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7/27/2019 Section 25 30
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7.33 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?
Assumptions
Reservoir >> suction pipeV1 0 flow is steady
Flow is turbulent 1 2 1 0= .
Given
Flow of water = const
1-D energy equation with a pump present:
hp V
gz
p V
gz hp lt+ + + = + + +
1
1
1
2
1
2
2
2
2
22 2
hp = pump head
p1g = 0, p2g = 100 kPa, z1 = 6m, z2 = 10m, = 9.81 kN/m3
hV
g
V
gVp = + + + = +
100
9 81 210 6 15
214 2
125
9 81
2
2
2
2
2
2
.( ) . .
.
.
- 84 -
Elevation = 10m
Elevation = 2m
Elevation = 6m
40 cm
D = 30 cm
p = 100 kPa
T = 10o C
water
1
2
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===
4
)3.0(
25.02
2
222A
QVAVQ
V m2
3537= . / sec
h mp = + =14 21 25
9 813537 15 79
2.
.
.( . ) .
pump power: . ( . )( . )W Q hp p= = 9 81 0 25 15 7 9
.W kWp = 3872
7.14 Water flow from a pressurized tank as shown. The pressure in the tank above the watersurface 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
h KV
gL L=
2
2
where KL is the head-loss coefficient. Find the value for KL.
Assumptions
Tank >> pipeV1 0
Flow is turbulent 1 2 10= .
1-D energy equation:
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Partly open valve
d
Air underpressure
water
1
2
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p V
gz
p V
gz hlt
1
1
1
2
1
2
2
2
2
22 2
+ + = + + +
p1g = 100 kPa, p2g = 0, z1 = 10m, = 9.81 kN/m3
100
9 8110
9
2 9 81
9
2 9 81
2 2
. . .+ =
+
KL
1981
9 81
81
2 9 811
.
. .( )=
+KL
12 1981
814 89 1 389+ =
= =K KL L
.. .
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
hV
gL = 10
2
2
where V is the velocity in the pipe. The cross-sectional area of the pipe is 5 cm2. Thehead 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.
Assumptions
Tank >> pipeV1 0
- 86 -
Partly open valve
d
Patm
water
1
2
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Flow is turbulent 1 2 10= .
1-D energy equation:
p V
gz
p V
gz hlt
11
1
2
12
22
2
22 2
+ + = + + +
p1g = p2g = 0, z1 = 10m, = 9.81 kN/m3
102
102
10 2 9 81 1122
2
2
2
2= + =V
g
V
gV.
V m s210 2 9 81
11
4 223= =. . /
Discharge: sec/1011.2105223.4334
22 mAVQ ===
7.36 A small-scale hydraulic power system is shown. The elevation difference between thereservoir 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 bythe turbine in kilowatts.
Assumptions
Reservoir >> pipeV1 0 flow is steady
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turbine
H
2
1
z
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Flow is turbulent 1 2 10= .
Given
Flow of water = const
1-D energy equation with turbine present:
p V
gz h
p V
gz hT lt
1
1
1
2
1
2
2
2
2
22 2
+ + = + + + +
hT = turbine head
p1g = p2g = 0, z1 = 10m, V = 5m / s2 , = 9.81 kN/m3
h zV
gmT = = =1
2
2 2
210
2 9 818726
(5)
..
Turbine power: . ( )(8. )W Q hT T= = 9 81 1 726
.W kWT = 85 6
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 atmosphericpressure = 100 kPa.
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z
D
B
A
C
WaterT = 20o C
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Assumptions
Reservoirs >> pipeV VA C= 0 Flow is steady
Flow is turbulent 1 2 1 0= .
Given
Flow of water = const
1-D energy equation:
p V
gz
p V
gz hA A
AA
BB
BB lt
+ + = + + +
2 2
2 2
pAa = 100 kPa, pBa = 1.23 kPa, zB = 10m, = 9.81 kN/m3
( )z
V
g
V
gV
A
B B
B=
+ + + = + +
123 100
9 81 210 2
21007 10
15
9 81
2 2
2.
..
.
.
z VA B= 15
9 810 07
2.
..
Continuity: 4
4
10
107
===
B
BBBA
QVAVQ
smVB /7=
( )z mA = =
15 7
9 810 07 7 42
2.
