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.
Rem = 5.5 x 106
Tm = 15o C = Tp  T = 1
pm = 10 atm, pp = 1 atm  p = 10
 p = 2 m ; L = 1/10
FD  f  ,  ,V ,
Choose MLT as fundamental units

ML-3
FD
MLT-2

ML-1T-1
V
LT-1

L
Dimensional matrix
FD


V

L
1
-3
-1
1
1
T
-2
0
-1
-1
0
M
1
1
1
0
0
FD ,  ,  produce
a 3 x 3 matrix with
a non-zero
determinant 
Rank = 3
3 repeaters are needed
choose  , V ,  as repeaters
Check on independence of the dimensions of the repeaters

L
-3
T
0
M
1
V
1
-1
0

1
0
0
 0  dimensions of  , V ,  are independent
- 110 -
Dimensionless groups:
1   1 V 1   1 FD  M 0 L0 T 0  (ML-3 )1 (LT -1 ) 1 (L)  1 MLT -2
M 0 L0T 0  M 1 1 L31  1  1 1T  1 2  1 + 1 = 0  1 = -1 ;
-1 - 2 = 0  1 = -2; -3(-1) + (-2) + 1 + 1 = 0  1 = -2
1 
FD
V 2 2
 2    2 V  2   2   M 0 L0 T 0  (ML-3 ) 2 (LT -1 )  2 (L)  2 ML-1T -1
M 0 L0T 0  M  2 1 L3 2   2  2 1T   2 1  2 + 1 = 0  2 = -1 ;
-2 - 1 = 0  2 = -1; -3(-1) + (-1) + 2 - 1 = 0  2 = -1
2 

V 
1 = f (2) 
  
FD
 f
  CD = f (Re)
2 2
V 
 V 
For dynamic similarity, Re = 1 
  V  
1

 = 1 since Tm = Tp = 15oC and  = f (T)   V 
m 
1
 

1
10

1
   10  
pp
pm
1013
.  10
1013
.

 12.3 kg / m 3 ;  p 

 123
. kg / m 3
RTm 287  288
RTp 287  288
  = 10
Vm  V p 

V = 1 
Re p   p
pp

55
.  10 6  1753
.
 10 5
 39.2 m / s
123
. 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
MLT-2 T-1
d
D
V


L
L
LT-1
ML-3 ML-1T-1
Dimensional matrix
FD

d
D
V


L
1
0
1
1
1
-3
-1
T
-2
-1
0
0
-1
0
-1
M
1
0
0
0
0
1
1
FD ,  , d
produce a 3 x 3
matrix with a nonzero determinant
 Rank = 3
3
repeaters are needed
choose  , D , V as repeaters
Check on independence of the dimensions of the repeaters

L
-3
T
0
M
1
D
1
0
0
V
1
-1
0
 0  dimensions of  , V , D are independent
- 112 -
Dimensionless groups:
1   1 V 1 D  1 FD  M 0 L0 T 0  (ML-3 )1 (LT -1 ) 1 (L)  1 MLT -2
M 0 L0T 0  M 1 1 L31  1  1 1T  1 2  1 + 1 = 0  1 = -1 ;
-1 - 2 = 0  1 = -2; -3(-1) + (-2) + 1 + 1 = 0  1 = -2
1 
FD
V 2 D2
 2    2 V  2 D  2   M 0 L0 T 0  (ML-3 ) 2 (LT -1 )  2 (L)  2 ML-1T -1
M 0 L0T 0  M  2 1 L3 2   2  2 1T   2 1  2 + 1 = 0  2 = -1 ;
-2 - 1 = 0  2 = -1; -3(-1) + (-1) + 2 - 1 = 0  2 = -1
2 

 VD
 3    3 V  3 D  3 d  3 = 3 = 0; 3 = -1 (by inspection)   3 
d
D
 4    4 V  4 D  4   4 = 0, 4 = -1; 4 = 1 (by inspection)   4 
1 = f (2, 3, 4) 
For dynamic similarity, Re = 1, St = 1
V 
  D
1
V
80 1

