Water is flowing at a rate of 0.25m3/s, and it is assumed that hL=1.5V2/2g from the
reservoir to the gage, where V is the velocity in the 30-cm pipe. What power must the
pump supply?
7.33
p = 100 kPa
Assumptions
Elevation =
10m
Reservoir >> suction
pipeV1  0  flow is
steady
Flow is turbulent 
 1   2  10
.
D = 30
cm
1
Elevation = 6m
water
o
T = 10 C
Given
40
cm
Elevation = 2m
Flow of water   = const
1-D energy equation with a
pump present:
V1 2
p2
V22
hp 
1
 z1 
2
 z2  hlt

2g

2g
p1
hp = pump head
p1g = 0, p2g = 100 kPa, z1 = 6m, z2 = 10m,  = 9.81 kN/m3
hp 
V2
100 V22
125
.

 (10  6)  15
. 2  14.2 
V22
9.81 2 g
2g
9.81
Q  V2 A2  V2 
Q
0.25

A2   (0.3) 2 


 4 
V2  3537
.
m / sec
h p  14.2 
2
125
.
(3537
. ) 2  15.79 m
9.81
- 84 -
pump power: W p   Q hp  9.81(0.25)(1579
. )
W p  38.72 kW
7.14
Water flow from a pressurized tank as shown. The pressure in the tank above the water
surface is 100 kPa gage, and the water surface level is 10m above the outlet. The water
exit velocity is 9m/s. The head loss in the systerm varies as
hL  K L
V2
2g
where KL is the head-loss coefficient. Find the value for KL.
Assumptions
Air
under
pressure
1
Tank >>
pipeV1  0
water
Flow is turbulent 
 1   2  10
.
d
Partly open
valve
1-D energy
equation:
V1 2
p2
V22
1
 z1 
2
 z2  hlt

2g

2g
p1
p1g = 100 kPa, p2g = 0, z1 = 10m,  = 9.81 kN/m3
100
92
92
 10 
 KL
9.81
2  9.81
2  9.81
1981
.
81

(1  K L )
9.81 2  9.81
1  KL 
2  1981
.

81
K L  4.89  1  389
.
- 85 -
2
7.19
In the figure for Probs. 7.14 and 7.15, suppose that the reservoir is open to the
atmosphere at the top. The valve is used to control the flow rate from the reservoir. The
head loss across the valve is given as
V2
hL  10
2g
where V is the velocity in the pipe. The cross-sectional area of the pipe is 5 cm2. The
head loss due to friction in the pipe is negligible. The elevation of the water level in the
reservoir above the pipe outlet is 10m. Find the discharge in the pipe.
Patm
Assumptions
Tank >> pipeV1  0
1
.
Flow is turbulent   1   2  10
water
d
2
Partly open
valve
1-D energy equation:
V1 2
p2
V22
1
 z1 
2
 z2  hlt

2g

2g
p1
p1g = p2g = 0, z1 = 10m,  = 9.81 kN/m3
10 
V2 
V22
V2
 10 2
2g
2g
 10  2  9.81  11V22
10  2  9.81
 4.223 m / s
11
Discharge: Q  V2 A2  4.223  5  10 4  2.11 10 3 m 3 / sec
7.36 A small-scale hydraulic power system is shown. The elevation difference between the
reservoir water surface and the pond water surface downstream of the reservoir, H, is 10 m. The
velocity of the water exhausting into the pond is 5 m/s, and the discharge through the system is
1m3/s. The head loss due to friction in the penstock is negligible. Find the power produced by
the turbine in kilowatts.
- 86 -
1
Assumptions
Reservoir >> pipeV1  0  flow is
steady
H
turbin
e
.
Flow is turbulent   1   2  10
z
Given
2
Flow of water   = const
1-D energy equation with turbine present:
p1

