Thermodynamics HW Solutions 347

Thermodynamics HW Solutions 347 - lambda_1_c=0.1704...

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Chapter 4 Transient Heat Conduction 4-74 "!PROBLEM 4-74" "GIVEN" D=0.08 "[m]" r_o=D/2 height=0.15 "[m]" L=height/2 T_i=150 "[C]" T_infinity=20 "[C]" h=40 "[W/m^2-C]" "time=15 [min], parameter to be varied" "PROPERTIES" k=110 "[W/m-C]" rho=8530 "[kg/m^3]" C_p=0.389 "[kJ/kg-C]" alpha=3.39E-5 "[m^2/s]" "ANALYSIS" "(a)" "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=0.2282 "w stands for wall" A_1_w=1.0060 tau_w=(alpha*time*Convert(min, s))/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-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"
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Unformatted text preview: lambda_1_c=0.1704 A_1_c=1.0038 tau_c=(alpha*time*Convert(min, s))/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 short cylinder" "(b)" theta_L_w=A_1_w*exp(-lambda_1_w^2*tau_w)*Cos(lambda_1_w*L/L) "theta_L_w=(T_L_w-T_infinity)/(T_i-T_infinity)" (T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c "center temperature of the top surface" "(c)" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_i-T_infinity) 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=0.0846 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer" 4-58...
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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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