10.34 – Fall 2006
Homework #8  Solutions
Problem 1 – Conduction/Convection (Beers’ text 6.B.1)
This problems involves heat conduction in a flowing Newtonian fluid with a know velocity
profile (incompressible flow in a tube). The governing equation for the system is the conduction
equation with convection in the zdirection:
2
∂
T
λ
⎛
1
∂
∂
⎛
T
⎞
∂
2
T
⎞
v
⋅∇
(
ρ
C T
)
=
λ
∇
T
⇒
v
=
r
+
P
z
P
⎜
⎝
r
r
⎜
∂
⎟
z
2
⎟
⎠
∂
z
ρ
C
∂
⎝
r
⎠
∂
⎛
r
2
⎞
∂
T
λ
1
∂
∂
2
T
∂
2
T
⎞
⎛
⎞
⎛
T
2
U
⎜
1
−
⎜
⎟
⎟
⋅
=
⎜
+
+
⎟
⎜
R
⎟
∂
z
ρ
C
r
r
∂
∂
r
2
∂
z
2
⎝
⎝
⎠
⎠
P
⎝
⎠
In this case, U is the average velocity of fluid and can be calculated from the given Reynolds
number and fluid properties.
The boundary conditions for this problem were given in the radial
part of the problem, but were somewhat ambiguous in the zdirection.
In the zdirection, a valid
boundary condition for upstream would be that T = T
0
at some distance upstream of the
temperature change at the wall. A BC for downstream could be that the gradient (dT/dz) far
downstream is zero, or that T = T
1
far downstream.
The zero derivative boundary condition is
probably better since it will be less likely to skew a solution where you do not go far enough in
the zdirection. The boundary conditions used in the problem were:
T r z
= −
⋅
0
R Pe
,
2
R Pe
=
T
T
r z
,
=
10
⋅
=
T
(
)
(
)
1
(
=
,
)
=
⎨
⎧
⎩
T
T
1
0
z
z
≥
<
0
0
∂
∂
T
r
(
r
=
0,
)
T r
R z
z
=
0
The boundary conditions in
z
are equivalent to starting 2 characteristic conduction times
upstream and going to 10 downstream of the temperature jump.
It is often useful to make problem nondimensional before solving in order to formulate the
problem in terms of as few parameters as possible.
For this problem, we choose the following
nondimensional variables:
η
≡
r
ξ
≡
z
θ
≡
T
T
−
0
where
:
Pe
=
2
UR
and
α
=
λ
R
R Pe
T
W
−
T
0
α
ρ
C
P
⋅
Carefully applying these nondimensional variables, one will find the following equation:
Flow
r
R
T
1
T
0
T
wall
z = 0
Figure by MIT OCW.
Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to
Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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z
,
1
1,
,
1,
,
1
i j
i
j
i j
i
j
i j
+
−
+
−
•
•
•
•
•
∂
θ
∂
2
θ
η
2
∂
2
θ
(
2
)
∂
θ
+
η
+
−
η
1
−
η
⋅
=
0
∂
η
∂
η
2
Pe
∂
ξ
2
∂
ξ
(
,
= −
2
)
=
0
,
=
10
)
=
1
θ η ξ
(
θ η ξ
θ η
1,
⎧
⎩
1
0
ξ
ξ
≥
<
0
0
∂
∂
η
θ
(
η
=
0
(
=
ξ
)
=
⎨
=
0,
ξ
)
You can see that the nondimensional form is much simpler, and the boundary conditions are
extremely simple, which is a nice aspect since the units in many problems can be tedious.
We
also multiply through by the radius term to avoid any potential problems at r = 0.
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 Fall '04
 ClarkColton
 Chemical reaction, Boundary value problem, MIT OpenCourseWare, Massachusetts Institute of Technology, Numerical Methods Applied

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