# Motor Example

The following example illustrates a complete investment analysis for a Motor Upgrade. Included are several steps needed to determine if the motor upgrade is a cost-effective. Investment calculations include simple payback, internal rate of return and savings to investment ratio.

Contact Energy Smart if you’d like us to assist with analyzing whether or not a motor upgrade is a good investment for your company.

**Motor Upgrade — Background**

Energy Smart staff identified a motor upgrade opportunity during an on-site energy consultation. Four existing general purpose motors had the following characteristics:

Horsepower (HP) = 100 HP

Full Load Efficiency (%FL) = 91.7%

Annual Operating Hours (AH) = 4550 hours

RPM Rating (RPM) = 1790 RPM

Voltage Rating (V) = 460 Volts

The following motor was proposed as a replacement for the existing model:

Horsepower (HP) = 100 HP

Full Load Efficiency (%FL) = 95.4%

Annual Operating Hours (AH) = 4550 hours

RPM Rating (RPM) = 1790 RPM

Voltage Rating (V) = 460 Volts

To determine the potential energy savings, the following equation is used:

Savings = HP * 0.746 * Cost (kWh) * AH * ((100/%FLExisting)- (100/%FLNew))

Savings = 100 * 0.746 * $0.07 *4550 * ((100/91.7)-(100/95.4))

**Savings = $1,004.81/year for each motor**

**Investment Analysis**

We now know the estimated savings for each motor ($1004.81). The cost for each new motor is $4,066 before rebate. The local utility company offers a $350 rebate for this type of motor upgrade. Energy costs in this area are $0.07 per kWh.

With this information in hand, the next step is to evaluate whether the motor upgrade project is an attractive investment. We’ll start with the most basic investment analysis and work our way toward determining if the project is acceptable base on your minimum attractive rate of return (sometimes known as your company’s minimum internal rate of return).

*Simple Payback Period (SPP)*

Simple Payback Period is generally the number of years required to recover the initial investment through project savings. This is a good first cut investment analysis tool, but it doesn’t take into account the time value of money. In this case we’ve taken the total project cost minus the rebates and divided that amount by the annual savings.

Using the example above, the total project cost is $16,264, the rebate amount is $1,400, and the annual savings amount is $4,019.

SPP = (Project Cost – Rebate)/Annual Savings

SPP = ($16,264-$1,400)/$4,019

**SPP = 3.70 years**

*Internal Rate of Return (IRR)*

Internal Rate of Return is designed to calculate the rate of return that is “internal” to the project. If you know what your company’s minimum attractive rate of return (MARR) is, an IRR that is higher than your MARR is an attractive project.

Again, using the example above, we know the total project cost after rebate to be $14,864 and the annual saving is $4,019. This company’s MARR is 15% and the new motors are expected to last 10 years. In order to determine the IRR the following equation should equal zero by using a trial and error approach.

The expression ‘(P|A,IRR,n)’ comes from Time Value of Money tables commonly used in economic analysis interest tables. In this case, P stands for the present value (to be determined), A stands for the annualized savings, IRR stands for the Internal Rate of Return for this project, and n stands for the number of years the project will last or the life span of the project.

Present Worth = -(Project Cost – Rebate) + Annual Savings * (P|A,IRR,n) = 0

PW = -$14,864 + $4,019*(P|A,IRR,10) = 0

Starting with an IRR of 15% …

PW = -$14,864 + $4,019*(P|A,15%,10)

PW = -$14,864 +$4,019*(5.0188)

PW = -$14,864 + $20,170

PW = $5,306 > 0

Let’s try an IRR of 25% …

PW = -$14,864 + $4,019*(P|A,25%,10)

PW = -$14,864 +$4,019*(3.5705)

**PW = -$514 < 0 so we know this project has an IRR of approximately 23% which is well above the MARR making this a very attractive project.**

Savings to Investment Ratio (SIR)

Savings to Investment Ratio is designed to determine how many times the project would pay for itself throughout its lifetime.

SIR = Total Savings/Investment

SIR = $4,019(P|A,15%,10)/$14,864

SIR = $4,019(5.0188)/$14,864

SIR = $20,170/$14,864

**SIR = 1.36 which is greater than 1.00 so this would be an attractive project.**