ME 353 Heat Transfer 1

M.M. Yovanovich

EX3P1.MWS

Example 3.1 of 4th edition of Incropera and DeWitt.

> restart:

> syspar:= (k[A]=0.15, k[B]=0.08, T[w]=400, T[a]=400, T[infinity]=25, T[so] = 50, h[i]=25, h[r]=25, h[o]=25);

Assumptions.

1. Steady-state conditions.

2. Conduction through window is one-dimensional.

3. Contact resistance is negligible.

4. Radiation absorption within the window is negligible.

5. Radiation exchange between window outer surface and surroundings is negligible.

6. Plastic materials are homogeneous and properties are constant.

Equivalent thermal circuit.

The equivalent thermal circuit consists of five resistors and six temperature nodes. The oven wall and air temperature are equal: , the inner surface temperature is and the outer surface temperature is . The outer air temperature is denoted as . The glass temperature is . The heat transfer from the interior of the oven is lost at the outer surface into the surroundings.

System heat transfer rate..

> Q[sys]:= (T[w] - T[so])/R[sys];

Heat loss from oven into surroundings.

> Q[loss]:= h[o]*A*(T[so] - T[infinity]);

Heat balance.

> HeatBalance:= Q[sys]=Q[loss];

System overall resistance.

> R[sys]:= R[A] + R[B] + 1/(1/R[fin] + 1/R[rad]);

Component resistances.

> Resistances:= (R[A] = L[A]/(k[A]*A), R[B] = L[B]/(k[B]*A), R[fout]=1/(h[o]*A), R[fin]=1/(h[i]*A), R[rad]=1/(h[r]*A));

> eq1:= subs(syspar, L[A] = 2*L[B], subs(Resistances, HeatBalance));

Now substitute and solve for the unknown thickness .

> eq2:= subs(A=1, eq1);

> LB:= solve(eq2, L[B]);

> LA:= 2*LB;

> L:= evalf((LA + LB)*1000, 4)*mm;