.. .
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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.
Assumptions
Tank >> pipeVA 0 Flow is steady
Flow is turbulent A C B= = 10.
Given
Flow of water = const
1-D energy equation with pump present:
p V
gz h
p V
gz hA A
AA p
BB
BB lt
+ + + = + + +
2 2
2 2
(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
Bernoullis equation along a streamline from C to B:
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A
water
D
C
B
nozzle
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p Vgz
p VgzC C
C
B B
B
+ + = + +2 2
2 2
( )V g z z V C B C C ms2 2 2 9 81 65 35 24 26= = =( ) . ( ) .
Continuity:
Q V A mC C= = =24 26 30 10 0 07284 3. . / sec
Pump power:
W Q hp p=
. ( . )( ) .W kWp = =9 81 0 0728 25 17 85
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 wateris 3.92 cfs. What horsepower is delivered by the pump to the flow?
Assumptions
Assume flow is steady
tm m min out = =0 &
Assume flow is uniform at inlet and outlet
Neglect friction uout = uin Q =0
z zout in
= const
Conservation of mass: m m V A V Ain out in in out out = =
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pump
pA pB
out
in
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= = =V A V A V ft in in out out . / sec3923
=
= =
=V ft V ft in out 392
1
4
4 99392
05
4
19962 2
.
( ). / sec;
.
( . ). / sec
Wss = 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.
Q W W W t
e dV ep
V dAss s other
V A
+ = + +
= +
W e
p
V As inlet
outlet
= +
W Q
p V p V s
B out A in
2 2
2 2
( ) =
+
.
.( . )
( ) ( . ) ( . )
..
Ws
624
322392
30 10 144 19 96 4 99
262 432 2
2 2
[ ] = + = = = , . .sec
, .sec
, . .W ft lbf ft lbf hp hps 11 289 6 1418 7 12 708 3 12 708 3550
231
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.
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z1 = 2 m
z2 = 4 m
D = 30 cmd = 15 cm
No head loss between B and outlet Bernoulli can be applied between B and the outlet:
p Vgz
p VgzB B B
out out
out
+ + = + +2 2
2 2
V VQ
Ap p zB out
pipe
out atm out = = = =; ; 0
pB = pout - gz1 = patm - 1000(9.81)(2)
pBg = -19.62 kPa
Apply Bernoulli between A and B:
p Vgz
p VgzA A
A
B B
B
+ + = + +2 2
2 2
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T = 10o C
water
A
B
machine
d
D
d
z2
z1 z
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( )V
Q
DA
ms=
=
=
2 2
4
0 06
0 30
4
0849.
..
( )V
Q
dBm
s=
=
= 2 2
4
0 06
015
4
3395
.
. .
( )[ ]
pg
z z zAg
= + +
19621000
3 395 0 849
2 10001 2 1
2 2
.( . ) ( . )
( )
[ ] kPapAg 5.53)1000(2
)849.0()395.3(10004
1000
)81.9(100062.19
22
=
++=
Neglect losses
Q=
0 uout = uin flow is uniform at inlet and outlet
Assumptions
reservoir >> delivery pipe Vin 0
flow is steady
t 0
= const
Energy equation for the c.v.
Q W W W t
e dV ep
V dAss s other
V A
+ = + +
Wss = 0 by choice of the c.v.
= + +
W Q
p Vgz
p Vgzs
out out
out
in in
in
2 2
2 2
= + + +
( . )( . )
. ( )Ws 1000 0 06 03395
20 0 0 9 81 2 4
2
= = . .W W kW s 31858 3 2
. ; W kW W s s= > 3 2 0 Machine is a turbine
Flow through a 90o reducing elbow
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Assumptions
Flow is steady
Fluid is incompressible
u & p are uniform at 1 and 2
Energy equation for the c.v.
Q W W W t
e dV ep
V dAss s other
V A
+ = + +
Wss = 0 by choice of the c.v.
Wother = 0
= + + + ( ) ( ) ( )Q W m u u mp p
mg z z V
V dAV
V dAsA A
2 1 2 1 2 1 2
2
2 2
2
1
2
1 1
12 2
= +
+
+ + ( ) W m
pgz
pgz
VV dA
VV dA m u u Qs
A A
2
2
1
1
2
2
2 2
2
1
2
1 1 2 1
12 2
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CVx
flow
g
y
z
1
2
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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.
