Manual De Calculos Del Ingeniero Mecanico

Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS

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⬚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 cylinder operates nonextraction. Steam leaves this cylinder 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. 5.1 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

FEEDWATER HEATING METHODS 5.2

POWER GENERATION

1,000,000 lb per hr 1,800 psia 1050°F H1 = 1,511.3

High-pressure cylinder

Reheater

450 psia 1000°F H3 = 1,521

500 psia 908,900 lb per hr 740°F H2 = 1,379.3 Heater

Intermediatepressure cylinder

91,100 lb per hr 1,000,000 lb per hr 2,000 psia 350°F H5 = 324.4

1,000,000 lb per hr 2,000 psia 430°F H6 = 409

908,900 lb per hr 200 psia 500°F H4 = 1,269

H7 = 449.4 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 ⫻ 106 Btu / 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 commercial 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 feedwater heater if 250,000 lb / h (31.5 kg / s) of water enters the heater at 100⬚F

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FEEDWATER HEATING METHODS FEEDWATER 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 to ⫽ tiww ⫹ 0.9wshg / (ww ⫹ 0.9ws), where to ⫽ outlet water temperature, ⬚F; ti ⫽ inlet water temperature, ⬚F; ww ⫽ weight of water flowing through heater, lb / h; 0.9 ⫽ heater efficiency, expressed as a decimal; ws ⫽ weight of steam flowing to the heater, lb / h; hg ⫽ 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), hg ⫽ 1160.6 Btu / lb (2599.6 kJ / kg), from the saturation pressure steam tables. Then to ⫽

100(250,000) ⫹ 0.9(25,000)(1160.6) ⫽ 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 (ho ⫺ hi) / ( hb ⫺ hi) where ho and hi ⫽ enthalpy of the water leaving and entering the heater, respectively, Btu / lb; hb ⫽ enthalpy of the steam at the boiler operating pressure, Btu / lb. For this plant from the steam tables ho ⫺ hi / (hb ⫺ hi) ⫽ 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 from v ⫽ W / 10,000, where v ⫽ heater internal volume, ft3. For this heater v ⫽ 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 temperature corresponding to the heating steam pressure [110 lb / in2 (abs) or 689.5 kPa], and 5⁄8-in (1.6-cm) OD admiralty metal tubes with a maximum length of 6 ft (1.8

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FEEDWATER HEATING METHODS 5.4

POWER GENERATION

m) 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 temperature 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), ts ⫽ 327.81⬚F (164.3⬚C). Then LMTD ⫽ tm ⫽

(ts ⫺ ti ) ⫺ (ts ⫺ to) ln [ts ⫺ ti / (ts ⫺ to)]

where the symbols are as defined in the previous calculation procedure. Thus, tm ⫽

(327.81 ⫺ 238) ⫺ (327.81 ⫺ 318) ln [327.81 ⫺ 238/(327.81 ⫺ 318)]

⫽ 36.5⬚F (20.3⬚C)

The average film temperature tf for any closed heater is then tf ⫽ ts ⫺ 0.8tm ⫽ 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 tf ⫽ 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 saturationtemperature steam table, hfi ⫽ 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, hfo ⫽ 288.20 Btu / lb (670.4 kJ / kg). Then the heater transferred Ht Btu / h is Ht ⫽ ww(hfo ⫺ hfi), where ww ⫽ feedwater flow rate, lb / h. Or, Ht ⫽ 37,640(288.20 ⫺ 206.32) ⫽ 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 to 1 in (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 The surface area required A ft2 ⫽ Ht / Utm. 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 ⫽ ww (passes) (ft3 / s per tube) / [v(ft2 per tube open area)], where ww ⫽ 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 ti and to, can be considered the average specific volume of the feedwater in the heater. From the saturation-pressure steam table, vf ⫽ 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 ⫽ (␲d 2 / 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; F1 and 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 ts ⫺ tm. For this heater, using correction factors from Fig. 3, ⌬ p ⫽ (0.136)(0.761)

冋

册

5.06(8)(14) 8 ⫹ 5.5(0.527) (8)(14) 0.5271.24

⫽ 14.6 lb/in2 (100.7 kPa)

