# Class38_Ex2 - Solution Reservoir to exit z = 0 exit V12 p2...

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Solution: Reservoir to exit ( 0 z = exit) 1 p γ 2 1 2 V g + 2 1 p p zh ++ = 2 2 2 2 f m V h g + + 22 12 4 pe b VL V z f h h gD g += + + 2 2 14 2 b z K K f ⎛⎞ +=+ + + + ⎜⎟ ⎝⎠ 2 2 2 10.18 1.611 2 0.5 4 QV Vf t s f t Ag π == = = × 0.5 e K = , ( ) ( ) 12 6 2 0.19 bb b KK r DK = = 5 5 44 2 Re 4.17 10 0.5 1.22 10 Q D νπ × = × ×× ×

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i.e., 0.0135 f = , Moody Diagram, 0,Re s k f D ⎛⎞ = ⎜⎟ ⎝⎠ 1700 30 1.611 1 0.5 4 0.19 0.0135 107.6 0.5 p hf t =+ × + + × + = 550 24.4 p pQ h h p γ == 2 2 mm
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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.

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Class38_Ex2 - Solution Reservoir to exit z = 0 exit V12 p2...

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