000 = m r s(i -i )

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50ENTH

ALPY - B

TU P

ER P

OUND OF

DRY AIR

SATU

RATION TE

MPERATURE - °

F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

35 40

40 45

45 50

5055

5560

60

65

65

70

70

75

75


80

85

15.5

16.0

16.5 VO

LUM

E - CU

.FT. P

ER

LB. DR

Y A

IR

17.0

17.5

18.0

HU

MID

ITY

RA

T IO

- P

OU

ND

S M

OI S

TUR

E P

ER P

OU

ND

DR

Y AI

Rr

0

css

Problem 3-39

90 (32)80 (27)

50 %59 (15)

62.5 (17)

0.8

R R




5000 FEET

0

1.0 1 .0

2.04 .08 .0

-8.0-4.0-2.0

-1 .0

-0 .5-0 .4- 0.3-0.2-0 .1

0 .1

0.2

0.3

0.4

0 .5

0 .6

0 .8-2000

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

Two unknowns & two equations Solve simultaneous:

fa n s a r c

a

a

W + q = m (i -i )

1 ,3 2 0 ,0 0 + 7 6 ,3 5 0m =(3 3 .7 -2 8 )

m = 2 4 4 ,9 7 4 lb a /h r


40

s r s ai = i - ( q m )

s1,320,000i = 33.7 - =28.3244,974

Btu/lba

Locate points on the condition line on Chart 1 Ha and point c is on cooler process line horz. to left of points. Read ts = 62.5 F, tc = 61.6F.

(a) s244,974Q = x16.2 = 66,143cfm

60

(b) 3

sQ 31.2 m= s

H

3-40 English Units –Tucson, Arizona, Elevation 2,556 ft. ; min 0i =i =31.1 Btu/lba and sat. air mint =64.5 F; PSYCH Shreveport, Louisiana, Elevation 259 ft. ; min 0i =i = 42.5 Btu/lba and sat. air mint 76.8 F; PSYC= SI Units – Tucson, Arizona ; min 0i =i 51.5 kJ/kga= mint =18.1 C; Chart 1b Shreveport, Louisiana ; min 0i =i =75.5 kJ/kga mint =24.8 C; Chart 1b


41

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50

ENTHALPY - B

TU PER P

OUND OF D

RY AIR

SATURATION T

EMPERATURE -

°F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

13.0

13.5

14.0 VOLU

ME - C

U.FT. P

ER LB

. DR

Y AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

SL

TLO

Problem 3-40Shreveport, LA

9576.8

R R




259 FEET

0

1.0 1 .0

2.04 .08 .0

-8.0-4.0-2.0

-1 .0

-0 .5-0 .4- 0.3-0 .2-0 .1

0 .1

0.2

0.3

0.4

0 .5

0 .6

0 .8-2 000

-1000

0

5 00

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW


42

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50

ENTHALPY - B

TU PER P

OUND OF D

RY AIR

SATURATIO

N TEMPERATU

RE - °F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.00 2

.00 4

.00 6

.00 8

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

35 40

40 45

45 50

50 55

5560

6065

65

70

70

75

75


80

85

85

14.0

14.5

15.0 VOLU

ME - C

U.FT. P

ER LB

. DR

Y AIR

15.5

16.0

16.5

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

TA

TLO

Problem 3-40Tucson, Arizona

10264.6

R R




2556 FEET

0

1.0 1 .0

2.04 .08 .0

-8.0-4.0-2.0

-1 .0

-0 .5-0 .4- 0.3-0 .2-0 .1

0 .1

0.2

0.3

0.4

0 .5

0 .6

0 .8-2 000

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-41 )ii(mq srs −= (a) = 1,319 lba/hr ton sm 12,000 /(28.2 19.1)= −

s1319 x 15.6Q 343 cfm/ton

60= =

o

s

m r1 13= 0.55 or 55%m 23.5r0

= =

(b) 3

sQ 0.046 m / s - kW≈ 0 sm /m 55%≈


43

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50

ENTHALPY - B

TU P

ER P

OUND OF

DRY AIR

SATU

RATION TEMPER

ATURE - °

F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

35 40

40 45

45 50

5055

5560

60

65

65

70

70

75

75


80

85

15.5

16.0

16.5 VO

LUM

E - CU

.FT. P

ER

LB. DR

Y A

IR

17.0

17.5

18.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y AI

R

r

s

0

1

Problem 3-41

100 (38)75 (24)50 (10)

10 %

40 %

0.7

R R




5000 FEET

0

1.0 1 .0

2.04 .08 .0

-8.0-4.0-2.0

-1 .0

-0 .5-0 .4-0.3-0 .2-0 .1

0.1

0.2

0.3

0.4

0 .5

0 .6

0 .8-2 000

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-42 2 2 3 2

500,000q m (i i ); m(41.1 21.9)

= − =−

lba/hr 042,26m2 = = 6315 cfm 14.55/60 x 26042Q2 = lba/hr 6511 26,042 x 25.0m0 == F 5.49/5.67t ;25.0m/m mix30 == Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

44

3-42 (cont’d) Preheat Coil: ph 0 p 4 0q = m c (t -t ) 6511 x 0.24 (60-6) 84,383 Btu/hr= = Heat Coil: h 2 5 1q = m (i -i ) 26,042 (28.4 - 20) 218,753 Btu/hr= = Humidifier: w 2 2 5m = m (W -W ) 26,042 (0.0144 - 0.0035) = 283.9 lbw/hr= (b) 3

2 ph hQ 2.98 m / s; q 24.7 kW; q 64.1 kW;= = =

kg/s 036.0mw =


45

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

3

2

4

1

5

2Problem 3-42

1153

70 (21)

30 %

105 (40)60 (16)

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 000

-100 0

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-43 Use Chart 1a; d a rq m (i is )= − or )ii/(qm srda −= (a) m a = 150 x 12,000 / (28.4-22) = 28,125 lbm/hr dQ 28,125 x 13.25/60 61,211 cfm= = m dQ 0.20 Q 1,242 cf= = m m mm = 1,242 x 60/13.5 5,521 lbm/hr [v assumed]= m r mi =i 1.8 x 12,000/5,521 24.5 Btu/lbm; t 62 / 57 F− = =


46

(b) 3 3d m mQ = 2.93 m /s; Q = .59 m /s; t = 17/14 C

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

rm

s

Problem 3-43

0.8

0.6

75 (24)60 (16)

62 (17)

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-44 (a) a15.0 x 12,000m 29,508 lba/hr(31.2 - 25.1)

= =

d mQ 29,508 x 16.0/60 7,869 cfm; Q = 0.2 x Q= = s 1,574 cfm= m mm =1,574 x 60/16.2 5,829 lba/hr (v assumed)= mi 35.7 1.8 x 12,000/5,829 27.5 Btu/lba;= − =


47

mt 62.5 /58 F= (b) 3 3

s m mQ =3.7 m / s; Q 0.74 m /s; t 17 /14.4 C= =

3-45 Use Chart 1a; r

1

m 100.8m 0r

= =

[Both design and min. load condition] is = ir - / mq sm

sr

ds ii

Qm

−= =

22.3)-(29.3512,000 x 50

conditions all for constant is m lba/hr; 106,85m ss = Btu/lba 25.83 10612,000/85, x 2535.29i 's =−= (a) From Chart 1a; F 64t 's = (b) s'bc'1bss i )mm(imim +=+

271.07.318.258.252.24

)ii()ii(

mm

'1's

'ss

c

b =−−

=−−

=

(b) From chart 1a; cases both for F 49td =


48

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

r

s

0

1

s

0'

1'

s

s'

Problem 3-45

0.9

95 (35)85 (29)77 (25)64 (18)55 (13)

50 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-100 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-46 Refer to problem 3-45. Results are similar. 3-47 (a) It is probably impossible to cool the air from 1 to 2 in one process. The extension of line 12 does not intersect the saturation curve. (b) Cool the air to state 1' and then heat to state 2.


49

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

ND

S M

OIS

TUR

E P

ER P

OU

ND

DR

Y A

IR

1

2

1'

2

Problem 3-47

80 (27)60 (16)52 (11)

67

54

90 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-48 (a) c

s

m sh= =.83m ch

7

h

s

m cs= 0.16m ch

= 3

c

h

m 0.837 5.14m 0.163

= =

s r sq m (i i= − )

s50 x 12,000m 93,750 lba/hr(28.2-21.8)

= =


50

sQ 93,750 x 13.2/60 20,625 cfm= = (b) 3

sQ 9.7 m /= s

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

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r

chs

Problem 3-48

90 (32)75 (24)52 (11)

90 %

0.65

20 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1 00 0

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-49 See diagram of problem 3-48

(a) c h

s s

m m36 10.10.9; 0.10m 46.3 m 46.3

= = = = ; c

h

m 0.9 9.0m 0.10

= =

s50 x 12,000m 83,333 lba/hr(30.1 - 22.9)

= =

sQ =83,333 x 15.67/60 21,763 cfm=Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

51

(b) 3

sQ =10.3 m /s 3-50 (a) See diagram for problem 3-48 c

c c r c c ss

m = 0.837; q = m (i -i ); m 0.714 x m 0.837 x 93,750m

= =

; cm 78,469 lba/hr= cQ 78,469 x 13.04/60 17,054 cfm= = cq 78,469 (28.2-20.6) 596,364 Btu/hr= = (b) 3

c cQ =8.1 m /s; q 175 kW= 3-51 SI Units

(a) On the basis of volume flow rate using Chart 1b:

2 313Q = Q 0.69 x 1.18 0.81512

= = m3/s

and m1 3 2Q = Q - Q = 1.18 0.815 0.365− = 3/s (b)

334 a3 4 3 4 3

3

34

Qq = m (i -i ) = (i -i )v

1.18q = (47.8-41.0) = 9.6 kW0.835


52

10 20 30 40 50

60

70

80

90

100

110

110

120

120


10

20

30

40

50

60

70

80

90

100

ENTHALPY - KJ P

ER KILOGRAM O

F DRY AIR

SATURATION T

EMPERATURE - °C

5 10 15 20 25

30 35 40 45 50

DR

Y B

ULB

TE

MP

ER

ATU

RE

- °C

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30


20%

30%

40%

50%

60%

70%

80%

90%

5

5

10

10

15

15

20

20

25

25


30

0.78

0.80

0.82

0.84

0.86 VO

LUM

E - CU

BIC

ME

TER

PER

kg DR

Y A

IR

0.88

0.90

0.92

0.94

HU

MID

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RA

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- G

RAM

S M

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TUR

E P

ER K

ILO

GR

AM D

RY

AIR

1

2

3 4

Problem 3-51

Problem 3-51

292417.212

50 %

11

14.7

R R




SEA LEVEL

0

1.0 1 .0

1 .52.0

4 .0

-4.0-2 .0-1.0

-0 .5

-0.2

0.1

0.2

0.3

0.4

0.5

0.60.7

0 .8-5.0

-2.0

0.0

1 .0

2.02.5

3.0

4.0

5.0

10.0

- SENSIB LE HEAT QsTOTAL HEAT Qt


hW

English Units (a) 1 34Q = 640 cfm; q = 33,684 Btu/hr

3-52 (a),(b)

From Chart 1b, states 1.4 and ADP are known. Based on approx. 11.8 C db, 11.2 C wb, and 90% RH locate state 2. Then for full load design condition air is cooled from 1 to 2 and the room process proceeds from 2 to 4.


53

For the high latent load condition, the air at 2 is reheated to state 3 where it enters the space and the process proceeds to state 4.

(c) 224 a 4 2 4 2

2

Qq = m (i -i ) = (i -i )v

2Q =35 x 0.817 (47.7-32) ; m2Q 1.8= 2 3/s

12 a 1 2

12

1.82q = m (i -i ) = (60.6-32)0.817

q = 63.7 kW

34 a 4 3

34

23 24 34

1.82q = m (i -i )= (47.7-39.4)0.817

q = 18.5 kWq = q - q = 35-18.5=16.5 kW


54

10 20 30 40 50

60

70

80

90

100

110

110

120

120


10

20

30

40

50

60

70

80

90

100

ENTHALPY - KJ P

ER KILOGRAM O

F DRY AIR

SATURATION T

EMPERATURE - °C

5 10 15 20 25

30 35 40 45 50

DR

Y B

ULB

TE

MP

ER

ATU

RE

- °C

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30


20%

30%

40%

50%

60%

70%

80%

90%

5

5

10

10

15

15

20

20

25

25


30

0.78

0.80

0.82

0.84

0.86 VO

LUM

E - CU

BIC

ME

TER

PER

kg DR

Y A

IR

0.88

0.90

0.92

0.94

HU

MID

ITY

RA

TIO

- G

RAM

S M

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TUR

E P

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ILO

GR

AM D

RY

AIR 1

ADP 2 3

4

3

Problem 3-52

Problem 3-52 21

2723

17

19

14

11.8

11

9

R R




SEA LEVEL

0

1.0 1 .0

1 .52.0

4 .0

-4.0-2 .0-1.0

-0 .5

-0.2

0 .1

0.2

0.3

0.4

0.5

0.60.7

0 .8-5.0

-2.0

0.0

1 .0

2.02.5

3.0

4.0

5.0

10.0



hW

3-52 English Units (a),(b) See above (c) Btu/hr 2Q = 4103cfm ; 12q =221,243 Btu/hr; 34q 67,49= 8 223q 52,50= Btu/hr 3-53 English Units (a) s r s sq=m (i -i ); m 5000 x 60/13.2 22,727 lba/hr= = (specific volume value of 13.2 ft3/lbm is assumed.) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

55

s r si = i - q /m =28.2 10 x 12,000 / 22,727 22.9 Btu/lba− = s o s ot = t = 57.5 F; W =W 0.0083 lbv/lba=

(b) r r

m s

m m0m= = 0.462m m0r

rm =0.462 x 22,727 10,500 lba/hr= om 22,727 10,500 12,227 lba/hr= − = rQ 10,500 x 13.68/60 2,394 cfm= = oQ 12,227 x 12.11/60 2,468 cfm= =

(c) r

m'

m 0'm'= =0.578m 0'r

r om =0.578 x 22,727 13,131 lba/hr; m 9,596 lba/hr= =' r o'Q =13,131 x 13.68/60 2,994 cfm; Q 9,596 x 13.48/60= = = 2,156 cfm (d) c s m' sq = m (i -i ) 22,727 (28.4 - 22.8) 127,271 Btu/hr= =


56

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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RA

TIO

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TUR

E P

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OU

ND

DR

Y A

IR

r

0

m

s

0'

r

m'

ADP

Problem 3-53

0.8

1150

75 (24)

70 (21)

65 (18)57.5 (14)40 (4)

43 (6)

90 %

50 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-1000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-53 SI Units (a) ts = 14.2C; Ws = 0.0083 kgv/kga (b) 3

rQ =1.13m s; 3oQ =1.17m s

(c) 3rQ =1.41m s; 3

o'Q =1.02m s (d) cq = 37.3 kW 3-54 (a) Any combination that will yield an enthalpy less than 57.0 kJ/kga or 33 Btu/lba Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

57

(b) rs mkga/s 95.584.0/5m ===

o

r

m mr= =0.36m 0r

om 0.36 x 5.95 2.14 kga/s= = 3

oQ = 2.14 x 0.852 = 1.82 m /s = 3,857cfm (c) adt 15.4 C or 60F= (d) (Essentially, no difference) o n m s r sq /q = (i -i )/(i -i ) = 1.0

10 20 30 40 50

60

70

80

90

100

110

110

120

120


10

20

30

40

50

60

70

80

90

100

ENTHALPY - KJ P

ER KILOGRAM O

F DRY AIR

SATURATION T

EMPERATURE - °C

5 10 15 20 25

30 35 40 45 50

DR

Y B

ULB

TE

MP

ER

ATU

RE

- °C

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30


20%

30%

40%

50%

60%

70%

80%

90%

5

5

10

10

15

15

20

20

25

25

30 WET BU LB TEMPERATURE - °C

30

0.78

0.80

0.82

0.84

0.86 VO

LUM

E - CU

BIC

ME

TER

PER

kg DR

Y A

IR

0.88

0.90

0.92

0.94

HU

MID

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RA

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AM D

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AIR

r

s

0m

s

Problem 3-54

25 (77)18 (64)

20 (68)

0.6

57

R R




SEA LEVEL

0

1.0 1 .0

1 .52.0

4 .0

-4.0-2 .0-1.0

-0 .5

-0.2

0 .1

0.2

0.3

0.4

0.5

0.60.7

0 .8-5.0

-2.0

0.0

1 .0

2.02.5

3.0

4.0

5.0

10.0



hW


58

3-55 424,000SHF530,000 424,000

−=

− = -4

Construct condition line on Chart 1a with preheat and mixing processes. (a) sen s p r sq = -424,000 = m c (t -t )

s424,000m 88,333 lba/hr

0.24 (75 95)−

= =−

3

sQ =88,333 x 14.07/60 20,714 cfm or 9.8 m /s=

(b) rr

m

m hm= =0.33; m 0.33 x 88,333 lba/hrm hr

=

r rm =29,150 lba/hr; Q 29,150 x 13.68/60= 36,646 cfm or 3.14 m /s=

hh

m

m =1 0.33 0.67; m 0.67 x 88,333m

− = =

h hm 59,183 lba/hr; Q 59,183 x 13.1/60= = 3

hQ 12,922 cfm or 6.1 m /s (at heated condition)= (c) ph h p h oq =m c (t -t ) = 59,183 x 0.24 (60-35) q= 355,098 Btu/hr or 104 kW (d) mq =88,333 x 0.24 (95 - 65) 635,998 Btu/hr or 186 kW= Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

59

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

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RA

TIO

- P

OU

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S M

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TUR

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OU

ND

DR

Y A

IR

r

0 h

m

ss

Problem 3-55

-4

95 (35)75 (24)

50 %

60 (16)35 (2)

20 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0.1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 000

-100 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-56 Refer to chart 1a. (a)

34 a3 4 3 3 4 33

34 33

4 3

33

60q = m (i -i ) = Q x (i -i )v

q v (1750 x 13.23)Q x =60(i -i ) 60(28.1-23)

Q = 75.7 or 76 cfm = 0.040 m /s


60

(b) t3db = 58.5 F and 80% RH or 15 C

(c) 32 3

31Q = ; Q = 0.754 x 75.7 = 57 cfm or 0.028 m /s12

3

1Q = 76 - 57 = 19 cfm or 0.012 m /s

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

- P

OU

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S M

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TUR

E P

ER P

OU

ND

DR

Y A

IR1

2

3

4

Problem 3-56

84

70

75

62

58.550

90 %

50 %

0.8

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2 00 0

-100 0

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW

3-57 (a) Refer to Chart 1

A reheat system is required. Process 1-2 is for the coil. Process 3-4 is defined by the SHF = 0.5

Process 2-3 represents the required heat. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

61

State 3 is defined by the intersection of the reheat and space condition lines.

(b)

334 a3 4 3 4 3

3

34 33

4 3

33

Q x 60q = m (i -i ) = (i -i )v

q v 100,000 x 13.4Q = =60(i -i ) 60(28.2-23.9)

Q = 5,194 cfm or 2.5 m /s

(c)

12 a 1 2

12

23

23

5,194 x 60q = m (i -i ) = (34.2-20.2)13.4

q = 325,594 Btu/hr or 95.4 kW

5,194 x 60q = (23.9-20.2)13.4

q =86,050 Btu/hr or 25.2 kW


62

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50


OUND OF D

RY AIR


F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

3540

40 45

45 50

50 55

55 60

6065

65

70

70

75

75

80

80


85

90

12.5

13.0

13.5

14.0 VO

LUM

E - CU

.FT. P

ER LB. D

RY

AIR

14.5

15.0

HU

MID

ITY

RA

TIO

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OU

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S M

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TUR

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ND

DR

Y A

IR 1

2 3

4

ADP

Problem 3-57

85

70

75

62

66

56

5145

50 %

R R




SEA LEVEL

0

1.0 1 .0

2.04 .08 .0

-8.0-4 .0-2.0

-1.0

-0 .5- 0.4-0.3-0 .2-0 .1

0 .10.2

0.3

0.4

0 .5

0.6

0 .8-2000

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIB LE HEAT Qs

TOTAL HEAT Qt


hW

3-58 Assume room temperature humidity of 50% and layout the state & processes on required from point c to s. Supply Air: = senq 120,000 x 0.5 60,000 Btu/hr= = s p s rm c (t -t )

s60,000m 53,192 lba/hr

0.24 (75-70.3)= =


63

3sQ =53,192 x 16.33/60 14,477 cfm or 6.8 m /s=

Mixed Air: om 53,192 x 0.333 17,703 lba/hr= = 3

oQ 17,713 x 17.2/60 5,078 cfm or 2.4 m /s= = rm 53,192 17,713 35,479 lba/hr= − = 3

rQ =35,479 x 16.5/60 9,757 cfm or 4.6 m /s= Reheat: rh c p s cq = m c (t -t ) 53,192 x 0.24 (70.3-55.2)= 192,768 Btu/hr or 56.5 kW= Coil: c m m cq =m (i -i ) 53,192 (34.4 - 24.2)= 542,558 Btu/hr or 159 kW=

= % 1.5412,200

100)109,190412,200(=

−


64

10 15 20 25

30

35

40

45

50

55

55

60

60


15

20

25

30

35

40

45

50

ENTHALP

Y - BTU

PER

POUND O

F DRY A

IR

SATU

RATION TE

MPERATURE - °

F

35 40 45 50 55 60

65 70 75

80

85

90

95 100

105

110

115

120

DR

Y B

UL

B T

EM

PE

RA

TU

RE

- °F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028


20%

30%

40%

50%

60%

70%

80%

90%

35

35 40

40 45

45 50

5055

5560

60

65

65

70

70

75

75


80

85

15.5

16.0

16.5 VO

LUM

E - CU

.FT. P

ER

LB. DR

Y A

IR

17.0

17.5

18.0

HU

MID

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RA

T IO

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TUR

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DR

Y AI

R

r

0

m

c ss

Problem 3-58

90 (32)

75

75 (24)

50 %

70 (21)

55 (13)

90 %

0.60.5

R R




5000 FEET

0

1.0 1 .0

2.04 .08 .0

-8.0-4.0-2.0

-1 .0

-0 .5-0 .4- 0.3-0.2-0 .1

0 .1

0.2

0.3

0.4

0 .5

0 .6

0 .8-2000

-1 000

0

500

1000

1500

2000

3000

5000

-SENSIBLE HEAT Qs

TOTAL HEAT Qt


hW


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Chapter 4

4-1 (a) comfortable

(b) too warm

(c) comfortabΙe

(d) too dry

4-2 (a) comfortable

(b) too warm

(c) comfortable

(d) too dry

4-3 (a) Assume sedentary dry bulb of 78 F , clo = o.5, met. = 1 .8,

using equation 4-4a, to,act = 75 - 5.4(1 + 0.5)(1.8 - 1 .2) = 71 F

Relative humidity should be less than 50%

(b) Should wear a S\Ι/eater or light jacket and slacks.

(clo = 0.8)

4-4 Use fig 4-1

(a) Summer, to = 76 F or 24 C; Winter, to = 72 F or 22 C

(b) Use equation 4-4a aS a guide, \Λ/ith clo = 0.2,

met = 3.0, tdb :76 F

t

o/

to =76 -5.4 (1+0.2)(3-1.2) = 64 F [winterorSummer]

4-5 From fig 4-3 temperature can rise about3.2 F.(j.g C)

t=68 +3.2=71.2 Fort=20+ 1.8= 21.8C

4-6 From fig 4-3 @200 fpm, temp rise ρ 5.3 F (2.9 C)

with t,,"-t _ 9 F (5 c), temp rise ε 6.5 F (3.6 c)

4-7 to = (t, +t^r)|2, then using Eq. 4-1

T,fn ='6* Cν\l2 σg _Tr) = (53s)4 +(O.103 x 109) (4o)1Ι2(78_74)

tmft:82For27'8C

to=(74+82)Ι2 = 78F or25.6C

4-8 Compute the operative temperature, to

Τ,xn = φ4q4 + (O.103 x 1o911eo11/'(εo _76)= 83.5 F or 28.6 C

to = (84 +76)12= 79.8 F or 26.5 C

From Fig 4-1, to = 79'8 F and 50 % R.Η. is out of the comfort

zone. Recommend lowering to to about 77 F or 25 C.

tu x72 F

4-9 Use Eq. 4-4 to estimate a value of the operative temperature

to, active, assuming to for sedentary activities is 78 F (25.6 C)

with met = 2.0. to, active = 78 - 5.4 (1 + 0.5) (2 - 1.2) = 71.5 F, (22C)

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Αs an approximation

Tmrt = 2To _Τ, and Tflx = Tno * ci1Ι21Tg _ Τ, ) Eq ' (4-1)

eliminating Tmrt between the 2 equations

2(Τo_T3)4 = Tno *CV1/21τn _Tr)

where all temperatures are absolute

Solve by trial and error with T, =72+ 460 = 532 R

and Te =(71.5+460)=531.5 R, C=0.103 x 1Oe, V=30

ta=85F(30C)

Cold surroundings require high ambient air temperature

for comfort, even with high activity level.

4-10 (a) Most occupants will be uncomfortable because the relative

humidity is more than 60%, even with trx = t,

(b) The lightest weight possible. Short sleeves, shorts,

open neck, etc.

(c) Lower relative humidity if possible by adjusting the cooling

system to remove more moisture. CouΙd also increase the

relative air motion to highest values, perhaps use fans.

4-11 (a) Even if the suit was heavy weight, many executives would be

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cool if sedentary.

(b) Would definitely be cold, especiaΙly hands and feet.

(c) Probably would be comfortable in typical work cΙothes

(d) Probably would be comfortable since they would keep

their coats on and would be walking around.

(e) Cold to very cold

4-12 Determine relative temperatures difference between inside and outside.

68 - 45 237 4 _ 45 = 29 Costs are79o/o of that for increased setting, or

74 - 45 2968 - 45 23

Costs are increased by 26o/o if thermostat is raised.

4-13 Too much air motion in the cold winter months tends to cause drafts and

make people uncomfortabΙe. Air velocity just sufficient to prevent large

temperature gradients from floor to ceiling is best for winter. Τhe opposite

is true for hot summer months. Higher air velocity tends to compensate

for high temperature and humidity.

4-14 (a) Raising the chiΙled water temperature will cause the cooling coil to

operate with a higher surface temperature and the relative humidity in

the space will tend to rise if the latent heat gain is signifΙcant such as

would be the case with many occupants, this could lead to

u ncomfortable cond itions.

(b) Yes, during the unoccupied hours the space load may be almost totalΙy

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SensibΙe heat gain and the load is much less than the design value. ln

this case the chiΙled water temperature may be increased.

4-15 Τhese fans may bring air down in the Summer, increasing the

velocity of air in the occupied zone and providing improved comfort.

ln the winter, air may be drawn upward, pushing the warm air at the

ceiling downward where it can increase the temperature in the

occupied zone without increasing significantly the air motion below

the fan.

4'16 (a) Τable 4-2 gives a minimum required amount of ventilation air

of 15 ft3 /min per occupant. this is the minimum amount of

outdoor air that should be used under any circumstances.

Therefore, (Qo)rin = 15(30) = 450 ft3/min

(b) on the basis of floor area, the occupancy wouΙd be 25 and the

minimum ventilation requirement would be

Q, = 15 (25) = 375 ft3 /min. lt would be better to design for

floor area if lowest air flow is desired. With 30 actuaΙ student air

flow is such a case wouΙd be insufficient.

4-17 Use Eq. 4-5, Solving for C,

Cs = (QtC" + N)/Qt = C" + (N/at)

= (2001196 + (O.25l9oο)

Ξ 478 x 1o-6 = 478 ppm

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or using Sl Units

c, = (2oo / 106)+ (0. 118 t 0.472x 9OO)

= (2OO / 106) + (278 t 106):478 ppm

4-18 n = number of people to occupy a room

N=n(5.Oml/s)

Solving Eq. 4-5 for Ν

N = Qt (C, - C") = n (5.0) ml/s-Person

n : Qt (C, - C") / (5.0)

: 2.8 (1000-280) / 5.ο

n = 403 persons or 0.0069 m3 /s - person

For English Units:

n = 6000 (1oOO - 28Ox 10-6) / O.O107

= 404 persons or 14.8 cfm/person

4-19 Use the M-100 media of fig. 4-8. From table 4-3, select a

12x24 x 8 unit; 650 cfm, ΔP = 0.4 in. wg

At ΔP = 0.25 in. wg. each unit will handle

Q = Ql |o'25 Ι o.40]1l2 = 650 [O.25 t o'4oJ1l2

:514 cfm/unit. Then the number of units

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-

;g μ=(2000 l 514) = 3.89 or 4. This is a satisfactory number.

4-20 Use the M-100 media from TabΙe 4-3 select a O.3 x O.6 x O.2 unit.This is rated at 0.3 m'/s with 1oo pa pressure drop.

Αt ΔP = 60 pa the alΙowabte flow rate for each unit would be

Q = (0.3) (60/1 OOf tz = 0.23 m3/s

1.OO m3/s wouΙd require 1'Oolo'23 = 4'34 units. This requires atΙeast 5 filter units, but since this is an odd number, recommendusing six units.

Trying the 0.6 x 0.6 x 0.2 filter the allowable flow per unit would be

Q = (0.62) (60/1}q1t2 = 0.48 This would require more than two

units of this size. Εconomies would determine the best choice.

4-21 Solving Εq' 4-1O for Q

Q = Qr [ΔP / ΔP,]1'2 = 9OO [o'1 l0'15]1|2 =735 cfm/module

N = ss00/235 = T.4g [must be integer] Use g modules

VeΙ = Q/Α = ΨΨ = 344fpm = 5.7 fps(2)(8) '|-

4-22 Solving Εq' 4-1O for Q

Q = Qr [ΔP / ΔP,J1l2 = (o'42) |24 l 37 '4]1t2 = 0.336

m=(2.8)/0.336=8.3

Use 9 modules, a 3 x 3 arrangement.

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/3

VelocitY =

4-23 M-200; 0.6x0.6

Use Eq. 4-10

ΔP = Δη ta / Qι.]' :1OO

0.4Velocity =

a (2.8)m3 /s=1.73m1s

FAcΕ AREA (0.3)(0.6)(e)m2

xO.2; O.4Om3 /s/module

a

lO,4OtO.42l2 =90.7 Pa

=2.22m1s

4-24

4-25

A (0,3)(o 6)

No solution exists due to the fixed air quantity for the unit. Thispart of the problem is intended to show the student that typicaldirect expansion equipment cannot be used in this \May. lt alsoshows that the load due to outdoor air is very large.

Γho : o'25 rh"; Locate point 1 on psychrometric Chart at82'4 F db and

66.8 F wb

it = 31.4 Btu / lbm and v1 = 13.9 ft3 /lbm

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by Sectiοns 107 οr ]08 ofthe t976 Llnited Stαtes Copyrilht Αctv,ithοut ιhe peιmissiοn ofιhe copyright oνner is unlανful'

exhaust

sHF= 0.7

74

Q1, = rhi (ii - is) = at /V (60) (i1 - is)

Φl = 35ο Ι 12,ooo Qls , = (350 l 12'000) (6ο / v1) (i1 _i.)

is = 31 . - ''''J8?rε;

'' = 23'46 Btu / lbm

Locate on psychrometric chart' ts = 65'6 F db' 55'5 F wb

Q.r = lil, (ir. - is) = 36'000; ir =27 '6 Btu / lbm

tr, = Γil1 = -ΨΨ ^ :8695'7 lb / hr(27 .6 - 23.46)

O, = rh, (vr, = Ψ(13.4) = 1940 cfm

Qt" = 8695.7 (31.4 - 23.46) = 69,000 Btu / hr = 5.75 tons

Qr = 5.75 (350) = 2014 cfm

(ο) Design filters for 2014 cfm, use M-200 media of fig 4-8.

Try the 24x24x8 units of table 4-3. 920 cfm @0.4 in. wg.

For max. ΔP of 0'125 in.wat.

Q = 920 [0. 125 tO.4O] 1t2 =514 cfm / module;

n = 2014 I 514 = 3'92, use 4 modules

4-26 Use the M-15 media, η = 93 % from fig' 4-3'

From table 4-2,60 cfm / person is required, outdoor air.

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--

A fresh air balance on the filter gives QrEt + Qo = Qs

where Q,. is recirculated air, Qs is outdoor air and

Q. is supply air.

8, = (60 - 20) 10.93 = 43.0; Q, = 43.0 +20 = 63.0 cfm / person

or the total amount of air supplied is

Qτ = 63.0 x 55 = 3465 cfm; Try the 12x24x8 unit of table 4-3

Q/unit=9oO[O'1 /O.35]1Ι2=481 cfm; n =4755 l481

= 7 .2 modules

Use 8 modules [Note: The M-24 media could also be used]

4-27 Q, = (25 - 15) / 0.S = 12.5 cfm / person

Φ. = 15 + 12.5 = 27 .5 cfm / person

4-28 Filter location is B, figure 4-9

Use Eq. 4-12, solve for RQ.. since

RQΓ = { -QoEv[C, _(1_Et)Co] + N}/ (EvEfcS)

RQr={-2OOxO.85[180-(1-0.8)0.0]+(10x150x35'32)]l

(0.85 x 0.8 x '180) where Co = 0.0

RQr = 185 ft3 / min or cfm

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by Sectiοns Ι 07 or 1 08 οf the Ι 976 tJnited Sιαtes Copyrιght Αcι ιiithout the permissiοn of ιhe copyrighι oνner is unlιτνν.ful'

75

76

Qo = 2OO cfm, Qs : (1S5 + 2OO) = 385 cfm

4-2g Solve Εq' 4-11 for RQ,

RQr = (_Qo)(Eu)(cr)+ N / ΕrEiC,

RQr = [ (-20) (0.65) (220) + (125) (35.32 ft3/m3)]

(0.65X0.7)(220)

RQ. = -Ψ9-*_!1!5^, = 15.53 cfm/person' (0.65)(0.7)(220)

4-3o For filter location A, use Εq. 4-1 1, solving for RQ,

RQr = (-QoEvCs + N) / (EvEfCs)

RQr = t (-2OO (0.85) 180) + (10 x 120 x 35.32 ft3/m3 )l t

(0.85 x 0.8 x '180 )

RQr = 183 cfm, Qo = 2OO cfm; d, = 383 cfm

4-31 (a) This type of space will require a high ventilation (supply air)

rate to handle the load, air cleanliness is not the main criterion.

Therefore, a low efficiency filter with low pressure drop is

acceptable. From table 4-2, assume occupancy will be about 30

persons / l OOO ft2. So the total design occupancy is 90

persons. Τhe design will be based on this occupancy although

the cooling requirements may dictate a larger supply air rate.

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77

A ''fresh air'' balance on the filter gives Φ, = (Q" _ Qo) / Ef

d," = (20 -15) / 0.5 = 10 cfm / person recirculation rate

4-31 (continued)

Φ, : 1O + 1 5 =25 cfm / person supply rate

Qτ = 25 x 90 = 2250 cfm total supply rate

Net face area, Αf = 2250 / 35o = 6'43 ft2

(b) A higher efficiency would reduce the total amount of air and

reduce the required face area. However this is not desirable in

this case. First the filter system would have to be enlarged to

handle the greater amount of air. A lower filter efficiency could

be used and still maintain the required air quality.

For example, suppose the load dictates 4000 cfm instead of

2250 cfm, then for 90 PeoPle

Φ, = 4ooo / 90 = 44'4 cfm / person

Using a minimum of 15 cfm / person of outdoor air.

Qr. = 44.4 - 15 = 29 '4 cfm / Person

8. = 29.4: (20 - 1s) / Er

Et = 5 Ι29'4 : 0'17 or 17oλ required

4-32 (a) Q = (Q, / v) 60 (i|. -ir)l5x

I

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studeπs enrolled in courses for which the textbook has been adopted. Αny oιher reprοduction ο

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225 people

75FRH=5ο%

125,ο0ο

78

Υ x13 ft3 / Ιba

Φ. = (125,oOo x13) Ι [ 60 x (28 - 1e.4 )]

Qs = 3,149 cfm

Φ, = Φo = 15 x225

Φ, = 3,375 cfm

Q. must be 3,375 cfm, find ne\Λ/ Supply air condition

125,000 = (3,375 I 13) 60 (28 - i.)

i' = 28 - (125,000 x 13 ) / ( 3,375x 60) = 20 Btu i lba

Locate new condition on chart aS Sho\Λ/n' Coil must cool oDA

down to this new condition.

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(b)

(c)

-D 50 52

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Chapter 5

k = CΔx =0.2 Χ4 =O.8 (Btu _ in) / ( hr - ft2 _ F)

k = 1 .14 x 0.1 = 0.114 W / (m-C)

C=k l Δx=o.3o / 5.5 = o.o55 Btu / 1ft2-nr_ 11

C = O.O43 / ο. 14 = .307 W/ (m2 _ C)

(a) R = 1lC= 1/0.055 = 18.3 (ft'- hr- F)/ Btu

R' = R l A= 1 l CA= 18.3/ 1ο0 = 0.183 ( hr-F)/Btu(b) R ='1 I .307 = 3.26(m'-C)/W

R'=3.26l9.3=0.35C/W

5-'1 (a)

(b)

5-2 (a)

(b)

5-3

5-4 R = ΣRi , Rgyp =1ΙC=1l3'1=o'32Rbtd = !0.33 = 3.03; Rair = 0.68

R = 0.68 + 0.32+ 3.03 + 0.32+ 0.68

R = 5.03 (hr- ft2- F) / Btu

R-0.68 R=0.68

R=0.32

5-5

tnb ιn2R'= Γ2 + η

2πk| 2πkoL

AssumeL=1ft

kι:0'2 Btu - in' t(ft2 _hr_F); kp =314 Btu-in tσe _hr_F)

81

lnside Surface (7 m/s )

overalI Τhermal Resis.

0120

= 0.652 m2clW

5-10

5-11

Outside Surface

4 in. Face Brick

Sheathing

lnsulation

2x4 stud

Gypsum board

lnside surface

Τotal

Between Frame

0.17

0.65

1.32

1 1.0

0.32

ο.68

14.14

Αt Framinq

0.17

0.65

1.32

4.27

0.32

0.68

7.41

UA:U1Α; +U1Α1, U = UiAi /Α + U1Α1/A

Αr =

14'5

^nd

A' =lΞanα υ = LA 16 A 16 R

, =ΓΨ " : -)- ΓΨ " +1= 0 o77Btu / (r',..-tt' -r)110 "14.14) 116 7.41J

An ordinary walt with ε = O.9 has a unit resistance of 0.68. A

highIy reflective wall, ε = O.05, has a unit resistance of 1'70.

Assume radiation heat transfer is zero for reflective wall. Τhen

the resistance due to convection alone is approximately

Rc=1.7; hc=1/Rc=0.59; h.*r:1/0'68 = 1'47

Frac. Conv. = ha lh. *, = 0.59 I 1'47 = 0'4

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*nιοh the texibook has been adopted' Αny οther reprοclucliοn o^:.:::Ei::::":!,::iχ:,i''Ψ'y''o"!:*! -, - - ,

82

5-12 Αssume 15 mph wind. Rr (2 x 6) RzQ x 4\

1. Outside surface, 15 mPh 0'17 0'17

2. Siding 0'79 0'79

3. Sheathing 1'32 1'32

4. lnsulation, 19'0 11'0

2x4 4'27

2x6 6.7

5. GYPsum wall board 0'32 0'32

6. lnside surface 0.68 0'68

Total 28.98 18'55

Ut = 0.035 Btu / (hr - ft2 - F)

υ2 __O.o54 Btu / 1ιlr - ft2 - F1

% DiffereΠCΘ = [o'O5-4r_0-035) ι'' ool = 35'2[ 0.0sη ) '

5-13 Air space will be near the indoor temperature with small

Δt across the air Space.

Use t."rn = 50 F and Δt = 10 F and read

R = 1.oz(rrr -f( -r) I Atu [Tabte S-3a] or 0.18 m2clW

5-14 Assume tr"rn = 50 F; Δt = 10 F

R = 3.55 (hr - ft2 _F) / Btu or 0.62 (.2 _c/W) [Τabte 5-3a]

5-15 qc/Α = U"Δt

Find U for highly reflective surfaces because radiation will be

minimal. This will give a good approximation for the convection

component. From Table 5-2a,l1orz', heat flow down

83

Uc=1/R = 1l(2x4.55) = 0'11

q./Α

or

U^ _ --l' = o.625; q. /Α = 0.625(63 - 43) = 12'5 W/m2" (2x0

Γ, _ '4 / a '.4lQ/A.. =σ'n l( |l l _[ l' 'l l'

'Lι1oO] _ι1οo]

-]'

for ε1= t2=O'9, E : 0.82, σ'

(q/A),. = O.1 713 x9.s2 t635 o _s.τol

m

Radiation heat transfer is about 10 times greater'

5-16 U* = O.O7 Btu / (hr - tt2 - 11

Ud = O.4O Btu / (hr - ft2 - F)

Uwin = O'81 Btu / (hr - ft2 - F)

Ad = 17 '78 ft2; Awin =25'0 ft2;

Α* = 117 '2 f(Parallel heat flow Paths

UΑ = U*A* + U6Α6 + U*;nΑ*in

ι ι (o.O7 x117'2) + (O.4 x 17.78) + (0'81 x25'0)I I_

117.2

= O.3O Btu /(hr - f( _F) or about 1 '72\Ν t(m' _ c)

5-17 q/Α = U(ti _to)

Exοerpts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis foι testing or instructiona] purposes onιy to

students enrolled in courses tbr wbich the texibook has been uJo|t.α. Αny οιher reprολucιion or ιrαnsιαtion of ιhis work beyond ιhαι peπnitted: : -'- -ι.41-^ ^^^'',.:/Ιa! ^''''aΔv;" "'-Ι6!$ιΙ

20'

g'

84

From Table 5-4b, construction 2, R = 8.90 (hr - ft2 - F) / Btu

Assume insulation does not fitl the airspace'

Remove R for metal bath and plaster of 0.47 (f''ι. - tt' - f)/Btu and\"' )

add R for acoustical tile and insulation'

Ceiling, R" = 1 / 0'8; insulation' R='1 1'00;

R1e61=20.68; U = 1/R = 0'048 Btu / (hr - ft2 -Fiq/Α = o'o48 (72 - 5) = 3'22 Bτυ / (hr _ ft2)

5-18 From Table 5-4a, Construction 1

Uw= 9!+=oe71wr(m2-c)0.1761 \ /

Ud = 2.27 Wl(m2 -a)' Table 5-8

Uwin = 4'62w1(m2 - ")'

Table 5-5b

Αw = 35 m2;Αwin =8m2;Αd = 2m2

UΑ = U*A* + U6 + Αα + UwinΑwin

u _ Q.e7 1x35) + (2.27x2) + (4'62x8) -_ 2.16 w I (m2 - c)

35

5-19

U = O.14 Btu / (hr _ ft2 _ F1τable5 _ 4a, Construction No. 2

R=1ΙO.14 = 7'14, Rn =7'14_(1 to'44) + (1/0.55) =6.69

Un = 0.15 Btu/(h r - ft2 - F) or about ο.85 W l 1m2-c1

5-20 Αssume Ηardwood, k = 1.25 (Btu-in) / (hr - ft2 - F)

Summer Winter

Rι = 0.68 Ri = 0.68

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students enτolled in courses for which the teΧtboοk has been adopted. Αny οther reproλuction or ιrαfisιαtion of ιhis ινork beyond thαι permiιιed

85

Rα = '1 '375 l 1.25 Rα = 1 '375 l 1'25

Ro = 0.25 Ro = 0.17

R, = 2.03 R* = '1.95

U, =0.49 Btu / (hr-ft2 -F) U* =0.51 Btu /(hr-ft2-F)

Both values are greater than the value given in

Τable 5-8 of O.39 Btu / (hr _ ft2 - F), but acceptable.

5-21 Computed: Ri = 0.68, Rs = 0.03 (estimate); Ro = 0.25

Ri +RgaRo=0.96=R

U = 1.04 Btu / (hr - tt2 -r); or 5.92 wl(m2 -c) computed

Utub : 1.O4 Btu / (hr - tι2 _r); Table 5-5a

or 5.91 \Ν l(m2 _c); Table 5_5b; Same result\ /'

5-22 (a) From Table 5-5 U=1.08 Btu / 1nr - ft2 - F)

(b) Αssume tr"rn : 50 F; Δt = 10 F

Ras = 1.ol (nr -f( -r) r atu

' ., = ++ 1 .O1 = 1'94, Un = O.52 Btu/1ιlr-ft2-F)Rn =U-R', 1.og

5-23 (a) Uw = o.o89 Btu /(hr - tt2 _ 11 or O.51 W/(m2 _ c)τaοle 5-9

l l- _ ο.o29 Btu / (hr - ft2 _F) or O'16 W/ (-'_c) τante 5-1o,.Jfl -ν'νLζ) lJιur\ιll -ιL _| l ιJlv'l\J vYl1ιιl -)

(b) Q=UΑ(ti _tg); tr= t"ur-A

Εxοerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in οourses for which the textbook has been adopted. Αny οther reproducιion or trαnslαtion οf ιhis νork beyond thαt permitιed

..-_''r-] ^_':'_^'- ! ^'.'.ι ^'.4'ι

86

tavg =35'8 F (2'44 C) Table 5-11}Chicaαo.lllinois

A :22 F (12 C) Figure 5-7 ) *

5-23 (continued)

tg__35'8-22 = 13.8; tι:72re2c)q* = 0.ο89 (4 x 20 x7) (72 - 1 3'8) = 2,900 Btu / hr or 0.85 kW

qn = 0.029 (20 x20) (72 - 13.8) = 675 Btu /hr or 0.2 kW

5-24 R" = R1, Un = 0.029 Table 5-10

Rrin = 1l 0.48 Table 5-1a (Fibrous Pad)

Re = #"+ 2.08 = 36.6

U" = 0.027 Btu / (hr - ft2 - F) or 0.16 Wl(m'z - C)

5-25 (a) R*=: +11+ (1 t g.1) = 22.6"0089 \

U* = 0.044 Btu/(hr-ftu-F) or 0.25 w1m2-c)

Rfι = -: + (1 tO.4s): 36.60029 \

Un = 0.027 Btu/(hr - tt2 -f) or 0.155 Wl(m'z-c)

(b) Refer to problem solution 5-23

Q* = 0.044 (4 x20 x7) (72 - 13.8) = 1434 Btu / hr or 0'42k\^Ι

Qn = 0 027 (20 x20) (72 - 13.8) = 629 Btu / hr or 0.18 kW

5-26 Rins = # = 4'1z (rrr _f( _r)l εtu

C = O.24Btu/(hr -'f( -F) or 1.36 Wl(m'z - c)

Excerpts fτom this work may be reproduced by instruοtors for distribution on a not-for-prοfit basis for testing οr instructional purposes only tostudentsenrolledincoursesforwhichthetextbοokhasbeenadopted. ΑnyοιherreproductionοrιrαnsΙαtionofιhisνοrkbeyondιhαιpermiιιedbνSectiοη.s1[]7ny ιnρn{t]ιo 1076'ΙΙb;l.)ζl-l-"/-^4'-')-ι"} 'l-''':''' :"'''.Ι^''Ω'Ι l λ / .*'

87

Then from Fig. 5-8, U' = 0.85 Btu/(hr -f( -F) or 1.47 W(m-C)

Q = U'P (ti - to) = 0.85 x 300172 -101 = 15,8'10 Btu/hr or 4.63 kW

5-27 tι = 72 F (22 C) Assumed

R" = R5 + R1η, Ub = o.ο52 Btu / 1nr -tt2 _ 11 Table 5-9

Rfi,.,, =R1 +Ru6*Rqyp=(5.0) + 0.0 +(1 12.22) = 7.22

R^- 1 +7.22=26.5- 0.052

U" = o.o38 Btu / (hr -ft2 _F) or o'22ννl(m'z_c)

5-28 Ub : 1.14 from Table 5-9

Rn = ++ O.7 + (1 t 12'6)= 1.66 (m2-c)Λ//, Un = 0.60 W(m2-1.14

c)

or Un = O'1OO Btu / 1nr -ft2 -F)This does not account for the walls above grade.

5-29 U = 0'16 Τable 5-'10 (no finish)

Rn:++(t69)+ (1 t4.6) =6.611m2-c1 lw0.16 \ '

Un = 0.15 W/(m2-C) or 0.027 Btu / (hr -ft2 - r)

5-30 q/A= Un(ti -ts)= (ti -tt)/Rt=(tt-t)lR2Rl=Rgyp+R1nr+R1, R1 = (112.6)+ 0.7 +0.12= 0.90

tl = ti _UnR1(ti _tg) =20 - [1.05 x 0.9 (2ο - 10)]

EΧcerpts frοm this work may be reproduced by instructors for distribution on a not-for_pΙofit basis fοr testing or instruοtional purposes only to

students enτolled in courses for which the textbook has been adopted . Αny οther reproduction or trαnsltιιion of ιhis νοrk beyοnd ιhαt permiιιedhιi 9λ"};.-" ιηa ^'' ]Λ9 ^.}L- ιoaA ι L-;.^) c'-''^' '.''-:-ι'ι ^'"-^-:' "'-ι5υ{ιi ' . . .

88

5-31

t1= 14.6 C or 58F

R2 = Rrr, +Ri = ( I 12.6) +0'12='20

tz = 20_ [O.60 x O'2 (2o-1ο)] = 18.8 c or 65.8 F

q/A=Un (ti _tg)=(tι _t'')/R1; Rl=Ri +Rc

= 0.12 + (1 I 4.6) -- 0.34

tι=20_ (O.15) (O.34) (20-10) = 19.5 c or 67 F

5-32 C = 0.2 Btui(hr-ft2-F); Figure 5-8

8 = U'P (t1 _to); Ui, = 0.81 Btu/ (hr-ft-F) or 1.a W(m-C)

U'ni = 1.37 Btu / (hr - ft -F)

(a) q/P = 0.81 (70 - 5) = 52'7

(b) q/P = 1.37 (70 - 5) = 89'1

5-33 Q = Δt / R' ; Eq. 5-25; L>>Ζ'

R'=

Btu / (hr - ft) or 50.7 W/m

Btu / (hr - ft) or 85.6 W/m

L = 100ft

. Γzoo x121Γ^ tn(12x1OO/2x3o)l

_'nL-o--@J_3"100Ι12)

R' = 8. 12 x 1O-3 thr - F) / Btu

of pipe wall.

Which neglects the resistance

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Students enΙollod in courses fbr ινh'iοh the textbook hu, υ"., uJo|i"a. Αny oιher reprολucιiοn οr ιrαnslαιioi οf this νork beyond ιhαt permitιed

L- Q.nl;n-" ιnη ^.' 'Λο

: '' ^ ' '1-"4'ι "'

2πkL

6- 70-42 - =3,4488tu / hr;' 8.12 x 1O-'

orq=1.01 kW; 9=33 1W/mL

q/L = 34.488tu/(hr-ft)

5-34 Q=Δt /R'

5-35

R'g = 2π (1.4) 100=4.98x10-3 C/W

Neglect resistance of the inside film and the tube wall'

. 60-5 =11.04kwO = ----------c' 4.98 x 10--

Moisturewillmovetowardtheinside.Locatethevapor

retardent on the outer side of the insulation'

The insulation will beοome wet if the retardent is placed on the

inside or left out entirely and the plywood would probably \Λ/arp

and rot.

