CHAPTER 7: ENERGY PRINCIPLE
7.1 Derivation of the Energy Equation
 Recall the first law of thermodynamics, i.e.
dE
 Q W
dt
(1)
where E, the energy of the system, is an extensive property. The corresponding intensive
property is given by e, which is made up of ek, ep, and u (kinetic, potential and internal
energy). Q is the rate of flow of heat into the system and W is the rate of work done by the
system on its surroundings.
 To convert the first law of thermodynamics to Eulerian form, we use the Reynolds transport
theorem (chapter 5). Here, B = E, and so b = e = ek + ep + u. Also, ek = V 2/2 and ep = gz.
Consequently,
V 2

V 2

d





  V. dA
Q W 

gz

u

d
V


gz

u

c.s.  2


d t c.v.  2



(2)
 For convenience of analysis, the work term is divided into shaft work Ws and flow work Wf.

Flow Work is the work done by pressure forces as the system moves through space.
The rate of flow work is given as
W f  cs p V. dA

Shaft Work is defined as work other than flow work. It is usually the work done
through a shaft and is commonly associated with a pump or turbine. The rate of shaft
work is given the symbol W s .
 Substituting for W in equation (2) and rearranging terms, the energy principle becomes
V 2

V 2
d
p 





Q  Ws 

gz

u

d
V


gz

u


c.s.  2

  V. dA
d t c.v.  2




(3)
 Noting that p/ + u is the specific enthalpy, h, the energy principle can be written as
V 2

V 2

d





  V. dA
Q  Ws 

gz

u

d
V


gz

h


c .v . 
c
.
s
.



dt  2

 2

(4)
7.2 Simplified Forms of the Energy Principle
 For steady flow, the energy accumulation term is zero. If the properties are uniformly
distributed across the inlet and outlet ports, equation (4) simplifies to
Q W s 
V 2

V 2







m

gz

h

m

gz

h
 o 2

i

 2

cs

 o cs 
i
(5)
 For a control volume with one inlet port and one outlet port, the energy principle for
incompressible flow (after substituting back h = p/ + u) can be written as
2
2
 p1
 p2
V
V
1
2


m  

Q Ws    g z1  u1 

   g z 2  u 2  2 m

2




(6)
where the inlet port is designated as 1 and the outlet port as 2.
 For nonuniform velocity distribution at the inlet and outlet ports (e.g. as in figure 7.2), the
kinetic energy correction factor is introduced to account for the velocity distribution. Equation
 ) as
(6) is then written (after dividing through by m
p1
V12
p2
V22
1  
Q Ws    g z1  u1  1   g z 2  u2   2
m

2

2
(7)
where V1 and V2 are the mean velocities at the inlet and outlet ports. The kinetic energy
correction factor, , is defined by
3
1 V 
  A   dA
A V 
(8)
Thus, a = 1 when the velocity is uniform across the section. For most cases of turbulent flow,
 ≈ 1.05 and therefore it is common practice in engineering applications to let  = 1.
 The shaft-work term in equation (7) is usually the result of a turbine or pump in the flow
system. It is convenient to represent the shaft-work term as
W s  Wt  W p
(9)
where W t and W p are the magnitudes of power delivered by a turbine or supplied to a pump.
 It is often conventional to express the term associated with a pump and a turbine as the pump
head (hp) and the turbine head (ht), respectively, defined as
hp 
W p
ht 
m g
Wt
m g
(10)
 Hence, using equations (9) & (10), equation (7) becomes (after re-arrangement of terms)
1
V12
p2
V22
Q 
 1
 z1  h p 
 2
 z 2  ht   u 2  u1  

2g

2g
m g 
g
p1
(11)
 The last term enclosed in brackets in equation (11) represents energy loss due to viscous
action between fluid particles and due to heat transfer. It is simply referred to as the head loss
and is represented by hL. Thus, the final form of the energy equation is
V12
p2
V22
 1
 z1  h p 
 2
 z 2  ht  hL

2g

2g
p1
(12)
 Both pumps and turbines lose energy due to factors such as mechanical friction, viscous
dissipation, and leakage. These losses are accounted for by the mechanical efficiencies,
defined as
p
W p

W p , act
W t , act
t 
W t


m g h p
W p , act
W t , act
m g ht

 Q hp 

W p , act 


W t , act 

 Q ht 
(13)
where W p, act and Wt , act are the actual power supplied to the pump and the actual power
delivered by the turbine, respectively.
 It should be noted that the energy equation is different from the Bernoulli equation.

