Banded radiation solver

A long cylinder is held at a temperature of T₁=1000 K in the long square duct shown. The edge dimension of the square duct is L=1 m, and three of the four walls are insulated. The fourth wall is held at a temperature of T₂=500 K and has an emissivity of ε₂=0.2. The cylinder has a diameter D=L/2 and an emissivity of ε₁=0.7. Treating the insulating walls of the duct (surface-3) as isothermal, determine the net rate of radiative heat loss from the cylinder Q_rad1 using network analysis. What is the temperature of the adiabatic walls T₃? Solve this problem using the banded radiation solver.

 

 

 

 

 

 

 

Solution:

The problem parameters are:

L=1;

D=L/2;

T1=1000;

T2=500;

e=[0.7; 0.2; 1.];

A=[pi*D*1; L*1; 3*L*1];

 

The view factors for radiative exchange are calculated from:

 

F=zeros(3,3);

F(1,2)=1/4;               % symmetry

F(1,3)=1-F(1,2);          % summation rule

 

F(2,1)=A(1)*F(1,2)/A(2);  % reciprocity

F(3,1)=A(1)*F(1,3)/A(3);  % reciprocity

 

F(2,3)=1-F(2,1);          % summation rule

F(3,2)=A(2)*F(2,3)/A(3);  % reciprocity

 

F(3,3)=1-F(3,1)-F(3,2);   % summation rule

 

A function that solves the banded radiation equations based on an array of guessed values for the unknown surface temperatures is defined. In this problem there is only one unknown temperature T₃, and the surfaces area gray (band=[0 Inf])

 

qrad = @(T3)SolveBandRadEqs([T1; T2; T3],F,e,[0 Inf]);

 

A function that evaluates the error in guessed temperatures is defined. For this problem, the condition that surface 3 is adiabatic is the appropriate error check:

 

RadBalEqs = @(qrad)[qrad(3)]; % qrad(3)=0

 

The final step is to employ a root finding function to determine the correct surface temperatures that satisfy the error condition.

 

T3=NSolve(@(G)RadBalEqs(qrad(G)),700)

T3 =  968.86

 

Therefore, the adiabatic wall temperature is found to be T₃=968.9 K. The radiative flux of heat from each surface is determined from the banded radiation equations solver:

 

q=SolveBandRadEqs([T1; T2; T3],F,e,[0 Inf]);

Q=q.*A

Q =

   9.6075e+03

  -9.6075e+03

   3.4985e-06

 

Therefore, the net radiative heat loss of heat from the cylinder is Q_rad1=9607.5 W. 

 

 

Supposed surface-1 has the spectral emissivity given by:

 

The radiation bands for this problem are defined by:

 

bands = [0 2 4 Inf];

 

and the spectral emissivities for the enclosure are defined by:

 

e=[0.0 0.8 0.4; ... % surface-1

0.2 0.2 0.2; ... % surface-2

1.0 1.0 1.0]; % surface-3

 

The function that solves the banded radiation equations is redefined:

 

qrad = @(T3)SolveBandRadEqs([T1; T2; T3],F,e,bands);

 

and solved:

 

T3=NSolve(@(G)RadBalEqs(qrad(G)),700)

T3 = 958.65

 

Therefore, the adiabatic wall temperature is now T₃=958.7 K. The radiative flux of heat from each surface is determined:

 

q=SolveBandRadEqs([T1; T2; T3],F,e,bands);

Q=q.*A

Q =

9.1784e+03

-9.1784e+03

2.3382e-06

 

Therefore, the net radiative heat loss of heat from the cylinder is now Q_rad1=9178.4 W.