Solutions Manual
for
Fundamentals of Thermal Fluid Sciences
5th Edition
Yunus A. Çengel, John M. Cimbala, Robert H. Turner
McGraw-Hill, 2017
Chapter 13
MOMENTUM ANALYSIS OF FLOW
SYSTEMS
PROPRIETARY AND CONFIDENTIAL
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13-2
Newton’s Laws and Conservation of Momentum
13-1C
Solution We are to express Newton’s three laws.
13-2C
Solution We are to discuss Newton’s second law for rotating bodies.
13-3C
Solution We are to discuss if momentum is a vector, and its direction.
13-4C
Solution We are to discuss the conservation of momentum principle.
13-5C
Solution We are to compare the reaction force on two fire hoses.
13-6C
Solution We are to discuss surface forces in a control volume analysis.
Analysis All surface forces arise as the control volume is isolated from its surroundings for analysis, and the
13-4
13-7C
Solution We are to discuss the importance of the RTT, and its relationship to the linear momentum equation.
13-8C
Solution We are to discuss the momentum flux correction factor, and its significance.
Analysis The momentum-flux correction factor
enables us to express the momentum flux in terms of the mass flow
13-9C
Solution We are to discuss the momentum equation for steady one-D flow with no external forces.
Analysis The momentum equation for steady flow for the case of no external forces is
out in
0F mV mV
where the left hand side is the net force acting on the control volume (which is zero here), the first term on the right hand
side is the incoming momentum flux, and the second term on the right is the outgoing momentum flux by mass.
Discussion This is a special simplified case of the more general momentum equation, since there are no forces. In this
case we can say that momentum is conserved.
13-10C
Solution We are to explain why we can usually work with gage pressure rather than absolute pressure.
13-11C
Solution We are to discuss if the upper limit of a rocket’s velocity is limited to V, its discharge velocity.
13-12C
Solution We are to describe how a helicopter can hover.
13-13C
Solution We are to discuss the power required for a helicopter to hover at various altitudes.
13-6
13-14C
Solution We are to discuss helicopter performance in summer versus winter.
13-15C
Solution We are to discuss if the force required to hold a plate stationary doubles when the jet velocity doubles.
Analysis No, the force will not double. In fact, the force required to hold the plate against the horizontal water
13–16C
Solution We are to describe and discuss body forces and surface forces.
13-17C
Solution We are to discuss the acceleration of a cart hit by a water jet.
13-18C
Solution We are to discuss the maximum possible velocity of a cart hit by a water jet.
13-8
13–19
Solution The velocity distribution for turbulent flow of water in a pipe is considered. The darg force exerted on the
pipe by water flow is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water is discharged to the atmosphere, and thus the gage
Daveave FApApdAuAUU
0
2211
1
Recalling that
max
816.0 UU ave
for turbulent flow, we write
7/17/1 /12255.1/1
U
ave
13–20
Solution A horizontal water jet is deflected by a stationary cone. The horizontal force needed to hold the cone
stationary is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water is discharged to the atmosphere, and thus the gage
Considering front view of the cone , we apply the conservation of mass,
A
c
j
jexitjhD
V
QdAVQ
2
0
h25.0
2
40
40025.0
4
2
mmmh 25.11025.1 3
Linear momentum equation in the x direction gives
x
FdAnVV
FdYDVVCosAVV c
h
jj
0
1
Since velocity is linear, it is in the form of
bYaV
. It is given that V=0 when Y=0, and V=Vj when
Y=h=1.25×10-3 m. Therefore we obtain
YV 32000
FdYYCos
3
1025.1
0
2222 32000)60(25.01000025.0
4
401000
N523 262785F
13–10
13–21
Solution A water jet of velocity V impinges on a plate moving toward the water jet with velocity ½V. The force
required to move the plate towards the jet is to be determined in terms of F acting on the stationary plate.
Assumptions 1 The flow is steady and incompressible. 2 The plate is vertical and the jet is normal to plate. 3 The pressure
Moving plate: (
)5.1(and 5.1 VAAVmVV iii
)
FAVVAFR25.225.2)5.1( 22
13-22
Solution A 90 elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The gage
pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Frictional effects are negligible in the calculation of the pressure
drop (so that the Bernoulli equation can be used). 3 The weight of the elbow and the water in it is negligible. 4 The water is
13–12
13-23
Solution A 180 elbow forces the flow to make a U-turn and discharges it to the atmosphere at a specified rate. The
gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Frictional effects are negligible in the calculation of the pressure
drop (so that the Bernoulli equation can be used). 3 The weight of the elbow and the water in it is negligible. 4 The water is
13-24E
Solution A horizontal water jet strikes a vertical stationary plate normally at a specified velocity. For a given
anchoring force needed to hold the plate in place, the flow rate of water is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in a plane normal to
the jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the
/s
3
ft 7.22 3
lbm/ft 62.4
lbm/s 450.8
m
V
Waterjet
13-25
Solution A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The
anchoring force needed to hold the elbow in place is to be determined.