1
V1 2
p
V2
 z1  hT  2   2 2  z2  hlt
2g

2g
hT = turbine head
p1g = p2g = 0, z1 = 10m, V2 = 5m / s ,  = 9.81 kN/m3
hT  z1 
V22
(5) 2
 10 
 8.726 m
2g
2  9.81
Turbine power: WT   Q hT  9.81(1)(8.726)
WT  85.6 kW
7.25
For this system, point B is 10m above the bottom of the upper reservoir. The head loss
from A to B is 2V2/2g, and the pipe area is 10-4m2. Assume a constant discharge of 7
x 10-4m3/s. For these conditions, what will be the depth of water in the upper reservoir
for which cavitation will begin at point B? Vapor pressure = 1.23 kPa and atmospheric
pressure = 100 kPa.
- 87 -
B
A
z
Water
T = 20o
C
C
D
Assumptions
Reservoirs >> pipeV A  VC  0  Flow is steady
.
Flow is turbulent   1   2  10
Given
Flow of water   = const
1-D energy equation:
pA

 A
V A2
p
V2
 z A  B   B B  z B  hlt
2g

2g
pAa = 100 kPa, pBa = 1.23 kPa, zB = 10m,  = 9.81 kN/m3
zA 
zA 
.  100 VB2
123

9.81
2g
 10  2
VB2
15
.
 10.07  10 
VB2
2g
9.81
15
.
V B2  0.07
9.81
- 88 -
Continuity: Q  VB AB  VB 
Q 7  10 4

AB
10  4
VB  7 m / s
zA 
15
. 7
2
9.81
 0.07  7.42 m
7.38 Neglecting head losses, detedmine what power the pump must deliver to produce the
flow as shown. Here the elevations at points A, B, C, and D are 40 m, 65 m, 35 m, and 30 m,
respectively. The nozzle area is 30 cm2.
B
Assumptions
Tank >> pipeV A  0  Flow is
steady
Flow is turbulent 
 A   C   B  10
.
A
water
Given
Flow of water   = const

 A
C
D
1-D energy equation with pump present:
pA
nozzle
V A2
p
V2
 z A  h p  B   B B  z B  hlt
2g

2g
(hp = pump head)
pAg = pBg = 0 ; zA = 40m, zB = 65m,  = 9.81 kN/m3 ;
VB = 0 (maximum height of fluid trajectory)
 hp = zB - zA = (65 - 40) m = 25 m
Bernoulli’s equation along a streamline from C to B:
VC2
p B VB2

 gzC 

 gz B

2

2
pC
- 89 -
VC2  2 g ( z B  zC )  VC  2 9.81 (65  35)  24.26 m s
Continuity:
Q  VC AC  24.26  30  10 4  0.0728 m3 / sec
Pump power:
W p   Q hp
W p  9.81(0.0728)(25)  17.85 kW
Note: 1 kW = 1.341 hp
Problem:
As shown in the figure, the pump supplies energy to the flow such that the upstream pressure
(12-in. pipe) is 10 psi and the downstream pressure (6-in. pipe) is 30 psi when the flow of water
is 3.92 cfs. What horsepower is delivered by the pump to the flow?
pA
pB
pump
out
in
Assumptions

 0 & m in  m out  m
t
Assume flow is uniform at inlet and outlet
Assume flow is steady 
Neglect friction  uout = uin  Q  0
z out  zin
 = const
Conservation of mass: m in  m out
  Vin Ain   Vout Aout
Vin Ain Vout Aout  V  3.92 ft 3 / sec
- 90 -
Vin 
3.92
  (1) 


 4 
2
 4.99 ft / sec; Vout 
3.92
 19.96 ft / sec
  (0.5) 2 


 4 
W ss  0 by choice of the c.v. [V is zero at walls and  = 0 at the inlet and outlet]
Energy equation for the c.v.


p  
Q  W ss  W s  Wother 
e

dV

e


A     V  dA
 t V
W s 

p  
e


    V  A
inlet 
outlet
p
V2
p
V2
W s   Q B  out  A  in 
2

2 

W s 
 30  10(144) (19.96) 2  (4.99) 2 
62.4
(3.92) 


