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)