Page 168 - Water Engineering Hydraulics, Distribution and Treatment

146
Chapter 5
Water Hydraulics, Transmission, and Appurtenances
or
2
1
f
2g
where subscript 1 refers to upstream and 2 refers to downstream.
Typical loss of head items can be listed as follows:
From tank to pipe (entrance loss) = K
2
v
1
From pipe to tank (exit loss) = K
2g
Expansion (enlargement) = K (v 1 −v 2 ) . 2 . v 2g 2 2 . h = K (v − v ) 2
2g
Contraction = K (v 2 ) 2 .
2g
Fittings, elbows, and valves = K v 2 .
2g
Ideally, K must be determined experimentally. If this is unavailable, the values of K can be obtained from Table 5.2 and
Appendixes 17 and 25.
EXAMPLE 5.42 FORM AND MINOR HEAD LOSS DETERMINATION
Determine the form (minor) resistance caused by a sudden contraction from a 12-in. (0.3048 m)-diameter pipe to a 6-in. (0.1504-m)-
◦
3
3
diameter pipe when water flow is 0.5 ft /s (0.01416 m /s) and angle of contraction = 180 .
Solution 1 (US Customary System):
D = 12 in. = 1ft.
1
D = 6in. = 0.5 ft.
2
◦
= 180 .
D /D = 0.5∕1 = 0.5.
2
1
From Appendix 25, K = 0.37.
c
From Table 5.2, K = 0.37.
From Appendix 17, equivalent length of 6 in. pipe = 5.5 ft.
Q = A v = Q = A v .
1 1
1
2 2
2
2 2
0.5 = ( ∕4)(1) v = ( ∕4)(0.5) v .
2
1
v = 0.637 ft∕s.
1
v = 2.548 ft∕s.
2
(v ) 2
h = K 2 .
f
2g
2
h = 0.37(2.548) ∕(2 × 32.2) = 0.037 ft = 0.44 in.
f
Solution 2 (SI System):
D = 0.3048 m.
1
D = 0.1504 m.
2
◦
= 180 .
D /D = 0.1504∕0.3048 = 0.5.
1
2