manual de calculos del ingeniero mecanico

8/14/2019 Manual de Calculos Del Ingeniero Mecanico

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5.1

SECTION 5

FEEDWATER HEATING

METHODS

Steam-Plant Feedwater-Heating cycleAnalysis 5.1

Direct-Contact Feedwater HeaterAnalysis 5.2

Closed Feedwater Heater Analysis andSelection 5.3

Power-Plant Heater Extraction-CycleAnalysis 5.8

Feedwater Heating with Diesel-EngineRepowering 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(565.6C). At exit from the cylinder the steam has a pressure of 500 psia (3445kPa) and a temperature of 740F (393.3C). A portion of this 500-psia (3445-kPa)steam is used in a closed feedwater heater to increase the temperature of 1,000,000lb per h (454,000 kg/h) of 2000-psia (13,780-kPa) feedwater from 350F (176.6C)to 430F (221.1C); the remainder passes through a reheater in the steam generatorand is admitted to the intermediate-pressure cylinder of the turbine at a pressure of450 psia (3101 kPa) and a temperature of 1000F (537.8C). The intermediate cyl-inder operates nonextraction. Steam leaves this cylinder at 200 psia (1378 kPa) and500F (260C). 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 theintermediate-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 enthalpyvalues using the data given and the steam tables. Write an equation for flow acrossthe feedwater heater, or (H

2 H

7) water (H

6 H

5). 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 10

6)(1511.3 1379.3)/3414 38,675.7 kW.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)Copyright 2006 The McGraw-Hill Companies. All rights reserved.

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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS


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5.2 POWER GENERATION

1,000,000 lb per hr1,800 psia 1050FH1 = 1,511.3

1,000,000 lb per hr2,000 psia 430FH6 = 409

Reheater

450 psia1000FH3 = 1,521

Intermediate-pressurecylinder

High-pressurecylinder

500 psia

740FH2 =

1,379.3

908,900 lb per hr200 psia 500FH4 = 1,269

908,900 lb per hr

91,100 lb per hr

1,000,000 lb per hr2,000 psia 350FH5 = 324.4

H7 = 449.4

Heater

1,000,000 lb/hr (454,000 kg/hr) 1800 psia (12,402 kPa) 1050F (565C)

500 psia (3445 kPa) 740F (393C) 1379.3 Btu/lb (3214 kJ/kg) 1511.3 Btu/lb (3521 k?

2000 psia (13,780 kPa) 430F (221C) 409 (953 kJ/kg) 350F (177C) 324.4 (756 kJ/kg)

450 psia (3101 kPa) 1000F (538C) 1521 Btu/lb (3544 kJ/kg) 500F (260C)

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 com-mercial steam power plants of all sizes. The method given can also be used forcombined-cycle plants using both a steam turbine and a gas turbine along with aheat-recovery steam generator (HRSG) in combination with one or more feedwaterheaters 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 water enters the heater at 100F



FEEDWATER HEATING METHODS


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FEEDWATER HEATING METHODS 5.3

(37.8C). Exhaust steam at 10.3 lb/in2 (gage) (71.0 kPa) saturated flows to theheater at the rate of 25,000 lb/h (31.5 kg/s). What saving is obtained by using thisheater if the boiler pressure is 250 lb/in2 (abs) (1723.8 kPa)?


1. Compute the water outlet temperatureAssume 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 asa decimal; ws weight of steam flowing to the heater, lb/h; hg enthalpy of thesteam 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

100(250,000) 0.9(25,000)(1160.6)t 187.5F (86.4C)o 250,000 0.9(25,000)

2. Compute the savings obtained by feed heatingThe percentage of saving, expressed as a decimal, obtained by heating feedwateris (ho hi) / ( hb hi) where ho and hi enthalpy of the water leaving and enteringthe heater, respectively, Btu/lb; hb enthalpy of the steam at the boiler operatingpressure, 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 11F (6.1C) rise in feedwatertemperature in a heater, there is approximately a 1 percent saving in the fuel thatwould otherwise be used to heat the feedwater. Checking the above calculation withthis rule of thumb shows reasonably good agreement.