240 3
Assume Tm = Tp ; pm = pp   = 1 ;  = 1
V  D  1 
V
 
FD
d  D

f
,
,

  CD = f (Re, St, d/D)
V 2 D2
  VD D V 
St : Strouhal number
  V  D
 1 and

D
1
D 1  D  3
3
Dm = D Dp = 3(1.68”) = 5.04”
- 113 -
 
V 1 1 1
  
D 3 3 9
 m   m p 
1
(9000)  1000 rpm
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,
Ca = 1 
( p  p )
v
 
2
V
1 
( p p )   V2
v
(see p. 295 for Cavitation Number)
Model & Prototype are tested in water with  = const   = 1
 ( p  pv )  V2
Also, Fr = 1 for dynamic similarity
V
1
2
1
1
2
g L
 1;  g  1 ( same location)   V   2L
  ( p  pv )   L 
p
p
m
 pvm
p
 pv p
1
( p  pv )  0.05
20
  0.05;

p p  14.7 psia ,
pv p  pv
 0.15 psia
T  45 F
pv m  pv
T 130 F
p m  0.05 p p  pv p   pv m  0.0514.7  0.15  2.22  2.95 psia
- 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
For dynamic similarity,  Re  1 
1
5
D  ;
  DV
1


 
   D V  1

v
vm  v water / T 20C ( assumed )  1  106 m2 / s
 v  0.069
v p  vair / T 15C  145
.  10 5
m2
s
 v 0.069

 0.345
1
D
V 
5
Vm   V V p  0.345 
100
 9.58m / s
3.6
Also for dynamic similarity, CD  
FD
1
V 2 D 2
2
F
D
  V2 2D
p 
 1   FD
2


9.58   1 

2
2
3
   V  D    
   4.758  10  

100
5

  
3.6 

101.3  10 3
kg
 1.23 3 ;
287  288
m
 m  1000kg / m 3    
F  4.758 10 3  0.813 103  3.87
D
- 115 -
1000
 0.813  10 3
1.23
Head loss and friction factor
1-D energy equation:
p1

 1
2
2
p

V1
V
 z1   2   2 2  z 2   hlT
2g
2g
 

T means ‘total’
For fully-developed flow through a constant-diameter pipe, V1  V2
p1  p2

  z2  z1   hl major
and if the pipe is horizontal, z1  z 2
p1  p2


p

 hl major (i)
Consider flow in a straight pipe:
Free-body diagram of fluid
Ff
mg
r
p1
d
x
Momentum equation in the x-direction:
 V2  V1 
p1 A  p2 A  Ff  m
  p1  p2  
Ff
V2
V1
x
A (ii)
wall shearing stress
dynamic heat
[ f’ = Fanning Friction Factor]
Drag coefficient: f ' 
- 116 -
Ff
p2
f '
w
2
V 2
(iii)
Hydraulic diameter: Dh 
Adx
4A
(iv)
4
dAw
dAw
dx
w means wetted
dAw
 P (wetted perimeter)
dx
For a circular duct of diameter D, and length L,
 
4  D 2
4A
4
Dh 
  
D
DL
P
L
For a rectangular duct of cross-section as shown:
h
4bh
2h
2h
Dh 


2b  h 1  h
1  AR
b
where AR (Aspect Ratio) =
b
h
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

with (ii)   p1  p2   4 f '  V 2 A
Dh A
 p1  p2  
f 
f 
V 2 dx
2 Dh
 p1  p2  
V 2 dx
2
Dh

hl major
V 2 dx
2
Dh

 ghl major
V 2 dx
2
Dh

ghl major
V 2 dx
2 Dh
- 117 -
For a pipe of length L and diameter D
ghl major
f 
2
V L
2 D

hl major
V2 L
2g D
(Darcy-Weisbach Equation)
hl major  hQ,  , D, L, v  


 h Re, 
D
V L

2g D
hl major
2
LAMINAR FLOW THROUGH PIPES
z
2
s
F2
Fs
1
F1
Fs

ds
ds
r
R

dz
dW
F1  p dA;

 p 
F2   p 
ds dA;
 s 

Fs   2r ds
 
dW   r 2 ds
If flow is steady and uniform at (1) & (2) then the momentum equation in the s-direction gives:
F
s
 V2  V1   F1  F2  Fs  dW cos  0
m
- 118 -

 p 
pdA   p 
ds dA   2r ds   r 2 ds cos  0
s 

 

 p
 dA 
dsdA    2  ds   dAds cos  0
 r 
s
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

