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2145-213 Fluid Mechanics for Aerospace Engineering

HW #4: Conservation of Linear Momentum, Conservation of Energy, Conservation of Angular Momentum and

Turbomachines, Bernoulli’s Equation

Due: Fri, April 22, 2011.

1

2145-213 Fluid Mechanics for Aerospace Engineering HW #4: Conservation of Linear Momentum, Conservation of Energy, Conservation of Angular Momentum and


Problem 1. Conservations of Mass and Linear Momentum for A Control Volume as observed from A

Stationary Frame of Reference [Adapted from Munson et al., 2002, Problem 5.41, p. 283.]

The hydraulic dredge is used to dredge sand from a river bottom, and the sand/water mixture is discharged as a

free-jet as shown below. The discharge has a cross sectional area A . The jet discharge speed is V , which is oriented at

an angle with respect to the horizontal. Assume that the bell-mouth suction has relative large cross sectional area

such that the sand/water mixture speed at the suction can be neglected. The specific gravity of the sand/water mixture is

SG and water density is .

1.1. Estimate the thrust needed from the propeller to hold the boat stationary.

1.2. Under this stationary operating condition, is the buoyancy force on the boat equal to the weight W of the boat

(including all the loads on the boat)? If not, how does the buoyancy force change from that when the hydraulic

dredge is not operated?

Solution

Control Volume: The control volume is stationary and non-deforming. It is the region occupied by the boat only.

Assumptions

1. Incompressible flow (steady and uniform field)

2. Steady V

field.

3. Uniform V

at each cross section.

4. The bell-mouth suction has relative large cross sectional area such that the sand/water mixture speed at the

suction can be neglected, VV 1

.

a. We assume that the magnitude of the velocity field around the boat is so small such that the pressure

distribution can be approximated by the hydrostatic pressure variation.

C- x -Mom:

CS

sfxCVxMV

BxSx AdVudt

dP

dt

dPFF )( /

,,

, V

xV dVuP )(, (A)

V

A

T

B

W

x

y

1

2

V

A





2

Unsteady term

CV

xCVdVu

dt

d

dt

dP)(

, :

0)0()(

)()(

,

CV

Vandsteady

CV

CVgmindefornonandstationary

CV

xCVdVdV

t

udVu

dt

d

dt

dP

Net Convection Efflux Term

CS

sf AdVu )( /

:

cos)(

)()()()()(

)(

0

121122

/1/2///

1

1

121221

Vmuumumum

AdVuAdVuAdVuAdVuAdVu

VV

u

MassC

A

sf

A

sf

AeachoveruniformisV

A

sf

A

sf

AACS

sf

o RTT part:

)1(cos

)( /,,

Vm

AdVudt

dP

dt

dP

CS

sfxCVxMV

Net Surface Force SxF

TFSx

Net Body Force BxF

0 BxF

Thus, C- x -Mom (A) becomes

cos)(cos 2AVSGVmT . ANS

C- y -Mom:

CS

sf

yCVyMV

BySy AdVvdt

dP

dt

dPFF )( /

,, ,

V

yV dVvP )(, (B)

Unsteady term

CV

yCVdVv

dt

d

dt

dP)(

, :

Similar to the case of x -momentum, we have

0,

dt

dP yCV.


CS

sf AdVv )( /

:

Similar to the case of x -momentum, we have

sin)()(1

21

12/ VmvvmAdVv

VV

AACS

sf

.

o RTT part:

sin

)( /

,,

Vm

AdVvdt

dP

dt

dP

CS

sf

yCVyMV

Thus, the RTT states that there is a time rate of change of linear y -momentum of the

coincident material volume MV .

According to the Newton’s second law, the net force in the y direction on the MV ,

hence also the CV , must not vanish.





3

Net Surface Force SyF

BFSy (Buoyancy is the resultant of pressure distribution over the surface of the

submerged body. Here, we also assume that the magnitude of the velocity field

around the boat is so small such that the pressure distribution can be approximated

by the hydrostatic pressure variation.)

Net Body Force ByF

WFBy

Thus, C- y -Mom (B) becomes

sin)(sin

sin

2AVSGWVmWB

VmWB

.

