( ) ( )iwsolutiontestbankfiles.com/wp-content/uploads/2018/05/...solutions chapter 2 ± 2.2.7 d pdvv...

Solutions Chapter 2

2–1

2.1.1

(a) N/mm or nm (nanometer) (b) °C/M/s (c) 100 kPa (d) 273.15 K (e) 1.50m, 45 kg (f) 250°C (g) J/s (h) 250 N 2.2.1

a. Basis: 1 mi3

1 mi 3 5280 ft 1 mi

⎛ ⎝

⎞ ⎠

3 12 in 1 ft

⎛ ⎝

⎞ ⎠

3 2 . 54 cm 1 in

⎛ ⎝

⎞ ⎠

3 1 m 100 cm ⎛ ⎝

⎞ ⎠

3

= 4.17 ×109 m3 b. Basis: 1 ft3/s

1 ft3

1 s60 s

1 min7.48 gal

1 ft 3 = 449 gal / min

2.2.2

a.

0.04 gmin( ) m3( )

60 min1 hr

12 in( )3

1 ft3

1 lbm

454g=

9.14

lbm

hr( ) ft3( )

b.

2Ls

3600 s1 hr

24 hr1 day

1 ft3

28.32L=

6.1×103 ft3

day

c.

6 (in.)(cm2 )(yr)(s)(lbm )(ft2 )

1 ft2

(12 in)2

(1 in)2

(2.54 cm)2

1 yr365 days

1 day24 hr

1 hr3600s

2.2 lbm

1 kg

1 ft12 in

0.3048 m1 ft

= 1.14 × 10–11 mkg( ) s2( )

Full file at http://testbankcart.eu/Solution-Manual-for-Basic-Principles-and-Calculations-in-Chemical-Engineering-8th-Edition-by-Himmelblau

Solutions Chapter 2

2–2

2.2.3

a. Basis: 60.0 mile/hr 60.0 mile 5280 ft 1 hr ft = 88

hr 1 mile 3600 sec sec

b. Basis: 50.0 lbm/(in)2

2 24m

2 2 2 2

50.0/lb 454 g 1 kg 1(in) (100 cm) kg = 3.52 10(in) 1 lb 1000 g (2.54 cm) (1 m) m

×

c. Basis: 6.20 cm/(hr)2

9 2

2 2 2

6.20 cm 1 m 10 nm 1(hr) nm = 4.79 (hr) 100 cm 1 m (3600 sec) sec

2.2.4

20 hp 0.7457 kW

1 hp= 14.91 kW

No , not enough power even at 100% efficiency; 68 kW = 91.2 hp.

2.2.5

1 hr525 mile

2200 gal1 hr

1000 mile = 4190.5 gal

1 hr475 mile

2000 gal1 hr

1000 mile = 4210 gal(20 gal)

None: 20 gal more are needed.

2.2.6

Let tA be the time for A to paint one house; tB for B A does a house in 5 hours, or 1 house/5 hr. B does one house in 3 hours, or 1 house/3 hr.

A B1 house hr 1 house hr+ = 1 house5 hr 3 hr

t t

Also tA = tB so that A A A3 5 8 + = 1 or =1

15 15 15t t t

A B15 = hr = = 1.875 hr or 112.5 min8

t t


Solutions Chapter 2

2–3

2.2.7

(a) mass, because masses are balanced (b) weight, because the force exerted on the mass pushes a spring

2.2.8

20.0g(m)(s)

1 lbm

453.6 g0.3048m

1 ft3600s1 hr

1 (lbf )(s2 )32.174(lbm )(ft)

1(hr)2

(3600)2s2

= 1.16×10−7 (lbf )(hr)ft2

2.2.9

1.0 Btu

(hr)(ft2 )o Fft

⎛⎝⎜

⎞⎠⎟

24 hrs1 day

1 ft2

(12 in)2

1 in2

(2.54 cm2 )(100 cm)2

1 m2

1.8oF1oC

2.54 cmin

252 cal1 Btu

12 in1 ft

4.184 J1 cal

1 kJ1000 J

= 1.49×104 kJ(day)(m2 )(°C / cm)