3. Determine the heater volumeWith a capacity of W lb/h of water, the volume of a direct-contact or open-typeheater 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 feedwatersupply is all makeup. With little or no makeup, the heater volume is chosen so thatthere 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 regenerativesteam-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 80F (44.4C)(from 238 to 318F or, 114.4 to 158.9C), bleed heating steam is at 100 lb / in2 (abs)(689.5 kPa) and 460F (237.8C), drains leave the heater at the saturation temper-ature corresponding to the heating steam pressure [110 lb/in2 (abs) or 689.5 kPa],and 58-in (1.6-cm) OD admiralty metal tubes with a maximum length of 6 ft (1.8





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m) are used. Use the Standards of the Bleeder Heater Manufacturers Association,Inc., when analyzing the heater.


1. Determine the LMTD across heaterWhen 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 theheating steam be used. At 100 lb/in2 (abs) (689.5 kPa), ts 327.81F (164.3C).Then

(t t ) (t t )s i s oLMTD t m ln [t t / (t t )]s i s 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)]

36.5F (20.3C)

The average film temperature tf for any closed heater is then

t t 0.8t f s m

327.81 29.2 298.6F (148.1C)

2. Determine the overall heat-transfer rateAssume a feedwater velocity of 8 ft/s (2.4 m/s) for this heater. This velocity valueis typical for smaller heaters handling less than 100,000-lb/h (12.6-kg/s) feedwaterflow. Enter Fig. 2 at 8 ft/s (2.4 m/s) on the lower horizontal scale, and projectvertically upward to the 250F (121.1C) average film temperature curve. This curveis used even though tf 298.6F (148.1C), because the standards recommend thatheat-transfer rates higher than those for a 250F (121.1C) film temperature not beused. So, from the 8-ft/s (2.4 m/s) intersection with the 250F (121.1C) curve in

Fig. 2, project to the left to read U the overall heat-transfer rate 910 Btu/(ft2 F h) [5.2 k]/m2 C s)].Next, check Table 1 for the correction factor for U. Assume that no. 18 BWG

58-in (1.6-cm) OD arsenical copper tubes are used in this exchanger. Then thecorrection factor from Table 1 is 1.00, and U

corr 910(1.00) 910. If no. 9 BWGtubes are chosen, U

corr 910(0.85) 773.5 Btu/(ft2

F h) [4.4 kJ/(m2 C s)],given the correction factor from Table 1 for arsenical copper tubes.

3. Compute the amount of heat transferred by the heaterThe enthalpy of the entering feedwater at 238F (114.4C) is, from the saturation-temperature steam table, hfi 206.32 Btu/lb (479.9 kJ/kg). The enthalpy of the

leaving feedwater at 318F (158.9C) 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), whereww feedwater flow rate, lb/h. Or, Ht 37,640(288.20 206.32) 3,080,000Btu/h (902.7 kW).





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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 58 to 1 in (1.6 to 2.5 cm) inclusive]





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4. Compute the surface area required in the exchangerThe 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 passAssume 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 ofpasses required. In a closed heater, number of tubes ww (passes) (ft

3 /s pertube)/[v(ft2 per tube open area)], where ww lb/h of feedwater passing throughheater; v feedwater velocity in tubes, ft/s.

Since the feedwater enters the heater at 238F (114.4C) and leaves at 318F(158.9C), its specific volume at 278F (136.7C), midway between ti and to, canbe considered the average specific volume of the feedwater in the heater. From thesaturation-pressure steam table, vf 0.01691 ft

3 /lb (0.0011 m3 /kg) at 278F(136.7C). Convert this to cubic feet per second per tube by dividing this specificvolume by 3600 (number of seconds in 1 h) and multiplying by the pounds perhour of feedwater per tube. Or, ft3 /s per tube (0.01691/3600)(lb/h per tube).

Since no. 18 BWG 58-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 tubeft2 (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 perft of tube length is obtained from a table of tube properties or computed from12(OD)/144 12(0.625)/155 0.1636 ft2 (0.015 m2).

7. Compute the actual number of passes and the actual tube lengthSince the tubes in this heater cannot exceed 6 ft (1.8 m) in length, the number ofpasses 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 aneven 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 heaterIn 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 theheater, lb/in2; F

1and F

2 correction factors from Fig. 3; L total lin ft of tubingdivided by the number of tube holes in one tube sheet; D tube ID; N numberof passes. In finding F

2, the average water temperature is taken as ts tm.

