Investing in Electrical Efficiency Two pumps capable of delivering 100 hp to an agricultural application are being evaluated in a present economy study. The selected pump will only be utilized for one year, and it will have no market value at the end of the year. Pertinent data are summarized as

Question

Investing in Electrical Efficiency
Two pumps capable of delivering 100 hp to an agricultural application are being evaluated in a present economy study. The selected pump will only be utilized for one year, and it will have no market value at the end of the year. Pertinent data are summarized as follows:

XYZ Pump	ABC Pump
$6,200	$2,900	Purchase price
$510	$170	Maintenance cost
90%	80%	Efficiency

If electric power costs $0.10 per kWh and the pump will be operated 4,000 hours per year, which pump should be chosen? Recall that 1 hp = 0.746 kW

Accepted Answer

The expense of electric power for the ABC pump is
(100 hp/0.80)(0.746 kW/hp)($0.10/kWh)(4,000 hours/yr) = $37,300.
For the XYZ Pump, the expense of electric power is
(100 hp/0.90)(0.746 kW/hp)($0.10/kWh)(4,000 hours/yr) = $33,156.

Thus, the total cost of owning and operating the ABC pump is $40,370, while the total cost of owning and operating the XYZ pump for one year is $39,866. Consequently, the more energy-efficient XYZ pump should be selected to minimize total cost. Notice the difference in energy expense ($4,144) that results from a 90% efficient pump relative to an 80% efficient pump. This cost reduction more than balances the extra $3,300 in capital investment and $340 in maintenance required for the XYZ pump.

Question 2-11: Investing in Electrical Efficiency Two pumps capable of deli...

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