Engg252 Engineering Fluid Mechanics
Week 9 Workshop SOLUTIONS
Ref: Cengel and Cimbala (2020)
Energy efficiency, energy equation and its applications
8.1 (Problem 5–19):
Solution:
Wind is blowing steadily at 10 m/s. The mechanical energy of air per unit mass, the
power generation potential, and the actual electric power generation are to be
determined.
Assumptions:
1. The wind is blowing steadily at
a constant uniform velocity.
2 The efficiency of the wind turbine
is independent of the wind speed.
3
Properties: The density of air is given to be ρ = 1.25 kg/m .
Analysis: Kinetic energy is the only form of mechanical energy the wind possesses,
and it can be converted to work entirely. Therefore, the power potential of the wind is
2
its kinetic energy, which is (V /2) per unit mass:
(2410 kW) (3sf)
The actual electric power generation is determined by multiplying the power
generation potential by the efficiency
Therefore, 722 kW of actual power can be generated by this wind turbine at the stated
conditions.
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Discussion: The power generation of a wind turbine can be shown to be proportional
to the cube of the wind velocity, and thus the power generation will change strongly
with the wind conditions.
8.2 (Problem 5–22):
Solution:
A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical
energy of the river water per unit mass, and the power generation potential of the
entire river are to be determined.
Assumptions:
1. The elevation given is the
elevation of the free surface of the
river.
2. The velocity given is the
average velocity of the river.
3. The mechanical energy of water
at the turbine exit is negligible
Properties: Take the density of water to be 1000 kg /m3.
Analysis:
The sum of the flow energy and the potential energy is constant for a given fluid body.
Then the total mechanical energy of the river water per unit mass becomes:
The power generation potential of the river water is obtained by multiplying the total
mechanical energy by the mass flow rate:
𝑚𝑚̇ = 𝜌𝜌 𝑄𝑄 = (1000 kg/m3) (5000 m3/s) = 500,000 kg/s
𝑊𝑊̇ max = 𝐸𝐸̇ mech
= 𝑚𝑚̇ emech
= (500,000 kg/s) (0.54405 kJ/kg) = 272,025 kW = 272 MW (3sf)
Therefore, 272 MW of power can be generated from this river as it discharges into
the lake if its power potential can be recovered completely.
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Discussion: Note that the kinetic energy of water is negligible compared to the
potential energy, and it can be ignored in the analysis. Also, the power output of an
actual turbine will be less than 272 MW because of losses and inefficiencies.
8.3 (Problem 5–73):
Solution:
The available head of a hydraulic turbine and its overall efficiency are given. The
electric power output of this turbine is to be determined.
Assumptions:
1. The flow is steady and incompressible.
2. The water levels at the reservoir and the
discharge site remain constant.
Eff.=78%
Turbine
Generator
Properties: The density of water = 1000 kg/m3.
Analysis:
When the turbine head is available, the corresponding power output is determined
from
[𝑉𝑉̇ =Q]
W turbine ,elec = η turbine − gen .W turbine ,e =η turbine − gen . m g hturbine ,e = η turbine − gen . ρ ∨ g hturbine ,e
Substitute values,
W
= 0.78 (1000kg / m 3 ) (1.3 m 3 / s ) (9.81 m / s 2 ) (50.0 m)
turbine ,elec
= 497367 W = 497 kW .
Discussion: The power output of the turbine is proportional to the available head and
the flow rate.
8.4 (Problem 5–74):
Solution:
A fan is to ventilate a bathroom by replacing the entire volume of air once every 15
minutes while air velocity remains below a specified value. The wattage of the fanmotor unit, the diameter of the fan casing, and the pressure difference across the fan
are to be determined.
Assumptions:
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1. The flow is steady and
incompressible.
2. Frictional losses along the flow (other
than those due to the fan-motor
inefficiency) are negligible.
3
4
3. The fan unit is horizontal so that z =
constant along the flow (or, the elevation
effects are negligible because of the low
density of air).
4. The effect of the kinetic energy
correction factors is negligible
α1 = α2 = 1.
3
Properties The density of air is given to be 1.25 kg/m .