 
Newton’s law of fluid viscosity:


0
dU r  d

   p   z 
dr
2  ds

dU 
U max
1
2
0
1

2
0  U max
U max 
 max  

R
 d

 ds  p   z  rdr
R
2
 d
r 


p  z  
 
 2 0
 ds
d

 ds  p   z 
R2
4
dU max
Rd

   p   z 
dR
2  ds

More generally,

0
U
dU 
1
2
1
0 U 
2

R
r
 d

 ds  p   z  rdr
R
2
 d
r 



p


z




 2 r
 ds
- 119 -
dU
dr
U
1
4
d
2
2 
 ds  p   z R  r 
Volumetric flow rate:
Q   U dA   U 2r dr 
R
A
Q
R
0
R
0
Q
2
4
2
Q
4
Q

 r2  d

  p   z 2r dr 
4  ds

2
d
 R 2
2
  p   z 0 R  r r dr
ds


R
2
r4 
d
 2 r
 
  p   z  R
2
4 0
 ds

4
2  d
R4 
 R


p


z




4  ds
4 
 2
R 4  d

Q
  p   z   V A
8  ds

V 
R 4
8
d

  p   z 
2

 ds
 R d 

p   z 

2
8  ds
R

For a horizontal pipe, dz  0  z  const
R 4 dp
and for a length, L, of the pipe,
Q
8 ds
Q
R 4  p D 4  p
(Hagen-Poiseuille equation)

8 L
128 L
p
128 L
128 L  D 2 

Q

 V 
D 4
D 4
 4 
- 120 -
p 
32  LV
 L  V
 32  
2
 D D
D
  
 L
 L
2  1 
 p  32  V 2 
  32  V 

 D
 D
 VD 
 Re D 


2
 L V
but  p   ghl major   gf  
 D 2 g
[ hl major
2
 L V
 Darcy-Weisbach Equation]
 f 
 D 2 g
2
 1 
 L V
 L
  gf  
 32  V 2 

 D 2 g
 D
 Re D 

f 

f
64
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  151
.  10 6 m2 / s (see Fig. A.3, p. 764, Text)
Re 
VD 0.1  0.02

 1324.5  Re  2000  flow is laminar
v
1.51  10 6
 f 
64
64

 0.0483
Re 1324.5
2
.
 L V
 30  01
 f 
 0.0483
 3.7  10 2 m of water

 D 2 g
 0.02  2  9.81
2
head loss: hl major
hl major  3.7cm of 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.
S .G.oil  0.92   oil  0.92  62.4  57.4lbf / ft 3
v  0.0205 ft 2 / s
Volumetric flow rate: Q 
m g W
50 tons / hr 50  2000 lbf / hr



 g  57.4 lbf / ft 3
57.4 lbf / ft 3
1742.16 ft 3
Q  1742.16 ft / hr 
 0.484 ft 3 / s
3600 s
3
2
9
 
2
D
Q 0.484
12
A
    0.4418 ft 2 ; V  
 1.096 ft / s
4
4
A 0.4418
9
1.096   
VD
 12   40.1  2000 
Re 

flow is laminar
v
0.0205
f 
64
64

 1596
.
Re 401
.
2
. 
5280 1096
 L V
 1596
.


 209.6 ft of oil
 f 
9
 D 2 g
2  32.2
12
2
head loss: hl major
Horsepower 
 Q hl major
550

57.4  0.484  209.6
 10.6 h. p.
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 
Vmax
 0.75 m / s
2
VCL = Vmax = 1.5
m/s
2
p
 p  D  2g
 L V

 f 
hl major  f  
 
 D 2 g

  L V 2
19
.  103
 0.3  2  9.81
f 
 0.0225


0.90  103  9.81  100  0.75 2
f 
64
DV 64

since flow is laminar  Re 
Re
v
f
2
DVf
0.3  0.75  0.0225
5 m
v

 7.91  10
64
64
s
TURBULENT FLOW IN PIPES
2
 L V
The Darcy-Weisbach equation: hl major  f  
can still be used to determine the head loss
 D 2 g
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  126
. (see Table A2, p.759, Text)
- 123 -
 Glycerin / T  20C  15
.
 H O / T  20C  998
2
Ns
(see Fig. A.2, p.763, Text)
m2
kg
(see Handout)   Glycerin/ T 20C  126
.  998  12575. kg / m3
3
m
3
litres
m3
3 m
Q  100
 100  10
 01
.
s
s
s
A
V 