Bernoulli equation expresses the relationship between velocity and piezometric
pressure along a streamline in a steady, incompressible, and inviscid flow.

The energy equation was developed for viscous, incompressible flow in a pipe with
additional energy being added through a pump or extracted through a turbine.

When the flow is inviscid, with uniform velocity at the inlet and exit ports and there
are no pumps or turbines, the energy equation yields the same relationship as the
Bernoulli equation. However, they have different meanings.
7.3 Application of the Energy, Momentum, and Continuity Principles in
Combination
 The energy, momentum, and continuity equations can be used together to solve a variety of
problems.
 The use of the three equations in combination is illustrated to obtain the head loss in flow
through an abrupt expansion on page 261 of the textbook.
Example: Problem 7.26
The discharge in the siphon is 2.80 cfs, D = 8 in., L1 = 3 ft, and L2 = 3 ft. Determine the head loss
between the reservoir surface and point C. Determine the pressure at point B if three-quarters of
the head loss (as found above) occurs between the reservoir and point B. Assume  = 1.0 at all
locations.
Solution: To be presented in class.
Example: Problem 7.53
Water flows in this bend at a rate of 5 m3/s, and the pressure at the inlet is 650 kPa. If the head
loss in the bend is 10 cm, what will the pressure be at the outlet of the bend? Also estimate the
force of the anchor block on the bend in the x direction required to hold the bend in place.
Assume  = 1.0 at all locations.
Solution: To be presented in class.
7.3 Concept of the Hydraulic and Energy Grade Lines
 Recall the energy equation
V12
p2
V22
 1
 z1  h p 
 2
 z 2  ht  hL

2g

2g
p1
 The terms of the equation above have dimensions of length. This has led to the conventional
use of the hydraulic grade line and the energy grade line.
 The hydraulic grade line (HGL) in a piping system is formed by the locus of points located a
distance p/ above the center of the pipe, or p/ + z above a pre-selected datum.
 The energy grade line (EGL) is formed by the locus of points a distance
the HGL, or the distance
 V 2 2 g above
 V 2 2 g  p   z above the datum.
 The HGL and the EGL are drawn for a piping system as shown in figure 7.4.
 The following are helpful hints for drawing HGL and EGL.

The EGL is positioned above the HGL by an amount V2/2g. Thus, the HGL and EGL
coincide if velocity is zero (e.g. at the liquid surface in a reservoir).

The HGL and EGL slope downward in the direction of flow due to the head loss in the
pipe. If the pipe has uniform physical characteristics (diameter, roughness, shape), the
HGL and EGL along the pipe will be straight lines with slope hL/L.

A jump occurs in the HGL and the EGL whenever useful energy is added to the fluid
such as occurs with a pump; a drop occurs if useful energy is extracted from the flow
as occurs with a turbine (see figs. 7.5 and 7.6).

In a pipe or channel where the pressure is zero, the HGL is coincident with the system
(because p/ = 0 at these points).

If a flow passage changes diameter, the velocity therein will also change. Hence, the
distance between EGL and HGL will change. The slope of the EGL will also change.

If the HGL falls below the pipe, then p/ is negative, indicating subatmospheric
pressure.
Example: Problem 7.71
What power is required to pump water at a rate of 3 m3/s from the lower to the upper reservoir?
Assume the pipe head loss is given by hL = 0.018(L/D)(V2/2g), where L is the length of pipe, D is
the pipe diameter, and V is the velocity in the pipe. The water temperature is 10oC, the water
surface elevation in the lower reservoir is 150 m, and the surface elevation in the upper reservoir
is 250 m. The pump elevation is 100 m, L1 = 100 m, L2 = 1000 m, D1 = 1 m, and D2 = 50 cm.
Assume the pump and motor efficiency is 74%. In your solution, include the head loss at the pipe
outlet and sketch the HGL and the EGL. Assume  = 1.0 at all locations.