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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 15% 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? 1 UU hp=14.6kVV; cost ot power to plant IS 2.2 cents per kVVh or kIlowatt-hour; Induced-dratt tans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|}{ Discrete Compounding; i=15%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } \\ \hline & Compound & & Compound & & Sinking & Capital \\ \hline & \begin{tabular}{c} Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{c} Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Fund \\ Factor \end{tabular} & \begin{tabular}{c} Recovery \\ Factor \end{tabular} \\ \hline & To Find F & To Find P & To Find F & To Find P & To Find A & To Find A \\ \hline & Given P & Given F & Given A & Given A & Given F & Given P \\ \hlineN & F/P & P/F & F/A & P/A & A/F & A/P \\ \hline 1 & 1.1500 & 0.8696 & 1.0000 & 0.8696 & 1.0000 & 1.1500 \\ \hline 2 & 1.3225 & 0.7561 & 2.1500 & 1.6257 & 0.4651 & 0.6151 \\ \hline 3 & 1.5209 & 0.6575 & 3.4725 & 2.2832 & 0.2880 & 0.4380 \\ \hline 4 & 1.7490 & 0.5718 & 4.9934 & 2.8550 & 0.2003 & 0.3503 \\ \hline 5 & 2.0114 & 0.4972 & 6.7424 & 3.3522 & 0.1483 & 0.2983 \\ \hline 6 & 2.3131 & 0.4323 & 8.7537 & 3.7845 & 0.1142 & 0.2642 \\ \hline 7 & 2.6600 & 0.3759 & 11.0668 & 4.1604 & 0.0904 & 0.2404 \\ \hline 8 & 3.0590 & 0.3269 & 13.7268 & 4.4873 & 0.0729 & 0.2229 \\ \hline 9 & 3.5179 & 0.2843 & 16.7858 & 4.7716 & 0.0596 & 0.2096 \\ \hline 10 & 4.0456 & 0.2472 & 20.3037 & 5.0188 & 0.0493 & 0.1993 \\ \hline 11 & 4.6524 & 0.2149 & 24.3493 & 5.2337 & 0.0411 & 0.1911 \\ \hline 12 & 5.3503 & 0.1869 & 29.0017 & 5.4206 & 0.0345 & 0.1845 \\ \hline 13 & 6.1528 & 0.1625 & 34.3519 & 5.5831 & 0.0291 & 0.1791 \\ \hline 14 & 7.0757 & 0.1413 & 40.5047 & 5.7245 & 0.0247 & 0.1747 \\ \hline 15 & 8.1371 & 0.1229 & 47.5804 & 5.8474 & 0.0210 & 0.1710 \\ \hline 16 & a 3576 & 16a & 557175 & 50512 & 170 & 167a \\ \hline \end{tabular} The AW of Wet Tower, Natural Draft is $ (Round to the nearest dollar.) The AW of Dry Tower, Mechanical Draft is $. (Round to the nearest dollar.) The AW of Dry Tower, Natural Draft is $ (Round to the nearest dollar.) What is the best alternative? Choose the correct answer below. A. Dry Tower, Natural Draft B. Dry Tower, Mechanical Draft C. Wet Tower, Natural Draft D. Wet Tower, Mechanical Draft 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 15% 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? 1 UU hp=14.6kVV; cost ot power to plant IS 2.2 cents per kVVh or kIlowatt-hour; Induced-dratt tans and pumps operate around the clock for 365 days/year (continuously). Assume that electric motors for pumps and fans are 90% efficient. \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|}{ Discrete Compounding; i=15%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } \\ \hline & Compound & & Compound & & Sinking & Capital \\ \hline & \begin{tabular}{c} Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{c} Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Fund \\ Factor \end{tabular} & \begin{tabular}{c} Recovery \\ Factor \end{tabular} \\ \hline & To Find F & To Find P & To Find F & To Find P & To Find A & To Find A \\ \hline & Given P & Given F & Given A & Given A & Given F & Given P \\ \hlineN & F/P & P/F & F/A & P/A & A/F & A/P \\ \hline 1 & 1.1500 & 0.8696 & 1.0000 & 0.8696 & 1.0000 & 1.1500 \\ \hline 2 & 1.3225 & 0.7561 & 2.1500 & 1.6257 & 0.4651 & 0.6151 \\ \hline 3 & 1.5209 & 0.6575 & 3.4725 & 2.2832 & 0.2880 & 0.4380 \\ \hline 4 & 1.7490 & 0.5718 & 4.9934 & 2.8550 & 0.2003 & 0.3503 \\ \hline 5 & 2.0114 & 0.4972 & 6.7424 & 3.3522 & 0.1483 & 0.2983 \\ \hline 6 & 2.3131 & 0.4323 & 8.7537 & 3.7845 & 0.1142 & 0.2642 \\ \hline 7 & 2.6600 & 0.3759 & 11.0668 & 4.1604 & 0.0904 & 0.2404 \\ \hline 8 & 3.0590 & 0.3269 & 13.7268 & 4.4873 & 0.0729 & 0.2229 \\ \hline 9 & 3.5179 & 0.2843 & 16.7858 & 4.7716 & 0.0596 & 0.2096 \\ \hline 10 & 4.0456 & 0.2472 & 20.3037 & 5.0188 & 0.0493 & 0.1993 \\ \hline 11 & 4.6524 & 0.2149 & 24.3493 & 5.2337 & 0.0411 & 0.1911 \\ \hline 12 & 5.3503 & 0.1869 & 29.0017 & 5.4206 & 0.0345 & 0.1845 \\ \hline 13 & 6.1528 & 0.1625 & 34.3519 & 5.5831 & 0.0291 & 0.1791 \\ \hline 14 & 7.0757 & 0.1413 & 40.5047 & 5.7245 & 0.0247 & 0.1747 \\ \hline 15 & 8.1371 & 0.1229 & 47.5804 & 5.8474 & 0.0210 & 0.1710 \\ \hline 16 & a 3576 & 16a & 557175 & 50512 & 170 & 167a \\ \hline \end{tabular} The AW of Wet Tower, Natural Draft is $ (Round to the nearest dollar.) The AW of Dry Tower, Mechanical Draft is $. (Round to the nearest dollar.) The AW of Dry Tower, Natural Draft is $ (Round to the nearest dollar.) What is the best alternative? Choose the correct answer below. A. Dry Tower, Natural Draft B. Dry Tower, Mechanical Draft C. Wet Tower, Natural Draft D. Wet Tower, Mechanical Draft