2.2.10

Basis: 1 lb H2O

a. 3 2

2 m ff

m

1 1 1 lb 3 ft (s )(lb )KE= mv = = 0.14(ft)(lb )2 2 s 32.174(ft)(lb )

⎛ ⎞⎜ ⎟⎝ ⎠

b. Let A = area of the pipe and v = water velocity. The flow rate is

q = Av = πD2

4⎛⎝⎜

⎞⎠⎟

(v)

π4

(2 in)2 (1 ft)2

(12 in)2

3 ftS

60 s1 min

7.48 gal1 ft3 = 29.37 gal/min

2.2.11

PE = 75 gal

min8.33 lbm

gal32.2 ftsec2

100ft

60 shr

s2-lbf

32.2 lbm -ftBtu

778 ft-lbf

= 4818 Btu/hr

2 hp 2545 BtuPump Work 5090 Btu/hrhp-hr

= =

Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr


Solutions Chapter 2

2–4

2.2.12

The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the force used to support the mass.

2.2.13

In American Engineering System Power = FV

5f f800 lb 300 ft (lb )(ft)= = 2.4 10 or 7.27 hp1 min min

×

In SI

Power = 4000 N 1.5 m 1 (watt)(s)

1 s 1(N)(m)

= 6000 watts

2.2.14

KE = 21 m v2

=

1 2

2300 kg 1 lb m 0 . 454 kg

10 . 0 ft s

⎛ ⎝

⎞ ⎠

2 1 32 . 2 lb m

lb f ft

sec 2

1 Btu 778 . 2 ft ( ) lb f ( )

= 10.11 Btu

2.2.15

Basis: 10 tons at 6 ft/s

KE = 2 2

2 m ff

m

1 1 20,000 lb 6 ft 1(s )(lb ) m v = = 11,200(ft)(lb )2 2 s 32.2(ft)(lb )

⎛ ⎞⎜ ⎟⎝ ⎠

2.2.16

Basis: 1 mRNA

1 nRNA 3x ribonucleotides

nRNA

1 active ribosome

264 ribonucleotides

1200 amino acids

min-active ribosome

protein

x amino acids=

13.6 protein molecules formed per min per nRNA


Solutions Chapter 2

2–5

2.3.1

a1. A is in g/cm3 a2. B is in g/(cm3)(°C) a3. Since the exponent of e must be dimensionless C is in atm-1

b1 A = 3 3

m m3 3 3

1 lb lb1.096 g (30.48) cm = 68.4 cm ft 454g ft

b2 B = 3 3

m3 3 3 o

lb0.00086 g (30.48) cm 1 lbm 1 C = 0.0298 (cm )( C) ft 454 g 1.8 R (ft )( R)

o

o o

b3 C = 2 2f f

0.000953 1 atm 1 = 0.0000648 atm 14.7 lb /in lb /in

2.3.2

Introduce the units. The net units are the same on both sides of the equation.

(ft)(ft)1.5 ft

s2

⎛⎝⎜

⎞⎠⎟

1/2

= ft3

s

2.3.3

No.

1/ 2

-3 32

22 2 2

(2) 9.8m 10 m 50×10 (kg)(m) 10.6(2m )s kg (s )(m ) 21-

5

q

⎡ ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥

⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

The net units on the right hand side of the equation are

1/23 3

24

m mms s

⎡ ⎤≠⎢ ⎥

⎣ ⎦

Consequently, the formula will not yield 80.8 m3/s, presumably in the formula the g should be gc for use in the AE system.


