14.2.1 Matrix Linear Equations
Apply method 2—the matrix method for N surfaces enclosure with participating gases. For case A problem, given temperatures to determine heat fluxes.
Let the right side of Equation 14.24 = the right side of Equation 14.27 to form
the matrix as before:
a 11 J 1 + a 12 J 2 + · · · + a 1N J N = c 1
a 21 J 1 + a 22 J 2 + · · · + a 2N J N = c 2
a 31 J 1 + a 32 J 2 + · · · + a 3N J N = c 3
a N1 J 1 + a N2 J 2 + · · · + a NN J N = c N
Therefore,
[A] [J] = [C]
[ J] = [A]
−1
[C]
In addition, combining Equations 14.23 and 14.26, we obtain J i = emission
from surface i + reflection from surface j
⎡
⎤
N
= ε i E bi + (1 − ε i )
F ij J j (1 − α g ) + ε i,g E bg
(14.28)
⎣
⎦
j=1
Similarly, Equation 14.28 can be used to form the matrix [A][J] = [C] as
follows:
J 1 = ε 1 E b1 + (1 − ε 1 )[F 11 J 1 (1 − α g ) + ε 1,g E bg + F 12 J 2 (1 − α g ) + ε 1,g E bg + · · · ]
J 2 = ε 2 E b2 + (1 − ε 2 )[F 21 J 1 (1 − α g ) + ε 2,g E bg + F 22 J 2 (1 − α g ) + ε 2,g E bg + · · · ]
. . .
.
. .
J N = ε N E bN + (1 − ε N )[F N1 J 1 (1 − α g ) + ε N,g E bg
+ F N2 J 2 (1 − α g ) + ε N,g E bg + · · · ]
Once matrix [ J] has been solved, then surface heat transfer rate can be
determined from Equation 14.24 as
E bi − J i
⇒ q i =
(1 − ε i )/(ε i A i )
284
Analytical Heat Transfer
Apply method 2—the matrix method for N surfaces enclosure with participating gases. For case A problem, given temperatures to determine heat fluxes.
Let the right side of Equation 14.24 = the right side of Equation 14.27 to form
the matrix as before:
a 11 J 1 + a 12 J 2 + · · · + a 1N J N = c 1
a 21 J 1 + a 22 J 2 + · · · + a 2N J N = c 2
a 31 J 1 + a 32 J 2 + · · · + a 3N J N = c 3
a N1 J 1 + a N2 J 2 + · · · + a NN J N = c N
Therefore,
[A] [J] = [C]
[ J] = [A]
−1
[C]
In addition, combining Equations 14.23 and 14.26, we obtain J i = emission
from surface i + reflection from surface j
⎡
⎤
N
= ε i E bi + (1 − ε i )
F ij J j (1 − α g ) + ε i,g E bg
(14.28)
⎣
⎦
j=1
Similarly, Equation 14.28 can be used to form the matrix [A][J] = [C] as
follows:
J 1 = ε 1 E b1 + (1 − ε 1 )[F 11 J 1 (1 − α g ) + ε 1,g E bg + F 12 J 2 (1 − α g ) + ε 1,g E bg + · · · ]
J 2 = ε 2 E b2 + (1 − ε 2 )[F 21 J 1 (1 − α g ) + ε 2,g E bg + F 22 J 2 (1 − α g ) + ε 2,g E bg + · · · ]
. . .
.
. .
J N = ε N E bN + (1 − ε N )[F N1 J 1 (1 − α g ) + ε N,g E bg
+ F N2 J 2 (1 − α g ) + ε N,g E bg + · · · ]
Once matrix [ J] has been solved, then surface heat transfer rate can be
determined from Equation 14.24 as
E bi − J i
⇒ q i =
(1 − ε i )/(ε i A i )
284
Analytical Heat Transfer