5-36 (a) Q/A = Uo(ti _to)=(tι _t1)/R1 =(ti _t)lR2

Ro = O'68 + O'45 +1 1 + 1'O +O'8 + O'17 = 14'1

Uo = O'o71 Btu /( ιrr - ft2 _F1

Rr = 0'68 + O'45 = 1'13( hr - f( -F) / Btu

R2 = O'68 + O'45 +11 -12'13(hr - ft2 -F) / Btu

tt=tι_R1Uo(tι_to)=7O_(1.13xO'O71)(7O-1O)=65.2F

Excerpιs Γrom ιhis \νοrk lnay be reproduced by insιructors for distribution οn a noι-for-proΓlι basis for ιesting or instructiona' purposes only to

students enrolled in courses foΙ ^γηι:ι lh. j:*tbook has *9,i'"Ιl'Ji."o"'i,-:';;;;;;;;;;;"Ζii"111i117; γ,:y:γ:'::'|"*o

ιyt.pern:ted

90

tz=70-(12'13 x 0.071) (70-10) = 18'3 F

(b) At 70 Fοο, 3O%o R.Η. and possible leakage of air to surfaces 1

or 2-

tdp=37F<65F-Πocondensationexpected

(c)Since|z=lS.3Fismuchlessthanthedewpoint,condensationwould oοcur'

Place vapor retardent at the location of interface 1'

5-37 Assume infiltration is negligible

ufAf (ti - t") = UwA*(t. - to)+ U'P(t. - to) = rilcp(tc - to)

* _ UrAtt, +(U*Α* +U'P+rhcp)to',ti=72F; to = 1o Fιc _

(UtAt + U*Α* + U'P + rhc, )

Αssume 1.5 in. of wood floor, Pine; Rwooο = '1'5l0'8 =1'88

U, =1; Rt =O'92+1'88+O'92 =3'72;Ut=0'27'Rf

UΛr= 0'27 x 30 x 60 = 484 Btu/(hr-F)

rr -a'R* =0.68+(6/15)+ O'17 =1'25',U* = O'80"*_Rw νv

U*A* = O.8O x 2(30 + 60)2 -- 288 Btu/(hr-F)

U'P=1.8x(30+60)2=324,h.p = 20 x 0.075 x 60 x o '24 = 21 '6 Btu/(hr-F)

+ -484 x72 + (288 + 324 + 21'6)10

= 36.85 Fιc_- 484+288+324+21'6

5-38 (a) Q/A = U(ti -to) = (tr-to)/R1;

Uz=0.112', Rr=O.17 + 0.33 + 4'17 + 2'22 = 6'89 (construction 2)

ExcΘrpts from this work may be rοproduced by instructors for distribution on a not-for-pτofrt basis foτ 1e^sting or lnstructional purposes only to '

students enτolled in courses fοr whiοh the textbo"t ι-'", υ"* "J"p

t.d. ,ι-ny otlr", ,rproEu.ιiοn οr ιrαnslαιioλ of ιhis νork beyond ιhαt permitιed

91

t1 = (6.89 x 0.112) (72-0) = 55.6 F

(b) U : 0.211 (construction 1)

Rr = 0.17 + 0.33 +2'22 = 2'72

ti =to+R1U(ti -to)=O + (2.72x0.211)(72-0) = 41.3F

(c) lf room air leaks into the air space for the case of no roof deck

insulation (b) there could be some condensation since t6p = 50

F at72 F and 45% RΗ. With the insulation, no condensation

would be exPected.

5-39 (ti -to) / Rr = (ti -ti) / R1

R, = 4'5 or O.79 1m2-cyw ; Τable 5-4a (Const. No.1)

Rτ = 0.68 + O.45 + O.94 =2.O7 1nr-tt2_F1/Btu or 0.365 1m2-cμru

Between Furring and block

tl = ti _

ft,' _ to) = 22_ ffiιr'+17) = 3.98 C or 39 F

tdp = 9.5 C, Assuming room air can diffuse into the air space,

condensation likely will form on the concrete block surface.

Therefore, place vapor retardant on inside surface of gypsum

board. Use foil backed retardent. Retardent must not touch

concrete blocks!

Exοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional purposΘS onιy to _

;;#;;;Ι;ln .o,,'",'to. *ni:| lhΘ j:;boοΙ has been adopted. o'l

"!!1: ,.,-r.':'!!i|',?! ":::::i:::::":!,*r::!'Ρ'n! 'y,'o*Ψ*o ,

92

5-40 U1A1(ti-tn) + UaΑ+(ti-tn) = 2U3Α3(tn-to) + lJzAzftn-to)

,^ _ (UlAl + UzΑz )ti + 2UgΑgto + UzΑzto'' 2υsΑg + UzAz * UlΑl * UqAι

UrAr = 0.09 x 8 x 20 = 14.4',

UzAz = 0.09 x 8.54 x20 = 15.4',

U3Α3 =0.09x3x8

UιAι=0.09x3x20=5'4

r _(4.4+15.4)70+(2 x 0.8 x o) + (15.a x o)L6= -S5.8F" (2 x 1.08) + 15.4 + 14.4 + 5.4

Place water pipes in this space with some caution.

Uf Αf (ti-tb ) = (U*Α* + UυtΑt )(tυ -tg ) ; ti = 72 F

tg = turg -Α - 37.6 _23 = 14'6 F or B C

1Ut =

& ; Rf = (2 x 0.92) + (1 5/0.8) + 2.1 = 5.82

Carpet and Fibrous pad assumed, Ur = 0.172 Btu/ (hr - ft2 - F)

U* =.164 Table 5 _ 9; Uot = 0.029 Τable 5 _ 1o

ltΑrti + ( U*Α* + Ubf Αf )tg

= 1.08

5-41

tb=U1Α1 + U*Α* *UυrΑr

r, _(0.172x400 x72) + (.1G4 x 80 x 7 + 0.029 x 4OO)1a.6l'b-

to = 30.3 F or -0.95 C

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Excerpts frοm this work may be reproduοed by instructοr9 jor distribution on a not-for-prοΓrt basis fοr

testing or instructiοnal purposes on1y to studeπs enrοlled in οourses for whiοh the textbook has been

adopted. Αny other reproduction or trαns,lαtion of this work beyond thαt permitted by Sections ]07 or

Ι0B οf the ]976 Unitei Stαtes Copyright Αct withλut the permission of the copyright owner is unlαwfuΙ'

Requests fοr permissioln or furthir"infοrmαtion should be'αddressed ti the Peimission Depαrtment' John

iiria sonλ, nr, ] 1 ] Riνer Street, Hoboken' NJ 07030'

6-1

CHAPTER 6

Refer to Table B-1 . The computer program PSYC may be used to find the

humidity ratio from t66 and assumed 100% RΗ'

Design relative humidity is determined by possible condensation on inside

of glass. Find glass surface temperature (which is the maximum dew-

poi;t temperature of the inside air allowed)'

q/A = U(t' - to) = Cr(tr - b)

tι= 72"F; t., = glass Surface temperature

U = 0.65 Btu/(hr-ft2-F)' Table 5-5a

111, hi = 1.46 Btu/inr-ft2-f)

c1 uhiCι = 1'172 Btυl (rrr-ft2-F1

6-2

WindDirection, deg.

CCW from N0.0

0.0

0.0

0.003

ο.0

0.003

140

290

10

340

360

20

6

13

4

12

I7

11

-2

-9

24

18

28

(a) Pendleton, OR

(b) Milwaukee, Wl

(c) Anchorage, ΑL

(d) Norfolk, VA

(e) Αlbuquerque, NM

(f) Charleston, SC

94

tr=Uti + to (Cr - U)

cl

6-3

* RH = 60 o/o would probably be uncomfortableRΗ = 40 to 50% would be more realistic

Assume that the weather strip does not change the conveοtive heat loss.

From Figure 6-2, Cp = 0.3. Using Eq. (6-7b) with the air density of 0 'F,the pressure difference due to wind is

ο: [o

.086ψ!\'( ls*ot * |.467 fi l'\ft')\ mph)

ΔP. =

z.(y.rrΙbm_ ft)ι lbf_s')

(o.rnr.o t"'tt )ι lbf l ft')

ΔP* = 0'037in'wg

Αssuming slight stack effect, ΔP ^y

0.04 in. water

Using Table 6-1 and Fig. 6-1,

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studentsοnrolledincοursestbrwhichthetextbookhasbeenadopted. ΑnyοtherreproducιionorιrαnslαιionofιhisνοrkbeyondthαtPermiιιedby Sections Ι07 or Ι08 ofthe ]976 United Stαtes Copyright Αctνιthout the permιssiοn ofιhe copyrighι oνner is unlaννful-

CitylndoorTου, oF

OutdoorΤου, oF

tr=top,

oF

Design orMax.

RH-%(a) Caribou, ME

(b) Birmingham, ΑL

(c) Cleveland, oΗ(d) Denver, CO

(e) San Francisco, CΑ

(g) Boise, lD

Rapid City

72

72

72

72

72

72

72

-10

23

b

3

39

-'16

I

35.5

50.2

42.6

41.3

57.3

32.8

44.0

26.2

46.1

34.6

32.9

59.9*

23.6

36.5

95

Loose fit with non-\Λ/eather-stripped, K - 6; a lL = O.75 cfm/ft

Loose fit with weather-stripped, K= 2; Q/L = O'24 cfmtft

Total length of crack, ι = [(3 x 3) + (2 x 5)] x 9 = 171 ftUsing Ll2for calculation, then

Q,, = 0.75 x 17112= 64.1 cfm, Q, = 0'24x17112= 2O'5 cfm

INow Q, = rh cr(t1 - to) = v cp(t; - tr)

Q,r-8,, -At-4, -Q,, Qt

ΔP., =

64.t-20.564.r

= 0.68

6-4

or a reduction of 68% in sensible heat loss.

Also, (Kl - K)lΚ1= β-2)16 = 0.67 or 670/o Reduction.

From Fig. 6-2, Cp = 0'52tor windward wind'

Assuming standard sea level air density, the pressure difference due to

the wind speed of 13 m/s is

_ 53.6Pα

z.( ι.okg _ *

t i/-s'

(a) From Table 6-2, K = 1 for tight-fitting.

Then, from Fig. 6-1, Q/L = 0.60 L/m-s

Q = 0.60 x (0.9 + 2'0) x2 = 3l8_LΔ

Αssuming that the wind speed and wind direction are the Same as the

given conditions for the bank at Rapid City, SD, the heating load (at -20'6

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students enτol1ed in οourses fbr which the textbook has been adopted. Αny oιher reprοiucιion or ιrαnslαιion οf ιhis ννοrk beyond ιhαt permitιed

by Sections 107 or ]08 ofthe Ι976 rJnited States Copyright ΑcιΙνilhouι ιhe permission οfιhe copyrighι owner is unlαwful'

Βry]s)

a, = (: + ε * ο. o ο 1

Τ ) (' .#l

ι r r,

* }r)rr, _ 1_zo.o1)" c = 393.tW

--σ6-.C outdoor temp. and 72'C indoor temp.) due to the door infiltration can

be calculating using Eq. (6-2b) as:

(b) From Table 6-2, Κ = 2 for average-fitting'

Then, from Fig. 6-1, Q/L = 1'25 L/m-s

Q = '1 .25x (0.9 + 2.0) x 2=7-25--Lls

α, _(', .zs* 0 OO'+) ('.rrfil ι rrrn?)o, _ e20.6))" C = 819 0W

(c) From Table 6-2, K = 6 for average-fitting'

Then, from Fig. 6-1, Q/L = 3.40 L/m-s

e = 3.40 x (0.9 + 2.0) x2 = 19f2-Lls

α, =(lν τ2* '001c) ('.#l ιrr, *}a)o'_e20.6))"C

_2227'6W

From Figure 6-2, Cρ = 0.52' Using Eq. (6-7b) with the standard air density,

the pressure difference due to wind is

r2

Λ p _ o'sz (o oτ οs#) (zz-en- ι κl #)' ( o.rnro :n:γ

r, ^\ι\ΓΙυ- ( b^_f) ι lbflft')

z.| ιz.17':::-_!- |

ι lbf _s'z)

ΔP' = 0't35in'wg

Neglecting stack effect and pressurization, ΔP - 0.135 in' water

From Table 6-1, K = 2 for average-fitting with non-weather-stripped.

From Fig. 6-1, Q/L = 0.60 cfm/ft.

L" = [(3 x2.5) + (2 x 4)]x3 = 46.5 ftExcerpts from this work may be reproduced by instruοtors for distribution on a not-for_profit basis for testing or instruοtronal puφoses only tο

studentsenrοlledinοourses1brwhiοhthetextbookhasbeenadopted. Αnyotherreproducι'ιonorιrαnsιαιionοfthisτνοrkbeyondιhαtpermiιted

by Sectiοns 107 or ]08 oftnn isri {Jniιed SιαιeS Copyrighι Αcιτνιιhοuι ιie permisiiοn ofιhe cοpyrighι oνner is unlαwful'

6-5

b-b

Q = 0.60 x 46.5 = 27.9 cfm.

(a) The wind effect is assumed to be independent of height and pressure

differences due to wind are the same as those given in Ex. 6-1.

3'd Floor: ΔP"/Cο = 0.037; ΔP, = 0.037 x 0.8 = 0.03 in. water

orientation ΔP, ΔP* ΔPτWindward 0.03 0.066 ο.ο96SidesLeeward

0.03 -0.066 -0.0360.03 -0.033 -0.003

gth Fιoor: ΔP./Co _ -0.100; ΔP, - -0.100 x 0.8 = -0.08 in. water

orientation ΔP. ΔP, ΔPτWindward -0.08 0.066 -0.ο14

SidesLeeward

-0 08 -0.066 -0.'146

-0.08 -0.033 -0.1 13

(b) For Bitlings, MT, design conditionS are to = -7oF, tι= 72"F, Φι= 28o/o.

From Table 6-3, K = 0.66 for conventional οurtain wall.

Αir will infiltrate on windward side only on 3'd floor.

Windward - 3E floorQiA = 0.15 cfm/ft2; Q = 0.15(120 x 10) = 180 cfmThen 9 "

= (1 80 x 60/1 2.4)(0.24)(72 - (-7)) = 16,514 Btu/hr

Q,. = (180 x 60/12.4)(0.005 - 0.000)1060 = 4,616 Btu/hr

Qt = Q, + 8r. = Ζ*1-3oBtu1hΙ [3'd Floor]

gth Floor - All exfiltration on this floor.

Qt = oοΞtuΔr 19th FlooηEXοerptS from this wοrk may be reproduced by instructors for distributioIr on a not-for-pro1it basis for testing or instruοtional puφoses only tο

students enrolled in courses tbr which the textboοk has been adopted. Αny οιher reprotluctiοn οr trαnslαtion of this νork beyond ιhαι permitted

by Secιions Ι07 or ] 08 of ιhe Ι 976 t'Jnited Sιαtes Copyrighι Αcι 1υiιhouι ιhe permission οf the cοpyrighι οlυner is unΙιrννful.

98-

6-7

(a) Windward Doors: Double vestibule type

ΔPτ = o.146 in' water, assume 1/8 in' οracks

Q/L = 16 cfm/ft tFιg. 6-7] , L -- 32 ft [Ex' 6-2]

Q = 16 x 32 x0'7 = 358 cfm

(Assume30o/oreductionforvestibuledoors)

6-7 (Cont.)

Side Doors: Double vestibule tYPe

ΔPτ = -o'o52 in' water' 1/8 in' cracks

Q = O.O (negative pres' dιff')

(b) Windward: ΔPτ = O146 in' water' K = 0'66 lΤable 6-3]

Q/A = o.2o cfm/ft2 [Fig. 6-6]

A=120x10=1200ft2Q = 0.29(12ο0) = 240 cfm

Sides: ΔPτ = -O'O52 in' water' K = O'66 [Table 6-2]

8 = O.O (negative pres' dιff')

UULeeward: ΔPτ = O'047 in' water, K = O'66 [Fig' 6-6]

Q/Α = O.1Oo cfm/ft2, A = 1200 ft2

Q = O.1οO(1200) = 120 cfm;

Total infiltration for the walls is

8* = 240 + O.O + 120 = 360 cfm

(c) Totat infiltration is sum for doors and walls'

From Εx.6-2'for leeward door' Q = 179 cfm'

Then the total door filtration is

Qo=358+179=537cfm(neglectinfil.duetotraffic).Andthetotalwallinfiltrationis360cfm,thenQτ=897cfm

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by Sectiοns 1 07 or ] 08 o7 ιn' to:ii'inιir-d Sιαιes Cοpyrιgh' 'ι'i"rii'i"'' 'i" per-ission of ιhe cοpyright oινner is unΙανfuΙ'

6-8

9V

For Charleston, WV: to = 1 1oF' ti = 70oF

q, = (897 x οοll \ 'τz)(o'z4)(70-11) = 65'ο25 Btu/hr

q. = (897 x 60/1 1'72)(1060)(O'OO5 - O OOO) = 24'338 Btu/hr

q = q" + 9. = 89,363 Btu/hr

(a) Assuming standard sea level air density, the pressure difference due

to the wind sPeed of 20 m/s is

( o.o,urtψ\'(zo*pt *1.461ιL:\ /

,,_ |υ.Ι9/+ tbfl fi,J =o 197in,water

UUWindward: ΔP*=O197xO25=ooo:

χ- } E:lι, n*l:Leeward: ΔP* = O"197 x (-0'5) = -0'099 in

AssumPtions:1) temperature differencΘ, tι _ to' = 40oF

zi tιle neutral pressure level is at floor 9'

3) the floor height is 12 ft', and

4) Cο = 0'80'

Then, from Fig.6-5,

Floor 1: h = 108 ft., ΔPr/Co = O.13, and ΔP" = O'13 x O'8O = 0'104 in' water

Floor 5: h = 60 ft., ΔP, = 0'065 x O'8O = 0'052 in' water

Floor 15: h = 72ft., ΔP, = -O'O85 x 0'8O = -0'068 in' water

Floor 20: h = 132ft., ΔPS = -Q'160 x O'80 = -0j28 in' water

Windward Leeward

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;}i::j;μιfii,Z'i'"fλ!"in:ii'ini,a s**, copy,ign,'a;,\iιiio"i''Ι, i,.ι'i-n of ιhe coρyrighι owner is unlαwful

& r *--,-,

Flnοr ΔP* ΔP" ΔPτ ΔP* ΔP" ΔPτ

1

5

1520

0.0490.0490.0490.049

0.1 ο40.052-0.068-0.128

0.1 530.101-0.019-0.ο79

-0.099-ο.099-0.ο99-0.099

0.1040.052-0.068-0.128

0.005-0.047-0.167-0.227

100

(b)

6-8 (Cont.)

-o.2 -ο.1 0.0 ο.1 Ε.2

ΔR. ilt. ιlrateι

*E*-lΛfiΠ$ffard ".-,t - LEθΛrard

lnfiltration - Windward Sides, from 1't to 13th Floor

Leeward Sides, 1't Floor onlY

Exfiltration _ Wind\Λ/ard Sides, from 14th to 2oth Fιoor

Leeward Sides, from 2nd to 2Oth Floor

(c) 1rt floor, lnfiltration on all sides - through doors, walls and fixed

windows

Windward Walls: from Table 6-3, K = 0'22 for tight fitting.

f rorn f ig. 6-6, Q/A = O.OB cfm/ft2'

A=(1οο+60)1 2=1920f(Q = O.O8 x 1920 = 154 cfm

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.* \λd& tl

#

1υ1

Leeward walls: from Table 6-3, K = 0'22 for tight fitting'

From Fig. 6-6, Q/A = O'OO5 cfmlft2'

A = (1OO + 60)12 = 1920 f(Q =0.005x1920= 10cfm

windward Doors: from Fig. 6-7, Q/L = 17 cfm/ft for 1/8 in' crack'

6-8 (Cont.)

For vestibule doors, assume a 35% reduction'

Q/L = 17 x 0.65 = 1 1'05 cfm/ft

L = (3 x 6.75) + (2 x 6) = 32'25 ft

Q = 11.05 x 32.25 = 356 cfm

Leeward Doors: from Fig. 6-7, Q/L = 1.5 cfm/ft for 1/8 in' crack'

Γor vestibule doors, assum e a 35o/o reduction'

Q/L = 1.5 x 0.65 = 0'975 cfm/ft

Φ = O.975x32'25 = 31 cfm

Then,totalinfiltration(neglectingtrafficeffect)isQtot = 154 + 1O + 356 + 31 = 551 cfm'

(d) and (e) lnfittration rate is zero due to negative pressure differentials for

the 1Sth and 2οth floors'

6-9

For Minneapolis, MN: to = -1 1oF' t1 = 70"F '

[Note:Δt=70-(-11)=81oFisinconsistentwithProblem6-8whereΔt = 40"F was used; however, an error is assumed to be minor]

(a) From Prob. 6-8, Qτ = 551 cfm for 1't floor'

q, = (5s1 , dotl 2'15)(0'24)(70 - (-11)) = 52'896 Btu/hr

o, = lbSt x 60/1 2.15)('1060)(0.O04 - o.ooo) = 11,537 Btu/hr

Qt= Qr* Qr. =64,433 Btuihr

(b) and (c) qt = O'O due to zero infiltration

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tie permιsiιon of the copyrighι owner is unlα:wful'

102

6-10

For Des Moines, lA: to = -4oF, ti = 70oF'

Transmission heat loss (negtecting infiltration) through windows, doors,

walls, and roofs can be determined by Eq. 5-19 as:q = UA(tι -to)

Windows: A = (3 xa)1 2= 144ft2;

From Table 5-5a, U = 0.55 Btu/(hr - ft2 -'F1;q = 0.55(144)(70 - (-4)) = 5,861 Btuihr

Doors: A = (3 x 6.75)1 2=243ft2',From Table 5-8, U = O'28 Btu/(hr - ft2 - 'F);

(assume panel with metal storm door)q = 0 28 (243)(70 - (-4)) = 5,035 Btu/hr

wails: A = 8[(36 + 64)21- 144- 60.75 = 1395.25ft2',

From Table 5-4a, U = 0.14 Btu/(h r -ftz - 'F);

Q = O. 14(1 395. 25)(70 - (-4)) = 14,455 Btu/hr

Roof/Ceilinq: A = 36 x 64 = 2304 ft2',

Ξrorn Example 5-3, U = O.83 Btu/(hr _ ft2 _ 'F1;

q = 0.083(2304)(70 - (-4)) = 14,151 Btu/hr

Transmission heat loss through the slat-on-grade floor can be determined

bY Eq 5-23 as:Q = U'P(ti -to)

Floor: p = (36 +64)2=200ft;u' = o.8o Btu/(hr - ft - F), from Fig' 5-8 (assume insulation R-value

of 5.4 (hr - ft2 -'F)/ Btu and d = 2ft)'q = 0.8(2OOX7O - (-4)) = 11'840 Btu/hr

Finally, total transmission heat loss is the sum of all heat losses;

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6-1 1

6-12

LP, -

ΔP* =

qt = 5l.ΞΖBtuΔt

From Figure 6-2, Cp= 0'52. Using Eq'

the pressure difference due to the wind

103

(6-7b) with the standard air density,

of 15 mph is

0.0765ψy\-fr' )

(0 5η[

9s =

Qr=

Qt=

z.(nlιlbry_ ι_ ι lbf _s2

0.058 in. water

For a low-rise building, neglect stack effect and pressurization' thus

ΔP1 = 0.058 in. water

From TabIes 6-'1 and 6-2, Κ = 1 for tight-fitting windows and doors'

From Fig. 6-1, Q/L = O.'13 cfm/ft'

L" = [(3 x 3) + (2 x a)]x3 + (3+6.75)x2x3 = 109'5 ft

e = 0.13 x 109.5 = 142 dm,

(14.2 x 60/1 2.15)(0'24)(70 - (-4)) = 1,245 Btu/hr

(ιι'zx60/12'15)(1ο6OXο.Oο5_0.00ο)=372BtulhrQ, + Qr. = 1,617 Btu/hr

For Ηalifax, Nova Scotia: to = 2oF, ti = 70oF'

Refer to Problem 6-10 for other data'

Windows: q = 0.55(1 44)(70 - 2) = 5'386 Btu/hr

Doors:q=0.28(243)(70-2)=4'627Btulhrwails: Q = 0.14(1 395. 25)(70 - 2) = 13,283 Btu/hr

κootrcuιιιno: q = O.083(23O4)(70 -2)= 13'004 Btu/hr

Γ*r, q = O.8(2OοX70 _2) = 10'880 Btu/hr

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students enrolled in οourses fbr which the textbοok t u, υ.", uαφi.J . Αny other reproλ3''ιo' or ιrαnslαtionω ιhis νork beyond thαι permitted

by Sectiοns ] 07 or 1 0B q 'n""i

/ii i"i.d Sror", copyrιsnt ii 'Ιiiiouι ιi" per*isiion of ιhe copyrighι οwner is unΙαwful'

104

6-13

6-14

6-15

6-16

Total: qt = 4Ζl€oBtu/hr

Memphis, TN; to = 21 oF; ti = 70"F

R*= 0.92+ 1.55 + 0.99 + 1.77 +0.17 = 5.4 (Tables 5-1 a'5-2a)

U* = 115.4 = O.'185 Btu/(hr - ft2 - 'F)Us = 0.81 Btu/(hr - f( - "F) (Table 5-5a)

Αs = 6xax3 = 54 ft2

n* = (40xl O)-54 = 346 ft2

q,= 0.185 x 346 x (70 - 21) = 3,136 Btu/hr

qs = 0.81 x 54 x (70 - 211= 2,143 Btu/hr

Qtotrl = 5zβ auk!

Concord, NH; to = -2F, ti = 70oF

R*= 5.4-0.99+ 3.0=7.41U* = ο.135 Btu/(hr _ ft2 - "F)

Us = 0.81 Btu/(hr -f( - 'F) (From problem 6-13)

q* = 0.135 x 346 x [70 - (-2)]= 3,363 Btu/hr

qn = 0.81 x 54 x (72) = 3,149 Btu/hr

Qtotrι = οβ1ΖBuer

lnstructor suPPlies solution.

(i' - iu)(a) q=9svs

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105

. qν" (280'00Ο)(14.6)Ο^= :-=:-_ =^251 cfm19(S _

i._iu β2.7 _21.8)60 v!

(b) q = rh cp(t, - t..) = 9ε

cp(ts - tr)vs

. qν, (250,00Ο)(14.6)Q' = ;r=δ= (O24X1 15_7O)ω = 5'633 cfm

6-17

SHF = Q./(9. + α") = -100'999 ===,

= -3.O34st (1 33, ooo - 1 oo, ooo)

Locate states, and οondition line and heating pΓocess on psychometric

chart.Q. = rh cp(t, _ tr) or Γh . = q

'/cr(t, _ t')

Γh. = 100,000/(0'24 x20) = 20,833 lbm/hr

Q. = Γh. X vr/6O = 20,833 x 14.05/60

Q" = 4,878 cfm or about 4,900 cfm

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106

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tffini

p

7-1

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Chapter 7

First, find longitude from Table B-1a

Then, convert Daylight Saving Time to Local Standard Time using Eq. 7-5

Next, determine the equation of time from Table 7-2

Finally, determine Local Solar Τime using Eq. 7-6

Τhe following table Summarizes the solutions of the problem.

Norfolk, VA 76.2Lincoln, NE 96.75

Casper, WY 106.47

Pendleton, OR 118.85London, UK 0.45

75 9:00:00 ΑM90 1:00:00 PM105 10:00:00 ΑM120 3:00:00 PM0 7:00:00 PM

Local^.-"":' Eouation LocalSolarδtanοarα : -'oτ llme llmeilme

8:00:00 ΑM -2'41 min 7:52:47 A|'Λ

12:00:00 PM -2.41min 11:30:35ΑM9:00:00 AM -2'41 min 8:51:43 ΑM2:00:00 PM -2'41min 2:02:11PΜ6:00:00 PM -2.41min 5:55:47 PM

Standard DaylightLocation Longltuαe' Meridian, Savingsoνν Τime

7-2Ηour angle (negative for morning and positive for afternoon) can bedetermined by

h:Ι5* (LST _l2)

(a) h = 15*(8.19 -'12:00) = 15*(-3.683) = -55.25 deg.

(b) h = 15*(10:03 - 12'.00) = '15*(-1 .950) = -29.25 deg.

(c) h = 15*(15:46- 12:00)= 15*(3.767)= 56.50deg.

107

(d) h = 15"(12.01 - 12:00) ='15*(0.017)= 0.25 deg'

7-3 Αt sunset and sunrise, β =0"; sin(B) = ρ

From Eq. 7-8; οos(/) ο οos(h). οos(δ) - _sin(/). sin(δ)

οos(h,,) = οos(ft,,) = _ tan(/)' tan(δ)

The following table summarizes the solutions of the problem.

Location Latitude,'N 'i"J;::',3" Cos(h) ^ffi::.

',|1:ff ',tι''iBillings, MT 45.8 20.6 -0.3865 112.7 4:29 AM 7:30 AM

orlando, FL 28.43 20.6 -0.2035 101.7 5:13 AM 6:46 AM

Anchorage, AL 61.17 20.6 -0.6829 133.1 3:07 ΑM 8:52 AM

Honolulu, Ηl 21.35 20.6 -0.1469 98.4 5:26 AM 6:33 AM

Note earlier sunrise at greater latitudes

7-4/ = 33.0 deg. N

h = 15.(9-12) = -45.0 deg.

on Sep 21, δ = 0.0 deg.

From Eq. 7-8; sin(B) : οos(/). οos(ft). οos(δ) + sin(/). sin(δ) = 0.593

β = 36'37 deg'

sin δοos/ _ cos δsin / οos hFrom Εq' 7-11 ; υυs ψ =

-"o,

βcos{ = -0.478

Φ = 118.57 deg. (clockwise from north)

Exοerpts from this work may be reDroduced bv instnlctοrq f'οr rliqtrih,lfinn nn a nnt-fnr_nrnfit hqqis fοr testinρ or instructional puφoses only to

108

7-5At sunris e, β _ 0"; sin(B) = g

From Eq. 7-8; οos(/) ' οos(h) ' οos(δ) _ _sin(/) ' sin(δ)

οos(h,. ) - _tan(/)' tan(δ)

(a) June 21'' δ-- 23'45 deg; / = 58 deg'

cos h = -O'6942; h = -133'96 deg' or -8'93 hours

Sunrise is at 3:04 AM (Solar Time)

7-6

= O.751

Φ = 41.33 deg. (clockwise from north)

(b) Dec21'' δ= -23'45 deg; / = 58 deg'

cos h = 0.6942; h = -46'04 deg' or -3'07 hours

Sunrise is at 8:55 AM (Solar Time)

. sinδcos/ _οosδsin lcoshFrom Εq'7-11; οosΦ=" =-0'751

Φ= 138.67 deg. (ctockwise from north)

Maximumsolaraltitudeangle,βwilloccuratsolarnoon,h=0

^^^ λ_ sin δ cos / - οos δ sin / οos h

From Eq. 7-11; ν"oo _ cos β

7-7

]09-

From Eq. 7-1O, β^u*:90 _ Min\, _ u,)

From Table 7-2, |δ'u"l = 23.45

(a) Denver, CO: I = 39.75 deg. N.

For north latitude, / is positive and greater than |δrrr| so we need largestpositive value of δ.

From Table 7-2, δ'u" = 23'45 deg. and hence β'", = 73.70 deg.

Therefore, maximum solar altitude angle occurs at solar noon on June 21.

(b) Lansing, Ml: l= 42.77 deg. N.

For north latitude, / is positive and greater than |δrrr| so we need largestpositive value of δ.

From Table 7-2, δ'u"= 23.45 deg. and hence β'u"= 70.68 deg.

Therefore, maximum solar altitude angle occurs at solar noon on June 21.

(c) Sydney, Αustralia: / = 33.95 deg. S.

For south latitude, / is negative and |/| is greater than lδ'u"l So \Λ/e need

largest negative rralue of δ.

From TabΙe 7-2, δ'u, = -23'45 deg. and hence β'", = 79.50 deg.

Therefore, maximum solar altitude angle occurs at solar noon on Dec 21.

Longitude'. Lt= 100 deg. W

Local Standard Time. LCT = 3:30 pm

on Nov 21, Eoτ = '13.8 min

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110

7-8

7-9

Using Eq. 7-6, LSf = 15.50 - (100-90).4 /60+ '13.8/60 = 15.063 Hr or 3:04pm.

Latitude: l = 37 '5 deg. Ν

Ηour angΙe: h = 15-('15.063-12) = 45.95 deg.

on Nov 21, δ = -19.8 deg.

Using Eq' 7-8 to calculate solar altitude, β= 21.36 deg.

Then using Εq.7-11 to calculate solar azimuth; Φ = 226'56 deg. (clockwisefrom north)

Surface azimuth; ψ = 12+180 = 192 deg. (clockwise from north)

Finally, using Εq.7-12 to calculate wall-solar azimuth

y= 1226.56-1921= 34.56 deg.

Using Εq' 7-13b to calculate angle of incidence for a vertical surface

θ = 39.92 deg.

Using Εq' 7-13a to caIculate angle of incidence for an inclined surface

For surface tilt = 70", θ= 32.30 deg'

For Ottawa, Ontario on July 21,

Longitude. Lt= 75.67 deg. W

Latitude: Ι = 45.32 deg. N

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Equation of Τime: ΕoT = -6.2 min

Dec|ination: δ= 20.6 deg.

(a) Eastern Daylight Savings Time: EDSI = 4:00 pm

Using Eq. 7-6, LS7- = 14.852 Hr or 2'.51 pm.

Hour angΙe: h = 15*(14'852-12) = 42.78 deg'

Using Eq. 7-8 to calculate solar altitude, β= 47"16 deg.

Using Εq'7-13c to calculate angle of incidence for a horizontal surface,

θ= cos-1(sin(47.16)) = 42'84 deg.

(b) At sunset, β = 0 and sinp = Q

cos(fr ): _tan(/). tan(δ)

Hour angle: h = 112.34 deg.

Solar time at sunset: LSf = 12 + hl15 = 19.49 hr or 7:29 pm.

Εastern Daylight Savings Τime can be calculated by

ΕDST = LST + (L' _ ΕSη(4min/deg-t4l) - ΕoT +L

EDSΓ = 19.49 +(75.67-75)-(4/60)-(-6.2/60)+1 = 20.638 hr or 8:38 pm.

7-10For Philadelphia, PA on July 21,

Longitude'' Lι= 75.25 deg. WLatitude: / = 39.88 deg. NEquation of Time: ΕoT = -6.2 minDeclination: δ= 20.6 deg.Eastern Daylight Savings Time: EDSI = 10:30 am

Εxοerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis fοr testing or instruοtional puφoses οnly tostudents enrol]ed in courses for which the textbook has been adopted. Αnyl other renro,l1ιft;Λh ^v

lνΛf.l^ιi^,^ ^{lL]d 'ιl^yL λo1'^') tι1nt ^trmιtted

Using Eq. 7-6, LS7Γ = 15.852 Hr or 3:51 pm.

Hour angle: h = 15-(15.852-12) = 57.78 deg.

Using Eq. 7-8 to calculate solar altitude, β = 49'42 deg.

Using Εq' 7-11 to find solar azimuth; Φ = 114.30 deg. (clockwise from north)

(a) Using Εq' 7-13c to calculate angle of incidence for a horizontal surface,

θ = cos-1(sin(49.42)) = 40.58 deg.

(b) For vertical surface facing southeast, Surface Tilt; α = 90 deg., and

Surface azimuth; ψ= 135 deg. (clockwise from north)'

Using Εq' 7-12 to calculate wall-solar azimuth , y= |114'3-135ι = 20.7 deg'

Using Εq' 7-13b to calculate angle of incidence for a vertical surface,

θ = cos-l(cos(49.42)cos(20 7ο)) = 52'52 deg'

(c) For inclined surface faοing south, Surface Tilt; α = (90-40) = 50 deg.,

and Surface azimuth', ψ= 180 deg. (clockwise from north)'

Using Εq' 7-12 to calculate wall-solar azimuth, y= l114.3-180| = 65.7 deg.

Using Εq.7-13a to calculate angle of incidence for an inclined surface,

d = cos-1 (cos(49.42)cos(65.70)sin(50)+sιn(+g.42)cos(50)) = 46.'t 1 deg.

7-11

112

7-12

7-13For Calibou, MΑ on July 21,

Exοerpts from this work may be reproduοed by instruοtors f,or distribution οn a not-Γor-profit basls for testing or instructional puφoses only to .

students enrolled in courses for which the texδook has been adοpted. Αny other reοrοιlu"',--:':::':::!*::ι::]'-υg:,oηd ιhαl permιιted

113

Longitude' Lt= 46.87 deg. WLatitude: l= 68.02 deg. NEquation of Time: EOT = -6.2 minDeclination: δ= 20.6 deg.Solar Parameters; Α = 346'4 Btu/hr-ft2 or 1093 Wm2, B = 0.186,

and C = 0.138Eastern Daylight Savings Time: ΕDSr = 2:00 pm

Surface Tilt; α = 60 deg.Surface azimuth, SW; ι/ = 225 deg' (clockwise from north)

Using Eq. 7-6, LSI = 14'72 Ηr

Hour angle: h = 15.(14.72-12) = 41 .58 deg.

Using Eq. 7-8 to calculate solar altitude, β = 36.04 deg.

Using Εq'7-11 to find solar azimuth; Φ__ 230.2 de9. (clockwise from north)

Using Εq.7-12 to calculate wall-solar azimuth, y= 5'2 deg'

Using Εq' 7-13a to calculate angle of incidence, θ = 7.45 deg.

Using Εq' 7-15 and clearness number of 1, Gryρ = 252'51 Btu/hr_ft2 or796.75 Wm2

Using Εq' 7-16a, Gρ = 250.28 Btu/hr-ft2 or 790'03 Wm2

Using Eqs, 7-18 and 7-2O, Gα= 26'13 Btu/hr-ft2 or 82'46 Wm2

Therefore, total clear sky irradiation is276.51 Btu/hr-ft2 or 872.49 Wm2

7-14Given lnformation:

Date: June 21Longitude' Lt= 96'0 deg' WLatitude: / = 36.0 deg. NΕquation of Τime: ΕoT = -1 .4 minDeclination: δ= 23.45 deg.

EΧοerpts fiom this wοrk may be reprοduοed by instruοtors for distribution on a not-for-profit basis Γor testing or instructional puφoses only to -

students enrolled in οourses fbτ whiοh the textbook has been adopted. Αny οther reρrολucιion οr ιrαJ'lsιalion οf ιhis wοrk beyond ιhαι permiιted

114

Solar Parameters; Α = 346.1 Btu/hr-ft2 or 1092 Wm" ' B = 0' 1 85,

and C = 0.137Central Daylight Savings Time: CDSI = 8:00 pm

Surfaοe Τilt; α = 90 deg.

Surface azimuth, SW; ι/ = 225 deg' (clockwise from north)

Reflectance from water; Ps = 0'25

Using Eq. 7-6, LSf = 18.58 Hr

Hour angle: h = 15-(18.58-12) = 98'65 deg'

Using Eq. 7-8 to calculate solar altitude, β = 7 '02 deg'

Using Εq' 7-15, G,νρ = 76'24 Btuihr-ft2 or 240'5 Wm2

lrradiation reflected from the ground can be determined by

Gn: PrF.r(sinβ +C)G'o

where F'ncan be determined from Εq'7-24'

ThereforΘ, Gκ = 2'47 Btu/hr-ft2 or 7'8 Wm'


Date: lιΛar 21

Latitude: / = 56.0 deg. NEquation of Time: ΕoΤ = -7'5 min

Declination: δ= 0.0 deg.Solar Parameters; Α = aoε.g Btu/hr-ft2 or 1164\Nlm', B = 0'149'

and C = 0.109Local Solar Time: LSf = 12:00 PmSurface Tilt; α = 90 deg'Surface Azimuth, S; ψ= 180 deg' (clockwise from north)

Clearness number; CN = 0'95Diffuse Reflectance from Sno\Λ/; ρn = 0'7

Hour angle: h = 0.0 deg.

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students enrolled in courses for which the textboοk has been adopted. 'qnν οthei ,eοroλucιion or ιrοnsιαtion of ιhis ιοrk beνοfrr] .hot nermiΙtρi

115

Using Eq. 7-8 to calcuΙate solar altitude, β= 34'0 deg'

Using Εq' 7-11 to find solar azimuth; Φ = 180.0 deg. (clockwise from north)

Using Εq' 7-12 to calculate wall-solar azimuth, 7r= 0.0 deg.

Using Εq' 7-13b to calculate angle of incidence, θ = 34.0 deg.

Using Εq.7-15, Gtvρ = 268.5 Btu/hr-ft2 or 847 '1 \Νlm2

Using Εq' 7 -16a, Gρ = 222'6 Btu/hr-ft2 or 702'3 Wm'

Using Εqs' 7-21 and 7-22, Gα = 33.O Btu/hr-ft' or 104. 1 \Νlm2

lrradiation reflected from the ground can be determined by

G^: PrF'r(slnβ+C)G'o

where F*n can be determined from Εq.7-24'

Therefore, GR = 62'8 Btu/hr-ft2 or 198.1 \Νlm'


Date: Aug 2'1

Latitude: I = 32.0 deg. NEquation of Time: ΕoT = -2'4 minDeclination: δ= 12.3 deg.Solar Parameters; Α = 350.9 Btu/hr-ft2 or 1107 \Λllm2, B = 0'182,

and C = 0.134Local Solar Time: LSf = 10:00 amSurface Tilt; α = 45 deg.Surface azimuth, SW; ι/ = 225 deg. (clockwise from north)Diffuse Reflectance from ground; ρn = 0.3

Hour angle: h = -30.0 deg.

Using Eq. 7-8 to calculate solar aΙtitude, β = 56.1 deg.

Exοerpts from this wοrk may be reproduοed by instruοtors for distribution on a not-for-pιofit basis for testing or instructional puφoses only tostudents enrolled in οourses for which the textbοok has been adopted. Αny οιher renrοdυcιiοn or trαnsΙαtion οf thi't τνοyk hcνοnι] thηl nPrmittρ/]

116

Using Εq.7-11 to find solar azimuth; Φ= 118.7 deg. (clockwise from north)

Using Εq.7-12 to calculate wall-solar azimuth, r= 106.3 deg.

Using Εq.7-13b to calculate angle of incidence, θ = 61.5 deg.

Using Εq' 7-15, Gruo = 281'8 Btu/hr-ft2 or 889'1 Wm2

Using Εq'7-16a, Gρ = 134'4 Btu/hr-ft2 or 424'0 Wm2

Using Eqs. 7-18 and 7-2o, Gα= 32'2 Btu/hr-ft2 or 10'1 '7 \ΝΙm"

Using Eqs. 7-23 and7-24, GR = 11'9 Btu/hr-ft2 or 37 '7 \'ΙΥlm2

Using Eqs. 7-25 , Gt= (34.4 + 32.2 + ',11.9) = 178.6 Btu/hr-ft2 or

= (424.0 + 101 .7 + 37.7) = 889.1 Wm'z

7-17The following results are determined from a computer program employing

equations in the book from Eqs. 7-6 to 7-26'

Following tables summarize input and output data calculated for southwest-

facing vertical window at32 deg. N latitude, 90 deg. W longitude, for all

daylιght hours of a clear day on July 21with ground reflectance of 0.2 and

clearness number of 1.

lnput DataLongitude 90 deg

Standard Meridian 90 degEOT -6.2 min

Latitude 32 deg

Declination 20.6 deg

Surf Azimuth 225 degSurf Tilt 90 deg

A 346.4 Btu/hr-ft2

B 0,186c 0,138cN1

RΗoG 0'2

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students enrοlled in courses tbr ιvhich the textbοοk has heen adonted. Αnν othcr fρn/^.h'"tinn nr trn'ζΙntinn n{1|1jq aιnrΙ' holnι'] th'ι Δ29Φ;fio)

117

Output Data

cDsr LsI h, o β,

. Φ,. Ψ, o θ, " Gruo* Go* Ga* Gρ* G,*

7.00 5.90 -91.55 9.50 71.57 153.43 151.90 112.19 O.OO 6.97 3.40 10.378.00 6.90 -76.55 21.78 78.63 146.37 140.64 209.84 o.OO 13.03 10.68 23.719.00 7.90 -61.55 34.38 85.69 139.31 128.74 249.18 o.OO 15.47 17.51 32.9810.00 8.90 -46.55 47.09 93.60 131.40 116.76 268.71 o.OO 16.69 23.39 40.071 1.00 9.90 -31.55 59.65 104.24 120.76 104.98 279.23 o.OO 17.34 27.g5 45.2s12.00 10.90 -16.55 71.33 123.59 101.41 93.63 284.65 o.OO 20.57 30.90 51.4613.00 1 1 .90 -1.55 78.52 172.69 52.31 83.01 286.52 34.88 24.03 32.03 90.9414.00 12.90 13.45 73.44 229.79 4.79 73.4s 285.30 81.05 27.54 91.28 139.8715.00 13.90 28.45 62.18 252.83 27.83 65.62 280.70 115.85 30.36 28.70 174.9116.00 14.90 43.45 49.71 264.52 39.52 60.08 271.44 135.41 31.69 24.45 1g1.5417.00 15.90 58.45 37.00 272.79 47.79 57.55 254.30 136.46 30.69 18.81 185.9718.00 16.90 73.45 24.37 279.93 54.93 58.44 220.69 115.51 26.33 12.15 153.9919.00 17.90 88.45 12.00 286.94 61.94 62.60 141.60 65.16 1s.97 4.90 86.03*Unit of lrradiation is Btu/hr-ft,

7-18Using the developed program, following tables summarize input and outputdata caΙculated for south-facing Surface tilted at 45 deg. on Apr 21 inLouisville, KY.

lnput DataLongitude

Standard MeridianEoΤ

LatitudeDeclination

Surf ΑzimuthSurf TiΙt

ABc

CNRHOG

tSI h, o β'.

Output Data

Φ," ψ," θ," Gruo*

18.8 161.2 164.4 0.035.7 144.3 150.0 0.049.8 130.2 135.4 0.0

85.73 deg90 deg1.1 min

38.18 deg1 1.6 deg180 deg45 deg

358.2 Btu/hr-ft20.1 640.12

1

0.2

1.02.03.0

-165.0 -38.3-150.0 -32.9-'135.0 -24.8

ιJD

0.00.00.ο

Gα*

0.00.00.0

^ * ^*ιJR ιra

0.0 0.00.0 0.00.0 0.0

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118

4.0 -120.0 _15.1 61 .55.0 -105.0 _4.3 71.66.0 -90.0 7.1 80.87 .0 -75.0 18.9 s9.98.0 -60.0 30.6 99.79.0 -45.0 42.0 111.310.0 -30.0 52.3 126.811 .0 -15.0 60.2 149.312.0 0.0 63.4 180.013.0 15.0 60.2 210.714.0 30.0 52.3 239.215.0 45.0 42.0 248.7'16.0 60.0 30.6 260.317 .0 75.0 18.9 270.118.0 90.0 7 .1 279.219.0 105.0 -4.3 288.420.0 120.0 -15.1 298.521 .0 135.0 _24.8 310.222.0 150.0 _32.9 324.323.0 165.0 -38.3 341.224.0 180.0 _40.2 360.0

*Unit of lrradiation is Btu/hr-ft,

118.5 120.7 o.o o.o o.o1ο8.4 106.0 o.o o.o o.o99.2 91.4 95.7 o.o 9.890.1 76.8 215.8 49.2 22.180.3 62.5 259.6 120.0 26.668.7 48.4 280.3 186..1 28.753.2 35.1 291.1 238.3 2g.830.7 23J 296.5 271.5 30.40'ο 18'4 298'2 282.9 3o.530.7 23.7 296.5 271.5 30.453.2 35.1 291.1 238.3 2g.868.7 48.4 2S0.3 186.1 28.780.3 62.5 259.6 120.0 26.690.1 76.8 215.8 49.2 22.199.2 91.4 95.7 o.o 9.8

108.4 106.0 o.o o.o o.o1 18.5 120.7 o.o o.o o.o130.2 135.4 o.o o.o o.o144.3 150.0 o.o o.o o.o161.2 164.4 o.o o.o o.o180.0 175.2 o.o o.o o.o

0.0 0.00.0 0.00.7 10.52.8 74.14.8 151.46.5 221.37.8 275.98.6 3,10.58.9 322.38.6 310.57.8 275.96.5 221.34.8 1s1.42.8 74.10.7 10.50.0 0.00.0 0.00.0 0.00.0 0.00.0 0.ο0.0 0.0

7-19Using the developed program,.following tables summarize input and outputdata calcuΙated for an east-facing windδw, 3 ft. wide by 5 ft. high, with noset baοk on a clea r Jul21 day in-Boise, lD.

Ιnput DataLongitude

Standard MeridianΕoT

LatitudeDeclination

Surf AzimuthSurf Titt

ABc

CNRHOG

LSr h, o β,.

5.0 -,105.0 3.86.0 -90.0 14.0

116.22 deg120 deg-6.2 min43.57 deg20.6 deg90 deg90 deg

346.4 Btu/hr-ft2ο.'1860.1 38

1

0.2

Φ,'65.074.8

ψ,"25.015.2

Output Data

θ, o Guo* Go" Gα*

25.3 21.5 19.5 3.620.6 160.9 .150.6

27.4

GR* Gt* e0.4 23.5 352.46.'1 184.1 2761 .3

Exοerpts from this work may be reproduced by instructors forilffi;;;;;l#;':il;:'r; rnich the texthn.ι. hqc h-.- ".1l'_'Ι:9*':' on a not-for-profit basis for testing or instruαionalρμmosesonl}, to

7.0 -75.0 24.7 84.4 5.68.0 -60.0 35.6 94.8 4.89.0 -45.0 46.2 106.9 16,910.0 -30.0 56.1 123.0 33.01 1.0 -15.0 63.8 146.7 56.712.0 0.0 67.0 18ο.0 90.013.0 15.0 63.8 213.3 123.314.0 30.0 56.1 237.0 147.015.0 45.0 46.2 253.1 163.116.0 60.0 35.6 265.2 175.217 .0 75.0 24.7 275.6 .185.6

18.0 90.0 14.0 285.2 195.219.0 105.0 3.8 295.0 205.0

*Unit of lrradiation is Btu/hr-ft2

25.3 222.0 200.7 36.835.8 251.6 203.9 38.548.6 267.7 177.2 36.162.1 276.8 129.6 31.476.0 281.6 68.2 26.290.0 283.0 0.0 21.5104.0 281.6 0.0 17 .5117 '9 276.8 ο.0 17 '2131.4 267.7 0.0 16.6144.2 251.6 0.0 .15.6

154.7 222.0 0.0 13.8159.4 160.9 0.0 10.0154.7 21.5 0.0 1.3

119

12.3 249.9 3747.918.1 260.6 3908.923.0 236.3 3544.726.8 187.8 2817.229.2 123.6 18s3.630.0 51.4 771.729.2 46.6 699.726.8 44.0 659.823.0 39.7 594.818.1 33.7 505.912.3 26.1 391.96.1 16.1 241.60.4 1.8 26.7

oE is the rate at which solar energy strike the window in Btu/hr

7 -20Given lnformation:

Latitude: l= 32.47 deg.Surface azimuth, S; ζz-=Windowwidth;W=4ft.Windowheight; H=6ft.Setbackdistance; b=1

N180 deg. (clockwise from north)

ft.

(a) On April2lDeclination: δ= 12.3 deg.Local Solar Τime: LSr= 9:00 am

Hour angle: h = 15.(9-12) = -45.0 deg.

Using Eq. 7-8 to calcu|ate solar altitude, β = 43'82 deg'

Using Εq' 7-1'1 to find solar azimuth; Φ= 106'27 deg. (clockwise from north)

Using Εq' 7-12 to calculate wall-soΙar azimuth, y= 73'73 deg.

Using Εqs. 7-28 to 7-30 to calcuΙate shaded dimensions,

Χ = (1 ft.)-tan(73.73) = 3.43 ft.y = (1 ft ).tan(43.82)/cos(73.73) = 3.42 ft.

v

6'Shaded area can be calculated by

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'-7

X

!tr-

4'

120

Α,h =W * H _ (W - x) * (H _ y) = 22'52 ft2

7-20 (Cont.)

Therefore, the percentage of the window that is shaded is 93"8%.

(b) On July 21Declination: δ= 20.6 deg.Local Solar Time. LSl. = 12:00 pm

Hour angle: h = 15*(12-12) = 0.0 deg.

Using Eq. 7-8 to calculate solar altitude, β = 78.13 deg.

Using Εq.7-11 to find solar azimuth; Φ= 180.0 deg. (clockwise from north)

Using Εq' 7-12 to calculate wall-solar azimuth, r= 0.0 deg.

