A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a highefficiency motor that has an efficiency of 95.4 percent. The motor operates 4368 hours a year at a load factor of 0.75. Taking the cost of electricity to be $0.12/kWh, determine the amount of energy and money saved as a result of installing the highefficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5449 and $5520, respectively.

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MrRoyal MrRoyal · Answer 1 · 2020-03-17T04:01:06+01:00

Answer:

1. Energy Savings = 9286.01Kwh/yr

2. Simple Payback Period = 0.0637 yr or 23 days

Explanation:

For we to calculate the energy savings, first we need to calculate the power savings.

Using

η, old = W,mech out ÷ W, elec in old

η, new = W,mech out ÷ W, elec in new

W, elec in old = W,mech out ÷ η,old

Where W, mech out = Shaft Power * Load Factor

W, elec in new = W,mech new ÷ η,new

Where W, mech in = Shaft Power * Load Factor

Power Savings = W,elec in old - W,elec in new

Power Savings = (Shaft Power)(Load Factor) * (1/η,old - 1/η,new)

Power Savings = 75hp * 0.75(1/0.91 - 1/0.954) ---- 1hp = 745.7 W

Power Savings = 75 * 745.7 * 0.75(1/0.91 - 1/0.954)

Power Savings = 2125.933029234916

Power Savings = 2125.93W ---. Approximated

Calculating Energy Savings

Energy Savings = Power Savings * Yearly Work Hours

Energy Savings = 2125.93 * 4368 hr/yr

Energy Savings = 9,286,062.24Whr/yr

Energy Savings = 9286.01Kwh/yr

2. Calculating the simple payback period.

First, we calculate the cost savings

Cost Savings = Energy Savings * Unit Cost of Electricity

Cost Savings = 9286.01 * 0.12

Cost Savings = $1,114.3212/yr

Simple Payback Period = Price Difference of Motors / Cost Savings

Simple Payback Period = (5520 - 5449)/1114.3212

Simple Payback Period = 0.063715919610970

Simple Payback Period = 0.0637 yr or 23 days