Page 150 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
But
L − x
.
Q
L
L − x
Q = Q
.
0
L
So
2
h = K
(L − x) dx.
L ∫
f
Q
2
2
h = K Q 2 0 2 2 0 Q 0 x x 0 L = L [(L − 2xL + x )dx].
f
L ∫
2
0
] L
Q 2 [ x 3
2
2
h = K 0 L x − x L + .
f 2
L 3 0
L
Q 2 [ L 3 ]
3
3
h = K 0 L − L + .
f L ∫ 3
2
0
2 L
h = KQ 0 3 .
f
Hence,
h = 1∕3(h ) .
f
f 0
(h ) = 3h .
f
f 0
That is, the head loss h in a pipe is equal to three times the head loss h in a perforated pipe.
f
f
EXAMPLE 5.23 HEAD LOSS BETWEEN TWO PRESSURE GAUGES
Water flows in a horizontal pipe of constant cross-sectional area. The pipe has a pressure gauge at location A with a pressure of 120
psig (832.8 kPa) and a pressure gauge at location B with a pressure of 90 psig (624.6 kPa). Determine the head loss between the two
pressure gauges.
Solution 1 (US Customary System):
Use the energy equations (5.5a and 5.6b):
2 2
(P ∕ + v ∕2g + Z ) + H = (P ∕ + v ∕2g + Z ) + h .
A A A a B B B f
Here Z = Z ,and H = 0.
A B a
2
2
Since the pipe size is constant, v ∕2g = v ∕2g.
B
A
The above energy equation reduces to
2
3
2
2
2
h = P ∕ − P ∕ = (120 lb∕in. − 90 lb∕in. )(144 in. ∕ft )∕(62.4lb∕ft ) = 69.23 ft.
B
f
A
Solution 2 (SI System):
2
1kPa = 1kN∕m .
Use the energy equations (5.5a and 5.6b),
2 2
(P ∕ + v ∕2g + Z ) + H = (P ∕ + v ∕2g + Z ) + h .
B
B
f
A
A
a
A
B
Here Z = Z ,and H = 0.
B
A
a
2
2
Since the pipe size is constant, v ∕2g = v ∕2g.
B
A
The above energy equation reduces to
2 2 3
h = P ∕ − P ∕ = (832.8kN∕m − 624.6kN∕m )∕(9.81 kN∕m ) = 21.22 m.
B
A
f