4.1 SECTION 4 STEAM GENERATION EQUIPMENT AND AUXILIARIES Determining Equipment Loading for Generating Steam Efficiently 4.2 Steam Conditions with Two Boilers Supplying the Same Line 4.6 Generating Saturated Steam by Desuperheating Superheated Steam 4.7 Determining Furnace-Wall Heat Loss 4.8 Converting Power-Generation Pollutants from Mass to Volumetric Units 4.10 Steam Boiler Heat Balance Determination 4.11 Steam Boiler, Economizer, and Air- Heater Efficiency 4.14 Fire-Tube Boiler Analysis and Selection 4.16 Safety-Valve Steam-Flow Capacity 4.18 Safety-Valve Selection for a Watertube Steam Boiler 4.19 Steam-Quality Determination with a Throttling Calorimeter 4.24 Steam Pressure Drop in a Boiler Superheater 4.25 Selection of a Steam Boiler for a Given Load 4.26 Selecting Boiler Forced- and Induced- Draft Fans 4.30 Power-Plant Fan Selection from Capacity Tables 4.33 Fan Analysis at Varying RPM, Pressure, and Air or Gas Capacity 4.35 Boiler Forced-Draft Fan Horsepower Determination 4.37 Effect of Boiler Relocation on Draft Fan Performance 4.38 Analysis of Boiler Air Ducts and Gas Uptakes 4.38 Determination of the Most Economical Fan Control 4.44 Smokestack Height and Diameter Determination 4.46 Power-Plant Coal-Dryer Analysis 4.48 Coal Storage Capacity of Piles and Bunkers 4.50 Properties of a Mixture of Gases 4.51 Steam Injection in Air Supply 4.52 Boiler Air-Heater Analysis and Selection 4.53 Evaluation of Boiler Blowdown, Deaeration, Steam and Water Quality 4.55 Heat-Rate Improvement Using Turbine- Driven Boiler Fans 4.56 Boiler Fuel Conversion from Oil or Gas to Coal 4.60 Energy Savings from Reduced Boiler Scale 4.64 Ground Area and Unloading Capacity Required for Coal Burning 4.66 Heat Recovery from Boiler Blowdown Systems 4.67 Boiler Blowdown Percentage 4.69 Sizing Flash Tanks to Conserve Energy 4.70 Flash Tank Output 4.71 Determining Waste-Heat Boiler Fuel Savings 4.74 Figuring Flue-Gas Reynolds Number by Shortcuts 4.75 Determining the Feasibility of Flue-Gas Recirculation for No Control in Packaged Boilers 4.77 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. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 4.2 POWER GENERATION DETERMINING EQUIPMENT LOADING FOR GENERATING STEAM EFFICIENTLY A plant has a steam generator capable of delivering up to 1000,000 lb/h (45,400 kg/h) of saturated steam at 400 lb / in 2 (gage) (2756 kPa). The plant also has an HRSG capable of generating up to 1000,000 lb/h (45,400 kg/h) of steam in the fired mode at the same pressure. How should each steam generator be loaded to generate a given quantity of steam most efficiently? Calculation Procedure: 1. Develop the HRSG characteristics In cogeneration and combined-cycle steam plants (gas turbine plus other prime movers), the main objective of supervising engineers is to generate a needed quan- tity of steam efficiently. Since there may be both HRSGs and steam boilers in the plant, the key to efficient operation is an understanding of the performance char- acteristics of each piece of equipment as a function of load. In this plant, the HRSG generates saturated steam at 400 lb/in 2 (gage) (2756 kPa) from the exhaust of a gas turbine. It can be supplementary-fired to generate additional steam. Using the HRSG simulation approach given in another calculation procedure in this handbook, the HRSG performance at different steam flow rates should be developed. This may be done manually or by using the HRSG software developed by the author. 