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Single Phase 26
1
A natural circulation test loop is heated at point “a”
and cooled at point “b” as illustrated below.
The total effective
loss coefficient for the cold leg is 0.8 (including friction).
The total effective loss coefficient for the hot leg is 1.0
(including friction).
Both the hot leg and cold leg have diameters of 1 ft.
The total piping length is 100 ft, evenly
split between the hot and cold legs.
If the loop contains 3395 lbm of water with a hot leg temperature of 620 F and
a cold leg temperature of 550 F, determine:
a)
The system pressure.
b)
The mass flow rate in the loop
You may assume a fluid state equation of the form
P
T
P
T
4
10
5906
.
2
0881
.
0
2847
.
95
)
,
(
−
×
−
−
=
ρ
where
T
is temperature in degrees F and
P
is pressure in psia.
30 ft
a
b
SOLUTION
System Pressure
The total system mass is given by
∫
=
V
dV
P
T
M
)
,
(
which for this system simplifies to
)
,
(
)
,
(
P
T
V
P
T
V
M
cl
cl
hl
hl
+
=
Since the hot leg and cold leg volumes are equal
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View Full DocumentSingle Phase 26
2
]
10
5906
.
2
2
)
(
0881
.
0
2847
.
95
2
[
)]
,
(
)
,
(
[
4
P
T
T
V
P
T
P
T
V
M
cl
hl
hl
cl
hl
hl
−
×
×
−
+
−
×
=
+
=
ρ
or solving for
P
4
10
5906
.
2
2
/
)
(
0881
.
0
2847
.
95
2
−
×
×
−
+
−
×
=
hl
cl
hl
V
M
T
T
P
The hot leg volume is
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 Fall '08
 Doster

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