When the dredge is operated, we have

sin)( 2AVSGWB

When it is not operated ( 0V ), we have

WB .

Thus,

under the operating condition, the buoyancy force: sin)( 2AVSGWB , is not equal to the weight

W of the boat;

the buoyancy force under the operating condition is more than when the dredge is not operated by the

amount of sin)( 2AVSG . ANS





4

Problem 2. Conservations of Linear Momentum for A Control Volume as observed from A Moving/Translating

Frame of Reference [Adapted from Fox et al., 2010, Problem 4.142, p. 155.]

A rocket sled, weighing 44,500 N and travelling 960 km/h, is to be braked by lowering a scoop into a water trough.

The scoop is w = 150 mm wide.

2.1. Determine the time required (after lowering the scoop to a depth of h = 75 mm into the water) to bring the sled

to a speed of 32 km/h.

2.2. Plot the sled speed as a function of time.

2.3. As a consequence of your current result, in terms of design, suggest three methods to reduce the time required.

Solution

Control Volume CV is stationary and non-deforming in the moving frame of reference (MFR). The frame MFR is

moving with the sled. The CV includes the sled and water in the scoop.

Observer An observer is in the MFR.

Assumptions

1. Incompressible flow (density field is both steady and uniform)

2. Neglect change in linear momentum in the CV.

3. Uniform velocity at each cross section.

4. No change in relative speed of water on the scoop from inlet to exit.

5. The scoop contains constant amount of water over time.

6. Neglect mass of water in the scoop, sledCV MM

7. Neglect any other drag and/or frictional forces.

8. Water density = 1,000 kg/m3.

Basic Equations

1. C-Mass:

)()(

/ )(,)()(

0

tV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

(A)

*** Note that here we use C-Mass for an observer in MFR and not in IFR, hence sfV /

and not sfV /

. ***

=

D

1

2

x

y

x

y

MFR

IFR W

itUV frˆ)(

=





5

2. C-Mom:

)()(

/ )()(,)(

tV

V

tCS

sfCVMV

frCVbs dVVtPAdVVdt

Pd

dt

PdaMFF

(B)

Velocities and Relative Velocities

Let iUV frˆ

. Then, i

dt

dU

dt

Vda

fr

frˆ

.

iUiUVVV frˆ)ˆ(011

UVV 12

[No change in relative speed of water on the scoop from inlet to exit.]

)ˆsinˆcos(2 jiUV

C-Mass

Unsteady Term dt

tdM CV )(

o Because 1) CV is stationary and non-deforming in MFR, 2) the flow of water is incompressible, and

3) the scoop contains constant amount of water over time, 0)(

dt

tdM CV .

The C-Mass, Eq. (A), then gives

mmm :12 , QQQ :12 .

The mass flowrate is given by

1/

1

UAAdVm

A

sf

, [incompressible flow, uniform velocity at each cross section,

iUVV sfˆ

11,/

]

C-x-Mom

0sxF . [Neglect any other drag and/or frictional forces.]

0bxF .

dt

dUMaM CVfrCV [ itUV fr

ˆ)(

. Thus, idt

dU

dt

Vda

fr

frˆ

]

0,

dt

Pd xCV

. [Neglect change in linear momentum in the CV.]

)cos1()()cos()()( 121122

)(

/ UmUUmVVmVmVmAdVV xxxx

tCS

sfx

[uniform velocity at each cross section, C-Mass]

C-Mom, Eq. (B), becomes

)1(1

1

,)cos1(

,1

)(

)cos1(11

)cos1(1

)cos1(1

)cos1(00

11

1

0

1

2

1

2

bU

Ut

hwAM

UAb

tb

UtU

tM

A

UU

dtM

AdU

U

M

A

dt

dU

U

Umdt

dUM

o

CV

oo

CVo

t

CV

U

U

CV

CV

o





6

1

2

2

31 234.1

/81.9

500,44

)30cos1(/

/

6.3

1960)15.0075.0(000,1

)cos1(

s

sm

N

hkm

sm

h

kmm

m

kg

M

UAb

o

CV

o

[Neglect mass of water in the scoop.]