62.4
32.2
2
32
.
2


W s  11,289.6  1418.7
ft  lbf
ft  lbf 12,708.3
 12,708.3

hp  231
. hp
sec
sec
550
1 hp = 550 ft-lbf/sec
If the discharge of water is Q = 0.06 m3/s, what are the pressures at A and B? Is the machine a
pump or a turbine? Neglect losses.
z1 = 2 m
z2 = 4 m
D = 30 cm
d = 15 cm
A
water
D
z2
d
T = 10o C
d
machin
e
- 91 -
z1
B
z
No head loss between B and outlet  Bernoulli can be applied between B and the outlet:
pB


VB2
p
V2
 gz B  out  out  gzout
2

2
VB  Vout 
Q
Apipe
pout  patm ; zout  0
;
pB = pout - gz1 = patm - 1000(9.81)(2)
pBg = -19.62 kPa
Apply Bernoulli between A and B:
pA


VA 
VB 
V A2
p
V2
 gz A  B  B  gz B
2

2
Q
 D 


 4 
2
Q
 d 


 4 
2


p Ag  19.62 
p Ag  19.62 
0.06
  0.30 2 




4


0.06
2
  015
.  




4


g
1000
 0.849 m s
 3.395 m s
  z1   z2
(3.395)
 z   
1
2
 (0.849) 2

2 (1000)


2
2
1000(9.81)
 4  1000 (3.395)  (0.849)  53.5 kPa
1000
2(1000)
Neglect losses  Q  0  uout = uin
 flow is uniform at inlet and outlet
Assumptions
reservoir >> delivery pipe  Vin  0

0
 flow is steady 
t
 = const
- 92 -
Energy equation for the c.v.


p  
Q  W ss  W s  Wother 
e

dV

e


A     V  dA
 t V
W ss  0 by choice of the c.v.
2
 pout Vout

pin Vin2

Ws   Q

 gzout 

 gzin 
2

2
 

 (3.395) 2

W s  1000 (0.06) 0 
 0  0  0  9.81(2  4)
2


W s  31858
. W  3.2 kW
W s  3.2 kW ; W s  0  Machine is a turbine
Flow through a 90o reducing elbow
z
2
Assumptions
g
y
x
C
V
Flow is steady
Fluid is incompressible
u & p are uniform at ‘1’ and ‘2’
flo
w
1
Energy equation for the c.v.


p  
Q  W ss  W s  Wother 
e

dV

e


A     V  dA
 t V
W ss  0 by choice of the c.v.
Wother  0
- 93 -
p2 p1
V22  
V12  


 ( z 2  z1 )  
 Q  Ws  m (u2  u1 )  m (  )  mg
 V2  dA2  
 V1  dA1


2
2
A2
A1
 p
 p

V2  
V2  
W s  m   2  gz2    1  gz1     2  V2  dA2   1  V1  dA1  m (u2  u1 )  Q
2
 
  A2 2
 
A1
Velocity is not uniform across ‘1’ and ‘2’; this is the case in all viscous flows. An average
velocity can, however, in conjunction with a KINETIC ENERGY FLUX COEFFICIENT ()
(also called KINETIC ENERGY CORRECTION FACTOR), be used.
V2  
V2  
V2
 V  dA   
 V  dA   m
2
2
2
A
A

 = 1 for uniform flow;  > 1 for non-uniform flow
 = 2 for fully developed laminar flow;   1.05 for turbulent flow.
 p
 p


V2
V2
Q 
W s  m   2   2 2  gz2    1   1 1  gz1    m  (u2  u1 )  
2
2
m 
 


 
 p
 
W s   p2
V2
V2
Q 
 
  2 2  gz2    1   1 1  gz1     (u2  u1 )  
m
2
2
m 
 
 
 
 p

W s  p1
V2
V2
    1 1  gz1    2   2 2  gz2   (u2  u1 )  q

m
2
2

  

 p
 1
W s  p2
V2
V2

  2 2  z2    1   1 1  z1   (u2  u1 )  q 

mg
2g
2g

 
 g
Pump
 p1

p

V2
V2
   1 1  z1   h p   2   2 2  z 2   hlT
2g
2g


 