Analysis
3
(a) The volume of air in the bathroom is V = 2 m × 3 m × 3 m = 18 m . Then the
volume and mass flow rates of air through the casing must be:
Q=
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
=
18 𝑚𝑚3
15𝑥𝑥60 𝑠𝑠
= 0.020 𝑚𝑚3 /𝑠𝑠
𝑚𝑚̇ = 𝜌𝜌 𝑄𝑄 = (1.25 kg/m3) (0.02m3/s) = 0.025 kg/s
Take points 1 and 2 on the inlet and exit sides of the fan, respectively.
Point 1 is sufficiently far from the fan so that P 1 = Patm and the flow velocity is negligible
(V1 = 0). Also, P2 = Patm. No turbine in the system. Then the steady flow energy
equation for this control volume between the points 1 and 2 reduces to:
P1
V12
P2
V22
+ α1
+ z1 + hfan, u =
+α2
+ z2 + hturbine, e + hL
ρg
ρg
2g
2g
0 + 0 + 0 + hfan, u
∴ hfan, u
V22
= 0 + 1 .0
+0+0+0
2g
V22
=
2g
𝑊𝑊̇ fan,u = 𝐸𝐸̇ mech
= 𝑚𝑚̇ g
= (0.025 kg/s) (9.81 m/s2)
72
𝑉𝑉22
2𝑔𝑔
2𝑥𝑥9.81
= 0.6125W
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𝑊𝑊̇ fan,elect
=
𝑊𝑊̇𝑓𝑓𝑓𝑓𝑓𝑓,𝑢𝑢
𝜂𝜂𝑓𝑓𝑓𝑓𝑓𝑓−𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
=
0.6125 𝑊𝑊
0.5
= 1.225 𝑊𝑊 (1.23 𝑊𝑊) (3𝑠𝑠𝑓𝑓)
Hence, the electric power rating of the fan/motor unit must be 1.23 W.
(b) For air mean velocity to remain below the specified value, the diameter of the
fan casing should be
𝜋𝜋
Q = A2 V2 = 𝐷𝐷22 V2
4
Hence D2 = �
4𝑄𝑄
𝜋𝜋 𝑉𝑉2
4𝑥𝑥 0.02 𝑚𝑚3/𝑠𝑠
=�
𝜋𝜋 𝑥𝑥 7.0
= 0.06031m (60.3 mm) (3sf)
(c) To determine the pressure difference across the fan unit, take points 3 and 4 to
be on the two sides of the fan on a horizontal line. Noting that z3 = z4 and V3 =
V4 since the fan is a narrow cross-section and neglecting flow losses (other than
the losses of the fan unit, which is accounted for by the efficiency), the energy
equation for the fan section reduces to:
2
3
=0
2
4
P3
V
P
V
+α3
+ z3 + hfan, u = 4 + α 4
+ z4 + hturbine, e + hL
ρg
ρg
2g
2g
P4 – P3 = (hfan,u)ρg =
𝑉𝑉22
2𝑔𝑔
ρg =ρ
𝑉𝑉22
2
= 1.25 x
72
2
=0
= 30.6 Pa
Therefore, the fan will raise the pressure of air by 30.6 Pa before discharging it.
Discussion: Note that only half of the electric energy consumed by the fan-motor
unit is converted to the mechanical energy of air while the remaining half is
converted to heat because of imperfections.
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8.5 (Problem 5–82): plus Excel solution
Solution:
Underground water is pumped to a pool at a given elevation. The maximum flow rate
and the pressures at the inlet and outlet of the pump are to be determined. Also plot
EGL and HGL for the flow between 1 and 2.
Assumptions:
1. The flow is steady and incompressible.
2. The elevation difference between the
inlet and the outlet of the pump is
negligible.
3. The frictional effects in piping are
negligible since we are after max. possible
flow. [hL or Emech loss = 0]
4
3
4 The effect of the kinetic energy
correction factors is negligible, α = 1. 0
Properties: The density of water = 1000 kg/m3.
Analysis: (a) The pump-motor draws 5-kW of power, and is 78% efficient. Then the
useful mechanical (shaft) power it delivers to the fluid is:
W pump,u = η pump motor W electric = 0.78 x 5 kW = 3.9 kW
Take point 1 at the free surface of underground water, which is also taken as the
reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both 1 and 2
are open to the atmosphere (P1 = P2 = Patm = 0,gauge pressure), the velocities are
negligible at both points (V1 ≅ V2 ≅ 0), and frictional losses in piping are neglected.
