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ME Heat Transfer 1

M.M. Yovanovich

FEF97P5.MWS

Solution for Problem 5 of Final Examination, Fall 1997.

Assumptions:

1) Steady-state heat transfer.

2) Negligible end effects.

3) Surfaces are gray and diffuse.

4) Room is much larger than the shield, or treat it as a black surface.

> restart:

System parameters

> sys:= (D1 = 20/1000, Ds = 60/1000, epsilon1 = 1, epsilon2i = 0.01,
epsilon2o = 0.1, epsilonw = 1, Tair = 27 + 273, Tw = 17 + 273,
ho = 10, Ts = 42 + 273, sigma = 5.67e-8);

(a) Sketch and label the interior radiative network and the exterior thermal circuit

The interior radiative network consists of a black surface and a gray, diffuse surface.

The nodes are Eb1 and Ebs and the radiative resistances are spatial radiative resistances R1s

and and the gray surface resistance at the inner surafce of the shield Rsi.

The thermal circuit for the exterior consists of two thermal resistances in parallel.

One resistance is the film resistance Rfo from the surface into the air at temperature

Tair = 27 C,

and the second resistance is the thermal radiative resistance Rro from the outer surface

of the shield to the surroundings at temperature Troom = 17 C.

The shield temperature is given, and the inner cylinder temperature T1 is unknown.

Calculation of heat transfer rates from cylinder to shield per unit length.

Net radiative exchange between cylinder and inner surface of the shield.

> Q1s:= (Eb1 - Ebs)/(Rs1 + R1s + Rsi);
Eb1:= sigma*T1^4; Ebs:= sigma*Ts^4;
Rs1:= (1 - epsilon1)/(A1*epsilon1);
Rsi:= (1 - epsilon2i)/(As*epsilon2i);
R1s:= 1/(A1*F1s);
A1:= Pi*D1; As:= Pi*Ds; F1s:= 1;

> Q1s:= evalf(subs(sys, Q1s), 4);

Heat transfer rate from the outer surface of the shield to the air and the walls.

Net radiative exchange between outer surface of the shield and the walls.

> Qso:= (Ebs - Ebw)/Rshield_wall:
Ebs:= sigma*Ts^4; Ebw:= sigma*Tw^4;
Rshield_wall:= Rsso + Rsw + Rswall:
Rsw:= 1/(As*Fsw): Rsso:= (1 - epsilon2o)/(As*epsilon2o):
Rswall:= (1 - epsilonw)/(Awall*epsilonw):
Rsw:= 1/(As*Fsw):
As:= Pi*Ds:
Awall:= 1e3: Fsw:= 1:
Qso:= evalf(subs(sys, Qso), 6);