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PPT05-solutions

# PPT05-solutions - Imperial College London ME4 Polymer...

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Imperial College London ME4 Polymer Processing Technology Solutions PSL 28 January 2008 Problem 5.1 Cooling law: T * = exp ± cFo ( ) . Cavity: flat plate, c = 2, Fo = 4 ± t H 2 . Sprue: cylinder, c = 4.4, Fo = ± t R 2 . If they are to cool in the same time, 2 ± 4 ² H 2 = 4.4 ² R 2 R = H 4.4 8 and with H = 2 mm, inlet diameter D i = 2 R = 2.97 mm . At outlet, D o = 2.97 + 30 ± 2 ± tan 2.5° = 5.59 mm. The conduction distance from point ‘X’ is greater than for either the cylinder or the plate approximations. It will freeze last, with a risk of sinking or voiding. Cooling time: T * = T E ± T B T P ± T B = 105 ± 60 240 ± 60 = 0.25 = exp ± 4.4 ² t R 2 ³ ´ µ · ¸ Hence cooling time t = D 2 4 ± ² ³ 1 4.4 ln0.25 = 5.59 ² 10 ³ 3 ( ) 2 4 ² 10 ³ 7 ² 4.4 ³ ln0.25 ( ) = 24.6 s

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Imperial College London ME4 Polymer Processing Technology Solutions PSL 28 January 2008 Problem 5.2 From Problem 4.4: longest flow path R = 110 mm Sprue size at cavity: R i = 2.8 mm Thickness H = 2 mm. Hence ± * = 110 2.8 ² 1 = 38.3 and C ± Gz n 1 + n ( ) = R i H ± * G ± * ( ) where G ± * ( ) = 1 + ± * ( ) n 1 + ± * ( ) 1 ² n ² 1 1 ² n ( ) ± * ³ ´ µ µ · ¸ ¸ ¹ º » ¼ » ½ ¾
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• Fall '10
• LEVERS
• Imperial College London, University College London, Polymer Processing Technology, Processing Technology Solutions

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PPT05-solutions - Imperial College London ME4 Polymer...

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