No turbine in the system. Then the energy equation for steady incompressible flow
through a control volume between these two points that includes the pump and the
pipes reduces to:
P1
V12
P2
V22
+ α1
+ z1 + hpump,u =
+ α2
+ z2 + hturbine,e + hL
ρg
2g
ρg
2g
0 + 0 + 0 + hpump,u = 0 + 0 + 30 + 0 + 0
∴ hpump,u = 30 m
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But , h pump ,u
W pump ,u
=
;
m g
∴ mass and volume flow rate are given by,
W pump ,u
3.9 x1000
=
=13.252 kg / s = 13.3 kg / s
m =
9.81 x 30
g h pump ,u
Q=
m
ρ
=
13.252 kg / s
= 13.252 x 10 −3 m 3 / s = 13.3 L / s
3
1000 kg / m
(b) Take points 3 and 4 at the inlet and the exit of the pump respectively.
We have,
Q
V3 =
A3
Q
V4 =
A4
=
=
13.252 x10 −3 m 3 /s
(π / 4 ) D32 m
13.252 x 10 −3 m 3 /s
(π / 4 ) D42 m
V32
=
= 3.443 m/s ;
= 0.604 m
2g
(π / 4 ) ( 0 .70 )2
13.252 x10 −3
V42
=
= 6 .749 m/s ;
= 2.32 m
2g
(π / 4 ) ( 0 .50 ) 2
13.252 x 10 −3
We take the pump as the control volume. Noting that z3 = z4, the energy equation for
this control volume reduces to:
V2
P3
V2
P
+ α 3 3 + z3 + hpump,u = 4 + α 4 4 + z 4 + hturbine,e + hL
2g
2g
ρg
ρg
∴ P4 − P3
ρ (α 3 V32 −−αα43 V42 )
+ ρ g hpump,u
2
1000 (1.0 x 3.443 2 − 1.0 x 6.749 2 )
=
+ 1000 * 9.81 * 30
2
= 277453 Pa = 277.5 x 10 3 Pa
= 278 kPa.
=
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Discussion: In an actual system, the flow rate of water will be less because of friction
in pipes. Also, the effect of flow velocities on the pressure change across the pump is
negligible in this case (under 6%) and it can be ignored.
Plots of EGL and HGL:
EGL
2.32m
NTS: Not to Scale
HGL
0.604m
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Prob 5-88
Pump-Motor wattage, W
5.0
Pump-motor efficiency
Density of water, kg/m3
Level difference between UG water
surface and pool, m
Diameter of pipe intake, mm
0.78
1000
30
Diamater of pipe discharge, mm
50.0
Acceleration due to gravity, g, m/s2
9.81
Kinetc energy correction factors for
inlet and outlet velocties, α
1.00
Elevation difference difference
between underground water
surface to pool level, m
15.0
30.0
45.0
60.0
75.0
90.0
105.0
120.0
135.0
150.0

m
70.0
h- pump, Maximum Intake
u (m) flow rate, Q, velocity,
L/s
v3, m/s
15.00
30.00
45.00
60.00
75.00
90.00
105.00
120.00
135.00
150.00
26.50
13.25
8.83
6.63
5.30
4.42
3.79
3.31
2.94
2.65
6.89
3.44
2.30
1.72
1.38
1.15
0.98
0.86
0.77
0.69
Intake
Discharge
velocity velocity, v4,
head, m m/s
(v3)^2/2g
2.417
0.604
0.269
0.151
0.097
0.067
0.049
0.038
0.030
0.024
Discharge
velocity
head, m
(v4)^2/2g
13.50
6.75
4.50
3.37
2.70
2.25
1.93
1.69
1.50
1.35
9.286
2.322
1.032
0.580
0.371
0.258
0.190
0.145
0.115
0.093
Pressure
difference,
kPa
(P4P3)
79.8
277.5
434.0
584.4
733.1
881.0
1028.7
1176.1
1323.5
1470.8
Since pipe friction losses were neglected, useful head supplied by the pump is the same as the water level difference.
For a given pump-motor assembly, as the head required to pump increases, the flow rate decreases.
This aspect is quite important for suitable pump-motor selection in practice.
The pressure difference across the pump however increases realtively linearly with the head difference.
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