2. Select the gas/steam temperature profile in the design mode Using a pinch point of 15 Њ F (8.33 Њ C) and approach point of 17 Њ F (9.44 Њ C), a tem- perature profile is developed as discussed in the procedure for HRSG simulation. The HRSG exit gas temperature is 319 Њ F (159.4 Њ C) while generating 25,000 lb/ h (11,350 kg/h) of steam at 400 lb /in 2 (gage) (2756 kPa) using 230 Њ F (110 Њ C) feed- water. 3. Prepare the gas/steam temperature profile in the fired mode A simple approach is to use the fact that supplementary firing is 100 percent effi- cient, as discussed in the procedure on HRSG simulation. All the fuel energy goes into generating steam in single-pressure HRSGs. Compute the duty of the HRSG—i.e., the energy absorbed by the steam—in the unfired mode, which is 25.4 MM Btu /h (7.44 MW). The energy required to gen- erate 50,000 lb/ h (22,700 kg /h) of steam is 50.8 MM Btu/ h (14.88 MW). Hence, the additional fuel required ϭ 50.8 Ϫ 25.4 ϭ 25.4 MM Btu/h (7.44 MW). If a manual or computer simulation is done on the HRSG, fuel consumption will be seen to be 24.5 MM Btu/ h (7.18 MW) on a Lower Heating Value (LHV) basis. Similarly, the performance at other steam flows is also computed and summarized in Table 1. Note that the exit gas temperature decreases as the steam flow increases. This aspect of an HRSG is discussed in the simulation procedure elsewhere in this handbook. 4. Develop the steam-generator characteristics Develop the performance of the steam generator at various loads. Steam-generator suppliers will gladly provide this information in great detail, including plots and tabulations of the boiler’s performance. As shown in Table 2, the exit-gas temper- 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.3 TABLE 1 Performance of HRSG Load, % 25 50 75 100 Steam generation, lb/ h (kg / h) 25,000 (11,350) 50,000 (22,700) 75,000 (34,050) 100,000 (45,400) Duty, MM Btu/ h (MW) 25.4 (7.4) 50.8 (14.9) 76.3 (22.4) 101.6 (29.8) Exhaust gas flow, lb/ h (kg / h) 152,000 (69,008) 153,140 (69,526) 154,330 (70,066) 155,570 (70,629) Exit gas temperature, Њ F( Њ C) 319 (159) 285 (141) 273 (134) 269 (132) Fuel fired, MM Btu / h LHV basis (MW) 0 (0) 24.50 (7.2) 50.00 (14.7) 76.50 (22.4) ASME PTC 4.4 efficiency, % 70.80 83.79 88.0 89.53 Boiler pressure ϭ 400 lb / in 2 (gage) (2756 kPa); feedwater temperature ϭ 230 Њ F (110 Њ C); blowdown ϭ 5 percent. Fuel used: natural gas; percent volume C 1 ϭ 97; C 2 ϭ 2; C 3 ϭ 1; HHV ϭ 1044 Btu / ft 3 (38.9 MJ / m 3 ); LHV ϭ 942 Btu / ft 3 (35.1 MJ / m 3 ). 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.4 TABLE 2 Performance of Steam Generator Load, % 25 50 75 100 Steam generation, lb/ h (kg / h) 25,000 (11,350) 50,000 (22,700) 75,000 (34,050) 100,000 (45,400) Duty, MM Btu/ h (MW) 25.4 (7.4) 50.8 (14.9) 76.3 (22.4) 101.6 (29.8) Excess air, % 30 10 10 10 Flue gas, lb / h (kg/ h) 30,140 (13,684) 50,600 (22,972) 76,150 (34,572) 101,750 (46,195) Exit gas temperature, Њ F( Њ C) 265 (129) 280 (138) 300 (149) 320 (160) Heat losses, % —Dry gas loss 3.93 3.56 3.91 4.27 —Air moisture loss 0.10 0.09 0.10 0.11 —Fuel moisture loss 10.43 10.49 10.58 10.66 —Radiation loss 2.00 1.00 0.70 0.50 Efficiency, % —Higher Heating Value basis 83.54 84.86 84.70 84.46 —Lower Heating Value basis 92.58 94.05 93.87 93.60 Fuel fired, MM Btu / h LHV basis (MW) 27.50 (8.06) 54.00 (15.8) 81.30 (23.8) 108.60 (31.8) Boiler pressure ϭ 400 lb / in 2 (gage) (2756 kPa); feedwater temperature ϭ 230 Њ F (110 Њ C); blowdown ϭ 5 percent. Fuel used: natural gas; percent volume C 1 ϭ 97; C 2 ϭ 2; C 3 ϭ 1; HHV ϭ 1044 Btu / ft 3 (38.9 MJ / m 3 ); LHV ϭ 942 Btu / ft 3 (35.1 MJ / m 3 ). 