quiz_1_04_sol - 2.25 ADVANCED FLUID MECHANICS QUIZ 1...

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± ² 2.25 ADVANCED FLUID MECHANICS QUIZ 1 FALL 2004 October 14, 2004 Problem 1 A ) This is a straightforward simple application of simple buoyancy ideas. 2 ³ i) Force balance gives F R + mg = ´ g ² µ R 2 h + 3 R 3 · ¸ ± ² or F R = ³ g R 2 h + 2 3 R 3 ¹ µ mg · ¸ ± ( ()( 1000 ) ² 1 2 ² 5 + 2 ² 1 3 ´ µ 15 1000 ) for specific values given (in cgs units) F R = 1 · ¸ 3 ¹ 2 ± ² ( 1000 ) = 2749 dynes F R = ³ ´ µ 3 ii) The ‘float’ is neutrally buoyant when F R · 0 then ³ m mg = gV new = g ± µ 2 3 R 3 + R 2 h new ¸ ± V new = = 15 cm 3 · ² If compression is slow then we can assume it is isothermal so P 1 V 1 = P 2 V 2 with P 1 ± 1 atm. 17 3 = 15 45 17 P 2 = V 1 = ² ³ ´ µ · P 1 V 2 2 The new height of the membrane is ± 1 3 + ± 1 2 ± h 2 = 15 ± h 2 = 4.107 cm 3 iii) As the pressure increases beyond this value, volume decreases further => the float valve sinks. This results in further compression of air and so it sinks further. Final resting place (labeled point ‘3’) is on bottom of tank. = 1.6 ± 10 5 Nm 2 P 3 = P 2 + gL = 1.5 P + gL = 1.6 P a a ± Volume of air at this level is given by expression for adiabatic compression P 3 V 3 11.4 = P 1 V 1 = 12.72 cm 3 17 1 ) ( ² ² ² ² V 3 = V 1 P P 3 a ³ V 3 = ´ µ ´ µ · · 3 1.6 final height is 2 ± 1 3 + ± 1 2 ± h 3 = 12.72 cm 3 h 3 = 3.38 cm ³ 3 B) Here we must be more careful as the buoyant force concept only applies to fully submerged fraction.
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This note was uploaded on 02/27/2012 for the course MECHANICAL 2.25 taught by Professor Garethmckinley during the Fall '05 term at MIT.

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quiz_1_04_sol - 2.25 ADVANCED FLUID MECHANICS QUIZ 1...

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