ME 353 Heat Transfer 1

M.M. Yovanovich

EX3P3.MWS

Example 3.3 of 4th edition of Incropera and DeWitt.

Steady conduction through a conical system of circular cross-section. The length is , and the diameters are and , and its thermal conductivity is . The lateral sides are adiabatic and the two ends are isothermal at temperatures and with . The local diameter is obtained from the relation , and the two ends are located at and . The geometric parameter .

1. Derive an expression for the one-dimensional temperature distribution .

2. Calculate the heat transfer rate given , , , , .

> restart:

Assumptions.

1. Steady-state conduction.

2. Source free system.

3. One-dimensional temperature distribution.

4. Constant properties.

System parameters.

> syspar:= (a = 0.25, x1 = 0.05, x2 = 0.25, T1 = 400, T2 = 600, k = 3.46);

Fourier's Law of Conduction.

> Fourier:= Q = k*A(x)*Diff(T(x),x);

Find temperature disrtribution.

Separate the variables and , then integrate between appropriate limits.

> LHS:= Int(Q/A(x),x=x1..x2);

Conduction Area.

> A(x):= Pi/4*D(x)^2; D(x):= a*x;

> LHS:= subs(D(x) = a*x, LHS);

> LHS:= value(%);

> RHS:= -Int(k, T=T1..T2); RHS:= value(%);

Heat Transfer Rate.

> Q:= solve(LHS = RHS, Q);

Alternative form of the relation for .

> Q:= Pi*a^2*k*(T1-T2)/(1/x1 - 1/x2)/4;

> Qval:= evalf(subs(syspar, Q), 4)*W;

Temperature distribution in conical system.

> restart:

To avoid confusing Maple let the variables be denoted as and .

> LHS:= Int(Q/(Pi*(a*x)^2/4),x=x1..xx); LHS:= value(%);

> RHS:= -Int(k, T=T1..TT); RHS:= value(%);

> TT:= solve(LHS = RHS, TT);

> Q:= Pi*a^2*k*(T1-T2)/(1/x1 - 1/x2)/4;

> T:= expand(TT);

Alternative form of temperature distribution.

> T:= T1 + (T1-T2)*(1/xx - 1/x1)/(1/x1 - 1/x2);

The temperature distribution gives at , and at .