Hence, to decelerate from oU = 960 km/h to )(tU = 32 km/h requires

sshkm

hkmt 5.23

234.1

11

/32

/9601

. ANS (3.1)

The plot of speed VS time is shown below. ANS (3.2)

In order to reduce the deceleration time, Eq. (1) suggests that we need to increase b . This can be achieved by

o decrease the angle , or

o increase 1A by

increase the depth h , or

increase the width w . ANS (3.3)

Since cos 30o is already equal to 0.87 (while cos 0

o = 1), the decrease in results in little change in

comparison to an increase in A1. The plot below shows the speed VS time for two values of h, one is twice the

other.

0

100

200

300

400

500

600

700

800

900

1000

0 5 10 15 20 25 30

t (s)

U (

km

/h)

h = 75 mm

h = 150 mm





7

Problem 3. Bernoulli’s Equation and Conservation of Linear Momentum [2145-213 – 2009, HW #6.]

A nozzle is attached to a vertical pipe and discharges water into the atmosphere as shown in the figure below. The

discharge rate is 0.1 m3/s. The nozzle has a weight of 200 N, and the volume of water in the nozzle is 0.012 m

3.

3.1. Find the gage pressure at the nozzle inlet.

3.2. Find the force vector jFiFF yxˆˆ

that is required to support the nozzle at the flange support.

Assume that the water in the nozzle is accelerated at a rate such that the frictional effect can be neglected.

Note: In the past, we have solved the conservation of linear momentum with pressure, hence pressure force, being given. With

the knowledge of the Bernoulli’s equation – under the assumption, among others, of invisicid flow, we can use the

Bernoulli’s equation to solve for the unknown pressure first, then use the pressure information to find the pressure force

for the C-Mom equation.

Solution

Control volume The control volume CV is stationary and non-deforming. It includes the water volume and the

nozzle as shown above.

Assumptions

1. All properties are steady.

2. Incompressible flow

3. Inviscid flow

4. Points 1 and 2 are on the same streamline.

5. All properties are uniform at each cross section. The velocity is axial and uniform at each cross section.

6. Water density = 1,000 kg/m3.

Basic Equations:

jVV ˆ11

)ˆsinˆ(cos22 jiVV

x

y

1

2

H

jFiFF yxˆˆ

waternoz WW 11 Ap g

0.25 m





8

C-Mass:

CS

sfCVMV AdV

dt

dM

dt

dM /0 ,

V

V dVM )(

(A)

C- x -Mom:

CS

sfxCVxMV

BxSx AdVudt

dP

dt

dPFF )( /

,,

, V

xV dVuP )(,

(B)

C- y -Mom:

CS

sf

yCVyMV

BySy AdVvdt

dP

dt

dPFF )( /

,, ,

V

yV dVvP )(,

(C)

Bernoulli’s Equation 2

2

1

2

2

1

2

1

gyVpgyVp

skgs

m

m

kgQm /1001.0000,1

3

31 ,

smm

sm

A

QV /5

02.0

/1.02

3

11 , sm

m

sm

A

QV /01

01.0

/1.02

3

22 .

Ns

mm

m

kgVgWwater 72.11781.9012.0000,1

2

3

3 .

2.1. Consider the streamline as shown in the figure. From Assumptions 1-4, we can apply the Bernoulli’s equation and

get

)1(:,1)/(2

1

1)/(2

1

2

1

2

1

2

1

2

1

22112

212

1

212

21121

22

2212

11

2

2

1

2

VAVAMassCgHAAV

gHVVVppp

gyVpgyVp

gyVpgyVp

g

Thus,

kPaPa

PaPa

ms

m

m

kg

s

m

m

kg

gHAAVppp g

0.405.952,39

5.452,2500,37

25.081.9000,1125000,12

1

1)/(2

1

23

2

2

22

3

221

21121

ANS

2.2.

C- x -Mom:

CS

sfxCVxMV

BxSx AdVudt

dP

dt

dPFF )( /

,,

, V

xV dVuP )(,

(B)

Unsteady term

CV

xCVdVu

dt

d

dt

dP)(

, :

0)0()(

)()(

,

CV

Vandsteady

CV


CV

xCVdVdV

t

udVu

dt

d

dt

dP


CS

sf AdVu )( /

:





9

)(

)()()()()(

121122

/1/2///

121221

uumumum

AdVuAdVuAdVuAdVuAdVu

MassC

A

sf

A

sf

AeachoveruniformisV

A

sf

A

sf

AACS

sf

o RTT part:

(2)cos)(

)(

212

/,,

Vmuum

AdVudt

dP

dt

dP

CS

sfxCVxMV

Net Surface Force SxF :

xSx FF [Except at 1, pressure is uniformly atmp throughout. Pressure on 1A does not

contribute to the force in the x direction.]

