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Fluid Mechanics 06
Energy, Work and Power
Work:- Work is force acting through a
distance when the force is parallel to the
direction of motion.
Energy:- Types of Work Stored like kinetic
energy, potential energy, thermal energy.
Power:- Power expresses a rate of work or
energy
Work and Energy Units
Common units include the joule (J), newtonmeter, watt-hour (Wh), foot-pound-force
(ft-lbf), calorie (cal), and the British thermal
unit (Btu). Where
Btu = 1055 J(It is the amount of energy needed
to heat one pound of water by one degree
Fahrenheit)
Cal = 4.2 J(it is the amount of thermal energy
needed to raise the temperature of 1 gram of
water by 1°C)
ft-lbf = 1.356 J
Power Units
Common units for power are the watt (W),
horsepower (hp), and the ft-lbf/s.
Where
1 hp = 550 ft-lbf/s.
1 hp = 745.7 watt
Energy Equation
First Law of thermodynamic

ʹ − ʹ =

Thermal energy is positive when there is an
addition of thermal energy to the system and
negative when there is a removal.
Work is positive when the system is doing work
on the environment and negative when work is
done on the system.
Pipe Flow Energy Equation
1
ɣ
+ 1
12
+
2∗
+ ℎ =
2
+
ɣ
2
22
+
2∗
+ ℎ + ℎ
The preview equation based on three main
assumptions:(a)the flow is steady.
(b) the control volume has one inlet port and
one exit port.
(c) the density of the flow is constant.
Example
A horizontal pipe carries cooling water at 10°C
for a thermal power plant from a reservoir as
shown. The head loss in the pipe is where L is
the length of the pipe from the reservoir to
the point in question, V is the mean velocity in
the pipe, and D is the diameter of the pipe. If
the pipe diameter is 20 cm and the rate of
flow is 0.06 m3/s, what is the pressure in the
pipe at L = 2000 m.
Solution
1
ɣ
+ 1
12
+
2∗
+ ℎ =
2
+
ɣ
2
Where
P1=Patm=0, v1=0, hp=0, ht=0
Z1=100m, z2=20m
2 =
ℎ =
2

=
0.02∗
0.06
= 1.910 m/s
Π
∗(0.22 )
4
2000
∗1.9122
0.2
2∗9.81
= 37.2 m
22
+
2∗
+ ℎ + ℎ
2
1.912
100 =
+ 20 +
+ 37.2
1000 ∗ 9.81
2 ∗ 9.81
From the Equation P2 = 418 KPa
Power
 = ʹ ∗  ∗ ℎ
There are loses in energy due to factors such
as mechanical friction and leakage.
Efficiency is the ratio of power output to
power input
Then
  =  ∗ ʹ ∗  ∗ ℎ
Example
A pipe 50 cm in diameter carries water
(10°C) at a rate of 0.5 m3/s. A pump in the
pipe is used to move the water from an
elevation of 30 m to 40 m. The pressure at
section 1 is 70 kPa gage and the pressure at
section 2 is 350 kPa gage. What power in
kilowatts and in horsepower must be
supplied to the flow by the pump? Assume
hL = 3 m of water.
Solution
1
ɣ
+ 1
12
+
2∗
+ ℎ =
2
+
ɣ
2
22
+
2∗
Where
P1=70000 Pa, P2=350000 Pa
z1=30 m, z2=40m
V1=v2, Ht=0, hL=3m
70000
+ 30 + ℎ
1000 ∗ 9.81
350000
=
+ 40 + 3
1000 ∗ 9.81
+ ℎ + ℎ
ℎ = 41.5
 = ʹ ∗  ∗ ℎ = 1000*0.5*9.81*41.5=204kW
Example
At the maximum rate of power generation, a small
hydroelectric power plant takes a discharge of
14.1 m3/s through an elevation drop of 61 m. The
head loss through the intakes, penstock, and
outlet works is 1.5 m. The combined efficiency of
the turbine and electrical generator is 87%. What
is the rate of power generation?
Solution
1
ɣ
+ 1
12
+
2∗
+ ℎ =
2
+
ɣ
2
22
+
2∗
+ ℎ + ℎ
Where
P1=P2=Patm=0
Z1=61m, z2=0
V1=v2=0
Hp=0, hL=1.5m
Ht=61-1.5=59.5m
 = ʹ ∗  ∗ ℎ=1000*9.81*14.1*59.5=8.23MW
  =  ∗  = 0.87 ∗ 8.23 = 7.16

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