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 10 .

Given

Flow of water = const

1-D energy equation with a pump present:

hp V

gz

p Vg

z hp lt 11

12

12

22

2

22 2

hp = pump head

p1g = 0, p2g = 100 kPa, z1 = 6m, z2 = 10m, = 9.81 kN/m3

hVg

Vg

Vp 1009 81 2

10 6 152

14 21259 81

22

22

22

.( ) . .

..

- 84 -

Elevation = 10m

Elevation = 2m

Elevation = 6m

40 cm

D = 30 cm

p = 100 kPa

T = 10o Cwater

1

2

V m2 3537 . / sec

h mp 14 2 1259 81

3537 15792. ..

( . ) .

pump power: . ( . )( . )W Q hp p 9 81 0 25 15 79

.W kWp 38 72

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

h K VgL L2

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 under

water

1

2

p Vg

zp V

gz hlt

11

12

12

22

2

22 2

p1g = 100 kPa, p2g = 0, z1 = 10m, = 9.81 kN/m3

1009 81

109

2 9 819

2 9 81

2 2

. . .

KL

19819 81

812 9 81

1.. .

( )

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

h VgL 10

2

2

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.

Assumptions

Tank >> pipeV1 0

- 86 -

Partly open valve

d

Patm

water

1

2

Flow is turbulent 1 2 10 .

1-D energy equation:

p Vg

zp V

gz hlt

11

12

12

22

2

22 2

p1g = p2g = 0, z1 = 10m, = 9.81 kN/m3

102

102

10 2 9 81 1122

22

22

Vg

Vg

V.

V m s210 2 9 81

114 223

. . /

Discharge:

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.

Assumptions

Reservoir >> pipeV1 0 flow is steady

- 87 -

turbine

H

2

1

z

Flow is turbulent 1 2 10 .

Given

Flow of water = const

1-D energy equation with turbine present:

p Vg

z hp V

gz hT lt

11

12

12

22

2

22 2

hT = turbine head

p1g = p2g = 0, z1 = 10m, V = 5m / s2 , = 9.81 kN/m3

h zVg

mT

12

2 2

210

2 9 818 726(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 atmospheric pressure = 100 kPa.

- 88 -

z

D

B

A

C

WaterT = 20o

Assumptions

Reservoirs >> pipeV VA C 0 Flow is steady

Flow is turbulent 1 2 10 .

Given

Flow of water = const

1-D energy equation:

p Vg

zp V

gz hA

AA

AB

BB

B lt

2 2

2 2

pAa = 100 kPa, pBa = 1.23 kPa, zB = 10m, = 9.81 kN/m3

z

Vg

Vg

VAB B

B

123 100

9 81 210 2

210 07 10 15

9 81

2 22.

.. .

.

z VA B 159 81

0 072..

.

Continuity:

z mA

15 79 81

0 07 7 422.

.. .

- 89 -

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 Vg

z hp V

gz hA

AA

A pB

BB

B 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

Bernoulli’s equation along a streamline from C to B:

- 90 -

Awater

D

C

B

nozzle

p Vgz

p VgzC C

CB B

B

2 2

2 2

V g z z VC B C Cms

2 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 water is 3.92 cfs. What horsepower is delivered by the pump to the flow?

Assumptions

Assume flow is steady t

m 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

outin

V A V A V ftin in out out . / sec392 3

V ft V ftin out3 92

14

4 993 92

0 54

19 962 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 Wt

e dV e p V dAss s otherV A

W e p V As

inlet

outlet

W Q

p V p Vs

B out A in

2 2

2 2

.

.( . )

( ) ( . ) ( . ).

.

Ws62 432 2

3 9230 10 144 19 96 4 99

262 432 2

2 2

, . .sec

, .sec

, ..W

ft lbf ft lbfhp 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 -

z1 = 2 mz2 = 4 mD = 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

Bout out

out

2 2

2 2

V VQ

Ap p zB out

pipeout 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

AB B

B

2 2

2 2

V Q

DAm

s

2 2

4

0 06

0 304

0849.

..

- 93 -

T = 10o C

water

A

Bmachine

d

D

dz2

z1z

V Q

dBms

2 2

4

0 06

0154

3 395.

..

p

gz z zAg

19 62

1000

3395 0 849

2 10001 2 1

2 2

.( . ) ( . )

( )

Neglect losses Q 0 uout = uin

flow is uniform at inlet and outlet

Assumptions

reservoir >> delivery pipe Vin 0

flow is steady t

0

= constEnergy equation for the c.v.

Q W W Wt

e dV e p V dAss s otherV A

Wss 0 by choice of the c.v.

W Q

p Vgz

p Vgzs

out outout

in inin

2 2

2 2

( . )

( . ). ( )Ws 1000 0 06 0

3 3952

0 0 0 9 81 2 42

. .W W kWs 31858 32

. ; W kW Ws s 3 2 0 Machine is a turbine

Flow through a 90 o reducing elbow

- 94 -

Assumptions

Flow is steadyFluid is incompressible

u & p are uniform at ‘1’ and ‘2’

Energy equation for the c.v.

Q W W Wt

e dV e p V dAss s otherV A

Wss 0 by choice of the c.v.

Wother 0

( ) ( ) ( )Q W m u u mp p

mg z zV

V dAV

V dAsA A

2 12 1

2 122

2 22

12

1 112 2

( ) W m

pgz

pgz

VV dA

VV dA m u u Qs

A A

22

11

22

2 22

12

1 1 2 112 2

- 95 -

CV

x

flow

gy

z

1

2

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.

