Chemistry Reference
In-Depth Information
1 + 1 + 4Bi / M 2
2
( T S
2 + M 2
T ) 2
q
λ
Bi
δ
T ) 2
2
( T 1
2
δ
1 + 1 + 4Bi / M 2
2
T ) 2
2
= M 2
δ
2 ( T 1
1 + 1 + 4Bi / M 2
2
2M 2
δ
Q
c
Q
2 c
( T S
.
T )
(1.27)
2
Substitution of Eq. (1.27) into Eq. (1.24) gives
1 + 1 + 4Bi / M 2
2
T ) 2
2 x = x 1
2
dT
dx
M 2
δ
2M 2 Q
δ
=
2 ( T 1
2 c
1 + 1 + 4Bi / M 2
2
(1.28)
Q
2 c
( T S
×
T )
exp
.
2 Qk 0 R T S
acE
E
R T S
±
Equation (1.7) with boundary conditions Eq. (1.4) and T
| x = x 1 = T 1 is the energy
equation for zone II. The solution of this equation is
1 + 1 + 4Bi / M 2
2
x
x 1
,
T = T +( T 1
T ) exp
M
(1.29)
δ
from which
1 + 1 + 4Bi / M 2
2
x = x 1
dT
dx
M
δ
=
( T 1
T ) .
If the temperature gradients on the border between the two zones are equal,
1 + 1 + 4Bi / M 2
2
1 + 1 + 4Bi / M 2
2
T ) 2
T ) 2
2
2
M 2
δ
= M 2
δ
2 ( T 1
2 ( T 1
1 + 1 + 4Bi / M 2
2
exp
.
2M 2
δ
2 Qk 0 R T S
acE
Q
c
Q
2 c
E
R T S
( T S
±
T )
2
(1.30)
 
Search WWH ::




Custom Search