5.1
SECTION 5
FEEDWATER HEATING
METHODS
Steam-Plant Feedwater-Heating cycle
Analysis
5.1
Direct-Contact Feedwater Heater
Analysis
5.2
Closed Feedwater Heater Analysis and
Selection
5.3
Power-Plant Heater Extraction-Cycle
Analysis
5.8
Feedwater Heating with Diesel-Engine
Repowering of a Steam Plant
5.13
STEAM-PLANT FEEDWATER-HEATING CYCLE
ANALYSIS
The high-pressure cylinder of a turbogenerator unit receives 1,000,000 lb per h
(454,000 kg /h) of steam at initial conditions of 1800 psia (12,402 kPa) and 1050
Њ
F
(565.6
Њ
C). At exit from the cylinder the steam has a pressure of 500 psia (3445
kPa) and a temperature of 740
Њ
F (393.3

values using the data given and the steam tables. Write an equation for flow across
the feedwater heater, or (H
2
Ϫ
H
7
)
ϭ
water (H
6
Ϫ
H
5
). Substituting using the
enthalpy data from the flow diagram, flow to heater
ϭ
(1
ϫ
10
6
)(409
Ϫ
324.4)/
(1379.3
Ϫ
449.4)
ϭ
90.977.5 lb/h (41,303.8 kg/h).
2. Determine the work done by the high-pressure cylinder
(b) The work done

H
6
= 409
Reheater
450 psia
1000°F
H
3
= 1,521
Intermediate-
pressure
cylinder
High-pressure
cylinder
500 psia
740°F
H
2
=
1,379.3
908,900 lb per hr
200 psia 500°F
H
4
= 1,269
908,900 lb per hr
91,100 lb per hr
1,000,000 lb per hr
2,000 psia 350°F
H

10
6
)(1521
Ϫ
1269)/3413
ϭ
67,118 kW.
4. Compute the heat added by the reheater
(d) Heat added by the reheater
ϭ
(steam flow through the reheater)(H
3
Ϫ
H
2
)
ϭ
(1
ϫ
10
6
Ϫ
90,977.5)(1521
Ϫ
1379.3)
ϭ
128.8
ϫ
10
6

Assume the heater is 90 percent efficient. Then t
o
ϭ
t
i
w
w
ϩ
0.9w
s
h
g
/(w
w
ϩ
0.9w
s
),
where t
o
ϭ
outlet water temperature,
Њ
F; t
i
ϭ
inlet water temperature,
Њ
F; w
w

ϭϭ
187.5
Њ
F (86.4
Њ
C)
o
250,000
ϩ
0.9(25,000)
2. Compute the savings obtained by feed heating
The percentage of saving, expressed as a decimal, obtained by heating feedwater
is (h
o
Ϫ
h
i
)/( h
b
Ϫ
h
i
) where h
o
and h
i
ϭ
enthalpy of the water leaving and entering
the heater, respectively, Btu /lb; h
b

With a capacity of W lb /h of water, the volume of a direct-contact or open-type
heater can be approximated from
v ϭ
W/10,000, where
v ϭ
heater internal volume,
ft
3
. For this heater
v ϭ
250,000/10,000
ϭ
25 ft
3
(0.71 m
3
).
Related Calculations. Most direct-contact or open feedwater heaters store in
2-min supply of feedwater when the boiler load is constant, and the feedwater
supply is all makeup. With little or no makeup, the heater volume is chosen so that
there is enough capacity to store 5 to 30 min feedwater for the boiler.
CLOSED FEEDWATER HEATER ANALYSIS AND
SELECTION
Analyze and select a closed feedwater heater for the third stage of a regenerative
steam-turbine cycle in which the feedwater flow rate is 37,640 lb/h (4.7 kg /s), the
desired temperature rise of the water during flow through the heater is 80
Њ
F (44.4
Њ
C)

