Choose MLT As Fundamental Units

7.31 The drag of an airfoil at zero angle of attack is a function of density, viscosity, and velocity, in addition to a length parameter. A 1/10th scale model of an airfoil was tested in a wind tunnel at a Reynolds number of 5.5 x 106 , based on chord length. Test conditions in the wind tunnel air stream were 15 C and 10 atm absolute pressure. The prototype airfoil has a chord length of 2 m, and it is to be flown in air at standard conditions. Determine the speed at which the wind tunnel model was tested, and the corresponding prototype speed.

6 Rem = 5.5 x 10 o Tm = 15 C = Tp  T = 1

pm = 10 atm, pp = 1 atm  p = 10

ℓp = 2 m ; L = 1/10

FD  f ,,V ,ℓ

Choose MLT as fundamental units

FD   V ℓ MLT-2 ML-3 ML-1T-1 LT-1 L

Dimensional matrix

L T M FD ,  ,  produce FD 1 -2 1 a 3 x 3 matrix with  -3 0 1 a non-zero  -1 -1 1 determinant  V 1 -1 0 Rank = 3 ℓ 1 0 0

3 repeaters are needed

choose  , V , ℓ as repeaters

Check on independence of the dimensions of the repeaters

L T M  -3 0 1

V 1 -1 0  0  dimensions of  , V , ℓ are independent

ℓ 1 0 0

- 110 - Dimensionless groups:

1 1  1 0 0 0 -3 1 -1 1  1 -2 1   V ℓ FD  M L T  (ML ) (LT ) (L) MLT

0 0 0  1 3   1  2 M L T  M 1 L 1 1 1 T 1  1 + 1 = 0  1 = -1 ;

-1 - 2 = 0  1 = -2; -3(-1) + (-2) + 1 + 1 = 0  1 = -2

F   D 1 V 2 ℓ2

 2  2  2 0 0 0 -3  2 -1  2  2 -1 -1  2   V ℓ   M L T  (ML ) (LT ) (L) ML T

0 0 0  1 3   1  1 M L T  M 2 L 2 2 2 T 2  2 + 1 = 0  2 = -1 ;

-2 - 1 = 0  2 = -1; -3(-1) + (-1) + 2 - 1 = 0  2 = -1

   2 V ℓ

F     = f ( )  D  f    C = f (Re) 1 2 V 2 ℓ2  V ℓ D

   V  ℓ For dynamic similarity, Re = 1   1  

1 1 10 o      = 1 since Tm = Tp = 15 C and  = f (T)  V 1    ℓ    10  

pm 101.3  10 3 p p 101.3 3  m    12.3 kg / m ;  p    1.23 kg / m RTm 287  288 RTp 287  288

  = 10  V = 1 

6 5 Re p   p 5.5  10  1.753  10 Vm  Vp    39.2 m / s  p ℓp 1.23  2

- 111 - 7.34 The fluid dynamic characteristics of a golf ball are to be tested using a model in a wind tunnel. Dependent parameters are the drag force, FD, and lift force, FL, on the ball. The independent parameters should include angular speed, , and dimple depth, d. Determine suitable dimensionless parameters and express the functional dependence among them. A golf pro can hit a ball at V = 240 ft/sec and  = 9000 rpm. To model these conditions in a wind tunnel with a maximum speed of 80 ft/sec, what diameter model should be used? How fast must the model rotate? (The diameter of a U.S. golf ball is 1.68 in.)

FD  f , d, D, V ,, 

Choose MLT as fundamental units

FD  d D V  

MLT-2 T-1 L L LT-1 ML-3 ML-1T-1

Dimensional matrix

FD 1 -2 1 FD ,  , d  0 -1 0 produce a 3 x 3 d 1 0 0 matrix with a non- D 1 0 0 zero determinant V 1 -1 0  Rank = 3  -3 0 1  -1 -1 1

3 repeaters are needed

choose  , D , V as repeaters

Check on independence of the dimensions of the repeaters

L T M  -3 0 1

D 1 0 0  0  dimensions of  , V , D are independent

V 1 -1 0

- 112 - Dimensionless groups:

1 1 1 0 0 0 -3 1 -1 1  1 -2 1   V D FD  M L T  (ML ) (LT ) (L) MLT

0 0 0  1 3   1  2 M L T  M 1 L 1 1 1 T 1  1 + 1 = 0  1 = -1 ;

-1 - 2 = 0  1 = -2; -3(-1) + (-2) + 1 + 1 = 0  1 = -2

F   D 1 V 2 D2

 2  2  2 0 0 0 -3 2 -1  2  2 -1 -1  2   V D   M L T  (ML ) (LT ) (L) ML T

0 0 0  1 3   1  1 M L T  M 2 L 2 2 2 T 2  2 + 1 = 0  2 = -1 ;

-2 - 1 = 0  2 = -1; -3(-1) + (-1) + 2 - 1 = 0  2 = -1

   2  VD d    3 V 3 D 3 d   =  = 0;  = -1 (by inspection)    3 3 3 3 3 D

 D     4 V 4 D 4    = 0,  = -1;  = 1 (by inspection)    4 4 4 4 4 V

F   d  D  = f ( ,  ,  )  D  f  , ,   C = f (Re, St, d/D) 1 2 3 4 V 2 D2   VD D V  D

St : Strouhal number

For dynamic similarity, Re = 1, St = 1

   V  D    1 and  D  1    V

80 1    V 240 3

Assume Tm = Tp ; pm = pp   = 1 ;  = 1

1    1    1    3 V D 3 D D

Dm = D Dp = 3(1.68”) = 5.04”

- 113 -  V 1 1 1        D 3 3 9

1      (9000)  1000 rpm m m p 9

7.36 A model hydrofoil is to be tested at 1:20 scale. The test speed is chosen to duplicate the Froude number corresponding to the 60 knot prototype speed. To model cavitation correctly, the cavitation number also must be duplicated. At what ambient pressure must the test be run? Water in the model test basin can be heated to 130 F, compared to 45 F for the prototype.

For dynamic similarity,

( p p ) v  1    2 Ca = 1  2  ( p pv )  V  V

(see p. 295 for Cavitation Number)

Model & Prototype are tested in water with  = const   = 1

   2 ( p pv ) V

Also, Fr = 1 for dynamic similarity

 1 V 2 1 1  1;  g  1 (same location)   V   L 2 2  g  L

1        0.05 ( p pv ) L 20 ( p pv )

 pm  pvm   0.05; p p  14.7 psia, pv p  pv  0.15psia T 45F  p p  pv p 

pvm  pv T 130F

pm  0.05p p  pv p  pvm  0.0514.7  0.15  2.22  2.95psia

- 114 - 7.41 An automobile is to travel through standard air at 100 km/hr. To determine the pressure distribution, a 1/5-scale model is to be tested in water. What factors must be considered to ensure kinematic similarity in the tests? Determine the water speed that should be used. What is the corresponding ratio of drag force between prototype and model flows? The lowers pressure coefficient is Cp = -1.4 at the location of the minumum static pressure on the surface. Estimate the minimum tunnel pressure required to avoid cavitation, if the onset of cavitation occurs at a cavitation number of 0.5.

Factors necessary to ensure kinematic similarity:

-model and prototype must be geometrically similar -model must be completely submerged to avoid surface effects -cavitation effects must be absent in model testing

   D  V For dynamic similarity,  Re  1   1  

1  D V D  ;     1 5  v

6 2 vm  vwater/T20C(assumed )  1 10 m / s

 v  0.069 m2 v  v  1.45  105 p air/T 15C s

 0.069   v   0.345 V  1 D 5

100 V   V  0.345   9.58m / s m V p 3.6

    1 Also for dynamic similarity, CD FD V 2D2

2    2 FD 2 2  9.58   1  3  1  F   V D      4.75810   2 2 D 100   5   V D  3.6 

101.3103 kg 1000    1.23 ;   1000kg / m3     0.813103 p 287  288 m3 m  1.23

  4.758103  0.813103  3.87 FD

- 115 - Head loss and friction factor

1-D energy equation: p V 2  p V 2  1  1  z   2  2  z   h 1 1  2 2  lT T means ‘total’  2g   2g 