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.5 TABLE 3 Fuel Consumption at Various Steam Loads Total steam HRSG steam SG steam HRSG fuel Sg fuel Total fuel lb/h lb/h lb/h MM Btu/h MM Btu/h MM Btu/h 150,000 50,000 100,000 24.50 108.60 133.10 150,000 75,000 75,000 50.00 81.30 131.30 150,000 100,000 50,000 76.50 54.00 130.50 100,000 0 100,000 0 108.60 108.60 100,000 25,000 75,000 0 81.30 81.30 100,000 50,000 50,000 24.50 54.00 78.50 100,000 75,000 25,000 50.00 27.50 77.50 100,000 100,000 0 75.60 0 76.50 50,000 0 50,000 0 54.00 54.00 50,000 25,000 25,000 0 27.50 27.50 50,000 50,000 0 24.50 0 24.50 kg/h kg/h SI Units kg/ h MW MW MW 68,100 22,700 45,400 7.2 31.8 38.9 68,100 34,050 34,050 14.7 23.8 38.5 68,100 45,400 45,400 22.4 15.8 38.2 45,400 0 45,400 0 31.8 31.8 45,400 11,350 34,050 0 23.8 23.8 45,400 22,700 22,700 7.2 15.8 23.0 45,400 34,050 11,350 14.7 8.1 22.7 45,400 45,400 0 22.4 0 22.4 22,700 0 22,700 0 15.8 15.8 22,700 11,350 11,350 0 8.1 8.1 22,700 11,350 0 7.2 0 7.2 ature decreases as the load on the steam generator declines. This is because the ratio of gas/steam is maintained at nearly unity, unlike in an HRSG where the gas flow remains constant and steam flow alone is varied. Further, the radiation losses in a steam boiler increase at lower duty, while the exit-gas losses decrease. However, the boiler’s efficiency falls within a narrow range. Table 3 also shows the steam generator’s fuel consumption at various loads. 5. Calculate steam vs. fuel data for combined operation of the equipment The next step is to develop, for combined operation of the HRSG and steam gen- erator, a steam flow vs. fuel table such as that in Table 3. For example, 150,000 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.6 POWER GENERATION lb/ h (68,100 kg /h) of steam could be generated in several ways—50,000 lb/h (22,700 kg /h) in the HRSG and 100,000 lb/h (45,400 kg / h) in the steam generator. Or each could generate 75,000 lb / h (34,050 kg /h); or 100,000 lb /h (45,400 kg / h) in the HRSG and the remainder in the steam generator. The table shows that max- imizing the HRSG output first is the most efficient way of generating steam because no fuel is required to generate up to 25,000 lb/h (11,350 kg/h) of steam. However, this may not always be possible because of the plant operating mode, equipment availability, steam temperature requirements, etc. Note also that the gas pressure drop in an HRSG does not vary significantly with load as the gas mass flow remains nearly constant. The gas pressure drop increases slightly as the firing temperature increases. On the other hand, the steam generator fan power consumption vs. load increases more in proportion to load. It is also seen that at higher steam capacities the difference in fuel consumption between the various modes of operation is small. At 150,000 lb/ h (68,100 kg/ h), the difference is about 3 MM Btu / h (0.88 MW), while at 100,000 lb/ h (45,400 kg/ h), the difference is 30 MM Btu/ h (8.79 MW). This difference should also be kept in mind while developing an