Page 224 - Bird R.B. Transport phenomena

208 Chapter 7 Macroscopic Balances for Isothermal Flow Systems

H* 120' Fig. 7.5-1. Pipeline flow
with friction losses be-
"f cause of fittings. Planes
20' 1 and 2 are just under
A, - - Plane 2 the surface of the liquid.
100'

-300'-
Pump

90° elbow
Plane 1

SOLUTION The average velocity in the pipe is

(7.5-11)
TTR 2 ТГ(1/6) 2
and the Reynolds number is

(7.5-12)

Hence the flow is turbulent.
The contribution to E from the various lengths of pipe will be
v

2(2.30) (0.0049) (5 + 300 + 100 + 120 + 20)
2
(1/3)
2
= (0.156)(545) = 85 ft /s 2 (7.5-13)
The contribution to E from the sudden contraction, the three 90° elbows, and the sudden ex-
v
pansion (see Table 7.5-1) will be
Е (^Ч)/ = ^(2.30) (0.45 + ф + 1) = 8 ft /s 2 (7.5-14)
З
2
2
г
Then from Eq. 7.5-10 we get
0 + (32.2X105 - 20) + 0 = W - 85 - 8 (7.5-15)
m
Solving for W we get
m
2
W = 2740 + 85 - 8 « 2830 ft /s 2 (7.5-16)
m
This is the work (per unit mass of fluid) done on the fluid in the pump. Hence the pump does
2
2830 ft /s 2 or 2830/32.2 = 88 ft lb /lb m of work on the fluid passing through the system. The
/
mass rate of flow is
w = (12/60X62.4) = 12.5 lb /s (7.5-17)
w
Consequently
W = w% = (12.5X88) = 1100 ft Ybf/s = 2 hp = 1.5 kW (7.5-18)
m
which is the power delivered by the pump.