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
Trang 1SECTION 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⬚F (565.6⬚C) At exit from the cylinder the steam has a pressure of 500 psia (3445 kPa) and a temperature of 740⬚F (393.3⬚C) A portion of this 500-psia (3445-kPa) steam is used in a closed feedwater heater to increase the temperature of 1,000,000
lb per h (454,000 kg / h) of 2000-psia (13,780-kPa) feedwater from 350⬚F (176.6⬚C)
to 430⬚F (221.1⬚C); the remainder passes through a reheater in the steam generator and is admitted to the intermediate-pressure cylinder of the turbine at a pressure of
450 psia (3101 kPa) and a temperature of 1000⬚F (537.8⬚C) The intermediate cyl-inder operates nonextraction Steam leaves this cylcyl-inder at 200 psia (1378 kPa) and
500⬚F (260⬚C) Find (a) flow rate to the feedwater heater, assuming no subcooling; (b) work done, in kW, by the high-pressure cylinder; (c) work done, in kW, by the intermediate-pressure cylinder; (d) heat added by the reheater
Calculation Procedure:
1. Find the flow rate to the feedwater heater
(a) Construct the flow diagram, Fig 1 Enter the pressure, temperature, and enthalpy values using the data given and the steam tables Write an equation for flow across the feedwater heater, or (H2 ⫺ H7) ⫽ water (H6 ⫺ H5) Substituting using the enthalpy data from the flow diagram, flow to heater ⫽ (1⫻ 106)(409 ⫺ 324.4) / (1379.3⫺ 449.4)⫽90.977.5 lb / h (41,303.8 kg / h)
2. Determine the work done by the high-pressure cylinder
(b) The work done⫽(steam flow rate, lb / h)(H1⫺H2) / 3413⫽(1⫻106)(1511.3
⫺1379.3) / 3414⫽ 38,675.7 kW
Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
Trang 21,000,000 lb per hr
1,800 psia 1050 °F
H1 = 1,511.3
1,000,000 lb per hr
2,000 psia 430 °F
H6 = 409
Reheater
450 psia
1000 °F
H3 = 1,521
Intermediate-pressure cylinder High-pressure
cylinder
500 psia
740 °F
H2 = 1,379.3
908,900 lb per hr
200 psia 500 °F
H4 = 1,269
908,900 lb per hr
91,100 lb per hr
1,000,000 lb per hr 2,000 psia 350 °F
H5 = 324.4
H7 = 449.4 Heater
1,000,000 lb/hr (454,000 kg/hr) 1800 psia (12,402 kPa) 1050 °F (565°C)
500 psia (3445 kPa) 740 °F (393°C) 1379.3 Btu/lb (3214 kJ/kg) 1511.3 Btu/lb (3521 k?
2000 psia (13,780 kPa) 430 °F (221°C) 409 (953 kJ/kg) 350°F (177°C) 324.4 (756 kJ/kg)
450 psia (3101 kPa) 1000 °F (538°C) 1521 Btu/lb (3544 kJ/kg) 500°F (260°C)
200 psia (1378 kPa) 1269 Btu/lb (2933 kJ/kg) 324.5 Btu/lb (756 kJ/kg)
908,900 lb/hr (412,641 kg/hr) 91,100 lb/hr (41,359 kg/hr) 324.4 Btu/lb (756 kJ/kg)
449.4 Btu/lb (1047 kJ/kg)
FIGURE 1 Feedwater heating flow diagram.