operational strategy. If a superheater is used, the performance of the superheater would have to be analyzed. Steam generators can generally maintain the steam temperature from 40 to 100 percent load, while in HRSGs the range is much larger as the steam tem- perature increases with firing temperature and can be controlled. Related Calculations. Developing the performance characteristics of each piece of equipment as a function of load is the key to determining the mode of operation and loading of each type of steam producer. For best results, develop a performance curve for the steam generator, including all operating costs such as fan power consumption, pump power consumption, and gas-turbine power output as a function of load. This gives more insight into the total costs in addition to fuel cost, which is the major cost. This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO Industries, Inc. The HRSG software mentioned in this procedure is available from Mr. Ganapathy. STEAM CONDITIONS WITH TWO BOILERS SUPPLYING THE SAME STEAM LINE Two closely adjacent steam boilers discharge equal amounts of steam into the same short steam main. Steam from boiler No. 1 is at 200 lb/ in 2 (1378 kPa) and 420 Њ F (215.6 Њ C) while steam from boiler No. 2 is at 200 lb/in 2 (1378 kPa) and 95 percent quality. (a) What is the equilibrium condition after missing of the steam? (b) What is the loss of entropy by the higher temperature steam? Assume negligible pressure drop in the short steam main connecting the boilers. Calculation Procedure: 1. Determine the enthalpy of the mixed steam Use the T-S diagram, Fig. 1, to plot the condition of the mixed steam. Then, since equal amounts of steam are mixed, the final enthalpy, H 3 ϭ (H 1 ϩ H 2 )/ 2. Substi- 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.7 800 ˚ F (471 ˚ C) 841.8 ˚ F (449.8 ˚ C) 200 psia (1378 kPa) FIGURE 1 T-S plot of conditions with two boilers on line. tuting, using date from the steam tables and Mollier chart, H 3 ϭ (1225 ϩ 1164)/ 2 ϭ 1194.5 Btu/lb (2783.2 kJ/kg). 2. Find the quality of the mixed steam Entering the steam tables at 200 lb / in 2 (1378 kPa), find the enthalpy of the liquid as 355.4 Btu/lb (828.1 kJ/ kg) and the enthalpy of vaporization as 843.3 Btu/lb (1964.9 kJ/kg). Then, using the equation for wet steam with the known enthalpy of the mixture from Step 1, 1194.5 ϭ H ƒ ϩ x 3 (H ƒg ) ϭ 355.4 ϩ x 3 (843.3); x 3 ϭ 0.995, or 99.5 percent quality. 3. Find the entropy loss by the higher pressure steam The entropy loss by the higher-temperature steam, referring to the Mollier chart plot, is S 1 Ϫ S 2 ϭ 1.575 Ϫ 1.541 ϭ 0.034 entropy units. The lower-temperature steam gains S 3 Ϫ S 2 ϭ 1.541 Ϫ 1.506 ϭ 0.035 units of entropy. Related Calculations. Use this general approach for any mixing of steam flows. Where different quantities of steam are being mixed, use the proportion of each quantity to the total in computing the enthalpy, quality, and entropy of the mixture. GENERATING SATURATED STEAM BY DESUPERHEATING SUPERHEATED STEAM Superheated steam generated at 1350 lb/ in 2 (abs) (9301.5 kPa) and 950 Њ F (510 Њ C) is to be used in a process as saturated steam at 1000 lb/ in 2 (abs) (6890 kPa). If the 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.8 POWER GENERATION superheated steam is desuperheated continuously by injecting water at 500 Њ F (260 Њ C), how many pounds (kg) of