Geoscience Reference
In-Depth Information
Solve the heat balance to determine
q
for cooling the superheated steam and condensing only:
q
= (Heat needed to cool steam to condensation temperature) + (Heat of condensation)
=
mC
p
∆
T + mH
v
The average specific heat (
C
p
) of steam at 250°F is roughly 0.45 Btu/lb-°F. The heat of vaporization
of steam at 212°F = 970.3 Btu/lb. Substituting into the equation,
q
= (660 lb/min)(0.45 Btu/lb-°F)(250 - 212°F) + (660 lb/min)(970.3 Btu/lb)
q
= 11,286 Btu/min + 640,398 Btu/min = 651,700 Btu/min
Now use Equation 16.24 to estimate the surface area for this part of the condenser:
A
=
q
/
U
∆
T
lm
For a countercurrent condenser, the log mean temperature is given by
(
)
−−
(
)
TT
−
TT
TT
TT
GL
1
2
GL
GL
GL
2
1
T
=
lm
−
−
1
2
ln
1
1
Remember that the desuperheater-condenser section is designed using the saturation temperature to
calculate the log mean temperature difference.
Gas entering temperature (
T
G
1
) = 212°F
Coolant leaving temperature (
T
L
2
) = 120°F
Gas leaving temperature (
T
G
2
) = 212°F
Coolant entering temperature (
T
L
1
) = 60°F
(
)
−−
(
)
212
−
120
212
60
T
lm
=
= 119 .°F
212
−
−
120
ln
212
60
The overall heat transfer coefficient (
U
) is assumed to be 100 Btu/°F⋅ft
2
⋅hr. Substituting the appro-
priate values into Equation 16.24:
(
651 700 Btu/min)(60 min/hr)
(100 Btu/°F ftthr)(119.5°F)
,
2
A
=
=
3272 ft
2
⋅
⋅
Estimate the total size of the condenser. Allow for subcooling of the water (212°F - 160°F). 160°F
is an assumed safe margin. Refiguring the heat balance for cooling the water:
q
=
UA
∆
T
m
where
m
= 660 lb/min (assuming all the steam is condensed)
q
= (660 lb/min)(1 Btu/lb-°F)(212 - 160)°F = 34,320 Btu/min
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