= =
VV dA
VV dA m
V
A A
2 2 2
2 2 2
= 1 for uniform flow; > 1 for non-uniform flow
= 2 for fully developed laminar flow; 1.05 for turbulent flow.
= + +
+ +
+
( )
W m
p Vgz
p Vgz m u u
Q
ms
2
2
2
2
2
1
1
1
2
1 2 12 2
= + +
+ +
+
( )
W
m
p Vgz
p Vgz u u
Q
m
s 2
2
2
2
2
1
1
1
2
1 2 12 2
+ + +
+ +
=
( )
W
m
p Vg z
p Vgz u u qs 1 1
1
2
12
22
2
2 2 12 2
[ ]
= + +
+ +
+
( )W
mg
p V
gz
p V
gz
gu u qs 2
2
2
2
2
1
1
1
2
1 2 12 2
1
Pump
lTphz
g
Vphz
g
Vp+
++=+
++
2
2
2
2
2
1
2
1
1
1
22
hp = pump head ; hlT = head loss term
gm
W
gm
Wh ssp
= (since Ws < 0 for pump)
W h Q h mg s
pumpp p= =
Turbine
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p V
gz h
p V
gz hT lT
11
1
2
12
22
2
22 2
+ +
= + + +
+
hT = turbine head
hW
mgT
s=
W h Q h mg s
turbineT T= =
Head loss in an abrupt expansion
Continuity: V A V A AV
VA1 1 2 2 1
2
1
2= =
Momentum equation: [ ]( )p p A Q V V1 2 2 1 2 = +
[ ] ( ) ( )( )p p Q
AV V
V A
AV V V V V 1 2
2
2 12 2
2
2 1 2 2 1
= = =
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1-D energy equation:
p V
gz
p V
gz hlt
11
1
2
12
22
2
22 2
+ + = + + +
z1 = z2
g
VVpphle
2
2
2
2
121 +
=
1 2
1
( ) ( )( )
( ) ( )
+
=+
=
22
21
2
12
21
12122 VVVg
VVVV
g
VV
g
VVVhle
( )g
VVhle
2
2
12 =
Abrupt Contraction
Bend in a pipe
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Vena Contracta
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Examples for the application of Rayleighs Theorem
1. Period of a simple pendulum
( )t f l g m= , ,
gmlconstt =
Choose LTM
( ) = T L M LT
2
Dimensional homogeneity: L M T L M T0 0 2
=+
= = = + = = =
= 0 2 11
20
1
2
1
2; ;
=t const l
g
the constant must be determined experimentally (const = 2)
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flow
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Note: Rayleighs 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
lf d V v= , , ,
vVdconstl
p=
Choose LTM
( ) ( ) ( ) 3121122
= MLTLLTLLLMLT
MTLTML ++ = 3222
Dimensional homogeneity:
= = = 1 2 2;( ) + + = + + = = 2 3 2 2 2 3 2 1
( ) ( )
d
V
Vd
vconstvVdconst
l
p 221
==
const must be determined experimentally and must be chosen arbitrarily
3. ( )p
lf d V v e= , , , , e = roughness
pl
const d V v e=
Choose LTM
( ) ( ) ( ) = = + + + ML T L LT L T ML L L T M2 2 1 2 1 3 2 3
Dimensional homogeneity:
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( )
= = =
+ + + = + + =
=
1 2 2
2 2 3 2 1
1
; ;
( ) ( )p
l
const d V v e= + + 1 2
pl
constv
dV
V
d
e
d=
2
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.
( )rlpfQ ,,, =
rlpconstQ =
Choose LTM
( ) ( )L T MLT L L ML T L3 1 2 2 1 1 =
Dimensional homogeneity: M L T M L T0 3 1 2 + + + =
+ = = = =
=
= + + = + + + = + =
0 2 11
2
1
1 3 1 1 3 3
; ;
;
= 3
==
t
lr
pconstrl
pconstQ 33
= chosen arbirtarilyconst 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
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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.
