Problem:
Recall from the lecture note the empirical relation for approximation of the wall heat flux
for turbulent flow in a circular tube for large Pr, using Reynolds analogy. Based on the
following approximation
(
)
(
)
0
0
3
3
0
0
*
*
1
Pr
14.5
1
Pr
14.5
R
dy
dy
q
q
yu
yu
ν
ν
∞
+
+
∫
∫
±
Show that
(
)
(
)
1/3
0
*
0
3 3
RePr
2
14.5
z
R
q D
u
u
k T
T
π
⎛
⎞
=
⎜
⎟
⎜
⎟
−
⎝
⎠
Answer:
The difference between the temperature at the wall (T
0
) and the time-averaged
temperature at the tube axis (
R
T
) is expressed as
(
)
(
)
0
0
3
0
*
(1)
1
Pr
14.5
R
dy
q
k T
yu
ν
∞
=
−
+
∫
T
The integral in Eq. (1) can be evaluated by making a change of variable. Let
Pr
.
/
*
1 3
145
yv
v
x
⎛
⎝
⎜
⎞
⎠
⎟ =
Then Eq. (1) becomes
1
1/3
*
0
3
0
Pr
14.5
1
o
R
v
dx
q
k
x
ν
−
−
∞
−
⎛
⎞
⎛
⎞
=
−
⎜
⎟
⎜
⎟
+
⎝
⎠
⎝
⎠
∫
T
T
The integral may be found in a table of integrals, where we find

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