Fluid Mechanics Chapter Five The Energy Equation
CHAPTER FIVE THE ENERGY EQUATION
Energy Transfer via Pump Head, Turbine Head, and Head loss:
The energy equation can be developed by modifying the original Bernoulli's equation to
include the effects of the energy transfer devices. This process can be done by identifying the
energy transferred by these as follows:
HP: is the energy added to a unit weight of fluid as it flows through a pump. This term is
called the Pump Head.
HT: is the energy removed from a unit weight of fluid as it flows through a turbine (or a
hydraulic motor). This term is called the Turbine Head.
HL: is the energy loss (due to frictional effects) from a unit weight of fluid as it flows
through a components such as pipes, fittings, valves, strainers, and filters. This term is
called the Head Loss.
Development of Energy Equation:
The energy equation can now be developed by referring to figure (1), and now can state
the energy equation as follows;
Fig. (1). System for developing the energy equation.
The total energy possessed by a unit weight of fluid at station 1, plus the energy added to
this fluid by a pump, minus the energy removed by a turbine, minus the energy loss due to
friction, equals the total energy possessed by this unit weight of fluid upon arriving at station 2
Stated in mathematical terms, the energy equation is,
P1 V12 P2 V2 2
z H P HT H L z (1)
2g 1 2g 2
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Fluid Mechanics Chapter Five The Energy Equation
Comments on Energy Equation:
The following five comments should be noted about the energy equation;
1. Each energy term is called a head and thus has units of length (ft.lb)/lb or (N.m)/N.
2. The Bernoulli's equation is a special case of the energy equation. If HP, HT, and HL are
set equal to zero, the two equations becomes identical.
3. The pump head term has a positive sign in front of because pumps add energy to the
fluid.
4. The turbine head and head loss terms have a negative sign in front because these terms
represent energy removed from the fluid.
5. If HP, HT, and HL are each set equal to zero, the result is Bernoulli's equation.
Example 1: Water flows from a large reservoir at a rate of (1.0 ft3/s), as shown in figure. In
this system, there are frictional losses in the sudden contraction from the reservoir, the
pipe, the two 90o bend, and the wide-open globe valve. Determine the head loss HL due to
these fittings, the pipe, and the valve.
Solution: Applying energy equation between (1) and (2) gives,
P1 V12 P2 V2 2
z H P HT H L z
2g 1 2g 2
Where P1=P2= atmospheric pressure.
HP = HT = 0 since there is no pump or
turbine.
V1 = 0 due to large reservoir.
Z1-Z2=18 ft.
V2 2 V2 2
H L z1 z2 18
2g 2g
4 1
From continuity equation, Q2 V2 . A2 V2 24.3 ft/s
2.75 12 2
HL 18
24.32
18 9.17 8.83 ft
2 32.2
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Fluid Mechanics Chapter Five The Energy Equation
Example 2: Determine the flow rate for the system of above example, if frictional losses are
negligible small.
Solution: From the energy equation, setting HL=0, we obtain,
V2 2 g z1 z2 2 32.2 18 34 ft/s
2.75
2
The flow rate is, Q A.V 34 1.4 ft /s
3
4 12
Thus the effect of the frictional losses calculated in example 1 is to reduce the flow rate
from 1.4 ft3/s to 1.0 ft3/s, a 29% reduction.
Example 3: A pump is inserted in the pipe line of the system of example 1 to increase the flow
rate from 1.0 ft3/s to 1.25 ft3/s. if the head loss due to friction increases to 12 ft due to the
increased flow rate, determine the pump head.
Solution:
P1 V12 P2 V2 2 V2 2
z H P HT H L z HP z2 z1 H L
2g 1 2g 2 2g
Q 4 1.25
V2 30.3 ft/s
A 2.75 12 2
HP
30.32 18 12 8.3 ft
2 32.2
Thus, the addition of a pump with a pump head 8.3 ft increases the flow rate from 1.0 ft 3/s
to 1.25 ft3/s, a 25% increase.
The Energy Diagram:
Figure (2) shows the energy diagram for a constant diameter horizontal pipe in which the
fluid flowing from left to right. The use of constant-diameter horizontal pipeline here allows us
to better illustrate the HP, HT, and HL head terms because there are no changes in elevation
head Z and velocity head V2/2g between any two stations in the pipeline. In this way, any
change in the total head is directly due to HP, HT, and HL which affect only the value of the
pressure head P/γ.
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Fluid Mechanics Chapter Five The Energy Equation
Thus the system gain or lose pressure from one location in this pipeline to another as a
result of the existence of components such as pumps, turbines, pipe sections, valves, fittings,
filters, or strainers.
