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14.4 Combined Collection of Wastewaters and Stormwaters 511

EXAMPLE 14.1 SIZING OF AN INVERTED SIPHON
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A siphon is to carry a minimum dry-weather flow of 1.0 ft /s (28.3 L/s or 0.0283 m /s), a maximum
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dry-weather flow of 3.0 ft /s (85 L/s or 0.085 m /s), and a storm flow of 45.0 ft /s (1,274 L/s or
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1.274 m /s) in three pipes. Select the proper diameters to ensure velocities of 3.0 ft/s (0.90 m/s) in
sanitary sewer pipes and 5.0 ft/s (1.5 m/s) in storm sewers.
Solution 1 (U.S. Customary System):
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1. For minimum dry-weather flow of 1.0 ft /s, the nearest standard diameter of pipe is calcu-
lated from:
A = Q>V
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pD >4 = Q>V
D = 24Q>pV
= 1224 * 1>p * 3.0 = 8in.
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Actual capacity AV (8) 3>(4 144) 1.05 ft /s
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2. For maximum dry-weather flow in excess of the minimum, namely, 3.0 1.05 2.0 ft /s,
the nearest standard diameter of pipe is:
D = 1224 * 2.0>p * 3.0 = 12 in.
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Actual capacity (1>4) 3 2.36 ft /s
3. For storm flows in excess of maximum dry-weather flow, namely, 45 (2.36 1.05)
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41.6 ft /s, the next lowest standard diameter of pipe is
D = 1224 * 41.6>p * 5.0 = 39 in; use 40 in., which will have to flow at a velocity of
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V (41.6 4) 144>( 40 ) 4.8 ft/s
Solution 2 (SI System):
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1. For minimum dry-weather flow of 0.0283 m /s, the nearest standard diameter of pipe is
calculated from:
D = 24Q>pV
D = 24 * 0.0283>p * 0.90 = 0.199 m; select 200 mm (or 8 in.)
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Actual capacity Q AV (0.785 D )(V) (0.785 0.200 )(0.90) 0.028 m /s 28 L/s
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2. For maximum dry-weather flow in excess of the minimum, namely, 0.085– 0.028 0.057 m /s,
the nearest standard diameter of pipe is:
D = 24 * 0.057>p * 0.90 = 0.284 m; select 300 mm (or 12 in.)
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Actual capacity Q AV (0.785 D )(V) (0.785 0.300 )(0.90) 0.064 m /s 64 L/s
3. For storm flows in excess of maximum dry-weather flow, namely, 1.274 (0.064
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0.028) 1.182 m /s, the next lowest standard diameter of pipe is
D = 24 * 1.182>p * 1.50 = 1.002 m; select 1,000 mm (or 40 in.)
which will have to flow at a velocity of
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V Q>A (1.182 m /s)>(0.785 D ) (1.182 m /s )>(0.785 1.000 ) m 1.51 m/s