saturated steam will be produced per pound (kg) of superheated steam? Calculation Procedure: 1. Using the steam tables, determine the steam and water properties Rounding off the enthalpy and temperature values we find that: Enthalpy of the superheated steam at 1350 lb/in 2 (abs) (9301.5 kPa) and 950 Њ F (510 Њ C) ϭ H 1 ϭ 1465 Btu /lb (3413.5 kJ/kg); Enthalpy of saturated steam at 1000 lb/ in 2 (abs) (6890 kPa) ϭ H 2 ϭ 1191 Btu/lb (2775 kJ /kg); Enthalpy of water at 500 Њ F (260 Њ C) ϭ (500 Ϫ 32) ϭ H 3 ϭ 488 Btu/lb (1137 kJ/kg). 2. Set up a heat-balance equation and solve it LX ϭ lb (kg) of water at 500 Њ F (260 Њ C) required to desuperheat the superheated steam. Then, using the symbols given above, H 1 ϩ X(H 3 ) ϭ (1 ϩ X)H 2 . Solving for X ϭ (H 1 Ϫ H 2 )/(H 2 Ϫ H 3 ) ϭ (1465 Ϫ 1191)/ (1191 Ϫ 488) ϭ 0.39. Then, 1.0 ϩ 0.39 ϭ 1.39 lb (0.63 kg) of saturated steam produced per lb (kg) of super- heated steam. Thus, if the process used 1000 lb (454 kg) of saturated steam at 1000 lb/in 2 (abs) (689 kPa), the amount of superheated steam needed to produce this saturated steam would be 1000/1.39 ϭ 719.4 lb (326.6 kg). Related Calculations. Desuperheating superheated steam for process and other use is popular because it can save purchase and installation of a separate steam generator for the lower pressure steam. While there is a small loss of energy in desuperheating (from heat losses in the piping and desuperheater), this loss is small compared to the savings made. That’s why you’ll find desuperheating being used in central stations, industrial, commercial and marine plants throughout the world. DETERMINING FURNACE-WALL HEAT LOSS A furnace wall consists of 9-in (22.9-cm) thick fire brick, 4.5-in (11.4-cm) Sil-O- Cel brick, 4-in (10.2-cm) red brick, and 0.25-in (0.64-cm) transite board. The ther- mal conductivity, k, values, Btu / (ft 2 )( Њ F)(ft) [kJ/(m 2 )( Њ C)(m)] are as follows: 0.82 at 1800 Њ F (982 Њ C) for fire brick; 0.125 at 1800 Њ F (982 Њ C) for Sil-O-Cel; 0.52 at 500 Њ F (260 Њ C) for transite. A temperature of 1800 Њ F (982 Њ C) exists on the inside wall of the furnace and 200 Њ F (93.3 Њ C) on the outside wall. Determine the heat loss per hour through each 10 ft 2 (0.929 m 2 ) of furnace wall. What is the temperature of the wall at the joint between the fire brick and Sil-O-Cel? Calculation Procedure: 1. Find the heat loss through a unit area of the furnace wall Use the relation Q ϭ ⌬ t/R, where Q ϭ heat transferred, Btu /h (W); ⌬ t ϭ temper- ature difference between the inside of the furnace wall and the outside, Њ F( Њ C); R ϭ resistance of the wall to heat flow ϭ L/(kXA), where L ϭ length of path 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.9 through which the heat flow, ft (m); k ϭ thermal conductivity, as defined above; A ϭ area of path of heat flow, ft 2 (m 2 ). Where there is more than one resistance to heat flow, add them to get the total resistance. Substituting, the above values for this furnace wall, remembering that there are three resistances in series and solving for the heat flow through one square ft (0.0.0929 m 2 ), Q ϭ (1800 Ϫ 200)/ {[(1 /0.82)(9/12)] ϩ [(1/ 0.125)(4.5 /12)] ϩ [(1 /052)(4 / 12)] ϩ [(1/ 0.23)(0.25 /12)]} ϭ 344 Btu / h ft 2 (1083.6 W /m 2 ), or 10 (344) ϭ 34400 Btu/h for 10 ft 2 (10,836 W/10 m 2 ). 