PS11_sol_S08 - Problem Set 11(2.006 Spring 08 Problem 1...

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Problem Set 11 (2.006, Spring 08) Problem 1 Conservation of energy for the control volume around the interface gives a balance between heat ‘generated’ in the control volume by condensation and heat conducted through the liquid layer to the cold surface. Considering an area into the page, A, and the rate of condensation, m & , A k T T h m Q Q l l surf sat fg conduction on condensati δ ) ( 0 = = & & & and dt d A m l ρ = & Substituting this in the previous equation, = t fg l surf sat l dt h T T k d 0 0 ) ( Solving, we get 2 / 1 ) ( 2 = t h T T k fg l surf sat l In one hour (3600 s), the layer thickness is =0.00672m . The mass condensed is 1.29×10 -3 kg. (T sat =100 o C, k l =0.679 W/mK, ρ l =958.35kg/m 3 , ρ v =0.59817kg/m 3 h fg =2256.4 kJ/kg, c p,l =4215.7 J/kgK, μ l =2.8174×10 -4 Pa·s) When the surface is placed vertically (condensation), the length L is m mm 0141 . 0 ) 200 ( 2 / 1 2 = . () kgK kJ h T T c h h fg surf sat l p fg fg / 7 . 2313 68 . 0 1 , ' = + = 4 . 331 ) ( ) ( 943 . 0 25 . 0 3 ' = = = surf sat l l fg v l l l L L T T k L h g k L h u N μ 15959 = L h ' ) ( fg surf sat L h T T A h m = & The condensation rate m & =2.76x10 -5 kg/s is constant in time, so the mass condensed is 9.94x10 -2 kg. This is more than an order of magnitude larger than for the horizontal configuration. For the vertical plate the condensed fluid flows downward with gravity, maintaining a thin condensate layer, but for the horizontal plate the conduction resistance increases as the liquid layer thickness increases.
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Problem 2 Water properties (at 77 o C = 350K) 34 973.36 / , 4194.7 / , 3.68364 10 , Pr 2.31314, 0.66805 / p kg m c J kg K Pa s k W m K ρμ == = × = = 41 6.2441 10 K β −− a) If the thermal resistances between the flowing water and the water being heated are neglected, then the temperature of all the components within the water heater are uniform and equal to each other. Since the temperature is uniform, we can apply the first law to the entire water heater in order to determine the time necessary to heat the water to 350 K. The temperature of the water exiting the tubes at any instant is the same as the temperature within the tank at that instant. T tank = T out () 22 tank tank tube tube 11 2 44 in out out vp o u t i n dE mh h dt dT cd h d L m c T T dt cc ρπ π =− ⎛⎞ +⋅ = ⎜⎟ ⎝⎠ & & The thermal mass of both the water to be heated and the water within the two heating tubes are included in the first law. Since the thickness of the copper jacket is not given in this problem, we neglect the thermal mass of all the copper material in our analysis. In evaluating the difference in enthalpy between the inlet and outlet flow, we neglect the effects of any pressure drop within the tubes. We now solve the above differential equation in order to calculate the desired time. All water properties are taken to be those of saturated liquid water at 350 K. The initial temperature of the water in the water heater is 300 K.
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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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PS11_sol_S08 - Problem Set 11(2.006 Spring 08 Problem 1...

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