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Unformatted text preview: V and Re and / D = (0.00085 ft)/(6/12 ft) = 0.0017, by using the equation (3), = 0.0227 (+2 points) The energy equation (1) and the head loss equation (2), along with p 1 = p 2 = p a , V 1 V 2 0, and h t = 0 (no turbine), yield an expression for pump head: = ( ) + 2 = (120 ft) + 0.0227 2000 ft 6 12 ft 15.3 ft s 2 32.2 ft s = 450 ft (+5 points) Thus, the pump power is = = 1.94 slug ft 3 32.2 ft s 3 ft 3 s (450 ft) = 84332 ft lbf s = 153 hp (+1 point)...
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This note was uploaded on 12/08/2011 for the course MECHANICS 57:020 taught by Professor Fredrickstern during the Fall '10 term at University of Iowa.
 Fall '10
 FredrickStern
 mechanics

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