2. Compute the temperature within the wall at the stated joint Use the relation, ( ⌬ t)/( ⌬ t 1 ) ϭ (R/R 1 ), where ⌬ t ϭ temperature difference across the wall, Њ F( Њ C); ⌬ t 1 ϭ temperature at the joint being considered, Њ F( Њ C); R ϭ total resistance of the wall; R 1 ϭ resistance of the first portion of the wall between the inside and the joint in question. Substituting, (1800 Ϫ 200)/ ( ⌬ t 1 ) ϭ 4.646/ 0.915); ⌬ t 1 ϭ 315 Њ F (157.2 Њ C). Then the interface temperature at the between the fire brick and the Sil-O-Cel is 1800 Ϫ 315 ϭ 1485 Њ F (807.2 Њ C) Related Calculations. The coefficient of thermal conductivity given here, Btu/ (ft 2 )( Њ F)(ft) is sometimes expressed in terms of per inch of thickness, instead of per foot. Either way, the conversion is simple. In SI units, this coefficient is expressed in kJ/ (m 2 )( Њ C)(m), or cm 2 and cm. The exterior temperature of a furnace wall is an important considered in boiler and process unit design from a human safety standpoint. Excess exterior tempera- tures can cause injury to plant workers. Further, the higher the exterior temperature of a furnace wall, the larger the heat loss from the fired vessel. Therefore, both safety and energy conservation considerations are important in furnace design. Typical interior furnace temperatures encountered in modern steam boilers range from 2400 Њ F (1316 Њ C) near the fuel burners to 1600 Њ F (871 Њ C) in the superheater interior. With today’s emphasis on congeneration and energy conservation, many different fuels are being burned in boilers. Thus, a plant in Louisiana burns rice to generate electricity while disposing of a process waste material. Rice hulls, which comprise 20 percent of harvested rice, are normally processed in a hammermill to increase their bulk from about 11 lb/ft 3 (176 kg/m 3 )to20 lb/ft 3 (320 kg /m 3 ). Then they are spread or piled on land adjacent to the rice mill. The hulls often smolder in the fields, like mine tailings from coal production. Con- tinuous, uncontrolled burning may result, creating an environmental hazard and problem. Burning rice hulls in a boiler furnace may create unexpected temperatures both inside and outside the furnace. Hence, it is important that the designer be able to analyze both the interior and exterior furnace temperatures using the procedure given here. Another modern application of waste usage for power generation is the burning of sludge in a heat-recovery boiler to generate electricity. Sludge from a wastewater plant is burned in a combustor to generate steam for a turbogenerator. Not only are fuel requirements for the boiler reduced, there is also significant savings of fuel used to incinerate the sludge in earlier plants. Again, the furnace temperature is an important element in designing such plants. The data present in these comments on new fuels for boilers is from Power magazine. 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.10 POWER GENERATION CONVERTING POWER-GENERATION POLLUTANTS FROM MASS TO VOLUMETRIC UNITS In the power-generation industry, emission levels of pollutants such as CO and NO x are often specified in mass units such as pounds per million Btu (kg per 1.055 MJ) and volumetric units such as ppm (parts per million) volume. Show how to relate these two measures for a gaseous fuel having this analysis: Methane ϭ 97 percent; Ethane ϭ 2 percent; Propane ϭ 1 percent by volume, and excess air ϭ 10 percent. Ambient air temperature during combustion ϭ 80 