R134a condenses at 313.15 K on a horizontal integral-rectangular-fin tube (1024 fpm) with 47=6 K. Compute...

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R134a condenses at 313.15 K on a horizontal integral-rectangular-fin tube (1024 fpm) with 47=6 K. Compute the condensation heat transfer coefficient for the tube using: 1. The Beatty and Katz gravity drained model. 2. The Kedzierski-Carr-Brown surface tension drained model accounting for condensate retention. Assume 100% fin efficiency for both cases. The R134a properties at 310.15 K are: p=1147 kg/m³, p = 50 kg/m², k₁=0.075 W/mK, o=0.006 N/m, λ=163020 J/kg, M=1.4 x 107 m/s, 161.5 × 10 kg/ms The tube geometry is: D. = 18.9 mm, D = 16.652 mm, 4, = 0.252 mm, 4 = 0.576 mm, s, = 0.4 mm, e = 1.124 mm, 1024 fpm (fins-per-meter) Please answer the following: 1. What fraction of the surface is condensate flooded (Cs)? 2. Compare the / calculated by the two methods to the measured data (h= 11000 W/m²K at 47,-6 K) and discuss. 3. Find the fpm that gives the maximum / by plotting at least ten points of h versus fpm. Plot the fpm from 700 fpm to 3000 fpm. Use the surface- tension drain model (model # 2) with all the parameters given above except set=4₁ = 0.252 mm and s, 1/fpm-1₁. Use the variation of C, and the relative contributions of him and h, to explain why a maximum is or is not observed. Note that you will not get the same answer for 1024 fpm that you got for the first part of the problem because you have simplified the problem by using th=t, resulting a different s, for 1024 fpm. R134a condenses at 313.15 K on a horizontal integral-rectangular-fin tube (1024 fpm) with 47=6 K. Compute the condensation heat transfer coefficient for the tube using: 1. The Beatty and Katz gravity drained model. 2. The Kedzierski-Carr-Brown surface tension drained model accounting for condensate retention. Assume 100% fin efficiency for both cases. The R134a properties at 310.15 K are: p=1147 kg/m³, p = 50 kg/m², k₁=0.075 W/mK, o=0.006 N/m, λ=163020 J/kg, M=1.4 x 107 m/s, 161.5 × 10 kg/ms The tube geometry is: D. = 18.9 mm, D = 16.652 mm, 4, = 0.252 mm, 4 = 0.576 mm, s, = 0.4 mm, e = 1.124 mm, 1024 fpm (fins-per-meter) Please answer the following: 1. What fraction of the surface is condensate flooded (Cs)? 2. Compare the / calculated by the two methods to the measured data (h= 11000 W/m²K at 47,-6 K) and discuss. 3. Find the fpm that gives the maximum / by plotting at least ten points of h versus fpm. Plot the fpm from 700 fpm to 3000 fpm. Use the surface- tension drain model (model # 2) with all the parameters given above except set=4₁ = 0.252 mm and s, 1/fpm-1₁. Use the variation of C, and the relative contributions of him and h, to explain why a maximum is or is not observed. Note that you will not get the same answer for 1024 fpm that you got for the first part of the problem because you have simplified the problem by using th=t, resulting a different s, for 1024 fpm.