Using Eqs. 7-28 to 7-30 to calculate shaded dimensions,

x = (1 ft.)*tan(0.0) = 0.0 ft.

y = (1 ft.)*tan(78.13)/cos(0.0) = 4.76 ft.

Shaded area can be calculated by

Α,h=W*H_(W _x)*(Ι-Ι_y) = 19.03ft2

Τherefore, the percentage of the window that is shaded is 79.3%.

(c) On Sep 21Declination: δ= 0.0 deg.Local Solar Time: LSl. = 5:00 9m

Hour angle: h = 15*(17-12) = 75.0 deg.

Using Eq. 7-8 to οalculate solar altitude, β= 12.61 deg.

Using Εq'7-11 tofind solarazimuth; Φ=261.81 deg. (clockwisefrom north)

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studentsenrolledincoursesforwhiοhthetextbοokhasbeenadopted. ΑnyotherreproducιionortrαnsΙαιionοfthisνοrkbeyondthαιpermιιtedby Secιions Ι07 or ]08 ofιhe Ι976 [JnitedιSιαιesCοnνriqhl Α'| \υithn1ιt thο hovh'i'CiΛ' n{tho '^---:nιot '1ιlv'- ;a ''"Ι^''t\'1

121

Using Εq' 7-12 to calculate wall-solar azimuth , y= 81.81 deg.

7 -20 (Cont.)

Using Εq.7-28 to calculate the horizontally shaded dimension, X,

x = (1 ft.).tan(81.81) = 6.95 ft.

Since x is greater than W, the window is completely shaded.

Therefore, the percentage of the window that is shaded is 100%'

7-21Given: Problem 7-20 with a tong 2 ft overhang located 2ft above the top of

the window.

For this problem, bo for overhang is the sum of the overhang depth and the

setback; henοe, bo = /+] = 3 ft.

(a) Τhe vertically shaded dimension on the window due to the overhang

can be calculated bY:

lo=botanβlcosy_!o-.

where 1rr-, is the distance of the overhang above the window. Therefore,

Υo = (3 ft.)*tan(43.82)/cos (73'73) - 2 _ 8'27 ft'

Sinοe η is greater than H, the window is completely shaded.

Therefore, the percentage of the window that is shaded is 100%.

(b) Similafly, yo= (3 ft.)- tan(78.13)/cos(0.0) - 2= 12'27 ft'

Since y, is greater than H, the window is completely shaded'

Therefore, the percentage of the window that is shaded is '100%'

Exοeφts from thls work may be reproduced by instruοtοrs for distribution on a not-for-profit basis 1br testing or instruοtional puφosΘs onιy to

students enrolled in courses fbr whiοh the texibook has been aJopted. Αny οιher reprολucJioι or ιrαnslαιiοn ο| this νork beyond ιhαt permiιιed

b,ySectiοns Ι07 οr Ι08of the 1q76τΙ-ito)-qf'''"?""";*13 l^'-!'1 ^ ' '1

122

(c) Since the window is completely shaded due to the setback' there is no

need to calculate Yo'

7 -22Given. Problem 7-2owith 6 in. setback instead of 1 ft' setback'

(a) Using Eqs. 7-28 to 7-30 to calculate shaded dimensions,

x = (0.5 ft.)*tan(73.73) = 1'71 fL

, = (o.s ft.)-tan(43.82)/οos(73'73) = 1'71 ft'

Shaded area can be calculated bY

Α,n =W * H _ (W _x)* (H _ y) = 14.19 fi2

Therefore, the percentage of the window that is shaded is 59'1%'

(b) Using Eqs. 7-28 to 7-30 to calculate shaded dimensions,

γ = (O.5 ft.)-tan(0'0) = ο'0 ft'

, = (ο.s ft.)*tan(78.13)/cos(0'0) _- 2'38 ft'


Α,l, =W * H _ (W _'T )

* (H _ y) = 9'52 ft2


(c)UsingEqs.T-28to7-3Otocalculateshadeddimensions,

x = (0.5 ft.).tan(81'81) = 3'48 ft'

, = (O.S ft.)-tan(12.61)/cos(81 'S1) = 0'79 ft'


Α,h =W * H _(W _.χ)* (H _ y) = 21'27 ft2

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studentsenrolledinοoursestoιwhichtnetexδookhasbeen ^l"i'r'i λiy'nrrrrproλur^ιιoλ-o,ιrαnslαιiοiofιhisνorkbeyondthαtρermitted

hιι'Sρrtinnrl07nrl0Rn{tι1o Ιo7AΙΙbi+^)c'-!"^^f''";..1'! ''',.'"''^""i'-''""'''''-"'^-^{i"^^^'^'';-ι^1 '''-''"iο"-ι^'{"l

123


7 -23on December 21, Declination: δ= -23'45 deg. Using the same procedure

as described in Problem 7-20, the following table summarizes the

calculated data.

:"^::j Hour Solar Solar Surface- '"Jlff#:"' ι"#::γ Shaded %Shaded:,ol1' Angle, Αltitude, Azimuth, Solar Dimension Dimension Area, ft2 ΑreaTime, ""; o o Azimuth,' "hr f\ZιlΙluιl '' 1x1, ft (y)' ft

8.OO -6O.O0 9.98 126.22 53 78 1'37 o'30 8'98 37 '4

9.OO -45.00 1g.4g ',136.52 43 48 o'95 o'49 7 '18 29'9

1o,oo -3o.oo 27.17 148.96 31'04 0'60 0'60 5'65 23',5

'1 1.OO -15.00 32.27 163.69 16'31 o'29 0'66 4'20 17 '5

12.00 O.OO 34.08 18o.OO O'OO o oo 0'68 2'71 1 ',1 ',3

13.00 15.00 32.27 196.3',1 16'31 0 29 0'66 4'20 17 '5

14.00 3O.OO 27.17 211'04 31'04 o 60 0'60 5 65 23'5

15.00 45.00 19.49 22g 48 43'48 0 95 0 49 7 18 29',9

16.00 60.0O 9.98 23g.78 53'78 1'37 0 30 8 98 37 '4

7 -24This problem is similar to ProblemT-21 but the overhang depth is 3 ft

instead of 2ft. Since the window in ProblemT-21 is completely shaded in

all cases. Τhe window in this problem is also completely shaded in all

cases since the overhang depth is greater in this problem'

7 -25

7 -26

7 -27First, we need to know angle of incidence and solar irradiation. Using Eqs'

7-8 to 7-26 (or a computeiprogrφ developed for previous problem),

Exοeφts 1iom this work may be reproduced by instructors for distτibution on a not-for-profit basis for testing or instructional puφoses only to

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124

incidence angle and solar irradiation on a southwest-facing window for

Boise, lD on a clear July 21 day at 3:00 pm solar time are

Angle of Incidence θ= 52.4 deg',Direct Solar lrradiation: Gp = 163'4 Btu/hr-ft2'

Diffuse Solar lrradiation: Ga + Gκ = 34'5 + 23'0 = 57 -3 Btu/hr-ft', and

Total Solar lrradiation: G1 = 163'4 + 57 '3 = 220'9 Btu/hr-ft2'

Then, the area of the glazing and of the frame is calculated to be 12'44 ft2

and 2.56 ft2, resPectivelY.

From Table 7-3, solar heat gain coefficients for the glazing system lD 21c

are

SHGGgo(52.4") = 0.548 and SHGGgα= 0'52'

From Table 5-2, the outside surface conductance may be estimated to be

4.0 Btu/hr-ft2-'F.

From Τable 5-6, the U-value for the fixed, double glazed window having

aluminum frame with thermal break utilizing metal spacers is 1.13 Btu/hr-

ft2-'F.

From Table 7-1, solar absorptance of the aluminum frame (assuming the

window is not a nev/ one) is 0.8.

Αssuming the window with no setback (Ar,u'" = Aruπ), the SHGC for the

frame can be calculated using Eq' 7-31 as:

SHGGr = 0.8*(1 '1314'0) = 0'226'

Then, using Εq.7-32, the total solar heat gain is

Qsnc = (0.548.1 2.44 + 0.226*2.56)*163'4+ (0.52*1 2.44 + 0.226*2.56)-57.5 = 1613'68 Btuihr'

7 -28From Table 7-3, the glazing transmittance and absorptances for the glazing

system lD 21c are

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students enrοl]ed in οourses for whiοh the tΘx;book has υ.", uJopt"α. Αny oιher reproλuction or ιrαnsιαtιοi of ιhιs wοrk beyond ιhαι permitted

125

7 -28 (Cont.)

Tρ6(52'4") = O.+1 56, Αtwοβ2Α) = O'140, Αfzυοφ2'4) = 0'1524'Ta= O'40, -Af ια= o.'13, and Αfza= O.15.

Using Eq. 7-35, total transmitted solar heat gain is

Qrru"', = (0'4156-163.4 + 0.4ο*57.5)-12'44 = 1130.9'1 Btu/hr.

Using Eq. 7-36, total solar heat gain absorbed by the glazing is

8 oroo,, = [1 63'4*(0"1 4+0' 1 524) + 57'5*(0"1 3+0' 1 5)]-1 2'44

= 794.64 Btu/hr.

From Table 5-5a, the U-value for the center of glass is 0.42 Btu/hr-ft2-"F.

Similar to the previous problem, the outside surface conductance may beestimated to be 4.0 Btu/hr-ft2-'F.

Then, the inward flowing fraction for glazing layer 1 can be calculated by:

Nt=0.42 14.0=0.105

From Table 5-2a, the inside surface conductance may be estimated to be1.46 Btu/hr-ft'-"F.

The conductance from the inner pane to the outdoor air can be calculatedby:

,11flo'2= 1 1=-1 1 =o'59Btu/hr-ft2-'F

U hi 0.42 ι.46

Then, the inward flowing fraction for glazing layer 2 can be calculated by:

Nz= 0.42 / 0.59 = 0.71

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126

7-28 (Cont.)

Using Εq. 7-38, the inward flowing fraction of the gΙazing system is

N = [1 63.4*(0.'1 0S*0. 14+0.7 1*0.1 524) +

57.5*(ο.'1 05*0. 13+0.71-0. 1 5)] l 220 '9= 0.122

Using Eq. 7-39 and the SHGGr calculated from the previous problem, thesolar irradiation absorbed by the frame is

Qoroo,f = (163.4 + 57.5)*2.56*0.226 = 127 .80 Btu/hr.

Using Εq' 7-40, the total absorbed solar heat gain of the fenestrationsystem is

Quruo,ur = 794.64*0.122 + 127 .80 = 224.75 Btu/hr.

The total solar heat gain is then

Qsrc = 1130.91 + 224.75 = 1355.66 Btu/hr.

7 -29From Table 7-4, lAC for a lighted-color Venetian blind installed on aresidential double-pane window is 0.66.

Using Εq' 7-41, the total solar heat gain is

Q suc = φ'226-2'56-220'9)+ [0.548*12.44*163.4 + 0.52* 12.44*57 .5]*0.66

= 1108.48 Btu/hr.

7 -30From Table 7-6, for a Ιighted-color Venetian blind, shade transmittance,reflectance, and absorptance are 0.05, 0.55, and 0.40, respectiveΙy.

Using Εq' 7-42, the transmitted solar heat gain is

127

7-30 (Cont.)

Qrroo = 0.05*1130.91 = 56.55 Btu/hr.

Using Εq' 7-43, the absorbed solar heat gaΙn is

4or"" = 224.75 + 0.40*1 130.91+ 0.55*1 130.91 *0.122*(0.'13+0.I S)

= 698.36 Btu/hr.

7-31From Table 7-3, solar heat gain coefficients for the glazing system lD 5bare

SHGGgo(52'4") = 0.6256 and SHG Ggα = 0.60.

Similar to Problem 7-27 , SHGGr = 0.226.

Then, using Εq'7-32, the total solar heat gain is

Qsμc = (0.6256-12.44 + 0'226*2.56)*1 63.4+ (0.60.12.44 + 0.226*2.56)*57.5 = 1828.64 Btu/hr.

7 -32From Table 7-3, the glazing transmittance and absorptances for the glazingsystem lD 5b are

TDθ(52'4") = 0.5332, 'Af ,οoβ2'4) = 0.1924, 1froθβ2'4) = O'12,Ta= 0.51, -Atro= 0.19, and -Arzd= 0.11.

Using Εq. 7-35, total transmitted solar heat gain is

Qrsac,g = (0.5332*'163.4 + 0.51*57.5)*1 2.44 = 1448.64 Btu/hr.

Using Eq. 7-36, total solar heat gain absorbed by the glazing is

λQ πllc'g = [1 63.4*(0.1924+0'12) + 57.5-(0. 1 9+0. 1 1)1-12.44

-===Ξ7 -32 (Cont.)

= 849.60 Btu/hr.

From Table 5-5a, the U-value for the center of gtass is 0.55 Btu/hr-ft2-"F.

similar to the previous problem, the outside surface conductance may beestimated to be 4.0 Btu/hr_ft2_.F.

Then' the inward flowing fraction for glazing layer 1 can be caΙculated by:

Nz=0.SS /4.0=0.1375

SimiΙar to the prevΙous probtem, the inside surface conductance may beestΙmated to be 1.46 Btυ/hr-ft2-"F.

The conductance from the inner pane to the outdoor air can be calcuΙatedby.

ho,z=t\ =T-] 1 = 0.88 Btu/hr-ft2-"F(]_τ O55 - 1-46

Τhen' the Ιnward flowing fraction for gtazing layer 2 canbe calcutated by:

/vz=0.55/0.gg=0.625

Using Εq. 7-38, the inward flowing fraction of the glazing system is

N = []-63 4-(O.1375-ο. 1924+o.625*0'12) +

= 'Jr';Δ'' 1375*0' 19+0'625*0' 1 1 )] l 22o'g

Τhe solar irradiation absorbed by the frame is the Same as the previousprobΙem, and is equaΙ to 127 .80 Btu/hr'

Using Εq' 7-4O, the totat absorbed solar heat gain of the fenestrationsystem is

7 -32 (Cont.)

Qo'"o'gt = 849'60*0'1OO + 127 '80 = 212j6 Btu/hr'

The total solar heat gain is then

Qwc = 1448'64 + 212J6 = 1661'4 Btu/hr'

B-1

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Chapter 8

The heat gain is generally gΓeater than the cooling load during themorning hours M/hen sunlight first strikes a building and the internalloads first begin. Heat is being stored in the building structure,furnishings, etc.Late at night when occupants are not present, lights and equipment areoff and solar radiation is zero, the building gives up stored heat to theair, which the equipment removes as cooling load. The heat gain maybe quite small, zero, or negative.At some time during the day, probably early evening, as heat gain isdecreasing, and equilibrium condition can be established when heatgain and cooling load are equal. Or, some interior zones, where thecooling load is driven only by internal heat gains may reach equilibrium ifthe heat gain remains constant for a number of hours.

MultipΙe design conditions should be checked, including peak dry bulbalong with mean coincident \Μet buΙb, and peak wet bulb along with meancoincident dry bulb temperature.

ΑSHRΑE 90.1 specified the2'5% design conditions, which roughlycorresponds to the 1% design conditions in the current Handbook ofFundamentals and the textbook.

a)

b)

c)

8-2

8-3

Location OutdoorDB.'F

OutdoorWB,'F

lndoorDB.'F

lndoorRH. %

Elevation, ft Latitude,ON

Norfolk, VA 91 76 75 50 30 36.90

8-4

select materials; some may need to be entered into the tayer library' The

resulting wall construction ,pp"rrt as shorrun here' (Note that not

everything is specified exacity, .o that a student using a density of 120

ιυit1yt"i υiicκ wiιι get a different set of CTF coefficients')

program in executefor this wall.

131

for room mode, we

B-5

After running the HvacloadExplorerobtain the following CTF coefficients

nxn,

Btu/h-ft2-"F

Yn,Btu/h-ft2-"F

zn,Btuih-ft2-"F

Φn

0 4.276507 0 ο0ο445 0.642344

1 -5.36497 0.ο1 1581 -0.98287 0.638772

2 1.141149 0.01 1845 0.376555 -0.02179

3 -0.027 59 0.001 134 -0.01 101

4 -7 .7Ε-05 0.0ο0017 -5E-06

This problem is solved in the Same \Λ/ay aS Problem 8-4,

13 insulation is changed to 5.5" thick R-19 insulation.except that the R-

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students enrolled in courses tor whiοh the teΧtbook t,u, υ"., uai-pt.α' iny oιh* up-iurrιλn-o, oαnsΙαtionΖf ιhis work beyοnd thαι permιιιed

by Secιions 107 or ] 08 o7 ιne ii;i' initid Sror^ copyrιg|rt 'a|ι rr"ιiiλuι ιn, prr^ι''ιλn of the cοpyrighι owner is unlανful'

-"..-"...-τns1] 130

0 5α] - "183 5nΟ 3.

8-6

132

The folΙowing CTF coefficients are obtained:

Αgain, this problem follows the procedure of the last two problems. Thethickness of the roll roofing must be estimated, and the conductivity chosento match the overall conductance. (k=thickness*conductance)

The following CΤF coefficients are obtained:

Εxceφts from this work may be reproduced by instruοtors for distτibution on a not-for-profit basis for testing or instruοtional puφoses only tostudents eι'ιrolled in courses for whiοh the textbook has been adopted. Αny oιher reproduction or ιrαnsιαιion of this ινork beyond ιhαι permiιtedby Secιiοns ] 07 οr 1 08 of ιhe Ι97 6 Uniιed Sιαtes Cοpyrighι Αcι w ithοuι ιhe permissiοn οf ιhe copyrighι oνner is unlανιful'

nxn,

Btu/h-ft2-"FYn,

Βtu/h-ft2-'FΖn,

Btu/h-ft2-"FΦn

0 4.277384 0.000071 0.644513

1 -5.95084 0.004622 -1.08666 0.779066

2 1.847897 0.008936 0.510931 -0.10021

3 -0.16027 0.001835 -0.05401 0.001435

4 0.001331 0.00004 0.000734

133

8-7

nxn,

Btu/h-ft2-'FYn,

Btu/h-ft2-'FΖn,

Btu/h-ft2-'F Φn

0 1.014657 0.006092 0.6445131 -1.09939 0.029838 -0.6816 0.1505942 0.126521 0.006044 0.0791043 0.000256 0.000071 0.00003

The following CΤF coefficients are obtained:

nxn,

Btu/h-ft2-'FYn,

Btu/h-ft2-'Fzn,

Btu/h-ft2-'F Φn

0 1.014651 0.00468 0.6544711 -1.12785 0.027234 -0.71129 0.1 781 592 0.1 51 609 0.00674 0.0955263 0.000351 ο.0001ο6 0.000053

ln this case, a reasonable value for the resistance of the air-space must beselected. For the air^-space, an R-value of 1 is chosen; thus conductivity isset to '12 Btu-in lhr- ft2- F, and the thickness Ιs set to 12 in Density anα bpare set to zero and 0.24, respectively.

Νntε: Ιayers listed fιnm tοp tο bοfiom :epr*s*ni {rα:n thg nutsiιJ* tο inside ot lhe sur{εce

Exc-erpts from this work may be reproduced by instruοtors for.distribution on a not-for_profit basis for testing or instruοtional puφoses only tostudents enrolled in courses Γor which the textbook has been adopted' Αny οther reproλcιion or trαnsl(]ιiοn of ιhis νοrk beyontl thαι permiιιedby Sectiοns ]07 οr ]08 ofιhe 1976 Uniιed StαιeS Cοpyrighι Αcιw'ithouι ιhi permissιλn οfιhe cοpyrighι ονner ιi unlαινful'

8-8

134

Again, reasonable values must be assumed for the density of theacoustical tile and the specific heat of the limestone concrete.

The foΙlowing CTF coefficients are obtained:

nxn,

Btu/h-ft2-'FYn,

Btu/h-ft2-'FZn'

Btu/h-ft2-'FΦn

CI 3.162792 0.002232 0.2851 '16

1 -3.76069 0.01895 -0.38995 0.71ο366

2 0.633425 0.007779 0.137 459 -0.01912

3 -0.00642 0.000149 -0.0ο352

First, app|y the exterior οonvective heat transfer correlation, Equation 8-18a, to determin Θ h"' Assume the 15 mph wind is windward on the surface,which results in h" = 2.3 Btu/(h-ft2-F).

Estimate the sky temperature as 10.8 R below the outdoor ambienttemperature = 546.87 R. Then estimate the effective sky temperature for avertical surface from Εquation 8-25

tsky,o= cos (90/2)λry +(1-cos(90/2))f, = 550.0 R

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by Sectiοns Ι 07 οr Ι 08 οf the 1976 tJnited Stαtes Copyrι1hι Αcι \νitlιout ιhe permission of the copyrighι oιιner is unlαινfuΙ.

8-9

136

DeclinationSurf Αzimuth

Surf TiltΑBc

CNRΗoG

20.6 deg270 deg90 deg

346.4 Btu/hr-ft20.1 860.1 38

1

0.2

MDST LSI h, o β, "

1 .00 23.79 176.83 -34.272.00 0.79 -168.17 -33.233.00 1.79 -153.17 -28.804.00 2.79 -138.17 -21.655.00 3.79 -123.17 -12.546.00 4.79 -108.17 -2.11

7.00 5.79 -93.17 9.198.00 6.79 -78.17 21.059.00 7.79 -63.17 33.2210.00 8.79 -48.17 45.4911.00 9.79 -33.17 57.5112.00 10.79 -18.17 68.4613.00 11.79 -3.17 75.2814.00 12.79 11.83 72.1915.00 13.79 26.83 62.3616.00 14.79 41 .83 50.6317.00 15.79 56.83 38.4118.00 16.79 71 .83 26.1719.00 17 .79 86.83 14.1520.00 18.79 101 .83 2.5821.00 19.79 1 16.83 -8.27

22.00 20.79 '131.83 -18.0023.00 21.79 146.83 -26.0624.00 22.79 161.83 -31.74*Unit of lrradiation is Βtu/hr-ft2

Output Data

Φ,' Ψ,o θ,o

356.41 86.41 87.0313.26 256.74 101.0628.82 241.18 114.9942.20 227.80 128.6353.39 216.61 141.5862.87 207.13 152.8071 .22 198.78 159.1779.02 190.98 156.3786.89 183.11 146.6595.77 174.23 134.23107.55 162.45 120.81127.34 142.66 106.97168.24 101.76 92.97218.87 51.13 78.94245.60 24.40 65.01259.78 10.22 51.37269.56 0.44 38.42277.71 7.71 27.20285.45 15.45 20.83293.49 23.49 23.63302.43 32.43 33.35312.83 42.83 45.77325.24 55.24 59.19339.93 69.93 73.03

* ^ * ^ * ^ * ^*ιr/vD ιJD ιrd ιJR ιrt

0.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.00

108.07 0.00 6.71 3.22 9.93206.38 0.00 12.82 10.26 23.08246.69 0.00 15.32 16.92 32.24266.87 0.00 16.57 22.71 39.29277.85 0.00 17.25 27.27 44.53283.62 0.00 17.61 30.29 47.91

285.80 0.00 20.83 31.59 52.42284.93 54.68 25.38 31.06 111.11

280.80 1 18.63 30.63 28.75 178.01272.32 170.01 35.51 24.81 230.32256.78 201.20 38.43 19.50 259.13227.20 202.07 37.19 13.16 252.42161 .85 151.27 27 .51 6.19 184.975.53 5.07 0.93 0.10 6.100.00 0.00 ο.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.00

The hourly dry-bulb temperature is calculated using Equation 8-2. Here,the hour nearest to the local solar time has been used to determine thetemperature. A spreadsheet is used to obtain the solution. lteration isaccomplished by simpΙy pasting the calculated values of Io" back into the" Io", estimated" column.

lnput Data

U-Value 0.1 Btui(h-ft2-F)

Solar absorotivitv 0.8

Ihermal emissivitν 0.9

Exceφts from this work may be reproduced by instruοtors for distrrbution on a not-for-proΓit basis for testing or instructional puφoses only tοstudents enrolled in courses for whiοh the textbook has bοen adopted. Αny other reproducιion or trαnsιαtion οf ιhis wοrk beyond thαι permittedby Secιions 107 or ]08 ofthe Ι976 United Stαtes Cοpyriqht Αctιliιhοuι ιhe permissiοn οfthe cοpyright oνner is unlανful.

Τis 72.O F

Peak temperature 96.0 FDailv Ranqe 25.4 F

Mean Wind Soeed 10.0 mph

8-11

137

This problem uses the same solution procedure as Problem 8-10.

lnput DataLongitude 116.22 deg


Latitude 43.57 degDeclination 20.6 deg

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students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproduction οr trαnsιαtion of this ινοrk beyond ιhαι permittedby Secιions ] 07 or ]08 of ιhe Ι 976 United Sιates Cοpyright Αcι ν,ithout ιhe permission of the cοpyrighι oνner is unlαwfuΙ.

Output Data

ClockTime

LocalSolarTime

OutdoorDry-bulbTemp.

(F)

skvΤemp.

(F)

Effectiveskv

Temp.(F)

To",estimated

(F)

hc(Btu/(h-ft'-F))

hrsky

(Btu/(h-fr-F))

hrg..o

(Btu/(h-ft'-F))

To",calculatedfrom 8-24

(F)

9conduction(Btu/(h-

tt'))

1.00 23.79 75.2 64.4 67.5 74.60 1.58 0.46 0.47 73.14 0.1 1

2.00 0.79 73.9 63.1 66.3 73.15 1.58 0.46 0.47 71.93 -0.01

3.00 1.79 72.6 61.8 65.0 71.94 '1.58 0.45 0.46 70.72 -0.13

4.00 2.79 71.6 60.8 64.0 70.72 1.58 0.45 0.46 69.75 -0.23

5.00 3.79 70.9 60.1 63.2 69.75 '1.58 ο.45 0.46 69.02 -0.306.00 4.79 70.6 59.8 63.0 69.03 1.58 0.45 0.46 68.78 -0.32

7.00 5.79 71.1 60.3 63.5 69.'18 1.58 0.45 0.46 72.34 0.038.00 6.79 72.4 61.6 64.7 74.55 1.58 0.46 0.47 77.55 0.569.00 7.79 74.7 63.9 67.0 78.99 1.58 0.47 0.48 82.47 1.0510.ο0 8.79 78.0 67.2 70.3 83.53 1.58 0.48 0.49 87.66 1.5711.00 9.79 8'1.8 71 .0 74.1 88.42 1.59 0.49 0.50 92.76 2.0812.00 10.79 86.1 75.3 78.5 93.24 '1.59 0.5ο ο.51 97.77 2.58

13.00 11.79 90.2 79.4 82.5 98.33 1.59 0.51 0.52 102.84 3.0814.00 12.79 93.2 82.4 85.6 109.05 1.59 0.53 0.54 122.43 5.0415.00 13.79 95.2 84.4 87.6 131.23 1.61 0.57 0.58 142.11 7.01

1 6.00 14.79 96.Ο 85.2 88.4 148.77 1.62 0.60 0.61 155.98 8.4017 0ο 15.79 95.2 84.4 87.6 159.92 1.62 0.62 ο.63 162.15 9.0218.00 16.79 o?ξ 82.7 85.8 162.49 1.62 0.62 ο.63 158.51 8.6519.00 17.79 90.7 79.9 83.0 153.00 1.62 0.60 0.6'1 138.39 6.6420.00 18.79 87.4 76.6 79.7 118.72 1.60 0.54 0.55 86.48 1.45

21.0ο 19.79 84.1 73.3 76.4 84.81 1.58 0.49 0.50 81.63 0.9622.00 20.79 81.3 70.5 73.6 81.65 1.58 0.48 0.49 78.96 0.7023.0ο 21.79 78.7 67,9 71 .1 78.97 1.58 0.47 0.48 76.54 0.45

24.00 22.79 76.7 65.9 69.1 76.55 1.58 0.47 0.48 74.60 0.26

138

Surf ΑzimuthSurf Tilt

ABc

CNRHOG

180 deg90 deg

346.4 Btu/hr-ft20.1 860.1 38

1

0.2

MDST LSr h, " β, "

1.00 23.15 167.23 -24.772.00 0.15 -177 .77 -25.803.00 1.15 -162.77 -23.914.00 2.15 -147.77 -19.345.00 3.15 -132.77 -12.596.00 4.15 -117.77 -4.217.00 5.'15 -102.77 5.318.00 6.15 -87 .77 15.609.00 7.15 -72.77 26.3210.00 8.15 -57 .77 37 .171 1 .0ο 9.15 -42'77 47 '7612.00 '1 0.15 -27 .77 57 .4213.00 11.15 -12.77 64.6814.00 12.15 2.23 66.9515.00 13.15 17.23 62.9116.00 14.15 32.23 54.7117.00 15.15 47 .23 44.6718.00 16.1 5 62.23 33.9519.00 17 .15 77 .23 23.1020.00 18.'15 92.23 12.4821.00 19.15 107 .23 2.3822.00 2015 122.23 -6.8523.00 21.15 137.23 -14.7924,00 22.15 152.23 -20.95*Unit of lrradiation is Btu/hr-ft'z

Output Data

Φ, " Ψ, o θ, " Grρ* Go* Gα*

346.83 166.83 152.14 0.00 0.00 0.002.32 177 '68 154.11 ο.00 0.00 0.0017.66 162.34 150.59 0.00 0.00 0.0031.94 148.06 143.20 0.00 0.00 0.0044.76 135.24 133.87 0.00 0.00 0.0056.15 123.85 123.75 0.00 0.00 0.0066.47 113.53 113.42 46.47 0.00 2.8976.20 103.80 103.29 173.45 0.00 10.7785.92 94.08 93.66 227.72 0.00 16.4596.42 83.58 84.88 254.61 22.70 20.78108.99 71.01 77.37 269.45 58.92 24.56125.92 54.08 71.58 277.79 87.76 27.57151 .07 28.93 68.02 281.98 105.54 29.47185.34 5.34 67.06 283.00 1 10.30 29.99217 .50 37.50 68.82 281.09 101.56 29.04239.78 59.78 73.09 275.81 80.22 26.78255.07 75.07 79.44 265.88 48.70 23.50266.84 86.84 87.38 248.28 11.34 19.55277.01 97.01 96.45 215.64 0.00 15.02286.67 106.67 106.26 146.49 0.00 9.10296,51 116.51 116.49 3.97 0.00 0.25307 .11 127.11 126.80 0.00 0.00 0.00318.90 138.90 136.76 0.00 0.00 0.00332.16 152.16 145.67 0.00 0.00 0.00

^ * ^*ιJR Lz1

0.00 0.000.00 0.000.00 0.000.00 0.000.00 0.000.00 0.001.07 3.967.06 17,8313.24 29.6918.90 62.3823.67 107.1527.24 142.5829.38 164.3929.95 170.2428.90 159.5026.32'133.3122.36 94.5717 .29 48.1811,44 26.465.19 14.280.07 0.320.00 0.000.00 0.000.00 0.00

Output Data

Exceφts from this work may be reproduοed by instructors fbr distribution on a not-for-profit basis for testing or instructional puφosΘs only tostudents enrolled in courses for which the textboοk has been adopted. Αny other reprοducιion or ιrαnslαtiοn of ιhis work beyond thαι permiιιedby Sectiοns ]07 οr Ι08 ofιhe 1976 Uniιed Stαtes Cοpyright Αctτιiιhout the permission ofιhe cοpyright oνner is unΙαwful.

lnput Data

U-Value 0.1 Btu/(h-ft2-F)

Solar absorptivitv 0.9Γhermal emissivih ο.9

Tis 72.0 FPeak temoerature 96.0 F

Dailv Ranqe 30.3 FVlean Wind Soeec 1 1.0 mph

139

ClockTime

LocalSolarTime

OutdoorDry-bulbTemp.

(F)

skvTemp.

(F)

EffectiveSkv

Temp.(F)

To"'estimated

(F)

h"(Btu/(h-ft'-F))

hrsky

(Btu/(h-ft'?-F))

hrgrα

(Btu/(h-tt'-ε))

To.,calculatedfromS-24

(F)

Qconduction(Btu/(h-

ft2))

1.00 23.15 73.0 62.2 65.3 71.15 1.72 0.45 0.46 71.15 -0.092.O0 0.15 71.2 60.4 63.5 69.41 1.72 0.45 0.46 69.41 -0.263.00 1.15 69.6 58.8 62.0 67.96 1.72 ο.45 0.46 67.96 -0.404.00 2.15 68.'1 57.3 60.5 66.51 1.72 0.44 0.45 66.51 -0.555,00 3.15 66.9 56.'1 ξoa 65.35 1.72 0.44 0.45 65.35 -0.676.00 4.15 66.0 55.2 58.4 64.48 1.72 0.44 0.45 64.48 -0.757.00 5.15 65.7 54.9 58.1 64.1 I 1.72 0.44 0.45 65.5'1 -0.658.00 6.15 66.3 55.5 58.7 68.60 1.72 0.44 0.45 70.67 -0.139.00 7.15 67.8 57.0 60.2 74.15 1.72 0.45 0.46 75.96 0.4010.0ο 8.15 70.5 59.7 62.9 81.31 1.73 0.46 0.47 89.08 1.71

1 1.0ο 9.15 74.5 63.7 66.9 99.67 1,74 0.49 0.50 106.51 3.4512.00 10.15 79.0 68.2 71.4 116.02 1.74 0.52 0.53 121.05 4.9013.00 11.15 84.2 73.4 76.5 129.01 1.75 0.55 0.56 131.75 5.9714.ο0 12.15 89.ο 78.2 81.4 137.21 1.75 0.57 0.58 137.47 6.5515.00 I 3.15 92.7 81.9 85.0 '139.67 1.75 0.58 0.59 137.47 6.551 6.00 14.15 95.1 84.3 87.5 136.52 1.75 0.58 0.59 132.07 6.0117.00 1 5.15 96.0 85.2 88.4 127.91 1,74 0.57 0.58 121.67 4.9718.00 16.15 95.1 84.3 87.5 114.37 1.73 0.54 0.55 107.09 3.5119.00 17.15 93.0 82.2 85.3 100.24 1.73 0.52 0.53 98.57 2.6620.ο0 18.15 89.6 78.8 82.0 93.64 1.72 0.51 0.52 91.61 1.9621.00 '19.15 85.7 74.9 78.1 85.63 1.72 0.49 0.50 83.43 11422.00 20.15 81.8 71 .0 74.1 79.56 1.72 0.48 0.49 79.56 0.7623.00 21 15 78.4 67.6 70.8 76.37 1.72 0.47 0.48 76.37 0.4424.00 22.15 75.4 64.6 67.8 73.47 1.72 0.46 0.47 73.47 0.15

B-12

This problem is solved in the same manner as Example 8-2. The results

Day 1

0.3120.4630.5080.4940.4540.4100.3870.4020.473

Day 2 Day 3

1.126 11260.954 0.9540.804 0.8040.673 0.6730.56't 0.5610.475 0.4750.426 0.4260.426 0.4260.487 0.487

(conduction heat fluxes for each hour in Btu/(hr-ft2)) may be summarized intabular form as:

Hour

1

2

345

67

B

9

Exceφts from this work may be reproduοed by instructors for distrrbution οn a nοt-for-profit basis for testing or instructional puφoses only tοstudents enrolled in courses for whiοh the textbook has been adοpted. Αny οther reproduction or trαnslαtion οf ιhis νork beyond ιhαι permiιιedby Secιiοns Ι07 or Ι08 οfιhe Ι976 Uniιed Stαιes Cοpyright Αct wiιhouι the permission ofthe copyrighι oνner is unlανful.

-

140

1011

12131415161718192021

222324

0.6120.8201.0891.3991.7151.9982.2232.3622.4052.3522.2152.0161.7861.5511.327

0.6200.8251.0921.4011 .7161.9992.2232.3622.4052.3532.2152.0161.7861.5511.327

0.6200.8251.0921.4011 .7161.9992.2232.3622.4052.3532.2152.0161.786'1.551

1.327

8-13

Because the wall is Ιightweight, the results converge rapidly.

This problem is soΙved in the Same V/ay aS the previous problem. Note thatthe additional insulation substantially reduces the conduction heat flux, asexpected. The resuΙts (conduοtion heat fΙuxes for each hour in Btu/(hr-ft2))may be summarized in tabular form as:

Hour1

2

34567o

I1011

12

131415161718192021

22

Day 1

0.2030.3290.3790.3800.3560.3230.2980.2960.3280.407ο.5350.7110.9251.1541.3731.5591.6911.7571.7541.6851.5621.406

Day 20.9150.7780.6590.5540.4640.3900.3400.3220.3440.4170.5420.7150.927'1 .1561.3741.5591.6921.7581.754'1.685

1.562'1.406

Day 30.9150.7780.6590.5540.4640.3900.3400.3220.3440.4170.5420,7150.9271 .1561.3741.5591.6921.7581.7541.6851.5621.406

Exc-erpts from this work may be reproduced by instructors for.distribution οn a not-for-profit basis for testing or instruοtional purposes only tostudents enrolled in οourses fοr which the textbook has been adopted. Αny other reprοdλCιion or ιrαnslαtion of ιhis lιork beyond thαι permiιιedby Sections ] 07 οr ] 08 οf ιhe 1 976 (Jniιed Sιαιes Cοpyright Αcι ιυithouι thi permissιλn ο7 ιhe copyrighι ονner ιi unlαwfut'

2324

141

1.2371.070

1.237 1.2371.070 1.070

8-14

B-'15

The solution to this problem is similar to that of Problem 8-9, except that toestimate the maximum possible surface temperature, the surface may beassumed to be adiabatic, and U is then zero. Also, the surface-to-groundradiation coefficient is zero, and no correction is necessary for the skytemperature, as the surface is assumed to be horizontal. Assume the windis windward, h"= 1.3 Btu/(h-ft2-F). Then, the final converged answer for thesurface temperature is:

hr,sky= 1.361 Btu/(h-ft2-F) fr"= 201.0 F

From Table 8-2, heat gain for occupants that are "Seated, very light work"have 245 Btulhr (72 W) sensible heat gain, and '155 Btu/hr (45 W) latentheat gain. The sensible portion is assumed to be 70% radiative/ 30%convective.

The sensible heat gain from people is72 Wperson x 30 people = 2160 W.

The radiative portion is 0.7 x 2160 = 1512νν '

Τhe convective portion is 0'3 x2160 = 648 W.

The latent heat gain from people is 45 Wperson x 30 peoPΙe = 1350 W.

The sensibΙe heat gain from lighting is '1 .5 ννft2 x 4OOO sq. ft. = 6000 W;20o/o is assumed to enter the plenum space directly, leaving 4800 W whichis assumed to be 59% radiative I 41% convective.

The radiative portion is 0.59 x 4800 = 2832\'tΥ.

The convective portion is 0.41 x 4800 = '1968 W.

The sensible heat gain from equipment is 1 ννfi( x 4OOO sq. ft. = 4O0O W,which is assumed to be 20o/o radiative I 80% convective. (Note this

Εxcerpts fiom this work may be reproduοed by instruοtοrs for distribution on a nοt-fbr_profit basis for testing or instruοtional puφoses only tostudents enrolΙed in courses for which the textbook has been adopted. Αny οιher reproductιοn οr trαnslαtiοn of this work beyοnd ιhαt permittedby Sectiοns l07 οr 108 ofιhe Ι976 Uniιed Sιαιes Copyrighι Αcιlνithoul the permission ofthe copyright οwner is unlαwfuΙ.

8-16

142

assumption is based on the assumption that most of the equipment is fan-cooled. Students are likely to make varying assumptions.)

Τhe radiative portion is 0'2x 4000 = 800 W.

The convective portΙon is 0.8 x 4000 = 3200 W.

The total sensible heat gain is 2160 + 4800 + 4000 = 10960 W.

The radiative portion is 1512 + 2832 + 800 = 5144 W.

The convective portion is 648 + 1968 + 3200 = 5816 W.

The total latent heat gain is 1350 W.

From Table 8-2, heat gain for occupants that are involved in "Sedentary\ι/ork" is275 Btu/hr (81 W) sensible heat gain, and 275 Btulhr (81 W) latentheat gain. The sensible portion is assumed to be 70o/o radiative/ 30%convective.

The sensible heat gain from people is 81 Wperson x 35 people = 2835 W.

Τhe radiative portion is 0'7 x 2835 = 1984.5 W.

The convective portion is 0.3 x 2835 = 850.5 W.

The latent heat gain from peopΙe is 81 Wperson x 35 peoPle = 2835 W.

Τhe sensible heat gain from lighting is '15 \,ΙΥlm2 x 75O m' = 11250 W; 50%is assumed to enter the plenum Space directly, Ιeaving 5625 W that isassumed to be 59% radiative I 41% convective.

The radiative portion is 0.59 x 5625 = 3319 W.

The convective portion is 0.4'l x 5625 = 2306 W.

The sensible heat gain from office equipment is 7000 W, which is assumedto be 20o/o radiative I 80% convective. (Note this assumption is based on

EXοeφtS from this work may be reprοduced by instruοtοrs for distribution on a not_for-profit basis for testing or instruοtional purpοses only tostudents enrolled in οourses fοr which the textbοok has been adopted' Αny oιher reproduction or ιrαnsιcιιiοn ofιhis work beyond ιhαι permittedbySectiοns Ι07 οr ]08οf the Ι976UniιedSιαιesCopyrighιΑcιy,ithoutιhepermissionof ιhecopyrighιοwnerisunlατνful.

143

the assumption that most of the equipment is fan-cooled. Students arelikely to make varying assumptions.)

The radiative portion is 0.2x 7000 = 1400 W.The convective portion is 0.8 x 7000 = 5600 W.

The total sensible heat gain is 2835 + 5625 + 7000 = 15460 W.

The radiative portion is 1984.5 + 3319 + 1400 = 6703.5 W.

The convective portion is 850.5 + 2306 + 5600 = 8756.5 W.

The total latent heat gain is 2835 W.

8-17

Heat gain to the space = 0.8 x 6000 W = 4800 W

Problem 8-18

At 4.00 p.m., 70 people are present. Assuming "seated, light offiοe \Mork",the sensible heat gain per person is245 Btu/hr (72νν) and the latent heatgain per person is 200 Btu/hr (59 W).

Sensible heat gain = 245 Btu/hr/person x 70 people = 17150 Btu/hr.

Latent heat gain = 200 Btu/hr/person x 70 people = '14000 Btu/hr.

Αt 6:00 p.m., no one is present; sensible and latent heat gains are O Btu/hr.

8-19

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144

First, compute the properties of the corresponding fictitious surfaces, usingEqns 8-35, 8-36, 8-37. ResuΙts are shown in the shaded table entries,beΙow.

Surface Area (ft') A-ε T(F) A-ε-T λ εl]f τ;lF)1 Νorth roof 639.7 0.9 575.8 122 70241.8 1983.7 0',9 , 1,10.62 South roof 639.7 0.9 575.8 143 82332.6 1983.7 0.9 1ο3.93 West wall 84.0 0.9 75.6 102 7711.2 2539.4 0.9 113,84 East wall 84.0 0.9 75.6 92 6955.2 2539;4, 0.9 114.15 Αttic floor 1176.0 0.9 1058.4 95 100548.0 1'4;47.',4' 0:9 128,4

Τhen, compute the radiant interchange factor and radiation heat transfercoefficient using Eqns. 8-38 and 8-39. Using Eqn. 8-40, estimate theradiative heat flux from each Surface (Q,.uα), then determine the radiativeheat transfer from each Surface (Q,"rα). Then, compute the total radiativeheat transfer from all surfaοes = -69,769.5 Btu/hr. Divide by the totalSurface area, 2623.4 ft2, to get the baΙancing factor, -26'6 Btu/(hr-ft'1, whichmust be subtracted from the previously caΙculated heat flux from eachsurface to determine the "balanced" radiation heat flux from each surface(q,rοlbal). Multiply by the area to determine the radiation heat transfer fromeach Surface (Q,"α/bal)' Cheοk to see that they noνv Sum to zero.

Suι"face Fit Trus (R) hriQraο

(Btu/(hr-ft2))

Q,uo(Btu/hQ

qr"6/bal(BtΨ

(hr-ft'))

Q,u6lbal(Btu/hr)

I North roof 0.872 576.0 11.4 129.7 82950.9 156.3 99964.12 South roof 0.872 583.1 11.9 463.7 296624.7 490.3 313637.93 West wall ο.897 567.6 11.2 -132.6 -11140.0 -'106.0 -8906.14 East wall 0.897 562.7 1 1.0 -242.5 -20368.1 -215.9 -18134.25 Αttic floor 0.832 571.4 10.6 -355.3 -417837.0 -328.7 -386561.8

8-20

First, compute the properties of the corresponding fictitious surfaces, usingEqns 8-35, 8-36, 8-37. Results are sho\Mn in the shaded table entries,below.

Surface Area(m2) A-ε T(C) Α-ε-Τ A, t1 Ti(e)

1 North roof 120.7 ο.9 108.7 43 4672.9 ' 372:7, 0.:,9, 38.3

Exοerpts from this work may be reproduοed by instruοtors for distribution on a not_for-profit basis for testing or instructional puφoses only tostudents enro]led in courses for which the textbook has been adopted' Αny other reproducιiοn or ιrαnslαιion of ιhis ινork beyond ιhαt permiιtedby Sectiοns l 07 οr Ι 08 of the Ι 976 Uniιed Sιαtes Copyrighι Αcι withοuι ιhe permission of ιhe copyrighι olυner is unlaνvfuΙ.

8-21

145

2 South roof 120.7 0.9 108.7 50 5433.6 372.7 0.9 36.02 West wall 18.0 0.9 16.2 36 583.2 475t.:5 ,0.9 39;64 East wall 18.0 0.9 16.2 38 615.6 4Ι5.5 0.;9 39.55 Attic floor 216.0 0.9 194.4 32 6220.8 277.5 ο.,9 45,3

Τhen, compute the radiant interchange factor and radiation heat transfercoefficient using Eqns. 8-38 and 8-39. Using Eqn. 8-40, estimate theradiative heat flux from each Surface (9,"rα), then determine the radiativeheat transfer from each Surface (Q,"α). Then, compute the total radiativeheat transfer from all surfaces = -3027.9 W. Divide by the total surfacearea, 493.5 m2, to get the balancing factor, -6.1 Wmz, which must besubtracted from the previously calculated heat flux from each surface todetermine the "balanced" radiation heat flux from each surface (q,u6lbal).Multiply by the area to determine the radiation heat transfer from eachsurface (Q,"α/bal). Check to see that they no\M Sum to zero.

Surface Fir Tuus (K) hrt Qraα ^(Wm'\ Q,"o (il4 qru6/baΙ

(Wm')Q,,ο/bal

(w)1 North roof 0.872 313.8 6.'1 28.6 3459.0 34.8 4199.92 South roof 0.872 316.2 b_J 87.2 10534.5 93.4 11275.4J West wall 0.897 310.9 6.1 -22.0 -395.4 -15.8 -285.04 East wall 0.897 31 1.9 6.2 -9.4 -168.4 -3.2 -58.05 Attic floor 0.835 31 1.8 57 -76.2 -16457.6 -70.1 -15132.3

The solution procedure is identicaΙ to that of Problem 8-19, except theemissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties areshown in the first table.

Surface Area (ft') ε A-ε T(F) A-ε-T At ε:1 T^,(F)1 North roof 639.7 0.1 64.0 122 7804.6 1983.7 0.6 97.62 South rool 639.7 0.1 64.0 143 9148.1 1983:7, ο.6 96.6J West wall 84.0 0.9 75.6 102 7711.2 2539;4 0'5 98.64 East wall 84.0 0.9 75.6 92 6955.2 2539.4 ο.5 99.25 Attiο floor 1176.0 0.9 1058.4 oξ 100548.0 14:47.'.4 t 0,2 11'3.3

Τhe total radiative heat transfer from a|l surfaces = -3476.1 Btu/hr. Thebalancing factor is -1. 3 Btu/(hr-ft2).

Surface Fir Τ*s (R) hri 9rao(Btu/

Q,rα(Btu/hr)

Qrrα/bal(Btu/

Q,,a/bal(Btu/hr)

Exceφts from this work may be reproduced by instruοtors for distributiοn on a not-for-profit basis for tΘsting or instructional puφoses only tοstudents enrolled in courses Γor which the textbook has been adοpted. Αny other reprοducιion or ιrαnslαιion of ιhis νork beyond thcιι peιmittedby Secιions ]07 or ]08 οfιhe Ι976 Uniιed Sιαιes Copyright Αctνiιhοut the permission ofιhe cοpyrighι οwner is unlcrνυful.

146

(hr-ft')) (hr-ft'))

1 North roof 0.098 569.5 1.2 30.3 19378.4 31.6 20226.02 South rooi 0.098 579.5 1.3 60.8 38903.3 62.1 39751.03 West wall 0.874 560'ο 10.5 35.5 2984.3 36.9 3095.64 Εast wall 0.874 555.3 10.3 -74.1 -6220.9 -72.7 -6109.65 Attic floor 0.222 563.8 2.7 -49.8 -58521.2 -48.4 -56963.0

Note that the radiative heat fluxes from surfaces 1 and 2, and to surface 5

are significantly lower. (The heat flux incident on surface 5 has beenreduced by s5%.) Τhe catch is that "in real life", everything else does notremain the same. ln particular, the temperatures would changesignificantly.

The solution procedure is identical to that of Problem 8-20, except theemissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties areshown in the first table.

SurfaceArea(m2) A-ε r(c) A-ε-T ,Λ ε',Ι Tτ(c)

1 North roof 120.7 0.1 12.1 43 519.2 372'7' ο.6 33;62 South roof 120.7 0.1 12.1 50 603.7 372,7 0'6 33ι'2

J West walΙ 18.0 0.9 16.2 Jb 583.2 .475.5 0,5 ,,33.9

4 East wall 18.0 0.9 16.2 38 615.6 ,475,,5' 0.5 33,85 Attic floor 216.0 0.9 194.4 32 6220.8 277,,5 o'.2 41.1

The total radiative heat transfer from all surfaces = -341.5 W. Thebalancing factor is -0.7 Wm2.

Surface Fy Trrs (K) hrl 9raα ^/Wm') Q,"ο (h4

qru6/bal(Wm')

Q.,a/bal(w)

1 North roof 0.098 311.4 0.7 6.3 765.2 7.0 848.7

2 South roof 0.098 314.8 0.7 117 1407.1 12.3 1490.7

J West wall 0.870 308.1 5.8 12.1 217.6 12.8 230.04 East wall 0.870 309.0 5.8 24.6 443.7 25,3 456.1

5 Attic floor 0.241 309.7 1.6 -14.7 -3175.0 -14.0 -3025.6

Exοerpts frοm this wοrk may be reproduced by instruοtors 1br distribution on a not_for-prοfit basis for testing or instructional puφoses οnΙy to

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8-22

8-23

147

Convective heat transfer coefficients are determined from Table 8-8.Coefficients for the pitched roof surfaces are based on the "Sloping - 45degrees" surface position. Α more sophisticated approach would involveinterpolation. The resuΙts are summarized below.

Surface Area (ft2) τ(F) SurfacePosition

Direction ofHeat Flow

h"Btu/(hr-ft2-F)

9""onu""tion(Btu/(hr-ft"))

1 North roof 639.7 122Sloping -

45 deqreesDownward o.42 15.54

2 South roof 639.7 143 Sloping -

45 deοreesDownward 0.42 24.36

3 West wall 84.0 102 Vertical Horizontal 0.56 9.52

4 East wall 84.0 92 Veftical Horizontal 0.56 3.92

5 Attic floor 1 176.ο 95 Horizontal Downward 0.18 1.80

Convective heat transfer coefficients are determined from Table 8-8.Coefficients for the pitched roof surfaces are based on the "Sloping - 45degrees" surface position. A more sophisticated approach \Mould involveinterpolation. The results are Summarιzed below.

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8-24

B-25

Surface Area (m2) r(c) SurfacePosition

Direction ofHeat Flow

hc(wm2-K)

Qt'"onu""1onrιΛ//m')

1 Νorth roof 120.7 43Sloping -

45 deqreesDownward 2.39 33.46

2 South roof 120.7 50Sloping -

45 deqreesDownward 2.39 50.1 9

J West wall 18.0 36 Vertical Horizontal 3.18 22.26

4 Εast wall 18.0 38 Vertical Horizontal 3.18 28.62

5 Attic floor 216.0 32 Horizontal Downward 1.02 3.06

r_

8-26

148

First, the solar irradiation on the window is obtained in the same manner asthe solution for Problem 7-17. The following tables show results for thewest-facing window.


