ME 353 Heat Transfer 1

M.M. Yovanovich

EX1P5.MWS

Example 1.5 of 4th edition of Incropera and DeWitt, 1996.

In the first case the temperature T1 is unknown, and in

the second case the temperature T2 is unknown.

Compute the heat transfer rates by conduction, convection and

radiation for the second case. Compare the relative magnitude

of Qconvection/Qconduction.

> restart:

Assume that the heat transfer area is 1 m^2.

Example 1.

> case1:= (A = 1*m^2, L = 0.15*m, k = 1.2*W/(m*K),
h = 20*W/(m^2*K), epsilon = 0.8, T1 = T1, T2 = 373*K, Tinfty = 298*K, Tsur = 298*K, sigma = 5.67*10^(-8)*W/(m^2*K^4));

Fourier's Law of Conduction.

> Qcond:= k*A*(T1 - T2)/L;

Newton's Law of Cooling.

> Qconv:= h*A*(T2 - Tinfty);

Stefan-Boltzmann Law of Radiation.

> Qrad:= epsilon*sigma*A*(T2^4 - Tsur^4);

Heat balance at the right boundary of the wall.

> HeatBalance:= Qcond - Qconv - Qrad = 0;

Find the temperature at the left boundary.

> T1:=
evalf(simplify(solve(subs(case1, HeatBalance), T1)), 5);

Example 2.

For the next case we will find T2 when all other parameters are given.

For this case I will not specify the units of the parameters.

> case2:=
(T1 = 350 + 273, T2 = T2, A = 1, L = 0.15, k = 1.2, h = 20, epsilon = 0.8, Tinfty = 25 + 273, Tsur = 25 + 273,
sigma = 5.67e-8);

Find the temperature at the right boundary.

> T2:= fsolve(subs(case2, HeatBalance), T2);

Take the positive value returned by the fsolve command.

> T2:= 372.4568;

Compute the heat transfer rates for the second case.

> Qcond:= evalf(subs(case2, Qcond), 5)*W;

> Qconv:= evalf(subs(case2, Qconv), 5)*W;

> Qrad:= evalf(subs(case2, Qrad), 5)*W;

> evalf(Qconv/Qcond, 4);

In case 2 the heat transfer rate by convection accounts for approximately

75 % of the overall heat transfer rate.