Motor Efficiency

The first important concept we need to keep in mind is "Small differences in motor efficiency can have a large effect when motor operates many hours". Let's look at some examples.

Example #1

A plant's ventilation system used 10, 40-hp motors on a 24/7 basis except for a 2-week shut down. The current motors are 85% efficient at rated/full load. They are considering updating the motors with 93%-efficiency motors at $2000 each. They pay 6 cents/kWh for electricity.  Assume load has been close to rated/full load.  What is the Payback and ROI?


Overall approach: We need to figure out the annual electricity cost of current and future motors. The difference would be the annual saving, then payback and return on investment calculation would be straightforward.

How to find out the electricity cost of motors? Two steps:

Step 1 – power demand



kW - is the power demand of the motor in kW

hp - is the power of the motor in horsepower (hp)

Eff = rated efficiency (e.g. 0.85)

LF = load factor – assume 1 when load is close to rated/full load

Step 2 – power consumption cost

kW = power demand (from step 1)

hrs - annual operating hours

$/kWh - electricity cost.

Now applying these two steps in Example 1,

we find for current motors:

Since there are 10 of such motors, total power demand is 10 x 35.12 kW = 351.2 kW

Using the same approach, for proposed motors:

Since there are 10 of such motors, total power demand is 10 x 32.1 kW = 321 kW


Annual savings = $177005 - $161784 = $15221

Capital cost = $2,000 x 10 = $20,000

Payback (yrs) = capital cost/annual savings = $20,000/$15221= 1.31 years

Return on Investment (ROI) = 1/payback = 1/1.31 = 76.33%

Example #1 continued

How does the above analysis change if the current motors are near the end of their lives and need replacing anyway? Assume replacement cost is $1,500/motor.

In this case, current motors need replacing at replacement cost of $1,500/motor. If we take this opportunity and spend additional $500/motor, we can upgrade to a 93% efficiency motor. Then:

Annual savings remains the same as above which is $177005 - $161784 = $15221

Capital cost however is only $500 x 10 = $5,000 since these motors need to be replaced anyway, only the additional cost of $500/motor should be considered capital investment for the better efficiency motors

Payback (yrs) = capital cost/annual savings = $5,000/$15221= 0.33

Return on Investment (ROI) = 1/payback = 1/0.33 = 303%

From the above example, one can see that the best time to upgrade the motors to higher efficiency is the time when current motors need to be replaced. Payback and ROI is just fabulous!

Now, let's look at another example.

Example #2

If the standard efficiency (85%) replacement motors are expected to last 10 years, what percent of their 10-year cost will be cost of motor? What percent will be the cost of the electricity?

The 10-year cost of a motor should be the cost of purchasing the motor and 10 year electricity cost of running the motor, so for a standard efficiency (85%) motor, based on analysis above,

power demand = 35.12 kW

!0 year electricity = $17700.5 x 10 = $177005

!0-year cost = purchasing cost + electricity cost (10 year) = $1,500 + $177005 = $178505

Percent of the 10-year cost that is cost of motor = $1500/178505 =0.84%

Percent of the 10-year cost that is cost of the electricity = $177005 /178505 = 99.16%

From the above analysis, you can see that initial purchasing cost of the motor is only a tiny fraction of the 10-year cost of the motor. It is true that $1,500 - $2,000/motor is a significant cost to any company, but we need to keep this in perspective when comparing to the cost running the motor for 24/7 for 10 years.

Example #3

A grinding operation used a central coolant system to pump coolant up to 50 grinding machines on the floor above. This requires 6, 60-hp motors (85% efficiency), operating 24/7 for 50 weeks/yr. Electricity costs 5 cents/kWh. They spend $100k/yr on coolant. What do they spend on electricity to pump it?

Since there are a total of 6 motors, so total power demand = 6 x 52.68 kW = 316.08 kW

As you can see, coolant is an expensive supply for this operation at $100,000/yr, but the electricity used to pump it cost even more!

Is it possible to do above calculations in EXCEL? You guessed it correctly! Yes, you can modify lighting worksheet anyway you like to do motor efficiency calculations. I have made one for your reference. It is called "Motor Efficiency Worksheet" and you can access it under Unit 3 learning resources. In the same location, you can also find a file called "Motor Efficiency Examples" that carried out the calculations for the above two examples in EXCEL spreadsheet.