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cooling tower designs that could be considered. Assuming that the cost of capital to the utility company is 12% per year, your job is to recommend the best alternative (ie, the least expensive during the servic life) Further, assume that each alternative is capable of satisfactorily removing waste heat from the condensers of a 200-MW power plant. What noneconomic factors can you identify that might also play a role the decision-making process? Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when /= 12% per year. The AW of Wet Tower, Mechanical Draft is $ (Round to the nearest dollar) Data table Alternative Initial cost Power for I.D. fans Power for pumps CI Dry Tower Mech. Draft Wet Tower Mech, Draft Wet Tower Natural Draft $8.8 million None $2.8 million. 40 200-hp I.D. fans 20 150-hp pumps $0.15 million Mechanical maintenance/year 20 150-hp pumps $0.13 million 30 years 0 30 years Service life 30 years 0 Market value 0 100 hp 74.6 kW, cost of power to plant is 2 2 cents per kWh or kilowatt-hour, induced-draft fans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. Dry Tower Natural Draft $4.8 million 20 200-hp I.D. fans. 40 100-hp pumps $0.17 million 30 years 0 $8.9 million None 40 100-hp pumps $0.13 million

cooling tower designs that could be considered. Assuming that the cost of capital to the utility company is 12% per year, your job is to recommend the best alternative (ie, the least expensive during the servic life) Further, assume that each alternative is capable of satisfactorily removing waste heat from the condensers of a 200-MW power plant. What noneconomic factors can you identify that might also play a role the decision-making process? Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when /= 12% per year. The AW of Wet Tower, Mechanical Draft is $ (Round to the nearest dollar) Data table Alternative Initial cost Power for I.D. fans Power for pumps CI Dry Tower Mech. Draft Wet Tower Mech, Draft Wet Tower Natural Draft $8.8 million None $2.8 million. 40 200-hp I.D. fans 20 150-hp pumps $0.15 million Mechanical maintenance/year 20 150-hp pumps $0.13 million 30 years 0 30 years Service life 30 years 0 Market value 0 100 hp 74.6 kW, cost of power to plant is 2 2 cents per kWh or kilowatt-hour, induced-draft fans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. Dry Tower Natural Draft $4.8 million 20 200-hp I.D. fans. 40 100-hp pumps $0.17 million 30 years 0 $8.9 million None 40 100-hp pumps $0.13 million

ENGR.ECONOMIC ANALYSIS
ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN:9780190931919
Author:NEWNAN
Publisher:NEWNAN
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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You have been asked to evaluate the economic implications of various methods for cooling condenser effluents from a 200-MW steamelectric plant. There are two basic types of cooling towers: wet and dry. Furthermore, heat may be removed from condenser water by (1) forcing (mechanically) air through the tower or (2) allowing heat transfer to occur by making use of natural draft. Consequently, there are four basic cooling tower designs that could be considered. Assuming that the cost of capital to the utility company is 12% per year, your job is to recommend the best alternative (i.e., the least expensive during the service life) Further, assume that each alternative is capable of satisfactorily removing waste heat from the condensers of a 200-MW power plant. What noneconomic factors can you identify that might also play a role in the decision-making process? Fir Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when i = 12% per year. The AW of Wet Tower, Mechanical Draft is $ (Round to the nearest dollar.) Data table Wet Tower Mech. Draft $2.8 million 40 200-hp I.D. fans 20 150-hp pumps $0.15 million Initial cost Power for I.D. fans. Power for pumps Alternative 30 years 0 Wet Tower Natural Draft $8.8 million None 20 150-hp pumps $0.13 million B Dry Tower Mech, Draft $4.8 million 20 200-hp I.D. fans 40 100-hp pumps $0.17 million 30 years 0 $8.9 million None Mechanical maintenance/year 40 100-hp pumps $0.13 million 30 years 0 Service life Market value 30 years 0 100 hp = 74.6 kW; cost of power to plant is 2.2 cents per kWh or kilowatt-hour, induced-draft fans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. Dry Tower Natural Draft X Clear all gla Check answer