Solutions Chapter 2

2–6

2.3.4

1.19 Q = 0.61S (2Δp)/ρ assume hole is open to atmosphere

Δp = 144 in2

1 ft2

23 lbf

in2 + 73 in gas. 0.703 H2O1 gas

1 ft12 in

14.7lbf

in2

33.91 ft H2O

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 3,579 2flb /ft

ρ = 0.703

ft3lb

1 lb/ft3H2O

62.4lbH2Oft3H2O =43.87

lbm

ft3

S = π 1

(4)12⎛⎝⎜

⎞⎠⎟

2

= 3.41×10-4ft2

Q = (3600)(0.61)(3.41×10−4 ) (2)(3579)gc / 43.87 = 54 ft3

hr


Solutions Chapter 2

2–7

2.3.5

a. Z = 1 + ρB + ρ2C + ρ3D Units B cm3 / g mol C (cm3/ g mol)2

D (cm3/ g mol)3 b. Z = 1 + ρ

*B*+(ρ*)2C*+(ρ*)3D* Units B* ft3 / lbm C* (ft3/ lbm)2

D* (ft3/ lbm)3

If B is the original coefficient, B* is obtained by multiplying B by conversion factors. Let MW is the molecular weight of the compound.

33 3*

m m

ft B cm 1 ft 1 g mol 454 g 0.016B = = B lb g mol 30.48 cm MW g 1 lb MW

⎛ ⎞⎜ ⎟⎝ ⎠

22 2 263 3 -4

*2

m m

ft cm 1 ft 1 g mol 454g 2.57 10C = C = C lb g mol 30.48 cm MWg 1 lb MW⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ×⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠

33 -6

*3

m

ft 4.096×10D = D lb MW⎛ ⎞⎜ ⎟⎝ ⎠


Solutions Chapter 2

2–8

2.3.6

u ms

⎡

⎣⎢

⎤

⎦⎥ = k[1] τ

ρ⎛⎝⎜

⎞⎠⎟

1/2Nm2

⎛⎝⎜

⎞⎠⎟

1/2m3

kg⎛⎝⎜

⎞⎠⎟

1/2⎡

⎣⎢⎢

⎤

⎦⎥⎥

To get u in ft/s, substitute for τ and for ρ , and multiply both sides of the equation by 3.281 ft/1 m (k is dimensionless).

τ N

m2 = ′τ lbf

ft2

3.281 ft1 m

⎛⎝⎜

⎞⎠⎟

21 N

0.2248 lbf

ρ kg

m3 = ρ′ lbm

ft3

3.281 ft1 m

⎛⎝⎜

⎞⎠⎟

31 kg

0.454 lbm

′u = 2.57 k ′τ

′ρ⎛⎝⎜

⎞⎠⎟

2.3.7

Place units for the symbols in the given equation, and equate the units on the left and right hand sides of the equation by assigning appropriate units to the coefficient 0.943. LHS RHS

Btu hr ( ) ft 2 ( ) Δ ° F ( )

? Btu hr ( ) ft ( ) Δ ° F ( )

⎛ ⎝ ⎜ ⎞

⎠ ⎟

3 lb m ft 3

⎛ ⎝

⎞ ⎠

2 ft hr ( ) 2

Btu lb m ft

hr ( ) ft ( ) lb m Δ ° F

⎡

⎣ ⎢

⎤

⎦ ⎥

1 4

The units are the same on the right and left so that 0.943 has no units associated with it.

2.3.8

η = numerator

denominator

Numerator = / /

c cx s b b eY Hγ −Δ

= mol cell C e equiv. energymole substrate C mol cell C mol e equiv.

−

−

⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

= energy

mol substrate C (1)


Solutions Chapter 2

2–9

Denominator = c

catHΔ

= energy

mol substrate C (2)

η = (1) energy (2) energy

=

There is no missing conversion factor What the author claimed about the units is correct.

2.3.9

For dimensional consistency:

B – absolute temperature in either ºR or K C – absolute temperature in either ºR or K

A – dimensionless In most cases the argument of a logarithm function should be dimensionless, but in this case it will not be. Therefore, the numerical values of A, B, and C will depend upon the units used for temperature and the units used for p*.

