ME 353 Heat Transfer 1

M.M. Yovanovich

FEF96P5.MWS

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Problem 5 of Final Examination, December 9, 1996.

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Consider two large, diffuse, gray, parallel surfaces separated

by a small space.

(a) If the surface emissivities are 0.8 and 0.9 respectively,

sketch the radiation circuit labelling all resistors and nodes.

Calculate the total radiation resistance for this system.

(b) A thin, high conductivity shield is placed between the

two surfaces.

The shield emissivities, epsilons, are the same for both sides

of the shield.

Sketch the radiation circuit labelling all resistors and nodes.

Calculate the total radiation resistance for this system.

(c) What value of the shield emissivity will reduce the net

radiation heat transfer by a factor of 20?

> restart:

(a) Total radiation resistance of the system without the radiation shield.

> Rgen1:=
(1 - epsilon1)/(A1*epsilon1) + 1/(A1*F12) +
(1 - epsilon2)/(A2*epsilon2);

> case1:=
(A1 = 1, A2 = 1, F12 = 1, epsilon1 = 0.8, epsilon2 = 0.9);

(a) The value of the total radiation resistance without the radiation shield.

> Rrad1:= evalf(subs(case1, Rgen1), 6);
#The units are 1/m^2.

(b) Total radiation resistance when a single shield is inserted

between the gray surfaces.

> Rgen2:=
(1 - epsilon1)/(A1*epsilon1) +
(1 - epsilon2)/(A2*epsilon2) +
1/(A1*F1s1) + (1 - epsilons1)/(As1*epsilons1) +
(1 - epsilons2)/(As2*epsilons2) + 1/(As2*Fs22);

> case2:=
(A1 = 1, A2 = 1, F1s1 = 1, Fs22 = 1,
epsilon1 = 0.8, epsilon2 = 0.9, epsilons1 = 0.1,
epsilons2 = 0.2, As1 = 1, As2 = 1);

(b) Value of the total radiation resistance when a single shield is present.

The total resistance depends on the emissivity of the shield which is

open at this point.

> Rrad2:= evalf(subs(case2, Rgen2), 6);
#The units are 1/m^2.

(c) Establish an equation that relates the total resistances of the two systems.

The total resistance of the system with the shield is required to be 20 times

greater than the resistance without the shield.

> case3:=
(A1 = 1, A2 = 1, F1s1 = 1, Fs22 = 1,
epsilon1 = 0.8, epsilon2 = 0.9,
epsilons1 = epsilons, epsilons2 = epsilons,
As1 = 1, As2 = 1);

> Rrad3:= evalf(subs(case3, Rgen2), 6);

> eq:= Rrad3 - 20*Rrad1 = 0;

(c) Calculate the value of the shield emissivity.

> epsilons:= evalf(solve(eq, epsilons), 3);

The above value of emissivity corresponds to a highly polished,

clean surface such as aluminum foil.