3. Find the work done by the intermediate-pressure cylinder
(c) The work done ⫽ (steam flow through the cylinder)(H3⫺ H4) / 3413⫽ (1 ⫻
106– 90.977.5⫻ 106)(1521⫺1269) / 3413 ⫽67,118 kW
4. Compute the heat added by the reheater
(d) Heat added by the reheater⫽(steam flow through the reheater)(H3⫺H2)⫽(1
⫻106⫺ 90,977.5)(1521⫺1379.3) ⫽128.8⫻ 106Btu / h (135.9 kJ / h)
Related Calculations. Use this general procedure to determine the flow through feedwater heaters and reheaters for utility, industrial, marine, and com-mercial steam power plants of all sizes The method given can also be used for combined-cycle plants using both a steam turbine and a gas turbine along with a heat-recovery steam generator (HRSG) in combination with one or more feedwater heaters and reheaters
DIRECT-CONTACT FEEDWATER HEATER
ANALYSIS
Determine the outlet temperature of water leaving a direct-contact open-type feed-water heater if 250,000 lb / h (31.5 kg / s) of feed-water enters the heater at 100⬚F
Trang 3FEEDWATER HEATING METHODS 5.3
(37.8⬚C) Exhaust steam at 10.3 lb / in2 (gage) (71.0 kPa) saturated flows to the heater at the rate of 25,000 lb / h (31.5 kg / s) What saving is obtained by using this heater if the boiler pressure is 250 lb / in2(abs) (1723.8 kPa)?
Calculation Procedure:
1. Compute the water outlet temperature
Assume the heater is 90 percent efficient Then t o⫽t i w w⫹0.9w s h g / (w w⫹0.9w s),
where t o ⫽ outlet water temperature, ⬚F; t i ⫽ inlet water temperature, ⬚F; w w ⫽ weight of water flowing through heater, lb / h; 0.9⫽heater efficiency, expressed as
a decimal; w s⫽weight of steam flowing to the heater, lb / h; h g⫽enthalpy of the steam flowing to the heater, Btu / lb
For saturated steam at 10.3 lb / in2(gage) (71.0 kPa), or 10.3⫹14.7 ⫽25 lb /
in2 (abs) (172.4 kPa), h g ⫽ 1160.6 Btu / lb (2599.6 kJ / kg), from the saturation pressure steam tables Then
100(250,000)⫹0.9(25,000)(1160.6)
t o⫽ ⫽187.5⬚F (86.4⬚C)
250,000⫹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⫺h i ) / ( h b⫺h i ) where h o and h i⫽enthalpy of the water leaving and entering
the heater, respectively, Btu / lb; h b⫽ enthalpy of the steam at the boiler operating
pressure, Btu / lb For this plant from the steam tables h o⫺h i / (h b⫺h i)⫽155.44
⫺67.97 / (1201.1⫺67.97)⫽0.077, or 7.7 percent
A popular rule of thumb states that for every 11⬚F (6.1⬚C) rise in feedwater temperature in a heater, there is approximately a 1 percent saving in the fuel that would otherwise be used to heat the feedwater Checking the above calculation with this rule of thumb shows reasonably good agreement
3. Determine the heater volume
With a capacity of W lb / h of water, the volume of a direct-contact or open-type
heater can be approximated fromv⫽W / 10,000, where v⫽heater internal volume,
ft3 For this heaterv⫽250,000 / 10,000⫽25 ft3(0.71 m3)
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⬚F (44.4⬚C) (from 238 to 318⬚F or, 114.4 to 158.9⬚C), bleed heating steam is at 100 lb / in2(abs) (689.5 kPa) and 460⬚F (237.8⬚C), drains leave the heater at the saturation temper-ature corresponding to the heating steam pressure [110 lb / in2(abs) or 689.5 kPa], and5⁄8-in (1.6-cm) OD admiralty metal tubes with a maximum length of 6 ft (1.8
FEEDWATER HEATING METHODS
Trang 4m) are used Use the Standards of the Bleeder Heater Manufacturers Association,
Inc., when analyzing the heater.
