sm_ch03

Water Resources Engineering

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Unformatted text preview: 3 – 1 Chapter 3 3.2.1. D 1 = 8 in V 1 = Q/A 1 = 2 1 2 1 D 4Q /4)D ( Q S S ; Similarly, 2 2 2 D 4Q V S From the requirement that V 2 d 2V 1 , 2 D Q/ 4 D Q/ 4 2 1 2 2 d S S , that is, 2 D D 2 2 2 1 d or 2 8 2 D D 1 2 t D 2 t 5.66 in. Hence, a pipe of 5 ¾ in diameter is appropriate. 3.3.1. D 1 = 3 in = 0.25 ft. Velocity head in the given pipe = 2g V 2 1 , V 1 = Q/A 1 , 4 D A 2 1 1 S , 2 1 1 D 4Q V S Thus, ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § g 2 1 D Q 4 2g V 2 2 1 2 1 S Similarly, velocity head in the second pipe = ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § g 2 1 D Q 4 2g V 2 2 2 2 2 S Since 2g V 2 2 = 2g V 4 2 1 , ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § g 2 1 D Q 4 2 2 2 S = ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § g 2 1 D Q 4 4 2 2 1 S 4 1 4 2 D 4 D 1 or D 2 = 0.707D 1 = 0.707(3) = 2.12 in. 3.4.1. Using Bernoulli’s equation: 2g V p 2g V p 2 2 2 2 1 1 ¡ ¡ J J , or 2g V V p p 2 1 2 2 2 1 ¢ ¢ J Q ) / 9 ( (8/12) 4Q V 2 1 S S , Q ) 529 / 9216 ( (5.75/12) 4Q V 2 2 S S Hence, » » ¼ º « « ¬ ª ¸ ¹ · ¨ © § ¢ ¸ ¹...
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sm_ch03 - 3 – 1 Chapter 3 3.2.1. D 1 = 8 in V 1 = Q/A 1 =...

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