How To Evaluate Potential Savings From Retrofitting Energy Efficient Motors In Existing Systems (2004)

By: Henry Manczyk, CPE, CEM

Manczyk Energy Consulting

A large portion of the electric energy consumption is attributed to electric motors. It is highly recommended that the existing, lower efficiency motors identified in a building survey be upgraded from standard to premium energy-efficient motors. Those that meet the minimum efficiency requirements set by the New York State Energy Research and Development Authority (NYSERDA) are eligible for incentive rebates offered to commercial, institutional, and industrial customers.

Energy-efficient motors pay for themselves in a few years, or sometimes even a few months, after which they will continue to accumulate savings valued many times their purchase cost for as long as they remain in service. That is another way of saying that operating costs, not just initial costs, are what we should look at when purchasing a new motor.

Electric motors are simply devices that convert electrical energy into mechanical energy. Like all electro-mechanical equipment, motors consume some extra energy in order to make the conversion. Efficiency is a measure of how much total energy a motor uses in relation to the rated power delivered to the shaft. There are several techniques to express motor efficiency, but the basic concept and the numerical results are the same. For example:

Efficiency = (746) (horsepower output) (100) / watts input

or = (watts output) (100) / watts input

Let's use an example with one motor that was selected for replacement in a typical local commercial office building:

Here are the existing motor versus new premium motor characteristics:

Existing New

Motor Hp = 15 Motor Hp = 15

Speed = 1,800 rpm Speed = 1,800

Efficiency = 87.5% Efficiency = 93%

Load = 75% Load = 75%

Voltage = 440 Volts Voltage = 440 Volts

Operating hours = 8,760 Operating hours = 8,760

· Cost per KWh at this particular office building is $0.075

· Cost per KW demand at this particular office building is $8.04

Total energy savings arise from reduction in power demand (KW) and energy consumption (KWh).

· KW saved = (Motor Hp)(Load)(0.746)[(1/Eff. Existing Motor) – (1/Eff. New Motor)] =

(15) (75) (0.746) [(1/87.5) – (1/93)] = 0.567 KW

· KWh savings = (Demand Savings)(Operating Hours) =

(0.567) (8,760) = 4,968.98 KWh

· Annual energy savings =

(Demand Savings)(12 months)(Demand Cost) + (Consumption Savings)(Cost per KWh)

= (0.567) (12) (8.04) + (4,968.98) (0.075) = $427.37 per year

Divide the motor replacement cost by the annual energy savings to determine the simple payback period.

Additional advantages of installing new, premium-efficiency motors are:

· Reduced operating cost

· Less heat loss (generated by the resistance)

· Extended winding lifespan

· Extended lubricating grease service

· Reduced noise levels

· Positive environment effects

For information about NYSERDA’s “Smart Equipment Choices” programs, visit their website at www.nyserda.org