kN 0.4905 N 490.5)m/s kg)(9.81 50(2 mgW
We take the elbow as the control volume, and designate the
entrance by 1 and the outlet by 2. We also designate the
horizontal coordinate by x (with the direction of flow as being
m/s 12
)m 0025.0)( kg/m(1000
kg/s30
m/s 0.2
)m 0150.0)( kg/m(1000
kg/s30
23
2
2
23
1
1
A
m
V
A
m
V
gage ,112
212
1
2
2
22 z
g
g
g
g
g
g
Substituting,
kN 1
m/s) 2(m/s) 12(
22
23
45
FRz
FRx
13–15
13-26
Solution A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The
anchoring force needed to hold the elbow in place is to be determined.
2
by 1 and the outlet by 2. We also designate the horizontal coordinate
by x (with the direction of flow as being the positive direction) and
m/s 12
)m 0025.0)( kg/m(1000
kg/s30
m/s 0.2
)m 0150.0)( kg/m(1000
kg/s30
23
2
2
23
1
1
A
m
V
A
m
V
Water
30 kg/s
W
1
13-27
Solution Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity.
The braking force and the power wasted by the brakes are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all directions in
the plane of the back surface. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the
splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Fiction during motion is
negligible. 5 There is no acceleration of the cart. 7 The motions of the water jet and the cart are horizontal. 6 Jet flow is
nearly uniform and thus the effect of the momentum-flux correction factor is negligible,
1.
Analysis We take the cart as the control volume, and the
direction of flow as the positive direction of x axis. The relative
velocity between the cart and the jet is
m/s 251035
cartjet VVVr
Therefore, we can view the cart as being stationary and the jet moving
with a velocity of 25 m/s. Noting that water leaves the nozzle at 20 m/s
and the corresponding mass flow rate relative to nozzle exit is 30 kg/s,
the mass flow rate of water striking the cart corresponding to a water jet
velocities in the negative x-direction. Substituting the given values,
N 536
N 8.535
m/skg 1
N 1
m/s) 52kg/s)( 43.21(2
brake rrVmF
35 m/s
Waterjet
10 m/s
FRx
13–17
13-28
Solution Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the
cart if the brakes fail is to be determined.
35 m/s
2
m/s 1.34
N 1
m/skg 1
kg 400
N 8.535 2
cart
m
F
a
Discussion This is the acceleration at the moment the brakes fail. The acceleration will decrease as the relative velocity
between the water jet and the cart (and thus the force) decreases.
Waterjet
10 m/s
400 kg
13-29E
Solution A water jet hits a stationary splitter, such that half of the flow is diverted upward at 45, and the other half is
directed down. The force required to hold the splitter in place is to be determined.
Assumptions 1 The flow is steady and incompressible. 2
lbm/s 6240/s)ft )(100lbm/ft 4.62(33 Vm
We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and
the outlet of either arm by 2 (both arms have the same velocity and mass flow rate). We also designate the horizontal
coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z.
The momentum equation for steady flow is
inout
VmVmF
. We let the x- and y- components of the
00)sin()sin(
)1(coscos)(2
2
2
1
2
2
1
12
2
1
VmVmF
VmVmVmF
Rz
Rx
18 ft/s
13–19
13–30E
rho=62.4 “lbm/ft3″
V_dot=100 “ft3/s”
V=20 “ft/s”
m_dot=rho*V_dot
F_R=-m_dot*V*(cos(theta)-1)/g “lbf”
,
m
, lbm/s
FR, lbf
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
6240
0
59
234
519
907
1384
1938
2550
3203
3876
4549
5201
5814
6367
6845
7232
7518
7693
7752
Discussion The force rises from zero at
= 0o to a maximum at
= 180o, as expected, but the relationship is not linear.
13–20
13–31
Solution A horizontal water jet impinges normally upon a vertical plate which is held on a frictionless track and is
initially stationary. The initial acceleration of the plate, the time it takes to reach a certain velocity, and the velocity at a
given time are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water always splatters in the plane of the retreating plate. 3
The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric
pressure which is disregarded since it acts on all surfaces. 4 The track is nearly frictionless, and thus fiction during motion
is negligible. 5 The motions of the water jet and the cart are horizontal. 6 The velocity of the jet relative to the plate remains
constant, Vr = Vjet = V. 7 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible,
1.
The momentum equation for steady flow in the x (flow) direction reduces in this case to