Quantity MLT FLT
p Pressure ML-1T-2 FL-2
Viscosity (dynamic) ML-1T-1 FL-2T
Viscosity (kinematic) L2T-1 L2T-1
surface tension MT-2 F/L
density ML-3 FT2L-4
c velocity LT-1 LT-1
a acceleration LT-2 LT-2
e roughness (absolute) L L
g acceleration (due to gravity) LT-2 LT-2
F Force MLT-2 F
shear stress ML-1T-2 FL-2
A Area L2 L2
V Volume L3 L3
specific weight ML-2T-2 F/L3
Q discharge L3T-1 L3T-1
(volumetric flow rate)m mass flow rate MT-1 FL-1T
hl head loss L L
N rpm T-1 T-1
angular speed T-1 T-1
T Torque ML2T-2 FL
H Impulse and Momentum MLT-1 FT
E Engergy and Work ML2T-2 FLP Power ML2T-3 FLT-1
E Modulus or elasticity ML-1T-2 F/L2
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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.
- 103 -
V
Moon surface
castinginstrumentpackage
cushion
liquid
D
V
hd
ball
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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 afunction 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 ofdimensionless parameters suitable for organizing experimental data.
Solution to #1
h = f (D, d, V, , , W)
h D d V W
MLT
h D d V W
L L L LT-1 ML-3 ML-1T-1 MLT
Dimensional matrix
L T M
- 104 -
At least one 3 x 3determinant is
nonzero Rank = 3
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h 1 0 0
D 1 0 0
d 1 0 0
V 1 -1 0
-3 0 1
-1 -1 3W 1 -2 1
Choose D, V, , as repeaters
Check on independence of the dimensions of the repeaters
-groups
D
hhVD == 11 111
(obtained by inspection)
D
ddVD == 22 222
3333 VD=
( ) ( ) 11131131000 33333333 ++ == MTLTMLMLLTLMTL
= = + = = 3 3 3 3
1 0 1 1 0 1;
( ) ( ) 3 3 3 3 3 33 1 0 1 3 1 3 1 1 1+ = = + = + =
VD
= 3
WVD 4444
=
( ) ( ) 231000 444 = MLTMLLTLMTL
1213000 44444 +++= MTLMTL
- 105 -
L T M
D 1 0 1
= V 1 -1 0 0 dimensions of the repeaters are independent
-3 0 1
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4 4 43 1 0+ + =
= = + = = 4 4 4 4
2 0 2 1 0 1,
( ) ( ) 4 4 43 1 3 1 2 1 2= = =
224 DV
W
=
( )4321 ,, = f
=
22,,
DV
W
VDD
df
D
h
2. a = f (, m, E, , V)
a m E V
Choose MLT
a m E VLT-2 L M ML-1T-2 ML-1T-1 LT-1
Dimensional Matrix
L T M
a 1 -2 0
l 1 0 0
m 0 0 1
E -1 -2 1
-1 -1 1
V 1 -1 0
Choose , m, V as repeaters
Check on the independence of the dimensions of the repeaters
- 106 -
At least a 3 x 3
determinant is
nonzero
Rank = 3
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-groups
aVm 1111
=
( ) 21000 111 = LTLTMLMTL 11111 2000 MTLMTL++
=
( ) = = = + + = = = = 1 1 1 1 1 1 10 2 0 2 1 0 1 2 1 1; ;
21 V
a=
EVml 2222=
( ) 211000 222 = TMLLTMLMTL 121000 2222 ++= MTLMTL
( ) 321101
;101;202
2222
2222
====+
==+==
mV
E3
2
=
3333 Vm=
( ) 111000 333 = TMLLTMLMTL 111000 3333 ++= MTLMTL
( )
= = + = = + =
= = =
3 3 3 3 3 3
3 3
1 0 1 1 0 1 1 0
1 1 1 2
; ;
mV
23
=
( )
==
mVmV
Ef
mV
af
23
321,,
- 107 -
L T M
1 0 1
= m 0 0 1 0 dimensions of the repeaters are independent
V 1 -1 0
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4. V = f (, , )
V
Choose MLT
V LT-1 MT-2 L ML-3
Dimensional Matrix
L T M
V 1 -1 0
0 -2 1
1 0 0
-3 0 1
