Question

Problem 6-28 (algorithmic) Festione Two 180 horsepower (HP) motors are being considered for installation at a municipal sewage r ent plant The 600 and has an operating ency of the second costs 5 000 and has an ency of 77% Both motors are projected to have zero salvage value therae of 10 years the charges, such as insurance manance et amount to a total of 0 of the cost of each mal and powers are fatte cents per kilowatthout how many minimum hours of load operation per year w a y to purchasing expensive for a cono cer you might find 1 75 7 Click the icon to view the interest factors for discrete compounding when per year A minimum of hours of load operation per year are necessary to justify purchasing the more expensive motor Round up to the nearest whole number

Answer 1

Let X be no. of hours of operation

EUAC of first motor = 6600 * (A/P,14%,10) + 0.1 * 6600 + 180 * 0.746 * 0.05 / 0.81 * X

= 6600 * 0.1917 + 0.1 * 6600 + 8.288889 * X

= 1925.22 + 8.288889 * X

EUAC of second motor = 5000 * (A/P,14%,10) + 0.1 * 5000 + 180 * 0.746 * 0.05 / 0.77 * X

= 1458.50 + 8.719480 * X

As per given condition

1925.22 + 8.288889 * X = 1458.50 + 8.719480 * X

1925.22 - 1458.50 = 8.719480 * X - 8.288889 * X

X = (1925.22 - 1458.50) / (8.719480 - 8.288889) = 1083.90 ~ 1084 hours