ME 353 Heat Transfer 1

M.M. Yovanovich

EXP13P6.MWS

Example 13.6 of 4th edition of Fundamentals of Heat Transfer and

Mass Transfer by Incropera and DeWitt, 1996.

Assumptions:

1. Steady-state.

2. All surfaces are opaque, diffuse and gray, and of uniform radiosity.

3. Convection effects are negligible.

4. End effects are negligible.

5. The reradiating surfaces is denoted as the R surface.

1. Find the net radiative heat transfer rate between the heated surfaces.

2. Calcualte the temperature of the reradiating surface.

> restart:

System parameters

> sys:= (A1 = 1, A2 = 1, AR = 1, epsilon1 = 0.8, epsilon2 = 0.4, epsilonR = 0.08, T1 = 1200, T2 = 500, sigma = 5.67e-8);

System heat transfer rate

> Qsys:= (Eb1 - Eb2)/Rsys;
Eb1:= sigma*T1^4; Eb2:= sigma*T2^4;

System radiative resistance with reradiating surface

The radiative resistances of the triangle net consist of three spatial resistances.

The resistance between surfaces 1 and 2 is in parallel with the two spatial resistances

which connect surfaces 1 and 2 via the reradiating surface.

> Rsys:= Rs1 + 1/(1/R12 + 1/(R1R + RR2)) + Rs2;

> Rs1:= (1 - epsilon1)/(A1*epsilon1);
Rs2:= (1 - epsilon2)/(A2*epsilon2);
R12:= 1/(A1*F12); R1R:= 1/(A1*F1R); RR2:= 1/(A2*FR2);
F12:= 0.5; F1R:= 0.5; FR2:= 0.5; #By symmetry

System heat transfer rate

> Qsys1:= evalf(subs(sys, Qsys), 5);

The net radiative heat transfer rate between surfaces 1 and 2 with refractory surface

present is approximately 37 kW.

Temperature of the reradiating surface

> TR:=
(T1^4 - (T1^4 - T2^4)/Rsys*(Rs1 + R12/(R12 + R1R + RR2)))^(1/4):

> TR1:= evalf(subs(sys, TR), 4);

The network consists of 5 nodes: Eb1, J1, J2, JR and Eb2 along with 5 resistors.

The net system heat transfer rate Q12 separates into two heat flow rates: QD and QID

where QD is the direct heat transfer rate from J1 to J2 and QID is the indirect heat

transfer rate between J1 and J2 via JR. The energy balance requires that Q12 = QD + QID.

There are several relationships which arise from the network.

> J1:= Eb1 - Q12*Rs1:

> J2:= Eb2 + Q12*Rs2:

> QD:= (J1 - J2)/R12:

> QID:= (J1 - J2)/(R1R + RR2):

> eq1:= (J1 - JR)/R1R:

> eq2:= (JR - J2)/RR2:

> J_R:= solve({eq1- eq2 = 0}, JR);

> evalf(subs(sys, Q12 = Qsys1, J_R), 5);

> TR:= evalf((83675/5.67e-8)^(1/4), 5);

The refractory surface temperature is approximately 1102 K and it lies

between the temperatures of the two heated surfaces.

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