Њ F (26.7 Њ C) and relative humid- ity ϭ 60 percent; fuel higher heating value, HHV ϭ 23,759 Btu/ lb (55,358 kJ/ kg); 100 moles of fuel gas is the basis of the flue gas analysis. Calculation Procedure: 1. Find the theoretical dry air required, and the moisture in the actual air The theoretical dry air requirements, in M moles, can be computed from the sum of (ft 3 of air per ft 3 of combustible gas)(percent of combustible in the fuel) using data from Ganapathy, Steam Plant Calculations Manual, Marcel Dekker, Inc. thus: M ϭ (9.528 ϫ 97) ϩ (16.675 ϫ 2) ϩ (23.821 ϫ 1) ϭ 981.4 moles. Then, with 10 percent excess air, excess air, EA ϭ 0.1(981.4) ϭ 98.1 moles. The excess oxygen, O 2 ϭ (98.1 moles)(0.21) ϭ 20.6 moles, where 0.21 ϭ moles of oxygen in 1 mole of air. The nitrogen, N 2 , produced by combustion ϭ (1.1 for excess air)(981.4 moles)(0.79 moles of nitrogen in 1 mole of air) ϭ 852.8 moles; round to 853 moles for additional calculations. The moisture in the air ϭ (981.4 ϩ 98.1)(29 ϫ 0.0142/ 18) ϭ 24.69, say 24.7 moles. In this computation the values 29 and 18 are the molecular weights of dry air and water vapor, respectively, while 0.0142 is the lb (0.0064 kg) moisture per lb of dry air as shown in the previous procedure. 2. Compute the flue gas analysis for the combustion Using the given data, CO 2 ϭ (1 ϫ 97) ϩ (2 ϫ 2) ϩ (3 ϫ 1) ϭ 104 moles. For H 2 O ϭ (2 ϫ 97) ϩ (3 ϫ 2) ϩ (4 ϫ 1) ϩ 24.7 ϭ 228.7 moles. From step 1, N 2 ϭ 853 moles; O 2 ϭ 20.6 moles. Now, the total moles ϭ 104 ϩ 228.7 ϩ 853 ϩ 20.6 ϭ 1206.3 moles. The percent volume of CO 2 ϭ (104/ 1206.3)(100) ϭ 8.6; the percent H 2 O ϭ (228.7/ 1206.1)(100) ϭ 18.96; the percent N 2 ϭ (853/ 1206.3)(100) ϭ 70.7; the percent O 2 ϭ (20.6/ 1206.3)(100) ϭ 1.71. 3. Find the amount of flue gas produced per million Btu (1.055 MJ) To relate the pounds per million Btu (1.055 MJ) of NO x or CO produced to ppmv, we must know the amount of flue gas produced per million Btu (1.055 MJ). From step 2, the molecular weight of the flue gases ϭ [(8.68 ϫ 44) ϩ (18.96 ϫ 18) ϩ (70.7 ϫ 28) ϩ (1.71 ϫ 32)]/ 100 ϭ 27.57. The molecular weight of the fuel ϭ [(97 ϫ 16) ϩ (2 ϫ 30) ϩ (1 ϫ 44)]/ 100 ϭ 16.56. Now the ratio of flue gases/ fuel ϭ (1206.3 ϫ 27.57)/ (100 ϫ 16.56) ϭ 20.08 lb flue gas/lb fuel (9.12 kg /kg). Hence, 1 million Btu fired produces (1,000,000)/ 23,789 ϭ 42 lb (19.1 kg) fuel ϭ (42)(20.08) ϭ 844 lb (383 kg) wet flue gases. 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. STEAM GENERATION EQUIPMENT AND AUXILIARIES [...]... CO)]}(Cb ϩ S / 2.67) ϩ S / 1.60; Mƒ ϭ lb of moisture per lb fuel burned; Ma ϭ lb of moisture per lb of dry air to furnace; Cb ϭ lb of carbon burned per lb of fuel burned ϭ C ϭ RCr; Cr ϭ lb of combustible per lb of refuse; R ϭ lb of refuse per lb of fuel; H2, N2, C, O2, S ϭ lb of each element per lb of fuel, as fired; CO2, CO, O2, N2 ϭ percentage parts of volumetric analysis of dry combustion gas entering the... pound of fuel as fired Therefore, input ϭ heating value of fuel ϭ 13,850 Btu / lb (32,215 kJ / kg) 2 Compute the output of the boiler The output of any boiler ϭ Btu / lb (kJ / kg) of fuel ϩ the losses In this step the first portion of the output, Btu / lb (kJ / kg) of fuel will be computed The losses will be computed in step 3 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)... Btu / lb (557 kJ / kg) of fuel 8 Compute the loss due to unconsumed carbon in the refuse