hp = pump head ; hlT = head loss term
- 94 -
hp 
 W s
W
 s
m g
m g

 h p  Q  h p mg
(since W s  0 for pump)  W s
pump
Turbine
 p1

p

V2
V2
   1 1  z1   hT   2   2 2  z2   hlT
2g
2g




hT = turbine head
hT 
W s

mg
 W s

 hT  Q  hT mg
turbine
Head loss in an abrupt expansion
Continuity:
V1 A1  V2 A2

A1 
V2
A2
V1
Momentum equation:
( p1  p2 ) A2   Q V1  V2

1
( p1  p2 )



2

V A
Q
V2  V1  2 2 V2  V1   V2 V2  V1 
A2
A2
1-D energy equation:
p1

1
V1 2
p
V2
 z1  2   2 2  z2  hlt
2g

2g
z1 = z2
- 95 -

 hle 
p1  p 2

V1 2  V22

2g
1  2  1
hle 
hle 
V2  V1  V  V   V2  V1  V  V1  V2  
V2 V2  V1 

1
2
 2

g
2g
g
2


V
 V1 
2g
2
2
Abrupt Contraction
Vena Contracta
Bend in a pipe
flo
w
- 96 -
Examples for the application of Rayleigh’s Theorem
1. Period of a simple pendulum
t  f  l , g , m
t  const  l  m  g 
Choose LTM

 T  L M  LT 2


Dimensional homogeneity: L0 M 0 T  L  M  T 2
1
 1 1
   0;  2  1     ;     0         
 2 2
2
l
g
the constant must be determined experimentally (const = 2)
 t  const
Note: Rayleigh’s method can be applied without difficulty when the number of independent
variables does not exceed the available number of fundamental units. However, when the
number of fundamental units, r, is less than the number of independent variables, p, then (p-r)
exponents must be chosen arbitrarily. See example #2
2. Consider pressure losses per unit length in pipes due to friction:
p
 f d ,V , v , 
l
 p
 const d  V  v   
l
Choose LTM

 MLT 2 L2 L1 L LT 1
 L T  ML 

2
1 
3 
ML2T 2 L    2 3 T    M 
- 97 -
Dimensional homogeneity:
  1;      2    2  
    2  3  2    2     2  3  2    1  

2
p
 v  V

 const d  1 V 2  v    const  
l
 Vd  d
const must be determined experimentally and  must be chosen arbitrarily
3.
p
 f  d , V , v ,  , e
l
e = roughness
p
 const d V  v    e 
l
Choose LTM

 ML2 T 2  L LT 1
  L T   ML 

1 
2
3 
L  L   2 3  T    M
Dimensional homogeneity:
  1;      2    2   ;
  2     2  3    2        1
   1    
p
 const d  1  V  2   v   e 
l
p
 const
l

2
 v  V


 dV  d
 e
 
d

const must be determined experimentally and  and  must be chosen arbitrarily
4. Rate of flow, Q, of a fluid of viscosity, , through a tube of radius, r, and length, l,
under a pressure difference, p.
Q  f  p, l ,  , r 
Q  const p  l    r 
Choose LTM
- 98 -

L3 T 1  MLT 2 L2



L ML1T 1


L
Dimensional homogeneity: M 0 L3T 1  M   L     T 2 
1 
     0     ;  2    1   
   1;
2
  1;          3  1    1    3      3
  3 
Q  const 
p

 l   r 3   const 
l
r3 
 t
p

 = chosen arbirtarily
const determined experimentally
Dimensional Analysis Procedure using the Buckingham Pi Theorem:
1. List all variables which influence a given problem
2. Choose a set of fundamental dimensions e.g. MLT or FLT
3. List the dimensions of all the variables in terms of the fundamental dimensions
4. Determine the rank of the dimensional matrix
5. Choose from the independent variables a number (equal to the rank of the dimensional
matrix) of repeating variables (also known as repeaters). Note that the dependent variable
cannot be chosen as a repeating variable
6. Check on the dimensional independence of the chosen repeating variables
7. Set up dimensional equations by combining the repeating variables with each of the
remaining variables, including the dependent one, in turn, to form dimensionless (or -) groups.
Dimensional homogeneity must be observed hereby.
p




