5 dimensional analysis

7/29/2019 5 Dimensional Analysis

1/12

Chapter 5Dimensional Analysis

Page 1 of 12

5_001 A gas having a molecular weight of 86 is flowing at a rate of 1.38 kg/min through a tube having an actual i.d.of 1.20 in. At a cross section where the absolute pressure is 330 atm and the temperature is 340 F, the true

density is 1.2 times that predicted by the perfect-gas law, and the viscosity is 2.0 x 10-6

force-pound xseconds per square foot. Calculate the numerical value of the dimensionless Reynolds number.

Solution:

Reynolds number:

NRe

= DG/

F

= 2.0 x 10-6 Force-pound x seconds per square foot

= Fg

c

= (2.0 x 10-6)(32.2)

= 6.44 x 10-5 lb matter / sec-ftD = 1.2 in = 0.1 ft

G = 4w/D2

w = 1.38 kg/min = 0.050715 lb/sec

G = 4(0.050715)/[(0.1)2] = 6.457234

NRe

= (0.1)(6.457234)/6.44 x 10-5

NRe

= 10,027 . . . Ans.

5_002 It is desired to investigate the drag force on geometrically similar solid objects placed in a free stream offlowing fluid. In the general case, the force Fexerted on the body is thought to depend on the characteristiclength of the object x, the approach velocity V, the acceleration due to gravity g, and the following properties

of the fluid: the density , the viscosity , the surface tension , and the velocity Va

at which sound waves

are propagated in the liquid.

a. What is the numbern

of algebraically independent dimensionless groups which may be used to

correlate the variables?

b. Determine a set of these n

dimensionless groups, using the force Fin only one group.

c. Calculate the numerical value of each of these groups except the one in which the force Fappers,for the condition in which x= 7 inches, g= 32.2 ft/(sec)(sec), V= 11 ft/sec, and the fluid is water at 68 F

(assume Va = 4900 ft/sec).

Solution:

Variables -

aV,,,g,V,x,F, Then:

( ) 0V,,,g,V,x,F, a = Dimension of Factors: MLF

Dimensions Factors

F x V g Va

M 1 1L 1 1 1 -3 1 -1 1

-1 -2 -1 -1

F 1 1

a. nf= no. of physical factors of importance

nf= 8

ni= nd = number of dimension

ni= 4


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Page 2 of 12

n

= no. of independent dimensionless group

n

= nf- n

i= 8 - 4 = 4

b. Incompatible factors

( ) ( ) ( )L

M&,T

1,L

M,Lx 3 ==

==

FF

x

x

x

L

M

xLM

xL

23

33

=

===

==

=

Remaining factors:

aV,g,V,

xV;

x

x

x

LV

F

x;

x

F

L

F

gx;

x

x

x

Lg

xV;

x

x

x

LV

a42a

3

2

32

232

2

22

2

12

====

===

==

==

====

c. Values:x = 7 inches = 0.583333 ftg = 32.2 ft/(sec)(sec)V = 11 ft/secV

a= 4900 ft/sec

at 68 F

= 62.282 lb/cu ft

= 6.75 x 10-4 lb/(sec)(ft)

= 72.7 Dynes/cm2

= 0.67534 N/ft2

( )( )( )

( ) ( ) ( )

( )( )( )263,737

106.75

62.28249000.583333

xV

36,730106.75

32.20.58333362.282

gx

592,063106.75

62.282110.583333

xV

4

a4

4

32

2

32

2

41

=

==

=

==

=

=

=

5_003 Below are some data from an early article on the performance of a packed regenerator. In heating runs, thepacked spheres were initially at an elevated uniform temperature, and air at room temperature entered atconstant mass rate. In cooling runs, the spheres were initially at room temperature, and air at approximately


