Thermodynamics HW Solutions 365

Thermodynamics HW Solutions 365 - T_infinity""For...

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Chapter 4 Transient Heat Conduction 4-85 "!PROBLEM 4-85" "GIVEN" 2*L=0.20 "[m]" 2*r_o=0.15 "[m]" T_i=20 "[C]" T_infinity=1200 "[C]" "T_o_o=300 [C], parameter to be varied" h=80 "[W/m^2-C]" "PROPERTIES" k=236 "[W/m-C]" rho=2702 "[kg/m^3]" C_p=0.896 "[kJ/kg-C]" alpha=9.75E-5 "[m^2/s]" "ANALYSIS" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.1439 "w stands for wall" A_1_w=1.0035 tau_w=(alpha*time)/L^2 theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-
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Unformatted text preview: T_infinity)" "For long cylinder" Bi_c=(h*r_o)/k "c stands for cylinder" "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1762 A_1_c=1.0040 tau_c=(alpha*time)/r_o^2 theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of cylinder" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_infinity-T_i) Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2*theta_o_c*J_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0876 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer" 4-76...
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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