2.3.10

The equation is

Δp = 4fρ (v2 / 2g) (L/D)⎡⎣ ⎤⎦ The units on the right hand side (with f dimensionless) in SI are

ρ kgm3

m2

s2

(kg)(m)s2

mm

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

→ 1m2

hence the equation is not dimensionally consistent because Δp has the units of N/m2. If g is replaced with g, the units would be correct.

2.4.1

Two because any numbers added to the right hand side of the decimal point in 10 are irrelevant.

2.4.2

The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an implied decimal point, the answer should be 1287 (no decimal point).


Solutions Chapter 2

2–10

2.4.3

The number of significant figures to the right of the decimal point is 1 (from 210.0m), hence the sum of 215.110 m should be rounded to 215.1 m.

2.4.4

A calculator gives 569.8269 000, but you should truncate to 4 significant figures, or 569.8 cm2.

2.4.5

Two significant figures (based on 6.3). Use 4.8×103.

2.4.6

Step 1: The product 1.3824 is rounded off to 1.4 Step 2: Calculate errors. For absolute error, the product 1.4 means 1.4 + 0.05

Thus 0.05 100% 3.6%1.4

× = error

Similarly 3.84 has 0.005 100% 0.13%3.84

× = error

and 0.36 has 0.005 100% 1.4%0.36

× = error

Total 2.7% error

2.6.1

a) =

b) =

c) = 321.26 10 lb mol N−×

(d) 2 6 2 62 6

2 6

3 lb C H O 1 lb mol C H O 454 g mol = 29.56 g mol C H O(46.07) lb C H O 1 lb mol

4 g mol mg Cl2 95.23( )g MgCl2gmol MgCl2

380.9 g

2 lb mol C3H8 44.09( )lb C3H8lb mol C3H8

454g C3H81 lb C3H8

4 ×104g C3H8

16 g N2 gmol N228.02( )g N2

1 lb mol N2454 gmol N2


Solutions Chapter 2

2–11

2.6.2

(a) 16.1 lb mol HCl 36.5 lb HCl = 588 lb HCl1 lb mol HCl

(b) 19.4 lb mol KCl 74.55 lb KCl = 1466 lb KCl1 lb mol KCl

(c) -3

3 33

3

11.9 g mol NaNO 85 g NaNO 2.20 10 lb = 2.23 lb NaNO1 g mol NaNO 1 g

×

(d) -3

2 22

2

164 g mol SiO 60.1 g SiO 2.20 10 lb = 21.7 lb SiO1 g mol SiO 1 g

×

2.6.3

Basis: 100 g of the compound comp. g MW g mol Ratio of Atoms C 42.11 12 3.51 1.09 O 51.46 16 3.22 1 H 6.43 1.008 6.38 2 13.11 Multiply by 11 to convert the ratios into integers The formula becomes C12O11H22 Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)


Solutions Chapter 2

2–12

2.6.4

Vitamin A, C20O H30, Mol Wt.: 286 Vitamin C, C6 H8O6, mol. wt: 176 Vitamin A:

a. Vitamin A = 2.00 g mol 286 g 1 lb = 1.26 lb

1 g mol 454 g

16 g 1 lb 0.0352 lb

454 g=

Vitamin C = 2.00 g mol 176 g 1 lb = 0.775 lb

g mol 454 g

16 g 1 lb 0.0352 lb

454 g=

b. Vitamin A = 1.00 lb mol 286 lb 454 g = 130,000 g

1 lb mol 1 lb

Vitamin C = 1.00 lb mol 176 lb 454 g = 79,900 g

1 lb mol 1 lb

For both 12 lb 454 g 5450 g

1 lb=

2.6.5

1 kg5.2 m3

1160 m3

kg mol = 223.1 kg/kg mol

2.6.6

Mass fraction to mole fraction:

x1 =

ω1

MW1

ω1

MW1

+(1−ω1)MW2

Mole fraction to mass fraction

ω1 =

x1MW1

(x1)(MW1) + (1− x1)(MW2 )


Solutions Chapter 2

2–13

2.6.7

Basis: 100 kg mol gas Comp. Mol % = mol MW kg CH4 30 16 480 H2 10 2 20 N2 60 28 1680 Total 100 2180

