ME 353 Heat Transfer 1

M.M. Yovanovich

EX1P2.MWS

Example 1.2 of 4th edition of Incropera and DeWitt.

Convection and radiation heat transfer from horizontal steam pipe.

> restart:

System parameters.

> syspar:= (Ts = 200, Tinfty = 25, Tsur = 25, h = 15, epsilon = 0.8, D = 70/1000, sigma = 5.67e-8);

Heat loss from pipe surface by convection and radiation.

> Qloss:= Qconv + Qrad;

Newton's Law of Cooling.

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

Stefan-Boltzmann's Law of Radiation Exchange between surface and its enclosure.

> Qrad:= sigma*epsilon*A*((Ts + 273)^4 - (Tsur + 273)^4);

Pipe surface area per unit length.

> A:= Pi*D*1;

Calculation of convection, radiation and total heat transfer rates per unit length.

> Qconv:= evalf(subs(syspar, Qconv), 5)*W/m;

> Qrad:= evalf(subs(syspar, Qrad), 5)*W/m;

> Qloss:= evalf(subs(syspar, Qloss), 5);

The heat transfer by convection is approximately of the total loss.