Electric motors: Repair or replace?

In brief:

• There are many software tools, spreadsheets and other freely available tools that one can use to make a sound judgment about the repair/replace decision for electric motors.
• The decision should be based on simple payback, the ready availability of a spare motor, relative reliability for a spare and the rewound motor, and other factors, taken separately or in combination.
• The reliability or maintenance engineer should provide input into the repair/replace decision to ensure considers the motor’s intended application.

In the 1990s it was popular to base repair-versus-replace decisions on energy consumption and reduced efficiency through the repair process. This meant considering the efficiency of a new energy- or premium-efficient motor, operating hours and load against the original machine so as to determine the simple payback. If the payback was less than a specified time, such as two years, the replacement was considered economically viable.

However, energy is only one aspect. Other considerations include:

• The availability of a replacement
• The number of times repaired
• What was repaired and how it was done
• Machine reliability
• General condition
• The effect of downtime on profitability
• The number of inventoried spares.

The first question usually asks which size is the cutoff for a motor repair. The answer: It depends.

Energy decisions

“The replacement or repair cost of a production-related motor that has no immediate spare often is far less than the lost production income.”

How do you weigh the relative values of repairing versus replacing a motor from the standpoint of energy consumption? There are software tools, spreadsheets and more resources dedicated to this type of decision-making, including the U.S. Department of Energy’s MotorMaster Plus software (www1.eere.energy.gov/industry/bestpractices/software.html). Each of these methods compares the efficiency of the original motor with energy assumptions following the repair process and the new, more efficient electric motor. The result normally is expressed as a simple payback for which the motor owner must decide if it’s of value. In the past, the normal threshold was two years or fewer.

The calculation is quite simple. You need to know motor horsepower, load (or average load) and the original efficiency, energy and demand costs. The numbers are plugged into a standard set of formulae, as follow:

To find the power difference, use Equation 1.

P = 0.746 * hp * L *(1/effo -1/effn)     (Eqn 1)

Where:
P = power difference (kW)
hp = horsepower
L = load (decimal fraction)
effo = original efficiency (decimal fraction)
effn = new motor efficiency (decimal fraction)

To find the demand charges, use Equation 2.

D = 0.746 * hp * 12 * C        (Eqn 2)

Where:
D = demand charge ($)
hp = horsepower
C = monthly charge ($/kW/mo)

Then, use Equation 3 to determine the usage charges:

U = 0.746 * hp * T * R       (Eqn 3)

Where:
U = usage charges ($)
hp = horsepower
T = hours of operation
R = utility rate ($/kWh)

Finally, use Equation 4 to find the simple payback.

PB = Diff / (D + U)    (Eqn 4)

Where:
PB = simple payback (yrs)
Diff = cost difference to repair or replace ($)
D = demand charge ($)
U = usage charges ($)

Practical example

Let’s assume a decision needs to be made about an older, standard-efficiency motor rated at 50 hp, 92.5% efficient, 80% loaded, with $0.12 usage and $12/kW demand charges for 4,000 hours per year to be replaced with a 95% efficient motor. Assume that the difference between the motor replacement cost and repair cost is $900.

According to the U.S. Department of Energy’s publications, you can expect an average of 0.5% efficiency reduction per rewind using standard motor repair practices. Alternate rebuild practices can avoid this reduction.

The power difference will be:

P = 0.746 * hp * L *(1/effo -1/effn)
  = 0.746 * 50 * 0.8 *(1/0.925 -1/0.95)
  = 29.84 * (1.0811 – 1.0526)
  = 29.84 * (0.0285)
  = 0.85044 kW

The demand charge will be:

D = P * 12 * C
  = 0.85044 *12 * 12
  = $122.46

The usage charge will be:

U = P * T * R
  = 0.85044 * 4,000 * $0.12
  = $408.21

The simple payback will be:

PB = Diff / (D + U)
   = $900 / ($122.46 + $408.21)
   = $900 / $530.67
   = 1.7 years

Because the replacement cost is less than the two-year threshold, this motor would be replaced. Also, if the motor efficiency is assumed to be reduced by 0.5% and the carbon output is 0.909 tons/MWh, then the energy and environment decision might be improved. Using the above, the original motor would be assumed at 92% efficient, or a total savings of $637 per year, or a simple payback of 1.4 years. The greenhouse gas emission reduction would be (1.02 kW * 4,000 hrs * (1 MWh/1,000 kWh) * 0.909 tons/MWh = 3.7 tons/year carbon. The combination of both numbers makes the replacement decision even more palatable.

Availability of replacement

On the other hand, if the motor isn’t available because of some special part of its design or the delivery lead time, one might opt to repair the motor, regardless of the energy effect. This would be classified as a production-related decision versus a decision concerned with energy or the environment. The replacement or repair cost of a production-related motor that has no immediate spare often is far less than the lost production income.