21.8 kgkg mol

2.6.8

Basis: 100 g mol gas Comp. mol = % MW g CO2 19.3 44 849.2 N2 72.1 28 2018.8 O2 6.5 32 208.0 H2O 2.1 18 37.8 100.0 3113.8

3113.8Avg. mol. wt = 31.138100.0

=

2.6.9

Basis: 100 lb mol Comp. % = mol MW lb CO2 16 44 2640 CO 10 28 280 N2 30 28 840 3760 Avg. MW = 37.6 lb/lb mol

2.7.1

(a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if use SI units). (b) A gas requires a convenient basis of 1 or 100 lb moles (if use AE units). (c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass is a convenient basis. (d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas. (e) Same answer as (e).


Solutions Chapter 2

2–14

2.7.2

Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.

2.7.3

Pick one day as a basis that is equivalent to what is given—two numbers: (a) 134.2 lb C1 (b) 10.7 ×106 gal water.

2.8.1

Basis: 1000 lb oil

= 129.5 gal

2.8.2

Basis: 10,010 lb

2.8.3

Basis: 1 g mol each compound g mol mw g ρ (g/cm3) V (cm3) Pb 1 207.21 207.21 11.33 18.3 Zn 1 65.38 65.38 7.14 9.16 C 1 12.01 12.01 2.26 5.31

3

mass (g)V̂ = density (g/cm )

Set 2 is the correct one

2.9.1

Component g mol mol fraction MW g mass fraction Na 1 0.20 23 23 0.22 C1 1 0.20 35.45 35.45 0.33 O 3 0.60 16 48 0.45 5 106.45 1.00

0.926 lb oilft3

1.00 lb H2Oft3

62.4 lb H2Oft3

= 57.78lb oilft3 oil

1000 lb oil 1 ft3 oil57.78 lb oil

7.48 gal1 ft3

10,010 lb 1 gal8.80 lb

0.134 ft3

gal=

152 ft3


Solutions Chapter 2

2–15

2.9.2

Basis: 1 gal of solution. Mass of solution:

= .

Mass fraction KOH = =

Mass fraction H2O = 1 – 0.09 =

2.9.3

Basis: 30 lb gas Comp. lb MW lb mol mol fr. CO2 20 44 0.455 0.56 N2 10 28 0.357 0.44 30 0.812 1.00

2.9.4

a) 1000 b) No c) Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases the ratio is in moles.

1.0824 lb solnft3 H2O

1.00 lb H2Oft3H2O

62.4lb H2Oft3H2O 1 ft3

7. 481 gal1 gal 9.03 lb soln

0.813 lb9.03 lb

0.09

0.911.00 Total


Solutions Chapter 2

2–16

2.9.5

On a paper free basis the total ppm are: Brand A: 6060 ppm Brand B: 405 ppm The respective mass fractions are: The other entries are similar Fe Cu Pb

Brand A:

1310 0.2166060

=

2000 0.3306060

=

2750 0.4546060

=

Brand B:

350 0.864405

=

50 0.123405

=

5 0.0123405

=

Ends

= 54.4 m3

Vs =5 m 30.2 m 200 mm 1 m

1000 mm2 sides

= 60.4 m3

Ve =5 m 27.2 m 200 mm 1 m

1000 mm2 ends


Solutions Chapter 2

2–17

Floor

Total volume = 281.392 m3

Mass of concrete = 3

3

281.392 m 2080 kg = 585,295m

Volume of displaced water required to float:

V = LWh

h = 0.703 m

2.9.6

Basis: 190,000 ppm

Vf =27.4 m 30.4 m 200 mm 1 m

1000 mm= 166.592 m3

586,543.36 kg 1 m3 H2O1000 kg H2O

= 586.543 m3

→ h =V

LW

586.54 m3

27.4m 30.4 m=

190,000 g PCB106 ×100 = 19%


Solutions Chapter 2

2–18

2.9.7

Basis: 100 g of the sample The biomass sample is g = % dry weight of cells C 50.2