For this heater, using correction factors from Fig. 3,

5.06(8)(14) 8p (0.136)(0.761) 5.5(0.527) 1.24(8)(14) 0.527

2 14.6 lb/in (100.7 kPa)





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FIGURE 3 Correction factors for closed feedwater heaters. (Standards of Bleeder Heater Manufacturers Association, Inc.)

9.

Find the heater shell outside diameterThe total number of tubes in the heater (number of passes)(tubes per pass) 8(14) 112 tubes. Assume that there is 38-in (1.0-cm) clearance between eachtube and the tube alongside, above, or below it. Then the pitch or center-to-centerdistance between the tubes pitch tube OD 38 58 1 in (2.5 cm).

The number of tubes per ft2 of tube sheet 166/(pitch)2, or 166/12 166tubes per ft2 (1786.8 per m2). Since the heater has 112 tubes, the area of the tubesheet 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 heatershell OD 11.1 2(0.25) 11.6 in (29.5 cm).

10. Compute the quantity of heating steam requiredSteam enters the heater at 100 lb/in2 (abs) (689.5 kPa) and 460F (237.8C). Theenthalpy 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 ascondensate at the saturation temperature corresponding to 100 lb/in2 (abs) (689.5kPa), or 327.81F (164.3C). The enthalpy of the saturated liquid at this temperatureis, 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 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 forselecting hot-water heaters for buildings, marine, and portable service. Various au-





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thorities recommend the following terminal difference (heater condensate temper-ature 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) (3171.7kPa), a turbine throttle pressure of 415 lb/ in2 (abs) (2861.4 kPa) and 725F(385.0C), and a turbine capacity of 10,000 kW (or 13,410 hp). The Rankine-cycleefficiency ratio (including generator losses) is: full load, 75.3 percent; three-quartersload, 74.75 percent; half load, 71.75 percent. The turbine exhaust pressure is 1inHg absolute (3.4 kPa); steam flow to the steam-jet air ejector is 1000 lb/h (0.13kg/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 secondstage a direct-contact or mixing heater. Use engineering-office methods in analyzingthe cycle.


1. Sketch the power-plant cycleFigure 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 extractionUse 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 725F (385.0C).

Assume an irreversible adiabatic expansion between throttle conditions and theexhaust pressure of 1 inHg (3.4 kPa). Compute the final enthalpy H

2s by the samemethod used in earlier calculation procedures by finding y

2s, the percentage ofmoisture at the exhaust conditions with 1-inHg absolute (3.4-kPa) exhaust pressure.Do this by setting up the ratio y

2s (sy S1) /sfg, where sg and sfg are entropies atthe exhaust pressure; S

1is entropy at throttle conditions. From the steam tables, y

2s

2.0387 1.6468/1.9473 0.201. Then H2s hg y2shfg, where hg and hfg areenthalpies at 1 inHg absolute (3.4 kPa). Substitute values from the steam table for1 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 differencebetween the throttle and exhaust conditions, or 1375.5 885.3 490.2 Btu/lb





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FIGURE 4 (a) Two stages of feedwater heating in a steam plant; ( b) Mollier chartof 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 theMollier chart to find, at 1.0 inHg absolute (3.4 kPa) and 1006.4 Btu/lb (2340.9kJ/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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traction throttle flow is (SR)(kW output) 9.25(10,000) 92,500 lb/h (11.7 kg/s).

3.Determine the heater extraction pressuresWith 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 forextraction.

Assume that the final feedwater temperature will be 212F (100.0C) and thatthe 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 at79.03F (26.1C). This feedwater is pumped through the air-ejector intercondensersand aftercondensers, where the condensate temperature will usually rise 5 to 15F(2.8 to 8.3C), depending on the turbine load. Assume that there is a 10F (5.6C)

rise in condensate temperature from 79 to 89F (26.1 to 31.7C). Then the temper-ature range for the two heaters is 212 89 123F (68.3C). The temperaturerise per heater is 123/2 61.5F (34.2C), since there are two heaters and eachwill have the same temperature rise. Since water enters the first-stage closed heaterat 89F (31.7C), the exit temperature from this heater is 89 61.5 150.5F(65.8C).