This loss is h7 Btu / lb of fuel ϭ Wc(14,150), where Wc ϭ lb of unconsumed carbon in refuse per lb of fuel fired With an ash and refuse of 9.42 percent of the dry coal and combustible in the ash and refuse of 32.3 percent, h7 ϭ (9.42 / 100) (32.3 / 100)(14,150) ϭ 430.2 Btu / lb (1006 kJ / kg) of fuel 9 Find the radiation... subject to the Terms of Use as given at the website STEAM GENERATION EQUIPMENT AND AUXILIARIES 4.12 POWER GENERATION First find Ws, lb of steam produced per lb of fuel fired Since 45,340 lb / h (20,403 kg / h) of steam is produced when 4370 lb / h (1967 kg / h) of fuel is fired, Ws ϭ 45,340 / 4370 ϭ 10.34 lb of steam per lb (4.65 kg / kg) of fuel Once Ws is known, the output h1 Btu / lb of fuel can be found... relieving capacity based on the pounds of steam generated per hour per square foot of boiler heating surface and waterwall heating surface In the edition of the Code used in preparing this handbook, the relieving requirement for oil-fired boilers was 10 lb / (ft2 ⅐ h) of steam [13.6 g / (m2 ⅐ s)] of boiler heating surface, and 16 lb / (ft2 ⅐ h) of steam [21.9 g / (m2 ⅐ s)] of waterwall surface Thus, the minimum... generated steam, Btu / lb; hƒ1 ϭ enthalpy of inlet feedwater; Sr ϭ reheated steam flow, lb / h (if any); hg3 ϭ outlet enthalpy of reheated steam; hg2 ϭ inlet enthalpy of reheated steam; B ϭ blowoff, lb / h; hƒ3 ϭ blowoff enthalpy, where all enthalpies are in Btu / lb Using the appropriate steam Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006... (0.014)(14.61) ϭ 0.2045 lb of moisture per lb (0.092 kg / kg) of air And h5 ϭ (0.46)(0.2045)(500 Ϫ 79) ϭ 39.6 Btu / lb (92.1 kJ / kg) of fuel 7 Compute the loss from incomplete combustion of C to CO2 in the stack This loss is h6 Btu / lb of fuel ϭ [CO / CO ϩ CO2)](C)(10.190), where CO and CO2 are the percent by volume of these compounds in the flue gas by Orsat analysis; C ϭ lb carbon per lb of coal With the... moisture, in terms of dry coal this is (1.61) / (100 Ϫ 1.61) ϭ 0.0164, or 1.64 percent Then h4 ϭ (1.64 / 100)(1089 Ϫ 79 ϩ 0.46 ϫ 500) ϭ 20.34 Btu / lb (47.3 kJ / kg) of fuel 6 Compute the loss from moisture in the air This loss is h5 Btu / lb of fuel ϭ 0.46Wma(Tg Ϫ Ta), Wma ϭ (lb of water per lb of dry air)(lb air supplied per lb fuel) From a psychrometric chart, the weight of moisture per lb of air at a... ft3 / min (52.1 m3 / s) In the engineering data used for this fan, the nearest capacity at 11-inH2O (2.7-kPa) static pressure is 110,467 ft3 / min (52.1 m3 / s), with an outlet velocity of 4400 ft / min (22.4 m / s), an outlet velocity pressure of 1.210 inH2O (0.30 kPa), a speed of 1222 r / min, and an input hp of 255.5 bhp (190.5 kW) The tabulation of these quantities is of the same form as that given... Find the total heating surface The total heating surface of any fire-tube boiler is the sum of the shell, tube, and head areas, or 297.0 ϩ 1583 ϩ 43.1 ϭ 1923 ft2 (178.7 m2), total heating surface 5 Compute the quantity of steam generated Since the boiler evaporates 34.5 lb / h of water per 12 ft2 [3.9 g / (m2 ⅐ s)] of heating surface, the quantity of steam generated ϭ 34.5 (total heating surface, ft2) . reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 4.2 POWER GENERATION DETERMINING. C ϭ RC r ; C r ϭ lb of combustible per lb of refuse; R ϭ lb of refuse per lb of fuel; H 2 ,N 2 ,C,O 2 ,S ϭ lb of each element per lb of fuel, as fired; CO