problems chapter 4

problems chapter 4 - Problem # 1 For the simple pipe system...

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Problem # 1 For the simple pipe system shown in the figure, the pressure are p1=14 kPa, p2=12.5 kPa, and p3=10kPa. Determine the head loss between 1 and 2 and the head loss between 1 and 3. The discharge is 7 l/s. The energy equation between 1 and 2 is: 2 1 L 2 2 2 2 1 2 1 1 h z g 2 V p z g 2 V p + + + = + + γ g 2 V g 2 V 2 2 2 1 = Where and z 1 =z 2 2 1 L 2 1 h p p + = γγ m 153 . 0 m kN 79 . 9 kPa 5 . 12 kPa 14 p p h 3 2 1 2 1 L = = =
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The energy equation between 1 and 3 is 3 1 L 3 2 3 3 1 2 1 1 h z g 2 V p z g 2 V p + + + = + + γ Now using the energy equation, we get: m 24 . 4 h 3 1 L = () () s / m 477 . 2 m 0028 . 0 s m 007 . 0 4 1000 mm 60 1000 s l 7 A Q V 2 3 2 1 1 = = = = π () () s / m 385 . 5 m 0013 . 0 s m 007 . 0 4 1000 mm 40 1000 s l 7 A Q V 2 3 2 3 3 = = = = () () () () 3 1 L 2 2 3 2 2 3 h 0 s m 81 . 9 2 s m 385 . 5 m kN 79 . 9 kPa 10 5 s m 81 . 9 2 s m 477 . 2 m kN 79 . 9 kPa 14 + + + = + +
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problems chapter 4 - Problem # 1 For the simple pipe system...

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