V V dA V V dA mV

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

ms2

22

2

21

11

2

1 2 12 2

( )

Wm

p Vgz

p Vgz u u Q

ms 2

22

2

21

11

2

1 2 12 2

( )

Wm

p Vgz

p Vgz u u qs 1

11

2

12

22

2

2 2 12 2

( )

Wmg

p Vg

zp V

gz

gu u qs 2

22

2

21

11

2

1 2 12 21

Pump

hp = pump head ; hlT = head loss term

(since Ws 0 for pump) W h Q h mgspump

p p

Turbine

p Vg

z hp V

gz hT lT

11

12

12

22

2

22 2

- 96 -

hT = turbine head

hWmgT

s

W h Q h mgsturbine

T T

Head loss in an abrupt expansion

Continuity: V A V A AVV

A1 1 2 2 12

12

Momentum equation: ( )p p A Q V V1 2 2 1 2

( )p p QA

V VV AA

V V V V V1 2

22 1

2 2

22 1 2 2 1

1-D energy equation:

p Vg

zp V

gz hlt

11

12

12

22

2

22 2

- 97 -

1 2

z1 = z2

1 2 1

Abrupt Contraction

Bend in a pipe

- 98 -

Vena Contracta

Examples for the application of Rayleigh’s Theorem

1. Period of a simple pendulum

t f l g m , ,

Choose LTM

T L M LT 2

Dimensional homogeneity: L M T L M T0 0 2

0 2 1 1

20 1

212

; ;

t const lg

the constant must be determined experimentally (const = 2)

- 99 -

flow

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:

pl

f d V v , , ,

Choose LTM

Dimensional homogeneity:

1 2 2; 2 3 2 2 2 3 2 1

const must be determined experimentally and must be chosen arbitrarily

3. pl

f 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 -

1 2 22 2 3 2 1

1

; ;

pl

const d V v e 1 2

pl

const vdV

Vd

ed

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.

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 1 12

1

1 3 1 1 3 3

; ;

; 3

= 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 -

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 Lg 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 LN rpm T-1 T-1

angular speed T-1 T-1

T Torque ML2T-2 FLH Impulse and Momentum MLT-1 FTE Engergy and Work ML2T-2 FLP Power ML2T-3 FLT-1

E Modulus or elasticity ML-1T-2 F/L2

- 102 -

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

castinginstrument

cushion

D

V

hd

ball

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 D d V W L L L LT-1 ML-3 ML-1T-1 MLT

Dimensional matrix

L T Mh 1 0 0

- 104 -

At least one 3 x 3 determinant is nonzero Rank = 3

D 1 0 0d 1 0 0V 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

(obtained by inspection)

3 3 3 31 0 1 1 0 1;

3 3 3 3 3 33 1 0 1 3 1 3 1 1 1

4 4 43 1 0

- 105 -

L T MD 1 0 1

= V 1 -1 0 0 dimensions of the repeaters are independent -3 0 1

4 4 4 42 0 2 1 0 1,

4 4 43 1 3 1 2 1 2

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 Ma 1 -2 0l 1 0 0m 0 0 1E -1 -2 1 -1 -1 1V 1 -1 0

Choose , m, V as repeatersCheck on the independence of the dimensions of the repeaters

- 106 -

At least a 3 x 3 determinant is nonzeroRank = 3

-groups

1 1 1 1 1 1 10 2 0 2 1 0 1 2 1 1; ;

3 3 3 3 3 3

3 3

1 0 1 1 0 1 1 01 1 1 2

; ;

4. V = f (, , )

- 107 -

L T M 1 0 1

= m 0 0 1 0 dimensions of the repeaters are independentV 1 -1 0

V

Choose MLT

V LT-1 MT-2 L ML-3

Dimensional Matrix

L T MV 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

2 1 01

20

12

3 1 0 31 1 1 1 1 1 1 1 1 1 ; ;

1 3 1

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

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 10 2 -1 0r 1 0 0 0 1 0v 2 -1 -

Choose r and as repeaters

Check on the independence of the dimensions of the repeaters

-groups

(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

2 2 2 22 0 2 1 0 1;

2 2 0r

(by inspection)

1 2 3 2 0 2

f fr r

v, ,

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

ddpd

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

p = f (V, , d, , , K, , g)

- 110 -

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 Mp -1 -2 1V 1 -1 0l 1 0 0d 1 0 0 -3 0 1 -1 -1 1K -1 -2 1 0 -2 1g 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

Number of dimensionless groups: N = (n-m) = 6

groups

1 1 1 1 1 1 11 0 1 2 0 2 3 1 0; ;

3 1 2 1 0 01 1

(EULER NUMBER, Epressure forcesinertia forcesu )

(by inspection)

-- implies that shape is a controlling factor

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, )

4 4 4 4 4 4 41 0 1 2 0 2 3 1 0; ;

- 112 -

3 1 2 1 0 04 4

4 2KV = (reciprocal of the MACH NUMBER, Ma

inertial forceselastic forces

)

3 1 2 0 15 5

= (reciprocal of the WEBER NUMBER, )

6 6 6 6 6 60 2 0 2 3 1 0; ;

3 0 2 1 0 16 6

= reciprocal of the FROUDE NUMBER, )

- 113 -