c
a
e
g
Quantity
Pressure
Viscosity (dynamic)
Viscosity (kinematic)
surface tension
density
velocity
acceleration
roughness (absolute)
acceleration (due to gravity)
MLT
ML-1T-2
ML-1T-1
L2T-1
MT-2
ML-3
LT-1
LT-2
L
LT-2
- 99 -
FLT
FL-2
FL-2T
L2T-1
F/L
FT2L-4
LT-1
LT-2
L
LT-2
F

A
V

Q
m
hl
N

T
H
E
P
E
Force
shear stress
Area
Volume
specific weight
discharge
(volumetric flow rate)
mass flow rate
head loss
rpm
angular speed
Torque
Impulse and Momentum
Engergy and Work
Power
Modulus or elasticity
MLT-2
ML-1T-2
L2
L3
ML-2T-2
L3T-1
F
FL-2
L2
L3
F/L3
L3T-1
MT-1
L
T-1
T-1
ML2T-2
MLT-1
ML2T-2
ML2T-3
ML-1T-2
FL-1T
L
T-1
T-1
FL
FT
FL
FLT-1
F/L2
ball
Problems
1. 7.19 The sketch shows an air jet discharging vertically.
Experiments show that a ball placed in the jet is suspended in a
stable position. The equilibrium height of the ball in the jet is
found to depend on D, d, V, , , and W, where W is the
weight of the ball. Dimensional analysis is suggested to
correlate experimental data. Find the Pi parameters that
characterize this situation.
2. The instrument package for a moon landing is
encased in a viscoelastic liquid as shown. The
acceleration, a, of the package is expected to depend
on , a dimension of the package, m, the mass of the
package, E, the modulus of elasticity of the liquid, ,
the liquid viscosity, and V, the impact speed.
Dimensional analysis is suggested to help design
suitable experiments. Determine the dimensionless
parameters that result.
- 100 -
h
d
V
instrume
nt
package
cushio
n
liquid
D
casting
V
Moon
surface
3. 7.6 Measurements of the liquid height upstream from an obstruction placed in an open
channel flow of a liquid can be used to determine volume flow rate. (Such obstruction,
designed and calibrated to measure rate of open-channel flow, are called weirs.) Assume the
volume flow rate over a weir, Q, is a function of upstream height, h, gravity, g, and channel
width, b. Use dimensional analysis to develop an expression for Q.
4. 7.8 Capillary waves are formed on a liquid free surface as a result of surface tension. They
have short wavelengths. The speed of a capillary wave depends on the surface tension, ,
wavelength, , and liquid density, . Use dimensional analysis to express the wave speed as a
function of these variables.
5. 7.13 The vorticity, , at a point in an axisymmetric flow field is thought to depend on the
initial circulation, 0, the radius, r, the time, , and the fluid kinematic viscosity, . Find a set of
dimensionless parameters suitable for organizing experimental data.
Solution to #1
h = f (D, d, V, , , W)
h D d V   W
MLT
h
L
D
L
d
L
V
LT-1


ML-3 ML-1T-1
W
MLT
Dimensional matrix
h
D
d
V


W
L
1
1
1
1
-3
-1
1
T
0
0
0
-1
0
-1
-2
M
0
0
0
0
1
3
1
At least one 3 x 3
determinant is
nonzero 
Rank = 3
Choose D, V, , as repeaters
Check on independence of the dimensions of the repeaters
- 101 -
L
1
1
-3
D
V

=
T
0
-1
0
M
1
0
1
 0  dimensions of the repeaters are independent
-groups
 1  D  1 V 1   1 h   1 
h
D
(obtained by inspection)
 2  D 2 V 2   2 d   2 
d
D
 3  D3V 3   3 