Body Force BxF :

0 BxF

Thus, Eq. (B) becomes

Ns

m

s

kgVmF o

x 86630cos10100cos2 . (3)

C- y -Mom:

CS

sf

yCVyMV

BySy AdVvdt

dP

dt

dPFF )( /

,, ,

V

yV dVvP )(,

(C)

Unsteady term

CV

yCVdVv

dt

d

dt

dP)(

, :

0)0()(

)()(

,

CV

Vandsteady

CV


CV

yCVdVdV

t

vdVv

dt

d

dt

dP


CS

sf AdVv )( /

:

)(

)()()()()(

121122

/1/2///

121221

vvmvmvm

AdVvAdVvAdVvAdVvAdVv

MassC

A

sf

A

sf

AeachoveruniformisV

A

sf

A

sf

AACS

sf

o RTT part:

(4)1sin)/()sin()(

)(

2111212

/

,,

AAVmVVmvvm

AdVvdt

dP

dt

dP

CS

sf

yCVyMV

Net Surface Force SyF :

11111 ApFApApFF gyatmySy

Body Force ByF :

)( waternozBy WWF

Thus, Eq. (C) becomes

)1sin)/()( 21111 AAVmWWApF waternozgy . (5)

and





10

N

N

NVm

ApWWAAVmF

o

gwaternozy

28.482

80072.3170

800)72.117200()130sin2(

)()1sin)/(

1

11211

Thus, we have the force vector F

that is required to support the nozzle at the flange support

NjiF ˆ3.482ˆ866

. ANS





11

Problem 4. Conservation of Energy [Fox et al., 2010, Problem 4.202, p. 160.]

Air enters a compressor at 96 kPa, 27 oC with negligible speed and is discharged at 480 kPa, 260

oC with a speed

of 152 m/s. If the power input is 2.38 MW and the flow rate is 9 kg/s, determine the rate of heat transfer.

Solution

Control Volume CV is stationary and non-deforming. It covers the compressor from inlet to exit.

Assumptions

1. Steady property fields.

2. Uniform properties at each cross section.

3. Neglect all other works, except shaft work.

4. Neglect velocity/kinetic energy at inlet.

5. Neglect change in potential energy.

6. Treat air as a perfect gas with constant pc = 1,005 J/kg-K, R = 287 J/kg-K.

Basic Equations

1. C-Mass:

CS

sfCVMV AdVdt

tdM

dt

tdM /

)()(0 ,

)(tV

V dVM . (A)

2. C-Energy

CS

sf

CV

AdVpedVedt

dWQ ))(v()( /

(B)

: gzVhpeWWWW othersshears 2

2

1v,

C-Mass

For steady property fields, from C-Mass we have mmm :12 .

C-Energy

?Q

sWW [Neglect all other works, except shaft work.]

0)(

CV

dVedt

d [Steady property fields.]

CS

sf AdVpe ))(v( /

m = 9 kg/s

1p = 96 kPa

1T = 27 oC

1V 0 m/s

2p = 480 kPa

2T = 260 oC

2V = 152 m/s

sW = 2.38 MW

+

Q W

Q = ?





12

gas]perfect [,2

1)(

energy.] potentialin changeneglect inlet,at energy kineticneglect [,2

1)(

mass.]-Csection, crosseach at properties uniform[,)2

1()

2

1())(v(

2212

2212

12

22

/

VmTTcm

Vmhhm

gzVhmgzVhmAdVpe

p

CS

sf

Thus, the C-Energy, Eq. (B), becomes

kW

MWsmKKkg

J

s

kg

WVTTcmQ

VTTcmWQ

sp

ps

55.168

38.2/1522

1)27260(005,19

2

1)(

2

1)(

222

2212

2212

Thus, there is heat transfer out at the rate of 168.6 kW. ANS





13

Problem 5. Idealized Machines: Is it a pump or a turbine?