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 212F(100.0C). Assume a 10 percent pressure drop between the turbine and heater steaminlet. This is a typical pressure loss for an extraction heater. Extraction pressure forthe second-stage heater is then 1.1(14.7) 16.2 lb/in2 (abs) (111.7 kPa).

Assume a 5F (2.8C) terminal difference for the first-stage heater. This is atypical terminal difference, as explained in an earlier calculation procedure. Thesaturated steam temperature in the heater equals the condensate temperature 150.5F (65.8C) exit temperature 5F (2.8C) terminal difference 155.5F(68.6C). From the saturation-temperature steam table, the pressure at 155.5F(68.6C) 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 enthalpiesTo establish the enthalpy of the extracted steam at each stage, the actual turbine-expansion line must be plotted. Two pointsthe throttle inlet conditions and theexhaust conditionsare known. Plot these on a Mollier chart, Fig. 4. Connect thesetwo points by a dashed straight line, Fig. 4.

Next, measure along the saturation curve 1 in (2.5 cm) from the intersectionpoint A back toward the enthalpy coordinate, and locate point B. Now draw agradually sloping line from the throttle conditions to point B; from B increase theslope to the exhaust conditions. The enthalpy of the steam at each extraction pointis read where the lines of constant pressure cross the expansion line. Thus, for thesecond-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, becausethe turbine stages using superheated steam (stages above the saturation curve) aremore efficient than stages using wet steam (stages below the saturation curve).





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5. Compute the extraction steam flowTo determine the extraction flow rates, two assumptions must be madecondensersteam flow rate and first-stage closed-heater extraction flow rate. The complete cyclewill 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 thenonextraction flow of 92,500 lb/h (11.7 kg/s). The reason is that extraction ofsteam will reduce flow to the condenser because the steam is bled from the turbineafter 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 thesecond assumption made. Since it will be checked later, an error in the assumptioncan be detected.

Assume a 2 percent heat radiation loss between the turbine and heater. This isa typical loss. Then

Compare the required extraction, 5950 lb/h (0.75 kg/s), with the assumed ex-traction, 5900 lb/h (0.74 kg/s). The difference is only 50 lb/h (0.006 kg/s), whichis less than 1 percent. Therefore, the assumed flow rate is satisfactory, becauseestimates within 1 percent are considered sufficiently accurate for all routine anal-yses.

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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FIGURE 5 Diagram of turbine-expansion line.

The computed extraction flow for the second-stage heater is not compared withan 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 com-puted. As Fig. 5 shows, the total enthalpy drop from the throttle to the exhaust is369 Btu/lb (389.3 kJ/kg). Each pound of extracted steam from the first- and sec-ond-stage bleed points causes a work loss of 75.7 Btu/lb (176.1 kJ/kg) and 129.7Btu/lb (301.7 kJ/kg), respectively. To carry the same load, 10,000 kW, with ex-tractions, it will be necessary to supply the following additional compensation steamto 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 addi-tional throttle flow heater extraction condenser flow. Set up a tabulation of

the flows as follows:





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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 calcu-lations, 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 cycleFor a nonextraction cycle operating in the same pressure range,

Heat chargeable to turbine (throttle flow air-ejector flow)(heat supplied byboiler) / (kW output of turbine) (92,500 1000)(1328.3)/10,000 12,410Btu/kWh (13,093.2 kJ/kWh), which is the actual heat rate HR of the nonextractioncycle.

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 (nonextractionHR 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 steamcycle resulting from additional feedwater heaters in the cycle, use the same pro-cedure as given above for three, four, five, six, or more heaters. Plot the percentimprovement vs. number of stages of extraction, Fig. 6, to observe the effect ofadditional heaters. A plot of this type shows the decreasing gains made by addi-tional heaters. Eventually the gains become so small that the added expenditure foran 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 thekW output of the turbine.

(3) Where a marine plant has more than two stages of feedwater heating, followthe procedure given in (1) above.

FEEDWATER HEATING WITH DIESEL-ENGINE

REPOWERING OF A STEAM PLANT

Show the economies and environmental advantages possible with Diesel-enginerepowering of steam boiler/turbine plants using feedwater heating as the entree.





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FIGURE 6 Percentage of improvement inturbine heat rate vs. stages of extraction.