L0T 0 M 0  L 3 LT 1
 ML 
3
3  3
ML1T 1  L 3   3 3 3 1T   3 1 M  3 1
   3  1  0   3  1;  3  1  0   3  1
 3   3  3 3  1  0   3  1  3 3   3  1  3 1   1  1
 3 

 VD
 4  D 4 V  4   4 W

L0T 0 M 0  L 4 LT 1
 ML 
4
3  4
MLT 2
L0T 0 M 0  L 4   4 3 4 1 T   4 2 M  4 1
 4   4  3 4  1  0
  4  2  0   4  2,  4  1  0   4  1
 4  3 4   4  1  3 1   2  1  2
4 
W
V 2D2
 1  f  2 ,  3 ,  4 
- 102 -
d
h

W 

 f  ,
,
2
2 
D
D

VD

V
D


2. a = f (, m, E, , V)
a  m E  V
Choose MLT

L
a
LT-2
m
M

ML-1T-1
E
ML-1T-2
V
LT-1
Dimensional Matrix
a
l
m
E

V
L
1
1
0
-1
-1
1
T
-2
0
0
-2
-1
-1
M
0
0
1
1
1
0
At least a 3 x 3
determinant is
nonzero
Rank = 3
repeaters
Choose , m, V as
Check on the independence of the dimensions of the repeaters
=

m
V
L
1
0
1
T
0
0
-1
M
1
1
0
 0  dimensions of the repeaters are independent
-groups
1  1 m 1V  1 a

L0T 0 M 0  L1 M 1 LT 1

1
LT 2  L0T 0 M 0  L1  1 1 T  1 2 M 1
  1  0;   1  2  0   1  2;  1   1  1  0   1   1  1   2  1  1
- 103 -
1 
a
V2
 2  l 2 m 2V  2 E



2
L0T 0 M 0  L 2 M  2 LT 1 ML1T 2  L0T 0 M 0  L 2  2 1T  2  2 M  2 1
  2  2  0   2  2;  2  1  0   2  1;
 2   2  1  0   2  1   2  1   2  3
3 E
2 
mV
 3  3 m 3V  3 

L0T 0 M 0  L 3 M  3 LT 1

3
ML1T 1  L0T 0 M 0  L 3  3 1T  3 1 M  3 1
  3  1  0   3  1;  3  1  0   3  1;  3   3  1  0 
 3  1   3  1   1  2
3  2

mV
1  f  2 ,  3  
 3 E 2  
a

 f 
,
mV
 mV mV 
4. V = f (, , )
V   
Choose MLT


MT-2 L
V
LT-1

ML-3
Dimensional Matrix
V



L
1
0
1
-3
T
-1
-2
0
0
M
0
1
0
1
At least a 3 x 3 determinant
is nonzero  Rank = 3
- 104 -
Choose ,  and  as repeaters
Check on the independence of the dimensions of the repeaters



=
L
0
1
-3
T
-2
0
0
M
1
0
1
 0  dimensions of the repeaters are independent
1   1 1   1V

L0T 0 M 0  MT 2

1
2 1  1  0   1 

L1 ML3

1
LT 1  L0T 0 M 0  L1 3 1 1T 21 1 M 1  1
1
1
;  1   1  0   1   1  ;  1  3 1  1  0   1  3 1
2
2
1
 1
  1  3   1 
 2
2

V

1 
 1  const  V 

 const or


 V  const

5.  = f ( 0, r, , )
 0 r  
Choose MLT

T-1
0
r
2 -1
LT L

T

L2T-1
Dimensional Matrix

0
r

v
L
0
2
1
0
2
T
-1
-1
0
1
-1
M
1
0
0
0
-
all 3 x 3 determinants are
zero
at least one 2 x 2
determinant is nonzero
 Rank = 2
- 105 -
Choose r and  as repeaters
Check on the independence of the dimensions of the repeaters
=
L
1
0
r