Transcribed Image Text:You have been asked to evaluate the economic implications of various methods for cooling condenser effluents from a 200-MW steamelectric plant. There are two basic types of cooling towers: wet and dry. Furthermore, heat may be removed from condenser water by (1) forcing (mechanically) air through the tower or (2) allowing heat transfer to occur by making use of natural draft. Consequently, there are four basic cooling tower designs that could be considered. Assuming that the cost of capital to the utility company is 12% per year, your job is to recommend the best alternative (i.e., the least expensive during the service life) Further, assume that each alternative is capable of satisfactorily removing waste heat from the condensers of a 200-MW power plant. What noneconomic factors can you identify that might also play a role in the decision-making process? Fir Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when i = 12% per year. The AW of Wet Tower, Mechanical Draft is $ (Round to the nearest dollar.) Data table Wet Tower Mech. Draft $2.8 million 40 200-hp I.D. fans 20 150-hp pumps $0.15 million Initial cost Power for I.D. fans. Power for pumps Alternative 30 years 0 Wet Tower Natural Draft $8.8 million None 20 150-hp pumps $0.13 million B Dry Tower Mech, Draft $4.8 million 20 200-hp I.D. fans 40 100-hp pumps $0.17 million 30 years 0 $8.9 million None Mechanical maintenance/year 40 100-hp pumps $0.13 million 30 years 0 Service life Market value 30 years 0 100 hp = 74.6 kW; cost of power to plant is 2.2 cents per kWh or kilowatt-hour, induced-draft fans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. Dry Tower Natural Draft X Clear all gla Check answer
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 35 40 45 50 Single Payment Compound Amount Factor To Find F Given P F/P 1.1200 1.2544 1.4049 1.5735 1.7623 1.9738 2.2107 2.4760 2.7731 3.1058 3.4785 3.8960 4.3635 4.8871 5.4736 6.1304 6.8660 7.6900 8.6128 9.6463 10.8038 12.1003 13.5523 15.1786 17.0001 29.9599 52.7996 93.0510 Discrete Compounding; /=12% Compound Amount Factor Present Worth Factor To Find P Given F P/F 163.9876 289.0022 0.8929 0.7972 0.7118 0.6355 0.5674 0.5066 0.4523 0.4039 0.3606 0.3220 0.2875 0.2567 0.2292 0,2046 0.1827 0.1631 0.1456 0.1300 0.1161 0.1037 0.0926 0.0826 0.0738 0.0659 0.0588 0.0334 0.0189 0.0107 0.0061 0.0035 To Find F Given A FIA 1.0000 2.1200 3.3744 4.7793 6.3528 8.1152 10.0890 12.2997 14.7757 17.5487 20.6546 24.1331 28.0291 32.3926 37.2797 42.7533 48.8837 55.7497 63.4397 72.0524 81.6987 92.5026 104.6029 118.1552 133.3339 241.3327 431.6635 767.0914 1358.2300 2400.0182 Uniform Series Present Worth Factor To Find P Given A P/A 0.8929 1.6901 2.4018 3.0373 3.6048 4.1114 4.5638 4.9676 5.3282 5.6502 5.9377 6.1944 6.4235 6.6282 6.8109 6.9740 7.1196 7.2497 7.3658 7.4694 7.5620 7.6446 7.7184 7.7843 7.8431 8.0552 8.1755 8.2438 8.2825 8.3045 Sinking Fund Factor To Find A Given F A/F 1.0000 0.4717 0.2963 0.2092 0.1574 0.1232 0.0991 0.0813 0.0677 0.0570 0.0484 0.0414 0.0357 0.0309 0.0268 0.0234 0.0205 0:0179 0.0158 0.0139 0.0122 0.0108 0.0096 0.0085 0.0075 0.0041 0.0023 0.0013 0.0007 0.0004 Capital Recovery Factor To Find A Given P A/P 1.1200 0.5917 0.4163 0.3292 0.2774 0.2432 0.2191 0.2013 0.1877 0.1770 0.1684 0.1614 0.1557 0.1509 0.1468 0.1434 0.1405 0.1379 0.1358 0.1339 0.1322 0.1308 0.1296 0.1285 0.1275 0.1241 0.1223 0.1213 0.1207 0.1204
Transcribed Image Text:N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 35 40 45 50 Single Payment Compound Amount Factor To Find F Given P F/P 1.1200 1.2544 1.4049 1.5735 1.7623 1.9738 2.2107 2.4760 2.7731 3.1058 3.4785 3.8960 4.3635 4.8871 5.4736 6.1304 6.8660 7.6900 8.6128 9.6463 10.8038 12.1003 13.5523 15.1786 17.0001 29.9599 52.7996 93.0510 Discrete Compounding; /=12% Compound Amount Factor Present Worth Factor To Find P Given F P/F 163.9876 289.0022 0.8929 0.7972 0.7118 0.6355 0.5674 0.5066 0.4523 0.4039 0.3606 0.3220 0.2875 0.2567 0.2292 0,2046 0.1827 0.1631 0.1456 0.1300 0.1161 0.1037 0.0926 0.0826 0.0738 0.0659 0.0588 0.0334 0.0189 0.0107 0.0061 0.0035 To Find F Given A FIA 1.0000 2.1200 3.3744 4.7793 6.3528 8.1152 10.0890 12.2997 14.7757 17.5487 20.6546 24.1331 28.0291 32.3926 37.2797 42.7533 48.8837 55.7497 63.4397 72.0524 81.6987 92.5026 104.6029 118.1552 133.3339 241.3327 431.6635 767.0914 1358.2300 2400.0182 Uniform Series Present Worth Factor To Find P Given A P/A 0.8929 1.6901 2.4018 3.0373 3.6048 4.1114 4.5638 4.9676 5.3282 5.6502 5.9377 6.1944 6.4235 6.6282 6.8109 6.9740 7.1196 7.2497 7.3658 7.4694 7.5620 7.6446 7.7184 7.7843 7.8431 8.0552 8.1755 8.2438 8.2825 8.3045 Sinking Fund Factor To Find A Given F A/F 1.0000 0.4717 0.2963 0.2092 0.1574 0.1232 0.0991 0.0813 0.0677 0.0570 0.0484 0.0414 0.0357 0.0309 0.0268 0.0234 0.0205 0:0179 0.0158 0.0139 0.0122 0.0108 0.0096 0.0085 0.0075 0.0041 0.0023 0.0013 0.0007 0.0004 Capital Recovery Factor To Find A Given P A/P 1.1200 0.5917 0.4163 0.3292 0.2774 0.2432 0.2191 0.2013 0.1877 0.1770 0.1684 0.1614 0.1557 0.1509 0.1468 0.1434 0.1405 0.1379 0.1358 0.1339 0.1322 0.1308 0.1296 0.1285 0.1275 0.1241 0.1223 0.1213 0.1207 0.1204
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