Thermodynamics HW Solutions 215

Thermodynamics HW Solutions 215 - R_conv_o=1/(h_o*A_o)

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Chapter 15 Steady Heat Conduction 3-80E "GIVEN" T_infinity_1=100 "[F]" T_infinity_2=70 "[F]" k_pipe=223 "[Btu/h-ft-F], parameter to be varied" D_i=0.4 "[in]" "D_o=0.6 [in], parameter to be varied" r_1=D_i/2 r_2=D_o/2 h_fg=1037 "[Btu/lbm]" h_o=1500 "[Btu/h-ft^2-F]" h_i=35 "[Btu/h-ft^2-F]" m_dot=120 "[lbm/h]" "ANALYSIS" L=1 "[ft], for 1 ft length of the tube" A_i=pi*(D_i*Convert(in, ft))*L A_o=pi*(D_o*Convert(in, ft))*L R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_pipe*L)
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Unformatted text preview: R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_conv_o Q_dot=(T_infinity_1-T_infinity_2)/R_total Q_dot_total=m_dot*h_fg L_tube=Q_dot_total/Q_dot k pipe [Btu/h.ft.F] L tube [ft] 10 1176 30.53 1158 51.05 1155 71.58 1153 92.11 1152 112.6 1152 133.2 1151 153.7 1151 174.2 1151 194.7 1151 215.3 1151 235.8 1150 256.3 1150 276.8 1150 297.4 1150 317.9 1150 338.4 1150 358.9 1150 379.5 1150 400 1150 3-58...
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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