Page 166 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
EXAMPLE 5.39 DETERMINATION OF PIPE DIAMETER USING A NOMOGRAM FOR SOLVING THE
HAZEN–WILLIAMS EQUATION
Given the following:
1. If a flow of 210 L/s is to be carried from point A to point B by a 3,300-m ductile iron pipeline (C = 100) without exceeding
a head loss of 43 m, what must be the pipe diameter?
2. If the elevation of point A is 580 m and the level of point B is 600 m and a minimum residual pressure of 3 bars (30 m of
water) must be maintained at point B, what then must be the minimum actual pressure at point A?
Solution:
1. Pipe diameter:
C = 100.
s = h ∕L = 43∕3,300 = 13 .
f
Q = 210 L∕s = 0.210 m 3∕s.
From the nomogram of Fig. 5.17, D = 380 mm.
Therefore, use a nominal pipe diameter of 400 mm so as not to exceed a head loss of 43 m.
2. Actual pressure at point A:
2
2
P ∕ + Z + (v ) ∕2g = P ∕ + Z + (v ) ∕2g + h .
A A A B B B f
Knowing that
v = v since no change in D or Q
A
B
3
the nomogram (C = 100) of Fig. 5.17 for D = 400 mm and Q = 0.210 m ∕swillgive s = 11 .
h = sL = 11 × 3,300 m = 36 m.
f
P ∕ + 580 = 30 + 600 + 36.
A
P ∕ = 86 m or 8.6 bars is the actual pressure at point A.
A
EXAMPLE 5.40 APPLICATIONS OF HAZEN–WILLIAMS FORMULA
Determine the expected velocity, full flow, and head loss for a 24 in. (609.6 mm) circular conduit with a Hazen–Williams coefficient
C = 100, a length L = 1,000 ft (304.8 m), and a hydraulic gradient s = 2.25%.
Solution 1 (US Customary System):
1. According to the nomogram shown in Fig. 5.17,
3
v = 3.3ft∕sand Q = 10 ft ∕s.
s = 0.0025 = h ∕L = h ∕1,000 ft.
f
f
h = 1,000 ft × 0.0025 = 2.5ft.
f
2. Using Eqs. (5.34) and (5.36),
v = 0.115Cd 0.63 0.54 = 0.115 × 100(24) 0.63 (0.0025) 0.54 = 3.3ft∕s.
s
s
v = 0.55CD 0.63 0.54 = 0.55 × 100(24∕12) 0.63 (0.0025) 0.54 = 3.3ft∕s.
3
s
Q 3 = 0.432CD 2.63 0.54 = 0.432 × 100(24∕12) 2.63 (0.0025) 0.54 = 10.5ft ∕s.
ft ∕s
h = 1,000 ft × 0.0025 = 2.5ft.
f
Solution 2 (SI System):
1. According to the nomogram shown in Fig. 5.17,
3
v = 1m/s andQ = 0.28 m ∕s.
h = 304.8m × 0.0025 = 0.762 m.
f
2. Using Eqs. (5.34) and (5.36),
v = 0.3545CD 0.63 0.54 = 0.3545 × 100(0.6096) 0.63 (0.0025) 0.54 = 1m∕s
s
3
= 0.278CD 2.63 0.54 = 0.278 × 100(0.6096) 2.63 (0.0025) 0.54 = 0.298 m ∕s.
s
m
Q 3 ∕s
h = 304.8m × 0.0025 = 0.762 m.
f