Standard Meridian 90 degEoτ -6.2 min


Surf Αzimuth 270 degSurf Tilt 90 deg

A 346.4 Btu/hr-ft2B 0.186c 0.138cN1

RHOG 0.2

Output Data

cDSr Lsr h, " β, " Φ, " Ψ, o θ, " G,νρ" Go* Ga* Gr* G,*

15.00 13.12 16.75 69.25 229.59 40.41 74.35 283.92 76.59 27.06 30.47 134.12

The layer absorptances of the double-pane \Mindo\Λ/ v/ith '1l8 in. sheet glass(lD5a) can be found from Τable 7-3 as:

GDirect,outer: 7f ,774 dοg) : O.13 (χdiffuse,outer: .fiyaφrn_ 0.11

σDirect,inner :'dr(7 4deg) : 0.06 acΙiffuse'inne, : .fz'aixur, : 0.07

Then, the solar radiation absorbed by each pane of the double-panewindow may be determined by (neglecting incident solar radiation from theinside):

Q"itob,o,b,d,outer, j,ρ: 0.13(76.5g) + O.11(57.53) : 16.29 εtu/(hr_ft^2)

Q"itob,o,b,d,n*',," i, θ :0.06(76.5g) + 0.07(57.53) : 8.62Βtνl(hr-ft2)

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149

First, the solar irradiation on the window is obtained in the same manner asthe solution for Problem 7-17 ' Τhe following tables show results for thewest-facing window.




Surf Αzimuth 270 degSurf Τilt 90 deg

A 346.4 Btu/hr-ft2B 0.186c 0.138cN1

RHOG 0.2

Output Data

MDST LSr h, " β,' Φ,' Ψ, o θ, o Grvo* Gρ" Gd* Gπ* Gt*

15.00 13.66 24.92 57.57 227.35 42.65 66.77 277.89 109.60 29.57 27.29 166.46

The layer absorptances of the double-pane \Λ/indo\M \Λ/ith 1/8 in. sheet glass(lD5a) can be found from Table 7-3 as:

&Direct,outer: .ir(67 deg) : 0.L27 ddffise'outer: -fι,aι1urr: 0.11

aDirect,inner: 7t1167 deg) : 0.073 σc]iffuse,inner: .lz,aιρr, : 0.07

Τhen, the Solar radiation absorbed by each pane of the double-panewindow may be determined by (neglecting incident solar radiation from theinside):

Q" it ob,o,bnd,outer' j, ο : 0.Ι27 (|0g.6) + 0. 1 1 (56.s6) : 20.1 7 Btu/(hr-ft2)

Q"ιt ob,o,bud,inner, j, θ : 0.073(109.6) + 0.07(56.86) : 1 1.98 Btu/(hr-ft2)

8-27

RεsULTs BY ΤΗΕ Ηts ΜΕΤΗΟD UsιΝG T}*|Γ ΗVΑCεXpLORΕRpRΟBRΜΑ ARΕ l-{lcl-lΕR Τι-.|ΑΝ RΕ$ULΤ$ βY ΤΗf; RΤS ΜETl*iΟDUsl ΝG Τl-tΕ spRΕΑD$ι*| HΕΤ.

8-28

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150

The RTS method is used to obtain the cooling load results for this problem.

The following table shows total cooling loads and cooling loads due to

window heaigains for both low-e and regular double-pane windo\Λ/S' Αιl

cooling loads due to other heat gains are the same as those shown in

Example 8-16.

Hour

Low-E Windows(from ExamPle 8-16)

Regular Windows

WindowConduοtion

(Btu,hr)

WindowSΗG

(Btu,&r)

Total(Btu/hr)

WindοwConduction

(Btu,&ιr)

WindowSΗG

(Btu,4lr)

Total(BtuAr)

1 186 364 4418 228 420 4516

2 146 299 3843 179 345 3921

2 110 246 3352 135 284 3414

4 79 203 2940 97 234 2989

5 58 167 2623 72 193 2662

6 52 tJo 2419 63 159 2452

7 61 254 2465 75 293 2518

8 92 465 2737 112 537 2829

9 145 710 8'190 178 820 8333

10 215 978 9562 toc 1129 9761

11 300 1247 1 0883 JOO 1437 11141

12 389 1492 12143 477 1720 12458

13 469 1694 13275 574 1 951 I 3637

14 533 1 833 14250 654 2111 14648

15 577 1897 1 5007 707 2185 15425

16 593 188'1 1 5486 726 2167 1 5905

17 585 1787 15701 717 2060 16105

1B 553 1624 1 0635 677 1 873 11008

19 503 139'1 9550 616 '1604 9877

20 444 1 089 8460 544 1256 8727

21 386 839 7477 472 968 7692

22 327 674 6588 401 777 6765

23 274 546 5777 336 629 5922

24 228 445 5057 279 513 5176

As shown in the above table, usιng the regular \Μindo\Μ Would resuιt in

slighily higher cooling loads than using the low-e window. The following

figure illustrates the 'hcrease in cooling loads due to changing the type of

w]ndow from the low-e window to the reguιar window.

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151

Cooling Load Comparison

αldΦocr)

:Ξooζ)

18000

'15000

12000

9000

6000

3000

0't0 '13 '16

Tirne, Hour

8-29

8-30

SolutΙon to be provided by an instructor.

First, the solar irradiation must be determined and is the same as thatshown for Problem 8-10. Τhen, the hourly dry bulb temperature iscalculated using Εquation 8-2. Here, the hour nearest to the local solartime has been used to determine the temperature. Finally, the sol-airtemperature is calculated using Εquation 8-63 with the thermaΙ radiationcorrection term being zero for a vertical surface.

Local lnsolation Outdoor Sol-airSolar (Btu/h- Drybutb Temp

Clock Time Time ft2) Τemp (F) (F)1.00 23.79 0.00 75.2 75.22.00 0.79 0.00 73.9 73.93.00 1.79 0.00 72.6 72.64.00 2.79 0.00 71.6 71.65.00 3.79 0.00 70.9 70.96.00 4.79 0.00 70.6 70.6

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152

7.008.009.0010.001 1.0012.0013.0014.0015.0016.00'17.00

'18.00

19.0020.0021.0022.0023.0024.00

ClockTime1.002.003.004.005.006.007.008.009.0010.0011.0012.0013.0014.0015.00'!6.00

17.0018.0019.0ο20.0021.0022.00

5.796.797.798.799.7910.7911.7912.7913.7914.7915.7916.7917.7918.7919.7920.7921.7922.79

9.9323.0832.2439.2944.5347.9152.42111 .11

178.01230.32259.13252.42184.976.'100.000.000.0ο0.00

73.177.081 .1

85.890.795.7100.6115.4130.8142.1147.1143.9127.788.684.1

81.378.776.7

Sol-airTemp (F)

73.071.269.668.166.966.066.369.072.379.990.6100.4108.8114.61 16.6115.1110.2102.396.991.885.781.8

71.172.474.778.08'1.8

86.'1

90.293.295.296.095.293.590.787.484.181.378.776.7

8-31

This problem uses the same solution procedure as Problem 8-30. Notethat the solar irradiation is the same as that shown for Problem 8-1 1.

LocalSolarTime23.150.1 5

1.152.153.154.155.156.157.158.1 5

9.1510.1511.1512.151 3.1514.1515.'1 5

16.1517.15'18.15

19.1520.15

OutdoorΙnsolation Drybulb(Btu/h-ft2) Temp (F)

0.00 73.00.00 71.20.00 69.60.00 68.10.00 66.90.00 66.03.96 65.717.83 66.329.69 67.862.38 70.5107.15 74.5142.58 79.0164.39 84.2170.24 89.0159.50 92.7133.31 95.194.57 96.048.18 95.126.46 93.014.28 89.60.32 85.70.00 81.8

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153

23.0024.00

21.1522.15

78.475.4

78.475.4

LongitudeStandard Meridian

EOTLatitude

DeclinationSurf Azimuth

Surf ΤiltAparBparCparCN

RHOG

0.000.00

106.62 deg105 deg-6.2 min35.05 deg20.6 deg0 deg0 deg

346.4 Btu/hr-ft20.1 860.1 38

1

0.2

8-32

This problem uses the simiΙar Solution procedure as Problem 8-3O. First,the solar irradiation is determined for the flat roof using the proceduredescribed in Chapter 7. The resuΙts are shown below.

Ιnput Data

MDST Lsr h, o β,.1.00 23.79 176.83 -34.272.00 0.79 -168.17 -33.233.00 1 .79 -153.17 -28.804.00 2.79 -138.17 -21.655.00 3.79 -123.17 -12.546.00 4.79 -108.17 -2.117.00 5.79 -93.17 9.'198.00 6.79 -78.17 21.059.00 7.79 -63.17 33.2210.00 8.79 -48.17 45.491 1.00 9.79 -33.17 57 51

12.00 10.79 -18.17 68.46'13.00 11.79 -3.17 75.2814.00 12.79 11.83 72.1915.00 13.79 26.83 62.3616.00 14.79 41.83 50.6317.00 15.79 56.83 38.4118.00 16.79 71.83 26.1719.00 17 .79 86.83 14.1520.00 18.79 101 .83 2.5821 .00 19.79 1 16.B3 -8.2722.00 20.79 131.83 -'18.0023.00 21.79 146.83 -26.06

Output Data

Φ, " Ψ, o θ, o Gruo* Go* Gα* Gπ* Gt*

356.41 356.41 124.27 0.00 0.00 0.00 0.00 0.0013.26 13.26 123.23 0.00 0.00 0.00 0.00 0.0028.82 28.82 1 18.80 0.00 0.00 0.00 0.00 0.0042.20 42.20 111.65 0.00 0.00 0.00 0.00 0.0053.39 53.39 102.54 0.00 0.00 0.00 0.00 0.0062'87 62'87 92'11 0.00 0.00 0.00 0.ο0 0.0071 .22 71.22 80.81 108.07 17 .26 14.91 0.00 32.1779.02 79.02 68.95 206.38 74.12 28.48 0.00 102.6086.89 86.89 56.78 246.69 135.17 34.04 0.00 169.2195.77 95.77 44.51 266.87 190.31 36.83 0.00 227.14107.55 107 .55 32.49 277 .85 234.37 38.34 0.00 272.71127.34 127.34 21.54 283.62 263.81 39.14 0.00 302.95168.24 168.24 14.72 285.80 276.42 39.44 0.00 315.86218.87 218.87 17.81 284.93 271.28 39.32 0.00 310.60245.60 245.60 27.64 280.80 248.75 38.75 0.00 287.50259.78 259.78 39.37 272.32 210'52 37.58 0.ο0 248'10269.56 269.56 51.59 256.78 159.55 35.44 0.00 194.98277 '71 277 '71 63.83 227 .20 10ο.20 31 .35 0.00 131 .56285.45 285.45 75.85 '161.85 39.s6 22.33 0.00 61.90293.49 293.49 87.42 5.53 0.25 0.76 0.00 1.01302.43 302.43 98.27 0.00 0.00 0.00 0.00 0.00312.83 312.83 108.00 0.00 0.00 0.00 0.00 0.00325.24 325.24 116.06 0.00 0.00 0.00 0.00 0.00

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154

24.00 22.79 161.83 -31.74 339.93 339.93 121.74 0.00*Unit of lrradiation is Βtu/hr-ft2

Then, the sol-air temperature is determined using Equation 8-63 with thethermal radiation correοtion term being 7 "F for a horizontal surface.

0.000.000.000.00

Clock Time1.002.003.004.005.006.007.008.0ο9.0010.0011.0012.0013.ο014.0015.ο016.0017.0018.0019.0020.0021.0022.0023.0024.00

Local lnsolationSolar (Btu/h-Time tt2)23.79 0.000.79 0.ο01.79 0.002.79 0.003.79 0.004.79 0.005.79 32.176.79 102.607.79 169.218.79 227.149.79 272.7110.79 302.9511.79 315.8612.79 310.6ο13.79 287.5014.79 248.1015.79 194.9816.79 131.5617.79 61.9018.79 1 .01

19,79 0.0020.79 0.0021.79 0.0022.79 0.00

Outdoor Sol-airDrybulb Temp

Τemp (F) (F)

75.2 68.273.9 66.972.6 65.671.6 64.670.9 63.970.6 63.671 .1 70.572.4 85.974.7 101.578.0 116.481.8 129.386.1 139.790.2 146.393.2 148.395.2 145.796.0 138.695.2 127.293.5 112.890.7 96.087.4 80.684.1 77.181.3 74.378.7 71.776.7 69.7

8-33

For hour 15, Equation 8-64 is used to find the conduction heat flux.

Q"conduction = .0052 x (151 '2-74) + .001 44 x (138'1-74) +

.00645 x (120.3-74)...

= 2.897 Btu/(hr-ft2)

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8-34

Equation 8-64 is used to find the conduction heat flux for each hour.

Hourq"

(Btu/(hr-ft2))

Hourq"

(Btu/(hr-ft2))

1 1.835 13 0.828

2 1.824 14 0.798

J 1.772 15 0.791

4 1.693 16 0.810

5 1.595 17 0.861

6 1.486 '18 0.948

7 1.372 19 1 .071

8 1.259 20 1.225ο 1.149 21 1.396

10 1.047 22 1.563

11 0.956 23 1.704

12 0.882 24 1.797

8-35

For hour 12, Εqυation 8-64 is used to find the conduction heat flux.

Q"conduction = 0.0061 92 x (143.9-72) + 0.044510 x (1 34.3-72) +

0.047321 x (1 21 .4-72)...

= 7 .028 Btu/(hr-ft2)

8-36

Equation 8-64 is used to find the conduction heat fΙux for each hour.

Ηour

q"(Btu/(hr-

ft2)) Hour

q"(Btu/(hr-

ft2))

1 0.674 13 1.050

2 0.401 14 1.544

3 0.1 99 15 2.0124 0.051 16 2.409

155

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156

5 -0.058 17 2.6946 -0.138 18 2.841

7 -0.197 19 2.8348 -0.232 20 2.671

9 -0.209 21 2.361

10 -0.075 22 1.936

11 0.194 23 1.466

12 0.583 24 1.031

Equation 8-64 is used to find the conduction heat flux for each hour withsoΙ-air temperatures calculated in Problem 8-31.

Hour

q"(Btu/(hr-

ft2)) Hour

q"(Btu/(hr-

ft2))

1 3.492 13 -0.055

2 3.147 14 0.1 65

3 2.758 15 0.563

4 2.348 16 1.112

5 1.937 17 1.754

6 1.536 18 2.417

7 1.154 19 3.026

I 0.796 20 3.510

9 0.472 21 3.823

10 0.1 99 22 3.958

11 0.00ο 23 3.931

12 -0.098 24 3.765

Using the simplified approach, the solution procedure is the same as that ofProbΙem 7-27. First, we need to know angΙe of incidence and Solarirradiation. Αssuming a west-facing window, the incidence angle and solarΙrradiation for Albuquerque, NM on a c|ear July 21 day at 3:OO pm solartime are (see solution in Problem 8-10 for reference)

Angle of lncidence θ = 65.0 deg.,Direct Solar lrradiation. Gρ = 1 18.6 Btu/hr-ft2,

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8-37

8-38

B-39

157

Diffuse Solar lrradiation: Ga a Gκ = 30.6 + 28.9 = 59'4 Btu/hr-ft2

Then, the area of the glazing and of the frame is calculated to be 27 .2 ft2and 4.8 ftz, respectively.

From Table 7-3, solar heat gain coefficients for the glazing system lD 5bare

SHGG9ο(65") = 0.515 and SHGGsα = 0.60.

From Τable 5-2, the outside surface conductance may be estimated to be4.0 Btu/hr-ft2-'F.

From Table 5-6, the U-value for the fixed, double glazed window havingaluminum-clad wood/vinyl frame with insulated spacers is 0.48 Btu/hr-ft2-"F.

From Τable 7-1, solar absorptance of the vinyl frame painted white is 0.26.

Αssuming the window with no setback (Ar,r'" = Asuπ), the SHGC for theframe can be calculated using Eq. 7-31 as:

SHGG1= 0.26*(0.4814.0) = 0.031.

For an unshaded window, the total solar heat gain is calculated using Eq.7 -32 as

Qsμc = (0.51 5*27 '2 + 0.031-4.8)-'1 '18.6

+ (0.60-27 '2 + 0.031*4.8)*59.4 = 2657 '2 Btυlhr'

This problem uses the same solution procedure as the previous problem.Assuming a south-facing window, the incidence angle and solar irradiationfor Boise, ]D on a clear Jυly 21 day at 3:00 pm solar time are (see solutionin Problem 8-1 '1 for reference)

Angle of lncidence' θ = 68.8 deg.,Direct Solar lrradiation: Gρ = 101.6 Btu/hr-ft2,Diffuse Solar lrradiation: Ga + Gκ = 29'0 + 28.9 = 57 '9 Btu/hr-ft2, and

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t '158

The area of the glazing and of the frame is the same as that calculated inProblem 8-38.

From Table 7-3, solar heat gain coefficients for the glazing system lD 29aare

SHGG,ο(68.8") = 0'408 and SHGGsα= 0.57.

From Table 5-6, the U-value for the fixed, triple glazed window havingaluminum-clad wood/vinyl frame with insulated spacers is 0.44 Btu/hr-ft2-"F.

The outside surface conductance and solar absorptance of the frame areassumed to be the same as those in Problem 8-38.

Assuming the window with no setback (Ar,u'" = Asuπ), the SHGC for theframe can be calculated using Eq. 7-3'1 as:

SHGGr = 0.26*(0 .4414.0) = 0.029.

For an unshaded window, the total solar heat gain is calculated using Eq.7-32 as.

Qsuc = φ'408-27 '2 + 0.029-4.8)*101.6+ (0.57*27.2 + 0.029*4.8)-57.9 = 2047.4 Btu/hr.

8-40

First, determine conduction heat gain by multiplying fluxes from Problem 8-33 by the surface area, 8OO ft2. Then, from Table 8-20, select theradiative/convective split to be 63%137o/o Apply the split to determine theconvective and radiative heat gains. Then, apply Equation 8-67 to theradiative heat gains to determine the radiative cooling load. Sum theradiative cooling load and the convective heat gain to get the cooling load.

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students enrolled in courses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnsΙαtion οf this τυork beyond ιhαt permiιιedby Secιions ] 07 or ] 08 of ιhe Ι 976 United StαιeS Cοpyrighι λcι νithοut ιhe permissiοn of the copyright owner is unlατνful.

HourConductionHeat Gain(Btu/h0

ConvectiveHG

RadiativeHG

RadiativeCoolingLoad

CoolingLoad

1 5462.3 2021.0 3441.2 2903.7 4924.72 4899.6 1812.8 3ο86.7 2813.1 4625.9

3 4334.5 1603.8 2730.7 2702.9 4306.7

4 3796.4 1404.7 2391.7 2583.0 3987.7

5 3300.5 1221 2 2079.3 2460.2 3681.4b 2854.1 1056.0 1798.1 2339.0 3395.07 2460.6 910.4 1550.2 2222.4 3132.9

8 2123.8 785.8 1338.0 2113.0 2898.99 1854.2 686.1 1168.2 2014.0 2700.1

10 1673.6 619.2 1054.4 '1930.6 2549.9

11 1598.0 591.3 1006.7 1868.0 2459.212 1629.0 602.7 1026.3 1829.3 2432.1

13 1759.4 651.0 1 108.4 '1815.8 2466.8

14 1983.0 733.7 1249.3 1827.8 2561.5

15 2318.0 857.6 1460.3 1868.9 2726.616 2803.7 1037.4 1766.3 1947.4 2984.817 3450.5 1276.7 2173.8 2068.9 3345.518 4215.9 1559.9 2656.0 2230.3 3790.2

19 5016.0 1855.9 3160.1 2419.2 4275.220 5741.3 2124.3 3617.0 2614.7 4739.0

21 6266.5 23'18.6 3947.9 2789.4 5108.022 6473.8 2395.3 4078.5 2913.3 5308.6ZC 6345.1 2347.7 3997.4 2969.4 5317.1

24 5971.2 2209.4 376'1.9 2961.2 5170.6

'159

Cooling Loads and Heat Gains

7ο00.0

60ο0.0

5000.0

4000.0

3000.0

2000.0

1 000.0

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Ιt

ΦΘG6)Ιoδ

ωoJ

0.0

16ο

8-41

First, determine conduction heat gain by multiplying fluxes from Problem 8-

35 by the surface area, 1OOO ft2. Then, from Τable 8-20, select the

radiative/convective split to be 84oλl16o/o' Apply the split to determine the

convective and radiative heat gains. Then, apply Εquation 8-67 to the

radiative heat gains to determine the radiative cooling load. Sum the

radiative cooling load and the convective heat gain to get the cooling load.

HourConductionΗeat Gain(Btuihr)

ConvectiveHG

RadiativeHG

RadiativeCooling

Load

CoolingLoad

1 2227.2 356.4 1870.8 4864.4 5220.7

2 1338.2 214.1 1124.1 4530.5 4744.6

J 627.7 100.4 527.3 4222.8 4323.2

4 57.3 9.2 48.2 3938.6 3947.7

5 -397.4 -63.6 -333.8 3676.2 3612.7

6 -746.0 119.4 -626.7 3436.0 3316.7

7 -935.8 -149.7 -786.0 3228.9 3079.2

o -610.8 -97.7 -513.1 3120.5 3022.8

9 507.2 81.2 426.1 3179.7 3260.8

10 2313.5 370.2 1943.3 3417.1 3787.2

11 4567.1 730.7 3836.3 3808.3 4539.0

12 7028.4 1124.5 5903.8 4316.0 5440.6

13 9455.0 1512.8 7942.2 4893.4 6406.2

14 116ο9.7 1857.6 9752.2 5486.8 7344.4

15 13293.3 2126.9 I 1 166.3 6042.8 8169.7

16 1 4350.1 2296.0 12054.1 651 1.5 8807.6

17 14672.7 2347.6 12325.1 6849.5 9197.1'18 14222.5 2275.6 I 1946.9 7024.8 9300.4

19 13ο18.3 2082.9 1Ο935.4 7018.0 9101.0

20 11142.9 1782.9 9360.0 6824.5 8607.3

21 8809.6 1 409.5 7400.1 6467.5 7877.0

22 6593.1 1ο54.9 s538.2 6042.8 7097.7

z3 4782.8 765.3 4017.6 5622.2 6387.5

24 3353.'1 536.5 2816.6 5228.0 5764.5

Exc€φts frοm this work may be reprοduοed by instructors for distribution οn a not-for-profit basis for testing or instructional puφoses only to

students enτo]led in courses for which the textbook has been adopted. Αny oιher reprοducιion or trαnsιCιιion of thιS work beyond ιhαt permitted

η s;"iι'n' ]07 οr ]08 οfthe Ι976 Uniιed Stαtes Copyrighι Αctwiιhout ιhe permission οfιhe cοpyright oνner is unlαwful.

161

cooling Loads and Ηeat Gains

L

flαl

(!(,(!Φ!οδ

ΦoJ

1 6000.0

1 4000.0

1 2000.0

'10000.0

8000.0

6000.0

--r-- Conduction Heat Gain(Btu/h0

-*x* Cooling Load

8-42

First, determine conduction heat gain by multiplying fΙuxes from Problem 8-36 by the surface area, 1200 ft2 ' Ther , from Τable 8-2o, seΙect theradiative/convective Split to be 84γoμe% Αpply the split to determine theconvective and radiative heat gains. Then, apply Equation 8-67 to theradiative heat gains to determine the radiative cooling load. Sum theradiative cooling load and the convective heat gain to get the cooling load.

Hour

ConductionHeat Gain(Btιl/hr)

ConvectiveHG

RadiativeΗG


CoolingLoad

1 809.2 129.5 679.7 1203.0 1332.42 481.0 77.0 404.1 1 105.3 1182.33 238.5 38.2 200.4 1018.3 1056.54 61.2 9.8 51.4 941.3 951.15 -69.3 11 .1 -58.2 872.8 861.76 -166.0 -26.6 -139.5 811.4 784.97 -237.0 -37.9 -199.0 756.2 718.3o -278.7 -44.6 -234.1 707.9 663.3I -250.9 -40.2 -210.8 674.0 633.810 -90.4 -14.5 -75.9 667.9 653.411 233.3 37.3 196,0 699.7 737.112 700.2 112.0 588.2 771.3 883.3

Excerpts Γrom this work may be reproduοed by instructοrs for distribution on a not_for_profit basis for testing or instruοtional puφoses only tostudents enrolled in οourses for which the textbook has been adopted. Αny other reproducιion or trαnslαιion of this work beyοnd ιhαt permiιtedby Secιions 107 οr 108 ofthe Ι976 United Stαtes Copyright Αcινηithοut the permissiοn ofthe copyrighι oνner is unlcrwful.

911131517192123

13 1260.3 201.6 1058.6 877.1 1078.8

14 1852.6 296.4 1556.2 1007.4 1303.8

15 2414.9 386.4 2028.5 1149.7 1 536.1

16 2890.5 462.5 2428.0 1290.4 1752.9

17 3233.3 517.3 2716.0 1416.6 1933.9

18 3409.7 545.6 2864.2 1516.5 2062.019 3401.3 544.2 2857.1 1580.7 2124.920 3205.0 512.8 2692.2 1602.7 2115.5

21 2833.6 453.4 2380.2 1579.7 2033.0

22 2323.3 371.7 1951 .6 1513.7 1885.5

23 1759.4 281.5 1477.9 1417.5 1699.0

24 1237.6 198.0 1039.6 1309.3 1507.3

162

Gooling Loads and Heat Gains

4000.0

3500.0

3000.0

2500.0

2000.0_--ο_ Conduοtion Heat Gain

(Btu/hr)

*_α_* Cooling Load1500.0

'1000.0

500.0

0.0

-500.0

First, the hourly Soιar heat gains are determined using the same solutionprocedure Shoν1/n in Problem 8-38. The results are Sho\Λ/n below. Notethat the SolaΓ irradiation on the window is the Same aS that shown in

Problem 8-10. Also, note that the calculated Soιar gain at 3:00 p.m. isslightly different from that shown in Problem 8-38 due to rounding errors.

αt

ΦΘ(τΦΙcδ

ιl1oJ

8-43

lnput DataGlass AreaFrame Αrea

27.2 ft',

4.8 ft2

Excerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional purposes οnly to

students enrolled in courses for whiοh thο textboοk has been adopted. Αny οιher reproducιiοn or ιrαfiSlαιiοn of this work beyond thαι permiιιed

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163

Diffuse SHGCAngular SHGC - 0'Angular SHGC - 40'Angular SHGC - 50'Angular SHGC - 60"Angular SHGC - 70"Αngular SHGc - 80"

Frame sΗGc

0.60.70.670.640.580.450.23

0.03'1

Output Datalnc. Dir lrradiation, Diff lrradiation, Solar Heat Gain,

Αngle, ' Btu/hr-ft2 Btu/hr-ft2 Btu/hr87.03 0.00 0.00 0.00101 .06 0.0ο 0.00 0.00114.99 0.00 0.00 0.00128.63 0.0ο 0.00 0'00141.58 0.00 0.00 0.00152.80 0.00 0.00 0.00159.17 0.00 9.93 163.51156.37 0.00 23.08 380.04146.65 0.00 32.24 530.96134.23 0.00 39.29 647.01120.81 0.00 44.53 733.28106.97 0.00 47 .91 788.9892.97 0.00 52.42 863.2878.94 54.68 56.44 1314.4465.01 1 18.63 59.38 2657 .0551.37 170.01 60.32 3940.0638.42 201.20 57.93 4657.1227.20 202.07 50.35 4594.5520.83 151.27 33.70 3393.4023.63 5.07 1.03 111.7433.35 0.0ο 0.00 0.0045.77 0.00 0.00 0.0059.19 0.00 0.00 0.0073.03 0.00 0.00 0.00

ClockTime1.002.003.004.005.006.007.008.009.0010.0011.0012.0013.0014,0015.0016.0017.0018.0ο19.0020.0021.0022.0023.0024.00

ln the original RΤS methodology, two types of radiant time faοtors v/ereutilized to convert So|ar heat gains into cooling Ιoads. The Solar.-RTS wasused to convert the beam transmitted solar gain whiΙe the NonsoΙar-RΤSV/aS used to convert all other Solar gains' However, to simpΙify thecalculations, only one RTS (Nonsolar-RTs) is used in this edition. Sincethe calculated Solar heat gains include both transmitleΞ aιd absorbed Solargains, the recommended radiative and convectivΞsplits shown in Table 8-20 would not be applicable. For this problem, it is assumed that theradiative fraction of the combined solar heat gain is about 0.9. Therefore,

Exοeφts from this wοrk may be reproduced by instructors fοr distribution on a not-for-profit basis for tΘSting or instructional purposes only tοstudentsenrolledincoursesforwhichthe textbookhasbeenadopted. Αnyοtherreproductionοrtrαnslαtionοfιhisνοrkbeyondιhαιpermiιιedby Sectiοns 107 or ]0B ofthe Ι976 Uniιed Stαtes Copyrighι Αcιwιιhοuι ιhe permissiοn ofιhe cοpyright oνner is unlατνfuΙ.

164

the radiative/convective split is 90%110%. Then, apply the split todetermine the convective and radiative heat gains and apply Equation 8-67to the radiative heat gains to determine the radiative cooling load. Αnd,finally, sum the radiative cooling load and the convective heat gain to getthe cooling load.

Hour

SolarHeat Gain

(Btu/hr)Convective

HGRadiative

HG

RadiativeCooling

LoadCooling

Load

1 0.0 0.0 0.0 141 0 141.0

2 0.0 0.0 ο.0 91.6 91.6

3 0.0 0.0 0.0 59.8 59.8

4 0.0 0.0 0.0 39.2 39.2

5 0.0 0.0 0.0 25.9 25.9

6 0.0 0.0 0.0 17.2 17,2

7 163.5 16.4 147.2 87.6 103.9

I 380.0 38.0 342.0 215.2 253.2

9 531.0 53.1 477.9 339.6 392.7

10 647.0 64.7 582.3 450.5 515.2

11 733.3 73.3 660.0 543.8 617.1

12 789.0 78.9 710.1 615.8 694.7

13 863.3 86.3 777.0 687.3 773.6

14 1314.4 131.4 1 183.0 932.5 1064.0

15 2657.1 265.7 2391.3 1661.8 1927.5

16 3940.1 394.0 3546.1 2566.6 2960.617 4657.1 465.7 4191.4 3304.7 3770.4

18 4594.6 459.5 4135.1 3630.8 4090.3

19 3393.4 339.3 3054.1 3261.0 3600.4

20 111.7 11.2 100.6 1624.6 1635.8

21 0.0 0.0 0.0 912.1 912.1

22 0.0 0.ο 0.0 549.4 549.4

23 0.0 0.0 0.0 343.2 343.2

24 0.0 0.0 ο.0 218.7 218.7

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by Secιions 1 07 or ] 08 of the 1 97 6 Uniιecl Sιαιes Copyrιghι Αcι |uithοuι the permissiοn οf ιhe copyright owner is unΙαινful.

--165

IΦ

(!ΘGΦJ.Φ(!oJ

5000.0

4500.0

4000.0

3500.0

3000.0

2500.0

2000.0

1500.0

1 000.0

500.0


OutPut DataDir lrrad, Btu/hr- Diff lrrad, Btu/hr-

ft2 f1'

0.00 0.000.00 0.000.00 0.00

_ο- Soiar Heat Gain (Btu/hr)

*x- Cooling Load

Solar Heat Gain,Btu/hr0.000.000.00

0.0

B-44

This problem uses the same solution procedures as Problem 8-43. Note

that the solar irradiation on the window is the same as that shown in

Problem 8-11. Also, note that the calculated Soιar gain at 3:00 p.m. is

slightly different from that shown in Problem 8-39 due to rounding errors.

ThΞ radiative/convective split of 90%l10% is also used for this problem.

lnPut DataGlass Area 27.2 ft'Frame Area 4.8 ft2

Diffuse SHGC 0.57Angular SΗGc - 0" 0.68Angular SHGC - 40' 0.65Αngular SHGO - 50' 0.62

Angular SHGC - 60' 0.54

Angular SHGC - 70' 0.39Angular SHGC - 80" 0.18

Frame SHGC 0.029

ClockTime1.002.003.00

lnc.Angle,'152.14154.11'150.59

EΧοeφtS from this work may be reprοduοed by instructors fbr distribution on a not-for-profit basis for testing or instructional puφoses only to

students enrolled in courses fbr which the textbook has been adopted. Αny other reproiucιιon or trαnslαιiοn οf ιhis νork beyond thαι Permiιιed

by Sectiοns Ι 07 or Ι 08 οf the ] 97 6 (]niιed Stαtes Copyright Αct wiιhouι the permission οf the copyrighι oινner is unlωυful.

166

4.005.006.007.008.009.0010.001 1.0012.00'13.00

14.0015.0016.0017.0018.0019.0020.0021.0022.0023.0024.00

143.20133.87123.75113.42103.2993.6684.8877.3771.5868.0267.0668.8273.0979.4487.3896.45106.26116.49126.80136.76145.67

0.000.000.000.000.000.0022.7058.9287.76105.541 10.30101.5680.2248.7011.340.000.000.000.000.000.00

0.000.000.003.9617.8329.6939.6848.2354.8158.8559.9457.9553.1 0

45.8636.8426.4614.280.320.0ο0.000.0ο

0.00ο.00ο.00

6'1.91278.89464.39680.731139.731721.362140.212255,352046.811551.04978.1 5592.46413.94223.454.970.000.000.00

Ηour

SolarHeat Gain(Btιl/hr\

ConvectiveHG

RadiativeΗG

RadiativeCooling

LoadCooling

Load1 0.0 ο.ο 0.0 356.3 356.32 0.0 0.0 0.0 341.8 341.83 0.0 0.0 0.0 328.6 328.64 0.0 0.0 0.0 316.4 316.45 0.0 0.0 0.0 305.0 305.0o 0.0 0.0 0.0 294.1 294.17 61.9 6.2 55.7 297.0 303.1I 278.9 27.9 251.0 338.9 366.8I 464.4 46.4 418.0 391.2 437.610 680.7 68.1 612.7 457.1 525.211 1139.7 114.0 1025.8 583.2 697.212 1721.4 172.1 1549.2 763.0 935.113 2140.2 214.0 1926.2 935.3 1149.314 2255.4 225.5 2029.8 1046.5 1272.015 2046.8 204.7 1842 1 1067.8 1272 4to 1551.0 1 55.1 1395.9 991 0 1146.117 978.1 97.8 880.3 853.6 951.418 592.5 59.2 533.2 723.9 783.119 413.9 41.4 372.6 633.6 675.020 223.4 22.3 201.1 551.8 574.121 5.0 0.5 4.5 465.4 465.922 0.0 0.0 0.0 421.3 421.323 0.0 ----ΑΔ 0.0 393.5 393.524 0.0 0.0

..--o.0373.0 373.0

Exceφts from this wοrk may be reproduοed by instructors Γor distribution on a not-for-profit basis for testing or instruοtional puφoses only tostudents enrolled in courses for whiοh the textbοok has been adoρted' Αny oιher reprοducιion or ιrαnslαιιon of ιhis νork beyond thαι permιιιedby Secιions ]07 or ]08 ofιhe Ι976 Uniιed Stαtes Cοpyrighι Αcιlνithout lhe permissiοn ofιhe copyright oνner is unΙcrννful'


L

Φ

ιEΘ(EΦΙΦΦoJ

2500.0

2000.0

I 500.0

1 000.0

500.0

0.0

--l_Solar Ηeat Gain (Btu/hr)

,--x- Cooling Load

8-45

167

Hourlnternal

Heat Gain(w)

ConvectiveHG

RadiativeHG


CoolingLoad(Vψ

I 200.0 100.0 100.0 125.3 225.32 200.0 100.0 100.0 116.6 216.63 200.0 100.0 100.0 111.O 211.04 200.0 100.0 100.0 107.3 207.35 20ο.0 100.0 100.0 105.0 205.06 200.0 100.0 100.0 103.4 203.47 200.0 100.0 100.0 102.4 202.4δ 2000.0 1ο00.0 1000.0 566.7 1566.7I 2000.0 1000.0 1000.0 753.8 1753.810 2000.0 1000.0 1000.0 85'1.1 1 851 .1

11 2000.ο 10οο.ο '10οο.ο 906.9 1906.912 2000.0 1000.0 1000.0 940.8 1940.813 2ο00.0 1000.0 1000.0 962.1 1962.1

14 2000.0 1ο00.0 1000.0 975.6 1975.615 2000.0 1000.0 1000.0 984.3 1984.316 2000.0 1000.0 1000.0 990.0 1990.017 2000.0 1000.0 1000.0 993.6 1993.618 2ο00.ο 1000.0 1000.0 996.0 1996.019 200.0 100.0 100.0 532.6 632.6

EXceφtS from this wοrk may be reprοduced by instructors fοr distribution on a not-fοr-profit basis for testing or instructional puφοses only tostudents enτolled in οourses for which thθ teΧtbook has been adopted. Αny οιher reproduction or trαnsΙαtion of this τνοrk beyond ιhαι permittedbySectiοns Ι07 or Ι08of ιhe ]976UniιedStαtesCοpyrightΑctwithοutthepermissiοnοf thecοpyrighιownerisunlcrννful.

20 200.0 100.0 100.0 346.2 446.221 200.0 100.0 100.ο 249.2 349.222 200.0 100.ο 1ο0.ο 193.6 293.623 200.0 100.0 100.0 159.9 259.924 200.0 100.0 100.0 138.7 238.7

'168


2500.0

2000.0

1500.0+- lnternal l-1eat Gain (W)

*x- Cooling Load (\Λ/)

Hour

lnternalHeat Gain

rw)Convective

HGRadiative

HG

RadiativeCooling

Load

CoolingLoad(w)

1 200.0 100.0 100.0 405.'1 505.12 200.ο 1οο.0 '100.0 392.9 492.93 2ο0.ο 100.0 100.0 381.7 481.74 200.0 100.0 100.0 371.2 471.25 200.0 100.0 100.0 361.3 461.36 200.0 100.0 100.0 352.1 452.17 200.0 100.0 100.0 343.3 443.3I 2000.0 1000.0 1000.0 518.9 15'18.9o 2000.0 1000.0 1000.0 562.8 1562.810 2000.0 1000.0 1ο00.0 59ο.6 1590.611 2000.0 1000.0 1000.0 612.3 1612.312 2000.0 1ο00.ο 10ο0.ο 630.8 1 630.8'13 2000.0 1000.0 1000.0 647.1 1647.1

14 2000.0 10οΟ.0 1000.0 661.9 1661.9

Εxοeφts from this work may be reproduced by instΙuctors for distribution on a not-for-profit basis for testing or instructional puφoses only tostudents enrolΙed in οourses for which the textbook has been adopted' Αny oιher reprοductιοn or ιrαnsl(tιion of ιhis work beyond thαt permiιιedby Sectiοns Ι 07 or ] 08 of the 1 976 Uniιed Stαtes Copyrιghι Αcι'ννiιhouι ιhe permissionbf the cοpyrighι οwner is unΙαlνful.

Ξ'6o(!(Σ)Ιoδ

(υoJ

8-46

Γ

'169

15 2000'ο 100ο.0 1000.0 675.5 1675.516 2000.0 1000.0 1000.0 688.2 '1688.2

17 2000.0 1000.0 1000.0 700.0 1700.018 2000.0 1000.0 1000.0 711 .1 1711 .1

19 200.0 100.0 100.0 537.6 637.620 200.0 100.0 100.ο 495.7 595.721 20ο.ο 100.0 1ο0.ο 469.9 569.922 200.0 100.0 100.0 449.9 549.923 200.0 100.0 100.0 433.2 533.224 200.0 100.0 100.ο 418.4 518.4

Gooling Loads and Heat Gains

2500.0

2ο00.0

1500.0

1000.0

500.0

0.0

---ο- lnternal Fleat Gain (W1

*-s* οooling Load (W)

&-x** *

; :. .::l L-.*

ΕΧcerpts tiom this ινork may be reproduοed by instructοrs for distributiοn on a not_for-profit basis fοr testing or instruοtional puφoses only tostudents enrolΙed in οouτses 1br whiοh the textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this work beyond ιhαι permiιιedby Secιiοns ]07 or ]08 οfιhe Ι976 LΙnited StαιeS Cοpyright Αcι νiιhouι ιhe permiSSiοn οfιhe copyrighι oνner is unlαwful'

ΞoΞ(!o

σ,:=oooτ,(τ,

aι'

'6Θ(τ,α)-t-

170

8-47

Gomparison of LW and Mνι/ 1 Zone Responses

2000.0

1500.0

1000.0

50ο.ο

Ι

ι*.\Η\ 5**+-

ι ξ--'. _l)*ι ':'*Ι^__x

--r-- lnternal Heat Gain β)_--α.* MW1 Zone Clg. Ld αv)-_.*- ΗW Zone Clg. Ld. (W)

232119171511 '13

Hour

As shown in the figure, there is a signifrcant difference in the response ofthe two Zones, with the ΗW zone having substantiatly more damping andtime delay.

Exceφts f?om this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only tostudents enrolled in courses fοr whiοh thΘ textbook has been adopted. Αny oιher reproclucιiοn or ιrαnsιαιionbf ιhιs work beyoλd ihαι permiιιedby Sections ]07 οr ]08 ofιhe }976 United Stαιes Cοpyright Αctwithοut the permissiοn οfιhe cοpyrighι οwner ιi unlανful'

171

8-48

Assumptions applied to each heat gain are discussed in the solution toProbΙem 8-15. The equipment heat gain is assumed to be continuous. Thetotal convective and radiative heat gains are determined in the folΙowingtable. Τhe Ιatent cooΙing Ιoads are equivalent to the latent heat gainsshown in the last column.

Name: ξ99p]9_ Liοhtino Eοuipment ΤotalTotal

RadiativeTotal

Convective

Latentfrom

PeopleRadiativeFraction: 0.7 0.59 0.2

lloυr

ΗeatGain(\Λ/)

ΗeatGain/w)

Heat Gain(w)

HeatGain (W)

HeatGain (W)

Heat GainrιΛΛ

HeatGain(w)

1 0 0 4000 4000 800 320C) 02 U 0 4000 4000 800 320ο 03 0 0 400ο 4000 800 3200 04 ο 0 4000 4ο00 800 3200 05 0 0 4000 4000 800 3200 06 0 0 4000 4000 800 3200 07 ο 0 4000 4000 800 3200 n

o 2160 4800 4000 1 0960 5144 581 6 1 350I 2160 4800 400ο 1 0960 5144 581 6 1 35010 2160 4800 4000 1 0960 5144 581 6 1 35011 2160 4800 4000 1 0960 5144 5816 1 35012 2160 4800 4000 1 0960 5144 581 6 1 35013 2160 4800 4000 1 0960 5144 581 6 1 35014 2160 4800 4000 1 0960 5144 581 6 1 35ο15 2160 48CI0 4000 1 ο960 5144 5816 1 35016 2160 4800 4000 1 0960 5144 581 6 1 35017 2160 4800 4000 1 0960 5144 58l 6 1 35018 0 4800 4000 8800 3632 5168 019 0 0 4000 4000 800 3200 020 U n 4000 40ο0 80ο 32ο0 021 ο 0 4000 400ο 800 3200 022 0 4οο0 400ο 800 3200 023 0 0 4000 4000 800 3200 024 0 0 4000 4000 800 32ο0 0

The sensible loads are then determined from the radiative and convectiveheat gains using Equation 8-67 and the radiant time factors from Τable 8-21, as shown in the next table.

ΕΧceφts liom this work may be reproduced by instructors 1br dlsιribιltlon οn a noιtbr-pτofit basis for testing or instructiona] purposes onΙy tosιuderrts en:-olled it-ι οοιlrses fοr ιhich the teΧιbook has bοen adοptοd. lπ;l οιΙιer reprοdιιction or trαnslcιιioιι cf ιhis ιι'ork beyond ιhαι ρerfrιiιtedby Sectiοlιs ] 07 οr Ι08 οf ιhe Ι 97 6 Uniιed StαιeS Copyrighι Αcι ινiιhouι ιlιe ρernιission οf the cοpyrighι ο'!'ner iS unlαιν.ful'

172

Hour

lnternalHeat Gain

(w)Convective

ΗGRadiative

HG

RadiativeCooling

Load

CoolingLoad(w)

1 4000.0 32ο0.0 800.0 2218.4 5418.4

2 4000.0 3200.0 800.0 2162.5 5362.5

3 4000.0 3200.0 800.0 2110.6 5310.6

4 40ο0.0 3200.0 800.0 2062.1 5262.1

5 4000.0 3200.0 800.0 2016.7 5216.7

6 4000.0 3200.0 800.0 1973.8 5173.8

7 4000.0 3200.0 800.0 1933.3 5133.3

8 10960.0 5816.0 5144.0 2782.4 8598.4

9 10960.0 5816.0 5144.0 2996.'1 8812.1

10 10960.0 5816.0 5144.0 3131.2 8947.2

11 10960.0 5816.0 5144.0 3237.5 9053.5

12 10960.0 5816.0 5144.0 3328.ο 9144.0'13 10960.0 5816.0 5144.0 3408.1 9224.1

14 1096ο.0 5816.0 5144.0 3480.7 9296.7

15 10960.0 5816.0 5144.0 3547.6 9363.6

16 '10960.0 58 1 6.0 5144.0 3609.6 9425.6

17 10960.0 5816.0 5144.0 JOO /. / 9483.7

18 88ο0'0 5168.0 3632.0 3413.3 8581.3'19 4000.0 3200.0 800.0 2799.1 5999.1

20 4000.0 320ο.0 8ο0.0 2625.5 5825.5

21 4000.0 3200.0 800.0 2512.4 5712.4

22 4000.0 3200.0 800.0 2422.9 5622.9

23 4000.0 3200.0 800.0 2346.8 5546.8

24 4000.0 3200.0 800.0 2279.4 5479.4


Ξ'6ΘΦΦΞoδ

π,oJ

120ο0.0

10000.0

8000.0

6000.0

4000.0

2000.0

0.0

ι

Εxοeφts fiom this woτk may be reproduced by instructofs for distrlbution on a not-for-profit basis for testing or instructional purposΘs only to

students enrolled in οouτses foτ which thΘ textbοok has been adopted' Αny other reprοducιion or trαnsιαιion of this νοrk beyond ιhαt permiιted

by Secιions Ι07 or Ι08 οfthe ]976 ιJnited Stαtel Cοpyrighι Αct\'ιithout the permissiοn ofιhe cοpyrighι owner is unΙαlυfuΙ.

8-49

173

Assumptions applied to each heat gain are discussed in the solution toProblem 8-'16. The equipment heat gain is assumed to be continuous; thelighting heat gain is assumed to occur from 8 a.m.-6 p.m. The totalconvective and radiative heat gains are determined in the following table.The latent cooling Ιoads are equivalent to the latent heat gains shown in thelast column.

Name: People Liohtinο Eουioment TotalΤotal

RadiativeTotal

Convective

Latentfrom

Peοοte

RadiativeFraction: 0.7 ο.59 0.2

Hour

HeatGain(w)

HeatGain(w)

Heat Gainrw)

HeatGain (W)

HeatGain (W)

Ηeat Gain/\ΛΛ

HeatGain(w)

1 0 0 70ο0 70οο 1400 5600 0

2 0 ο 7000 7000 1400 5600 0

3 0 0 7000 7ο00 1400 5600 0

4 0 0 7000 7000 1400 5600 0

5 U 0 7000 70ο0 1400 5600 0

6 0 0 7000 7000 1400 560ο 0

7 0 0 7000 7ο00 140ο 5600 0

I 2835 5625 7000 1 5460 6703.25 8756.75 28359 2835 5625 7000 1 5460 6703.25 8756.75 2835't0 2835 5625 7000 1 5460 6703.25 8756.75 283511 2835 5625 700ο 1 5460 6703.25 8756.75 283512 2835 5625 7000 1 5460 6703.25 8756.75 283513 2835 5625 7000 1 5460 6703.25 8756.75 283514 2835 5625 7000 1 5460 6703.25 8756.75 2835

15 2835 5625 7000 1 5460 6703.25 8756.75 2835'16 2835 5625 7000 1 5460 6703,25 8756.75 283517 2835 5625 7000 '15460 6703.25 8756.75 2835

18 0 5625 70ο0 12625 4718.75 7906.25 0

19 0 0 7000 7000 1 400 5600 0

20 0 0 7000 7000 1400 5600 0

21 0 0 7000 7000 1400 5600 0

22 0 ο 7000 7000 1 400 5600 0

23 0 ο 7000 700ο 1400 5600 ο

24 0 0 7000 7000 1400 5600 0

The sensible loads are then determined from the radiative and convectiveheat gains using Εquation 8-67 and the radiant time factors from Τable 8-21 , as shown in the next table.

EXceφtS fτοm this work may be reproduοed by instruοtors for distribution on a not_for_prοfit basis Γor testing or instructional puφoses only to

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174

Hour

lnternalHeat Gain

(w)Convective

ΗGRadiative

HG

RadiativeCooling

Load

CoolingLoad(w)

1 7000.0 5600.0 1400.0 2888.2 8488.2

2 7000.ο 5600.0 1400.0 2830.6 8430.6

3 7000.0 5600.0 1 400.0 2777.5 8377,5

4 7000.0 5600.0 14ο0.0 2727.8 8327.8

5 7000.ο 5600.0 14ο0.0 2680.7 8280.7

6 7000.0 5600.0 1 400.0 2635.8 8235.8

7 7000.ο 5600.0 140ο.0 2592.7 8192.7

8 15460.0 8756.8 6703.3 3814.6 12571.3

9 15460.0 8756.8 6703.3 4292.6 13049.3

10 15460.0 8756.8 6703.3 4539.8 13296.5

11 15460.0 8756.8 6703.3 4694.4 13451.2

12 15460.0 8756.8 6703.3 4806.9 13563.6

'13 1546ο.0 8756.8 6703.3 4897.4 13654.2

14 15460.0 8756.8 6703.3 4975.3 13732.1

15 15460.0 8756.8 6703.3 5045.3 13802.0

ιo 15460.0 8756.8 6703.3 5109.7 13866.5

17 15460.0 8756.8 6703.3 5170.2 '13926.9

'18 12625.0 7906.3 4718.8 4754.8 12661.0

19 7000.0 5600.0 1400.0 3825.2 9425.2

20 7000.0 5600.0 1400.0 3446.7 9046.7

21 7000.0 5600.0 1400.0 3246.7 8846.7

22 7000.0 5600.0 1400.0 31 19.9 8719.9

23 7000.0 5600.0 1400.0 3027.2 8627.2

24 7000.0 5600.0 1400.0 2952.4 8552.4


Ξ'6Θ(!q)ΞoE

IEoJ

18000.0

16000.0

14000.0

'12000.0

10000.0

8000.0

6ο00.0

4000.0

2000.0

0.0

---G- lnternal Fteat Gain (W;

*x* Cooling Load (νγ)

Εxcerpts tioιτ this work may be reproduced by instΙuctors fοτ distribution οn a not-for-profit basis for tΘsting or instruοtional puφoses only to

students enrolled in οourses tbr which the textbook has been adopted. Αny other reprodλc'tiοn or trαnsιαιιon of thιs νork beyond ιhαt permitted

iysr,itιon, 107 οr ]0Bοf the 1976ΙJnitedStαtesCopyrighιΑctlνithoutthipermissiοnof thecopyrightoνnerisunΙαινful'

8-50

175

Heat gain to the space = 0.8 x 6000 W = 4800 W, assumed 59% radiative,41o/o convective, from Τable 8-20' The sensible loads are then determinedfrom the radiative and convective heat gains using Equation 8-67 and theradiant time factors from Table 8-21, as shown below. Τhere are no latentcooling loads.

