Energy Consideration in Steady Flow
Lecture - 6
Introduction



Up till now we have studied the motion of liquid
particle without taking into consideration any force
or energy causing the flow.
This lecture deals with the motion of liquids and
the forces causing the flow from a viewpoint of
energy considerations.
The first law of thermodynamics tells us that
energy can neither be created nor destroyed. But it
can of course, be changed from one form to other.
It follows that all forms of energy are equivalent.
Energy



The energy, in general, may be defined as:
“The capacity to do work.”
Or
“Quantity that is often understood as the ability of
a physical system to do work on other physical
systems.”
Since work is defined as a force acting through a
distance (a length of space), energy is always
equivalent to the ability to exert pulls or pushes
against the basic forces of nature, along a path of a
certain length.
Energies of a flowing fluid

1.
2.
3.
Though the energy exists in many forms, yet the
following are important from the subject point of
view.
Potential Energy
Kinetic Energy
Pressure Energy
1. Potential Energy

Energy of an object or a system due to the position
of the body or the arrangement of the particles of
the system.
PE in terms of Fluid





“It is the energy possessed by a liquid particle by
virtue of its position.”
A fluid particle of weight W situated a distance z
above datum possesses a potential energy of Wz.
Thus its potential energy per unit weight is z,
measured in units ft.lb/lb=ft or N.m/N=m.
The particle's potential energy per unit mass is gz,
measured in units of ft2/sec2 or m2/s2;
ts potential energy per unit volume is pgz, measured
in units of lb/ft2 or N/m2.
2. Kinetic Energy




Anything that is moving contains kinetic energy.
The kinetic energy of an object is
the energy which it possesses due to its motion.
It is defined as the work needed to accelerate a
body of a given mass from rest to its
stated velocity.
E.g. Wind, waves,
falling rocks
KE in terms of Fluid



A body of mass m when moving at a velocity V possesses a
kinetic energy, KE = ½ m V 2.
Thus if a fluid were flowing with all particles moving at the
same velocity, its kinetic energy would also be ½ m V2;
for unit weight of the fluid we can write this as:
1
KE
Weight

 2
mV

2
1
 2
( p  )V
pg 
2

V
2
(5.1a)
2g
where  represents the volume of the fluid mass. In BG units
we express V2/2g in ft-lb/lb = ft and in SI units as N .m/N = m.
KE in terms of Fluid

Similarly,
1
KE
KE
Volume

 2
mV

2
 2
Mass
1
mV
2

V
2
m
2
1
2
 2
( p  )V

(5.1b)

pV
2
(5.1c)
2
The units of V2/2 of course are ft2/sec2 in BG units or m2/s2 in
SI units. The units of pV2/2 are lb/ft2 or N/m2, which are units
of pressure.
True KE


In most situations the velocities of the different fluid particles
crossing a section are not the same, so it is necessary to
integrate all portions of the stream to obtain the true value of
the kinetic energy.
It is convenient to express the true value in terms of the mean
velocity V and a factor a (alpha), known as the kinetic-energy
correction factor or coriolis coefficient. Then,
True KE
Weight

V
2
2g
(5.2)
Correction Factor

In order to obtain an expression for alpha, consider the case
where the axial components of the velocity vary across a
section, as in Fig.

If u is the local axial velocity component at a point, the mass
flow per unit of time through an elementary area dA is:
r dQ = r udA
Correction Factor


Thus the true flow of kinetic energy per unit of time across
area dA is:
1/2(r udA)u2 = 1/2 r u3 dA
The weight rate of flow through dA is  dQ = rg udA. Thus,
for the entire section,
1
TrueKE / time

Weight / time

trueKE
weight

 udA
3
p u dA
 2
pg


3
u dA

2 g udA
Comparing Eq. (5.3) with Eqs. (5.2) and (4.3
get,
3
 
1
V
2
 u dA
 udA

1
AV
3

(5.3)
3
u dA
 udA
(5.4)
 A.V
), we
Correction Factor

and we get the same result if we use True KE/Mass, where
mass flow rate is:


p dQ  p udA

or if we use True KE/Volume, where the volume flow rate is.
 dQ   udA .

As the average of cubes is always greater than the cube of the
average, the value of  will always be more than 1. The
greater the variation in velocity across the section, the larger
will be the value of . For laminar flow in a circular pipe, 2
(Problem 5.1), for turbulent flow in pipes, ranges from 1.01 to
1.15, but it is usually between 1.03 and 1.06.
Correction Factor



In some instances it is very desirable to use the
proper value of , but in most cases the error made in
neglecting its divergence from 1.0 is negligible.
As precise values of  are seldom known, it is
customary in the case of turbulent flow to assume
that = 1, i.e., that the kinetic energy is V2/2g per unit
weight of fluid, measured in units of ft.lb/lb=ft or
N.m/N=m.
In laminar flow the velocity is usually so small that
the kinetic energy per unit weight of fluid is
negligible.
3. Pressure Energy