For fully-developed flow through a constant-diameter pipe, V1  V2 p  p 1 2  z  z   h  2 1 lmajor

and if the pipe is horizontal, z1  z2 p  p  p 1 2   h (i)   lmajor

Consider flow in a straight pipe:

Free-body diagram of fluid

x V1 V2

p1 F p2 dx f

Momentum equation in the x-direction:

p1 A  p2 A  Ff  m˙ V2 V1 

F  p  p  f (ii)  1 2  A

wall shearing stress Drag coefficient: f ' dynamic heat [ f’ = Fanning Friction Factor]

- 116 -  f ' w V 2 (iii)  2

4 A Adx D   4 Hydraulic diameter: h (iv) dAw dAw dx w means wetted dA w  P (wetted perimeter) dx For a circular duct of diameter D, and length L,

  4 D 2 4A 4 D      D h P DL L

For a rectangular duct of cross-section as shown: h 4bh 2h 2h Dh    2b  h 1 h 1 AR b b h where AR (Aspect Ratio) = b

 V 2  Adx Ff   w  dAw  f '  4  2  Dh

 V 2  dx Ff  4 f '   A  2  Dh Darcy Friction Factor  f  4 f '

2 dx 1  p  p  4 f ' V A  with (ii)  1 2   2  Dh A

V 2 dx p1  p2   f  2 Dh

p  p   hlmajor  ghlmajor ghlmajor f  1 2    V 2 dx V 2 dx V 2 dx V 2 dx

2 Dh 2 Dh 2 Dh 2 Dh

- 117 - For a pipe of length L and diameter D

gh h f  lmajor  lmajor V 2 L V 2 L (Darcy-Weisbach Equation) 2 D 2g D

hlmajor    hlmajor  hQ,, D, L,v   hRe,  V 2 L  D  2g D

LAMINAR FLOW THROUGH PIPES

1 Fs 

F1 ds r ds  dz R dW

  p    F1  p dA; F2   p  dsdA; Fs   2rds   s  dW   r 2 ds

If flow is steady and uniform at (1) & (2) then the momentum equation in the s-direction gives:

˙  Fs  mV2 V1   F1  F2  Fs  dW cos  0

- 118 -   p  2 pdA   p  dsdA  2rds   r ds cos  0   s 

 p  dA  dsdA  2 ds   dAdscos  0  s  r  dA  r 2  r  dA r

 p  dz   2    0  s r ds

r   p dz  r  d          p   z since p varies only in the s-direction 2   s ds  2  ds 

dU Newton’s law of fluid viscosity:    dr

dU r  d      p   z dr 2  ds 

0 1 R  d  dU   p   z rdr U 0   max 2  ds 

R 1  d  r 2  0 U max   p   z  2  ds  2  0

R 2  d  U max  p   z 4 ds 

dU max R  d   max    p   z dR 2 ds 

More generally,

0 1 R  d  dU   p   z rdr U 2 r  ds 

R 1  d  r 2  0 U   p   z  2  ds  2  r

- 119 - 1  d  U  p   zR 2  r 2  4 ds 

Volumetric flow rate:

R Q  U dA  U 2rdr A 0

2 2 R R  r  d  Q   p   z2r dr 0 4 ds 

R 2  d  2 2 Q   p   z R  r r dr 4 ds 0

2 4 R 2  d  2 r r  Q   p   zR   4 ds  2 4  0

2  d  R 4 R 4  Q   p   z   4 ds  2 4 

R 4  d  Q   p   z  VA 8 ds 

R 4  d   p   z 8 ds  R 2  d   V    p   z R 2 8 ds 

For a horizontal pipe, dz  0  z  const

R 4 dp  Q  and for a length, L, of the pipe, 8 ds

R 4  p D 4  p Q   (Hagen-Poiseuille equation) 8 L 128 L

128 L 128 L D 2    p  Q  V   4 4   D D  4 

- 120 - 32 LV  L  V  p   32   D2  D D

 L      L   1   p  32  V 2    32 V 2    D  VD  D  Re D 

 L  V 2 but  p  gh  gf   lmajor  D 2g

 L  V 2 [ h  f    Darcy-Weisbach Equation] lmajor  D 2g

 L  V 2  L   1   gf    32 V 2    D 2g  D  Re D  f

64 f  Re D

Example Re

Water flows in a pipe 2 cm in diameter and 30 m long; the pipe is running full. Find the head loss when the temperature is 5C and the velocity is 10 cm/s. v  1.51 106 m2 / s (see Fig. A.3, p. 764, Text)