Ηour

lnternalHeat Gain

(w)Convective

ΗGRadiative

HG

RadiativeCooling

Load

CoolingLoad(w)

1 0.0 0.0 0.0 79.8 79.8

2 0.0 0.0 0.0 52.4 52.4

3 0.0 0.0 0.0 34.7 34.7

4 0.0 0.0 0.0 23.1 23.1

5 0.0 0.0 0.0 '15.6 15.6

6 4800.0 1968.0 2832.0 1473.9 3441.9

7 4800.0 1968.0 2832.4 2060.5 4028.5

8 4800.0 1968.0 2832.0 2365.5 4333.5

9 4800.0 1968.0 2832.0 2540.5 4508.5

10 4800.0 1968,0 2832.0 2646.7 4614.7

11 4800.0 1968.0 2832.0 2713.2 4681.2

12 4800.0 '1968.0 2832.0 2755.6 4723.6

13 4800.0 1968.0 2832.0 2782.9 4750.9

14 4800.0 1968.0 2832.0 2800.5 4768.5

15 4800.0 1968.0 2832.0 2812.0 4780.0

16 4800.0 1968.0 2832.0 2819.5 4787.5

17 4800.0 1968.0 2832.0 2824.3 4792.3

18 48ο0.0 1968.0 2832.0 2827.5 4795.5't9 0.0 0.0 0.0 1366.5 1366.5

20 0.0 0.0 0.0 777.9 777.9

21 0.0 0.0 0.0 471.8 471.8

22 0.0 0.0 0.0 296.0 296.0

23 0.0 0.0 0.0 189.3 189.3

24 0.0 0.0 0.0 122.5 122.5

A plot showing the lighting heat gain and resuιting cooling loads follows.

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by Secιionι 107 οr ]08 ofthe Ι976 United Sιaιes Copyri?hι Αcι1|,ithouι the permission ofιhe copyrighι olνner is unlανful.

176

t-6000.0

5000.0

4000.0

3000.0

Ξ,6

Θ(υΦΙoδ

ιl,oJ

--{- |nternal Fleat Gain (\Λl)

___x* Cooling Load (\Λ/)

57 9 11 13 15 17 19 21 23

Hour

ι-

8-5'1

The schedule described in problem 8-18 is reduced to the number of

peopιe present per hour in the table below. Assuming "Seated, light office

*ork'', the sensible heat gain per person is 245 Btu/hr (72νν) and the latent

heat gain per person is 2bO Btu/hr (59 W). lnternal heat gains from

occupants are assumed to be 70o/o radiative. The latent cooling loads are

equivalent to the latent heat gains shown in the table.

The sensible loads are then determined from the radiative and convective

heat gains using Equation 8-67 and the radiant time factors from Table 8-

21 , as shown below.

As is readily evident from the plot, the heavyweight zone significantly

damps the response to the heat gains'

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177

HourPeopleοresent

lnternalHeat Gain

(w)Convective

HGRadiative

HG

RadiativeCooling

Load

CoolingLoad(w)

Latentheatgain(\Λ/)

1 0 0.0 0.0 0.0 20.6 20.6 0

2 0 0.0 0.0 0.0 12.5 12.5 ο

3 0 0.0 0.0 0.0 7.8 7.8 0

4 0 0.0 0.0 ο.0 5.0 5.0 0

5 0 0.0 0.0 0.0 3.4 3.4 0

6 0 0.0 0.0 0.0 2.5 2.5 0

7 0 0.0 0.0 0.0 2.0 2.0 0

I 0 0.0 0.0 0.0 1.6 1.6 0

9 40 2880.0 864.0 2016.0 1021.8 1885.8 2360

10 40 2880.0 864.0 2016.0 1484.5 2348.5 2360

11 60 4320.0 1296.0 3024.0 2233.7 3529.7 3540

12 60 4320.0 1296.0 3024.0 2593.8 3889.8 3540

13 60 4320.0 '1296.0 3024.0 2784.5 4080.5 3540

14 70 5040.0 1512.0 3528.0 3143.9 4655.9 4't30'15 70 5040.0 1512.0 3528.0 33'17.9 4829.9 4130

16 70 5040.0 1512.0 3528.0 3410.8 4922.8 4130

17 10 720.0 216.0 504.0 1931 .3 2147.3 590

18 0 0.0 ο.0 0.0 1010.7 1010.7 0

19 0 0.0 0.0 0.0 552.7 552.7 0

20 0 0.0 ο.0 0.0 309.2 309.2 0

21 0 0.0 0.0 0.0 175.8 175.8 0

22 0 0.0 0.0 0.0 101.1 101.1 0

23 0 0.0 0.0 0.0 58.8 58.8 0

24 0 ο.0 0.0 0.0 34.6 34.6 0


Ξ'ΞΘπ,ΦΞοδ

π,oJ

6000.0

5000.0

4000.0

3000.0

2ο00.0

1000.0

--*- lnternal Heat Gain (W1

*-s* Cooling Load (W)

Exceφts from this work may be reproduοed by instructors fοr distribution on a not-for-profit basis for testing or instruοtional puφoses only to

students enrolled in course s for which the textbook has been adopted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhιs νork beyond ιhαt permitιed

by Secιiοns ] 07 or ] 08 οf the Ι 97 6 (Jnitecl Sιαιes Copyrιghι Αcι Ι|iιhοuι the permissiοn οf ιhe copyright οwner is unlανfuΙ.

178

Solution to be provided by an instructor.

Excerpts from this wοrk may be reprοduced by instru:lor:.fo, distribution on a not-for-profit basis for

testing or instruοtional purposes on1y ω stuαeλs enrο1led i, "or.r"s for which the textboοk has been

adopted. Αny other repiodλction "y γ""ititi"λ

o}thιs ινork beyo'nd thαt permitted by Sections ]07 or 10B

of the 1976 (Jnitecl itotus Copyright Αct 'ιiinλ, the permission-o/'λu copyrιgit owner is unlαwful'

h'uouurt, for permission or further ιnfοrmαtionilrouιa i" oaarrrruiΙo ιhe Piimiision Depαrtment, John

'';i,;;;'';:"{,-i", il l Riνir Street' Hoboken' NJ 07030'

CHAPTER 9

From Table 9-1, the number of average degree day is 6283'

From Fig. 9-1, Co = 0'60Using Εq' 9-2,

f-

Or F = 438.7 mcf of natural gas

= 438,727 std ft3

= 102,867 kw-hr

(24hr l dαy)(6283' F _ dαy)\2-^2-^1000Β tu l hr)(}'60)

(0 η(7ο_ 12"tr)(10ο \Βtu l stdff)

f- (24hrldαy)(6283'F_dαy)(Ζ-^?5,000Βtulhr)(0'60)(1 J) ql o _Γz' r)βaL2Βtu l kW _ hr)

$Elec = 102,867(0.1 O) = $1 0'287

$Gas = 438.7(4.5) = $1,974

SΕΙec_$Gαs |0287_ι974= 4.2

$Gαs ι914

or the electric cost is about 5'2 times as much'

(1 02,s 67 k-t4l _ hDβ a lΨ !u_

l kY-' hr)

@at)(000stdff lmΦSource energy using eιec' =

= 1063'6 mcf

Source energy using gas = 438'7 mcf

>_Ξ-

9-3

9-4

ΕSΕ,_ΕSG 1063.6 _438'7

"o,"ff=-tiU - = 1.42

That is eleο. heat uSeS 242% more Source energy'

The following are information for Washington' DU'

From Table g-l,1he number of average degree day is 4224'

From Fig. 9-1 , CD= O'62'

From Table B-1a, the outdoor temperature is 20 "F'

Forenergyefficientfurnace,assume85o/oeffiοiencyfactor,Using Εq' 9-2,

04hr l d αν)(4224" F _ dαy)(|20'000Βt u l hr)(0'62\,-, - , 'o'=177,468stdft3" =

-10.85X7O_

20"trX1O00Βtu l stdft3)

Or F = 177 .5 mcf of natural gas

Qro" = ato + 5

120,OO0=a(20)+bg=2(gQ)+b

12ο,ooQ =(20 _60)aa = -3OOO, b = 180,000

Q uo" = 180,000 - 3'000 to

8o. = Qrn - Qιnt = 18O'oOO - 3'Oo0 to - 20'0ο0

ξo" = 160,000 _ 3,000 to

Exοeφts from this work may be reprοduced by instruοtοrs for distribution on a not-'br-pro'it basis for testing or instructional puφosΘS only to

studοnts enrolled in courses to. wι,λι-' the tοxtbook r-,* υ*. "α"pi.λ.

-

Α-ny oιher *p-a""riλλ"]r'i,o^ιoιι*7 *is wor-k beyond ιhαt permiιted

by Secιions ]07 or 108 o7 m, l'oii'iλιi'rλi,'λ,r, copyrιgn, 'a}iΙ:i';;u' iλ'! irrλ'*iλ" of ιhe copyright oνner is unlανful'

Load Profiles

1 80000

- 160000

ξ tηooooαt_ l z000ο€ lοοοoο6j sooooE' 6oooo

3 4ooooΞ 2OoOo

030 4o

Outdoor TemP, "F

9-5

186

9-6

Group

Sunday

Monday

Tuesday

WednesdayThursday

Friday

Saturday

lll lιl lVVVl

shift 2 shift 1

1481012162024Hour

Αssumes Sunday and Saturday in shift 2

See Table 9-2

EΧceφts from ιhis wοrk may be reproduοed by instruοtors for distribution οn a not-for-profit basis for testing or instructional puφoses only tο

students enrolled in courses for whrch the textbook has been adopted- Αny other reprodiction or ιrcιnsιαιion οf this wοrk beyond ιhαt permitιed

bySecιiοns ]07 οr ]08of the Ι976(JnitedStαιeSCopyrightΑcινιthoutthepermissionof ιhecopyrighιowneriSunlανfuΙ.

Groupshift 1

hrs inea. qp

shift 1

Days in

ea. qp.

Totalshift 1

hrs ea. qp

Totalhrs in

ea. οp.

Frac. ofshift 1

hrs ea. gp

Frac. ofhrs in

ea. qp.

I

il

Πt

IV

V

VI

0

0

2

4

4

4

0

0

5

5

5

5

0

0

1ο

20

20

20

28

28

28

28

28

28

0.0

0.0

0.36

0.71

0.71

0.71

'1.0

1.0

0.64

0.29

0.29

0.29

lo/

9-6 (Cont.)

9-9

9-7

9-8

The procedure is the same as Problem 9-6. Use appropriate bin data fromΑpp. B in last step as per Table 9-3.

Refer to Εxample 9-2, insert shift hours of Problem 9-6 in column 2 and 3of Table 9-5 and recalculate.

Reconstruct Τable-9-3 for the appropriate city to obtain Shift A and Shift Bhours. lnsert the hours in columns 2 and 3 of Table 9-5 and recalculate.

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students enτo11ed in courses for whiοh the textbook has been adopted. Αny other reproduction or ιrαnslαtion of this work beyond ιhαι permιttedby Secιiοns ] 07 or 1 08 of ιhe 1 976 Unιιed SιαιeS Cοpyrighι Αct ινithοut ιhe permission of the copyrighι οlνner is unlανful.

Bin.Temp

Shift t hrs in each GrouplII III IVVVI

shift I

hrsshift 2

hrs6257524742373227221712

0

0

0

0

ο

0

00

0

0

0

0

0

0

0

0

0

00

00

0

3438493635322710

I61

9677674748382817121

0

68829462655436171640

69588687997566282211

1

2672552962322471991577258222

3753463883374204223471571137316

TotaΙ: 1807 2994

188

9-10

Reconstruct Table 9-3 using the shift hour fractions from Problem 9-6 andbin hours and temperatures for the appropriate city. lnsert the shift hoursin column 2 and 3 of Table 9-5 and bin temperatures in column 1 andrecalculate.

9-1 1

Solution furnished by an instructor.

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students enrolled in courses for which the textbook has been adopted. Αny οther reproduction οr trαnsιαtion of this 'ιιοrk beyond ιhαt permitted

by Secιions ] 07 or 108 ofιhe Ι976 [Jniιed Sιαιes Copyrighι Αcι ινithout the permission ofthe copyrighι oνner is unlανful.

Exοeφts from this work may be reproduced by instructors for distributiοn on a not-for-prοfit basis fοrtesting or instruοtiοnal puφoses only to students enrolled in courses fοr which the textbook has been

adοpted. Αny other reproduction or trαnslαtion of this wοrk beyond thαt permitted by Sections 107 orΙ 0B of the 1976 ΓJnited Stαtes Copyright Αct without the permission οf the copyright ov)ner is unlαwful.

Requests for permission or further informαtion should be αddressed to the Permission Depαrtment,

John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hoboken, NJ 07030'

CHAPTER 1O

1O-1. (a) ',* o+ + ρΖι_Pz+'Ψ + ρZz+ ρWp + ρ tt, Vl = VziZι=Ζz

neglect (! iιz; Wp = -Η, g/g/" = -80 (ft - lbf)/lbm

Pz = Pl + ρWp = 20 + 62.4(80)1144 = 54.7 psig

Ps = Pz - PΖs_ U:)rg= 54.7 _ (62'4 X 50)/1 44 _ (20 x 62'41144)

Pg= 54.7 _ 3ο.3 = 24'4 psig

Pι=Pz- PZι-((izg_ (1't)g+ =54.7 ffiX25) (#)(20+ '15) =

28.7 psig

(b) Neglecting the pump, the pressure or head required for this pipe is:

ΔP = 28'7 - 20 psi or ΔP = 8.7 psi

Δ Η 20.1 ft.

Note:20 1

This is thecharacteristicfor only partof the totalsystem.

150

(8 x 2.31) - 250 + 30 + 300 =

I87IΓlot-lοlσlο)|ΞΞ

0L_ oa

10-2. Ηl = Hz * Ηο + (. trZz=

190

'10-3.

10-4.

10-5. (a)Ηl=alΦι+ZιHz=ΔzQz+Ζz

Series ConnectionQl=Qz;SumΗ

Ηl_Ηz=Hp+ !,t+Ζz=0

H^ = - ('-2, = -25- 300 = -325ft: H^ = 325ft of head

= 970 kPa

100

Ηl = 98.5 ft of waterPι = 42'6 psig ε 294 kPa Jt

H2

z2Hq

Ξ(σοΞ5o

75

ιLI

U(σο.)-c.

zaa H=Ηlr

Εxcerpts from this work may be reproduced by instructors 1br distribution on a not-for-proΓit basis for testing or instruοtiοnal puφoses only tostudents enτolled in οourses for ιvhiοh the textbook has been adoρted' Αny οιher reproducιion or ιrαnsΙαιion of ιhis νοrk beyond ιhαιpermitted by Sections ]07 or Ι08 ofιhe ]976 United Sιαιes CopyrighιΑctwithouι the permissiοn οfιhe copyrighι oνner is unΙανfuΙ.

(Ι +ΙΙ )+ 12 + 1b

191

H=Hl *Hz

Η=Qz(aι+Az)+Gι+zz)

Parallel ConnectionHl = Ηz, Ζι=ZzorZ=0Sum Q

a

4.02610-6.

12

Ql= ^E,Φz= ^EYar YazQ = Q l + Q r= JH_Ζ(Jil q * ^[-ν η) .

6ra*M)'+Ζ

t t=f!Y1 ; V = 6. 3ft/sec, L = 3OO ft, D =' D2g

Referring to Figures 10-2a and 1 0-2b

Re = ρνD _ 62'4(1 .04)6'30(4'026 l12) = 75,696

μ (2'7 11490)

ε = 0.00015 ft; Table 10-1

ε/d=o.ooo1 5" 12

4.026

ε/D = 0.00045; f = 0'022 Fig. 10-1

/3OO) (6.30)2( , = 0.022 \' = 12.1 ft of 30o/o E.G.sol.(4.026 t 12) 2(32.17)

= 12.6 ft of water : 38 kPa

-'.--Ξ-Excerpts frοm this work may be reproduοed by instructors for djstribution on a nοt-for-profit basis for testing or instruοtional puφoses only to

students enrolled in courses Γor which the textbook has been adopΙed' Αny οther reproduction or ιrαnsιαtion of ιhis νork beyond ιhαιpermitted by Secιions ] 07 or 1 08 of the ] 976 United Sιαtes Cοpyright Αct νithouι ιhe permission of ιhe copyright οwner is unlανful.

192

10-7. (a) so

Ξ(σ

925J-

Qn = 48 gpm;

28ft

Qn = 60 gpmQs = 32 gpmQc = 41 gpm

100

gpm

Qs = 24 gpm; Qc = 32 gpm;

125 gpm

a2

125 150

(b)

(c)

(d)

10-8.

} ο_

Refer to Problem 10-7; Ζι = Ζz _ 0

(a) Hn= uoQi, θn= 9= + =O.9O8Oai 50'

Hs = auQ3, Θs = 9= =

= o'o278a6 30'

Hc = a.Q3, θc = +=+ =o.O'148aa 45'

H- 02

(^n/o!o8 + JΤt ozτa + J.roιua)2 645.06= 0.001 55Q2

(b) Q = 100 gpm; H = O.OO155 x (1OO), - 1S.5 ftExcerpts from this work may be reproduced by instruοtors for distribution on a nοt-for-prοfit basis for testing οr instructional purposes only tostudents enro]Ιed in courses for which the textbook has been adopted' Αny other reproduction or ιrαnslαtion of this work beyoncl ιhαιpermiιιedbySectiοns Ι07 or l08 of ιhe ]976LlniιedSιalesCοpyrightΑctιι]iιhouιιhepermissionοf ιhe copyrighιονnerisunlανfuΙ.

193Hn = Hs = Ηc = 'l5.5 ft

Qn= JΠ/rΑ =.,/l55/O.OO8 =44gpm

Q g = .'m lo'o2?8 = 23'6 οpm

Qc = Jss/oJl4g = 32.4 gom

(c) From (a) above. H = 0.001 55(12q2 = 24.2 tt

Qn= =55qpm

Qg = Jπ2loβ278 = 29.5 gpm

Qc = J%2toβ148 = 4O.4 qpm

e = 125 gpm

1o-9. Q = CαA r|2g"(P, _Pr)l'''' D2

=84.8 = O.55'--L ρ -] 'D1 154.1 v'vv

aSSume Cα = 0.638 using Fig. 10-9;

Ar= ζ (O.os4s)2 = O.00565 m2-4Pι_Pz

= O.O98(13.55)9.s = 13.O13 J/kgρ

Q = 0.63S(0.0056 5)t2(13,013)11/2 = 0.0184 m3/s x 292 gpm

V z = 3.26 m/s; Re - 999(3'26)(0:085) = 1.98 x'105

1.4x10-3

C6 ^:

0.638 From Fig. 10-9.

Τherefore the original assumption is satisfactory.

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ι

194

1o-1o. (a)V.. = Γ2g" (Po, _ρ,ll"' _|z x s?'ιτ x o.os x

L-" [ ρ )] L ι0 +91x 144)144

1t2

=3.97 ft/sec

Table A-1a; ρ = 59'83 lbm/ft3

(b)rir = ρVA(O. 821= 59.83 x 3.97 r+][qΨ''l2 * o.εzι4rι 12 )rh = 39 lbm/sec or 140,674 lbm/hr

10-11 (a)

Read from Fig. 10-11a at 35 ft and '125 gpm, W. = '1.6 HP

Q = 180 gpm, H, = 20 ft; 1.8 ΗP

Τhis is actually out of the operating range of the pump and the

efficiency is very low. ln situations like this there is a danger of

overloading the pump motor; however, that does not appear to be

a problem in this case since the motor is probably a 2HP model.

Εxcerpts from this work may be reprodιιced by instruοtors for distribution on a not_fοr-profit basts for testing or instruοtional purpοses only tostudents enrοlled in οourses for whiοh the textbοok has been adοpted. Αny other reprodιιctiοn οr ιrαnslαιion οf ιhis work beyond thαtpermiιιed by Sections ]07 or Ι08 οfιhe 1976 Unιιed Stαιes CopyrightΑct1|iιhout ιhe permissiοn οfthe copyrighι oινner is unlcrlυful.

40

#35Ξ

(σΦT2s

(b)

(c)

(d)

195

10-12. NPSHR =

NPSHA =

to20='uΧ

12

NPSHΑ = 20 ft (Figure 10_1 1b)

[e"e") -2,-r,- f&9"]\ ρg / ι ρg)

( l'soτxιt_\ az.z13.55x62.4 -Ζ"_2_ 44

62.2

Ζ"=32'85_20_2_1'17 = 9.68 ft; (2.)rr, = 9.68 ft

'10-13. (a) 231 gpm, ξp = 73'4%W, = 12* tlP

(b) 225 gpm ηo = 73 '3o/o

W, = 12- ΗP

10-14. η, = 73.3oλ; W, = 11. 5 HP

225 gpm; '185 ft of head

ηr=73.5%;Ws=14ΗP

(c) ηo= 73o/o, W. = 14 ΗP 225 231

gpm

10-15 From Problem 10-14b, the original system defined by 225 gpm and 149ft of head and would operate at 242 gpm, 173 ft of head and require14.4 HP with the 7in impellor. Τhen,

EΧcerpts 1iom this work may be reprοduced by Ιnstruοtors for distribution on a not-for-profit basis for testing or instruοtiοnal puφoses οnly tostudents enrolled in οοurses for which the textbook has been adopled' Αny οιher reproducιion or ιrαnsιαιion οf ιhis wοrk beyοnd thαιpermiιted by Sections ] 07 or Ι 08 of ιhe Ι 97 6 United Sιαtes Copyrighι Αcι ινithouι ιhe permission of ιhe cοpyright οlυner is unlαwful.

_.ι _Ι

200

140

100

185

149

(a)

(b)

196

ΓPln= 35οo[+.l =3o37\242 )

H^ = Π3 (go37 )' = l3O ft" ι3500/

w^ 1 4.4(s037)' = e.4 HP" ι3500,The Εfficiency wouΙd not Chan ge,74.2oλ

1o-16 Dn = "r(#)=

7(o e7) = 6 8 in

Hn = *, (Ψ)' = 173 (o'g44)= 163 ft

η 1. .[Ψ)u = 14'4(o'g17) = 13 2 ΗP

The Εfficiency would not Chan ge,74.2oλ

10-17. Uses Fig. 1 0-20 or program PIPE

(a) 25 gρm; 1 % in., V < 4 ftlsec,2in dia. or less(b) 40 gpm; 2in. V < 4 fVsec,2in dia. or less(c) 15 gpm; 1Υιin., v < 4 fUsec, 2in dia. or less(d) 60 gρm;2% in', '(,'1< 4 pVsec; dia > 2 in.

(e) 2OO gpm; g %in., .(,'1< 4 ftllOO pUsec; dia > 2in.(η 2ooo gpm; 8 in., /1 slightly > 4 ftl1oo ft

Exοerpts 1iom this work may be reprοduced by instructors fbr distribution on a not-fοr-profit basis for testing or instructional puφosos only tostudents enrolΙed in courses tbr which the textbook has been adopted' Αny οιher reproduction or trαnsιαιion of this νork beyond ιhαιpermiιιed by Secιions ] 07 or Ι 08 of ιhe ] 97 6 Unιted Sιαιes Copyrighι Αcι ν ithοuι the permission of ιhe cοpγ,ι'ighι ονner is unlαwful.

1971ο-18. (a) K = 30 ft, ft = 0.019; K = 0.57 (Table 10-2; Figure 10-22a)

V = 3.82 fVsec; ! r = 0'57(3'822lβ2.2x2)= O.13 ft

(b) K = 340 ft, ft= 0.017; K = 5.78V = 5.0 f/sec; ! r = 5.78 x 5.02t132.2 x2) = 2.24 ft

(c) K= 60ft, ft= 0.018; K = 1.08V = 6.5 ft/sec; ! r= 1.08 x 6.52t(2x32.2) = 0.71 ft

'10-1e. ! r = 2.31 (#)' = 10.8 ft of water or 4.7 psi.

10-20. Assume com. stl. pipe

Q = O.O3 mt/s = 108 m3/hr, size pipe for about 4 mllOO m

From Fig. 10-20, use 5 inch pipe, lD = 130 mm

nt - a,λ' f _ ..25 ml100 m; [1= (3'251100)200 = 6.5 m of water or 63.7 kPa

ΔPg = 35 kPaΓ o-99-1'

= ι7 '27 kPaFor strainer. ΔP" - | '" 1 0.00722 J

Then for the pump:

ΔPp = 63.7 + 35 + '17 .3 + 3(1000)(9 .807)11000 = 145'4 kPa

Ηp = 145.419'807 = 14.8 m

Q = O.O3 mu/s - 30 L/s

10-21. Size the pipe using Fig. 10-20 or program PIPE. Fitting equivalentlengths found using Fig. 10-22a; 10-22b and Τable 10-2' ProgramPιPE could be used to solve the complete problem including fitting

losses. Data for hard calculations are summarized below:

Εxcerpts from this work may be reproduced by instruοtors for distribution on a not-for-pro1it basis for testing or instruοtional puφoses οnly tostudents enrolΙed in οourses Γor whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsΙαtion of ιhis νork beyond ιhαtpermitted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed SιζlιeS Cοpyright Αcιlνiιhοuι ιhe permission ofιhe cοpyrighι oνner is unΙανful.

L

ro(3) I o(3)

lO(3) Θ/o\

1ot3] Θ \:-/

198

] 5ι5)

Sec.No.

qpm itft./'100 ft

Left.

ιtft.

Con.Valve ft.

Coilft.

Τotalft.stze tn.

1120

3.38 45 1.5 1.53

570

3.64 '15 0.6 0.62.5

640

3.'1 24 0.7 11.4 12.12

740

3.'1 13 0.4 12.0 12.42

490

5.84 27 1.6 1.62.5

10120

3.38 42 1.4 1.43

25ο

4.7 22 1 10.0 11.02

350

4.7 26 1.2 10.0 11.22

I 306.3 28 1.8 14.4 16.21.5

930

6.3 13 0.8 15.0 15.81.5ch 120 20

Exοerpts lrom this work may be reproduced by instruοtors for distribution on a not-1br-profit basis for testing or instructional purposes only tostudents enrol]ed in courses Γor whiοh the tΘXtbook has been adopted' Αny οιher reproducιiοn or trαislαtiοn of ιhis νori beyoncl ιhαιpermitted by Secιions ] 07 or Ι 08 of ιhe ] 976 (Ιnιted SιαιeS Copyrighι Αct wiιhοut the peimission οf ιhe cοpyrighι owλer is unΙcnνful.

199

The head losses forsame.

the three parallel runs are approximately the

For run (1-5-6-Z-4-10), Hp = 49.6 ftFor run (1-5-S-g-10), Hp = 55.5 ftFor run (1-2-3-4-10), Hp = 46J ft

Therefore, a pump should be selected to provide about 56 ft of head at120 gpm.

10-22. 500 gpm, Use 5 inch pipe; !'f = 4.17 fil1OO ft

V = 8.0 ftlsec

Length of pipe = 160 + 3O + 12 = 202 ft

6-5 in elbows = ,lS ft (Figure 10_22)

3-5 in gate valve = 12 ft

1-5 in gΙobe valve = 130 ft; Total equivalent length = 419 ft

. 4.17(41e\/, = -Ι_1}-J "/ = 17 '5 ft of water

For strainer: ! "=

2.31 [#)' = e.24ft of water

For cond | ! "o

= 20 ft of water

Τhen Hp = 17 '5 + 9.24 + 20 + (3o _ 12) = 64.7 ft at 5OO gpm

10-23'Use Εq. 10-33

L

Exοerpts frοm this work may be reproduοed by instructors for distribution on a not-Γor-profit basis fbr testing or instructional puφosοs only tostudents enrolled in οourses fοr which the textbook has been adopted Αny οιher Ιeprοdactιon or ιrq'nsιαιion of ιhis νork beyοnd thαιpermiιιed by Sections ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyright Αct νiιhλut ιhe peimission of the cοpyτ.ighι owner is unlανfuΙ.

,,_ _ 6oott*:;*g -,l) - 3x6 sxlo-u (1,0 -oull

". - = 19.4 9a1. = 74 L

I gz.οgο og.οgο ]

οoo[[9 Ψξg _,'l _ 3x6.5x1 o_u (1,, o _ ou1l10-24' Use Eq. 10-34 v, = L[ 0'ο1ο0zz ) '-___-_-J

ι '_ 69:ο%,Vr-_8.Ζ-gal = 33 L

200

10-25. Use Eq. 10-33tl = 60oF, P2= 50 psig, P1 = 20 psig, v1 = O.O16ο53 ft3/lbm

vz = O.O1 6772 ft3/lbm, tz = 220"F

Vτ= = _11_5_gal. - 435 L

10-26(a) Use Eq. '10-16

P,+ PτPzι= Pzgzz +ρνv+ρg'g"9cι

'' = ff (zz-zι)+ ρw +

Exοerpts from this work may be reproduced by instruοtors for distribution on a nοt_for-profit basis for testing οr instructional purposes only tostudents enrolled in οourses for which the textbook has been adopted,' Αny other reproductιon or ffαλιαιion of ιhis wori beyond-ιhαιpermiιιed by Secιiοns ]07 οr ]08 οfιhe Ι976 L]niιed Sιαιes CopyrightΑctνΙιhouι ιhe peimission οfthe copyrighι owλer is untιιυful'

lL9c

Τ24o tΙ,

ΙP9n-

LE

9cl

-

(b)

= # e4o) #(60) + # es1=Be psis or 61 2 kpa201

P,'+ρg!= P2+ρg29c - --gc

Pz=Pι- t (zι-zz)=89.o -effiPz = -15 psig = -i03 kpa or about o absorute

(c) No, makeup water is not available to overcome a pressure of ggpsig. However, the domestic water system probabry has abooster pump.

10-27 (a) Pι = Pz * o* (zz-zι) + ρνν * ρ9ι,9c gc

= 5 + 62.4(240)144

=Pz+ !99c

Pl = 109 psig

62.4(60) . 62.4(25)144 144 'n0.8

Pr = 93.8 psig or 647 kpa

(b) η (zz-zι) = 5 *62|(,2ra0)

or 752 kPa

= 5 + 104 =.t09 psig

Excerpts from this work may be repτoduced by instructors for distributiοn οn a not-for-profit basis for testing or instructiοnal puφoses only tostudents enrolled in courses for which the iextbook t-,u. υ".n 1dγνωj Ατ-ιl λrir, Ιrpr"a""rιon or ιrαnslαιion οf ιhis work beyond ιhαιpermiιιed by Sections ] 07 or Ι 08 of the ] 976 United Stαιes copyrighi Αcι withλuι ιhe p,e1miιsιon of ιhe copyright ονner is unlαινful.

202

10-28

(c) This location is at Ιeast workable. Ηowever the pressure at thepump is still very high. The domestic service waterpressure would have to be boosted to a higher pressure at the 2othfloor.

(oo , sz) +

Qo *Q. =

x +o)= (ao x ιτ)100

(a,

Qα=

Solve Simultaneous

57Qb + (O x 1OO) - 4OQb = 100 x47

= Ψ = 41.2 say 41gpm17

Q. = Q.. = 1OO - 41 = 59 gpm

Size all pipe for 100 gpm

D = 3 in. from Fig. '1 0-20 or PIPE

10-29. (a) Each chiller requires 600 gpm. Since chiller 2 is partially loaded it

must have the full flow of 600 gpm.Therefore, Q"p = 1200 - 750 = 450 gpm

(b) (150 - 60) + (450 x 42) = 600 ts, ts = 46.5 F

(c) LR = 150/600 = 0.25

(d) Main pipe to and from sec. Circuits: D = 8 in. com. stl.

Excerpts from this work ιnay be reproduοed by instructors for distribution on a not-for_profit basis fbr testing or instructional puφoses only tostudents enτοlled in courses for whiοh the textbook has been adopted' Αny οther reproductiοn or trαnslαιion of this work beyond ιhαιpermitιed by Sectiοns ]07 or Ι08 οfιhe ]976 United Snιes CοpyrighιΑcιιυiιhοuι the permissiοn ofιhe copyright oνner is unlαννfuΙ'

ab

(b)

(c)

203Dns = Dco = O in. com. sfl. (S in. a litile small)

D"p = 6 in. com. sfl.

Dsc = DRo = 6 Ιn. com. stl. ( could be 5 in. but easier to make all 6in.)

(e) Rpm, = ΓPIΠl 050l12oo) = 35OO(75o/12o0) = 2188

(η ιW = *#=1-ff =,l_[ffiJ' =, (##) = O 756

or 75o/o

10-30. (a) Q"n = 12OO -750 = 450 gpm

(b) Qrtr + Qztz = Qsts; ,. = (450x42)+ (750x60)

= 53.3 F1200

Both chillers receive the same temp. water

(c) Load ratios are the same:

LR= ##=0628or63%10-31.

Εxcerpts from this work may be reproduοed by instructors fοr distribution on a not-for_profit basis fοr testing or instruοtionaΙ puφoses only tostudents enrolled in courses for whiοh the textbook has been adopted. Αny οιher )eproclucιion or ιrαnslαιiοn of this ιυοrk beyoncl thαιpermitted by Sectiοn's Ι 07 or 108 of ιhe ] 976 ΙJnitecl Stαtes Cοpyrιghι Αcι wιthλuι ιhe peλιssion of the copyright ονner is untωνful.

204

10 (13) 2 2Ο(6) 3 *{s)T'r^;n ^ !ι }iJJυα'

**ntΓ*if Φr

u?

ι$

BaΙ*n*sve,v*

{typi*aii

Note: Piping is type L copper

10-31.

Αll |engths are total equivalent lengths

1*{S}

Ω*{s} s 2Ο(s) 8

**rnnr** pip*

CoilFΙow rateopm (L/s)

Lost head ft (m)

Coil Con. valveABc

40(2.5)40(2.5)50(3.2)

12(3.7)15(4.6)18(5.5)

10(3)12(3.7)15(4.6)

(continued)

SectionNo. gpm

Dia.in.

irftl1ο0'

L"ft

!.rft

1-2

2-33-4^,^il C

1309050

3

2%2

3.74.85.0

602030

2.2

r; ) 355

' he reproduced by instruοtors for distribution on a not_Γor-profit basis fortiting or instructionai puφoses onΙy tohiοh the textbook has been adopted' Αny oιher λeprοduction οr ιrαislαιiοn of ιhis wori beyond -ιhαι

''/ United Sιαιes Cοpyrighι Αcι ιιithouι ιhe permission of ιhe copyv.ight oνλer is unlωυful.

Con. C4-5Com. oi

1300

33

3.7 40151.50.0 39.2 ft (tote

2-6Coil ACon. A

40 2 3.4 30 1.012

10 23 ft (total)3-7Coil BCon. B7-8

40

80

2

2%

3.4

3.9

10

20

0.515120.8 28.3 ft (totat)

205

Circuit 1-2-3-4-5-1 is the path of greatest lost head. From Fig. 1O-1 1 choose at40 ftof head and 130 gpm the 7 in., 1750 rpm model which froduces about 43ft of head.

rfι*{}ιJ

Ξp^q 2ΟΟ 2ΟΟlbu] 2(rΟ)3(rΟ}

Ιs*ianc*vxlγ*

{typi*ai}

*οntr*i. valv*{ typi*xi}

10-32.Notes: PΙpe is schedule 40, commercial steel.

Αll lengths are total equivalent lengths exc|uding control valves.

Circuit Flow rategpm (L/s)

Control valvehead loss ft (m)

Α 60(3.8) 40(12)B 70(4.4\ 5ο(15)c 70(4.4\ 50(15)

t p,ih-*:y +ο* 4QPumP {12)l t1i

*hill*rp ιl Π]$} $

'' be reproduοed by instruοtors for distribution on a not-for_profit basis for testing or instructional puφoses on1y to"hiοh the textboοk has been adopted. Αny oιh-er )eproducιiοn or ιrαλhιiοn o7 tnπ 'orκ beyond thαι

'976 United Stαtes Cοpyrighι Αcι ννithouι the peλission of ιhe copyright ονner is unlαwful.

206

SectionNo. gpm

Dia.in.

ifft/'l00'

L"ft

(.t

ft

1-22-33-4Con.4-1

20014070

200

432%

4

2.44.23.5

2.4

200200240

400

4.88.48.4509.6 81 .2 ft (total)

2-5Con. A5-4

60

140

2%

3

2.5

4.2

240

200

o408.4 54.4 tt (total)

3-6Con. B

70 2% 3.5 40 1.450 51.4 ft (total)

Circuit 1-2-3-4-1 has the largest head loss of alΙ paths. Select pump for 2OOgpm at 81 ft of head. From Fig. 10-1 1, use: 5Υ' in., 3500 rpm model. Willoperate at 96 ft at 200 gpm.

10-33 (a) qst = 20 x 12,000 x2= 480,000 Btu

/1 _ gst 480000= 512.8 ft3

ρc, (t,.-t.) 62.4(1) (60-45)

orQ=3,8369a1

(b) Vol = 513 ft3 ora Space 8 ftx 8 ftx 8 ft ora cyΙindrical tank 8 ftdia. x10.2 ft

10-33. (continued) Solution - Sl:

(a) Qg1 = (352_280) (2)= 144k\^l-hr= mc, (trt.) _ Qρcp(t-t.)

q = Qst - 144 x3600 = 14 m3

ρο, (t1ts ) 980(4. 184) (1 6-7)

' 1le reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instruοtional purposes only to'.'hiοh the textboοk has been adopted. Αny oιher reproduction οr trαnslαtion of ιhis wοrk beyond thαt

^76 (Ιnιιed Sιαιes Copyrighι Αcινιιhout ιhe permission οfthe cοpyrighι oιυner is unlανful'

207

(b) Vol. = 2Αmx2.4mx2.4m

10-34. Solutions may/can Vary. Α typical solution is:

(a) Use 2 chillers of '15 tons total capacity in a reverse return systemsimilar to figure 10-32. The piping would be routed overhead aroundthe complex with supply and return running parallel, starting andreturning to the equipment room.

(b) Total flow rate is

Qτ =16 x2'25 = 36 gpm Using PIPE or Fig 10-21; Dia. = 2in'

(c) Estimated length =225x4x2= 1s00ft. Τotal Eq. Length=2x 18OO =3600 ft

Assuming an average Ιoss ofabout 2.5 ft1100 ft; The pump head required would be:

Hp = 2.5 x 3600/100 = g0 ft with flow rate of 36 gpm

'10-35 Solutions may vary

(a)Figure 10-34 is a schematic of what the system wouΙd be.

However, there would be 3 chillers and the secondary piping would

be routed in a square fashion around the outside of the parking

garage in reverse return.

(b)Τhe primary system would appear as in Figure 10-34 with the

common pipe as shown because of the expected variable and light

load at night.

(c)The tertiary circuits would be as shown in Figure 10-34 and piped in

a reverse return manner.

(d)For each building:Excerpts Γrom this work may be reproduοed by instructors Γor distribution on a not-fοr-profit basis for testing or instruοtiοnal puφoses only tostudents enrolΙed in οourses for whiοh the textboοk has been adopted' Αny other reprοductiοn οr ιrαislαιion of ιhis νork' beyοnd thαtpeιmitted by Sections Ι 07 or Ι 08 of the ] 976 tJnited Sιαιes Cοpyrighι Αcι 1ψiιhouι the permission of ιhe cοpyrighι owner is unΙα:wful.

208

Qi =1500 x 't2000

= 600 gpm4x500(60-45)QΤ=4x600 =2400 gpm

(e) Dia. = 10 in., Figure 10-20 or plpE

10-36.

10-37 ' Αssume boiler pressure of 2'O psig with ΔP/L = 2'O oz or 0'125 psi/1OOft' (TabΙe 10-4a). Τhen, ΔP = o'125x 175l1OO = 3'5 ozor 0'22 psi ΔPis about Υzthe alΙowabΙe from Table 10-4a'

Assume boiler pressure of 1.o psig with ΔP/L Ξ 0'125 psi/1Oo ft.(TabΙe 10-4a)' Then ΔP = 0'125 x 175l1OO = 0'22 psi which is nearthe maximum in Τable 10-4a' Either boiler pressure could be used,but select2.0 psig to be conservative.

Εxc-erpts fiom this work may be reproduced by instructors fοr distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in cοurses for whiοh the textboοk has been adοpted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhis wοrk beyoncl thαtpermiιted by Sectiοns Ι 07 or 108 of ιhe ] 97 6 (]nitect Stαιes Cοpyrigh} Αcι νith'οuι ιhe peimission of ιhe cοpyright ονλer is untωνfuΙ.

;ControΙ valve (Typical)IY φ a_Air Vent (Typical)

,+9*Π4 Heating Device (Typical)

T<-TypicatTrapr-€

IEι o r--5erξι--] Possible! -ξ2 < Vacuum Breaker on each;- - - -<-'-' Heatinq Device

10-39. Refer to Table '10-5a.

The available head is = 2 x'100/110 = 1'82ftl1ο0 ft. Then at 850 lb/hr

of condensate flow, D = 1 in. nominal is adequate.

10-40 (a)

q = rhcp(t,-t,) = ga##(1)(6s - 42)

^ - 1200x 1200x7'48 - λ ^F^ -..-ι'ι = 60 X 624(654η = 1'250 gpm

209

From Figure 10-48a at 850 lb/hr; ΔP/L = 0.'125 ρsil100 ft, and boilerpressure of 2.0 psig: Pipe diameter = 4 in., with steam velocity of4,00ο ftlmin at zero psig. Correct velocity to 2'0 psig (Fig. 10-49a)V = 3,800 ftlmin

10-38. For each unit at full load:ft. = 283 lb/hr

Pipe size depends on slope of line, Τable 10-5a. For slope of 1/8 to Υιin./ft, D = 1 in. nominal specify slope of % in./ft (conservative).

(b) Αssuming no changes in the temperatures, the total flow rate wouldbe:

Φ., = Ξ9t 1l25O) = 937.5 or 938 gpm-P 1200 \ '/

Τhe chillers could share the flow:

^-938(J, = -=469gpm'2

and be above their minimum flow of 70o/o.

jι - 469 = 0.75 or or Ts%

Qmin 625EΧcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only tostudents enrolΙed in οourses for whiοh the textbook has been adopted. Αny οther reproducιiοn or ιrαnslαιion οf ιhis νork beyond thαιpermitted by Secιions ]07 οr Ι08 οfthe E976 LΙniιed Stαtes Copyright Αct\νithout ιhe pe'ιmission οfιhe copyrighι οwner is unΙαlvful.

210

Τhis is probably the best way to operate. Τhere would be no bypassflow and the parallel pumps couΙd operate at:

RPM, = *x 35OO = 2,626625

(c) At 6ο% full load, again assuming no change in temperature, the

totaΙ flow rate would be:

Qp = 0.6(1 250) = 750 gpm

This is too much flow for one chilΙer and not enough for two chillers

at minimum flow of 875 gpm or 438 gpm each. Τherefore, both

chilΙers will have to operate at least at 438 gpm each and some ftow

bypassed equal to:

Φυ, = 875 - 75O = 125 gpm

The pumps could both be slowed to:

RPMP = ,9'? ι35oο) = 2'450' 1250

(d) Αt 25% of fuΙΙ capacity, again assuming the temperatures do not

change, the flow rate for the load is:

d, = 0.25(1250) = 313 gpm

which is less than the minimum flow rate for even one chiller.

Τherefore, shut down one chiller and operate the other at least its

minimum flow of 438 gpm. The bypass flow would be:

Qop = 438 - 3]3 = 125 gpm

One pump would be shut down and the speed reduced for the other

pump to:

RPM, = *(3sOO) =2,453' 625Excerpts 1iom thrs work may be reproduced by instructors for distributiοn on a not-for-profit basis for testing or instructional puφoses only tostudents enrolled in οοurses for whiοh the textbook has been adopted. Αny other reproclucιion οr ιrαislαιion of this work beyοnd'thαtpermitted by Sections ] 07 or Ι 08 of the ]976 Uniιed Sιαtes Cοpyrighι Αcι withοuι ιhe permission of ιhe copyright owner is unlανfut.

Excerpts from this work may be reproduοed by instructors for distribution on a not-fοr-prοfit basis fortesting or instructional pulposes only to students enrolled in courses for which the textbοok has beenadopted. Αny other reproduction or trαηslαtion of this work beyond thαt permitted by Sections ]07 orΙ0B of the Ι976 United Stαtes Copyright Αct without the permission of the copyright iwner is unΙαιυful.Requests for permission οr further informαtion should be αddressed to thi permission Depαrtm'ent,John Wiley & Sons, Ιnc, ] Ι ] Riνer Street, Hobοken' NJ 07030.

GHAPTΕR 1 1

11-1 . (a) Using Eq. 1 '1-1b

, = 1=L959ll-,

Ao = Ψ= O.353 ft2; Αssumed K = 6ζJΑ" '- 'ν s5ο

X5o = ' 1: -=ug-^o)

= 68.5 ft; x1ρρ = 34.2ft; xlso = 22.8ft50.,/0.353

(b) Q, = CQov"/v,, C = 2; Εq. 11-2a

(Q,)so = 2(3OO)850/SO = 10,2OO ft3/min

(Q,)loo = 6Oο(85O)/1OO = 5,1O0 ft3/min

(Q,)rso = 6OO(850)/150 = 3,400 ft3/min

11_2' Using Εq. 11-3

t, _ t, = 0.8(to _ tι.) (V,/Vo), Δt, = 0.8(100-75) V,ll'tοO

(Δt,)so = 0.8(28)50l1100 = 1'02 F

(Δt*)loo = 2'04 F

(Δt )lso = 3.06 F

11-3. 50 ftlmin throw = 24 - 6 = 18 ft

212

11-4. Q, =

Vo=

V,=

11-5.

Qo_ Xζ Q"λ;=1l3Kro'ffi = *v* - 18 x 5o

= 132.71.13K '1 .13x6

From Eq. '1 1-'1 and Qo = Vo x Αo; Assume K = 6

Αny combination shown would be

acceptable. The size would depend

on the available total volume flow

rate of air and the size of the space.

Q oC V "lV, , Eg. 11=2a

Q o/Ao = ιzstl L- r9)'l= οsο fVminL4 \ιz1

1

V"(1.13)Kl lx ; K= 6, x = 12ft

vr=

Qr=

r-636 x 6(1 .i q

^l+x(O.S)2 t12 = 159 ft/min' 'γ4

125x2x6361159 = 1000 cfm

A ceiling type diffuser system has the ability to handle large

quantities of air because the air is discharged radially and

diffuses the high velocity jet in a short distance.

11-6. (a) A perimeter type system would be necessary to achieve a

satisfactory heating performance. Αny other type of system

would lead to a cold and drafty floor.

ExοeΙpts Γrom this wοrk may be reproduced by instructors for distribution on a not-fοr_profΙt basis for testing or instruοtional puφoses only tostudents enrolled in οοurses for which the textbοok has been adopted. Αny other reprοduction or trαnsι{ιιion οf this work beyοnd ιhαιpermiιted by Sections ]07 οr l08 ofthe Ι976 tJnited Stαtes CopyrighιΑctνiιhoutιhe permission ofιhe copyrighιονneris unlαιllfuΙ.

D^

δ ln. ft.

29 aJ 0.2539 4 0.3349 5 0.4t759 6 0.500118 t2 1.00

213(b) An overhead type system would be preferred because of thegreater need for cooling during the summer and less needfor heat durΙng the winter.

11-7 ' Α perimeter type system would be the best choice. This typesystem is required to do a good job of heating. Α spreading jetshould be used when heating and a nonspreading jet shouldbe used when cooling.

11-8' Some kind of overhead system wouΙd be preferable sincecooΙing would be the dominant mode of operation. However,ceiling diffusers with radial discharge woutd not be required dueto a low volume of circuΙated air' Α high side walt type of systemor ceiling diffusers with discharge in only one or two directionswith a Ιarge throw would be preferred. This would give themaximum air motion with a smaΙl amount of circulated air.

11-9. 10 in. round diffuser, TabΙe 11-4;650 cfm

lnterpolation between 600 & zoo cfm is required

NC=0.5x(21 -17)+17=19

x5o = 0.5(1 1-10) + 10 = 10.5 ft

p=oo62(ffi)'=oo73in wq

11-10.

For 150 cfm/ft, ΔPo = O.08 x (150116712 = O.ο65 in' wg.Excerpts from this work may be reproduced by instructors for distributiol on a not-for-profit basis fοr testing or instructionaΙ puφoses only tostudents enrolled in courses for which the textbοοk has been adopted Αny oιh,er Ιλaurtιon or ιrαnslαιion οf this wοrk beyοnd thαιpermitted by Secιions ] 07 or Ι 08 of ιhe ] 976 ('|nited Sιαιes Cοpyright Αct w iιhλuι ιhe p,e{m|ssιon of ιhe copyrighι oνner is untιnνfut.

214Throw values are for a 4 ft active length then

X5ο = 21 _ 0.6(4) = '18.6 ft;

Τhe uncorrected NC for a 10 ft length is NC = 23 _ 0.6(5) = 20.

For a length of 6 ft the correction is -2.

Corrected: NC = 20 -2 = 18

11-11. Model 28, 4-48 T-Bar; Table '1 1-6, 270 cfmlnterpolate:

NC = 0.7 (36 - 32) + 32 = 34.8 or 35

xso = 0.7(11 -10) * l0 = 10'7 or 11 ft

, ^-^'t2p=O.r[ +l =o.13in.wqι245 )

-11-12' From Table 11-1, L = 12 ft' Then from Τable 11-2 at

Q = 40 Btu/(hr-ft21, lxuo/L1rr*

=1.3 and the range is1.2 - 1.8, and X5s = 1.3 x12= 15.6ft

A good solution would be to use the 4 in. size with

'150 cfm/ft. with uncorrected throw of 18 ft and NC = 19.

The corrected throw is:

Xso ='18 x 0.85 = 15.3 ft and NC = 19 - 4= 15

P = 0.057rΨ)' = O.066 in. wqι139 ) -

11-13. (a) Room char. Length = 14 ft, Table 1 1-'1

(x5ρ/L)rr, = 0.8, Table 11-2

Range of xso/L = 0.5 to 1.5; xso = 0.8 x 14= 11.2ft

EΧοerpts fτom this work may be reproduced by instΙuctoΙS fοτ distribution on a not-for-profιt basis for testing or instruοtional purposes only to

students enrolled in courses for which the textbook has been adopted. Αny other reprοducιιon or ιrαnsιαιion οf this ιυork beyond thαt

permiιted by Secιions ] 07 or Ι 08 οf the ] 976 ιJniιed Stαtes Copyrighι Αcι lυ iιhouι the permission οf the cοpyrighι οwner is unlαwful.

215The best choice would be a 12 in. size with 600 cfm

(b) xso = 13. :9 Q) = 14.3ft;x5e/L - 14'3 = 1.02(in the range)80'ι --'-'--νν'-

14

ΔPo = o.o81 (ffi)' = 0.0g6 in. wg., NC = 22+ffιol =24.5

11-14. Room char. Length is 26 ft,Table 11-2

(a) (x5ρ/L),,, = 1.6 (Table 11-2); range of (x5g/L) = 1'2 - 2'3

Xso = 1.6 x 26 = 41.6 ft; Q/diff = 60012 = 3OO οfm

From Τable 11-5, the 18 x 4, 14 x 5, or

12 x 6 sizes may be acceptable aΙthough the throw is

less than desired. Xso = 31 ft

Xso/L = 31.6126 = 1.2 (barely in the acceptable range)

(b) Xso = 31- ft (zero defΙection)

NC = 22*, ^po:

o.o6e (#)'= 0.065 in. ws.

11-15 lt is good practice to keep the core veΙocity below 5OO ftlmin. Asolution is the 18 x '1 2; Table 11-7

ΔPo- -o.O45 [,Ψ)' = -O.O57 in. wgι535/

NC= 21 + 65 fZl =24135' ι

Note that static pressure and ΔPo are negative.

11-16. Guidelines:

1-Place diffusers under or between double windows.

2-Select throw using the ΑDPl procedure. Characteristic length

Εxοerpts from this work may be reprοduοed by instΓuctoΙS for distribution on a not-for-proflt basis for testing or instruοtional puφoses only tostudents enrolled in courses for which the textbook has been adopted' Αny oιher reproduction or ιrαislαtiοn of ιhis wori beyond'thαtpermiιιed by Secliοns ] 07 or Ι 08 οf ιhe Ι 976 Uniιed Stαtes Copyrighι Αct νithout ιhe permission οf ιhe cοpyrιghι owλer is unlαwful.