O 20.1 N 14.0 H 8.2 P 3.0 95.5 other 4.5 Total 100.0 10.5 g cells 50.2 g C 1 g mol C = 0.439 g mol C/g mol ATPg mol ATP 100 g cells 12 g C

2.9.8

2MMM(s) NN(s) 3CO (g)→ +

a. 7 6

10

2×10 disintegrations 1 min 1 curie 10 µ curie = 11µ curie

min 60 3×10 disintegrations/s 1 curie

b. 7

72 10 disin tegrations 0.80 1.6 10 cpmmin

× = ×

2.9.9

The relation to use is 1/ 2t n(2) /(k)(OH )−= l with 6(OH ) 1.5 10− = × k 1/ 2t (seconds)

Methanol 120.15 10−× 530.8 10× Ethanol 121 10−× 54.6 10× MTBE 120.60 10−× 57.7 10× The order is in increasing persistence ethanol, MTBE, and methanol.


Solutions Chapter 2

2–19

2.9.10

Yes. Bases are first entries.

34 4 4

9 34 4

4800 g mol CC1 10 g mol air kg mol air 154 g CC1 103 mg CC1 =10 g mol air kg mol air 22.8 m g mol CC1 g CC1

3432.4 mg CC1 /m which exceeds the NIOSH standards

2.9.11

a.

b.

c.

For d. and e. assume that (P)outflow remains unchanged. d. P retained/yr = 2,240 + 6,740 + 0.7 19,090 – 4,500 = 17,843 ton/yr. P conc. in ppb =

= 92.6 ppb

This is greater than 10 ppb.

e. P retained/yr = 2,240 + 6,740 + 0.3 19,090 – 4,500 = 10,207

P conc. in ppb =

25,600 ton Pyr

2.6 yr 11.2× 1014 gal

1 gal3.785 L

2000 lb1 ton

454 g1 lb

106µ g1 g

= 133.1µ g / L

19,090 lb P( )municipal30,100 lb P( )total

= 63. 4%

19,090 lb P( )municipal30,100 lb P( )total

0.70 lb P( )det.1 lb P( )municipal

100 lb P( )tota l30,100 lb P( )tota l

= 44.4%

×

17,843 tonyr

2.6 yr 11.2 ×1014 gal

2000 lbton

1 g / cm3

8.345 lb / gal109 g

1billion g

Eutrophication will not be reduced.

×tonyr

10,207 tonyr

2.6 yr1.2×1014 gal

2000 lbton

1 g / cc8.345 lb / gal

109 g1 billion g

= 53.2 ppb

This will not help.


Solutions Chapter 2

2–20

2.9.12

Basis: 106 g mol gas

2 2 26

2 2 2

350 g mol H S 1 g mol gas 34g H S 1 g mol CO10 g mol gas 1 g mol CO 1 g mol H O 44 g CO

2 2

6 62

270 g H S 270 g H S= = 10 g CO 10 g total liquid

mass fraction H2S = 42.70 10−×

2.9.13 (a)

442 2

6

1 mol O 10 mol O or 10 ppm100 mol gas 10 mol gas

⇒

(b) Basis: 100 mol gas answer Comp. % = mol mol fr. or mol % SO3 55 0.932 93.2 SO2 3 0.051 5.1 O2 1 0.017 1.7 Total 59 1.000 100.0

2.9.14

MW CaCO3: 100.06

Ca 40.05 Mg 24.3 C 12.01 O 16.00

3 3 3

3

100.06 g CaCO 1 g mol CaCO g CaCO1 g mol Ca = 2.50 g mol Ca CO 1 g mol Ca 40.05 g Ca g Ca

Similarly 3 3g CaCO g CaCO = 4.118 g Mg g Mg

Total alkalinity = 2.50 (56.4) + 4.118 (8.8) = 3mg CaCO177 L


Solutions Chapter 2

2–21

2.9.15

No. 1 molecule in 1023 or more is not 13-20 ppb

2.9.16

On a mol basis, the carbon dioxide concentration in air is about 350 parts per million (ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to 353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at present.

