Two 150-horsepower (HP) motors are being considered for installation at a munici

Question

Two 150-horsepower (HP) motors are being considered for installation at a municipal sewage-treatment plant. The first costs $4,500 and has an operating efficiency of 83%. The second costs $3,600 and has an efficiency of 80%. Both motors are projected to have zero salvage value after a life of 10 years. If all the annual charges, such as insurance, maintenance, etc., amount to a total of 15% of the original cost of each motor, and if power costs are a flat 5 cents per kilowatthour, how many minimum hours of full-load operation per year are necessary to justify purchasing the more expensive motor at i = 6%? (A conversion factor you might find useful is 1 HP = 746 watts = 0.746 kilowatts.)

Explanation / Answer

a)Let "a" be no of minimum hours to be operated

Horse power to KW conversion :

=150*.746=111.9 KW

calculate the annual payment to be made if we buy $4500 at 6% for 10 years

=pmt(rate,nper,pv,fv,type)

=pmt(6%,10,4500,,0)=$611.41

annual Manintenace cost=(.15*4500)+((111.9*a)/.83)*.05 (Here divideb ny .83 since it is 83% efficient)

=6.741a+675

b)Simiarly in second case

calculate the annual payment to be made if we buy $3600 at 6% for 10 years

=pmt(rate,nper,pv,fv,type)

=pmt(6%,10,3600,,0)=$489.12

annual Manintenace cost=(.15*3600)+((111.9*a)/.8)*.05 (Here divideb ny .83 since it is 83% efficient)

=6.994a+540

Equating total annual cost in both cases

611.41+6.741a+675=489.12+6.9938a+540

a=1017.8 hours per year

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