Fig. (2). Energy diagram for a constant-diameter horizontal pipeline.
Unlike the case for Bernoulli's equation, the total energy is not constant and hence the
TEL is not horizontal. Rather, in the direction of flow, the TEL increases a cross pumps,
decreases a cross turbines, decreases a cross valves, fittings, and decreases over the length of a
section of pipe.
Power Rate of Pumps and Hydraulic Motor:
In this section, we determine the amount of
fluid power delivered to a fluid by a pump. We
also determine the amount of fluid power delivered
to a hydraulic motor (or turbine) by the fluid.
Figure (3) shows the fluid power system for
performing this function where Fload is the external
load that equals the weight to be lifted.
The upper velocity of the piston can be
Q
calculated as, V Fig. (3). Pump driving a hydraulic cylinder
A against a load force.
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Fluid Mechanics Chapter Five The Energy Equation
Where, Q is the oil flow rate delivers by the pump to the hydraulic cylinder, and A is the
area of the piston.
Weight = Fload = P.A
The work done or energy expanded in the lifting the weight through a distance S can be
found as,
Work = energy = force time distance traveled by force.
Thus, energy = Fload × S = (P.A) × S
Since power is the rate of doing work or the rate at which energy is expanded.
energy PAS S
power V
time t t
power P.A.V F .V (2)
Equation (2) states that mechanical power (in this case delivered by a hydraulic cylinder)
equals the product of force and velocity. Checking the units for this product, we obtain power
units,
Power = N. m/s = J/s = watt S.I. units
Power = lb. ft/s = ft-lb/s U.S. units
1 h.p = 550 ft-lb/s or 1 h.p = 746 watt
Q A.V
Fluid Power P.Q (3)
Equation (3) states that fluid power equals the product of pressure and volume flow rate.
Observe the following power analogy between mechanical, electrical, and fluid systems:
Mechanical power = force × linear velocity = F.V
Electrical power = voltage × amperage = V.I
Fluid power = pressure × volume flow rate = P.Q
Fluid Power Added to Fluid by Pump:
For the test setup in figure (4), the following equation can be used for determining the
power delivered by a pump to a fluid, where ΔP is the pressure rise across the pump.
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Fluid Mechanics Chapter Five The Energy Equation
Therefore, ΔP = P2 – P1
Thus, Fluid Power P Q
Also, since the change in pressure (ΔP) across a
pump equals a product of the pump head (HP) and the
specific weight (γ) of the fluid,
Fluid Power .H P .Q (4)
Equation (4) is a key equation to use in conjunction
with the energy equation since both equations contain the
pump head parameter. Also, since that the product of
specific weight γ and volume flow rate Q equals weight
Fig. (4). Test setup for power delivered
flow rate W, equation (4) can be written as;
to a fluid by a pump.
Fluid Power W.H P (5)
Where W is the weight flow rate in units of (N/s) or (lb/s).
Fluid Power Delivered by Fluid to Hydraulic Motor/Turbine:
Figure (5) shows the test setup used when examining the
power delivered by the fluid to a hydraulic motor/turbine.
ΔP = P2 – P1
Fluid Power P Q .H T .Q W .H T
Example 1: What fluid power must a pump deliver to the
Fig. (5). Test setup for power delivered
water in a pipeline if the pump must increase the by a fluid to a turbine.
pressure by 500 kPa and maintain a flow rate of 0.05
m3/s?
Solution: Fluid Power P Q 500 10 3 0.05 25000 watt 25 kW
Example 2: A hydroelectric turbine receives water from a large reservoir and discharges the
water into the atmosphere through an 18 in. diameter pipe, as shown. If the velocity in the
pipe is 20 ft/s, determine the fluid power delivered to the turbine by the water.
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Fluid Mechanics Chapter Five The Energy Equation
Solution: writing the energy equation between (1) and (2) yields,
P1 V12 P2 V2 2
z H P HT H L z
2g 1 2g 2
Where,
P1 = P2 = atmospheric pressure,
V1 = 0 since we have a large reservoir,
HP = 0 there is no pump in the system,
HL = 0 since frictional losses are being neglected,
Z2 = 0 selected to be a reference (zero elevation).