T
0
1
0  dimensions of the repeaters are independent
-groups
1  r 1 1    
(by inspection)
 2  r  2   2 0
L0T 0  L 2 T  2 L2T 1  L0T 0  L 2  2T  2 1
  2  2  0   2  2;  2  1  0   2  1
2 

r2
0
 3  r  3  3 v 

r2
v (by inspection)
 

 1  f   2 ,  3     f  2 0 ,
v
r
r2 
DIMENSIONAL ANALYSIS OF A GENERAL FLOW PROBLEM
1. Variables should include such fluid properties as:
density
surface tension
compressibility
viscosity
gravitaional effect
compressibility is most conveniently expressed in terms of its inverse:
K

 dp
dp

dV
 d 
V
 

(Bulk modulus of elasticity)
- 106 -
2. Variables should also include the geometry
two linear dimensions are used :
 (length of pipe in pipe flow, or chord width in flow around an airfoil)
d (diameter of pipe or thickness of airfoil)
3. The velocity is used to characterize the mass flow rate or volumetric flow rate
4. Main performance (i.e., dependent) variable
p-pipe flow; drag (or resistance) or lift in external flows
p = f (V, , d, , , K, , g)
Number of variables: n = 9
Choose MLT as fundamental units
p
ML-1T-2
V
LT-1

L
d
L


-3
ML ML-1T-1

g
MT-2 LT-2
Dimensional Matrix
p
V
l
d


K

g
L
-1
1
1
1
-3
-1
-1
0
1
T
-2
-1
0
0
0
-1
-2
-2
-2
M
1
0
0
0
1
1
1
1
0
At least one 3 x 3
determinant is nonzero 
Rank = 3 = m
Choose v, d,  as repeaters
- 107 -
K
ML-1T-2
Check on the independence of the dimensions of the repeaters
=
V
d

L
1
1
-3
T
-1
0
0
M
0
0
1
0  Dimensions of the repeaters are independent
Number of dimensionless groups: N = (n-m) = 6
 groups

1   1 d 1V  1 p  M 0 L0T 0  ML3

1

L1 LT 1

1
ML1T 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   1   1  1  0
3 1   1  2  1  0   1  0
1 
pressure forces
p
 (EULER NUMBER, E u 
)
2
inertia forces
V
 2    2 d  2 V  2    2   2  0,  2  1 (by inspection)
2 

d
-- implies that shape is a controlling factor

 3    3 d  3V  3   M 0 L0T 0  ML3

3

L 3 LT 1

3
ML1T 1
M 0 L0T 0  M 3 1 L33  3  3 1T  3 1
  3  1  0   3  1;   3  1  0   3  1;  3 3   3   3  1  0
3 1   3  1  1  0   3  1
3 

 Vd
= (reciprocal of the REYNOLDS NUMBER, Re 
- 108 -
inertial forces
)
viscous forces


4
 4    4 d  4 V  4 K  M 0 L0T 0  ML3


4
L 4 LT 1
ML1T 2
M 0 L0T 0  M  4 1 L3 4   4  4 1T  4 2
  4  1  0   4  1;   4  2  0   4  2;  3 4   4   4  1  0
3 1   4  2  1  0   4  0
4 
inertial forces
K
= (reciprocal of the MACH NUMBER, Ma 
)
2
elastic forces
V

 5    5 d  5 V  5   M 0 L0T 0  ML3

5

L 5 LT 1

5
MT 2
M 0 L0T 0  M 5 1 L35  5  5 T  5 2
  5  1  0   5  1;   5  2  0   5  2;  3 5   5   5  0
3 1   5  2  0   5  1
5 

d V
2
= (reciprocal of the WEBER NUMBER, We 

 6    6 d  6 V  6 g  M 0 L0T 0  ML3

6

L 6 LT 1

6
inertial forces
)
surfacetension forces
LT 2
M 0 L0T 0  M  6 L3 6   6  6 1T  6 2
  6  0;   6  2  0   6  2;  3 6   6   6  1  0
30   6  2  1  0   6  1
6 
dg
inertia forces
= reciprocal of the FROUDE NUMBER, Fr 
)
2
gravity forces
V
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