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Boiling Heat Transfer 17
1
The hot channel in a PWR operates under the conditions given below.
Compute and compare the outer clad
temperature distribution assuming the JensLottes and Thom correlation in the nucleate boiling region and the
Bergles and Rosenhow correlation in the mixed boiling region.
You may assume an axial heat flux profile of the
form
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
′
′
=
′
′
e
H
z
q
z
q
)
(
sin
)
(
0
λ
π
Problem Parameters
Maximum channel heat flux
474,500 Btu/hrft
2
Channel Mass Flux
2.48 x 10
6
lbm/hrft
2
Inlet Temperature
552 F
Channel Pressure
2250 psia
Rod Pitch
0.496 inches
Rod Diameter
0.374 inches
Rod Height
144 inches
Axial Peak to Average Ratio
1.5
SOLUTION
Heat Flux
The heat flux profile is in terms of two unknown parameters, the extrapolation distance
and the amplitude
0
q
′
′
.
The extrapolation distance is determined by the axial peak to average ratio.
The amplitude sets the magnitude of the
heat flux.
Extrapolation Distance
The axial peak to average ratio is defined to be
q
z
q
F
z
′
′
′
′
≡
)
(
max
where
max
z
is the position of maximum heat flux in a particular channel, and
q
′
′
is the axially averaged heat flux in
the same channel.
Note, that since for any given channel
)
(
max
z
q
′
′
and
q
′
′
both contain the amplitude
0
q
′
′
, this
parameter cancels and the axial peak to average ratio is only a function of shape.
The position of maximum heat
flux is that location such that
0
max
=
′
′
z
q
dz
d
For this heat flux profile, the maximum heat flux occurs at
2
H
, such that
500
,
474
)
(
max
max
0
=
′
′
=
′
′
=
′
′
q
z
q
q
Btu/hrft
2
.
The axially averaged heat flux is defined to be
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View Full DocumentBoiling Heat Transfer 17
2
dz
H
z
q
H
dz
z
q
H
q
e
H
H
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
′
′
=
′
′
≡
′
′
∫
∫
λ
π
0
0
0
sin
1
)
(
1
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
′
′
=
′
′
e
e
e
H
H
H
H
H
q
q
)
(
cos
cos
0
The axial peaking factor is then
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
e
e
e
z
H
H
H
H
H
F
)
(
cos
cos
For
2
+
≡
H
H
e
, this expression is transcendental in
and must be solved iteratively.
Iterating on
gives
feet.
3009
.
0
=
Coolant Enthalpy and Temperature Distributions
The coolant enthalpy distribution is given by
∫
′
′
′
′
+
=
z
z
Dd
z
q
m
h
z
h
0
)
(
1
)
0
(
)
(
±
which for the heat flux profile given here yields
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
′
′
+
=
e
e
e
H
z
H
m
D
H
q
h
z
h
)
(
cos
cos
)
0
(
)
(
0
±
and is valid over the entire channel height.
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 Fall '08
 Doster

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