c) Is the project cconomically viable? Problem 4 25 points An area can be izmiga

Question

c) Is the project cconomically viable? Problem 4 25 points An area can be izmigated by pumping water from a nearby river. Two competing installations are being considered The MARR is 12% pd year At what level of operation (hours per year) would you be indifferent between the two pumping systems. The table below gives the different estimates needed to do the calculations Pump A 6730 75% 45 12 0.08Acwh Initial cost, Electrical efficiency 3522 175 0.08kwh 10 load on motor18 hp Electric power costs, S lhp-0.746 kW Years of usags,n 10 a) At what level of operation (hours per year) would you be indifferent between the two pumping systems b) Which pump will you sclect if the desire hours operation per year is 670 hours

Explanation / Answer

The level of operation at which the two pumping systems can be indifferent based on equivalent-worth equation (EUAC) Note= EAC = equivalent annual cost O&M cost = operating and maintenance cost P = asset initial cost S = asset salvage value end of life EAC for pump capital cost = P(A/P, i %, n) - S(A/F, i %, n) + Other Costs Initial Cost P(A/P,i%,N) + Annual energy cost - Salvage value (A/F,i%,N) The cost of the electric power for the pump A and B $0.08 per kWh. The annual energy cost as per the calculation : Annual Energy Cost= (Operating hours/year) ($0.08/kWh) ((operating load on motor) (0.746kW/hp)/ Motor efficiency) Assumed the operating hours per year fpr pump A and pump B = X The Annual energy cost of pump A= X($0.08/kwh) ((18hp)* (0.746kw/hp)/0.60))= $1.7904X The Annual energy cost of pump A= X($0.08/kwh) ((12hp)* (0.746kw/hp)/0.75))= $0.9549X Now, the calculation of EAUC with both the values of Annual Energy Cost: EUAC of Pump A (12%)= $3522(A/P,12%,10) + $1.7904X - $175(A/F,12%,10) =$3522*(0.1769)+$1.7904X-$175(0.0569) =$623.04+$1.7904X-$9.9575 = $613.08+1.7904X EUAC of Pump B (12%)= $6730(A/P,12%,10) + $0.9549X - $45(A/F,12%,10) =$6730*(0.1769)+$0.9549X-$45(0.0569) =$1190.54+$0.9549X-$2.5605= $1187.98+0.9549X Equation between EAUC of Pump A and EAUC of Pump B : To find out the level of operation so that one is indiffrent between the two pumping system $613.08+1.7904X = $1187.98+0.9549X 1.7904X-0.9549X=$1187.98-$613.08 0.8355X=574.90 X=688 a) the level of operation so that one is indiffrent between the two pumping system is 688hours per year. b) If Pumping system is expected to operate 670hours per year, I would be recommend for PUMP A as per the minimize EUAC. EAUC of Pump A if operating hours is 670 hours per year X= 670 $613.08+1.7904*670= $1812.65 EAUC of Pump B if operating hours is 670 hours per year X= 670 $1187.98+0.9549*670= $1827.76 Since the EUAC of PUMP A is lesser than the PUMP B.

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