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


1. Determine the output ranges possible with todays diesel enginesMedium-speed Diesel engines are available in sizes exceeding 16 MW. While thiscapacity may seem small when compared to gas turbines, it is appropriate forrepowering of steam plants up to 600 MW via boiler feedwater heating.

Modern Diesel engines can attain simple cycle efficiencies of over 47 percentburning natural gas or heavy fuel oil (HFO). The ability to burn natural gas inDiesels 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 SO2emissions on a lb/million Btu-fired basis (kg/MJ), a boiler feedwater heating sys-tem that can help make these reductions while simultaneously improving overallplant efficiency is attractive. Diesel engines offer these reductions when used inrepowering and feedwater heating.

Today Diesel engines convert about 45 percent of mechanical energy to elec-tricity; 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 generallynot 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 aregenerative feedwater heating system. The heaters progressively increase the con-densate temperature until it approaches the steam saturation temperature. Conden-sate then enters the final economizer and evaporator sections of the boiler.

Using the waste heat from Diesel engines to partially replace the feedwaterheaters is almost completely non-intrusive to the operation of the existing system,





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

but causes several significant alterations in the cycle. Two particular cycle altera-

tions are: (1) Jacket water temperature from a Diesel engine is available at about195F (91C). The lube-oil cooling system produces water at about 170F (77C).These temperatures are appropriate for partial displacement of the boilers low-temperature feedwater heaters.

(2) A gas/Diesel engine can operate on about 97 percent natural gas/3 percentHFO and has an exhaust temperature of 680F (360C). The exhaust gas can beducted through an economizer that is equipped with selective catalytic reduction(SCR) and has heat-transfer sections that can adjust the exit temperature to matchthe preheated-burner-windbox air temperature. The SCR reduces NOx emissionsfrom the engine to about 25 ppm on leaving the economizer. This exhaust econo-mizer, 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 heatingloopExhaust 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-NOxburner. In the installation in Fig. 7, the two Diesel generators produce 351,600lb/h (159,626 kg/h) of exhaust gas. Most of this gas is ducted to the boiler windboxto achieve a 17.5 percent O

2level needed for the low NOx burners. The balance

enters the boiler as overfire air.

4. Determine the heat-rate improvement possibleDiesel engines are highly efficient on a simple-cycle basis. When combined with asteam turbine, as described, the cycle efficiency reaches about 56 percent on anincremental basis. In the example here, the incremental heat rate of the enginecombined 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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be averaged with the heat rate of the associated plant. Total system heat rate maybe 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 whilemaintaining superheat and reheat temperatures. Adding Diesel feedwater heat in-creases system output by about 25 percent. More important, the system is almostcompletely non-intrusive, and can return to normal operation when the Diesel out-put is not required. Thus, the total turndown of the plant is increased from 40 to52 percent, making plant operation more flexible.

6. Compare diesels vs. gas turbines for feedwater heatingComparing Diesels vs. gas turbines (GT) in this application, it appears that themajor differences are in the temperature of the exhaust gas and the quantities ofexhaust gas that must be introduced to the boiler. Most GTs have fairly high ex-haust-gas rates on a per-kilowatt basis, varying from 25 to over 30 lb/kW (9 to13.6 kg/kW). GT exhaust may contain from 14.5 to 15.5 percent O

2.

Conversely, Diesels have exhaust-gas rates of 15 to 16 lb/kW (6.8 to 7.3 kg/kW). The O

2concentrations for Diesels vary between 11 percent for spark-ignited

gas engines up to 13 percent for gas/Diesels or HFO-fired Diesels. Thus, whenproviding inlet gases to the boiler and adjusting the windbox concentrations to 17.5percent O

2, the volume of gas has to be even further increased with GTs.

7. Evaluate the cost of this type of feedwater heatingCapital 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 plantconnected with an existing power system, net output of the existing system isincreased, as noted, because of increasing flow to the steam turbines condenser.This increased output offsets the cost of interconnection to the boiler.

Related Calculations. The data and procedure given here represent a new ap-proach to feedwater heating and repowering. Because three function areservednamely feedwater heating, repowering, and environmental compliance, theapproach is unique. Calculation of the variables is simple because basic heat-transfer relationscovered elsewhere in this handbookare 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.


manual de calculos del ingeniero mecanico

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