HW2_solution

# HW2_solution - Minxue(Kevin He Homework#2 Solution Set...

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Minxue (Kevin) He CEE151 W08 Homework #2 Solution Set Page - 1 - of 2 1. Solution: According to the problem, the normal depth of flow d could be determined as, (2 ) (cos25 ) 1.813 d ft ft ° = × = Since the width 1 w ft = , the cross-section area A could be computed as, 2 1.813 A dw ft = = Thus the want discharge is, 2 (1.813 )(3 / ) 5.439 Q AV ft ft s cfs = = = 2. Solution: According to Table 2.1.2 in the textbook, at 20 C ° , 6 2 1.007 10 / m s υ - = × . The diameter of the pipe 10 0.01 D mm m = = . The cross-section area of the pipe A can then be computed as, 2 5 2 7.85 10 4 D A m π - = = × For the flow to be laminar, 2000 e R . To have the maximum discharge for which a laminar flow may be expected, e R should be set to 2000. Therefore, max max 6 2 max max 2000 2000( ) / 2000(1.007 10 / ) / 0.01 0.201 / e V D R V D V m s m V m s υ υ - = = = = × = The wanted maximum discharge is, 5 2 5 3 max max (0.201 / ) (7.85 10 ) 1.58 10 / Q V A m s m m s - - = = × × = × 3. Solution: Using Equation (3.5.8) in the textbook, b h x

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Minxue (Kevin) He CEE151 W08 Homework #2 Solution Set Page - 2 - of 2 3 2 2 36 ( ) 2 3 ( ) 2 3 2 3 6(3 2 ) cp c c I y y yA bh x h bh x h h x h x h = + = + + + = + + + 4.
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