ME 353 Heat Transfer 1

M.M. Yovanovich

P2MT97.MS

Solution of Problem 2 of Midterm Exam, October 1997.

> restart:

System Parameters

> sys:= (D = 6/1000, L = 100/1000, k = 25, h = 65);

(a) System consists of two parts:

1) the insulated part which a rod of length L1 = f L,

2) the convective part which is a fin of length L2 = (1 - f) L.

The resistance of the rod is equal to the resistance of the fin.

(b) The equivalent thermal circuit consists of three temperature

nodes: T0, T1 and Tf where T1 is the temperature at the point

where the rod is in contact with the air. The two resistors which

are in seies are the rod resistance and the fin resistance.

(c) Rrod = Rfin

Define the rod and fin resistances, and the fin parameters.

> eq:= Rrod - Rfin = 0;

> Rrod:= L1/(k*A);

> Rfin:= 1/(sqrt(h*P*k*A)*tanh(m*L2));

> m:= sqrt((h*P)/(k*A));

> L1:= f*L; L2:= (1 - f)*L;

> P:= Pi*D; A:= Pi/4*D^2;

Examine the equation again. This equation is transcendental

because f appears on both sides and it cannot be solved

analytically. Numerical methods are required.

> eq:= eq;

(d) Substitute the system parameters to obtain a numerical

relation which contains the unknown parameter f.

> eq_num:= evalf(subs(sys, eq), 8);

This is a transcendental equation which requires a numerical

solution such as Newton-Raphson. The problem does not require

a numerical result.

Maple will be used to find f.

> f1:= fsolve(eq_num, f, 0..1);

The insulated portion is approximately 24% of the total length.