Calculation Procedure:
1. Determine the LMTD across heater
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 / in2(abs) (689.5 kPa), t s ⫽327.81⬚F (164.3⬚C) Then
(t s⫺t ) i ⫺(t s⫺t ) o
LMTD⫽t m⫽
ln [t s⫺t / (t i s⫺t )] o
where the symbols are as defined in the previous calculation procedure Thus,
(327.81⫺238)⫺(327.81⫺318)
t m⫽
ln [327.81⫺238/(327.81⫺318)]
⫽36.5⬚F (20.3⬚C)
The average film temperature t ffor any closed heater is then
t f⫽t s⫺0.8t m
⫽327.81⫺29.2⫽298.6⬚F (148.1⬚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⬚F (121.1⬚C) average film temperature curve This curve
is used even though t f⫽298.6⬚F (148.1⬚C), because the standards recommend that heat-transfer rates higher than those for a 250⬚F (121.1⬚C) film temperature not be used So, from the 8-ft / s (2.4 m / s) intersection with the 250⬚F (121.1⬚C) curve in
Fig 2, project to the left to read U⫽ the overall heat-transfer rate ⫽ 910 Btu / (ft2䡠 ⬚F䡠h) [5.2 k] / m2䡠 ⬚C䡠s)]
Next, check Table 1 for the correction factor for U Assume that no 18 BWG
5⁄8-in (1.6-cm) OD arsenical copper tubes are used in this exchanger Then the
correction factor from Table 1 is 1.00, and Ucorr⫽910(1.00)⫽910 If no 9 BWG
tubes are chosen, Ucorr ⫽910(0.85) ⫽ 773.5 Btu / (ft2䡠 ⬚F䡠h) [4.4 kJ / (m2䡠 ⬚C䡠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⬚F (114.4⬚C) is, from the
saturation-temperature steam table, h fi ⫽ 206.32 Btu / lb (479.9 kJ / kg) The enthalpy of the leaving feedwater at 318⬚F (158.9⬚C) is, from the same table, h fo⫽288.20 Btu / lb
(670.4 kJ / kg) Then the heater transferred H t Btu / h is H t⫽ w w (h fo ⫺ h fi), where
w w⫽ feedwater flow rate, lb / h Or, H t⫽ 37,640(288.20⫺ 206.32) ⫽3,080,000 Btu / h (902.7 kW)
Trang 5FEEDWATER 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 to 1 in (1.6 to 2.5 cm) inclusive]
FEEDWATER HEATING METHODS
Trang 64. Compute the surface area required in the exchanger
The surface area required A ft2 ⫽ H t / Ut m Then A ⫽ 3,080,000 / [910)(36.5)] ⫽ 92.7 ft2(8.6 m2)
5. Determine the number of tubes per pass
Assume the heater has only one pass, and compute the number of tubes required Once the number of tubes is known, a decision can be made about the number of passes required In a closed heater, number of tubes ⫽ w w (passes) (ft3/ s per tube) / [v(ft2per tube open area)], where w w ⫽ lb / h of feedwater passing through heater;v⫽feedwater velocity in tubes, ft / s
Since the feedwater enters the heater at 238⬚F (114.4⬚C) and leaves at 318⬚F (158.9⬚C), its specific volume at 278⬚F (136.7⬚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 ⫽ 0.01691 ft3/ lb (0.0011 m3/ kg) at 278⬚F (136.7⬚C) Convert this to cubic feet per second per tube by dividing this specific volume by 3600 (number of seconds in 1 h) and multiplying by the pounds per hour of feedwater per tube Or, ft3/ s per tube⫽(0.01691 / 3600)(lb / h per tube) Since no 18 BWG 5⁄8-in (1.6-cm) OD tubes are being used, ID ⫽ 0.625 ⫺ 2(thickness) ⫽ 0.625 ⫺ 2(0.049) ⫽ 0.527 in (1.3 cm) Then, open area per tube