V2 2 (20) 2
H T Z1 200 200 6.21 193.8 ft of water
2g 2 32.2
18
2
Q A.V 20 35.3 ft /s
3
4 12
Fluid Power .H T .Q 62.4 193.8 35.3 427,000 ft - lb/s 550 776 h.p
Efficiencies of Pumps and Hydraulic Motor:
Not all the input mechanical power to pump is converted into fluid power. Similarly, not
all the fluid power from a fluid is converted into output mechanical power of a hydraulic
motor/turbine. Energy is lost due to mechanical friction in the bearings and between other
mating parts of a pump or hydraulic motor/turbine.
The equation for the efficiency of a pump is,
Pump output fluid power
P 100 % (6)
pump input mechanical power
Due to the frictional energy losses, the output mechanical power of a hydraulic motor is
always less than the motor input fluid power.
The equation for the efficiency of a hydraulic motor/turbine is,
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Fluid Mechanics Chapter Five The Energy Equation
Turbine/Mo tor output mechanical power
T 100 % (7)
Turbine/Mo tor input fluid power
Due to frictional energy losses, the output mechanical power of a hydraulic motor is
always less than the motor input fluid power.
Example 3: For the system of example 1, determine the mechanical power that an electrical
motor must deliver to the pump if its efficiency is 80%.
Solution: using equation (6), we have
Pump output fluid power 25
P 100 % 80 100
pump input mechanical power mech. power
100
pump input mech. power 25 KW 31.3 KW
80
Example 4: For the system of example 2, determine the mechanical power delivered to the
electric generator by the hydroelectric turbine if its efficiency is 90%?
Solution: using equation (7), gives
Turbine/Mo tor output mechanical power
T 100 %
Turbine/Mo tor input fluid power
turbine output mechanical power
90 100
776 h.p
turbine output mech. power 0.9 776 698 h.p
Example 5: A ventilation fan that receives 0.5 KW of
power from an electric motor pulls air out of a large room
and into a square duct of side dimension 500 mm, as
illustrated in figure shown. If the velocity in the duct is 14
m/s, find the efficiency of the fan. Frictional losses in the
duct are negligibly small. Assume that the atmospheric
pressure and temperature remain constant.
Solution: select point (1) inside of the large room and point (2) downstream of the motor to
establish the CV for the energy equation. Thus,
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Fluid Mechanics Chapter Five The Energy Equation
P1
V1 2
z1 H P H T H L
P2 V2 2
V
z2 H P 2
14 9.99 m
2 2
2g 2g 2 g 2 9.81
Fan output fluid power .H P .Q
F 100 100
Fan input mechanical power 0.5 KW
Where γair = 12.0 N/m3 for standard atmospheric air
Q A.V 0.52 14 3.5 m3 /s
12 9.99 3.5
F 100 83.9 %
0.5 KW
Hence, 16.1 % of the electrical motor input shaft power is lost due to the friction and
turbulence of the air in the area of the fan blades.
Example 6: Water flows from the large reservoir through a (6 in)
diameter pipe and valve and discharges into the atmosphere, as
shown in figure. Even when the valve is wide open, the flow rate
is considered inadequate. It has been decided to add a pump in
the line to increase the flow rate by 50%. Neglecting frictional
losses in the pipe and valve, determine the horse power
requirement of an electric motor if the pump efficiency is 75%.
Solution: First, we determine the flow rate without a pump using the energy equation between
points (1) and (2),
P1 V12 P2 V2 2 V2 2
z H P HT H L z2 z1 z2
2g 1 2g 2g
V2 2 g ( z1 z2 ) 2 32.2 40 50.8 ft/s
The existing flow rate without a pump is found next;
6
2
Q A.V 50.8 9.97 ft /s
3
4 12
With a pump, the flow rate is to increase by 50%. Thus, the new flow rate is;
Qnew 1.5 9.97 15 ft 3 /s
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Fluid Mechanics Chapter Five The Energy Equation
Qnew 4 15
The new velocity can now be found, Vnew 76.4 ft/s
A 6 / 12 2
Applying the energy equation and solving for the pump head yields,
Vnew 2 (76.4) 2
HP z2 z1 (40) 50.6 ft of water
2g 2 32.2
Pump output fluid power .H P .Qnew
P 100 % 100
pump input mechanical power mech. or elect. power
62.4 50.6 15
mech. or elect. power 100 63150 ft - lb/s 115 h.p
75
Example 7: A pump delivers water at 0.1 m3/s from a lake
to a factory, as shown in figure. Frictional losses between
stations (1) and (2) can be represented by HL = 10V2/2g and
P2 is the atmospheric pressure. Determine the power
required to drive the pump if it has an efficiency of 75%.
