Question

Two pumps are being considered for service in Africa. Each can draw water from a depth of 39 meters and both pumps deliver 60 horsepower​ (output) to the pumping operation. One of the pumps must be selected. It is expected that the pump will be in use 600 hours per year. The following data are available. LOADING... Click the icon to view the alternatives description.

If electricity costs ​$0.16 per​ kilowatt-hour, which pump should be selected if the MARR is ​7% per​ year? Recall that 1 hp=0.746 kW. What is the basic tradeoff being made in this​ problem?

LOADING... Click the icon to view the interest and annuity table for discrete compounding when the MARR is ​% per year. The AW of Pump A is ​$ nothing. ​(Round to the nearest​ dollar.)

                                                                                                                                                                                                                                                                                                                                                                                        

	Pump A	Pump B
Capital investment	​$2,000	​$1,500
Electrical efficiency	0.93	0.88
Annual maintenance cost	​$340	​$190
Useful life	8 years	4 years

Answer 1

R = 7%

Annual operation cost for pump A = 60*.746*600*.16/.93

Annual operation cost for pump A= $4620.39

So,

AW of pump A = annualized initial investment + annual operations cost + annual maintenance cost

AW of pump A = 2000*(A/P, 7%, 8) + 4620.39 + 340

AW of pump A = 2000*.1675 + 4620.39 + 340

AW of pump A = 5295.39  ( it is an annual total cost)

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Annual operation cost for pump B = 60*.746*600*.16/.88

Annual operation cost for pump B = $4882.91

AW of pump B = annualized initial investment + annual operations cost + annual maintenance cost

AW of pump B = 1500*(A/P, 7%, 4) + 4882.91+ 190

AW of pump B = 1500*.2952 + 4882.91+ 190

AW of pump B = 5515.71   ( it is an annual total cost)

So, pump A will be selected, because it is more cost effective than that of pump B.

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The basic tradeoff is the difference in efficiency of the motor. A higher efficiency, causes lower annual operations cost that makes pump to be more cost effective in nature. Other than this, lower annualized initial cost is also responsible for the lower annual cost for the pump A.