ft2⫽ (d2/ 4) / 144 ⫽ 0.7854(0.527)2/ 144⫽ 0.001525 ft2(0.00014 m2) 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 ⫽ [(37,640)(1)(0.01681 / 3600)] / [(8)(0.001525)]⫽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 ⫽A / (number of tubes per pass)(passes)(area per ft of tube) Or,
tube length for 1 pass⫽ 92.7 / [(14)(1)(0.1636)]⫽40.6 ft (12.4 m) The area per
ft of tube length is obtained from a table of tube properties or computed from
12(OD) / 144⫽12(0.625) / 155⫽0.1636 ft2(0.015 m2)
7. Compute the actual number of passes and the actual tube length
Since the tubes in this heater cannot exceed 6 ft (1.8 m) in length, the number of passes required⫽(length for one pass, ft) / (maximum allowable tube length, ft)⫽ 40.6 / 6⫽ 6.77 passes Since a fractional number of passes cannot be used and an even number of passes permit a more convenient layout of the heater, choose eight passes
From the same equation for tube length as in step 6, l⫽ tube length⫽ 92.7 / [(14)(8)(0.1636)]⫽5.06 ft (1.5 m)
8. Determine the feedwater pressure drop through heater
In any closed feedwater heater, the pressure loss ⌬p lb / in2 is ⌬p ⫽ F1F2(L ⫹
5.5D)N / D1.24, where ⌬p ⫽ pressure drop in the feedwater passing through the heater, lb / in2; F1and F2⫽correction factors from Fig 3; L⫽total lin ft of tubing
divided by the number of tube holes in one tube sheet; D⫽tube ID; N⫽number
of passes In finding F2, the average water temperature is taken as t s⫺ t m
For this heater, using correction factors from Fig 3,
⌬p⫽(0.136)(0.761)冋 (8)(14) ⫹5.5(0.527)册0.5271.24
2
⫽14.6 lb/in (100.7 kPa)
Trang 7FEEDWATER HEATING METHODS 5.7
FIGURE 3 Correction factors for closed feedwater heaters (Standards of
Bleeder Heater Manufacturers Association, Inc.)
9. Find the heater shell outside diameter
The total number of tubes in the heater⫽ (number of passes)(tubes per pass) ⫽ 8(14) ⫽ 112 tubes Assume that there is 3⁄8-in (1.0-cm) clearance between each tube and the tube alongside, above, or below it Then the pitch or center-to-center distance between the tubes⫽pitch⫹tube OD⫽ 3⁄8⫹5⁄8⫽1 in (2.5 cm) The number of tubes per ft2 of tube sheet ⫽ 166 / (pitch)2, or 166 / 12 ⫽ 166 tubes per ft2(1786.8 per m2) Since the heater has 112 tubes, the area of the tube sheet⫽112 / 166⫽0.675 ft2, or 97 in2(625.8 cm2)
The inside diameter of the heater shell ⫽ (tube sheet area, in2/ 0.7854)0.5 ⫽ (97 / 0.7854)0.5⫽11.1 in (28.2 cm) With a 0.25-in (0.6-cm) thick shell, the heater shell OD⫽11.1⫹2(0.25)⫽ 11.6 in (29.5 cm)
10. Compute the quantity of heating steam required
Steam enters the heater at 100 lb / in2(abs) (689.5 kPa) and 460⬚F (237.8⬚C) The
enthalpy at this pressure and temperature is, from the superheated steam table, h g
⫽ 1258.8 Btu / lb (2928.0 kJ / kg) The steam condenses in the heater, leaving as condensate at the saturation temperature corresponding to 100 lb / in2(abs) (689.5 kPa), or 327.81⬚F (164.3⬚C) The enthalpy of the saturated liquid at this temperature
is, from the steam tables, h f⫽298.4 Btu / lb (694.1 kJ / kg)
The heater steam consumption for any closed-type feedwater heater is W, lb /
