chandra_324_hw8_soln

# chandra_324_hw8_soln - Solution for ME324 HW#8 1. Flow Rate...

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Solution for ME324 HW#8 1. Flow Rate Q = AV = A 1 V 1 = A 2 V 2 Given: Q =10 6 mm 3 /s , A 1 = 800 mm 2 , h = h 1 - h 2 =175 mm . Thus: v 1 = Q / A 1 =1250 mm/s Based on Eq. Assumption : p 1 = p 2 , Then we get Thus : A 2 = Q / v 2 = 447.6 mm 2 Thus : Selecting area A 2 at the base of the sprue could avoid aspiration of molten metals 2. Solidification Time t = C (Volume / Surface area) 2 . Given: the same casting conditions. Thus: the same C value. Given: the same length and cross-section area. Thus: the same Volume. Given: l = 500mm , r = 50mm . The Surface Area of the round casting , is S round = 2 ± rl + 2 ± r 2 = 172,787 mm 2 . Given: ab = ± r 2 , and a = 2b , Thus b = 62.7 ,so the perimeter of the rectangular casting is 376 mm. The Surface Area of the rectangular casting, is S rectangular = (376)(500) + 2ab= 203,710 mm 2 . Thus : t rectangular / t round = (S round / S rectangular ) 2 = 0.72 3.For the cylindrical part, given : r = 4 in , h = 2 in V = ± r 2 h = 100.5 in 3 A = 2pr 2 + 2 prh = 150.8 in 2 t cylindrical = C(V / A) 2 =0.444C h 1 p 1 ² g ³´ µ¶ 1 2 2g µ ++ h 2 p 2 ² g 2 2 µ = 2 1 2 h 1 h 2 · + 2234 mm s µ == h 1 2

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For a square plate, given : the same volume with the cylindrical, h = 2 in.
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## This note was uploaded on 10/13/2008 for the course ME 324 taught by Professor Boylan during the Spring '08 term at Iowa State.

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chandra_324_hw8_soln - Solution for ME324 HW#8 1. Flow Rate...

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