216= floor to ceiling.

3-Noise criteria (Nc) should usuaΙly be less than 30.4-Be Sure that the totaΙ pressure required is compatiblewith the pressure characteristics of the system. For example,a smalΙ commercial system may have a fan that produces onlyabout 0.6 in. wg. total pressure while a large commerciat systemmay operated at2-5 in. wg. total pressure. The diffusertotalpressure Ιosses should be no more than abo ut 10% of thefan total pressure.

5- Use data from Table 1 1-3

11-17 ' GuideΙines:

1-center diffusers in square or nearly square spaces. Dividelarge or irregular spaces into imaginary square spaces andpΙace a diffuser in each Space. Select throw using ΑDPl procedure.

2-Τry to obtain a balance between many small diffusers versusa few very rarge diffusers to be cost effective.

?_vι

4_ t See Problem 11-16

5- Use data from Table 11-4

11-18. Guidelines:

1-Locate diffusers about 12 in. below ceiling on inside waΙls. Setectthrow using ADpl procedure.

2-Τhe jet may be spread with this type diffuser. However, morethan one diffuser should be used where the room width is atleast two times the room depth.

3-ζ.I See problem 11-16

Exοerpts frorn this work may be reproduοed by instructors for distributiοn on a not-fοr-prοfit basis for testing or instruοtional purposes only tοstudents enroΙ]ed in courses for which the textbook has been adopted Αny oιl'r, Ιrpioauαιοn or ιrαnslαriοn of this νοrk beyond ιhαιpeιmiιιed by SeCιionS Ι07 or Ι08 οfιhe ]976 (]niιed Sιqtes Copyrigh'ι Αcιlνiιhλuι ιhe peimιssiοn οfιhe copyrighι ονner is untανυful.

v&

2174-

5- Use data from Table 1 '1-S

Excerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instruοtional puφoses onιy tostudents enrolled in courses for whiοh the textbοok has been adopted' Αny other reproduction or trαλlωion of this work beyond'thαιpermitted by Secιions ]07 or Ι08 ofthe ]976 (]niιed Stαtes CopyrighιΑcιιυithouι the permission οfιhe cop7righι owλer is unlανful.

-ffi

218

11-19. Guidelines:'1-Locate grilles in ceiling near the inside wall.

2-Noise criteria (NC) should be less than 30.

3-The negative static pressure should be held to minimum,

especially for light commercial systems with small fans.core velocities of less than 5oo ftlmin will usually yield aquiet system with a reasonably low-pressure loss. Highervelocities and pressure ross may be tolerated with heavycommercial applications.

4- Use data from Tale 11-7.

11-20. (a) H-24, Mod 28at 89 cfm each

2Ψ'-windows

(b) X1oo= 3ft; L =7 ft; X19ρ/L =3l7 = O.43; o.3 < (x19ρ/L) < 1.o89 cfm/diffuser; ΔPo - 0.06 in. V/g.; ΝC = 20Νote: other acceptable soΙutions also exist.

11-21. SimiΙar to 11-2O- Diffusers shouΙd throw air towards the windows- Arrange to obtain uniform air motion- Might use diffusers with short throw around exposed walls with

larger units in the interΙor.

11-22. L = 9 ft; Table'l i-1Exοeιpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only tostιιdents enrolled in courses for whiοh the textbook has been adopled,' Αny othe, )eprocturtιon or trαnsιαtion of ιhis νork beyond ιhαιpeιmiΙιed by Secιions ] 07 or Ι 08 οf the ] 97 6 tJniιezι Sιαιes Copyrιgh'ι Αct νιthλut ιhe peλission οf the cοpyrighι oνner is unlαwful.

+l+ <-l-+ +l-+

*l-+ +t-e +l-r

€l+ +l-, €Ι_+

219X59/L = 0.9; Τable 11=2, straight Vanes ( Assume light load for a

secondary system)xso=0.9x9=8.'1 ft

Α solution: 9-4 ft length diffusers with 50 cfm/ft, 2 in. size,Τable 11-3, x = 8.5 ft (no correction required); NC = 15 - 4 = 11ΔPo = β0l4q2 x O.036 = O'047 in. wg.Place 3 diffusers on each exposed wall

11-23. Use 4-'12 in. size from Τable 11-4

650 cfm/diffuser; L = 20 ft 80

Room Load = 18 Btu =(hr _ ftΖ )

x5sil = 0.8, Table 11-2

x = 16 ft (desired)

(650 - 630)Xactual =

F5 _ 63οi (7 _ '15) + 15 = 15.5 ft

xact_15.5_.,-fl : ^

= 0.78 (in acceptable range)

NC = 27; ΔPo=O 105 rΨ]'= O.112 in. wg.ι630/

11-24' Use 14-H-48, Model 28 diffusers from Τable 11-6;

229 cfmldif. as shown. L = 20 ft, xlse/L

= 0.3 and acceptable range is 0.3 to 1.0.

Εxcerpts 1iom this work may be reproduced by instruοtors for distribution on a not-for-prοfit basis fοr testing or instructional puφoses only tostudents enrolled in οourses for which the textbook has been adopted' Αny οther reproductiοn or trαnsιαιion of ιhis νork beyond-thαιpermitιed by Secιions ] 07 or Ι 08 of ιhe ] 976 United Sιαtes Copyrighι Αcι νithouι ιhe permissiοn of ιhe copyrighι oνner is unlιrνιful'

78

220Desired throw: xloo = 0.3 x 20 = 6 ft.

Actual throw: Xact = 6.5, TabΙe '1 '1-6

Xr"t/L = 6'5Ι20 = 0.33, o.K. 80

NC = 29, ΔPo = 0.095 in. wg. ltLr78

11-25. Refer to Problem 11-23, Q = 2600 cfm; refer to Table 11-7.

Αssume a Ιay-in ceiling with 2 ftx 4 ft tiles. Τo assure a quiet

return, limit NC to about 20. Use 2-24 in. x24 in. grilles with

1300 cfm each. Nc < 25, ΔPo = -0.048 in. wg.

11-26. Refer to Problem 11 -24,3200 cfm.

Assume a2ftx4 ft lay-in ceiling.

Use 24 in. x 24 in. size from Τable 11-7 ' Using three units,

cfm/grille = 320013 = 1067; Nc < 2ο

ΔPo = -0.033 + 0.006 = -0'027 in. wg.

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Exοerpts from this wοrk may be reproduced by instructors fοr distribution on a not-for-profit basis fortesting or instruοtional puφoses οnly to students enrοlled in courses fοr whiοh the textbook has been

adopted. Αny other reproduction or trαnslαtion of this work beyond thαt permitted by Sections ]07 or

Ι 0B of the t 976 United Stαtes Copyright Αct without the permission of the copyright owner is unlαwful.

Requests for permissiοη or further informαtion should be αddressed to the Permission Depαrtment,

John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hοboken, NJ 07030.

CHAPTER 1 2

(a) W. = rt (Pl _ Pz)lρ Αssume standard air

= Q(Pl -P2) _ 2ooο(1.9)

= o.60 HP - O.45 kW

12-1.

6350 6350

(b) r,= ξffi =#ffi+ = O 54 or 54oλ

(c) V = 2Ooo/O '84=2,381ft/min, Pu = Q381l4ooq2 = ο.35 in wg

P. = 1.9 - 0.35 = 1.55 in wg

W" =2000x'1.55

= 0.496350

Ιs = Ws/Wrr, = 0'4911'1 = 0'44 or 44o/o

(d) From (c) abovet P, = 1.55 in wg

12-2 Qz = a, #ffi = 2ooo (ffi#) =24oocrm = 1,133 L/s

ΔP,z = ΔP.l [Hffi)'

= 1.55 (ffi#)' = 223in wg ρ 555 Pa

ΔPoz = ΔΓ (κpv")2 ι o( -*Ψ9\ =z'τιin wg =682Pa".[ffi)= -ι1οOo/

W, = \iν, [RPM, ]' = l '1 (12o0lu = l.9 HP = 1'42k\ΝιRPM1

' ι1000/

12-3. (a,b) Qz = Q (750i900) = 0.833 Q r

Poz= PO1(75ol9OO)2 = ο'694 Pο1

HPz- ΗP1(75oi9oo)3 = 0.579 HPl

12-3

2.4

2.0

'1.6

1.2

0.8

0.4

221

(d) Po = 1-30 in. wg

Φ = 9625 cfm

ΗP = 2.34(LΞ

σJΞ.Ξ

I

ΦΞΦU)ΦαEΞot-

6

5

4

3

2

1

00.002 468111214cfm x 10-3

12-4. Since pressure in in. wg. is plotted on the ordinate instead of head the

pressure must be adjusted to reflect the barometric pressure at 5280 ft

elevation.

Po = (Po)rtο(ρ/P.tα) = (Po).tα(Po/Po,'tα)

also, W = W.tα(ρ/Ps*o)= Wr16(P6/P5'r16)

Po,,tα = 14.696 psia; Po = O.491(29'42 _ 0.0009 x 5280); Eq. 3-4

Excerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφosοs only to

students enrolled in courses for wtriοh the tixtboοk has been adopted. Αny other reproduction or trαnslαιion of this work beyοnd ιhαt

permiιιed by Sectiοns ] 07 οr Ι 08 of ιhe Ι976 LΙnitec] Sιαιes Cοpyrigh} Αcι ν)ithouι the permissiοn οf ιhe copyrighι oνner is unΙανful'

80ο RPM 7OO RPM

a Po HP a Po HP

6,000 2.3 2.75 5,250 1.76 1.84

10,000 1.87 3.5 8,750 1.43 2.34

14,000 1.15 3.45 12,250 0.88 2.31

222Po = 12.112 psia

Τhen in Denver, Co the ne\M characteristics may be obtained by

computing Po and W at various volume flow rates from Fig. 12-8.

Po = (Po),ro(2.112t14.696) = 0'824(P6)916 ?Πd W - 0-824 Wstο.

Q cfm

6,000 '10,000 14,000

Po W Po W Po W

Sea Level 2.3 2.75 1.87 3.5 1.15 3.45

Denver 1.9ο 2.27 1.54 2.88 0.95 2.84

. :. (w",0 - w)1oo _ 3.5 - 2.88(b) ΔW=ff= 35

ΔW = 18% (decrease)

12-5. Refer to Problem 12-4 for explanation.

P6 = (99.436 - 0.10 x 1618) = 83.256 kPa

Po = (Po)stο = (83.256/101 '325) = 0.822(Po).16

\Λ/ = Wstο x0'822

(a)

Q m3/min

125 155 180

Po W Po W Po W

Sea Level 400 1 350 320 1 600 260 2000

Αlbuquerque 329 1110 263 1315 214 1644

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students enrolled in οourses fοr whiοh the textbook has been adopted. Αny oιher reproducιion or ιrαnsιαιiοn of ιhis wοrk beyond ιhαι

permiιted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed Stαιes Copyrighι Αctwiιhout ιhe permission ofιhe copyright oνner is unlαινful.

223

(b) ΔW = (1600--1315)

10O = 17'ro/odecrease1 600

12-6. (a) This is at the limit of the good selection range. lt would be

better to choose a different fan.

(b) Α near perfect match with the fan capable of producing about

1.85 in. V/g. totaΙ pressure at 10,000 cfm.

(c) A bad application and out of the recommended range.

Would probably be unstable.

12-7 [From Fig. 12-9]

(a) No, fan is too small.

(b) No, not a good application, fan is too large.

(c) YeS, near perfect appΙication; moderate fan speed,

high efficiency.

12-8. 150 m3/min, 4OO Pa [From Fig 1 2-10J

The fan would be acceptable and is reasonable.

η1 = 55%; RPM = 85O; W, = 185o W

12-9. (a) At 1418 cfm = 1420 cfm, Ve = 2OOO ft/min

P" = [+Ψl' = o.zs in. wg., Po = P, + P, = O.88 in. wg.ι4005,

P. = 0.88 - 0.25 = 0,63 in. wg. - 518 in. wg.EΧcerpts from this work may be reproduced by instructors 1br distribution οn a not-for-profit basis for testing or instructional puφoses only tostudents enrolled in οourses for whiοh the textbook has been adopted. Αny οther reproducιion or ιrαnslαιion of this νοrk beyοnd thαιpermitted by Sections ] 07 οr Ι 08 of the ] 976 United Stαtes Copyright Αcι νithοuι ιhe permission of ιhe copyright ονner is unlαιιful.

224

(b) From Τable 12-1 in col. For 5/8 in. wg.

The rpm is 1092 and power is 0.39 ΗP

1240 1420cfm

12-10' (a) ΔPo = 3.0 + 0.3 + 0'20 = 3.50 in. wg.

0.94O.BB

0.80

(b)

3.

3.CI

. 't3,500 15,00ο

Q cfm

12-11.

(c) 13,500 to 14,000 cfm

/2istem, actual

\o10e2 rPm

desired fanr t.ι ,' svstem with svstem{z' P eifect factor

'osystem wlthout systemι effect faοtor

iffu" setection withoutI system effect factor

βε,,f-οr whiιih tlοr ]08 οfthe ]9

225

System eff. Factor = 610 - 430 = .1gO pa

12-12' D" = (4 x 12 x 16lπ)112 = 15.6 in

Αssume blast area ratio = 0.7, Table 12-3\/V

" = 4ο00(1 2 x 161144) = 5333 fVmin

One eff. Duct length = 5.3 diameters, table 12_2

or L" = 5.3 x '15.6 = 83 in.

% Εff ' Duct length = 100 x 30/83 = 36

Elbow in position C, Fig. 12-13

EΙbow loss factor = 0'79, Τable 12-5

ΔPo = 0.79(5333l4o0q2 = 1.40 in. wg.

12-13' V ι = 4OO 0l |π x 142l14 x 1aa)] = 4,276 ftlmin

Duct length = 28 in.; R/D = 10.5114 = o.7s; L/D = 29114 = 2.0

Elbow and duct loss factor = 1.O, Τable 12-6

ΔPo - 1'0(4276l4ooq2 = 1'14 in. wg.

12-14' Blast area ratio = 0.7, TabΙe 12-3

D" = (4 x 20 x 2Olπ)1l2 = 22'6 in.

V = 1O,OOO x 1441(20 x20) = 3,600 fVmin

L" = 3.6 dia., Table 12-2

L/Le - (0122.6)13.6 = 0.12

Co = 0.4, Table 12-4

ΔPo = 0. 4(3600l4ooq2 = O.32in. wg.

12-15' D" = (4 x 12 x 12lπ)1l2 = 13.5 in.

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226V" = 25001(12x 121144) = 2500 ft/min

One eff. Duct length = 2.5 diameters,Table 12-2

L = 2.5 x 13.5 = 33.9 or 34 in.

12-16. From Problem 12-15, Vr= Ve = 2500 ft/min (assumed)

ΔPo = C"(v /4005)2; Co = O.16/(250O t400q2 = O'41

From Τable 12-6, LlH = 4'3

Length = 4.3 x 12 = 51.5 in.

12-17. (a) Τhe design condition and the observed condition are on

nearly the same system characteristic. Therefore, it is

probable that the fan is not running at the desired speed

of about 920 rpm but at a lower speed of about 6'10 rpm.

(b) The fan is operating near the 920 rpm characteristic

but something related to the duct system has changed.

Possibly a damper is closed, a duct has collapsed or some

other obstruction is present.

(c) Both the system and the fan characteristic have

changed. The duct system has probably

become fouled or slightly damaged is some way while

the fan speed has decreased slightly due to \Μear and tear.

12-18' Wsn,l = 16 HP; Wsh,2 Ξ 1.5x5000= 1.62

6350x0.73

% Diff = rco(16_,!62) = +οO% [decrease from 1 to 2!ι 16 )

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227

12-19. (a) Assume 15,000 cfm is an equivalent value for the day.

Forfull load point 1: Wr = 16 x0.746x24=286.5 kwh

For part load cond.: Wp = 6.7 x 0]46 x24 = 120.Okwh

ιW = (28q'9_ ]20) x '1OO = 58o/o (decrease)

286.5

(b) No, the fan would be forced to operate to the left of the maximum

pressure and would probably be unstable.

12-20' W l = 28.5 ΗP; W z= 17 '5 HP (static po\Μer used)

ΔW = (28'?:17 '5) 1OO = 39o/o (decrease)

28.5

12-21. (a) Wr = 28.5 x 0.746 x24 = 510 kwh

W, = 27 '0 xO'746 x24 = 483 kwh (vanes assumed "λ open)

ιW = (510_:183)

1OO = 5'3% (decrease)510

(b) W, = 27 x0.746x24 = 483 kwh

ΔW = (510_l_483)

1OO = 5'3% (decrease)51ο

(a) and (b) essentiaΙly the same.

12-22' The actual inside dimensions are 10 x 8 in. or D" = 9.8 ιn., Table 12-7

For duct, unlined, ΔPo/L = 1 .8 in. wg./100 ft (Fig. 12-21)

O - 2ooox144 = 3600 fvmin

1 0x8

From Fig. 12-23, roughness corr. Factor = 1.51, then for the

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228lined duct, ΔPo/L = 1 .8(1 .5'11 = 2'72 in' wg./100 ft. and

ΔPo - 50 x2'7211ο0 = '1.36 in. wg. or about 338 Pa

12-23. ΔPo = (ΔPo)rl X P/P.l, ρ _]!}-Psι Pο,rz

Pυ = 0.491(29'42 - 0.0009 x 5000) = 12.236 psia

ΔPo = 1.36(12'236114.696) = 1'13 in. wg. or about 282 Pa

12-24.

Q = 600 cfm

tl|

-=t'

()Θ

Dz = Ds = 10 in'', A2lA1= 0.6 = Α3/Aa

ΔPd/L = 0.185 in. wg./100 ft; Fig' 12-21

ΔPzs = 0.1 85 x20l100 = ο.037 in. wg.

For contraction, AzlAι = 0.6, Coz= 0'21 ] Table 12-gAFor expansion, A+/As = 1'67, Cρa = O.80 'Υz= Vs= 600 - = 11OOfUmin; Υι=Vgx ff =660ft/min

ι(ιo1'4112 )

ΔPιz= 0'21(11}ol4oO5)2 = O.O16 in. wg.

ΔPοo = 0.80(660/4o05)2 = a'O22 in. wg.

ΔPo = 0.037 + 0.016 + 0'022 = 0.075 in. wq.

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22912-25' (a) Bellmouth, Co = 0'2i Αbrupt, Co = 0.5; Table 12-109 & 10Α

ΔPos = O'2(1ooo/4oo5)2 = O.o125 in' W9. - 3.1 Pa

ΔPon = O.5(1oOo/4oO5)2 = O.0313 in. wg. x7'8Pa

o/o Diff . - (0'0313 - 0'0125) (1oo) = lsoo/o0 0125

(b) ΔPoa = O'2(40O0l40oq2 = O.2O in. wg. ε 50 Pa

ΔPon = O.5(4OOol4ooq2 = O.50 in. wg' x 124 Pa

o/o Diff. - (0.5 - 0.2) (1oo) = 15Oo/oo.2\/

12-26. Table 12-8a, Co = 0.25

V o = 1200l[(πla)x(1 4ln)2! = 1122'5 ftlmin

ΔPo = O.25(1122.5l4oO5)' _ O.O2 in. wq.

also

Vo = 0'6l|@lξ(0.35)2] = 6'24 mls

ΔPo - 0.25(6.2411.2q2 = 5.8 Pa

12-27. (a) Co = 0.15, Table 12-8b

Vo = 25OO x 1441(16 x 16) = '1406 ftlmin

ΔPo - o.15(1 40614005)2 = O.O'185 in. wq.

or Vo = 1.21(0.4 x 0.4) = 7.5 m/s

ΔPo = 0'15(7 '5l1'2q2 = 5.1 Pa

(b) Co = 1 '2 Τable 12-8C

ΔPo = 1.2(140614005)2 = O.148 in. wq.

or ΔPo = 1'2(7.5l1'29)'= 40.6 Pa

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230

12-28. a/a c= 25O18OO = 0.3125

Aυ/Ac=(6112)2=O.25

Co = 0.345, Τable 12-11A

Vυ = 25Ol|(πl4)(6t2)2]= 1273 ft/min

Vυ = 0'12[(ila)(ο.15)2] = 6.8 m/s

ΔPoο = 0.345(1 273t4o05)2 = O.O35 in. wg.

ΔPoο = 0.345(6 '8t1'2q2 = 9.6 Pa

Qr/Qο=55O/80O=0.6875

ΑS/Ac = (0112)'= 0.694

C" = 0.135, Table 12-11A

V. = 550l|(πla)(0l12)'! = 1οo8 ftlmin

V, = o.26ll(ila)(0.25)21 = 5.3 m/s

ΔPo" = O.135(1oO8/4οο5)2 = O.OO9 in. wq.

ΔPo" = O.135(5.3/1 2972 = 2'3 Pa

12-29. From Problem 12-28

a/aο = O.31 25; A/A"= 0'25

Vo = 1273 fVmin or 6.8 m/s

Cο = 0.93, Τable 12-11F

ΔPoo = 0.93(1 273l4oo5)2 = O.O94 in. wq.

or ΔPoο = O.93(6.811'2q2 = 25.8 Pa

Qr/Qc = 0.6875; Αr/A" = 0.694

C, = 0.135; Table 12-118

V, = lOOB fVmin or 5.3 m/s

ΔPo, = O.135(1oOs/4oo5)2 = O.OO9 in. wq.

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or

or

or

or

or ΔPo, = o.135(5.3/1 29)2 = 2'3 Pa231

12-30' (a) Ao/Al = 6.0, θ = 180 deg., Co = 37'4, Table 12-98

vr = o?9-0"'^!o

= 2,ooo ft/min(18x18)

ΑoVo = ΑlVl, Vo = 1, x 2,O0O = 2000/6 = 333 fVminΑo

ΔPo = 37.4(333l4o0q2 = 0.260 in. wq.

(b) Co = 14.35 (Table 12-98)

ΔPo = 14.35(333t4ooq2 = O.O99 in. wq.

12-31. ao/Q" = 5OO/1OOO = O.5o

Αο/A" = (8Ι12)'= o'444

ΑS/Αc = (8/1 2)' = 0'444

Q,/Qc = 500/1OOO = O'5

(a) Cυ = 0.755, Table 12-12A 1 lnterpolation required or use

C" = O.2'15, Table 12-12A l ASHRΑE Duct Fitting Data

Vο = 50Ot|πt4)(8t2)2]= 1432 ft/min - v,or Vυ= O.24l|(πl4χo.2)'] =7'64mls

ΔPoυ = 0.755(1 4g2l4o05)2 = O.O97 in. wο.

or ΔPoo = O'755(7 .64112972 = 26'5 Pa

ΔPo, = O'215(1432I4OO5)2 = O.O28 in. wq.

or ΔPo, = O'215(7.641129)2 = 7 '54 Pa

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1

2

3

55

72

20

0.50

0.16

2.0

1020

630

550

23335

14

13

0.'135

0.055

0.050

0.07 4

0'ο40

0.010

2(0.26)

2(0.26)

0.17

0.032

0.004

0.038

0.034

0.013

0.003

0.140

0.057

0.051

12-33' L = D x Co/f; f = 0.019, ΤabΙe 12-13

Bellmouth: L = 1 x0.210.019 = 10.5 ft

Αbrupt Int.: L = 1 x 0.5/0.019 = 26.3 ft

Q = 1OOO x πl4 = 785 cfm; ΔPo/L = O.12in'wg'l1oo ft, Fig' 12-21

ΔPog = 0.12x 10'5l100 = 0.0126 in. wg. or about 3.'1 Pa

ΔPon = 0.'12x26'3l100 = 0.0316 in. wg. or about 7'9 Pa

12-34' From ProbΙem 12-26' Co = 0.25, D = 14 in'

Lu = DxC olfi f = 0.017 , Table 12-13

γ'= lx 0'25 = 17 '2ft- 12 0.017

Q = 12OO cfm; ΔPο/L = 0.'13 in. wg./1OO ft

ΔPo = 0.13 x 17 '2l100 = 0.022 in. wg. or about 5.6 Pa

Note: Most of following duct sizing problems can be solved with the computer

program, DUCT.

12-35. From Figure 12-36.

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234

Estimate Tota] Εquivalent Length of Run 1-2-3 to be approximately

132 ft, Τable 12-14' Then ΔPo/Le = (0.13 x 1 00)1132

= 0.'10 in. wg./100 ft size ducts using Figure 12-21 and record the

actual ΔPo/L from Figure 12-21'

Section

No.

Le

ft

acfm

D

in.

ΔP/L

in. wq.

100 ft

ΔPn

in. wg.

1

2

3

4

5

45

16

71

55

55

300

220

100

80

120

II6

5

6

0.084

ο.090

0.083

0.14

0.125

0.ο38

0.014

0.059

0.077

0.069

Run 1-2-5 actually has the greatest lost pressure.

ΔPlη = 0.038 + 0'077 = 0.115 in. wg.;

ΔPιzs = 0.038 + 0.014 + 0.059 = 0.121 in. wg.

ΔPlzs = 0.038 + 0.014 + 0.059 = 0.1'1 '1 in. \Λ/g.

12-36. The design pressure loss is (0.25 - 0.1 ) = 0.1 5 in. wg. (for supply ducts)

Assume the run with the largest equivalent length is:

1-2-3-4-5, Le = 185 ft

Τhen for design: ΔPo/Le - (0'15 _ 0'03) x 1OO = 0.065 in. wg./'1oO ft185

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235

Section Θ has a total flow of 845 cfm. Τherefore, the maximum

velocity in section Θ wjl! be about 800 fVmin if a 14 in. duct is used.

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\

236

12-36. (continued)

(a)

Section

No.

Le

ft

acfm

D

in.

ΔP/L

in. wο.

100 ft

ΔPn

in. wg.

1

2

3

4

5

6

7

II

88

18

16

17

46

51

43

49

49

845

595

395

275

125

250

200

120

150

14

12

12

9

7

I8

7

I

0.065

0.07ο

0.034

0.065

0.065

0.060

0.072

0.065

0.040

0.ο57

0.013

0.0054

0.0'1 1

0.030

0.031

0.031

0.032

0.020

With the equal friction method, every branch should have a damper for

balancing purposes.

Αctual total pressure loss:

ΔPo - ΔP1 + ΔP2 + ΔP3 + ΔPa + ΔPu + ΔPοs

ΔPo = 0'146 in' wg'

Note that run 1-2-3-4-7 actuaΙly has the greatest loss in total pressure

but the difference is not significant. Use ΔPo = 0.15 in. wg.

(b) Sizing of the longest run, 1 -2-3-4-5, is the same as (a) above where

ΔPo/L = 0.065 in. wg./100 ft. Construct a new table as follows:

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----------------

237MAIN DUCT RUN BRΑNCH DUcτS

(1)

Sec

No.

(2)

Le

ft.

(3)

cfm

(4)

DJwxh

in.

(5)

ΔP

L

(6)

vfpm

(7)

ΔPo

(2)(5)

100

(8)

ΣΔPo

Σ(7)

(e)

Br.

Seο

No.

(10)

ΔPι

ΔPoot

-(8)*

ΔPα

(1 1)

Le

ft.

(12)

ΔPi

L

('10)100

(11)

(1 3)

οfm

(14)

D"/wxh

in.

(1 5)

vfpm

I BB 845 14 0.065 800 0.057 0.057 f) 0.ο39 51 0.076 250 9 550

2 '18 595 12 0.070 760 0.013 0.070 7 0.036 43 0.084 200 I 570

J 16 ?oξ 12 0.034 500 0.005 ο.ο75 8 0.035 49 0.071 120 7 500

4 17 275 I 0.065 600 ο.01'1 ο.086 9 0.020 49 0.041 150 I 420

5 46 125 7 ο.065 500 0.ο3 0.116

ffuser G 0.030 0,146

The left 8 columns are the same as (a) above. The branches, 6-7-8-9,

are sized to balance in the right hand 7 columns.

(c) Equal Friction Method

-- Design Procedure --Sysιem type: Supp}y

Duct Sizing Method: Equal FrictionRounding Method: Round Nearest

-- Ean Selection --Knoι\Ιn Fan Parameter: F'an TotaΙ Ρressure : 0 . 250 in. wg

Fan Αlrflow: B45.0 cfmFan or Externa1 Total Pressure: Ο.25Ο in. wg

Coi1 Lost PreSsure: Ο . ΟΟ0 in. wgΕi1ter Lost Pressure: 0 . 000 in. wgMisc. Lost PreSSure: 0.ΟΟΟ in. wα

ΑΗU External Total Pressure: Ο.25Ο in. wg

AΗU Pressure for Supply System: 0.150 in. wg - or 60.0 %

AHU Pressure for Return System: 0.1Ο0 in. wg - or 40.0 z

-- Lost Pressure from Αir HandJ_ιng Unit to Diffuser --Diffuser ΙD Q Tota1 Delta P

(cfm) (1n. wg)

71 125.0 A.L20Excerpts from this work may be reproduced by instruοtοrs for distribution οn a not-for-prοfit basis for testing or instruοtional puφoses only tostudents etrrolled in οourses for which the textbook has been adopted. Αny other reproducιion or ιrαnsιαιion οf this wοrk beyond thαtpermitted by Sections ] 07 οr 108 οf the ] 976 United Stαιes Copyrighι Αcι ''νiιhout ιhe permission οf ιhe cοpyrighι oνner is unlανful.

23822 150.0

120.a 0.200.0 0.250.A 0.

0.154130t25Ι21

t63034

Total B45. Ο

ΙD Fitt1ng Type

1 Air Ηandling Unit2 Straight Duct3 ConicaΙ Contraction4 Εlbow5 Elbow6 Tee / νΙΥe main

branchcommon

7 Straight DuctB Tee / wye main

branchcoτnmon

9 Straight Duct1Ο Tee / wye main

branchcoΙτιmon

11 Straight Duct12 Tee / ,νΙye il?}l.n

coΙnmon13 Straight Duct14 Εlbow15 Straight Duct16 Elbow17 Diffuser / GrilΙe18 Straight Duct19 Elbow20 Straight Duct21 EΙbow22 Diffυser / GrΙΙ23 Elbow24 Straight Duct25 ΕΙbow26 Diffuser,/ Gri11e21 Ε]-boν'τ28 Straight Duct29 Elbow30 Diffuser / GritΙ31 Elbow32 Straight Duct33 EΙbow34 Dlffuser / GrilΙ

-- Calculated Fitting VaΙues --

Dia.( in)

0.014. Ο

14. Ο

1Δ a

14. Ο

72 .4oΔ

14.072 .010.08.0

72 .010.09.01 .0

10.09.01.0?n9.07.0'7.01 .07.0

1.0'1 .01 .07.0

?n7.Ο7.0

8.08.0

Q Velocity(cfm) (ftlmin)

845.0 0.0845.0 '79Ο.4845.0 19A.4845.0 '790.4845.0 19A.4595.Ο '75'7.6250.0 565. 9B45.Ο 19a.4595.0 151 .6395.0 124.220Ο.0 573.0595.0 151 .6395.0 124.2215.0 622.5L20.0 449.0395.0 '724.2215.0 622.5L25 .0 461 .1150.0 561.3215.0 622.5125 .0 461 .1L25 .0 461 .1L25 .0 461 .'7L25 .0 461 .1] 2η n

15Ο.0 561.31s0.0 561.315Ο.0 561.315Ο.0 561.315Ο.0L24.0 449.0ι20.0 449.a120.0 449.0120 .02Ο0.0 573.02AA. 573.020Ο.0 573.02Ο0.0250.0 565.9250.0 565.9250.0 565.9250.0

Delta P(in. wg)

Ο.0000.0130. Ο130. Ο060.0060. Ο050.018

0. ΟΟ70.0040.01_7

0.0070.0030. Ο16

0.0070.0020.009

0.0080.0020.0020.0030.0300.0060.0050.0160.0050.0400.0020. Ο110. Ο03Ο.036Ο.0Ο30.0100.0050.0400.0030.0130.0040. Ο50

AP /L(in. wg)

0.06415

0.011_61

0.08259

0.07138

0.05817

Ο.05B17

Ο. Ο8082

Ο. ΟB0B2

0.05405

0. ο7106

0.060049.09.09.0

(c) Balanced Capacity Method

-- Design Procedure --

EΧceφts from this wοrk may be reproduced by instruοtors for distribution on a not_for-profit basis for testing οr instruοtional puφoses only tostudents enτolled in οourses for whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsιαtion of ιhis νοrk beyond ιhαιpermiιΙed by Sectiοns ] 07 or l 08 οf the 1976 United Stαιes Copyright Αcι withοuι ιhe permissiοn of ιhe cοpyright oνner is unlανful.

____

System type: SupplyDuct Sizing Method: Balanced Capacity

Rounding Method: Round Nearest

-- Fan Selection ---

239 -

Known Fan Parameter: Fan Total Pressure : 0.250 in. wg

Fan Αirflow: B45.0 cfmFan or External Total Ρressure: Ο.250 in. wg

CoiΙ Lost Pressure: 0.Ο00 in. wgE'iΙter Lost Ρressure: 0.000 in. wgMisc. Lost PreSSure : 0 . 0Ο0 in. ι^rq

ΑHU Εxterna1 Total Pressure: Ο.25Ο in. wg

AΗU Pressure for SuppΙy System: 0'150 in. wg - or 60.0 %

ΑHU Pressure for Return System: 0.1Ο0 in. wg - or 40.0 %

-- LoSt Pressure from Αir Ηand1ing Unit to Diffuser _-

Diffuser ΙD Q Tοtaι Delta P(cfm) (in. wq)

L] 125.0 0.13822 150.0 Ο.15426 720.0 0.14530 200.0 0.14034 250.0 0.141

Total 845.0

-- Cal cu-Lated Fitting Values --

ΙD Fitting Type Dia. Q Velocity DeΙta P ΔP/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 Air Handling Unit 0. Ο 845.0 0.0 0.002 Straight Duct 14.0 845.0 '790.4 0.013 Ο.064153 Conicaf Contraction 14.0 845.0 190.4 0.0134 E}bow 14.0 845.0 190.4 Ο.0065 trlbow 14.0 845.0 '190.4 0.0066 Tee / Wye main 1'2 .0 595 . 0 '7 57 .6 0 . 005

branch 8.0 250.0 116.2 0.016common 14.Ο 845.Ο '790.4

7 Straight Duct 72.a 595.0 '751 .6 Ο.0Ο7 0.071678 Tee / \NΥe main 10.0 395.0 124.2 0.004

branch 1 .0 200.0 1 48.4 0.016common 72.0 595.0 15'1 .6

9 Straight Duct 10.0 395.0 -124.2 0.007 0.0825910 Tee / Wye main 9.0 215.0 622'5 0.0Ο3

branch 6.0 720.0 6lL.2 Ο.013common 10.0 395.0 124.2

11 Straight Duct 9. Ο 215 '0 622.5 0.0Ο7 0.07138L2 Tee / Wye main 6.0 \25.0 636.6 0.003

branch 1 .0 150.0 561.3 0.009coΙnmon 9.0 21 5.0 622.5

Exceφts from this rνork may be reproduced by instructors for distribution οn a nοt-for-profit basis for testing or instruοtional puφοses only tostudents enrolled in courses for which the textbook has been adopted. Αny οther reproductiol1 or trαnsιatιon οf ιhis work beyond thαt

permitted by Sections ] 07 οr Ι 08 οf the t 97 6 Uniιed Sιαtes Copyrighι Αct ||1ιthout ιhe peιmissiοn of the copyrighι oνner is unΙcτwful.

24013 Straight Duct

0.72304L4 Elbow15 Sιraight Duct76 Εlbοwl1 Diffuser / Grille

1 8 Sιraight Duct19 Elbow2a Straighι Duct27 Elbow22 Diffuser / Grille23 Elbow24 Straight Duct25 Elbow26 Diffuser / GrilΙe2"7 El-bow28 Straighι Duct29 Elbow30 Diffuser / Grille31 Εlbow32 Strarght Duct33 Elbow34 Diffuser,/ Grille

12-37.

(a)

6.0

6.06.06.0

7.0/.u7.01.0

6.06.Ο6.0

1.41.07.0

8.08.08.0

125 .0

L25 .0L25 .01rη n

1,25 .0

150.0150.0

15Ο.0150.01"20 .0120.0t2a .0120.0200. Ο

200.0200.0200. Ο

250.025Ο.0250.0250.0

636.6

636.6636 .6636 .6

561.3561.3561.3561.3

6tt .26L7.26L1.2

148.4'1 ΔQ Δ

"7 48 .4

'716.21L6 .2'Ι 16 .2

0.016

0.0040.0050.007Ο.030

0.0060.00s0.0160.0050.0400.0040.0230.0060.036Ο.005n nT q

0.0090. Ο400.005^ ^410.0070.050

0.72344

0.08082

Ο.08082

0.ΙL427

0.13629

0.10661

ΔPos + ΔPon = ο.70 _ 0.35 = 0.35 in. wg.

ΔPos = 0.65(0.35) = 0.23 in. wg.

ΔPoκ ^y

0.35 _0'23 = 0.12 in. wg.

Τhe method of Solution is similar to Problem 12-36' An

acceptable solution follows:

Longest run - 1 -2-3-4-5-6-7-8-9-1 1 -13

The summation of equivalent lengths may vary v/ith designers.

ΔPo/Le =(0.23 - 0.03)

100 = 0.092 in. wg./'100 ft217

Size all Suppιy ducts for this pressure loss per unit length.

EΧcΘrpts from this work may be reproduced by instructοrs for distτibution on a not-for-profit basis for testing or instructional puφoses only tο

students enrolled in courses for whiοh the tΘXtbook has been adopted' Αny other reproductiοn or ιrαnslαtion οf ιhis work beyond ιhαι

pe,rmιιed by Sections ] 07 οr l δε o7 ,ιr, 1976 Uniιed Stαtes Copyrigh't Αct νiιhout ιhe peimission of ιhe copyright oνner is unlατιful'

241

150

100

100

75

100

200

16

14

14

12

12

12

10

I7

5

6

5

6

5

1200

1 050

850

750

650

550

475

225

175

50

125

50

75

50

1

2

3

4

5

b

7

8

9

10

11

12

13

14

Εxοerpts frοm this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only ω

students enrolled in courses for wt'iοh the textbook has been adopted. Αny oιher reproducιion or ιrαnsl(tιion of ιhis work beyond thαι

permiιιed by Secιiοns ] 07 or l δε i7 in, Ι976 United Stαιes Cοpyrigh't Αct ιlithouι the peimissiοn ofιhe copyrighι oνner is unΙανful'

----=-

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12-42. (continued)

-ο

-

12-37. (continued)

ΔPo for the longest run will be about 0'22 in. wg' for the above sizes"

Τherefore, size the return system for a pressure loss of (0.35 _0'22)

or 0.13 in. wg.

For the return system:

(L")r* x 230 ft, ('1 r - 2r - 3r)

then ΔPo lL"= r0'13-__0'05'] ''oo = O.O35 in. wg./'1oo ftι230)

Using the equal friction method:

Section

No.

acfm

D

in.

L"

ft

ΔP"/L ΔPo

in. wg.

1r

2r

3r

4r

5r

1200

800

400

400

400

18

'16

12

12

12

115

70

44

14

28

0.038

0.033

0.ο36

0.036

0.036

0.044

0.023

0.0'16

0.005

0.010

Return system is the same for parts (a) and (b).

ΔPo for return = 0.133 in. \Mg.

(c) Equal Friction Method

-- Desiqn Procedure --

System type: SupplyDuct Sizing Method: Equal- Γriction

Rounding Method: Round Nearesι

-- Ean SeΙection --

Known Εan Parameter: Εan Total Pressure : 0.700 in. wg

244

Εan Airflow: 1000.0 cfmFan or ExternaΙ Total Pressure: 0.700 in. wg

Coil Lost Pressure: 0.25Ο in. wgF'iΙter Lost Pressure: 0 . 10Ο in. wgΜisc. Lost ΡresSure: 0.0Ο0 in. wg

ΑHU Externa1 Total Pressure: 0.350 in. ι^rg

AΗU Pressure for SuppΙy System: 0.228 in. wg - or 65.0 %

AHU Pressure for Return System: 0.123 in. wg - or 35.0 %

-- Lost Ρressure from Αir Ηand1ing Unlt to Diffuser --Diffuser ΙD Q Total DeΙta Ρ

(cfm) (in. wg)

2'7 75.Ο 0.22130 75. Ο 0.20134 75.0 0.2Ι138 s0.0 0.27256 75.0 0.17160 100.0 0.15863 100.0 0.17161 100.0 0.L4212 200.0 0.2a215 150.0 0.131

Total 1000. Ο

-_ Calculated Fittinα Values --

ΙD Εitting Type Dia. Q VeΙocity Delta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 Αir Ηandling Unlt Ο.0 1000.0 0. Ο 0.0002 Conical- Contraction 14.0 1000.0 935.4 0.0113 Straight Duct 14.0 1Ο00.0 935.4 0.007 0.087454 Tee / wye main l2.a 850. Ο 082.3 O. O1O

branch 1 .0 15Ο. Ο 561.3 0.061coΙτΙnon 14.0 1000.0 935.4

5 Straight Duct 12.0 850.0 L082.3 0.003 0.138196 Tee / Wye main 12 .0 650 . 0 82'7 .6 0 . 008*<10>

branch 1 '0 2a0.0 1 48.4 Ο.058coΙnmon L2 '0 B50.0 1082.3

7 Straiqht Duct l2.0 650.0 B2'7 .6 0.005 0.08429B Tee / vfrze main 10.0 550.0 1008.4 Ο.0Ο9

branch 6.O 1OO.O 5O9.3 Ο.Ο37common L2.0 650.0 821.6

9 Straight Duct 10.0 550.0 1008.4 0. ΟΟ6 0.1516410 Tee ,/ Wye main 10.0 450.0 825.! O. O1O*<10>

branch 6.0 100.0 509.3 0.064coΙπnon 10.0 550. Ο 1008.4

11 Straight Duct 10.0 450.0 825.7 0.004 0.10485L2 Tee ,/ Wye main 9. 0 350 . Ο 192 .2 O . ΟΟ5

branch 6.0 100.Ο 509.3 a.024conτnon 10.0 450.0 825.L

13 Straight Duct 9. 0 350 . 0 1 92.2 0 . 014 0. 11082

Excerpts from this work may be reproducοd by instruοtors for distribution on a not_for-profit basis for testing or instructional puφoses only tostudents enrolled in οοurses for whiοh the textbook has been adopted. Αny οther reproductiοn of trαnslαtion of ιhis νork beyοnd thαιpermiιιed by Secιions Ι07 or l08 οfthe Ι976 Uniιed Stqιes Copyright Αcι'|ι)iιhout the permission ofthe cοpyrighι oνner is unlcrwful.

24514 Tee ,/ Wye main 8. Ο 215.a "781 .8 Ο. Ο05

branch 5.0 75.Ο 550.0 0.019common 9.0 35ο.Ο 192.2

15 Straight Duct B.0 215.0 '7B1 .8 0.0Ο6 0.7268'7L7 ΕΙbow B.0 215.0 1B1.B 0.0Ο818 Straight Duct 8.0 215.0 181 .B 0.013 0.L268'719 Tee ,i Wye main B.0 225.a 644.6 Ο.0Ο9*<10>

branch 4.0 5Ο. Ο 573.0 a .021common 8.0 215.4 l1i .B

20 Straight Duct B.0 225.a 644.6 0. Ο07 0.0880Ο27 Tee / Wye main 7.0 150.0 561.3 0.003

branch 5.0 75.0 550. Ο 0.011coΙnmon B.0 225.0 644.6

22 Straight Duct 1 .0 150.0 561.3 0.018 0.080822? Tρρ / Ιn1lzo main 5.0 75.0 550.0 0.002f νν / νγli ν

branch 5.Ο 75.0 550.0 0.007coΙnmon 1.a 150.0 561.3

24 Εlbow 5. Ο 75.0 550.0 0.00625 Straight Duct 5. Ο 75.0 550.0 0.072 Ο.1186926 Rectangular Transition 5. Ο 75.0 210.a 0.0Ο52'l Diffuser / GrilΙe 75.0 0.03028 Straight Duct 5.0 75.0 550.0 0.0Ο7 0.1186929 Rectangular Transition 5. Ο 75.0 210.a 0.0Ο53Ο Diffuser / Grille 75.0 0.02531 Elbow 5.0 75. Ο 55Ο.0 0.0Ο632 Straiqht Duct 5. Ο 75.0 550.0 Ο.018 0.1186933 RectanguΙar Transition 5.0 75.0 21a.0 0.00534 Dif fuser / Gril-le 75.0 A.02535 Elbow 4.0 50.0 573.0 0.00736 Strarght Duct 4.0 50.0 573.0 0.017 0.1691631 Rectangular Transition 4.0 5Ο.0 180.0 Ο.01038 Dif fuser ,/ cril1e 50. Ο 0.02053 Elboιv 5.0 75.0 550.0 0.00654 Straiqht Duct 5.0 75.0 550. Ο 0.01B 0.1186955 RectanguΙar Transition 5.0 75.0 210.0 0.00556 Diffuser / cril1e 75.0 0.03051 Ε1bow 6.0 100.0 509.3 0.00458 Straight Duct 6.Ο 100.0 509.3 0.008 a.0122l59 Rectangular Transition 6.0 1Ο0.0 360.0 0.0036Ο Diffuser / cri]ιe 100.0 0.0456L Straight Duct 6. Ο 100.0 5Ο9.3 0.072 a.0B221'62 Rectangular Transition 6. Ο 100.0 36Ο.0 Ο.00363 Diffuser / Gri1Ιe 1Ο0.0 0.03264 E1bow 6.0 100.0 509.3 0.0Ο465 Straight Duct 6.0 1Ο0.0 509.3 0.008 a.08221-66 Rectangular Transition 6.0 100.0 360.0 0.00361 Diffuser / Gril]_e 1Ο0.0 0. Ο456B Straight Duct '7 .A 200.0 148.4 0.02'7 0.l.362969 Elbow 1.0 200.0 148.4 0.01370 Straight Duct 1 .0 20Ο. Ο "7 48 ' 4 0 .a21 0.1-362911 Rectangu1ar Transition 1 .0 20Ο. ο 120.0 0.00112 Diffuser / Gritle 20Ο. Ο 0.04513 Straight Duct 1 .0 150.0 561.3 0.032 0.080821 4 Rectangular Transition 7.0 150.0 54Ο.0 0.00115 Dif fuser ,/ Grille 150.0 Ο.02016 Elbow 8.0 215.0 181 .B 0.005

* - De}ta P ιfas computed using the fitt1ng equivalent lenqth

Balanced Capacity Method

Exοeφts from this wοrk may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enτοlled in οourses Γor whiοh the textbook has beοn adopted. Αny οther reproduction or trαnslαιion of ιhis νοrk beyond thαtpernιiιιed by Secιions ]07 or l 08 οf the ]976 Uniιed Stαtes Copyright Αcι νιιhout ιhe permissiοn οf ιhe cοpyrighι owner is unlcrνιful.

246

-- Desiqn Procedure --

Note that almost alΙ branch ducts need a damper to increase the diameter andreduce ve1οcity.



-- Fan Selection --Known F'an Parameter: Ean Tοta1 Pressure : 0.700 in. wg

Εan Αirflow: 100ο.Ο cfmFan or Εxternal Total Pressure: Ο.7ΟΟ in. wg

CoiΙ Lost Pressure: 0.25Ο in. wgFilter LoSt Pressure: 0 . 1Ο0 in. wgMisc. Lost Pressure: 0.000 in. wq

ΑHU External Total Ρressure: 0.350 in. wg

AHU Pressure for Supply System: 0.228 in. wg - or 65.0 ?ΑHU Pressure for Return System: 0.L23 in. wg - or 35.0 ?

-- Lost Pressure from Αir Handling Unit to Diffuser --

Diffuser lD Q TotaΙ Delta P(cfm) (in. wq)

2'7 75.0 0.22130 75.0 0.24334 75.0 0.2713B 50.0 0.2L256 75.0 0.23260 100.0 0.21963 100. Ο 0.28561 100.0 0.25'712 2Ο0.0 0.20215 150.Ο 0.17Ο

TotaΙ 10Ο0.0

-- CaΙcuΙaιed Γiιting Vafues --

ΙD Εltting Type Dia. Q VeΙocrty DeΙta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 Αir Ηand1ing Unit Ο.0 10Ο0.0 0.0 0.0002 Conical Contraction 14 .0 1000.0 935.4 0.0113 Stralght Duct 14. Ο 1000.0 935.4 0.007 0.08745/ .Γaa / τlτ.,o main 72.a 850.0 L082.3 0.010

branch 6. Ο 15O . O 1 63 .9 O. O57coττunon 14.0 1000.0 935.4

5 Straight Duct l2.0 850. Ο 1-082.3 0.0Ο3 0.138196 Tee / wye main !2.a 650.0 821.6 Ο.OΟ8*<1O>

branch 1.a 20Ο.Ο '748.4 0.058

Excerpts from this work may be reproduοed by instruοtors for distribution on a nοt_for-profit basis fοr testing or instruοtional puφoses only tostudents enrοlΙed in οourses fοr whiοh the textbook has been adopted. Αny οther reProduction or ιrαnsΙαιiοn of ιhis work beyond ιhαιpermiιιed by Secιiοns Ι07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighl Αct νiιhouι ιhe peιmission of ιhe cοpyrighι olνner is unΙcrννful.

coΙnmon7 Straight DuctB Tee / wye main

branchcoΙnmon

9 Straight Duct10 Tee / Wye main

branchcoΙnmon

11 Strarght Duct'r ^^ / r^ir7^ malnΙ99 / νlyν

branchcoΙτιΙnon


branch

ι5L1LBL9

202L

2223

coΙnmonStraight DuctElbowStraight DuctTee / Wye main

branchcoτnmon

Straiqht Duct/ l1lrΣ^ ηa1nιγ ] ν

branchcommon

Straight DuctTee / Wye main

branchcoτΙυnon

24 Elbow25 Straight Duct26 Rectangufar Transition21 Diffuser / crille28 Straight Ducι29 Rectangular Transition30 Diffuser / Gritle31 tr1bow32 Straight Ducc33 Rectangular Transition34 Diffuser / Grille35 Elbow36 Straight Ducι31 Rectangular Transition38 Diffuser / Grille53 Elbow54 Straighι Duct55 Recιangufar Trans1tion56 Diffuser / GriΙ1e51 ΕΙbow58 Sιraighι Duct59 Rectangular Transition60 Diffuser / GriΙ]e67 Straight Ducι62 Rectanqular Transition63 Diffuser / Grille64 ΕΙbow65 Scraight Duct66 Rectangular Transition61 Diffuser / crille68 Straight Ducι

72 .01_2 .010.0Δo

L2 .010.010.04.Ο

10.01Ο.09.04.0

10.09.08.04.0on8.08.08.08.04.ο8.08.07.05.Οo.u1.44.05.0?n5.Ο5.05.0

4.04.Ο

ηn5.ΟtrΛ

4.04.0ΔΓ\

4.44.04.0

85Ο.0650. Ο

10Ο.0650.0550.0450.0100.055Ο. Ο

450.0350.01Ο0.0450.035ο. ο275.075.0

350.0t1\ aο' tr Λ

215.0225.050.0

215 .0aa tr Δ

150.075.0

225.015Ο.075.Ο75.Ο

150.075.075.0?tr Λ

75.075.075.075.ο7η n

75.075.075.Ο5Ο. Ο

50. Ο

50.050.075.075.075.Ο'tr

Λ

1Ο0.01Ο0.010Ο. Ο

100.01ο0.010Ο.0100.0100.010Ο.0100.0100.02ο0. Ο

ι082.3821 .6

1008.4ι145.9821 .6

10ΟB.4825 .7

1145. 91ο0B.4825.L192.2

1145. 9825.1-7q2 2

181 .8859.41A) )181 .8181 .8181 .8644 .6573.0'7 81 .8644 .6561.3550.0644 .6561.3859.4550.0561.355Ο.0550.0210 .0

859.4214 .0

550. Ο

550.0214 .0

573.0573.0180.0

0.0050.0090.039

0.006

247

0.08429

0.15164

Ο.104B5

0.11082

0.010*<10>a .062

0.0040.0050.031

0.0140.0050.017

0.0060.0080.013

4.04.44.0

4.04.Ο

Δol4.Ο4.0

7.Ο

859.4859.4210 .0

1145.91145.9360.0

L1_45 .9360.0

1145. 97L45 .9360. Ο

148.4

0.L2681

0.!268'70.009*<10>0 .021

0.0Ο7 Ο. Ο88Ο00.0030.011

0.018 0.080820.0070.007

Ο.0060.at2 0.118690.00s0.0309.02! 0.352660 .0240.025Ο.0060.018 0.11869Ο.0Ο50. Ο250. ΟΟ70.017 Ο.169160.0100.0200.015Ο.053 0.352660 .4240.03Ο0 .0210.060 0.591440.0420.0450.090 0.591440 .0420.032Ο a2'10.06Ο 0.59'7440.0420.045a.021 0.!3629

Exοerpts from this work may be reprοduced by instructors for distribution on a not-fοr-profit basis for testing or instruοtional puφoses only to

students enrolled in courses for which the textboοk has been adopted. Αny other reprοducιiοn or ιrαnsιαtion οf this νork beyοnd thαιpermitted by Sections ] 07 οr l 08 οf the 1 976 United StαιeS Copyrighι Αcι'withouι ιhe permission of the cοpyrighι owner is unlανυful.