2.10.1

TK = –10 + 273 = 263K T°F = –10 (1.8) + 32 = 14°F T°R = 14 + 460 = 474°R

2.10.2

Yes, if the temperature scale is a linear relative one (°C, °F), or a logarithmic scale (ln 1° is zero). No, if the scale is absolute, but read J. Wisniak, J. Chem. Educ., 77, 518-522 (2000) for a different conclusion.

2.10.3

C

p = 8.41 + 2.4346×10-5TK

2.4346×10-5 J(gmol)(K)2

⎛

⎝⎜⎜

⎞

⎠⎟⎟(TK )

Substitute 1.8 TK = T°R

Cp =

= o5

R8.41 1.353 10 T−+ ×

8.41 + 2.4346× 10œ5 Jgmol( ) K( )2

T°R1.8

⎛ ⎝

⎞ ⎠


Solutions Chapter 2

2–22

2.10.4

a) + 32 =

b) + 32 + 460°R =

c)

−25oF – 32oF1.8

1.0oC1.8oF

+ 273K = 241.3K

d) =

2.10.5 First multiply the RHS of the equation so that Btu 1054.8J 1 lb mol 1.8°F 1°C (lbmol)(°F) 1 Btu 454 gmol 1.0°C 1K

= 4.182

and substitute T°F = 1.8T°C + 32

Cp = [8.488+ 0.5757 ×10−2(1.8T°C + 32)− 0.2159×10−5(1.8T°C + 32)2 +

0.3059×10−9(1.8T°C + 32)3]4.182 J(gmol)(°K)

Simplifying, Cp = 36.05 + 0.0447T – 0.2874 10–4 T2 + 0.7424 10–8 T3

2.10.6

The instrument does not contain mercury, but has to contain a fluid that responds at –76°C and can be calibrated to measure temperature.

10°C 1.8°F1.0°C 50°F

10°C 1.8°F1.0°C 510°R

150K 1.8°R1.0K 270°R

Jgmol( ) °K( )

× ×


Solutions Chapter 2

2–23

2.11.1

Basis: 15cm3 water

a. 3

1000 kg 10 m 10 m 0.15 mm = ρV= = 15,000 kgm

b. 2

2 2

F m g 15,000 kg 9.80 m 1 N 1 Pa = = = 1470 Pa (kg)(m) NA A s 10 m 10 m 1s m

= 1.47 kPa

1.470 kPa 1 atm 14.7 psi = 0.21 psi101.3 kPa atm

or 2

2

15 cm H O 1 in. 1 ft 14.696 psi = 0.212.54 cm 12 in. 33.91 ft H O

2.11.2

ρconcrete = 2080 kg/m3

ρwater = 1000 kg/m3 Volume of concrete:

Sides

Ends

= 54.4 m3

Floor

Vs =5 m 30.2 m 200 mm 1 m

1000 mm2 sides

= 60.4 m3

Ve =5 m 27.2 m 200 mm 1 m

1000 mm2 ends

Vf =27.4 m 30.4 m 200 mm 1 m

1000 mm= 166.592 m3


Solutions Chapter 2

2–24

Total volume = 281.392 m3

Mass of concrete = 3

3

281.392 m 2080 kg = 585,295m

Volume of displaced water required to float:

V = LWh

h = 0.703 m

2.11.3

The pressure is a gauge pressure. Basis: 50.0 psig a.

(difference)

b. No. Insufficient height. Alternate solutions can be applying ∆p =

2.11.4

2

ff m2

lb in. = lb lb in the AE systemin.

→

so the procedure is ok, although the unit conversion is ignored.