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FEEDWATER HEATING METHODS FEEDWATER 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, hg ⫽ 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, hf ⫽ 298.4 Btu / lb (694.1 kJ / kg). The heater steam consumption for any closed-type feedwater heater is W, lb / h ⫽ ww(⌬t)(hg ⫺ hf), 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 feedwater heaters in stationary and marine service. A similar procedure is used for selecting hot-water heaters for buildings, marine, and portable service. Various au-

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FEEDWATER HEATING METHODS 5.8

POWER GENERATION

thorities recommend the following terminal difference (heater condensate temperature minus the outlet feedwater temperature) 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 extraction 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 hf ⫽ 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 H2s by the same method used in earlier calculation procedures by finding y2s, the percentage of moisture at the exhaust conditions with 1-inHg absolute (3.4-kPa) exhaust pressure. Do this by setting up the ratio y2s ⫽ (sy ⫺ S1) / sfg, where sg and sfg are entropies at the exhaust pressure; S1 is entropy at throttle conditions. From the steam tables, y2s ⫽ 2.0387 ⫺ 1.6468 / 1.9473 ⫽ 0.201. Then H2s ⫽ hg ⫺ y2shfg, where hg and hfg are enthalpies at 1 inHg absolute (3.4 kPa). Substitute values from the steam table for 1 inHg absolute (3.4 kPa); or, H2s ⫽ 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

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FEEDWATER HEATING METHODS FEEDWATER 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 efficiency)(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 fullload 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-

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FEEDWATER HEATING METHODS 5.10

POWER GENERATION

traction 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 practice 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 temperature 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 extraction 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 turbineexpansion 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), hg ⫽ 1136 Btu / lb (2642.3 kJ / kg). For the first-stage closed heater where p ⫽ 4.7 lb / in2 (abs) (32.4 kPa), hg ⫽ 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).

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FEEDWATER HEATING METHODS FEEDWATER HEATING METHODS

5.11

5. Compute the extraction steam flow To determine the extraction flow rates, two assumptions must be made—condenser steam flow rate and first-stage closed-heater extraction flow rate. The complete cycle will be analyzed, and the assumption checked. If the assumptions are incorrect, new values will be assumed, and the cycle analyzed again. Assume that the condenser steam flow from the turbine is 84,000 lb / h (10.6 kg / s) when it is operating with extraction. Note that this value is less than the nonextraction flow of 92,500 lb / h (11.7 kg / s). The reason is that extraction of steam will reduce flow to the condenser because the steam is bled from the turbine after passage through the throttle but before the condenser inlet. Then, for the first-stage closed heater, condensate flow is as follows:

The value of 5900 lb / h (0.74 kg / s) of condensate from the first-stage heater is the second assumption made. Since it will be checked later, an error in the assumption can be detected. Assume a 2 percent heat radiation loss between the turbine and heater. This is a typical loss. Then

Compare the required extraction, 5950 lb / h (0.75 kg / s), with the assumed extraction, 5900 lb / h (0.74 kg / s). The difference is only 50 lb / h (0.006 kg / s), which is less than 1 percent. Therefore, the assumed flow rate is satisfactory, because estimates within 1 percent are considered sufficiently accurate for all routine analyses. For the second-stage direct-contact heater, condensate flow, lb / h is as follows:

The required extraction, calculated in the same way as for the first-stage heater, is (90,900)(61.7 / 932.2) ⫽ 6050 lb / h (0.8 kg / s).

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FEEDWATER HEATING METHODS 5.12

POWER GENERATION

The computed extraction flow for the second-stage heater is not compared with an assumed value because an assumption was not necessary. 6. Compare the actual condenser steam flow Sketch a vertical line diagram, Fig. 5, showing the enthalpies at the throttle, heaters, and exhaust. From this diagram, the work lost by the extracted steam can be computed. As Fig. 5 shows, the total enthalpy drop from the throttle to the exhaust is 369 Btu / lb (389.3 kJ / kg). Each pound of extracted steam from the first- and second-stage bleed points causes a work loss of 75.7 Btu / lb (176.1 kJ / kg) and 129.7 Btu / lb (301.7 kJ / kg), respectively. To carry the same load, 10,000 kW, with extractions, it will be necessary to supply the following additional compensation steam to the turbine throttle: (heater flow, lb / h)(work loss, Btu / h) / (total work, Btu / h). Then

Check the assumed condenser flow using nonextraction throttle flow ⫹ additional throttle flow ⫺ heater extraction ⫽ condenser flow. Set up a tabulation of the flows as follows:

FIGURE 5 Diagram of turbine-expansion line.