24869 EΙbow 1.0 200.a ']48.4 0.013

70 Straight Duct 1 .0 200.0 148.4 0.02'1 0.1-3629lt RectanguΙar Transition 1 .0 2Ο0.0 120.0 0.00112 Diffuser / Gril-le 200.0 0.04513 Straight Duct 6. Ο 150.0 '7 63.9 0.069 0.171391 4 Rectangular Transition 6. Ο 150.0 540.0 0.00715 Diffuser / Griι1e 150.0 Ο.02016 EΙbow B.0 2'75.a '7B'7.B 0.0Ο5

* - Delta Ρ was computed using the fitting equivalent Ιength

Return Ducts, Equal Friction Method

-- nηοj ^_ D:ocedure __vνυΙYlΙ L ]

System type: ReturnDuct Sizing Method: Εqual Friction


-- Fan SeΙection --

Known Εan Parameter: Εan Tota1 Pressure : 0.700 in. wg

Fan ΑirfΙow: 1200,0 cfmFan or External Tοtal Ρressure: Ο.7ΟΟ in' wg

Coil Lost Pressure: Ο.25Ο in. wqΕiΙter LoSt Pressure: 0.100 in. wgMisc. Lost Pressure: 0.000 in. wg

AΗU Εxternal Tota1 Pressure: 0.350 in. wq

AΗU Pressure for SuppΙy System: 0.228 in. wg - or 65.0 %

ΑΗU Pressure for Return System: 0.123 in. wg - or 35.0 %

-- Lost Ρressure from Αir Ηandling Unit to Diffuser --

Diffuser ΙD Q TotaΙ Delta P(cfm) (in. wg)

11 400.0 0.12514 400.0 0.113l"7 400.Ο 0.096

Tota1 120Ο.0

_- CaΙcu1ated F'itting VaΙues -_

TD Fitting Type Dia. Q Velocity Delta P LP/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 Αir Handllng Unit 0. Ο 1200.0 Ο.0 Ο. ΟΟ0

2 Rectangular Transition 18.0 72A0.0 679.1 0.0013 Straight Ducι 1B.0 L200.a 619.1 0.002 0.035124 Tee / \ηye main 1'2.a 400.0 509.3 0.026

branch 16.0 800.0 573.0 0.026coπlmon 18.0 1200.0 679.1

5 Straight Duct 16. Ο 800.0 573 ' 0 0.006 0.03a246 Tee / wye main 12.0 4ΟΟ.0 509.3 0.0]-5

branch \2.0 400.0 509.3 0.013coΙnmon 16.0 800.0 573.0

Exceφts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instructiοnal puφoses only to

students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnslαιion of ιhis νοrk beyond thαιpermitιed by Sectiοns ] 07 or Ι 08 of ιhe 1976 United Sιαtes Cοpyright Αcι νithouι the permission οf ιhe copyright oνner is unΙανfuΙ.

7 Stralght Duct0. Ο3476 B ElbowΟ. Ο039 Straight Duct10 Rectangular Transition11 Diffuser / criΙle72 Straight DucL13 Rectangular TransitionL4 Diffuser / GrilΙe15 Straight Dυct16 Rectangular Transition11 Diffuser / Gri-ι1e

12 .0

L2 .0L2 .0

L2 .0L2 .0

L2.AL2 .0

400.012 .0

4Οο.04Ο0.04Ο0.0400.040Ο.0400.0400.0400.0400.0

0.007509.3

0.0010.0140.0500.0010.0140.0500.0030.0140.05Ο

249

0 .034'7 6

0.03476

0.03476

509.3400.0

s09.3l-00.0

5Ο9.3100.0

509.3100. Ο

12-38. The three branches from the plenum

possibιe for the Same preSSUre ]oss.

to be more extensive than Α or c.

For B: ΔPo/Le

For Α: ΔPo/Le

must be designed as close as

Start \Ι/ith B since it appears

=[

=(

0.18 - 0.025

104

'100 = 0.107 in. wg./100 ft

'100 = 0.109 in. wg./10ο ft

100 = 0.149 in. wg./100 ft

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only tostudents enrolled in courses for whiοh the textbook has been adopted. Αny other reproductiοn or trαnsιαtion of ιhis wοrk beyond ιhαιpermiιted by SecιioιιS ] 07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighι Αct νithouι the peιmission of ιhe cop1',ι"ighι oνner is unlcrνful'

025

145

0.18-0.

025

142

0.18 - 0.For C: ιP"/L" = (

B

Α

c

25012-38. (continued)

Note that the resulting total pressures losses turn out to be:

(ΔPo)a = 0.144 in.wg., (ΔPo)n = 0.157 in. wg.; (ΔPo)c = 0.161 in. \Μg.

Within the accuracy of the calculation these are appΓoximately equal'

It may be necessary to use a damper in branch B, sec. 8.

12-38.

Branch A, Balanced Capacity Method__ Γ)Αq i α- pηocedure --ννJf Y11 ! l


Rounding Method: Round Nearesι

__ Εan Sefection -_

Known Εan Parameter: Plenum Tota1 Ρressure : 0.1B0 in. wg

Fan Αirflow: 4Ο0.0 cfm

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students enrolled in οourses 1br whiοh the textbook has been adopted' Αny other reproduction or trαnsιαιion of this νork beyοnd ιhαt

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MA|N DUcτ RUN BRΑNCΗ DUCTS

(1)

Seο.

No.

(2)

Le

ft.

(3)

cfm

(4)

D"

in.

(5)

ΔP

L

actual

(6)

vfpm

*r"

1

(7)

ΔPo

(2)(5)

100

(8)

ΣΔPo

Σ(7)

(e)

Br.

Sec

Νo.

(1 0)

ΔPi

ΔPo6+

-(8)+

-ΔPα

(11)

Le

ft.

(12)

ΔPi

L

(1 0)1 00

(1 1)

(13)

cfm

(14)

De

in.

(1 5)

vfpm

8 44 500 12 .057 650 .025 025 14 0.094 55 0.171 125 b 660

o 22 375 10 .085 700 .019 .044 10 0.ο75 52 0.144 200 7 760

11 25 175 7 0.1 1 630 o28 .072 12 0.047 38 0.124 75 5 550

IJ 54 100 o .ο87 520 .047 0.119

Tot 145 ΔPο .025 0 144

1 50 40ο 10 .095 760 ο.048 0.048 7 0.084 38 0.221 100 ξ 750

2 19 300 .ο92 700 0.018 .ubb b 0.ο66 48 0.1 38 100 b 510

25 200 ο8 59ο 0.02 .086 4 0.046 q7 0.ο8'1 '100 o 500

48 100 o .095 530 0.046 . t5z

Tot. 142 ΔPο 0.025 157

15 56 225 7 17 850 0.095 .095 to 0.041 38 0.1 ο8 125 ο 610

17 48 100^

.085 510 0.04'1 '136

Tot. 104 ΔPο 0.025 161

Fan or External TotalCoil Lost

E'iΙter LostMisc. Lost

AΗU External Total Pressure:

ΑHU Pressure for Supply SystemΑΗU Pressure for Return System

Diffuser ΙD

PressureΡreSSurePressurePressure

0 .269 1n.0.000 in.0.000 in.0.Ο00 in.

0.269 in.

0.180 in.0.089 in.

Dia.(in )

0.Ο9.ο9.09.Οqn8.05.0on8.01.4trΛ

8.0'7.0ξn5.01.05.05.05.0

5.0ζn5.0

5.0trΛ

5.0

5.05.Ο5.Ο

a(cfm)

4ΟΟ. Ο

40Ο. Ο

400.040Ο. Ο

400.0300.0100.0400.03Ο0.0200.0100.0300.0200.0100.0100.0200. Ο

100.0100.01ΟΟ.010Ο.010Ο. Ο

100.0100.0100.0100.01ο0. Ο

100.0100.0100.01Ο0. Ο

10Ο.Ο100.0

Velocity( ftlmin)

Ο.0905.4905.4905.4905.4Qξo /

733.4onξ /

859.4t48.47?? Ia ξo 1

148.4133.41)) Λ

148.4133.41aa Λ

1)) Λ

360. Ο

7?? Δ

a 1a Λ

36Ο.0

??? Δ

aa1 Λ

360.0

111 Λ

1" 1 Λ

360.0

0.0000.0110.0030.0110.0140.0060.039

251

0 . 14157

0.14157

0.1_3629

0.1_9911

0 .1997'7

0.Ι99'71

0.L9911

0.3-991"7

wg1^rg

wg\^rg

1^rg

wgwg

or 6'7or 33

0%0%

Lost Ρre$-sure from Αir Handling Unit to Diffuser --1

15192329

Total 400.0

ΙD Εitting Type

1 Αir Handllng Unit2 Conlcal Contraction3 Straight Duct4 Elbow5 Stralght Duct6 Tee / wye main

branchcoΙnmon

7 Straight Duct8 Tee / wye main

branchcoΙnmon

9 Straiqht Duct10 Tee ,/ Wye main

branchcoτnmon

11 Straighι Duct72 E1bοw13 Straight Duct1'4 RecιanguΙar Transition15 Diffuser / Grille16 EΙbowt'7 Sιraight Duct18 RectanquΙar Transition79 Diffuser / cr1lιe2A Efbow2L Straiqht Duct22 RectanguΙar Transition23 Diffuser / cri]_ιe24 Sιraight Duct2-Ι ButterfΙy Damper28 RectanguJ-ar Transition

^ r^-^1 n^l -^ nν ΙOιd-L ιreΙLd r

(cfm) (in. wg)

100. Ο 0.171100.0 0.155100. Ο 0.141100.0 0.187

-- Calculated Fittlng Values --

Delta P ^P/L(in. wg) (in. wg)

0.018 0.148780.0050.020

0.0160.0040. Ο]_3

0.0200 .0!20.0160.0100.0250.0Ο70.0160.0100 .4250.0070.0160. Ο10a.0250.0160.0580. Ο10

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25229 Diffuser / Grille

Branch B, Balanced Capacity Method/t"

-\.-- Desιgn Procedure _-

System tyρe: SupplyDuct Sizing Method: Balanced Capacity


-- Εan Sefection __

Κnown Ean Parameter: Plenum Total PressurΘ : 0.180 in. wg

Fan Αirflow: 500.0 cfmFan or Εxternal Total Pressure: Ο.18Ο in. wg

Coil Lost Ρressure: 0 . 000 in. wgΓi1ter Lost Pressure: 0.000 in. wgMisc. Lost Pressure: 0.Ο00 in. Wg

AHU External- Total Pressure: 0.180 in. wg

AΗU Pressure for SuppJ_y System: Ο.18Ο in. wg - or 1OO.O %

AHU Pressure f or Return System: 0 . 000 in. \^Ιg - or . ο %

-- Lost Ρressure from Αir Ηandling Unit to Diffuser --Diffuser ΙD Q TotaΙ De]ta P

(crm1 (1n. wg)

13 10Ο.0 0.185L6 75.0 0.17919 200.0 0.\4223 125.a Ο.15B

Total 5Ο0.0

-- Calculated Εittinα VaΙues --

ΙD Εitting Type Dia. Q Velocity Delta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 A1r Ηandling Unrt 0. Ο 500.0 0.0 Ο. OOO2 ConicaΙ Contraction 10.0 5Ο0.0 916.1 Ο. O113 Straight Duct 10.0 500.0 976.1 0. Ο18 0.72'723Λ 'Γaa / τι;l',o main 9.0 375.0 848.8 O.Ο06

branch 6.O Ι25.O 636.6 O.o42coπιrrion 1Ο.0 500.0 916.7

5 Straight Duct 9.0 375.0 B4B.B 0.019 0.725"756 Tee / wye main 6.ο 2ΟΟ.0 1O18.6 Ο.OO9

branch 6.0 175.Ο 891.3 0.019coΙnmon 9.Ο 375.0 84B.B

7 Elbow 6.0 175.0 891.3 0.0088 Straight Duct 6.0 175.0 891.3 O.014 0.221759 Tee / wye main 4.0 75.Ο 859.4 0.OO6

branch 5.0 1ΟΟ.0 '733.4 0.020coΙnmon 6.0 175.Ο B91.3

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100.0 0.025

10 El-bow11 Straight Duct\2 RectanguJ-ar Transition13 Diffuser,/ Gril1eL4 Straight Duct15 Rectangular Transition16 Diffuser / GriΙ1e71 Straight Duct18 RectanguΙar Transition19 Diffuser,/ Griιle20 Straight Duct27 Butterfly Damper22 Rectangular Transition23 Diffuser / critle

1-41B

5.0ξn5.0

4.04.0

6.06.0

b.u6.06.0

100.0100.0100.010Ο.075.075.075.0

200.0200.0200 .0125 .41-25.01oξ Λ

L2s.A

'733.41?2' Δ

360.0

859.4210 .0

1018 . 6120 .0

636.6636.6450.0

0.0070.0280.0100 .0250 .0280 .0250 .0250.0410.0140.0250 .0L20.0440.005a .02s

2530.1_99'71

0.35266

0 .29022

0.72304

AP /L(1n. wg)

-- Design Ρrocedure --System type: Supply

Duct Sizing Method: Balanced CapacityRounding Method: Round Nearest

-- Εan Se1ecιion --

Knoιtn Ean Parameter: Plenum Total Pressure : 0.180 in. wg

Branchffilanced Capacity Method

Εan ΑirfΙοw:F.an or External Total Pressure:

ColΙ Lost Pressure:Filter Lost Pressure:M1sc. Lost Ρressure:

AΗU ΕXterna1 Tota1 Pressure:

ΑHU Ρressure for Supply System:AHU Pressure fοr Return System:

225.0 cfm0. 180 in. wgΟ . 0Ο0 in. wg0 . 000 in. \^Ιg

0.000 in. wg

0 . 18Ο in. wq

0.1800. Ο0Ο

or 10Ο.0or .C)

ln. \.^/g

in. wgzz

-- Lost Pressure from Αir Ηandlinq Unit to DiffuserDiffuser ΙD a Totat DeΙta

(cfm) (in. wg)

10Ο.0 Ο.191725.0 0.186

Total aatr Λ

-- Ca1cu1ated Fittinα Values --

ΙD Fitting Type Dia.(in )

a(cfm)

Velocity( ftlmin)

Delta P(in. wq)

1 Air Handling Unit 0.0 225.A O. O O. OOO2 Conical Contraction 8 . Ο 225 .0 644 .6 O . OO5Excerpts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instructional puφoses only tostudents enrolled in courses Γor which the textboοk has been adopted. Αny οther reproducιion οr ιrαnslαιion of this νori beyond-ιhαιpermiιιed by Secιions ] 07 or Ι 08 of ιhe 1 976 ιJnited Stαtes Copyrighι Αct ιιiιhout ιhe permissiοn of the copyrighι oινner is untαwfuΙ'

2543 Stralght Duct

0.088004 Butterfly Damper5 Straight Duct6 Elbow7 Straight Duct8 Elbow9 Straight Duct1Ο Tee / Wye main

branchcoΙτ'ιmon

11 Straight Duct72 Efbow13 RectanguΙar Transitj-on74 Diffuser / Grille15 Sιraight Duct71 Rectangular Transition18 Diffuser / GriΙle

8.0

8.08.08.08.08.08.06.06.08.Ο6.06.Ο6.0

6.06.0

225 .0

22\ Λna tr n

225 .0225 .0C1ζ

^1_25 .01Ο0.0225.01ΟΟ. Ο

100.010Ο.0100.014tr

^

L25 .0L25 .0

644 .6

644 .6644 .6644 .6644 .6644 .6644 .6636.6509.3644 .6509.3509.3360.0

636.6450.0

0.003

0.!120.003 0.08800Ο.0060.004 0.088000.006Ο.005 0.088000.0030.011

o. oo7 o. 0822L0.0030.0030.02s0.010 0.123040.00s0 .025

12-39. Solution follows Example 12-14 closely.

12-40 Solution follows Exampte 12-14 closely

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-

25512-41.

12-42.

1

Po=o

1

2

2

S

Supply fan: ΔPo = 4 in. wg.

Return fan: ΔPo = 1.75 in. wg.

1

Pυ=0

-1

-2

4

2

SF

S Space Pressure

A

le

S

Fan, ΔPo = 5.75 in. wg.

Exοorpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional puφoses only tostudents enτolled in οourses for which the textbook has been adopted. Αny other reproduction or trαnsιαti()n οf this work beyond ιhαιpermitιed b) Sections ] 07 or Ι08 of the 1 976 Uniιed Stαtes Copyright Αct lνiιhout ιhe permission ofthe copyrighι owner is unΙawfuΙ.

SF

F

M c E

A

8,,

25712-43.

1

Pυ=0

-1

-2

Supply fan: ΔPo = 4 in. wg.

Return fan: ΔPo = 1.75 in. wg.

12-44.

1

Pυ=0

-1

-2

Excerpts from this work may be reproduced by instructors for distribution on a not-for-prοfit basis for testing or instructional purposes onΙy to

students enrolled in οοurses for whiοh thΘ tΘxtbook has been adopted. Αny other reprοduction or ιrαnslαιion οf ιhis νοrk beyond thαιpeιmiιted by Secιions ] 07 οr ] 08 of ιhe Ι 976 Uniιed Stαtes Copyright Αcι νithout ιhe permission of the cοpyrighι oινner is unΙαw/ul'

4

2

6

4

R F

M c Ε

S

SI

258Fan, ΔPo = 5'75 in. wg.

12-45.

(a) Αssume a reasonable duct velocity of about '1200 fpm. ΔPo/L = 0.095

in. wg./100 ft. and D" = 18 in. (may be converted to 20x14 in. for

example)

For the duct: ΔPο - (0.095 x 40)1100 = 0.038 in. wg.

For elbows: Co = O.15; ΔP" = 2x o.15(118O/4OO5)2 = 0'026 in. wg.

For damper: Co = O.52, ΔPα = 0'52(118o/4OO5)2 = 0.045 in. wg.

For grille: ΔP, = 0'25 in. wg

For expansion: Vo = V.'(A.'/Αo) = 118Ol2 =59O fpm

ΔPu - 1'2(59ol4oo5)2 = 0.026 in. wg.

overall: ΔPo = 0.038 + 0.026 + 0.045 + 0'25 + 0.026 =

ΔPo - ο.385 in. wg.

(b) For 18 in. duct with 1,ο00 cfm, ΔP/L = 0'027 in. wg./10ο ft

For duct: ΔP6 = 0'027 x 401100 = 0.01 1 in. wg.

For elbows: ΔP" = 2x0'15(59ol4og5)2 = 0.006 in. wg.

For griΙle: ΔP, = O'25(1Qoo/2oοο)2 = 0.063

For expansion: Vo = 59012 = 295 fpm

ΔP" = 1'2(295l4oO5)2 = O.OO7 in. wg.

Εxοerpts frοm this work may be reproduced by instruοtors Γor distribution on a nοt-for-profit basis for testing or instruοtional puφoses only tostudents enrolled in courses for which the textbook has been adopted- Αny oιher reproduction or trαnsιαιion of this νork beyond ιhαιpermitted by Sections Ι07 οr 108 οfthe ]976 Uniled Sιαtes Copyrighι Αcιwiιhοuι ιhe permissiοn οfthe copyrighι oνner is unlωνful.

259For damper: ΔP6 = 0.385 _ (0.01 '1 + 0.007 + ο.063 + 0.007) =

ΔP6 = 0'297 in' wg. = Co"(59O lAooqz

(c) Co" = 0.29710.022 = 13.7

12-46. Equal Friction Method

Note that a damper has been inserted in duct 6 (Nο. 34 betow) tocause an increase in duct diameter from 8 to 9 1n. with a consequentdecrease in velocity to an acceptable 1evel.

-- Deslgn Procedure --System type: Supply

Duοt Sizing Method: trqua1 Εr1ctionRounding Method: Round Nearest

-- Fan Selection --Known Fan Parameter: E'an Tota1 Pressure : ο.900 in. wg

Fan Αirflow: 845.0 cfmΕan or Ext.ernaΙ Total Pressure: 0.9ΟΟ in. wg

Coil Lost Pressure: Ο.500 in. wqΕilter LoSt Pressure: 0.100 in. wgMisc. Lοst Pressure: 0.050 in. wα

AΗU External Total Pressure: 0.250 in. wg

AΗU Pressure for Supply System: 0.150 in. wg - or 60.0 %

AΗU Pressure for Return System: Ο.10Ο in. wg - or 40.0 %

-- Lost Ρressure from Αir Ηand1ing Unit to Diffuser --Diffuser ID Q TotaΙ Delta Ρ

(cfm) (in. wq)

19 150.0 Ο.14124 L25.0 0.12828 120.0 a.12332 200.0 0.11s38 25Ο.0 0.143

Total 845.0

-- Ca1cu1ated Fittinα Values --

TD Εitting Type Dia. Q Velocity Delta P ^P/L(in) (cfm) (ftlmin) (in. \^/g) (in. wg)

EΧοerpts from this work may be reproduced by instruοtors for distributiοn on a not_fοr_profit basis for testing or instruοtional puφoses only tostudents enrolled in courses fοr which the textbοok has been adopted. Αny oιher reproducιiοn or ιrαnslαιion of this work beyond ιhαιpermitted by Secliοns ] 07 or 1 08 of the ] 976 Uniιed Sιαιes Copyrighι Αct wiιhοuι the permissiοn οf ιhe cοpyrighι ονner is unlωυful.

260

1 Αir ΗandΙing Unit2 Conical Contraction3 Straight Duct4 Elbow5 Straight Duct6 Elbow

7 Straight Duct8 Tee / vlye main

branchcoΙτιmon


branchco]πnon

11 Straight Duct:L2 Tee ,/ Wye main

branchcoΙτιmon

13 Stralght Duct1-4 Tee / Wye main

branchcoΙττnon

15 Straight Duct16 ΕΙbow11 Strai ght Duct18 Rectangular TransitionL9 Diffuser,/ Grltle20 Sιraight Duct21- EΙbow22 Straight Ducι23 Rectanqular Transition24 Diffuser / Gri1Ιe25 EΙbow26 Straight Ducc21 Rectangular Transition28 Diffuser,/ crill_e29 ΕlbowJU SιraΙqhι ι]uct31 Rectaigular Transition32 Diffuser,/ Grille3 3 E]_bοw34 Stralght Ducι35 Butterfly Damper36 Straight Duct3'/ RectanguΙar Transition38 Diffuser / Gri1le

0.0L4 .014.014.014.014.0

14.0ι2 .0on

74 .012.A10. Ο

8.012 .010. Ο

9.07.0

10.09.Ο1.01.09.01.01.01.41.0

1.41.07.01.0

1.07.07.0

845.0845.0845.0845.0845.0845.0

845.0595.0250 .0845.0ξoζ n

395.0200.0595.0395.0215 .012A.A395.Ο215.0150.0L25 .0

15Ο.015Ο.0150.0150.0150. Ο

125 .0L25 .01atr ΛrZJ . υ12η nT rξ n

L20 .072A.A720 .0120.0200.020Ο. Ο

240.A200.0250.0250.0250.0

250 .0250. Ο

0.0190 .4190 .4190.4190 .4190 .4

190.415'7 . 656η q

190.4151 .61'Δ

'573.0-151 .6'1

'Δ 'G22 \449 .0124 .2622 .5561.346'7 .'7

561.3561.3561.3450.0

46'7 .1461 .1Δ61 1

375.0

449.A449 .036Ο.0

573.0573.0600.0

565.9565.9ξ6η q

565.9750.0

0.0000.0080.0030.0Ο60.005Ο.006

0.0060. Ο050.018

0. Ο070.0040.017

0.0Ο70.0030.016

0.0Ο70. ΟΟ30.011

0.0060.0050.0160.0040.0400.013Ο.0Ο20.0020.002Ο.0300.0020.0110 .002Ο.0360.0030.0100.0000.04Ο0. ΟΟ3Ο. Ο130 .0240.0010.0010. Ο5Ο

0.06415

0. Ο6415

Ο. Ο6415

0 .01 L61

0.08259

0.07138

0.08082

0.0BΟ82

0.05817

0.05817

0.05405

0.07106

0.06Ο04

0.06004

8.08.0B.Ο

9.09.09.0oΛ9.Ο

Balanced Capacity Method

Note that dampers have been inserted in ducts 6 and 7 (No. 31 and36 below) to cause an increase in duct diameter and a consequentdecrease in velocity.

-- Design Procedure --System type: Supply

Duct Sizing Method: Balanced CapacityRounding Method: Round Nearest

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261

_- Εan Selection --Known Γan Parameter: Fan Tota1 Pressure : 0.9OO in. wg

Εan Airflow: 845.0 cfmFan or Εxternal Total Pressure: 0.900 in. wg

Coil Lost Pressure: Ο . 500 in. wgΕi1ter Lost Pressure: 0.100 in. wgMisc. Lost Pressure: 0.Ο50 in. wα

ΑΗU Externa1 Total Ρressure: 0.250 in. wg

AΗU Pressure for Supply System: 0.150 in. wg _ or 60.O ?AΗU Pressure for Return System: Ο.100 in. wg - or 4O.O %

_- Lost Pressure from Αir Ηandling Unit to Diffuser --Diffuser ΙD Q Total De]ta P

(cfm) (in. wg)

19 150.0 0.14124 725.0 0.72828 720.0 0.14033 200. Ο 0.14038 2s0.0 0.131

Total 845.0

-- Calculated Fittlng Values --

ΙD Εitting Type Dia. Q VeΙocity De1ta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)

1 Α1r ΗandΙing Unit 0. Ο B45. O O. O O. OOO2 Conical Contraction 14. Ο B45. O 19O.4 Ο. OΟ83 Straight Duct 14.0 845. O '79a.4 O. OO3 O. Ο64154 Elbow 14 . 0 845. 0 '7 90 .4 O . OO65 straight Duct 14.0 845.0 i90.4 o.oo5 0.064156 Ειbow 14.0 B45.O 190.4 O.OO67 Straight Duct 14.ο 845.0 19O.4 O.OO6 O.O64158 Tee / wye main 72.A 595. O j5j .6 O. OO5

branch 10.0 250.0 458.4 Ο.020coΙnmon 14.0 845.Ο 190.4

9 Stra1ght Duct 72.0 595.Ο 151 .6 O.ΟΟ7 0.0'716'710 Tee / Wye main 10.0 395. O j24.2 O. OO4branch 8.0 20Ο.0 573.0 O. O17coΙτunon ι2.0 595.0 151 .6

11 Straight Duct 10.0 395. Ο 124.2 O. OO7 O. O825912 'Tοο / τιι',o main 9.0 21 5.O 622.5 O.OO3branch 6. O L2a.O 6Ιl.2 O. O13coπ]Ιnon 10.0 395.0 '724.2

13 Stralqht Duct 9. Ο 2']5.O 622.5 O. OO7 O. O713B\4 τee / Wye main 1 .0 15O. Ο 561.3 O. OO3branch '7 .0 L25.0 46i .i 0.011co]nmοn 9.Ο 2'75.0 622.5

15 Straight Duct 1 .a 150.0 561.3 O. OO6 Ο. O8Οs2

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2621-6 Elbow j .0

1-1 Stralght Duct i .01B Rectangular Transj-tion i.019 Diffuser / Grilte20 Straiqht Duct j.A2ι Ε1bow 1.O22 Straight Duct 1.023 Rectangular Transition i.024 Diffuser / eriΙle25 El-bow 6. 026 Straight Duct 6. 021 Rectangular Transition 6.028 Diffuser / Grill_e29 Elbow 8.030 Butterfly Damper 8.031 Straight Duct 8.032 Rectangu1ar Transitiοn 8. Ο

33 Diffuser / Gri11e34 E1bow 1Ο.035 Butterfly Damper 10. Ο

36 Straight Duct 10.03-l Reclangular Transit1on 1Ο.038 Diffuser,/ critle

15ο.0150.0150.0150.012η n

125.01t\ cl

725 .0125.0120 .0120 .0120.0120 .0200 .0200. Ο

200 .0200.0200.0250.0250.0250.A250. Ο

250 .0

561.3561.3450.0

461 .'7461 .'7461 ."7

375. Ο

677 .2G1 1 2

36Ο.0

573.0573.0ξ?? n

600.0

458 .4458 .4458.4750. Ο

0.0ο50.016 0.080820.0040.0400.013 0.058170.0020.002 Ο.058170.002Ο.0300.0040.023 0.1\42-70.0070.0360.0030 .0240.010 0.071060.00Ο0.0400 .0020.0L60.008 0.035990.0010.050

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CHAPTER 13

13-1. From Ξq' 13-2

h-= tr* = lb* = ft'Α (C* _ c- ) r.,,. _ t2 (lο* , n, ) ft2 _hr

Now C and W are related by Eq 13-14

C = Wρ" ψ* |% = lb*/ft3

lba fto

Τhe density of dry air must be used. Then from Εq. ,1 3-17

hο = hrP" = -ft3 X9 = ]b,' ftz _ hr-- ft3 ft2 _hr

Consider Eq. 13-13 which is dimensionless

h Btu ft3 lb,F ft2 - hr- η-zι- ^-Ξ-^---;- - |

P3Cp,hη ftz _ hr _ F |b, Btu ft3

Clearly dimensionless when C* is used.

13-2. using Eq. 13-'18,

h = Le2t3 = 1; hα = -η- = ,,'9. = 41'7lba/(hr - ftr)cpahα Cp, 0.24

also h, = hα/ρ" = 41 .7lO'O75 = 555.6 ft3/1hr _ ft')

h6 : 0.057 kgal(m'- s)

13-3. hd = 0.61s p"oa7

k

2610,075x100x60x(1 t14_

= g52ΚΞo = oβ44

k = 0.0147 Btul(ft-hr-F) (Τable A-4a)

n = Ot.O,::!,

x 0.61 5(852)047 = 2.S9Btu/(hr-ft2-F)(1t12) \ '

h6 = h/cpa =2.5910.24= iO.B lba/(ft2-h11

h, = h6/ρ" = 1O.8/0 .O75 = 144 ff lσt'-hr)

13-4. Nu = 0.023 Re0.8 pro 3 0r h = 0.023 (k/D) Reo.s pro 3

and h6 = h/cpa, assuming Le = '1

Re = ρ V olμ; V = 600/ (il$ = 471 ft/min or 28,260 ftthr

νL= 0.044lbmlft-hr; ρ = 0.075 (TabΙe A-4a)

Re = 0'075Ι?8'260x1

= 48,1700.044

Pr = 0.7; k = 0.01 47 btulhr-ft-F (TabΙe A-4a)

h = O.O23 ΨΨ (48,17o)os(O.7)o., = 1.7 Btu/(hr-ft2-F)1

hα = 1'710'24 = 7'1 lba/(ft2-hr)

h* = h6/ρ^=7'1l0'075 = 95 ft3/(ft'-hr)

13-5. 43,560 ft2 = 1 acre; Γi'l * = hdΑ(Wr, - W-)

Use j factor anaΙogy, h/crh6 = Le2l3

Αssume: Le = 0.85; Cρ + 0'24E

Thenho- ' =23.221bal(ft2-hr)o.24(o.8q2t3

Using chart 1: W- = 0.013 Ιbu/lba

W* - 0.0223lb"ilba (assume sat. air at 80 F)

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__.Ξ_262

Γh, = (23'22)1ooo(43,56OXo.o223 - o.o1 3)fr,,, = 9,415,ooO lbr/hr Ξ 19 gpm/acre

13-6. Use analogy of Eq. 13_1g

Ql= hαΑ(W- - Wr)irgi Q. = hΑ(t- - t*)

q = qr + Q., W, = 0.00765; W- = O.O'1 10; Chart 1

ho= h 144Ε\- 9(1.15)=,- =49'23lba/(ft2-hηcoLe2l 3 \ l' | ν'il

024(σ82τπ =

Q/Α = 49.23(0.o 11-0.00765)1o65 = 176 Btu/(hr _ ft')Q./A = 9(1 .1 5)(75-50) = 25g Btu/(hr-ft2)

q/Α = 435 Btui(hr _ ft2) = 1'37 kWlm2

13-7. Qr= rh,(i*-i,,)

Γh, = hdΑ(WV, - W-)

h6 = h/(cr"Lezu); cp, = 0.24 Btul(lba _ F); Le2t3 = 1

hα = '1 '510'24 = 6'25lbal(hr - ft')W, = 0'0223lb,/lba W* = o.0096 lb,/Ιba

i- = 28.4 Btu/lba; Chart 1

Γh, = 6.25(300 x 15O)(O. 0223 - 0.0096)Γh, = 3,572lbWhr

Q1 = 3,572(1,05o) = 3,75O,600 Btu/hr oΓ x 1,o99 kWΑny water on the deck and occupants neglected.

13-8' lt is assumed that the blanket is folded in half over the clothes line withone side exposed to air.

ho = h Le-2t3 = #(0.g3)-2l3 = 18.g7tbai(hr -ft )cp

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Γh* =

Γhr, =

hdΑ(Wb - W,); Wn = 0.0312;W^= 0'0152

} 'ΔΘ=m*/[hdΑ(Wv_W")]

263

12. A. = rh r/G,

4000 x 0.071

6O/1OOO = 17 ft2

or instruοtional puφoses only to|αιion οf ιhis wοrk beyond thαtghι οwner is unlαwful.

Δθ=(16 - 4)

= 0.71 hrI 8. 87(56)(0. 03 1 2 - 0.01 52)

Δθ = 42'6 min Say 45 min.

13-9. The procedure is the same as example 13-1 except that the

energy balance line A-B will have a positive slope

and tl ι=75"F, tlz = 90"F

Ans: 68162"F; 17 .4 ft2:4.8 ft

13-'10. The solutions to this problem closely follows example 13-2.

Ans: 77169"F; 17.4 ft2; 4.8 ft

Αns. 31126 C; '1.6 m2; 3 m

13-11. The procedure is the Same aS example 13-2 exοept that the energy

balance line A-B will have a negative slope and the inlet and outlet

water temps. are reversed.

Αns: 71t69"F; 17 '4 ft2;5.5 ft

13- 50

45

40

35

30

25

20

Xhαam =5510.24=229.2

h1/h6 = -3.05 =#,

t

Ε_ο

J

οο|ι

o-τΙ_ιlJ

'lI

-J-I

t2tι

YΕxοerpts fromstudents enrοl]permiιted by Sι

9060 70 80 100

264Υ : 2'7, Then

L = Gry/hdθm = 10ο0 x2'71229'2 = 11 .8 ft

13-13. The solution to this problem closely follows example 13-3

13-14. Ans: 1.4 to 1.5

13-15. Solution of this problem follows example '13-4 closely.

'13-'16. Αns: 5OO ft2; 12'2 ft

13-17. Extrapolate the 72 F wb curve in Fig. 13-9. The largest cooling

tower modeI ''M'', iS not Ιarge enough to handle 2ο00 gpm.

Therefore use two towers of 1000 gpm each. Select the

model "L" which ls rated at about 1100 gpm.

13-'18 See example 13-3; the cooΙing tower must be larger.

13-19. See example '13-3; the cooling tower must be larger.

13-20. (a) Model B or C using Fig. '13-9

(b) Cooling Range = t 1 -t. 2= 100-85 = 15 FExοerpts Γrom this work may be reproduοed by instructors for distribution on a not-for-profit basis fοr testing or instruοtional purposes only tostudents enrolled in courses for which the textbook has been adopted' Αny οther reprοduction or trαnsιαtion of ιhis νοrk beyοnd thαιpeιmitted by Secιiοns ]07 or l 08 of ιhe ] 976 United Sιαιes Cοpyright Αcι

'yiιhouι the permission ofιhe copyrighι owner is unlαινful'

265Αpproach = t. z-twol = 85 _ 76 = 9 F

Tower capacity = Q

q = (200 x 60 x 8.33)(1)(15) = 1,499,400 Btu/hr

13-21 q - 5OO qpm x Δt; qpm = '=??o,o^o = 50500x1 0

qpm/ton= 50 -^^. =3.0

(250,000 / 1 5,000)

Note: !n this case, '1 ton = 15,000 Btu/hr

Cold water temperature; t" = 70 - 10 = 60 F

From Fig. 13-7; t*o = 42 F

13-22. Αlbuquerque, NM; tψ6 = 64 F (Τable B-1a)

(a) From Fig. 13-7; cold water temperature = 73 F;

gpm/ton = 2'5i \Λ/arm water temperature Ξ (73 + 10) 83 F

(b) Charleston, SC; tψ6 = 79 F (Table B-1a)

From Fig. 13-7; cold water temperature = 84 F,

gpm = ΨΨ x2'5= 83 gpm (a & b)15,000

13-24. Model G, nominal rating - 600 gpm & 250 tons (Table 13-2).

Using Figure 13-9; assume gpm is constant.

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13-23' (a) tons = ''?9?99o = 80, gpm/ton = 24OΙ8O= 3.O, maximum two = 72F15,000

(b) gpm/ton = Ψ = 4'o',max. t,5 = 65 F80

266With cooting range of (97 - 85) = 12

Max. two = 76 F (Figure 13-9)

With cooling range = 15, tu, = 100 F

Max. two = 74 F (Figure 13-9)

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-------_-=\ ---

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Requests for permission or furtλer ffirmαtion should be αddressed to the Permission Depαrtment,

Join Wiley & Sons, Ιnc, ] Ι ] Riνer Street' Hoboken' ΝJ 07030'

CHAPTER 14

14-1. (a) P - 120-60= 0.43

200 - 60

* = 200-180

= 0.33120 - 60

F = 0.985 {Fig. 14-191

LMTD _ (1so - 6_q)_:(299_ 120)(180 - 60)

ln

LMTD = 98.7oF

(200 -120)

Btu/hr-F275(7.48)16'500

=2η5ft3/hr; Q - = 34 gpm

(b) C" = (fr "r)",,

= 50oo *uo (29'92x0'491x144) φ'24)

53.35(520)

= 5490 Btu/hr-Fcn = cc(1,2 - tu1)/(t*z - t*r) = 5490(1 20 -60y(200 - 180)

= '16,500 Btu/hr-F

(rh cp)* = (Q pcp)* = 16,500(c) Cn =

Q=(6OlΧ1) 60

(d) q = UAF(LMTD)

UA = Cι-., (t*z--t*l) _

F(LMTD)

16,500(2ο0 _ 180)

0.985(e8.7)

UA

= 3390 Btu/hr-F

UA UΑ(e) NTU =

NTU =

Cc Cair

= 0.62

Cmin

33905490

^-267

(f) "=

120-60 0.43\ / 200-60

26814-2' (a) q = UΑF(LMTD) = (rh cp),i,(1 10-50)

Γh, = 4000 x 14'7 x 144l(53.35 x 510) = 311'2 lbΙminor 18,672|b/hrQ = 18,672(0.24)(110-50) = 268,874 Btu/hrq = (rh cr,,)('l80 - tr.,o) = Q5 x 8.33)(1)(180 - tho)60

tι.o = 1 80 - =?9y+ = 158.5 F or 159 F25x8.33x60

p= 110-50 =0.46: *='180-159 =0.35

180 -50 1 10 - 50F = 0.98; Fig. 14-1

LMrD=S# =88

't[ ^JA = 9/(UF x LMΤD) ='!u!!!^o== = 312 ft2' 10x0.98x88

(b) Cair = 18,672(0'24) = 4481 = Cmin

Cwat = 25 x (60.817 .48) x 60 = 12,193 = Cr*, =

110 - 5o = 0.401'

Cmin - 4481 = o.3T

180 - 50 crr, 12,193

NTU = 0.7, Fig. 1 4-18

UA/C,1. = O'7i μ= 0'7\!481 = 314 ft2, ,10

14-3. (a) Γh air = 32οo x 60 x ::'::'!!- = 13,726lb/hr53.35x555

Cair=Cmin=3294

NTU - 1ox3oo = 0.91 125

3294ε = 0.615, Fig. 14-18 T

at C.inlCrr, = 0

l refriο. l

atr.

+

125

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students enrolled in courses for whiοh the textbook has been adopted. Αny oιher reprοduction or trαnslαtiοn of this νork beyond ιhαιpermitted bν Secιιons ] 07 or Ι 08 οf ιhe Ι 97 6 LΙniιed Sιαιes Cοp1τ'ighι Αct Ιψiιhouι the permissiοn of ιhe copyrighι oνner is unlcτwfuΙ.

r_.-

(b) ,., = taηh(mrl)

, ο = [E -,][.'+ O.35rnBj = 1.243\mrφ) \r )\ r/mrΦ - (18.26)(0 '5112)1.243 = 0.9454

,_ tanh(0.9454) =0.7g' 0.9454

(c) Within readability of Fig. 14-4 the answers are the same


0.61s = ,':"=-'^? , tco = t"o = 0.6'1 s(125- gs) + g5125 - 95

t,o = 113'5'F

(b) Q = Cui,(tao - tai) - rh,.irg = 3294(11 3.5 - 95). 3294(1 13.5 - 95)

= 928 tb/hrlrl,. = ------------.6s.s

14-4' (a) ,= ΓηlL κy -] L 90(0.00 8 l 12) J

lm = (1.0 - 0.5)

18.26= 0.7612

R/r = 1l0.5 = 2.0i η = ο.8, Fig' 14-4

14-5. ξs = 1 + (1 - η); η = O.78 from proplem 14-4A

ξs = 1-0.9(1-0.78) = 0.80

1 1 Δx 1 1 (0.01 5112)'1 Λ^

Uo ho?ro k(Α, /Αo ) hi(Αi /Αo ) 10x0.8 (100x1)

1J- = 0.17; Assumes Α; = Αo and k.opp",. = 100

200x(1/9)

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270The second term may be neglected

Uo = 5.9 Btu/(hr-ft2-F)

14-7. '1 =! =0.133Uo 10x0.8 1 100(1/9)

Uo = 7.5 Btu/(hr-ft2-F)

14-8. , - tanh(m/)' m!.

, = Γeη-l''' =| z"sτ _1''' =64.18 m-1L κy ] [ l za1o.'t 6x1 o_3 l]

v

mI = 64.18(6 x 1O-') = 0.385; η = 0.953

14-g' ns=1 *(1 -η)_1-o.s5(1 _O.95)

η. = 0.96

Ar = 2ΗLWP, mm2; Α, = LW mm2; A = Ar + (LW _ tLWPS)

Where P" = fin pitch in fins/in. and L = W = 1

Δ = 2HLWP' + LW _ tP, _2x6x0'47 +1_0'16x'47 - '1 '1AΛ'

Αo - ,ΗL\Λ/P' 2x6x'47 l' lo+

+=\:γ--J^ = O.O19; U = 52.3 W(m' _ c)u 1400 57(0.96)

Excerpts from this rνork may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only tostudents enrolled in οourses for whiοh the textbook has been adopted. Αny οther reprοducιion or trαnsιαιiοn οf ιhis work beyond thαιpermitted by Secιions ] 07 or l 0B of ιhe ] 976 United Stαtes Cοpyrighι Αcι lνiιhout ιhe permission of the cοpyright owner is unlαwful.

14-11 .

tan h(m r/ )

mrΦ

R" = 1.2s ψ(β - 0'2)'''; m =

r

Γ znl1l2

Lπ]*=},β=;;L>M

(a) Diml =;'=Ψ = 0.56 in

Dim, = rΓrg)' *or1'''' 2Lι2l l

'Φ= [}_,][,+O35.(})]

e0(0.01/ 12)

η=

271

= 16.33 ft-1

= jιιo.uu)'* (1'35)'1'''= O.73

Then L = Dimz = 0.73in.; M = Dimr = 0.56 in.

ψ= 0'56 =1'75:3=0'73 =1.3

(0.64 t2) 0.56R"

= 1.27(1.Ts)(L3-0.3)1 t2=2.22r

φ = (2.22-- 1)t1 + O.35ln( 2.22)!= 1'56; . = ΓL

mrΦ - 16.33(0.32112)1.56 = 0.631

n_ tanh(0.762) =0.869' 0.762

(b) Dim.-a-12.5mm'2Dim2 = ! lZZt + 12.5'ltt2 = 12.65 mm' 2'L = Dimz = 12.65 mmM = Dimr = 12.5 mm

t=12'5 =2.s:g= 12'65

=1.0125 12.5R"

= 1.27(2.s)(1.01 2 - 0.g)1t2 = 2.69r

f'''2x10

Excerpts fiom this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses on1y tostudents enτolled in οourses for whiοh the textbook has been adopted' Αny οther reproducιιon or ιrαnsιαtion ofιhis lνοrk beyond thαιpermitted by Sections ] 07 οr ] 08 οf ιhe Ι 97 6 United Sιαιes Copyrighι Αcι y,ithοuι the permission of ιhe copyright owner is unlωνful.

b = 1.35 in

b=22mm

272

Φ = 2.69 - 1)t1 + 0.35 ln(2.69)] = 2.26Γ zxoε l _^ -_m=l E'\vv l=66.67m-11170(0.00018)_j

mrQ = 66.67(0.005)2.26 = 0.753_ tanh(O.753)

11 =' 0.753

η = 0'85

14-12. + = -1-, ; neslecting tube wall resistanceUo hoΠo h; (Α, 7no ) ι - v-_ _----{

(a) 1o=1 *(1 -η)-1_o.9(1 _O.s4)=0.86

1 1 :^ = 0.120; Uo = 8.60 Btu/(hr-ft2-F)%

= 1o10€6

* οoo(l o)

(b) ro = 1 -0.9(1 -0.81) = 0.831 1

^4^ = 17.8; Uo = 0.056 kW(m2 - c)% =

0,.068ro s3 *

1a.+rl o1

14-13. (a) Rct = 2.222 x 10-6ο.010

Rct = 4.15 x 10-a 1hr-ft2-r1/εtu

(b) '12 fins/in = 0.472 fin/mm

Rct=3.913x10-7

14-14' Re= ρνD;ρ=60.6 lbm/ft3μ

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0.18

]""'

1o

uott=

1 .oe3 x 1o-41m2 - cyw

l

0 64( 1t12 - 1)'

ι 0.010 )

rc(1t0 72_ 1)'ι 0.18 )

273Dr = 0.545112 = 0.0454 ft; Table C-2

V = e lA= 2'5 x+ =3.44fVsec' ι 7 '48^ 60(π t4)(o'o454)2

μ = o.93 lbm/(ft - hr) = 2'58 x 1O-a lbm/(ft-sec) Τable A-1a

Re = 60.6(3.44)0'0454

= 36,6g3i Re = 36,7002.58x10-a

(L/D),in = 410'0454 = 88 ftPr = 2'43 (Pr = crμ/k)ιlo'',u = O.ο23 κ"Bu Pro'; k = 0.383 Btu/(hr-ft-F)k

h = O.O23 ,Ψf''='J. (36,7oo)o'12.431o' = 1,136 Btu/(hr-ft2-F1(0.0454)

14-15' ρ = 1'o1(62.4) = 63.02 LBM/FT2 [Fig. 1 0-2a]

μ= 0.7l1490 = 4'7 x '1o-4 lbm/ft-sec [Fig. 1O-2b]

Cp = 0.93 Btu/lbm-F [Fig. 14-ε];

K = 0.93 Btu/lbm-F [Fig. 14-9]

Υ = 3.44 fVsec [ProbΙem M-1aJ;

D = 0.0454 ft [Problem 14-14]

63.02(3.4 4)0 '045ΔRe=ffi=20,940(L/D),1n = 88 ft [Problem 14-14]

o _Cpβ _4'7x1o41sοoo)o.93 _ΕΕ.)lr---ν''νia- k ο.285

h = O.O23 ,!o:',u=u], (2ο,94O)ou15'521o'= 690 Btu/(hr-ft2-F)(0.0454)

14-16' ρ = 1.O45 x 62.4 = 65'21 lbm/ft2 ;

μ = 1 '3l1490 = 8'725 x 1O-a lbm/(ft-sec)

co = 0.81 ;k= 0.22; V = 3.44 ft/seC, D = 0.ο454 ft

Re=#=11,670

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274

,," _ 0.81x8.725x1 0-43600 =

.1 ..16

0.22

h = 0.023 !:?3=). (i 1 ,670)0.(.1 ..t6)0. = 209 Btu/(hr-ft2-F10.0454 '

14-17. Use hydraulic dia. for rectangular channel

Dη = 4rn = a(AJP) = 4(3/8) = 1.5 in. = 0.125 tt

ρ = 62.4lbm/ft3 [Table A-1a];

μ = 3.45 lbm/(ft-hr) [Table A-1a]

Re = u':o!!_),Ψ

-'-?? = 32,556(3.45l3600)

cp = 1.003 Btu/lbm-F [Table a-1a]

k = 0.338 Btu/(ft-hr-F); Pr = 3.45 x 1 .003/0.338 = '10

(a) For cooΙin9, h = O.O23 ! R"o'Pro'D

h = O.O23 9Ψ (32,556)0r11o.21o. = 5O9 Btu/(hr-ft2-F)0.125

(b) For heating;

h = O.O23 9Ψ (32,556)0s11O.21o. = 642Btu/(hr-ft2-F)0.125

14-18. Dr, = 0.125 ft [From problem 14-171

ρ = 62.4(1.o45) = 65'2lbm/ft3 [Fig. 1 O-2a]

μ = 3.5/149O = 2.35 x 1o_3 lbm/ft-sec [Fig ' 10-2b]

Re=

c, = 0.89 btu/(lbm-F) [Fig. 14-8]

k = 0.28 Btu/(ft-hr-F) [Fig. 1a-g];

Pr = 2.35 x 1 o-3136oο)(o.s9) l0 '28 = 26.9

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275

(a) Cooling

h = 0.023 # ('13,900)0t126.910t = 28s Btu/(hr-ft2-F)

(b) Heating: h = 285 (26'9):1

= 396 Btu/(hr-ft2-F)'"" (26'910'e

14-1g' (a) Re = ρΥD _ 99ο'2(1 '5)(0'012) _ ^'''μ o'uffi =29'905

Assume L/D > 60 ,

Then Φ = o.o23 Reo.8Pro4, Pr = [o'sgοxlo3x+'lεzl = e.glk ι 6373 )

π = 9 9?: (O.637X31 ,157)o s(3.91)o 4

0.o12h = 8287 W(m2 - C) = 8.29 kw(m',- C)

Data from Figures 10-2a, 10-2b, 14-8 and 14-9.