0.22
4
Q
0.1

 3.185 m / s
A 0.0314
Re D 
f 
 0.0314 m 2
 V D 1257.5  3.185  0.2

 534  2000  Flow is laminar

1.5
64
64

 012
.
Re 534
2
 100  3.185
 L V
 0.12
 31.0 m of Glycerin
hl major  f  

 D 2 g
 0.2  2  9.81
2
If water is the working fluid, then
Ns
 H 2O / T  20C  1  10 3 2 (see Fig. A.2, p.763, Text)
m
Re D 
 V D 998  3.185  0.2

 635726  2000  flow is turbulent

1  10 3
D  20cm  7.874" 
e
 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


. 
LV2
 100   3185
 f
 0.021  
 5.43m of water

 0.2  2  9.81
D 2g
2
hl major
- 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
hl major  f
LV2
D 2g
 V 
2 gDhl major
L
where c1 
2 gDhl major
Re  V D
 Re  c 2V where c2 
f

1
2

c1
f
L
v
D
v
Solution Procedure:
(i) Determine e
D
(ii) Assume a value for f using the Moody diagram. (Locate e
D
and move horizontally across
the diagram to the f axis)
(iii) Using f calculate V from V  c1
f
(iv) Use V to determine Re from Re  c2V
(v) Using e
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
6
2
v  1  10 m / 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
hl major  f
LV2
VD
, Re 
D 2g
v
- 125 -
[2 equations with 3 unknowns (Re, V , f )  iteration is needed]
V 
c1
f , Re  c2V
2 gDhl major
c1 
L
 V  0.462
c2 

2  9.81  0.2  5.43
 0.462m / s
100
f
D
0.2
s

 2  105
6
v 1  10
m
 Re  2  105V
From the Moody diagram, with
V 
e
 0.0013  f  0.021
D
0.462
 319
. m/ s
0.021


Re  2  105 319
.   6.38  105


f  F 0.0013, 6.38  105  0.021  f assumed  convergence is attained
 Q  AV 
 0.22
4
3.19  0.1 m
3
s
Re, V , f and D are the unknowns
hl major
LV2
Q 4Q
 f
, V  
D 2g
A D 2
 hl major
 L   4Q  1
 f   2 
 D   D  2 g
 8LQ 2 
 f  c3 f
 D 
 hl major g 2 


5
c3
c3 is known
- 126 -
Re 
4Q
V D 4Q D 4Q 1 c4
where c 4 

 
 
2
2
v
v
D v D D D
c4 is known
Solution Procedure:
(i) Assume a value for f
(ii) Calculate D from D 5  c3 f
(iii) Calculate e
D
(iv) Calculate Re from Re  c4
D
(v) Using e
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 m 3 / s
hl major  5.43 m ;
v  10
.  10 6 m2 / s
L  100m ;
8LQ 2
8  100  0.1

 0.01522
2
hl major g
5.43  9.81   2
2
 c3 
 D5  c3 f  0.01522 f
c4 
4Q
4  0.1

 1.27  10 5
6
v   1  10
 Re 
c4 127
.  105

D
D
- 127 -
Iteration #1
Assume f = 0.01
 D  0.015220.01
1
5
 0172
. m  6.772"
e
127
.  105
 1453
.
 10 3 ; Re 
 7.38  105
D
0172
.
e

f  F  , Re  F 1453
.
 10 3 , 7.38  105  f  0.022
D



Iteration #2
Assume f = 0.021
 D  0.015220.021
Re 
1
5
 0.2m  7.874"
127
.  105
 6.35  105 ;
0.2
e
 0.0013
D
e

f  F  , Re  F 0.0013, 6.35  105  0.021  f assumed
D



 Convergence is attained  D = 0.2m
- 128 -