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

7/27/2019 Section 25 30

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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:

- 85 -

Partly open valve

d

Air underpressure

water

1

2

7/27/2019 Section 25 30

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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

7/27/2019 Section 25 30

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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

- 87 -

turbine

H

2

1

z

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Given


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.

- 88 -

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



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.

.. .

- 89 -

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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


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:

- 90 -

A

water

D

C

B

nozzle

7/27/2019 Section 25 30

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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 = =

- 91 -

pump

pA pB

out

in

7/27/2019 Section 25 30

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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.

- 92 -

7/27/2019 Section 25 30

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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

- 93 -

T = 10o C

water

A

B

machine

d

D

d

z2

z1 z

7/27/2019 Section 25 30

11/30

( )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


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

- 94 -

7/27/2019 Section 25 30

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Assumptions

Flow is steady

Fluid is incompressible

u & p are uniform at 1 and 2


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

- 95 -

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

- 96 -

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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

= = =

- 97 -

1 2

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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

- 98 -

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)

- 99 -

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:

- 100 -

7/27/2019 Section 25 30

18/30

( )

= = =

+ + + = + + =

=

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

- 101 -

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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

- 102 -

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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

7/27/2019 Section 25 30

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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

7/27/2019 Section 25 30

22/30

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

7/27/2019 Section 25 30

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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


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

-3 0 1

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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


-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

7/27/2019 Section 25 30

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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


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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 = )

Section 25 30

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Transcript of Section 25 30