Solution: writing energy equation between stations (1) and
(2), gives
P1 V12 P2 V2 2
z H P HT H L z2
2g 1 2g
V2 2 V2 2 V2 2
H P 10 z2 11 z2
2g 2g 2g
4 0.1
From continuity equation, Q A2 .V2 V2 12.7 m/s
(0.1) 2
(12.7) 2
H P 11 25 116 .4 m of water
2 9.81
Then, pump fluid power = .H P .Q 9800 116 .4 0.1 114 ,000 watt 114 KW
fluid power 114
P 100 % 75 100
mech. power mech. power
114
The power required to drive the pump = Mechanical power = 100 152 KW
75
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Fluid Mechanics Chapter Five The Energy Equation
Example 8: For steady, incompressible flow of water from the
tank through the pipe and nozzle, determine the discharge (Q)
and the pressure at point (3). Assume energy losses are (4V32/2g)
from section (1) to section (3) and (0.05V22/2g) from section (3)
to section (2).
Solution: Applying energy equation between points (1) and (2),
gives
P1 V12 P2 V2 2
z H P H T H L (1 2) z2
2g 1 2g
V2 2 V3 2 V2 2
z1 H L (1 2) z1 ( H L (1 3) H L (3 2) ) 25 4 0.05
2g 2g 2g
2
A D 1
From continuity equation, A3 .V3 A2 .V2 V3 V2 2 V2 2 V3 V2
A3 D3 9
V2 2 4 V2 2 V2 2 V2 2 V2 2
25 0.05 25 0.1 1.1 25
2g 81 2 g 2g 2g 2g
V2 38.25 ft/s and V3 4.25 ft/s
2
2
Q 38.25 0.8345 ft /s
3
4 12
To determine P3, written energy equation between points (1) and (3), gives
P1 V12 P3 V3 2
z H P H T H L (1 3) z3
2g 1 2g
P3 V3 2 V3 2 V3 2
25 4 25 5 23.6 ft of water
2g 2g 2g
P3 62.4 23.6 1472 .49 psfg 10.225 psig
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Fluid Mechanics Chapter Five The Energy Equation
Example 9: In a fire fighting system, a pipeline with a pump heads to a nozzle as shown in
figure. Find the flow rate when the pump heads to develop a head of 80 ft. assume that the
head loss in the (6 in) diameter pipe may be expressed by HL = 5V62/2g, while the head loss in
the (4 in) diameter pipe is HL = 12V42/2g. Sketch the energy line and hydraulic grade line, and
find the pressure head at the suction side of the pump.
Solution: Select the datum as the elevation of the water surface in the reservoir. If V3 is the jet
velocity, from continuity equation,
2
3
V6 . A6 V3 . A3 V6 V3 0.25V3
6
2
3
V4 . A4 V3 . A3 V4 V3 0.563V3
4
Applying energy equation from the surface of the reservoir (point 1) to the jet (point 3),
P1 V12 P3 V3 2
z H P HT H L z3
2g 1 2g
V32 V6 2 V4 2 V3
2
80 H L (6) H L ( 4) (80 70) 80 5 12 10
2g 2g 2g 2g
5(0.25V3 ) 2 12(0.563V3 ) 2 V3 2
80 10 V3 29.7 ft/s
2g 2g 2g
3
2
Q A3 .V3 29.7 1.458 ft /s
3
4 12
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Fluid Mechanics Chapter Five The Energy Equation
V6 2 5(0.25V3 ) 2
Head loss in suction pipe is, H L ( 6) 5 4.28 ft
2g 2g
V4 2 12(0.563V3 ) 2
Head loss in discharge pipe is, H L ( 4) 12 52 ft
2g 2g
V3 2 V4 2 V6 2
13.7 ft 4.33 ft 0.856 ft
2g 2g 2g
Inspection of the figure shows that the pressure head on the suction side of the pump is,
PB
70 50 4.28 0.856 14.86 ft
Example 10: A water pump of 25 h.p with 75% overall efficiency draws water from a sump
2.5m below its axis and delivers it to a storage tank 17.5 m above the axis. If the loss of head
in the system is 2.0 m of water due to frictional effect, find the discharge. Neglect the velocity
head.
Solution: Applying energy equation, taking datum at the axis of the pump.
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Fluid Mechanics Chapter Five The Energy Equation
2
P1 V1 P2 V2 2
z1 H P H T H L z2
2g 2g
H P 2.5 2 17.5 H P 22 m
fluid power .H P .Q
P 100 % 75 100 (i.e, 1 h.p =746 watts)
mech. power 25 746
25 75 746 13987 .5
.H P .Q 13987 .5 Q 0.0468 m3 /s
100 9810 22
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