Mechanics of Fluids

104 The principles governing fluids in motion
Then for a constant-density fluid with no other heat transfer and no shaft
work performed the Steady-Flow Energy Equation reduces to
p1
ϱg + u2 1
2g + z1 − hf = p2
ϱg + u2 2
+ z2
(3.16)
2g
Here u1 and u2 represent mean velocities over the cross-sections (1) and (2)
respectively and the kinetic energy correction factor α is taken as unity. If
we assume further that the flow occurs in a horizontal pipe of uniform crosssection
then u1 = u2 and z1 = z2 and so (p1 − p2)/ϱg = h f. That is, the
displacement work done on the fluid in the pipe is entirely used in overcoming
friction.
For an incompressible fluid the values of (e2−e1) and the heat transfer resulting
from friction are in themselves rarely of interest, and so combining the
magnitudes of these quantities into the single term h f is a useful simplification.
We can see that the head loss h f represents, not the entire disappearance
of an amount of energy, but the conversion of mechanical energy into thermal
energy. This thermal energy, however, cannot normally be recovered as
mechanical energy, and so h f refers to a loss of useful energy. For a compressible
fluid, on the other hand, that statement would not, in general, be
true since the internal energy is then included in the total of useful energy.
We shall consider the flow of compressible fluids in more detail in
Chapter 11. For the moment we look more particularly at the behaviour
of incompressible fluids.
Example 3.2 A pump delivers water through a pipe 150 mm in diameter.
At the pump inlet A, which is 225 mm diameter, the mean
velocity is 1.35 m · s −1 and the pressure 150 mmHg vacuum. The
pump outlet B is 600 mm above A and is 150 mm diameter. At a
section C of the pipe, 5 m above B, the gauge pressure is 35 kPa. If
friction in the pipe BC dissipates energy at the rate of 2.5 kW and the
power required to drive the pump is 12.7 kW, calculate the overall
efficiency of the pump. (Relative density of mercury = 13.56)
Solution
Mean velocity at A = uA = 1.35 m · s −1
∴ by continuity,
uB = uC = uA × (Area)A
(Area)B,C
� �2 225
= 1.35 m · s
150
−1 = 3.038 m · s −1
Steady-Flow Energy Equation:
pA
ϱ
1
+
2 u2 A + gzA
Energy added by pump/time
+
Mass/time