Choose , and as repeaters
Check on the independence of the dimensions of the repeaters
V1111 =
( ) ( ) 11111111 1213000132000 ++ == MTLMTLLTMLLMTMTL
= =
+ = = = + = =2 1 01
20
1
23 1 0 31 1 1 1 1 1 1 1 1 1 ; ;
=
=1 31
21
1
2
- 108 -
At least a 3 x 3 determinant
is nonzero Rank = 3
L T M
0 -2 1
= 1 0 0 0 dimensions of the repeaters are independent
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constVorconstVconst
V
===
=
1
1
5. = f (0, r, ,)
0 r
Choose MLT
0 r
T-1 L2T-1 L T L2T-1
Dimensional Matrix
L T M
0 -1 1
0 2 -1 0
r 1 0 0
0 1 0
v 2 -1 -
Choose r and as repeaters
Check on the independence of the dimensions of the repeaters
-groups
== 111
r (by inspection)
- 109 -
all 3 x 3 determinants are
zeroat least one 2 x 2
determinant is nonzero
Rank = 2
L T
= r 1 0 0 dimensions of the repeaters are independent
0 1
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0222 = r
12001200 2222 + == TLTLTLTLTL
+ = = = = 2 2 2 22 0 2 1 0 1;
2 2 0=
r
vr
vr23
33
== (by inspection)
( ) 1 2 3 2 0 2
= =
f f
r rv, ,
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
d
dp
d=
=
VV
(Bulk modulus of elasticity)
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
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p = f (V, , d, , , K, , g)
Number of variables: n = 9
Choose MLT as fundamental units
p V d KML-1T-2 LT-1 L L ML-3 ML-1T-1 ML-1T-2
gMT-2 LT-2
Dimensional Matrix
L T M
p -1 -2 1
V 1 -1 0
l 1 0 0
d 1 0 0
-3 0 1
-1 -1 1
K -1 -2 1
0 -2 1
g 1 -2 0
Choose v, d, as repeaters
Check on the independence of the dimensions of the repeaters
- 111 -
At least one 3 x 3
determinant is nonzero
Rank = 3 = m
L T MV 1 -1 0
= d 1 0 0 0 Dimensions of the repeaters are independent
-3 0 1
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Number of dimensionless groups: N = (n-m) = 6
groups
( ) ( )2113000
1111111 == TMLLTLMLTLMpVd
2131000 11111 +++= TLMTLM
+ = = = = + + = 1 1 1 1 1 1 1
1 0 1 2 0 2 3 1 0; ;
( ) + = =3 1 2 1 0 01 1
=
=21 V
p
(EULER NUMBER, E
pressure forces
inertia forcesu
= )
1,02222
222 ==== Vd (by inspection)
d
=2 -- implies that shape is a controlling factor
( ) ( ) 11313330003333== TMLLTLMLTLMVd
1131000 33333 +++= TLMTLM
+ = = = = + + = 3 3 3 3 3 3 31 0 1 1 0 1 3 1 0; ;
( ) + = = 3 1 1 1 0 13 3
3 =
Vd
= (reciprocal of the REYNOLDS NUMBER,forcesviscous
forcesinertial=Re )
( ) ( ) 211300044
44
444 == TMLLTLMLTLMKVd
2131000 44444 +++= TLMTLM
+ = = = = + + = 4 4 4 4 4 4 4
1 0 1 2 0 2 3 1 0; ;
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( ) + = =3 1 2 1 0 04 4
4 2=K
V= (reciprocal of the MACH NUMBER, Ma
inertial forces
elastic forces= )
( ) ( ) 2130005555555 == MTLTLMLTLMVd
231000 55555 +++= TLMTLM
03;202;101 5555555 =++====+
( ) + = = 3 1 2 0 15 5
25 Vd
= = (reciprocal of the WEBER NUMBER, forcestensionsurfaceforcesinertial
We =
)
( ) ( ) 21300066
66
666 == LTLTLMLTLMgVd
213000 66666 +++= TLMTLM
= = = + + + = 6 6 6 6 6 6
0 2 0 2 3 1 0; ;
( ) + + = =3 0 2 1 0 16 6
26 V
gd= = reciprocal of the FROUDE NUMBER,
forcesgravity
forcesinertiaFr = )