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, = 4.4, = R 2 . If they are to cool in the same time, 2 ± 4 ² 2 = 4.4 2 = 4.4 8 and with = 2 mm, inlet diameter D i = 2 = 2.97 mm . At outlet, 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: * = E ± B P ± B = 105 ± 60 240 ± 60 = 0.25 = exp ± 4.4 2 ³ ´ µ · ¸ Hence cooling time = 2 4 ² ³ 1 4.4 ln 0.25 = 5.59 ² 10 ³ 3 2 4 ² 10 ³ 7 ² 4.4 ³ ln 0.25 = 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: i = 2.8 mm Thickness H = 2 mm. Hence ± * = 110 2.8 ² 1 = 38.3 and C Gz n 1 + () = i * G * where * = 1 + * 1 + * 1 ² ² 1 1 ² * ³ ´ µ µ · ¸ ¸ ¹ º » ¼ » ½ ¾ » ¿ » 11 + .
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This note was uploaded on 04/07/2010 for the course MECH ENG 207 taught by Professor Levers during the Fall '10 term at Imperial College.

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

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