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FEEDWATER HEATING METHODS FEEDWATER HEATING METHODS

5.13

Compare this actual flow, 83,840 lb / h (10.6 kg / s), with the assumed flow, 84,000 lb / h (10.6 kg / s). The difference, 160 lb / h (0.02 kg / s), is less than 1 percent. Since an accuracy within 1 percent is sufficient for all normal power-plant calculations, it is not necessary to recompute the cycle. Had the difference been greater than 1 percent, a new condenser flow would be assumed and the cycle recomputed. Follow this procedure until a difference of less than 1 percent is obtained. 7. Determine the economy of the extraction cycle For a nonextraction cycle operating in the same pressure range,

Heat chargeable to turbine ⫽ (throttle flow ⫹ air-ejector flow)(heat supplied by boiler) / (kW output of turbine) ⫽ (92,500 ⫹ 1000)(1328.3) / 10,000 ⫽ 12,410 Btu / kWh (13,093.2 kJ / kWh), which is the actual heat rate HR of the nonextraction cycle. For the extraction cycle using two heaters,

As before, heat chargeable to turbine ⫽ (95,840 ⫹ 1000)(1195.3) / 10,000 ⫽ 11,580 Btu / kWh (12,217.5 kJ / kWh). Therefore, the improvement ⫽ (nonextraction HR ⫺ extraction HR) / nonextraction HR ⫽ (12,410 ⫺ 11,580) / 12,410 ⫽ 0.0662, or 6.62 percent. Related Calculations. (1) To determine the percent improvement in a steam cycle resulting from additional feedwater heaters in the cycle, use the same procedure as given above for three, four, five, six, or more heaters. Plot the percent improvement vs. number of stages of extraction, Fig. 6, to observe the effect of additional heaters. A plot of this type shows the decreasing gains made by additional heaters. Eventually the gains become so small that the added expenditure for an additional heater cannot be justified. (2) Many simple marine steam plants use only two stages of feedwater heating. To analyze such a cycle, use the procedure given, substituting the hp output for the kW output of the turbine. (3) Where a marine plant has more than two stages of feedwater heating, follow the procedure given in (1) above.

FEEDWATER HEATING WITH DIESEL-ENGINE REPOWERING OF A STEAM PLANT Show the economies and environmental advantages possible with Diesel-engine repowering of steam boiler / turbine plants using feedwater heating as the entree.

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FEEDWATER HEATING METHODS 5.14

POWER GENERATION

FIGURE 6 Percentage of improvement in turbine heat rate vs. stages of extraction.

Give the typical temperatures and flow rates encountered in such installations using gas and / or oil fuels.

Calculation Procedure:

1. Determine the output ranges possible with today’s diesel engines Medium-speed Diesel engines are available in sizes exceeding 16 MW. While this capacity may seem small when compared to gas turbines, it is appropriate for repowering of steam plants up to 600 MW via boiler feedwater heating. Modern Diesel engines can attain simple cycle efficiencies of over 47 percent burning natural gas or heavy fuel oil (HFO). The ability to burn natural gas in Diesels is a key factor when coupled with coal-fired boilers. Since the Clean Air Act Amendments of 1990 (CAA) require these boilers to reduce both NOx and SO2 emissions on a lb / million Btu-fired basis (kg / MJ), a boiler feedwater heating system that can help make these reductions while simultaneously improving overall plant efficiency is attractive. Diesel engines offer these reductions when used in repowering and feedwater heating. Today Diesel engines convert about 45 percent of mechanical energy to electricity; 30 percent becomes exhaust-gas heat; 12 percent is lost to jacket-water heat; and 6 percent is used to cool the lube oil. The remaining energy lost is generally not recoverable. 2. Show how the diesel engine can be used in the feedwater heating cycle Modern steam-turbine reheat cycles, Fig. 7, use an array of feedwater heaters in a regenerative feedwater heating system. The heaters progressively increase the condensate temperature until it approaches the steam saturation temperature. Condensate then enters the final economizer and evaporator sections of the boiler. Using the waste heat from Diesel engines to partially replace the feedwater heaters is almost completely non-intrusive to the operation of the existing system,