VD 0.1 0.02 Re    1324.5  Re  2000  flow is laminar v 1.51106

64 64  f    0.0483 Re 1324.5

2 2  L  V  30  0.1 2 head loss: hlmajor  f    0.0483   3.7  10 m of water  D 2g  0.02 2  9.81

hlmajor  3.7cmof water

- 121 - Oil at 60F has a specific gravity of 0.92 and kinematic viscosity of 0.0205 ft2/s. Find the horsepower required to pump 50 tons of oil per hour along a pipeline 9 in diameter and one mile long.

3 S.G.oil  0.92   oil  0.92 62.4  57.4lbf / ft v  0.0205 ft 2 / s

m˙ g W˙ 50 tons / hr 50  2000 lbf / hr Volumetric flow rate: Q      g  57.4 lbf / ft 3 57.4 lbf / ft 3

1742.16 ft 3 Q  1742.16 ft 3 / hr   0.484 ft 3 / s 3600 s

2  9     D 2 12 Q 0.484 A      0.4418 ft 2 ; V    1.096 ft / s 4 4 A 0.4418

 9  1.096   VD 12 flow is laminar Re      40.1  2000  v 0.0205

64 64 f    1.596 Re 40.1

2  L  V 2 5280 1.096 head loss: h  f    1.596    209.6 ft of oil lmajor  D 2g 9 2  32.2 12

 Q hlmajor 57.4 0.484 209.6 Horsepower    10.6 h.p. 550 550

Note: 1 h.p.  550 ft lbf / s

Example

Velocity measurements are made in a 30-cm pipe. The velocity at the centre is found to be 1.5m/s, and the velocity distribution is observed to be parabolic. If the pressure drop is found to be 1.9 kPa per 100m of pipe, what is the kinematic viscosity  of the fluid? Assume that the fluid’s specific gravity is 0.90.

- 122 - Velocity distribution is parabolic

flow is fully developed and laminar

V  V  max  0.75 m / s 2

VCL = Vmax = 1.5 m/s

 L  V 2  p  p  D  2g h  f     f    lmajor  D 2g    L V 2

1.9  103  0.3  2  9.81 f     0.0225 0.90  103  9.81100 0.752

64 DV 64 f  since flow is laminar  Re   Re v f

DVf 0.3  0.75  0.0225 m2 v    7.91 105 64 64 s

TURBULENT FLOW IN PIPES

 L  V 2 The Darcy-Weisbach equation: h  f   can still be used to determine the head loss lmajor  D 2g due to friction but f now depends upon the Reynolds number (Re) and the relative roughness  e  (e/D); f  F Re,  see the MOODY DIAGRAM  D

Example

Determine the head loss due to flow of 100 litres/s of glycerine at 20C through a 100m length of 20cm diameter pipe made of cast iron. Rework the problem for the case when the working fluid is water.

S.G.Glycerin  1.26 (see Table A2, p.759, Text)

- 123 - N   1.5 s (see Fig. A.2, p.763, Text) Glycerin/T20C m2

kg 3   998 (see Handout)   Glycerin/T20C  1.26  998  12575.kg / m H 2O / T 20C m3

litres m3 m3 Q  100  100  103  0.1 s s s

 A  0.22  0.0314 m2 4

Q 0.1 V    3.185 m / s A 0.0314

VD 1257.5 3.185 0.2 Re    534  2000  Flow is laminar D  1.5

64 64 f    0.12 Re 534

 L  V 2 100  3.1852 hlmajor  f    0.12   31.0 m of Glycerin  D 2g  0.2  2  9.81

If water is the working fluid, then Ns   1103 (see Fig. A.2, p.763, Text) H 2O / T 20C m2

 VD 998 3.185 0.2 Re    635726  2000  flow is turbulent D  1103

e D  20cm  7.874"  0.0013(see Fig. 8.15 for Cast Iron, p.351) D

 e  f  F Re,   F 6.36  105 , 0.0013  f  0.021(see Moody diagram)  D  

2 L V 2 100 3.185 h  f  0.021     5.43m of water lmajor D 2g  0.2  2  9.81

- 124 - Determine the discharge of water at 20C through a pipe, made of cast iron, with a diameter of 20cm if the head loss in 100m length of pipe is equal to 5.43m.