When heat-transfer rates in feedwater heaters are computed, the average film tem-
perature of the feedwater is used. In computing this the Standards of the Bleeder
Heater Manufacturers Association specify that the saturation temperature of the
heating steam be used. At 100 lb/ in
2
(abs) (689.5 kPa), t
s
ϭ
327.81
Њ
F (164.3
Њ
C).
Then
(t
Ϫ
t )
Ϫ
(t
Ϫ
t )
si so
LMTD
ϭ
t
ϭ
m
ln [t
Ϫ
t /(t

t
Ϫ
0.8t
fs m
ϭ
327.81
Ϫ
29.2
ϭ
298.6
Њ
F (148.1
Њ
C)
2. Determine the overall heat-transfer rate
Assume a feedwater velocity of 8 ft/ s (2.4 m /s) for this heater. This velocity value
is typical for smaller heaters handling less than 100,000-lb/ h (12.6-kg /s) feedwater
flow. Enter Fig. 2 at 8 ft/s (2.4 m/s) on the lower horizontal scale, and project
vertically upward to the 250
Њ
F (121.1
Њ
C) average film temperature curve. This curve
is used even though t
f
ϭ
298.6
Њ
F (148.1
Њ

8
-in (1.6-cm) OD arsenical copper tubes are used in this exchanger. Then the
correction factor from Table 1 is 1.00, and U
corr
ϭ
910(1.00)
ϭ
910. If no. 9 BWG
tubes are chosen, U
corr
ϭ
910(0.85)
ϭ
773.5 Btu/ (ft
2
⅐ Њ
F
⅐
h) [4.4 kJ /(m
2
⅐ Њ
C
⅐
s)],
given the correction factor from Table 1 for arsenical copper tubes.
3. Compute the amount of heat transferred by the heater
The enthalpy of the entering feedwater at 238
Њ
F (114.4
Њ

t
ϭ
37,640(288.20
Ϫ
206.32)
ϭ
3,080,000
Btu/h (902.7 kW).
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FEEDWATER HEATING METHODS
FEEDWATER HEATING METHODS
5.5
FIGURE 2 Heat-transfer rates for closed feedwater heaters. (Standards of
Bleeder Heater Manufacturers Association, Inc.)
TABLE 1
Multipliers for Base Heat-Transfer Rates
[For tube OD
5
⁄
8
to1in(1.6 to 2.5 cm) inclusive]
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FEEDWATER HEATING METHODS
5.6
POWER GENERATION
4. Compute the surface area required in the exchanger

per tube open area)], where w
w
ϭ
lb/h of feedwater passing through
heater;
v ϭ
feedwater velocity in tubes, ft /s.
Since the feedwater enters the heater at 238
Њ
F (114.4
Њ
C) and leaves at 318
Њ
F
(158.9
Њ
C), its specific volume at 278
Њ
F (136.7
Њ
C), midway between t
i
and t
o
, can
be considered the average specific volume of the feedwater in the heater. From the
saturation-pressure steam table,
v
f
ϭ

0.527 in (1.3 cm). Then, open area per tube
ft
2
ϭ
(
␲
d
2
/4)/144
ϭ
0.7854(0.527)
2
/144
ϭ
0.001525 ft
2
(0.00014 m
2
) per tube.
Alternatively, this area could be obtained from a table of tube properties.
With these data, compute the total number of tubes from number of tubes
ϭ
[(37,640)(1)(0.01681/3600)] /[(8)(0.001525)]
ϭ
14.29 tubes.
6. Compute the required tube length
Assume that 14 tubes are used, since the number required is less than 14.5. Then,
tube length l,ft
ϭ
A/(number of tubes per pass)(passes)(area per ft of tube). Or,

From the same equation for tube length as in step 6, l
ϭ
tube length
ϭ
92.7/
[(14)(8)(0.1636)]
ϭ
5.06 ft (1.5 m).
8. Determine the feedwater pressure drop through heater
In any closed feedwater heater, the pressure loss
⌬
p lb/in
2
is
⌬
p
ϭ
F
1
F
2
(L
ϩ
5.5D)N/ D
1.24
, where
⌬
p
ϭ
pressure drop in the feedwater passing through the

5.5(0.527)
ͫͬ
1.24
(8)(14) 0.527
2
ϭ
14.6 lb/in (100.7 kPa)
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FEEDWATER HEATING METHODS