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Page 3 of 12

200 F entered at constant mass rate. In the data given below the relation between the temperature ofentering air (t

1), the temperature of exit air (t

2), and initial uniform temperature (t

0) of the bed of spheres is

given by the dimensionless term y= (t2

- t0)/(t

1- t

0). The heat capacity of the vertical cylindrical column was

small compared with that of the spheres used for packing, and the column was well insulated.Correlate all the values ofygiven below, plotted as ordinates against some suitable abscissa:

Material Steel Steel Steel Steel Lead Glass

Bed depth, inches ..... 5.125 10.25 5.125 2.56 9.55 9.71

Bed diam, in .............. 4 8 4 2 8 8

Particle diam, in ......... 0.125 0.25 0.125 0.0625 0.233 0.237

Sp ht .......................... 0.111 0.111 0.111 0.111 0.0347 0.168

Sp gr ......................... 7.83 7.83 7.83 7.83 11.34 2.6

k, Btu/(hr)(ft)(deg F) .. 26 26 26 26 19.5 0.4

Air superficial mass

vel., lb/(sec)(sq ft) ... .. 0.15 0.15 0.3 0.3 0.3 0.3

y , , , , , ,

sec sec sec sec sec sec

0.1 300 600 140 . . . 113 127

0.2 360 680 160 . . . 133 149

0.3 380 740 190 . . . 147 166

0.4 410 810 205 100 162 182

0.5 450 890 220 . . . 176 199

0.6 480 970 240 . . . 191 215

0.7 510 1050 250 . . . 206 232

0.8 570 1140 300 140 222 250

0.9 650 1300 330 170 248 280

The quantity is the elapsed time at which the indicated ywas measured.

Solution:

L - ftD

t- ft

Dp

- ft

cp

- Btu/lb-ft

- lb/cu ftk - Btu/sec-fr-deg FG

o- lb/sec-sq ft

- time, sec

MLTFH System:

Hoppt Kq,,Gk,,,c,D,DL,y = p

Hnm

oife

pc

pb

ta KGkcDDLy =

( ) ( ) ( ) ( )p

2

2n

m

2

if

3

ecba

H

ML

L

M

LT

H

L

M

MT

HLLL1

=


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Page 4 of 12

(1) L: a + b + c - 3f - i - 2m + 2p = 0

(2) L: e + i - p = 0

(3) M: -e + f + m + p = 0

(4) T: -e - i = 0

(5) : -i - m + n - 2p = 0

(4) i = -e(5) e + i - p = 0

e - e - p = 0p = 0

(3) -e + f + m + p = 0-e + m + n = 0e = m - n

Equating:e = f + m = m - n

(1) a + b + c - 3f - i - 2m + 2p = 0a + b + c - 3f + e -2m = 0a + b + c - 3f + e - 2(e - f) = 0a + b + c - 3f + e - 2e + 2f = 0a + b + c -f -e = 0

a + b + c = f + ef = a + b + c -e

Then:

p = 0a = ab = bc = ce = ef = a + b + c -ei = -em = e -f = 2e - a - b -c

n = -f = e - a - b -c

Then:cbaecba2e

oeecbae

pc

pb

ta GkcDDLy

++=

e2

op

c

o

pb

o

t

a

o rk

Gc

G

D

G

D

G

Ly

=

Note: Strouhal Numbers:

G

LN

oLSl, =

GDN

o

tDSl, t

=

G

DN

o

pDSl, p

=

Ratio of Peclet Number to Strouhal Number:


5/12


Page 5 of 12

k

Gc

N

N2

op

Sl

Pe =

Steel # 1:

L = 5.125 in = 0.4271 ftD

t= 4 in = 0.3333 ft

Dp

= 0.125 in = 0.01042 ft

cp

= 0.111 Btu/lb-F

= 7.83 x 62.4 = 488.6 lb/cu ft

k = 26 Btu/hr-ft-deg F = 7.2222 x 10-3

Btu/sec-ft-deg FG

o= 0.15 lb/sec-sq ft

( )( )

1391.2

0.15

488.60.4271

G

LN

oLSl, ===

( )( )