Re, V and f are unknown

2 1 2gDh  L V lmajor 2 c1 hlmajor  f  V   f  D 2g L f

2gDh where c  lmajor 1 L

D Re  VD  Re  c V where c  v 2 2 v

Solution Procedure:

e (i) Determine D e (ii) Assume a value for f using the Moody diagram. (Locate D and move horizontally across the diagram to the f axis) c V  1 (iii) Using f calculate V from f

(iv) Use V to determine Re from Re  c2V

e (v) Using D and Re, determine f

(vi) Compare f obtained in (v) with f assumed in (ii). If they are the same, STOP. If not, repeat steps (iii) through (vi) until convergence is attained D 2 (vii) After convergence calculate the discharge using Q  V 4 v  1 106 m2 / s (see Fig. A.3, p.764, Text) e  0.0013 for Cast Iron with D  20cm  7.874" (see Fig. 8.15, p.351, Text) D

L = 100 m, D = 20 cm = 0.2m, hlmajor = 5.43m

L V 2 VD h  f , Re  lmajor D 2g v

- 125 - [2 equations with 3 unknowns (Re, V , f )  iteration is needed]

c V  1 f , Re  c2V

2gDhlmajor 2  9.81 0.2  5.43 c    0.462m / s 1 L 100

 V  0.462 f

D 0.2 s c    2  105 2 v 1 106 m

 Re  2  105V

e From the Moody diagram, with  0.0013  f  0.021 D

0.462  V   3.19m / s 0.021

Re  2  105 3.19  6.38  105

5 f  F0.0013, 6.38  10   0.021  f assumed  convergence is attained

 0.22 m3  Q  AV  3.19  0.1 4 s

Re, V , f and D are the unknowns

L V 2 Q 4Q h  f , V   lmajor D 2g A D 2

 L   4Q  1  8LQ 2   h  f       D5    f  c f lmajor D 2 2g  2  3   D   hlmajor g 

c3 c3 is known

- 126 - VD 4Q D 4Q 1 c 4Q Re       4 where c  v D 2 v D 2 D D 4 v c4 is known

Solution Procedure:

(i) Assume a value for f

5 (ii) Calculate D from D  c3 f

e (iii) Calculate D

c (iv) Calculate Re from 4 Re  D

e (v) Using D and Re, determine f

(vi) Compare f obtained in (v) with f assumed in (i). If the two values are approximately the same, STOP. If not, repeat steps (ii) through (v) until convergence is attained Example

Determine the size of a cast iron pipe required to convey 100 litres/s of water at 20C with a head loss of 5.43m in a 100m length of pipe.

Q  100 litres / s  0.1 m3 / s

6 2 hlmajor  5.43 m ; L  100m ; v  1.0  10 m / s

8LQ 2 8100 0.12  c3  2  2  0.01522 hlmajor g 5.43 9.81

5  D  c3 f  0.01522 f

4Q 4 0.1 c    1.27 105 4 v  1106

c 1.27  105  Re  4  D D

Iteration #1

- 127 - Assume f = 0.01

1  D  0.015220.01 5  0.172m  6.772" e 1.27  105  1.453  103 ; Re   7.38  105 D 0.172

 e  f  F , Re  F 1.453 103 , 7.38  105  f  0.022  D   

Iteration #2

Assume f = 0.021

1  D  0.015220.021 5  0.2m  7.874"

1.27  105 e Re   6.35  105 ;  0.0013 0.2 D

 e  f  F , Re  F 0.0013, 6.35 105  0.021  f  D    assumed

 Convergence is attained  D = 0.2m

- 128 -