5.1. Idealized Axial-Flow Turbomachines For the given blade angles,

1. qualitatively sketch the blade shape,

2. qualitatively sketch the inlet and exit velocity diagrams,

3. use appropriate governing equations to show whether the design is suitable for a pump or a turbine.

4. Also, state necessary assumptions you make in order to arrive at these results.

5.2. Idealized Radial-Flow Turbomachines [Adapted from Munson et al., 2002, Problem 12.3, p. 816.]

The rotor shown below has straight, though backwardly inclined, blades and constant passage width from

inlet to exit. It rotates with an angular velocity of 2,000 rpm. Assume that the fluid enters the impeller with the

absolute flow velocity purely in the radial direction and the relative flow velocity is tangent to the blades

across the entire rotor. Use

1) appropriate velocity diagram and

2) governing equations

to show whether the device is a pump or a turbine. Relevant velocity components and angles should be

calculated in order to show clearly that this device is a pump or a turbine.

U

flow

a. oo 70,20 21

U

flow

b. oo 20,70 21

flow

U

z





14

Solution 5.1: Idealized Axial-Flow Turbomachines

ANS (5.1/1, 2)

U

1V

1 12

12,2 V

12, V

12

1rbV

12, V

e

12,2 rbV

12,2 V

12,2 rbV

flow

nV

U

1

nV

U

2

1

2

U

12 : pump

concave side leading forward

U

1V

1 12

1rbV

V

e

2rbV

2V

1

flow

nV

U

nV

U

2

1

2

U

12 : turbine

convex side leading forward

U

1V

1 12

2V

V

1rbV

e

2rbV

a. oo 70,20 21 ( 12 )

b. oo 20,70 21 ( 12 )





15

Control Volume: For both axial- and radial-flow machines, control volumes are stationary and non-deforming as

shown above. It includes the rotor/impeller and cuts through the solid shaft. sT

is the shaft

torque.

Assumptions

1. The flow is incompressible ( is both steady and uniform).

2. The velocity field is steady in mean. Use and evaluate the mean properties.

3. Neglect all other moments/torques (e.g., moments due to other surface forces (friction/shear and pressure),

frictional torque at bearings, moment due to body force mg), except shaft torque.

4. ( Vr

) is uniform over each cross section.

For axial flow machines: evaluate properties/velocities at mean radius.

For radial flow machines: velocity (and the corresponding velocity components) is uniform with

respect to the zr coordinates.

For further details, see footnote.1

5. Shockless entry/exit condition. That is, the relative flow (i.e., the velocity of fluid relative to the rotating blade

rbV

) enters and leaves the rotor at the geometric blade angles ’s. (Use in the constructions of the velocity

diagrams.)

ANS

(5.1/4)

Basic Equations

1. C-Mass:

)()(

/ )(,)()(

0

tV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

. (A)

2. C-Angular Momentum:

)(

)(

,

)(

/,,

)(:

))(()(,))(()()(

tCV

Ssc

tV

cV

tCS

sfcCVcMV

c

dVgrFrTM

dVVrtHAdVVrdt

tHd

dt

tHdM

(B)

1 Assumption: At each inlet/exit cross section, )( Vr

is uniform over the cross section.

sectioninletforsection,exitfor),())(( / VrmAdVVr

A

sf

.

zzrrzzrrzr erVerVzVezVeVeVeVezerVr ˆˆˆ)ˆˆˆ()ˆˆ(

rrsfrn

zzsfznnzzrrsf

dAVAdVee

dAVAdVeeedAeVeVeVAdV

/

//

,(radial)ˆˆ

,(axial)ˆˆ,ˆˆˆˆ

Given that we are interested in the torque of the shaft, which is aligned in the z direction, the above assumption implies that

Axial-machine: rV does not depend on zdA ; or, rV is uniform over the cross section.

