ME 353 Heat Transfer 1

M.M. Yovanovich

P3MT97.MWS

Solution of Problem 3 of Midterm Exam, October 1997.

> restart:

System Parameters

> sys:=
(rho = 7830, k = 51.2, cp = 540, D = 150/1000,
L = 75/1000, Ti = 300, Tf = 1200, h = 80);

Biot numbers

> Bi1:= h*D/(2*k);

> Bi2:= h*L/k;

> Bi3:= h*V/(k*S);
V:= (Pi/4)*D^2*2*L;
S:= 2*Pi/4*D^2 + Pi*D*2*L;

Part (a) Calculation of Biot numbers

> Bi1:= evalf(subs(sys, Bi1), 4);

> Bi2:= evalf(subs(sys, Bi2), 4);

> Bi3:= evalf(subs(sys, Bi3), 4);

Since all Biot numbers are sufficiently small, the lumped capacitance model

can be applied.

Part (b) Characteristic time of the system

> tc:= (rho*cp*V)/(h*S);
tc_sec:= evalf(subs(sys, tc), 4);

Solution of lumped capacitance model

> T:= Tf - (Tf - Ti)*exp(-h*S*t/(rho*cp*V));

> T:= evalf(subs(sys, T));

Part (c) Calculation of the time for system to reach 800 K

> tsec:= evalf(solve(T = 800, t), 5);

Part (d) Calculate the time to cool from 800 K to 320 K

> restart:

> sys2:=
(rho = 7830, k = 51.2, cp = 540, D = 150/1000,
L = 75/1000, Ti = 800, Tf = 300, T = 320, h = 25);

> t2:= - 1/m*ln((T - Tf)/(Ti - Tf));

> m:= (h*S)/(rho*cp*V);

> V:= (Pi/4)*D^2*2*L;

> S:= 2*Pi/4*D^2 + Pi*D*2*L;

> t2_sec:= evalf(subs(sys2, t2), 4);

> t2_hrs:= evalf(t2_sec/3600, 4);