586,543.36 kg 1 m3 H2O1000 kg H2O

= 586.543 m3

→ h =V

LW

586.54 m3

27.4m 30.4 m=

50.0 psig 33.91 ft H2O14.7 psia

= 115 ft

ρΔhg

148 ft

60°F


Solutions Chapter 2

2–25

2.11.5

Basis: Dept = 1000 m p = 1 atm +

=

Alternative solution:

42 2

2 2

1000 m sea H O 1.024 g H O 3.28 ft 101.3 kPa101.3 kPa + = 1.003 10

1.00 g sea H O 1 m 33.91 ft H O

×

4

4 + 0.01 10

= 1.013 10

×

×

2.11.6

ABS Gauge atm(a) p = p + p

2 2

2ff2 2 2

22.4 lb 144 in. 28.6 in. Hg 14.7 psia 144 in.+ = 5250 lb /ft

1 in. 1 ft 1 29.92 in. Hg 1 ft⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

(b)

22.4 psig 29.92 in. Hg14.7 psia

⎛⎝⎜

⎞⎠⎟

+ 28.6 in. Hg = 74.2 in. Hg

(c)

22.4 psig1

1.013×105N/m2

14.7 psia

⎛⎝⎜

⎞⎠⎟

+ 28.6 in. Hg

11.013×105N/m2

27.92 in. Hg

⎛⎝⎜

⎞⎠⎟

= 2.51×105N/m2

22 222.4 psig 33.91 ft H O 28.6 in. Hg 33.91 ft H O

(d) + = 84.1 ft H O1 14.7 psia 1 29.92 in. Hg

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2.11.7

Neither John is necessarily right. The pressure at the top of Pikes Peak is continually changing.

p = po +ρgh

1000 m 1.024 gcm3

9.8 ms2

1 kg103 g

100 cm1 m

⎛ ⎝

⎞ ⎠

3 1 Nkg( ) m( )

s2

1 kPa1N

kg( ) m( )s2

1 atm1.013×105 N / m2

100.1 atm 101.3 kPa1 atm

= 1.014× 104 kPa


Solutions Chapter 2

2–26

2.11.8

a.

b.

Δp =ρgh

=1 gcm3

9.8 ms2

13.1 m 1 kg1000 g

100 cm( )3

m3Pa

1 kgm s2

1 kPa1000 Pa

= 128. 4 kPa

FA

=mg

ABottom=

ρVgA Bottom

=ρ S.A.( ) t( )g

ABottom

ABottom =π

4D2 =

π

430.5 m( )2

= 730.62 m2

ATop = 730.62 m2

Aside = πDh = π 30.5 m( ) 13.1m( ) = 1255.2 m2

S.A.= ATop + ABottom + ASide

= 730.62 + 730.62 +1255.22( ) m2 = 2716.46 m2

ρ S.A.( ) t( )gA

=

7.86 gcm3

100 cm3( )m3

2,716.46 m2 9.35×10œ3 m730.6 m2

9.8 ms2

1 kg1000 g

1 Pa1 kg / ms2

1 kPa1000 Pa

= 2.68 kPa


Solutions Chapter 2

2–27

2.11.9

A B 1 1 2 1 1 1p -p = ρ gh +ρ gΔh ρ gh ρ gΔ h− −

2 1 1 2 1= ρ gΔh ρ gΔh = gΔh(ρ ρ )− −

=18.4 mm Hg

A 1 1 2 D

B 1 1 2 D

p + ρ gh + ρ gΔh = Pp + ρ gh + ρ gΔh = P

The pressure at A is higher than the pressure at B.

Alternate solution:

Δp = (0.78 in.) (13.546− 0.91)

13.546760 mm Hg29.92 in. Hg

= 18.5 mm Hg

=9.8 m

s20.78 in 1 m

3.28 ft13.546 œ0.91g

cm21 kg

1000 g

1 ft12 in

100 cm1 m

⎛ ⎝

⎞ ⎠

31 Nm2

1 kgm( ) s( )2

1 Pa1 Nm2

760 mm H2101.3×103Pa


( ) ( )iwsolutiontestbankfiles.com/wp-content/uploads/2018/05/...solutions chapter 2 ± 2.2.7 d pdvv...

Documents