1085.7

0.15

488.60.3333

G

DN

o

tDSl, t

===

( )( )

33.9414

0.15

488.60.01042

G

DN

o

pDSl, p

===

( )( )

( )( )107.07754

107.2222488.6

0.150.111

k

Gc

N

N 43

22op

Sl

Pe

=

==

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4NN

Sl

Pe =

y X1 X2 X3 X4

0.1 300 4.637333 3.619000 0.113138 0.212326

0.2 360 3.864444 3.015833 0.094282 0.254791

0.3 380 3.661053 2.857105 0.089319 0.268947

0.4 410 3.393171 2.648049 0.082784 0.290179

0.5 450 3.091556 2.412667 0.075425 0.318489

0.6 480 2.898333 2.261875 0.070711 0.339722

0.7 510 2.727843 2.128824 0.066552 0.360955

0.8 570 2.440702 1.904737 0.059546 0.403420

0.9 650 2.140308 1.670308 0.052218 0.460040

Steel # 2:

L = 10.25 in = 0.8542 ftD

t= 8 in = 0.6667 ft

Dp

= 0.25 in = 0.02083 ft

cp

= 0.111 Btu/lb-F

= 7.83 x 62.4 = 488.6 lb/cu ft

k = 26 Btu/hr-ft-deg F = 7.2222 x 10-3

Btu/sec-ft-deg F


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Page 6 of 12

Go

= 0.15 lb/sec-sq ft

( )( )

2782.4

0.15

488.60.8542

G

LN

oLSl, ===

( )( )

2171.7

0.15

488.60.6667

G

DN

o

tDSl, t

===

( )( )

67.8503

0.15

488.60.02083

G

DN

o

pDSl, p

===

( )( )

( )( )107.07754

107.2222488.6

0.150.111

k

Gc

N

N 43

22op

Sl

Pe

=

==

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4NN

Sl

Pe =

y X1 X2 X3 X4

0.1 600 4.637333 3.619500 0.113084 0.424652

0.2 680 4.091765 3.193676 0.099780 0.481273

0.3 740 3.760000 2.934730 0.091690 0.523738

0.4 810 3.435062 2.681111 0.083766 0.573281

0.5 890 3.126292 2.440112 0.076236 0.629901

0.6 970 2.868454 2.238866 0.069949 0.686521

0.7 1050 2.649905 2.068286 0.064619 0.743142

0.8 1140 2.440702 1.905000 0.059518 0.806840

0.9 1300 2.140308 1.670538 0.052193 0.920080

Steel # 3:

L = 5.125 in = 0.4271 ftD

t= 4 in = 0.3333 ft

Dp

= 0.125 in = 0.01042 ft

cp

= 0.111 Btu/lb-F

= 7.83 x 62.4 = 488.6 lb/cu ft

k = 26 Btu/hr-ft-deg F = 7.2222 x 10-3

Btu/sec-ft-deg FG


( )( )

695.6

0.30

488.60.4217

G

LN

o

LSl, ===

( )( )

542.8

0.30

488.60.3333

G

DN

o

tDSl, t

===

( )( )

16.97

0.30

488.60.01042

G

DN

o

pDSl, p

===

( )( )

( )( )102.83102

107.2222488.6

0.300.111

k

Gc

N

N3

22op

Sl

Pe 3

=

==


7/12


Page 7 of 12

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4NN

Sl

Pe =

y X1 X2 X3 X4

0.1 140 4.968571 3.877143 0.121214 0.396343

0.2 160 4.347500 3.392500 0.106063 0.452963

0.3 190 3.661053 2.856842 0.089316 0.537894

0.4 205 3.393171 2.647805 0.082780 0.580359

0.5 220 3.161818 2.467273 0.077136 0.622824

0.6 240 2.898333 2.261667 0.070708 0.679445

0.7 250 2.782400 2.171200 0.067880 0.707755

0.8 300 2.318667 1.809333 0.056567 0.849306

0.9 330 2.107879 1.644848 0.051424 0.934237

Steel # 4:

L = 2.56 in = 0.2134 ftD

t= 2 in = 0.1667 ft

Dp

= 0.0625 in = 0.00521 ft

cp

= 0.111 Btu/lb-F

= 7.83 x 62.4 = 488.6 lb/cu ft

k = 26 Btu/hr-ft-deg F = 7.2222 x 10-3

Btu/sec-ft-deg FG


( )( )

347.6

0.30

488.60.2134

G

LN

o

LSl, ===

( )( )

271.5

0.30

488.60.1667

G

DN

o

tDSl, t

===

( )( )

8.485

0.30

488.60.00521

G

DN

o

pDSl, p

===

( )( )

( )( )102.83102

107.2222488.6

0.300.111

k

Gc

N

N3

22op

Sl

Pe 3

=

==

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4N

N

Sl

Pe =

y X1 X2 X3 X4

0.4 100 3.476000 2.715000 0.084850 0.283102

0.8 140 2.482857 1.939286 0.060607 0.396343

0.9 170 2.044706 1.597059 0.049912 0.481273


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Page 8 of 12

Lead:

L = 9.55 in = 0.7958 ftD

t= 8 in = 0.6667 ft

Dp

= 0.233 in = 0.01942 ft

cp = 0.0347 Btu/lb-F = 11.34 x 62.4 = 707.6 lb/cu ft

k = 19.5 Btu/hr-ft-deg F = 5.4167 x 10-3

Btu/sec-ft-deg FG


( )( )

1877

0.30

707.60.7958

G

LN

oLSl, ===

( )( )

1572.5

0.30

707.60.6667

G

DN

o

tDSl, t

===

( )( )

45.81

0.30

707.60.01942

G

DN

o

pDSl, p

===

( )( )

( )( )108.14797

105.4167707.6

0.300.0347

k

Gc

N

N3

22op

Sl

Pe 4

=

==

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4N

N

Sl

Pe =

y X1 X2 X3 X4

0.1 113 16.610619 13.915929 0.405398 0.0920720.2 133 14.112782 11.823308 0.344436 0.108368

0.3 147 12.768707 10.697279 0.311633 0.119775

0.4 162 11.586420 9.706790 0.282778 0.131997

0.5 176 10.664773 8.934659 0.260284 0.143404

0.6 191 9.827225 8.232984 0.239843 0.155626

0.7 206 9.111650 7.633495 0.222379 0.167848

0.8 222 8.454955 7.083333 0.206351 0.180885

0.9 248 7.568548 6.340726 0.184718 0.202070 Glass:

L = 9.71 in = 0.8092 ftD

t= 8 in = 0.6667 ft

Dp

= 0.237 in = 0.01975 ft

cp

= 0.168 Btu/lb-F

= 2.60 x 62.4 = 162.24 lb/cu ft

k = 0.4 Btu/hr-ft-deg F = 1.1111 x 10-4

Btu/sec-ft-deg FG


( )( )

437.6

0.30

162.240.8092

G

LN

oLSl, ===


9/12


Page 9 of 12

( )( )

360.6

0.30

162.240.6667

G

DN

o

tDSl, t

===

( )( )

10.68

0.30

162.240.01975

G

DN

o

pDSl, p

===

( )( )

( )( ) 0.838766101.1111162.240.300.168

k

Gc

N

N4

22op

Sl

Pe

===

Table No. 1

X1N LSl, = X2N

tDSl,=

X3N

pDSl,=

X4N

N

Sl

Pe =

y X1 X2 X3 X4

0.1 127 3.445669 2.839370 0.084094 106.523282

0.2 149 2.936913 2.420134 0.071678 124.976134

0.3 166 2.636145 2.172289 0.064337 139.235156

0.4 182 2.404396 1.981319 0.058681 152.655412

0.5 199 2.198995 1.812060 0.053668 166.914434

0.6 215 2.035349 1.677209 0.049674 180.334690

0.7 232 1.886207 1.554310 0.046034 194.593712

0.8 250 1.750400 1.442400 0.042720 209.691500

0.9 280 1.562857 1.287857 0.038143 234.854480

By curve fitting:

e2

op

c

o

pb

o

t

a

o rk

Gc

G

D

G

D

G

Ly

=

= 0.003593a = -29.130232b = 30.947910c = -3.944493e = -0.394509

Therefore:-0.394509

2op

-3.944493

o

p30.947910

o

t

-29.130232

o rk

Gc

G

D

G

D

G

L0.003593y

=

5_004 Two metals, A and B, are being considered for use in constructing tuning forks which are to produce

vibrations of a given frequency. What would be the ratio of the costs of the metal required for the two tuningforks, expressed as a function of the frequency, the physical properties of the metals, and the cost per unitmass of each metal (assumed independent of the quantity purchased)? The tuning forks are to begeometrically similar.

Solution:


10/12


Page 10 of 12

Let R = ratio of the cost of the metals.f = frequency, HzU

1, U

2= cost per unit mass.

Other properties:

1,

2= densities of material

V1, V

2= sound velocities

( )212121 U,U,V,V,,f,R = h

2g

1e

2d

1c

2b

1a UUVVfR =

hgedc

3

b

3

a

M

1

M

1

L

L

L

M

L

M

11

=

(1) : -a - d - e = 0

(2) M: b + c - g - h = 0

(3) L: -3b - 3c + d + e = 0

(1) d + e = -a(2) -3b - 3c - a =

b + c = -a / 3(3) b + c - g - h = 0

g + h = -a / 3

Then:

ga2

g1

da2

d1

ba2

b1

a 31

31

UUVVfR

=

g

2

1

d

2

1

b

2

1

a

222U

U

V

V

UVr

fR

31

31

=

g

2

1

d

2

1

b

2

13a

23

22

3

U

U

V

V

UVr

fR

=

Therefore:

=

2

1

2

1

2

1

23

22

3

U

U

V

V

UVr

fR ,,,


11/12


Page 11 of 12

5_005 It is agreed to assume that the thermal conductivities of gases and vapors depend on the following factors:

cp, specific heat at constant pressure; c

v, specific heat at constant volume; , viscosity; T, absolute

temperature; = /(p)T, coefficient of compressibility; b = /(T)

p, coefficient of thermal expansion.

List all conclusions which may be drawn from dimensional consideration. If one additional physical variablewere to be included, what should it be?

Solution:

unitsc

p- Btu/lb-R

cv

- Btu/lb-R

- sq ft / lb

- 1/R

additional variable is the viscositym - Btu/sec-ft-R

MLTF system

Hvp K,,,,c,ck = f

Hedcb

va

p1 Kcck = f

2edc

2ba

H

ML

L

M

T

1

M

L

MT

H

MT

H

LT

H

=

(1) H: a + b - f = 1

(2) : -e - f = -1e + f = 1

(3) L: 2c - e + 2f = -1

(4) T: -a - b - d = -1

a + b + d = 1(5) M: -a - b - c + e + f = 0

(1) and (4)-f = df = -d

(5) -a - b - c + e + f = 0a + b + c - e - f = 0a + b + c -1 = 0a + b + c = 1

(2) e = 1 - fe = 1 + de = 1 + c

but c = 1 - a - b

Then:a = ab = bc = 1 - a - bd = 1 - a - be = 1 + c = 2 - a - bf = a + b -1


12/12


Page 12 of 12

-1baH

b-a-2b-a-1b-a-1bv

ap1 Kcck

+=

=

H

b

Hv

a

Hp1

K

Kc

Kck

KH

= Jgc

= (778)(32.2) = 25,000 (lb matter)(ft)2 / (Btu)(sec)2

Final results.

=

Kc

Kc

K

k Hv

Hp

H

,

- end -

5 dimensional analysis

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