h⫽ w w(⌬t)(h g⫺ h f), where⌬t⫽temperature rise of feedwater in heater,⬚F, c⫽ specific heat of feedwater, Btu / (lb䡠 ⬚F) Assume c⫽1.00 for the temperature range
in this heater, and W⫽ (37,640)(318⫺ 238)(1.00) / (1258.8⫺ 298.4)⫽3140 lb /
h (0.40 kg / s)
Related Calculations. The procedure used here can be applied to closed feed-water heaters in stationary and marine service A similar procedure is used for selecting hot-water heaters for buildings, marine, and portable service Various
au-FEEDWATER HEATING METHODS
Trang 8thorities recommend the following terminal difference (heater condensate temper-ature minus the outlet feedwater tempertemper-ature) for closed feedwater heaters:
POWER-PLANT HEATER EXTRACTION-CYCLE
ANALYSIS
A steam power plant operates at a boiler-drum pressure of 460 lb / in2(abs) (3171.7 kPa), a turbine throttle pressure of 415 lb / in2 (abs) (2861.4 kPa) and 725⬚F (385.0⬚C), and a turbine capacity of 10,000 kW (or 13,410 hp) The Rankine-cycle efficiency ratio (including generator losses) is: full load, 75.3 percent; three-quarters load, 74.75 percent; half load, 71.75 percent The turbine exhaust pressure is 1 inHg absolute (3.4 kPa); steam flow to the steam-jet air ejector is 1000 lb / h (0.13
kg / s) Analyze this cycle to determine the possible gains from two stages of ex-traction for feedwater heating, with the first stage a closed heater and the second stage a direct-contact or mixing heater Use engineering-office methods in analyzing the cycle
Calculation Procedure:
1. Sketch the power-plant cycle
Figure 4a shows the plant with one closed heater and one direct-contact heater Values marked on Fig 4a will be computed as part of this calculation procedure.
Enter each value on the diagram as soon as it is computed
2. Compute the throttle flow without feedwater heating extraction
Use the superheated steam tables to find the throttle enthalpy h f⫽1375.5 Btu / lb (3199.4 kJ / kg) at 415 lb / in2(abs) (2861.4 kPa) and 725⬚F (385.0⬚C)
Assume an irreversible adiabatic expansion between throttle conditions and the
exhaust pressure of 1 inHg (3.4 kPa) Compute the final enthalpy H 2sby the same
method used in earlier calculation procedures by finding y 2s, the percentage of moisture at the exhaust conditions with 1-inHg absolute (3.4-kPa) exhaust pressure
Do this by setting up the ratio y 2s ⫽(s y⫺S1) / s fg , where s g and s fgare entropies at
the exhaust pressure; S1is entropy at throttle conditions From the steam tables, y 2s
⫽2.0387⫺1.6468 / 1.9473⫽0.201 Then H 2s ⫽h g⫺y 2s h fg , where h g and h fgare enthalpies at 1 inHg absolute (3.4 kPa) Substitute values from the steam table for
1 inHg absolute (3.4 kPa); or, H 2s ⫽ 1096.3 ⫺ 0.201(1049.2) ⫽ 885.3 Btu / lb (2059.2 kJ / kg)
The available energy in this irreversible adiabatic expansion is the difference between the throttle and exhaust conditions, or 1375.5 ⫺ 885.3 ⫽ 490.2 Btu / lb
Trang 9FEEDWATER HEATING METHODS 5.9
FIGURE 4 (a) Two stages of feedwater heating in a steam plant; (b) Mollier chart
of the cycle in (a).
