You have been asked to evaluate the economic implications of various methods for

Question

You have been asked to evaluate the economic implications of various methods for cooling condenser effluents from a 200-MW ...

a. The AW of Wet Tower, Mechanical Draft is $

b. The AW of Wet Tower, Natural Draft is $

c. The AW of Dry Tower Mech. Draft is $

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 (.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? Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when 1-12% per year. of Wet Tower, Mechanical Draft is s (Round to the nearest dola) (Round to the nearest dollar.) Data Table Alternative Wet Tower Mech. Draft Wet Tower Natural Draft Dry Tower Natural Draft Dry Tower Mech. Draft Initial cost Power for I.D. fans Power for pumps Mechanical maintenance/year Service life Market value $3.1 million 40 200-hp I.D. fans 20 150-hp pumps $0.13 million 30 years 0 $9.1 million None 20 150-hp pumps $0.09 million 30 years 0 $5.1 million 20 200-hp I.D. fans 40 100-hp pumps $0.14 million 30 years 0 $9.1 million None 40 100-hp pumps $0.13 million 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. PrintDone

Explanation / Answer

Annual Worth Analysis (AW) is an important parameter of selection of project is function of {P, SV, i%, and n }  

AW = - P(A|P, i, n) + SV(A|F, i, n) - A

where , P = intial investment , SV = salvage value , i = rate of interest , n =time , A = annual operation cost.

Here, Salvage value = 0 for all 4 options

so,  AW = - P(A|P, i, n) - A

at cost of capital 12% for 30years , (A|P, 12, 30) = 0.1241

efficiency 90% = 0.90

Cost of power($) = (Hp * .746kwh * 365 * 24 hr * 2.2 cents) * 0.9 / 100  

AW ofwet tower mech draft = -7.56 million

100 hp 74.6 Kwh Operation =365*24 =8760 hrs Cost 1kwh 2.2 cents

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