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FEEDWATER HEATING METHODS FEEDWATER HEATING METHODS

5.15

FIGURE 7 In repowering, Diesel exhaust is adjusted in temperature to the same levels expected from feedwater heaters in the existing plant. (Power.)

but causes several significant alterations in the cycle. Two particular cycle alterations are: (1) Jacket water temperature from a Diesel engine is available at about 195⬚F (91⬚C). The lube-oil cooling system produces water at about 170⬚F (77⬚C). These temperatures are appropriate for partial displacement of the boiler’s lowtemperature feedwater heaters. (2) A gas / Diesel engine can operate on about 97 percent natural gas / 3 percent HFO and has an exhaust temperature of 680⬚F (360⬚C). The exhaust gas can be ducted through an economizer that is equipped with selective catalytic reduction (SCR) and has heat-transfer sections that can adjust the exit temperature to match the preheated-burner-windbox air temperature. The SCR reduces NOx emissions from the engine to about 25 ppm on leaving the economizer. This exhaust economizer, Fig. 7, also elevates the temperature of the feedwater after it leaves the deaerator. 3. Explain the environmental impact of using diesels in the feedwater heating loop Exhaust gas from the economizer sections, Fig. 7, is ducted to the boiler windbox. This gas serves the same function as flue-gas recirculation (FGR) in a low-NOx burner. In the installation in Fig. 7, the two Diesel generators produce 351,600 lb / h (159,626 kg / h) of exhaust gas. Most of this gas is ducted to the boiler windbox to achieve a 17.5 percent O2 level needed for the low NOx burners. The balance enters the boiler as overfire air. 4. Determine the heat-rate improvement possible Diesel engines are highly efficient on a simple-cycle basis. When combined with a steam turbine, as described, the cycle efficiency reaches about 56 percent on an incremental basis. In the example here, the incremental heat rate of the engine combined with the additional output from the turbine is 6060 Btu / kWh (6393 kJ / kWh). This heat rate represents about 25 percent of the total system power and can

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FEEDWATER HEATING METHODS 5.16

POWER GENERATION

be averaged with the heat rate of the associated plant. Total system heat rate may be improved by as much as 10 percent as a result of repowering in this fashion. 5. Evaluate system turndown possible with this type of feedwater heating Typically, a coal-fired boiler can be turned down to about 60 percent load while maintaining superheat and reheat temperatures. Adding Diesel feedwater heat increases system output by about 25 percent. More important, the system is almost completely non-intrusive, and can return to normal operation when the Diesel output is not required. Thus, the total turndown of the plant is increased from 40 to 52 percent, making plant operation more flexible. 6. Compare diesels vs. gas turbines for feedwater heating Comparing Diesels vs. gas turbines (GT) in this application, it appears that the major differences are in the temperature of the exhaust gas and the quantities of exhaust gas that must be introduced to the boiler. Most GTs have fairly high exhaust-gas rates on a per-kilowatt basis, varying from 25 to over 30 lb / kW (9 to 13.6 kg / kW). GT exhaust may contain from 14.5 to 15.5 percent O2. Conversely, Diesels have exhaust-gas rates of 15 to 16 lb / kW (6.8 to 7.3 kg / kW). The O2 concentrations for Diesels vary between 11 percent for spark-ignited gas engines up to 13 percent for gas / Diesels or HFO-fired Diesels. Thus, when providing inlet gases to the boiler and adjusting the windbox concentrations to 17.5 percent O2, the volume of gas has to be even further increased with GTs. 7. Evaluate the cost of this type of feedwater heating Capital cost for modifying the boiler is largely dependent on the site and boiler. Cost for a turnkey-installed Diesel facility is about $850 / kW. For a Diesel plant connected with an existing power system, net output of the existing system is increased, as noted, because of increasing flow to the steam turbine’s condenser. This increased output offsets the cost of interconnection to the boiler. Related Calculations. The data and procedure given here represent a new approach to feedwater heating and repowering. Because three function are served—namely feedwater heating, repowering, and environmental compliance, the approach is unique. Calculation of the variables is simple because basic heattransfer relations—covered elsewhere in this handbook—are used. The date and methods given in this procedure are the work of F. Mack Shelor, Wartsila Diesel Inc., as reported in Power magazine (June 1995). SI values were added by the handbook editor.

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