PS10_sol_S08

PS10_sol_S08 - Problem Set 10(2.006 Spring 08 Problem 1 dP...

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Problem Set 10 (2.006, Spring 08) Problem 1 By fitting the data, we found K Pa dT dP sat / 10 64 . 8 4 × = (Note that 1mmHg = 133.3224Pa. Refer to the last chapter of the 2.006 data book) Get saturated vapor pressure at 4.00K by interpolating the data from 650mmHg and 600mmHg. P = 618.99mmHg = 82525Pa. Using the ideal gas model, kg m Pa mol kg K molK J MP RT g / 10067 . 0 ) 82525 )( / 004003 . 0 ( ) 4 )( / 314 . 8 ( 3 = = = ν Hence, by using Clausius-Clapeyron relation, we can find g J kg J K Pa kg m kg m K dT dP T h sat fg fg / 1 . 32 / 32112 ) / 10 64 . 8 )( / 129 1 / 10067 . 0 )( 4 ( 4 3 3 = = × = = This value is bigger than the experimental value of 21.9 J/g. The ideal gas model has overestimated the heat of vaporization. 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 5 2 4 6 8 10 12 14 16 x 10 4 T (K) P (Pa) y = 8.64e+004*x - 2.61e+005 data 1 linear

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Problem 2 a), b) Given the contact angle = 90°, and the cavity of 10 μ m with 4° included angle, we found the initial size of bubble, m m R i μ 3492 . 0 ) 2 / 4 tan( 10 = × = o Using the properties at P sat = 10 bar, the nucleation liquid superheat is (ignore the effect of water level), K kg J m K kg m m N h R T T T T fg i sat fg sat s 7 . 14 ) / 10 6 . 2014 )( 10 3492 . 0 ( ) 03 . 453 )( / 1932 . 0 )( / 0589 . 0 ( 2 2 3 6 3 = × × = = = Δ σν c) If the pressure becomes 3bar, the new liquid superheat is K kg J m K kg m m N h R T T T T fg i sat fg sat s 3 . 38 ) / 10 5 . 2163 )( 10 3492 . 0 ( ) 67 . 406 )( / 6047 . 0 )( / 0589 . 0 ( 2 2 3 6 3 = × × = = = Δ Where we assume the surface tension remains the same. d) The net force which the embryo bubble attached to cavity, ) ( 3 2 2 3 f g i i buoyancy gravity Tension surface net g R R F F F F ρ π σ + = + = Note that density of saturated vapor is evaluated at 453.03K + 14.7K = 467.73K (P>10bar), and the density of superheated liquid water can be approximately from saturated water density at 10bar. 3 / 0386 . 7 m kg g = , 3 3 / 15 . 887 ) / 0011272 . 0 /( 1 m kg kg m f = = ) ( 129 . 0 / ) 15 . 887 0386 . 7 )( / 81 . 9 ( ) 10 3492 . 0 ( 3 2 ) / 0589 . 0 )( 10 3492 . 0 ( 2 3 2 3 6 6 downward N m kg s m m m N m F net = × + × = Note that the force holding the bubble is much larger than the buoyancy force on the bubble. The bubble should keep growing such that the direction of the force by surface tension becomes smaller (due to angle change) and the bubble buoyancy gets larger. e) Increase in internal energy by vaporization: fg fg vap u dm dU = Word done by the expansion of the bubble: ( ) i i sat sat dR R P dV P W 2 exp 2 = = Energy stored by surface tension: ( ) ( ) ( ) i i i i i surf dR R R dR R dA dU 4 2 2 2 2 = + = = The first two together gives simply ( ) fg i i g fg fg sat vap vap h dR R h dm dV P dU dH 2 2 = = + =
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This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

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PS10_sol_S08 - Problem Set 10(2.006 Spring 08 Problem 1 dP...

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