Mechanics of Fluids

General energy equation for steady flow of any fluid 101
the mass flow rate ϱuδA. The rate at which kinetic energy passes through
the element is 1
2 (ϱuδA)u2 and consequently the rate at which kinetic energy
passes through the whole cross-section is � 1
2ϱu3dA. We may also integrate the mass flow rate through an element of crosssection
so as to obtain the total mass flow rate = � ϱudA. The rate at which
kinetic energy passes through the element divided by the rate of mass flow
is then:
�
1
2ϱu3dA �
ϱudA
For a fluid of constant density this reduces to
�
u3dA 2 � udA
(3.14)
Now unless u is constant over the entire cross-section this expression does
not correspond to u2 /2.
The factor by which the term u2 /2 should be multiplied to give the true
rate at which kinetic energy passes through the element divided by the rate
of mass flow is often known as the kinetic energy correction factor, α. Hence
for a fluid of constant density we may write
α = 1
A
�
A
�
u
�3 dA
u
The value of α can never be less than unity because the mean of different
cubes is always greater than the cube of the mean.
Example 3.1 Consider fully developed turbulent flow in a circular
pipe (to be discussed more fully in Chapter 7). The velocity over
the cross-section of the pipe varies approximately in accordance with
Prandtl’s one-seventh power law
u
�
y
�1/7 =
(3.15)
umax R
where R represents the radius of the pipe and u the velocity of the
fluid at a distance y from the wall of the pipe. The maximum velocity
umax occurs at the centre of the pipe where y = R. Calculate the kinetic
energy divided by mass.
Solution
Assuming the variation of density over the cross-section is negligible,
the integrals in the expression 3.14 may be evaluated by the use of
eqn 3.15. Because of axial symmetry the element of area δA may be
taken as an annulus of radius r and area 2πrδr (see Fig. 3.6). Then
�
u 3 �
dA = u 3 �
y
�3/7 max dA =
R
u3max R3/7 � R
y 3/7 2πrdr
0