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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
Turbomachines, Bernoulli’s Equation
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
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.
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.
Net Convection Efflux Term
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.
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.
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
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.
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)(
=
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.
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
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.
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
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.
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
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.
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
CVgmindefornonandstationary
CV
xCVdVdV
t
udVu
dt
d
dt
dP
Net Convection Efflux Term
CS
sf AdVu )( /
:
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.
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
CVgmindefornonandstationary
CV
yCVdVdV
t
vdVv
dt
d
dt
dP
Net Convection Efflux Term
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
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.
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
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.
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 = ?
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.
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
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.
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
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.
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 )
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.
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
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.
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
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.
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)
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.
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,
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.
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
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.
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
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.
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
CVgmindefornonandstationary
CV
CCVdVdV
t
VrdVVr
dt
d
dt
tHd
Net Convection Efflux Term
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)
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.
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