Radial-machine: rV does not depend on rdA ; or, rV is uniform over the cross section.

sT

,

1

2

c

flow

sT

1

2

c

U

Axial-flow machines Radial-flow machines





16

C-Angular Momentum:

Unsteady Term dt

tHd cCV )(,

]0

)( stedy, areand[0

]deforming.-non and stationary is CV[)(

))(()(,

t

VrV

dVt

Vr

dVVrdt

d

dt

tHd

CV

CV

cCV

Net convection efflux term

CS

sf AdVVr ))(( /

12

12

12

)(:,)(:),()(

section] croseach over uniform is)[()()()()(

))(())(())((

/1/2111222

/11/22

///

A

sf

A

sf

A

sf

A

sf

A

sf

A

sf

CS

sf

AdVmAdVmVrmVrm

VrAdVVrAdVVr

AdVVrAdVVrAdVVr

o RTT part:

)()(

))((

111222

/,,

VrmVrm

AdVVrdt

Hd

dt

Hd

CS

sfcCVcMV

Net external moment about c , cM

ueshaft torqexcept moments,other allNegelect ,

)(

)(

s

tCV

Ssc

T

dVgrFrTM

Then, Eq. (B) becomes the Euler’s Turbomachine equation,

Euler’s turbomachine equation: ) torquehydraulic:()()( 111222 hs TVrmVrmT

(C)

and we also have the associated power equation, Eq. (C) ,

The Associated Power Equation: )()( 111222 VUmVUmTW s

. (D)

Note that from C-Mass, we also have mmm :12





17

Idealized axial-flow machines

Because UUU

:12 , the associated power equation, Eq. (D), becomes

1212 :,)( VVVVUmVUmVVUmTW s

. (1)

Thus,

if V

is in the same direction as U

,

0 VUmW

and it is a pump (energy being input into the system);

if V

is in the opposite direction to U

,

0 VUmW

and it is a turbine (energy being extracted from the system).

The velocity diagram shown above therefore indicates that the condition for the angles 1 and 2 under which the

machine changes type is as follows:

12 0 VUmW

pump. Hence, (a) is a pump.

12 0 VUmW

turbine. Hence, (b) is a turbine. ANS

(5.1/3)





18

Solution 5.2: Idealized Radial-Flow Turbomachines

Because 1V

is purely radial (while 1U

is purely tangential), 011 VU

and the associated power equation, Eq.

(D), becomes

22 VUmTW s

. (2)

Thus,

if 2V

is in the same direction as 2U

,

022 VUmW

and it is a pump (energy being input into the system);

if 2V

is in the opposite direction to 2U

,

022 VUmW

and it is a turbine (energy being extracted from the system).

Inlet Velocity Diagram

sftftsrev

radrevrU /7.1045.0

60

min2

min000,211

sftVV n /2011

o

sft

sft

U

V8.10

/7.104

/20tantan 1

1

111

Exit Velocity Diagram

sftftsrev

radrevrU /5.1889.0

60

min2

min000,222

C-Mass

)(/1.11/209.0

5.0

)width,passageconstant(,2

20

1112

11

2

12

1212

11

22

111

2

121122

VVsftsftft

ftV

r

rV

r

rV

bbVr

rV

br

brV

A

AVAVAV

nnn

nnnnnn

on

sft

sft

U

V37.3

/5.188

/1.11tantan 1

2

212

Since for a straight, backwardly-inclined blade, 12 (see note below).

Thus, oo 37.38.10 212 .

The velocity diagram above then shows that, for 22 , 2V

is in the same direction as 2U

and, therefore,

022 VUmW

.

Hence, the machine is a pump. ANS (5.2)

Note The relation between the blade angles, 1 and 2 , for straight-bladed radial-flow machines.

2U

(188.5 ft/s)

2 (3.37o)

2rbV

nV2

(11.1 ft/s) nV

eU ˆ,

12 = 10.8o

2V

2V

1U

(104.7 ft/s)

1 (10.8o)

nV

1rbV

1V

(20 ft/s)

eU ˆ,





19

For a simple straight blade geometry, we have the following relations.

Backwardly-inclined blade

)(

2/

2/

1212

22

21

Forwardly-inclined blade

)(

2/

)2/(

1221

22

21

In both cases, we see that when 0 , 2/12 and we have a radial blade, i.e., the blade is aligned in

the radial direction.

1

2

1 2

1

2

2

Backwardly-inclined blade Forwardly-inclined blade





20

Problem 6. Radial-Flow Machine and The Effect of The Exit Blade Angle [2145-213 – 2009, HW #6.]

