Question

A Company uses a 10-hp motor for 16 hours per day, 5 days per week, 50 weeks per year in its flexible work cell. This motor is 85% efficient, and it is near the end of its useful life. The company is considering buying a new high efficiency motor (91% efficient) to replace the old one instead of buying a standard efficiency motor (86.4% efficient). The high efficiency motor cost $70 more than the standard model, and should have a 14 year life. The company pays $7 per kW per month and $0.06 per kWh. The company has set a discount rate of 10% for their use in comparing projects.
Determine;
a) SPP ( simple Payback Period),
b) IRR (Internal Rate of Return), and
c) BCR (benefit/cost ratio) for this project
Assume 60% load factor the motor.

Answer 1

Given,Assume the load factor (If) is 60%.

Now,

where,

DR = Demand reduction

Pm = Power rating of the motor,10-hp

effs = Efficiency of the standard efficiency motor, 86.4%

effh = Efficiency of the high efficiency motor, 91%

Therefore,

Now,

where,

DCR=Demand cost reduction

DC = Demand cost,$7/kW/mo

therefore,

Now,

where,

ECS=Energy cost savings

EC=Energy cost,$0.06 / kWh  

Therefore ,the annual cost savings (ACS) can be calculated as follows:

SPP = 0.825 yrs Answer.

Additionally,the ROI can be found with using the following factor in inerest rate tables:

% or IRR=121.2% Answer.

=$85.85/yr[P|A,10%,14yr]

=$84.85 * 7.3667

=$625

BCR=$625/$70

BCR=8.93 Answer.