(b) Re = (1 'ο28)999(1 '5)(0'012)

= 15,4OO1.2x10-3

1 .2x10-3x3.7x10-3,r=ffi=8.9:- o.o23h = Ξ'Ξ]Ξ (ο.5o)(15,4oO)08(8.9)oo = 5140 W(m2 _ c)

0.012= 5.1 4 k\Νl(m2 - c)

14-20. (a) Re = 62'4(0'5)(0'3!112)

= 922< 2soo(3.45l3600)

hD =

.1.86tReP. D .,lls [ra)o 'o : Αssume ( n)o'o

= ,τ - l.vvιl.-,, L, ιr',,l,,1oΦLΙl,,- ιλJ

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276

,,=ξ1# =1O'4

π = sff# p22(1o.ol(ffi),"' = 66 Btu/(hr-ft2-F)

(b) Same procedure as part a using data for 30o/o ethyleneglycol from Figures 10-2a, 10-2b,14-8 and 14-9.

14_21. Re - 992.2(0.1.0)x103 = 1519

653There is a question about the flow regime. lt is probably

in the transition region. Assume it is laminar and use

7 10.14

Eq. 1 4-24 and assu me | -U- I = 1.

[ρ'J

Pr - o.653X'1ο_3(4.182) = 4'34

0.63τ- 1 89φΞ3)

1519(4'34)rΨ']]1/3 - 328 W(m'- c)n=-O.O1 ι'--_\- 'ι 3 )'

14-22. Use average values for Gu and G. and Eq. 14-26.

G. =:o;(rh. )"us= 0.912;(Gr)",s = ln't,,u =237.81bm/(ft2-hr)' Α.'' "9'v z(o.sε9)'4\12)

(rh,),us - (1 + 0.1)12 = 0.55 lbm/hr

(Gu),us = 0.55/4" = 290'6 lbm/(ft2-hr)

DG, _ r0.589)r?ΞΞ) = 12 3lιt ι 12 ,ι 0.95 /

DGu _r0 589)rΞ9ψ) = 15

βι ι 12 ,ι 0.95 /

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permiιιed by Sections ] 07 or Ι 08 οf the ] 97 6 Uniιed Sιαtes Copyrighι Αct νilhout ιhe permission οf the copy'ighι oνner is unΙαινfuΙ.

H= = 138(pr.,,. [#j"'L? E)"102

277

pρ'( e")"' = .,u.o1, 61 o )"' = 1oo8

βι lρ") ι0.0135/

Pr _ 0.95(1.001) =2.48' 0.384

i,, = 1001 Btu/lbmΔt ρ 80 = 1'160 _ s0); twall Ξ 80'F (Using water outside the tubes)

6 = 13.8 0.384 - '4n' Γ 1φ]-l"u ,,,oo8]o

2

- - ο.58 g t12

QΑ8) '' L,1 ,(uη] "

Uυδl'

h = 888 Btu/(hr-ft2-F)

14-23. Use average values of G. and

A"= ζ (O.o15)'= 1'767 x 1O-a"4(G. )"us =

(Gu)"ug =

(ο.s8)o. 126x1o_3 Ιz1.767x10-a

0.126x10-3 (1 + O. 1 2) t 2

1.767x10-a

Gu and Eq. 14-26

mt

= 0.314 kg/(m2 - s)

= 0.399 kg/(m2 - s)

DGr _ ο 015(0.311) = 12'1

βι 0.390x10_"

DGu ( ρr)"' =

O.o15(0.39_9 [ gzο

l1

Ι2 = 1024

βι |ο" ) O.39Ox1o_3 L0'219J

Pr. - O'39Ox1 9jΙ1' 19x1 03 = 2'46; ig = 2326kJ/kg' 0.665

Δt = 45C = (73 _ 28); liquid water assumed outside tubes

h=13gφj95)(2'46)1t3lffi].,u..o24)o2=5022W(m2-c)

h = 5.02 kw(m2 - c)

14-24' Use Εq. 14-28

R_22 1 At inlet x : 0.20; at outlet 10oF superheat

Excerpts from this wοrk may ,l"rΓj],' ξ,ο,Jhm/fu p,ειJ*FsnRιε 1"Q,psi?i k,':. Ω if,'$-"t.nω puφoses only to

students enrolled in οourses'for which tbe textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this wοrk beyοnd ιhαιpermιtted by Sections ]07 or Ι08 ofιhe 1976 United Sιαιes CοpyrightΑcι\υithout ιhe permission ofιhe copyright ονner is unlαννful.

278

Since X" Σ 1.O; Cl = 8'2 x 1O-3; n = O.4

Αssume tube wall thickness of 0.016 in.

Τhen D; = 0.375 - 2(0.0161 = 0.343 in.

Ai= π-g|

= 6'417 x 1o-4 ft2'4rh 80

rJ = -:- =

-

= 124,700 lbm/(ft2 - hr)A j 6.417x10-a

μ. = 0'52lbm/(ft-hr) at 30'F (sat. temp. at 70 psia)GD

= 124,700(0.343 I 12)

= 6855βι 0.52

k = 0.056 Btu/(hr-ft-F) at 30'F (sat. temp.); Table Α-3ai1g = 88'5 Btu/lbm

h = 8.2x 1O-3 (0.o56)

Γrοεssl, (ττaφ'qaa'sβz'ιτ))lo o

(O.343, r)L'ooccΓ ι 5(32l 7) J.]

h = 779 btu/(hr-ft2-F)

14-25. Use Equation '1 4-28

R-22;G = 2OO kg/(m2 - s); Dr = 8.5 mm; L = 2 m; P,= 210 kPa

ξ = 30%, Xe = 100%

Tsat = -24C a|210 kPa abs. Pres. [Table A-3b]

μ. = O.27O x 1O-3 N-s/m2 [Table Α-3b]; extrapolate

k. = 0.107 W(m - c) [Table A-3b]

iβ= 223 kJ/kg [Table A-3b]

GD=200(0.0085_) =6296βι 0 '270x10_3

Cι = 8.2 x '1O-3; n = O.4

h = 4106 W(m2 _ C) = 4.11 kννlm',_ c

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students enrolled in οoursοs for which the textbook has been adopted. Αny other reprοducιion or ιrαnsιαιion οf ιhis work beyond thαιpermitlecl by Sectiοns ] 07 or Ι08 οf ιhe ] 976 tJnited Stαtes Cορyright Αct \νiιhout ιhe permissiοn οf the copyrighι oνner is unΙπwfuΙ.

6 = 8'2x 1O-3 (o'1O7)Γ

β296\20.0085 L'

. r0.

J]

0007

7(223)1

2x9.80Ie

279

14-26. lr, = f !!1 , Assume isothermat" D2g

Re = 36,700; ProbΙem 14-14; smooth tubes

f =0.022, Fig. 10-1;L=(6x6)+(5xl)=41 ft

Dr = 0.0454 ft; V = 3.44 fUsec, Problem 14-14

lr, =o.o2z* -!1- *lt!!l,t =3.65ft0.0454 2x32.17

14-27' l ι. = f !ζ, Αssume isothermal" D2g

V =O.5ft/sec; f =64/Re =641922=O.OOg

L = (10 x 10) + (9 x 1.5) = 113.5 ft; Di = 0.34 in.

lι. = 0.06 n*11?^'?x12, - Ψ', = 1'07 ft0.34 2x32.17

14-28. Refer to Fig. 14-10

Load/circuit = 10 x 1 2,000110 = 1 2,000 Btu/hr

Length/οircuit = (6 X 5) + (5 x 0.75) = 33.75 ft

(a) ΔP. /L = 0.10 psi/ft; CF = 1.25 Fig. 14-10

ΔP. = 0.10 x 33.75 x 1'25 = 4.22 psi

(b) ΔP. /L = 0.04 psi/ft

CF = 1.25

ΔP. = 0.04 x 33.75 x 1'25 = 1.7 psi

14-29. Gt,^ = 18OO lbm/(hr-ft'); t, = 7O"F,tz= 120"F

(a) Figure 14-12

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280

Re, = G'Xo

, Gc = G"

- ΨΨ = 3214 lbm/(ft2-hr)μ σ 0.56 ''_'-''\-

ψ = 4.6 x 10-2 lbm/(ft-hr) at t = 95oF (Table A-4a)

Re, 3214(1 .083 t12)= 6306

4.6x10-2

j = 0.0091;f = 0.021 Fig. 14-12

I = jG" c, Pr-2l3 = O.OO91 (3214)0.24(O.T)-2t3

= 8.91 Btu/(hr-ft2-F)

(b) Re6 = 6306 x 0.52511.083 = 3057; assumes expanded tubes

plus fin collars.

A 4xoxrα _ 4 1.25x1.083x0.56 = 1O.O8At πD6D π 0'0152x0 '525x12

JP = (3057)-04(1O.OS)-015 = 0.0285 (Eq. 14-39)

j= 9. 1x1ο-3 Fig. 14-14; h = O.OO91(3214)(0.24)(0.7γzl!

= 8.9 Btu/(hr-ft2-F)

D*=0.525x10 08(1.25 - 0.525) = 0.904

116.7

Using Eq. 14-44.

1+

( 0.525,0 " I(3o57)-0 2'[αoo+.1

t

1.25 - 0.525

]"

| ι 'zs _,,l_o u

_10.e04 I -FP=

4( L- o o06)\67 )

0.173

f = 4.2 x 1O-2 or f = 0.042 (Note that f may be in error up to

!35o/o (Figure 14-15)

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281

14-30. (a) G" = 4.5 kg/(m'-.); tp = 2OC; Re = Gc(0'0275)

'

μ

μ = 1 8'2 x 10-6 Ν-s/m2 1Table A-4b]

Re = 4'5(0'027Ξ)

= 6800; Cp = 1.OO5 kJ/kg-C18.2x10-o

j = 0.009; f = 0.020 [Fig. 14-121

h = jG" copr-2t3 = o.oog0(4.5)1.005(0.7)-2t3 - 0.051 kJ/(m2-s-c)

6 = 0.051 kw(m2-c)

(b) See problem 14-29b for prcedure

14-31. Use Eq. 14-42 or 14-45

l - el-Γr,r * o2{ eι_,l*'+4l,n - 2gρ-ρ1L\,*" )lo''_' )_, η λ-,

ρ, = #!S#L = 0.O75 lbm/ft3; ρ2 = O.068 lbm/ft3' 'i 53.35(530)

P,, = (ρl + ρ2)l2= 0'072lbmlft3

A _ αΥ _147(1.083112)5 = 118'45Aο σAf, (1)0.56

Where V = Αt. xL; A1," = 1 ft2;L = 5 x'1.083/1 2 = 0'451

l h = β214)2 _Γ(, * ιo.u6), )r9Ψ _ l)*'' 2(32'17)(0'072)(0.075)(3600)Ζ L\ /\0.068 )

o 042(1 1s 5)(0 075)lo 072 -l

h = 12.2ft of air

ΔPo = Ω, (1i::?)Α;'z = 0 18 in. wg

Εxοerpts from this work may be reproduced by instruοtors for distribution οn a not_for-profit basis for testing or instruοtional puφoses only tostudents enrolled in courses Γor whiοh the textboοk has been adopted. Αny other reproducιion or ιrαnsιαιion οf ιhis wοrk beyond thαιpermιtted by Sections ] 07 or Ι 08 of the ]976 United Stαtes Copyright Αcι νiιhout ιhe permissiοn οf the cοpyrighι owner is unΙανful.

28214-32. Use Equation 1 4-42 or 14_45

ti = 10C; t2 = 30C, ffΑ" = 1 18.5 from problem 14-31

1O1x103oι = ffi1 = 1.244 kg/m3; Pz= 1.161 kg/m3;

pm = 1.203 kg/m3

t' = ffi[(,'. (0 56)2 )(#- 1).

o.O19(1 18.5) 1μ!1\ / 1.203 )

I r, = 1.67 m of air

ΔPo = 1 6, (#lι',rool = 2'O8mm of wg.

or ΔPo = 1'67(1'244)9'807 = 20.4 Pa

14-g3. Re = G.Dr,

, Dι' = O.O101 ftμ

Αt 65.F, μ = 4.39 x 1O-2 Ιbm/(ft-hr) [Table A-4a]2700(0.0101)κe=lffi =621

From Fig. 14-16;j = 0.013, f = 0.053

h = G" co j Pr-2t3 = 2700(0.24)O.013(0.72)-2t3 = 1 O. S Btu/(h r-ft2-F)

Where co= 0'24 Btu/(lbm-F); Pr = O.72 From Τable A-4a

14-34' Use Eq. 14-33; tυ = 65 F

14.6x144o Ξ -ff = 0'074lbmlft3; ρ2= 0.077 lbmlft3' 53.35(535)

Pm = (ρl + ρ")l2 = 0.076 lbmift3

Εxοerpts from this work may be reproduced by instructors for distribution on a not-for_profit basis for testing or instructional puφoses only tostudents enrolΙed in οοurses fοr which the texδook has been adοpted. ,lny otlrr', iφr:oΙλ'-"ιiλn o, ιronrtotιon of ιhis νork beyoncl ιhαtpermiιιed by SeCιionS ] 07 or Ι 08 of ιhe Ι 97 6 Uniιed Sιαιes Copyrιghι Αcι νιιnout inte p,eλλsion of the cοpyrighι oτι,ner is untcιwful.

28s

A 4L -ψI2 = 132π=π= oοlοl

Assume a contraction ratio of 0'5

Then Κi -- 0 '32', Κ" = 0 '27 Fig ' 1 4-17

ΔPo _ e7oq2 _ t(o.32 * l- os2)

π _

2β2.17 )1 4'6(1 44)(0.07 4)(3600)'

+ 2 (#- 1) - o 053(1 3r)ιffi#] _ r _ (o 5)2 _'''1#'\

ΔPo/Pg1 =4'126x10-a

ΔPo = 4'126x 1o-4114.6)(1 44)(12)t62'4 = 0'17 in' wg'

14-35. (a) Coil DescriPtion'

Type of coil = Refrigerant condenser

Tube pattern = Staggered plate-fin-tube coil

Material = Αluminum fins with copper tubes

Refrigerant type = Refrigerant 134

Finned side fluid = air

Finned side air pressure = 29'92 inches of Hg

Face area -- 4'44 square feet

Height of heat exchanger = 20'0 inches

Width of heat exchanger = 32'0 inches

Numberofrowsoftubesintheairflowdirection=4

Numberoftubesperro\Λ/=16circuitsontubeside=4

Fin pitch = 8 fins/inch Fin thickness = '006 inches

Vertical tube spacing = 1'250 inches

ΕxcerDts from this work may be reproduced by instructors for distribution on a not-fbr-profit basis for testing or instructional puφoses only to

sιudenιs enτolled in courses ι", *-L1.λ ,ι-'. ιexibook ιr, υ..nrjopi.α*' λny- o,rnr, ,rρroλur'ioλir-ιron,Ιαιιoi of ιhis νork beyond ιhαι

permiιιed by Sections ] 07 * ,oi'iiiλ)jili'bnii i,o*, ciiilii, iri ''ιhout

ih" prr^ιr,rio' oj'n" copyrighι oνner is unlωυful'

284Horizontal tube spacing = 1'083 inches

Tube outside diameter = .500 inches

Tube wall thickness = .016 inches

lnside tube fouling factor = 'OOOO BTU-HR-SQFT-F

'/6C

14-95. (continued)

Total heat transfer rate = -48783'2 Btu/hr

Sensible heat transfer rate = -48783'2 Btu/hr

Entering air conditions:

Dry bulb temPerature = 95'0 F

Face velocitY = 650'00 FPM

Air volume flow rate = 2888'9 CFM

Leaving air conditions:


Tube side conditions:

Refrigerant saturation temperature = 125'0 F

Air pressure loss = '393 inches of water

Tube side Pressure loss = 1'07 PSI

Fin efficiency = '3gg Surface effectiveness = '824

Tubesideheattransfercoefficient=388.88tu/hr-SQFT-F

Finnedsideheattransfercoefficient=12'5Btu/hr-SQFT-F

Mean temperature difference = -20'7 F

(b) Yes

14-36. Coil DescriPtion'

Type of coil = water or brine solution

Tube pattern = staggered plate-fin-tube coil

Material = aluminum fins with copper tubes

Tube side fluid = water

:i:.J:fJ:f;':T:J$E::',i:;i1:i1i:'*iJ$ιι'{ij:{..Ti ι:;i:i'b:;,ff:;;i!:'ii:i1r,*"Ιir;?j':ii,:!;;iiii:i!jΙ:'i#ii!*''

ρermiιιed by Secιiοns Ι 07 "''iοi')i 'i i ii6 t'Jniιed Sιαιes Copyrighι Αcι w ilhouι ιhe permι

286Finned side fluid = air

Finned side air pressure = 29.92 iches of Hg

Face area = 5.56 square feet

Height of heat exchanger = 20.0 inches

Width of heat exchanger = 40.0 inches

Number or ro\Λ/S of tubes in the air flow direction = 2

Number of tubes per ro\Μ = 16 Circuits on tube side = 4

Fin pitch = 7 fins/inch Fin thickness Ξ .008 inches

Vertical tube spacing = 1.250 inches

Ηorizontal tube spacing = 1.083 inches


Τube wall thickness = .016 inches

lnside tube fouling factor = .0000 Btu-hr-SQFT-F

Diameter of inlet pipe/header = 1.0 inch(s)

Total heat transfer rate = -95759.1 Btu/hr

Sensible heat transfer rate = -95759.1 Btu/hr


Dry bulb temperature Ξ 7ο.ο F

Face velocity = 650.00 FPM

Air volume flow rate = 3611.1 CFM


Dry bulb temperature = 94.6 F


Entering fluid temperature = 150.0 F

Leaving fluid temperature = 128.2 FExcerpts from this work may be reproduced by instructors for distribution on a not-for_prοfit basis for testing or instructional purposes only to

students enrolled in courses for whiοh the textbook has been adopted . Αny oιher reprοduction οr trαnsΙαtion of ιhis 'work beyond thαι

permitted by Sections l 07 οr Ι 08 of ιhe ] 976 t]niιed Stαιes Copyrighι Αct wιthouι ιhe permissiοn οf ιhe copyrighι oιυner is unΙαwful'

287Τube side fΙuid velocity = 4.00 FPSCooling or heating liquid flow rate = 9.0 GPM

Air pressure loss = .187 inches of water

Tube side head loss = 8.20 feet or water

Fin effiοiency - .831 Surface effectiveness = .846

Tube side heat transfer coefficient = '1368.8 Btu/hr,SQFT-F

Finned side heat transfer coefficient = 14.4 Btu/hr-SQFT-F

Mean temperature difference = -56.8 F

14-37. From problem 14-29, Re = 6306 (based on xp)

and h5 = 8.9'1 btu/(hr-ft2-F)

jn _.1: = r -1280 Nr(Re)-1 2 (Eq. 14-42)jι

i

then Ψ _ 1_ 1280 x 5 (Re)-1'2 = 1_ 6400 Re_1 '2

iι

i

and Jξ_ 1_(8x12so)(6306)_1 1 = o 872, ^ _ ιJ.ιr, gjs 1 _ (5x1 2s0)(63o 6l_ι 'z

now hε/hs = jε/js = 0'872

hε = 0'872(8.91)

ha = 7.77 Btu/(hr-ft2-F)

14-38. From problem 14-30, Re = 6800 and h5 = 51W(m'- C)

jn - 1- 128oNr(Re)l 2

[From sotution to 14-371js 1- 64OORe-1 2

jο _ 1_(6x128ο)(680)] 2 = 0.96js 1- 64OO(6801-r z

Excerpts from this work may be reproduοed by instructors for distribution on a nοt_for_profit basis for testing or instructional puφoses only tostudents enrolled in cοurses for which the textbook has been adopted. Αny οther reprοducιion or ιrαnslαιion of ιhis τνork beyond thαtpermιιιed by Sections ]07 οr Ι08 ofthe ]976 United Stαtes CοpyrighιΑcιwithοuι the permιSsion ofιhe copyright o'ννner is unlαwful.

288ιl ur rr s = jo/js = 0.96

hο = 0'96 x 51 = 49W(m2 - C) or o.o49 kW(m2 - c)

14-39' Re, = ρ%xo = =o':1:= rΨ (1.os3/12) x 60 = 8225

μ 0.0445 0.54

j = 0.0095 (Fig. 14-12)

ιl - jcco Pr-2l3 = O.OO85 x

h = 10.5 Btu/(hr-ft2-F)

ο.o73, ffi x60x o-24(o'η-2t3

c.. _ 0.0123 - 0.0092 =

.1 .033 x 1O-a,) Using Chart 1 orv ι

85 _ 55

.^^ - 0.0123-0.0063 = 1.5 x 1O-a

) PSYCνz- -- 85_q5

Cavg = 1'27 x1O-a; Use Eq. 1 4-7o &14-73',k = 1zε*s}η;

From Table 5-1a.

M2 = ##l, -Ψ] = u, 3; M = 22'7 ft-1

Ψ = 1'27 ψ(β- o.3)"' = # (1 - o.3)1t2 = 1'265r

* = [E'_ l)[l + O.35ιn&l'ιr /ι r)Φ = (1 '265 _ 1)[1 + 0.35 ln(1.265)] = 0'287

MrΦ = 22.7 xΨ xO'287 = 0'285

- tanh(mrl) = 0.g74 or g7.4o/oΠm-_--'-' mrΦ

ξ,s = 1 * (1 - η,) = f _O.94(1 _ o'g74)= O.98 or 98%

14-40' For 80/67oF; tοp = 60oF

surface temperature must be equal to or less than 60"F'

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students enrolled in οourses for whiοh the texδook has bee, uJopt.α. Αny οιher reproλucιion οr trαnsιαιiοn of this wοrk beyond thαι

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289

Moisture would condense at the base of the fin on the

tube outer surface if it condenses at all'

Let t* be this temPerature.

9 = UiAi(t,- tr) = h6η9A(i, - ir)

where 1 =-1' *,1*,Αr=Ai--"-'_ UιAi hiAι kΑ,

,nα 1 =!*41= :_- '9o]:^

= O.OO1""- U, hi ' k 1000 12x190

Ui = l OOO Btu/(hr-ft2-F); where k"opp", = 190 Btu/(ft-hr-F)

t* = tr . ffi(i,- i*) = 50 + ffi (31.7 - i*)

Αssume a value for tr, read i* from chart 1 and compute t* to

check assumption' Assume t* = 55'8'F then i' = 23'7 Btu/lba

and the calculated t* checks O.K. Therefore moisture will

condense at the base of the fin and on some portion of the fin.

There will probably be no condensation near the outer edge of the

fin.

14-41. For 27Ι19 g, tοο = 15 C

Solution is similar to problem 14-40

UiAi(t, _ tr) = h6η6(i3 _ i*)

1 =

1 +0'0005 = O.O1888; Uι= 52.98ui 53 58

t*=ti - ffi (i"_i*) =14'3-,g4,,* β4.2_i,) *H

For t* = 16, i* = 45. Checks O.K.

Εxcerpts fτorτ this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instruοtiοnal puφoses only to

students enrοlΙed in οourses 1br which the textiook has been adopted. Αny oιher reprοduction or ιrαnsιαιion οf ιhis νοrk beyond thαl

prr^ιt'rca υy srctiοns ] 07 or l οε ij rni l ozο LΙniιed Sιαtes cip)rιgnt 'ιci lνιthοuι ihe permissiοn ξ ιhe copyright ονner is unlαwful'

290There will be no condensation because the tube outside wall is

greaterthanthedewpointtemperatureoftheair.

14-42" This problem is intended for computer solution because

considerable iteration is required'

Coil DescriPtion:

Type of coil = water or brine solution

Tube pattern = staggered plate-fin-tube coil

Material = aluminum fins with copper tubes

Tube side fluid = water





Width of heat exchanger = 60'0 inches; W = 2H

Number of rows of tubes in the air flow direction = 5

Number of tubes per ro\Λ/ = 24 Circuits on tube side = '12

Finpitch=12fins/inchFinthickness=.008inchesVertical tube spacing = 1'250 inches

Horizontal tube spacing = 1'083 inches

Tube outside diameter = '500 inches

Tube wall thickness = '016 inches

lnsidetubefoulingfactor=,OOO0Btu-hr-SQFT-F

Diameter of inlet pipeihead er = 2'5 inch(s)

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students enrolled in courses fbιwhich the texibook has υ*,' "J"pi"α. 'q

iy oιη" "p'oλucιιon

or trαnsιαιion of ιhis work beyond thαι

permitιed by Sectiοns ] 07 * l οε'"iiλr''i ii'6-inι,"d S,o'r, coiir|ιgn, 'a"i 'ιthout ihe permission οf the copyrighι owner is unlανful'

29114.42. (continued)

Total heat transfer rate = 232885.0 Btu/hr

Sensible heat transfer rate = 164919.4 Btu/hr



Wet bulb temperature Ξ 68.0 F

Enthalpy = 32'3 Btu/LBMΑ

Humidity ratio = 83.3 grains/LBMA



Comment: coil is 34.3 percent dry



Wet bulb temPerature = 57.1 F

EnthalPY = 24'4 Btu/LBMA

Ηumidity ratio = 68.7 Grains/LBMA



Leaving fluid temperature = 62.4 F

Τube side fluid velocity = 4.00 FPS

Cooling or heating liquid flow rate = 26.9 GPM


Τube side head loss = 14.09 feet of water

Fin efficiency = .gg9 Surface effectiveness = .819

Tube side heat transfer coefficient = 822.3 Btu/hr-SQFT-F

Finned side heat transfer coefficient = 10.0 Btuihr-SQFT-FExοeΙpts from this work may be reproduced by instructors 1br drstribution on a not-for-pro1lt basis for testing or instructional purposes only to

students enrolled in courses fοr whiοh the textbook has bοen adopted. Αny oιher reproducιion or τrαnsιαtion of ιhis wοrk beyond thαι

permitled by Sections ] 07 οr Ι 08 of ιhe ] 976 tJniιed Sιαtes Copyright Αct without ιhe permissiοn οf the cοpyrighι ονner is unΙανfuΙ.

292

14-43. Coil DescriPtion.

Τype of coil = Direct expansion

Tube pattern = Staggered circular-fin-tube coil

Material = Aluminum fins with copper tubes

Refrigerant tYPe = refrigerant22


Finned side air pressure = 29.92 inches of Hg





Number of tubes per ro\Μ = 16 Circuits on tube side = 16

Fin pitch = 12 Fins/inch Fin thickness Ξ .014 inches


Horizontal tube spacing = 1.300 inches



lnside tube fouling factor = .0000 Btu-HR-SQFT-F


Sensible heat transfer rate = 162201'8 Btu/hr


Dry bulb temperatuΓe = 82.0 FEΧceφts fiοm this work may be rοprοduced by instructors for drstribution on a not-for-profit basis for testing or instιuοtional purposΘS only to

students enrolled in courses for which the texibook has been adopted. Αny oιher reproducιion or trαnsιαtiοn of this νork beyοnd thαt

permitted by Secιiοns ] 07 or Ι 08 οf ιhe ] 976 (Ιnited Sιαιes Cοpyrιght Αct \υιιhouι the permissiοn of ιhe copyrighι oνner is unlανful'

293Wet bulb temPerature = 67.0 F

Enthalpy = 31.4 Btu/LBMA

Ηumidity ratio = 74'9 Grains/LBMΑ

Face velocity = 500'00 FPM


Comment: Coil is .0 Percent drY




Enthalpy = 20.6 Btu/LBMA

Humidity ratio = 52.1 Grains/LBMA

Tube side conditions.

Refrigerant saturation temperature = 35.0 F


Tube side pressure loss = 1.36 PSI

Refrigerant quality entering/leaving evaporator = '29

Enthalpy change in evaporator = 62'75 Btu/LBM

Fin efficiency = .679 Surface effectiveness = '885

Tube side heat transfer coefficient = 375.5 Btu/hr-sQFT-F

Finned side heat transfer coefficient = 9.3 Btu/hr-SQFΤ-F

14-44. Coil DescriPtion:

TYPe of coil = Steam

Tube pattern = Triangular plate-fin-tube coil


Τube side fluid = Steam

Finned side fluid = Air


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Students enroιled in courses for whiοh the texibook has been adopted. Αny οther reprοducιion or trαnslαιiοfi ofthiswοrkbeyοnd ιhαt

permitted by Sectiοns Ι 07 οr l οε iiin, ] 9t7 6 (Jnitecl Stαtes Copjrιght ,ιci τιithout ihe permission of ιhe copyrighι owner is unlιrννful'

294





Number of tubes per ro\M = 16 Circuits on tube side = 16

Finpitch=8Fins/inch Fin thickness = .006 inches


Ηorizontal tube spacing = 1'299 inches

Τube outside diameter = '625 inches

Tube wall thickness = .0'18 inches


Total heat transfer rate = -554503'6 Btu/hr

Sensible heat transfer rate = -5545ο3'6 Btu/hr


Dry bulb temperature Ξ 60.0 F

Face velocitY = 750'00 FPM

Αir volume flow rate = 9000.0 cFM


Dry bulb temperature Ξ 116'1 F


Steam temPerature = 227 '1 F

Steam saturation pressure = 5'000 PSIG

Air pressure loss = '269 inches of water

Fin efficiency = .738 surface effectiveness = '756

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students enrolled in οourses tbr w*ch the texibook has b..n uJo|t"d. Αny οιher reprολucιion or ιrαnslαtion οf ιhis τυork beyond thαι

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isizο LΙniteιl Sιαιes cοpyrιglιt ,ιci νιthout the permission of ιhe copyright ονner is unΙανful'

295

14-44. (continued)

Τube side heat transfer coefficient = 2051.7 Btu/hr-SQFT-F

Finned side heat transfer coefficient = 14.6 Btu/hr-SQFT-F

14-45. Coil DescriPtion:

Type of coil = Water or brine solution

Tube pattern = Staggered plate-fin-tube coil


Τubesidefluid=3Oo/oethyleneglycolsolution

Finned side fluid = Αir






Number of tubes per ro\Λ/ = 16 Circuits on tube side = 4

Fin pitch = 7 Fins/inch Fin thickness = .008 inches

Vertical tube spacing = 1'250 inches

Horizontal tube spacing = 1'083 inches



Insidetubefoulingfactor=.OOO0Btu-hr-SQFT-F

Diameter of inlet pipe/header = '1'0 inch(s)

Total heat transfer rate = -9ο610'1 Btu/hr

Sensible heat transfer rate = -90610'1 Btu/hr

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students enrolled in οourses fbr which the texibook has been aJopteα. Αny other reproλucιion or ιrcιnsιαιion of ιhis νork beyond ιhαt

permiιted by Secιions ] 07 or l οε^"i 'n,

i szο (Jniιed SιαιeS cip)rιgΙιt 'ιci νιιhοuι ihe permission of ιhe copyrighι oνner is unlωυfuΙ'

296Entering air conditions:


Face velocity = 650.00 FPMΑir volume flow rate = 3611.1 cFMLeaving air conditions:



Entering fluΙd temperature = 150.O F

Leaving fluid temperature = 128.4 F

Tube side fluid velocity = 4.00 FPSCooling or heating Ιiquid flow rate = 9.O GPM

Αir pressure loss = '186 inches of water

Τube side head loss ='10.13 feet of water

FΙn efficiency - .83'1 Surface effectiveness = .846

Tube side heat transfer coefficient = 796.O Btu/hr-SQFτ-FFinned side heat transfer coefficient = 14.4 Btu/hr-seFT-FMean temperature difference = -57.6 F

Τhere is a 5 percent reduction in capacity and increased pressure losson the tube side.

14-46. Coil Description:

Type of coil = Water or brine solution

Τube pattern = Staggered plate-fin-tube coil

Material = Αluminum fins with copper tubes

Tube side fluid = 30oλ ethylene glycol solutionExc€φts from this work may be reproduced by instruοtors for distribution on a nοt-for-profit basis Γor testing or instruοtional purposes only tostudentsenroΙIedincoursesfοrwhiοhthetextbookhasbeenadopted. Αnyοtherreprολucιionοrιrαnsιαtioiofιhislυorkbeyλnithαιpermitιed by Sectiοns ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αct ιυ ithοuι ihe permission of ιhe copyright olνner is unlανful.

297Finned side fluid = Air

Finned side air pressure = 29.92 inches of Hg





Number of tubes per ro\Λ/ = 24 Circuits on tube side = 12

Fin pitch = 12 Fins/inch Fin thickness Ξ .008 inches


Horizontal tube spacing = 1.083 inches

Τube outside diameter = .500 inches

Τube wall thickness = .016 inches


Diameter of inΙet pipe/header = 2'5 inch(s)


Sensible heat transfer rate = 155955.9 Btu/hr


Dry bulb temperature Ξ 80.ο F


EnthalPY = 32'3 Btu/LBMA

Ηumidity ratio = 83.3 Grains/LBMA


Αir volume flow rate = 6875.0 cFMComment: Coil is 43.6 percent dry

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Dry bulb temperature Ξ 58.6 F


Enthalpy = 25'1 Btu/LBMA

Humidity ratio = 71.4 Grains/LBMΑ



Leaving fluid temPerature = 62.0 F

Τube side fluid velocity = 4'00 FPS

Cooling or heating liquid flow rate = 26'9 GPM

Αir pressure loss = .756 inches of water

Tube side head loss = 18.13 feet of water

Fin efficiency = .699 Surface effectiveness = '819

Tube side heat transfer coefficient = 476.4 Btu/hr-sQFT-F

Finned side heat transfer coefficient = 10.0 Btu-hr-SQFT-F

The capacity is reduced by about 9 percent, the pressure loss on the

tube side is increased and the leaving air temperatures have increased

by about 1 degree.

14-47. Check Examples 14-1 through 14-5

Coil Description:

Type of Coil = Water or Brine Solution

Τube Pattern = Staggered Plate_Fin-Tube Coil

Material = Αluminum Fins With Copper Tubes

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299Tube Side Fluid = Water

Finned Side Fluid = Air

Finned Side Air Pressure = 0. FT. of Εlevation

Face Area = 2.17 Square Feet

Ηeight of Heat Exchanger = 12.5 ]nches

Width of Heat Exchanger = 25.0 linches

Number of Rows of Tubes in the Αir Flow Direction = 5

Number of Tubes Per Row = 10 Circuits on Τube Side = 5

Fin PΙtch = 8 Fins/lnch Fin Thickness = .006 lnches

Vertical Tube Spacing = 1.250lnches

Horizontal Tube Spacing = 1.083 lnches

Tube Outside Diameter = .525 lnches

Tube Wall Thickness = .015 lnches

lnside Tube Fouling Factor = .0000 ΗR-FTΛ2_F/Btu

Diameter of Inlet Pipe/Ηeader = 1.3 lnch(s)

Total Heat Τransfer Rate = -133026.9 Btu/HR

Sensible Heat Transfer Rate = -133026.9 BtuiHR

Entering Αir Conditions:

Dry Bulb Τemperature = 50.0 F

Face Velocity = 950.00 FPM

Αir Volume Flow Rate= 2061.6 cFMLeaving Air Conditions:

Dry Bulb Temperature = 107.6 F

Tube Side Conditions:

Entering Fluid Temperature = 150.0 FEXcerpts fiοm this work may be reproduced by instruοtοrs Γor drstribution on a not-for-profit basis for testing or instruοtional purpοses only tostudents enrolled in courses for whiοh the textboοk has been adopted. Αny οther reproducιιon or trαnsιαιion of this ιιork beyond thαιpermitιed by Sections ] 07 οr Ι 08 of ιhe Ι 976 United Sιaιes Cοpyrighι Αct withοuι ιhe permission of ιhe copyright oνner is unlα:ινful.

300Leaving Fluid Temperature = 128.3 F

Τube Side FΙuid Velocity = 4.00 FPS

Cooling or Heating Liquid Flow Rate = 12.5 GPM

Αir Pressure Loss = 1.13'1 lnches of Water

Tube Side Head Loss = 6.77 Feet of Water

Fin EfficieΠCy = '750 Surface Effectiveness = '771

Tube Side Ηeat Transfer Coefficient = '1 354.2 Btu/hr-SQFΤ-F

Finned Side Ηeat Τransfer Coefficient = '19.5 Btu/hr-SQFΤ-F

Mean Temperature Difference = -58.5 F

Τhe above results show that a 5 row coil would easily satisfy

the specified requirements. Τhe manual calculation of the

examples are very conservative.

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15-1

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o7 ilιe lιzο (]nited Smtes Copyright Αct without the permission of the copyright owner is unlαwful.

iequests for permission or furtier-idormαtion should be αddressed to the Permission Depαrtment, Jοhn

Wiley & Sons, Ιnc, 11Ι Riνer Street, Hoboken, ]\ΙJ 07030.

Chapter 15

COP : qe/\rv; UseP-idiagram

_νV = i+_ig =119.5_105.5

= - 14.0 Btu / lbm p

Q"=il -iι=42'5-119'5

= -77 Btu / lbm

g"=-Q"+\M=77-14

= 63 Btu / lbm

COP=63 114=4.5

CoP"".n.1 = 50ο l (570- 5οο) = 7'14

14 ooFtlooF

4 ooF

(a)

(b)

4.50r= = 0.63 or 63 %

7.14

(c)

ψ= Q" =coP

10 (12,000) _26,667 Btu lhr = 1ο.5 HP N 7.8 kW4.5

then HP =

10'5 = 1.O5 and Ψ =O.22ton 10 kW

(d) Qe = ΓhQe oΓ Γh =10 (12,000)

= 1905 lbm / hr = 0.24 kg /s63

300

15-3

(e) W = 10.5 HP from (c) above = 7.8 kW

(f) PD = rhv. = 1905_ι0'68)

= 21'6 ft3 / min x O.O1 m3 / s60

15-2 W = 2.5 kW; rh = 0.05 kg / s

(a) 8c=il_iη, _νv - i+_iο

i+ =ig _W / rir : 398.4 + 2'5 / 0.ο5 = 448'4 kJ / kg

Qc = i1 -i4:260'3 - 448'4 = -188'1 kJ / kg

w = i3 -i4 = 398.4 - 448'4: -50 kJ / kg

9e : 9c -\M - 188.1-50 : 138.1 kJ / kg

(b) COP = 138.1 I 50 = 2.76

( coP )carnot = #*:3.86iι = 448'4 kJ / kg from part (a)

0r = 2.76 / 3.86 = 0 '72 or 72 o/o

R- 134a; Energy Balanοe

rh1i1 + mηiι =m2i2+ rh5i5 , rh1 = fi2= rh3 = rhη = rhs

is = i.l +iι _iz = 44'94 + 101 '54 _ 37.98 : 108'5 Btu / Ibm

Ps = P+ :16.6 Psia

w=i6-is,So=Ss

! 100 F sat. liquidP1: 138.8 Pjsla

-r0 Fsat. vapor 80 F

Excerptsfromthiswοrkmaybereproduοedbyinstructorsfordistributionona.not-for-profitbasisfortestingorinstructionalpurposesonlyto _,l -_...__^ll^r;^^^llfaAqfnru,lιiοhthetextbοοkhasbeenadonteιl. Αnνotherreυroducιionortrαnslαιiono{thiswοrkb?νοbdthηι^2v":1'^)

(c)

(d)

5

301

io :'127 Btu / lbm lChart 3]

\Ι/ : 127 - 108.5 = 18.5 Btu / lbm

Qe = i+ _i3 : iι _iz = 1o1 '54 _ 37'98 = 63'6

15-3 (continued)

15-4

(a)

(b)

HP^t TT8ton rhge

HP =

(18.5) 778 (12,ooo)=1.37

132

Pt9

r80P

2

0.75

0.90 = 0.15

ton (63.6) 60 (33,000)

Γ. ^ rr..1l"' v30ν =| 1+ C _ c

IL \PoJl vb

n = k = 1.17; 1 ln= 0.855

P./P6:180120:9

v3 = ?

-1I: o.9o [chart 4 tabv6 2.38

nu =|ι+ O.O3 - o.03 (9)o uuu]

,.. : ftua or ]1ι1 =

0u lb / ft3'v PD PD v3

rh 0.75PD 2.14

Πν = Γ,

* o.15_ o.15 (91o'εss

le Α3a]

0.90 =

Exοemtsfrοmthiqrνnrl-tra\,L--__-^j,'^^lL-.:__}_'.^a^.^.--f,j_+-:!_,'+:^-_-___^}ε^--.nfi'+λ-"iafΛf tροt;fσΛrinctrlotinnql nllrnnsεsnnlvt^

15-5

15-6

GπlΦ() a β00)1728

= 52.36 ft3 / min

rh / PD = 0.15 t 2-14 = O'07 lb / ft3

(e\ fr, - tro = 1- o'07

= o.8o or 80 %\v" rha 0.35

(d) Power is directly proportional to the mass flow rate

therefore, Power compares as in (c) above'

0ν =o'7o

4 cyl - 3" bore, 4" stroke, 800 rPm

Pι = 49] psia (chart 3)

Pι = 138.8 psia (chart 3)

Πν = frΥ2, / PD; Υ2a = 1 'o4 ft3 / Ibm;

Table Α-2a @ 55 F / 52 psia

Qιz = rh (i2 _ η)

PD=

iι __iι = 46 Btu / tbm

iz = 112 Btu / lbm

1ιz=ηPt" (iz_,.,l = Ξ#f4 (.ε'46) = 2g26 Btu / minΥ2a

or 912=139,560 Btu/hr = 11'6tons

R-22, assume suPerheat = 20 F

2 qsoF

_. Ξ-- __^fi+ L-.]. f^f tea1inσ nr instnlctiοnaι nurooses onlv to

303

Subcooling = 10 F

t. = 130 F, te = 45 F

(a) 3t

90

'15-6 (Continued)

tl:130-10=120F

iι:iz:46 Btu / lb

(b) Q" = 144,000 Btu / hr (Fig. 15-7)

Wc = 14.8 kW = 50,498 Btu / hr

(c) te = 32.5 F (Fig. 1 5-τ); W = 13.3 kW

15-7 Refer to Fig ' 15-7 cΑP = 133,000 Btu/hr

-3\PΦ

P b*Ρ;.}

4\'d

ι|1 35

te= 47 F

120 E

^ l1 .'jL''L:^-.^_-^^+f^r-nrnfithasisfοrtestinsorinstructional purposesonΙyto

Ξ-Ξι_ΦΞoo-

t-

Ξφα)o()o

o(dο-(τ,

ζ)

304

15-8

Excerpts from this work maystudents enrolled in courses

orator Tι

ΞΥ13.7 ;

=o-c(ι)

Ξoα

-cΞΦooo

'δωo-(σo

only toιhis work

Evapl a),( c)

( b) (a)( c)io

Design Pointo -o(

Measured OperatingPoints

by Secιions ]07 οr ]08 οf ιhe 1976 United Sι*eΝwff#{&ψeff&Etrβwib,gΦeruεgffi'"'^''n''' -ιhαι permiιιed

--305

Suction valve, ΔP:2 psia

Discharge valve, ΔP = 4 psia

'10 F S.H. in intake man. and cyl.

Piston clearance=5oλ'

Γ ι r"']"nl γ"ηu=| 1+C_.ι ]1 rPoJ.] uo

15-9

Pe = 69 psia; te = 30 F (Τable A-4)

(a) Τhe condensing temperature is still about '1 '15 F, but the

evaporating temperature is low, about 30 to 31 F.

(b) (qα _q) / aα = 1_*=0.36 or 36 % low305

(c) lt appears that the evaporator is not loading the compressor.

Check for proper air fΙow over evaporator. Fan speed may be

low or an obstruction exists.

Glzτ s

sοsia,-Ι

sat R-22 νao450F '

P=90.73psia

ExcΘrpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolΙed in cοtlrses fnr whinh tLΔ +__}L^^l'- L^^ L^^- ^j^_ι' l

V3 = Vg at 45 F; vg = 0.604 ft3 / lbm (Τable Α-3a)

vo = 0.66 ft3 / lbm (Chart 4 at 55 F / 89 psia)

Pc=275+4= 279 psia; Pb = 90.73 -2= 88.73 psia

n:1.16, C=0.05

(a) o" =| 1+ O.O5 _ o.O5 (Ξ+\1/1 16l /o οo+\

L "_v'vι'ιs&rcJ ] ι .* .,l

+ 0'εeε

N 0.2 kg /s

res. loss

(b)ri'r -

(c)

15-9 (cont

'15-10 (a)

0.604 : 27.75|bmlmin

.n-1 IP. I

n -11Pυ)

]l, ,oi6 I

.66) l ( 2'79 )l 'ο _.' I'|(8εzs)

1

31 kJ/kg

1.0 HP x 8.2 kW

)'''o] ι*l>f k assumed and 2 psi p

ry, (PD) / v. : (ο.838) 20 /

ψ=!Ψ, W=*o",[[

* = 1J9(ss.73) (144) (o0.16'

inued)

= 1 0,466 ft - lbf /lbm x

w _27.75(10,4666) _ 1.0.80(33,ο00)

Γ

ou =|l+ O.04 _ o.o4 (Ψ'ν L 'u'Note: Αn average value c

Excerpts from this work may be reproduοed by instruοtors for distribution on a not_for-profit basis for testing or instruοtional purposes only toctrr.lan+. ^--^ir-

i

307

assumed in suction header and valve.

Tv = 0.90, Γh = (PD) ηu lνz 19

= 9'4 (ο.90) l 0'74

rh ='l 1.44 lbm / min

β*p=zooirδ oort"II

\l60.oF

91( ft _ lbη / lbm

Btu / min

o6o'55

Γ - 1'4_1

w = -_1!- (53) 1 44 (o'77) l [Ψ']

*(1'4_1) \ / \ /Lι53/

vιl = fr*

= 9491 (11.44) / (0.9 x778) =0m

ι6.I

=94

55 I

tz

1t

15-10 (Continued)

QH=ψ+Qι=(155x60) + 3O,OOO = 39,3ooBtu/hr

or qH = 655 Btu / min

9491 =125.2Btu / ΙbmW23 =i2 -i3, iZ =iZt W = 133 +

j

778

9Η = ia _i3i iη = 9μ +iu = _#+ +125'2= 68 Btu / lbm11.44

(b) lteration is required

P3 will decrease with the lighter load but Pz is also lower and

(P3 / P2) t'n will be about the same as part (a);

ν2 l ν6 will be about constant. Then Tv : Constant.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis flοr testinρ οr l'nstnlctiοnaΙ nιlrnοses onlv to-+,,f,^-+-.-.---rl

j

308

However , Υ2= O.85 and rh : (PD)ην l νe = (9.4) 0.87 / 0.85

= 9.62 lbm / min.

ΠoWW =9491 (48)(1 03) : 14.78 Btu / tbm

53 (0.77) 778

W = (1 4.78) (9.62) (601 = 8530 Btu/hr;

Qn = 8530 + 24,000 = 32,530 Btuihr

Which assumes Ps I Pz is constant and 2 psi pres. loss in the

valve.

ie = iz t w :83.5 + 14'78 --94'3 Btu / lbm

iι =iz_ 9ιz = 111_24,000 / 9'62 (60) = 69'4 Btu / lbm

,, = # (50) = 188 psia 18

Ps

15-11 Reduced air flow reduces the load on the evaporator. Without

suction pressure control the evaporator pressure will decrease

until condensate will freeze and completely block the

evaporator air flow. Liquid refrigerant will return to the

Εxοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to

41.9 .83.5 9 4.3t

309

15-12

15-'13

15-14 (a)

compressor and eventually cause the compressor suction

valve to faiΙ.

lnstall an evaporator pressure regulator set to maintain a

pressure such that the temperature of the evaporator surface

will not decrease below the freezing point for water.

lnstall a suction pressure regulator on the compressor inlet.

Τhe regulator shouΙd be set to limit the suction pressure to a

level compatible with the compressor capacity.

Using chart 2 with the construction shown,

15-15

the final temperature is 9ο F

(b) 'v - L=!=0.18s orm (ν 20 i

18.5 o/o vapor (Use chart 2)

Using chart 2

X3 = 0.495; ts = 125 F

28ooFl8ooF

0 0.25 .495χ+

Excerpts from this rπork may be reproduced by instructors for distribution on a not-for_profit basis for testing or instruοtionaΙ purposes only tostudents enrolled in courses fbr whiοh the textbook has been adopted. Αny other reproduction or trαnslαιiοn of this work beνond ιhαΙ nemilto)by Sectiοns ]07 or ]08 οfιhe !Q76 [Ιnifa'] Slnlo" r'^'.'-':'-ι'' ι ''

tb ammonia/lb sol.

TzΞτ7 l6TT-sia

310

15-16 (coP)ma, = Ψ:fd: T^ = 180 + 460 : 64OR;\ _ - /Ιlιaλ Tg(To _ ξ )' '9 -

Te=75 + 460 = 535 R; To =]00 + 460 = 560 R

(COP)ma x = 2.675

15-17 Refer to Chart 5 for saturated vapor at 10 mm hg.

Vapor must first be condensed to sat. liquid at '10 mm hg.

Q" = irs, Table Α-1a;

i'n = 1ο64.8 Btu/lbmu for 1 lbm of vapor oΓ 9. = 1064.8 Btu

at 50 F, P = 0.178 psia or

15-17 (continued)

P = 10 mm hg

Locate point I at x = 0; P = 10 mm hg

Locate point s atx = 0.6 ; P = 10 mm hg

ffru=1;m.=5

ms5mvmm6ms

5- Elv = ; tΠS = i x 45'5 Ξ 37.g (depends on scale used)ob

(a) x = 0.50

(b) Q, = i, - io = -50 - (-70) = 20 Btu / lbm of solution

Exοerpts Γrom this work may be reproduced by instruοtors for distribution on a not-fοr-profit basis for testing or instruοtional purposes only tostudents enτolΙed in οourses for which the textbook has been adopted. Αny oιher reproducιion or ιrαnslαιion of ιhis νοrk beyond thαι pemιιιedby Sectiοns Ι07 or ]08 ofιhe ]976 tΙnited Sιοιρs Canιιriohι Δ?r1υ!a!"^''a n'^ ' '

311

Qtot : 1064.8 + (6 x 20) = 1 ,184.8 Btu

i

70δh9

0.5 0.6χ+

0.8

Excerpts from this work may be reproduοed by instruοtors for distribution on a not_for_profit basis foτ testing or instructional purposes only tostudents enτolled in οourses for whiοh the textbook has been adopted. Αny other reproduction or trαnslαtion of ιhis ιιork beyond thαι permitιedbν Sections ]07 or ]08 οfιhe ]976 [kιitad Stntρc Γnh1'';-L} ι^l''':'1''' ' ''

lm

ib

\m-1

311

ADDENDUMto

Solutions ManuaΙ fοr McQuiston, ΙΙVAC 6e

ProbΙem 6-10

For the floor, it is unοlear what2 in. vertiοal edge insulation means (whether 2 in. is the

thiοkness of insulation or the depth of the edge insulated).. The solution assumes that the insulation has R-value of 5.4 hr-ft2-oF/Btu and the

depth of the edge is 2 ft.

For the door, Table 5-8 in the 6th edition does not have U-value for the wood storm doorand there are three types of the wood door with 1 % in. thickness.

ο The solution assumes that the doors are panel doors with metal storm doοr; henοe,

its U-value is 0.28 But/hr-ft'-'F.

ΡrobΙem 7-9

The standard time zone for ottawa, ontario is Εastern Standard Time instead of CentralStandard Time.

. The solution uses Eastern Standard Time.

Prοblem 7-14

For the specified loοatiοn, the sunset oοοurs before 9:00 p.m. CDST on June 21.

ο The solution uses 8:Ο0 p.m. CDST instοad of 9:00 p.m.

Prοblems 8-25 and 8-26

Both problems do not specifu the window orientation.. The solutions assume the west-facing window for both prοblems.

Tabte 8-20

Reοommended radiative and conveοtive fraοtions for solar heat gains should be revisedsinοe the 6th edition uses the SHGC values in the calοulation of the (οombined) solar heatgain for the RΤS methοd.

Example 8-16

Τhe example actually uses 90%/10% of radiative/οonveοtive split of the cοmbined solarheatgain. Ηowever,thetext (page270) says 100%/0Υoforthetransmittedsolarheatgainand 630Λ1370Λ for the absorbed sοlar heat gain.

L--**-

312

ProbΙems 8_25 and 8-26

The solutions for both problems use 90%11,0% for the combined solar heat gain.

ExampΙe 9_1

The οalοulation for this example should be

" =

('1Ι11?2Ψoi19.']!o:9Φ _ Ι22'606(0.ssx70- 0)(1000)

(Changing 13 to24 and 122790 to 122606).

mcquiston heating ventilating air conditioning 6th solutions

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