A centrifugal radial water pump has the dimensions shown in the figure below. The volume rate of flow is 0.25

ft3/s, and the absolute inlet velocity is directed radially outward. The angular velocity of the impeller is 960 rpm. The

exit velocity as seen from a coordinate system attached to the impeller can be assumed to be tangent to the vane at its

trailing edge. Calculate the (ideal) power required to drive the pump.

Solution

Control Volume: Control volume is stationary and non-deforming as shown below. It includes the rotor/impeller

and cuts through the solid shaft. sT

is the shaft torque.

Assumptions

1. Incompressible flow (steady and uniform density field).

2. Velocity field is steady in mean and evaluate the mean property.

3. Neglect frictional torque and torque due to other surface forces, except at shaft. .

4. Neglect torque by body force.

5. ( Vr

) is uniform over each cross section.

For radial flow machines: velocity (and the corresponding velocity components) is uniform with

respect to the zr coordinates.

6. Shockless entry/exit condition. (Use in the constructions of the velocity diagrams.)

Basic Equations:

C-Mass:

)(

/ )(,)()(

0

tV

V

CS

sfCVMV dVtMAdVdt

tdM

dt

tdM

.

(A)

z re

e





21

C-Angular Momentum:

V

CV

CS

CCVCMV

CV

sS dVVrtHAdVVrdt

tHd

dt

tHddVgrTFr ))(()(,))((

)()()( ,

,,

(B)

C-Angular Momentum:

V

CV

CS

CCVCMV

CV

sS dVVrtHAdVVrdt

tHd

dt

tHddVgrTFr ))(()(,))((

)()()( ,

,,

(B)

Unsteady term

CV

CCVdVVr

dt

d

dt

tHd))((

)(,

:

0)0()(

))(()(,

CV

Vandmeaninsteady

CV


CV

CCVdVdV

t

VrdVVr

dt

d

dt

tHd


CS

sf AdVVr ))(( /

:

12

1122

/1/2

,)(

///

)()(

)()(

)()()()(

))(())(())((

12

1221

VrVrm

VrmVrm

AdVVrAdVVr

AdVVrAdVVrAdVVr

MassC

A

sf

A

sf

radiusmeanatevaluateorAeachoveruniformisVr

A

sf

A

sf

AACS

sf

o RTT part:

)1(

))((

1122

/,,

VrVrm

AdVVrdt

Hd

dt

Hd

CS

sfCCVCMV

Torque due to surface force (except at shaft): SFr

o 0

SFr [Neglect frictional torque and torque due to other surface forces, except at

shaft.]

Torque due to body force:

CV

dVgr )(

o 0)(

CV

dVgr [Neglect torque by body force.]

Shaft Torque: sT

Then, Eq. (B) becomes the Euler’s Turbomachine equation,

Euler’s turbomachine equation: )( 1122 VrVrmTs

. (C)

and we also have the associated power equation

The Associated Power Equation: )( 1122 VUVUmTW s

. (D)





22

Idealized radial-flow turbomachines

Because 1V

is purely radial (while 1U

is purely tangential), 011 VU

and the associated power equation, Eq.

(D), becomes

22 VUmTW s

. (2)

3

3

3

22

2

3

22

2222

94.1,/485.025.094.1

/08.4612

11

2

153.100

/53.100min

60

1

min960)2(

60

)2(

/389.118.0

125.0

180.012

75.0

12

11

2

122

ft

slugsslug

s

ft

ft

slugQm

sftfts

radrU

srads

rev

rev

radN

sftfts

ft

A

QV

ftftftbrA

r

Velocity Diagram

2rbV : sftsft

VV

VVVorb

rrbrrb /696.1

55sin

/389.1

sinsin 2

2

22222

2V : sftsftsftVVUVUV orrb /10.4555cot/389.1/08.46cotcos 22222222

Thus, from Eq. (2), we have the ideal shaft power

hp

sftlbf

hp

s

ftlbf

s

ftlbf

s

ft

s

ft

s

slug

VUmVUmTW s

83.1/550

11008100810.4508.46485.0

2222

. ANS

2rbV

2U

2rV

2

2V

2V

e

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