Page 135 - Water Engineering Hydraulics, Distribution and Treatment

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A = (6∕12) × 0.785 ft = 0.196 ft .
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Q = 1.224 ft /s = (A )(v ) = (A )(v ).
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1.224 ft /s = (0.785 ft )(v ) = (0.196 ft )(v ).
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v = 1.6 ft/s.
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v = 6.2 ft/s.
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Solution 2 (SI System):
Q = (A )(v ) = (A )(v ) = 2,082 Lpm = 0.0347 m /s.
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A = (30.7∕100) × 0.785 m = 0.0725 m .
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A = (15.24∕100) × 0.785 m = 0.0182 m . 3 5.1 Fluid Mechanics, Hydraulics, and Water Transmission 113
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Q = 0.0347 m /s = (A )(v ) = (A )(v ).
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0.0347 m /s = (0.0725 m )(v ) = (0.0182 m )(v ).
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v = 0.479 m/s.
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v = 1.907 m/s.
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EXAMPLE 5.6 DETERMINATION OF HYDRAULIC PRESSURE IN A WATER PIPE
Water flows through the nozzle shown in Fig. 5.1 and deflects the mercury in the U-tube gauge at a treatment plant. Determine the
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value of h if the pressure at point D is 20 psi (138.8 kN/m ), and m is 2.75 ft (0.84 m).
Nozzle
D
m
A
h
B C
Figure 5.1 Fluid flow through a nozzle.
Solution 1 (US Customary System):
Method A. Pressure balance method:
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20 lb/in. + [(62.4 lb/ft )(2.75 + h) ft]/(144 in. /ft ) = [(13.57 × 62.4 lb/ft )(h) ft]/(144 in. /ft ) + P .
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P = 0psi = 0 lb/in. .
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144 × 20 + 62.4 (2.75 + h) = 13.57 × 62.4h.
h = 3.9 ft.
Method B. Water column length method:
1psi = 2.307 ft of water.
1ftHg = 13.57 ft of water.
20 × 2.307 + (2.75 + h) = 13.57h.
h = 3.9 ft.