It is the energy, possessed by a liquid particle, by
virtue of its existing pressure.
A particle of fluid has energy due to its pressure
above datum, most usually its pressure above
atmospheric, although we normally do not refer to
this as pressure energy.
From Eq. (3.4) this pressure is p = h, and so the
depth of liquid that would produce this pressure, or
the "pressure head", is h=p/.
4. Internal Energy

Internal energy is stored energy that is associated
with the molecular, or internal state of matter; it
may be stored in many forms, including thermal,
nuclear, chemical, and electrostatic.
Problem:

In Laminar flow through a circular pipe the velocity profile is
a parabola (Fig), the equation of which is:
2

 r  
 
u  u m 1  

r0  





Where U is the velocity at any radius r, and Um is the
maximum velocity in the center of the pipe where r = 0 and ro
is the radius to the wall of the pipe. Find  ?
Solution:
EQUATION FOR STEADY MOTION OF AN IDEAL
FLUID ALONG A STREAMLINE, AND BERNOULLI'S
THEOREM



Referring to Fig., let us consider frictionless steady flow of an
ideal fluid along the streamline. We shall consider the forces
acting in the direction of the streamline on a small element of
the fluid in the stream tube, and we shall apply Newton's
second law, that is F = ma.
The cross-sectional area of the
element at right angles to the
streamline may have any shape and
varies from A to A + dA.
Recalling that in steady flow the
velocity does not vary at a point
(local acceleration = 0), but that it
may vary with position (convective
acceleration 0).
Bernoulli's Theorem:

The mass of the fluid element is m = r ds(A + 1/2dA) = r dsA
when we neglect second order terms. The forces tending to
accelerate or decelerate this mass along s are:
(a) the pressure forces:
1


pA   p  dp  dA  ( p  dp )( A  dA )   dpA
2


(b) the weight component in the direction of motion:
1
dz


   A  dA  cos    pgdsA
  pgAdz
2
ds



Applying ƐF = ma along the streamline, we get,
 dpA  pgAdz  ( pdsA ) a
Bernoulli's Theorem:

Dividing by the volume dsA,

dp
 pg
ds

 pa
ds
This states that the pressure gradient along the streamline
combined with the weight component in that direction causes
the acceleration a of the element. Recalling that a = V(dV/ds)
for steady flow (Equation 4.24), we get:

dp
ds

dz
 pg
dz
 pV
ds
dV
ds
Multiplying by ds/p and rearranging,
dp
p
 gdz  VdV  0
(5.5)
Bernoulli's Theorem:



We commonly refer to this equation as the one-dimensional
Euler3 equation, because Leonhard Euler (1707-1783), a
Swiss mathematician, first derived it in about 1750.
It applies to both compressible and incompressible flow, since
the variation of p over the elemental length ds is small.
Dividing through by g, we can also express Eq. (5.5) as:
dp

 dz  d
V
2
2g
0
(5.6)
Compressible Fluid:

For the case of a compressible fluid, since Sp. weight
() is not constant, we must introduce an equation
relating  (or r) to p and T before integrating Eq.
(5.5) or (5.6).
Incompressible Fluid:

For the case of an incompressible fluid (=constant),
we can integrate Eq. (5.6) to give:

Energy per unit weight:
p


z
V
2
 constant (along a streamline )
(5.7)
2g
We know this famous equation as Bernoulli's theorem, in
honor of Daniel Bernoulli (1700-1782), the Swiss physicist
who presented this theorem in 1738. If we multiply each term
first by g and then by r, we obtain the alternate forms:
Incompressible Fluid:

Energy per unit mass:
p
 gz 
p

V
2
 constant (along a streamline )
(5.8)
2
Energy per unit volume:
p  z 
1
pV
2
 constant (along a streamline )
(5.9)
2

The constant (of integration) is known as the Bernoulli
constant.
Assumptions:
1. It assumes viscous (friction) effects are negligible
2. It assumes the flow is steady
3. The equation applies along a streamline
4. It assumes the fluid to be incompressible
5. It assumes no energy is added to or removed from
the fluid along the streamline
Problem:

Glycerin (specific gravity 1.26) in a processing plant flows in
a pipe at a rate of 700 L/s. At a point where the pipe diameter
is 600 mm, the pressure is 300 kPa. Find the pressure at a
second point where the pipe diameter is 300 mm if the second
point is 1.0 m lower than the first point, neglect the head loss.
Solution:
Assignment:
1.
Assume frictionless flow in a long, horizontal, conical pipe, which has a
diameter of 3.6 ft at entrance and 2.4 ft at exit. The pressure head at the
smaller end is 15 ft of water. If water flows through this cone at the rate
of 95 cfs, find the velocities at the two ends and the pressure head at the
larger end.
2.
Assume the flow to be frictionless in the siphon shown in Fig., where a =
3 ft, b = 12 ft. Find the rate of discharge in cfs and the pressure head at B
if the pipe has a uniform diameter of 3 in.
3.
Application of Bernoulli’s Theorem in real world…