PPT03-solutions

# PPT03-solutions - Imperial College London ME4 Polymer...

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Imperial College London ME4 Polymer Processing Technology Solutions PSL 5 April 2008 Problem 3.1 For a parallel circular channel: Q = ± ² n 3 + 1 1 2 ³ 1 ± p L ´ µ · ¸ ¹ 1 R 3 + 1 () , and by the lubrication approximation, at ANY z , dp dz = ± 2 1 3 + 1 ´ µ · ¸ ¹ r ± 3 + 1 and at inlet d d ± ² ³ ´ µ i = · 2 ¸ 1 3 + 1 ¹ º » ¼ ½ ¾ ¿ 1 · 3 + 1 , so that in general = P 1 1 ± ² ³ ´ µ · 3 + 1 . Integrating: 2 ± 1 = d d ² ³ ´ µ · 1 1 3 + 1 ± 3 + 1 1 2 ¸ For a linear taper, = const. so that 2 ± 1 = d d ² ³ ´ µ · 1 1 3 + 1 dr ± 3 + 1 1 2 ¸ = ± 1 3 ² ³ ´ µ · d d ² ³ ´ µ · 1 1 3 + 1 1 2 ± 3 ¹ º » ¼ = ± 1 3 ² ³ ´ µ · d d ² ³ ´ µ · 1 1 2 1 ² ³ ´ µ · ± 3 ± 1 ¹ º ½ ½ » ¼ ¾ ¾ Now = 1 2 D 2 · 1 ± ² ³ ´ µ , so that 2 ± 1 = ± 1 3 ² ³ ´ µ · d d ² ³ ´ µ · 1 1 2 ± 1 1 ± 2 1 ² ³ ´ µ · ± 3 ¸ ¹ º º » ¼ ½ ½ = d d ² ³ ´ µ · 1 1 3 1 ± 2 1 ² ³ ´ µ · ± 3 ¸ ¹ º º » ¼ ½ ½ 2 1 ± 1 » ¼ ½ QED. NB 2 ± 1 must be ± 1 for the lubrication approximation to be valid. For ± 1 = 16 ± 10 3 , = 0.3, 1 = 3 ± 10 –3 , 2 = 6 ± 10 –3 , = 0.06, = 1 ± 10 –5 , 1 = ± 32 ² 10 3 ² 1.9 0.3 ² 1 ² 10 ± 5 ´ µ · ¸ ¹ 0.3 ² 1.5 ² 10 ± 3 ± 1.9 = 290 MPa m –1 2 ± 1 = 0.06 ² 2.897 ² 10 8 ² 1 0.9 1 ± 2 ± 0.9 ³ ´ µ 2 ± 1 [] = 8.96 MPa

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Imperial College London ME4 Polymer Processing Technology Solutions PSL 5 April 2008 Problem 3.2 Total rate of work done on fluid: dW dt = wx , y , z 1 () A 1 ± p 1 dA ² , , 2 2 ± 2 = 1 Q 1 ² 2 2 = Qp 1 ² 2 assuming incompressibility. For adiabatic conditions: ± QC p ± T = 1 ² 2 hence ± = 1 ± 2 ² C p . For a flat channel, flow rate per unit width is D = 1 2 VH under drag flow P = 1 12 P 1 ± 2 L H 3 under pressure flow.
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• Fall '10
• LEVERS
• Trigraph, G protein coupled receptors, Imperial College London, University College London, Polymer Processing Technology

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PPT03-solutions - Imperial College London ME4 Polymer...

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