Page 164 - Water Engineering Hydraulics, Distribution and Treatment

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Water Hydraulics, Transmission, and Appurtenances
Chapter 5
EXAMPLE 5.36 HEAD LOSS DETERMINATION USING HAZEN–WILLIAMS FORMULA
Determine the loss of head in ft∕1,000 ft (m∕1,000 m) for a 12 in. (0.3048 m) pipe with a flow of 1 MGD (3.785 MLD) and a C=100.
Assume that the pipe is flowing full.
Solution 1 (US Customary System):
Method 1—use Fig. 5.16a.
When Q = 1 MGD, d = 12 in., C = 100
h = 2.07 ft∕1,000 ft.
f
Method 2—use Fig. 5.17.
When Q = 1MGD = 1.547 ft /s, d = 12 in., C = 100
h = 2.07 ft∕1,000 ft. 3
f
Method 3—use Appendix 14.
When Q = 1MGD = 684.4 gpm, d = 12 in., C = 100
h = 2.07 ft∕1,000 ft.
f
Method 4—use Eq. (5.37).
h = 10.6(Q MGD ∕C) 1.85 L∕D 4.87 (5.37)
f
h = 10.6(1∕100) 1.85 1,000∕(12∕12) 4.87 .
f
s = 2.07 ft∕1,000 ft.
Solution 2 (SI System):
Method 1—use Fig. 5.16a.
When Q = 3.785 MLD, D = 0.3048 m, C = 100, after all metric units are converted to the US customary units
h = 2.07 ft∕1,000 ft = 2.07 m∕1,000 m.
f
Method 2—use Fig. 5.17.
Read Solution 1 after SI units are converted to the US customary units.
h = 2.07 m∕1,000 m.
f
Method 3—use Appendix 14.
3
When Q = 0.0438 m /s = 43.8 L/s, D = 0.3048 m = 30.48 cm, and C = 100
h = 2.07 m∕1,000 m.
f
Method 4—use Eq. (5.37).
3
where Q = 0.0438 m /s, D = 0.3048 m, L = 1,000 m and C = 100
∕C) 1.85 L∕D 4.87 (5.37)
m
h = 10.67 (Q 3 ∕s
f
h = 10.67(0.0438∕100) 1.85 1,000∕(0.3048) 4.87 .
f
s = 2.17 m∕1,000 m.
EXAMPLE 5.37 MATHEMATICAL AND GRAPHICAL BASIS OF THE PIPE FLOW DIAGRAM
The pipe flow diagram (Fig. 5.16) establishes the numerical relationships between Q, v, d,and s for a value of C = 100. Conversion to
other magnitudes of C is simple because both v and Q vary directly as C. Show the mathematical and graphical basis of this diagram.
Solution:
s
1. Written in logarithmic form, Eq. (5.36), Q = 405Cd 2.63 0.54 ,is
gpd
(a) log Q = 4.61 + 2.63 log d + 0.54 log s,or
gpd
(b) log s =−8.54 − 4.87 log d + 1.85 log Q .
gpd