(1140.2 kJ / kg) The work at full load on the turbine is: (Rankine-cycle effi-ciency)(adiabatic available energy) ⫽ (0.753)(490.2) ⫽ 369.1 Btu / lb (858.5 kJ / kg) Enthalpy at the exhaust of the actual turbine ⫽ throttle enthalpy minus full-load actual work, or 1375.5 ⫺ 369.1 ⫽ 1006.4 Btu / lb (2340.9 kJ / kg) Use the Mollier chart to find, at 1.0 inHg absolute (3.4 kPa) and 1006.4 Btu / lb (2340.9
kJ / kg), that the exhaust steam contains 9.5 percent moisture
Now the turbine steam rate SR ⫽ 3413(actual work output, Btu) Or, SR ⫽
3413 / 369.1⫽9.25 lb / kWh (4.2 kg / kWh) With the steam rate known, the
nonex-FEEDWATER HEATING METHODS
Trang 10traction throttle flow is (SR)(kW output)⫽9.25(10,000) ⫽92,500 lb / h (11.7 kg / s)
3. Determine the heater extraction pressures
With steam extraction from the turbine for feedwater heating, the steam flow to the main condenser will be reduced, even with added throttle flow to compensate for extraction
Assume that the final feedwater temperature will be 212⬚F (100.0⬚C) and that the heating range for each heater is equal Both assumptions represent typical prac-tice for a moderate-pressure cycle of the type being considered
Feedwater leaving the condenser hotwell at 1 inHg absolute (3.4 kPa) is at 79.03⬚F (26.1⬚C) This feedwater is pumped through the air-ejector intercondensers and aftercondensers, where the condensate temperature will usually rise 5 to 15⬚F (2.8 to 8.3⬚C), depending on the turbine load Assume that there is a 10⬚F (5.6⬚C) rise in condensate temperature from 79 to 89⬚F (26.1 to 31.7⬚C) Then the temper-ature range for the two heaters is 212 ⫺ 89 ⫽ 123⬚F (68.3⬚C) The temperature rise per heater is 123 / 2⫽ 61.5⬚F (34.2⬚C), since there are two heaters and each will have the same temperature rise Since water enters the first-stage closed heater
at 89⬚F (31.7⬚C), the exit temperature from this heater is 89 ⫹ 61.5 ⫽ 150.5⬚F (65.8⬚C)
The second-stage heater is a direct-contact unit operating at 14.7 lb / in2(abs) (101.4 kPa), because this is the saturation pressure at an outlet temperature of 212⬚F (100.0⬚C) Assume a 10 percent pressure drop between the turbine and heater steam inlet This is a typical pressure loss for an extraction heater Extraction pressure for the second-stage heater is then 1.1(14.7)⫽16.2 lb / in2(abs) (111.7 kPa)
Assume a 5⬚F (2.8⬚C) terminal difference for the first-stage heater This is a typical terminal difference, as explained in an earlier calculation procedure The saturated steam temperature in the heater equals the condensate temperature ⫽ 150.5⬚F (65.8⬚C) exit temperature ⫹ 5⬚F (2.8⬚C) terminal difference ⫽ 155.5⬚F (68.6⬚C) From the saturation-temperature steam table, the pressure at 155.5⬚F (68.6⬚C) is 4.3 lb / in2(abs) (29.6 kPa) With a 10 percent pressure loss, the extrac-tion pressure⫽1.1(4.3)⫽4.73 lb / in2(abs) (32.6 kPa)
4. Determine the extraction enthalpies
To establish the enthalpy of the extracted steam at each stage, the actual turbine-expansion line must be plotted Two points—the throttle inlet conditions and the exhaust conditions—are known Plot these on a Mollier chart, Fig 4 Connect these two points by a dashed straight line, Fig 4
Next, measure along the saturation curve 1 in (2.5 cm) from the intersection
point A back toward the enthalpy coordinate, and locate point B Now draw a gradually sloping line from the throttle conditions to point B; from B increase the
slope to the exhaust conditions The enthalpy of the steam at each extraction point
is read where the lines of constant pressure cross the expansion line Thus, for the
second-stage direct-contact heater where p⫽ 16.2 lb / in2(abs) (111.7 kPa), h g⫽
1136 Btu / lb (2642.3 kJ / kg) For the first-stage closed heater where p⫽4.7 lb / in2
(abs) (32.4 kPa), h g⫽1082 Btu / lb (2516.7 kJ / kg)
When the actual expansion curve is plotted, a steeper slope is used between the throttle super-heat conditions and the saturation curve of the Mollier chart, because the turbine stages using superheated steam (stages above the saturation curve) are more efficient than stages using wet steam (stages below the saturation curve)