# The Motor System (Including Upstream Components)

Take the example of two 25 hp, off-the-shelf, 460 V, 3-phase, TEFC motors, 250 feet from the load center, running at 75% load, 6,000 hours per year, served by a No. 8 THHN circuit in steel conduit. This circuit is one of many supplied by a 1,500-kVA dry-type transformer. Let’s look at the motor efficiency decision isolated from other elements:

## Calculation of Energy Savings and Payback of High-Efficiency Motor

Motor No. 1: STANDARD EFFICIENCY, 90.2% NEMA efficiency, cost \$859.50

 Input Power (kW) = 25hp x 0.746 x 75% 0.902 efficiency = 15.51 kW Energy Usage (kWh) = 15.51 kW x 6,000 hr. = 93,060 kWh/yr

Motor No. 2: PREMIUM EFFICIENCY, 93.6% NEMA efficiency, cost \$1,043.00

 Input Power (kW) = 25hp x 0.746 x 75% 0.936 efficiency = 14.94 kW Energy Usage (kWh) = 14.94 kW x 6,000 hr. = 89,640 kWh/yr Savings = (93,060 - 89,640) = 3,420 kWh/yr
at \$0.09 per kWhat \$0.07 per kWh
Value of Annual Savings \$307.80 \$239.40
Simple Payback = added cost (\$183.50)
savings
0.6 yr 0.8 yr

Now, let’s figure in the additional savings related to the wiring of the motor circuit, as a result of reduced I2R losses from the motor load alone.

Remember, in computing kW and kWh losses or savings, the wire doesn't know the phase angle of the current it's seeing, so we can reasonably take 75% of the rated full load amps (FLA) to calculate watts loss.

 For Motor No. 1: I=30.6 FLA x 0.75 = 22.95 amps per phase For Motor No. 2: I=30.0 FLA x 0.75 = 22.5 amps per phase

## Calculation of Energy Savings in the Wire

 For No. 8 AWG THHN Wire (from the NEC, Chap. 9, Table 9): R = 0.78 Ω per 1,000 ft @ 75°C To correct resistance to 30°C, use NEC Table 8 footnote: R2 = R1 [1 + α (T2- 75)] where α = 0.00323 No. 8 THHN @ 30°C, R = 0.667 Ω /1,000 ft., or, dividing by 4, 0.167 Ω / 250 ft. For Motor No. 1: 3-Phase Watts loss = I²R = (22.95)² x (0.167) x 3 = 263.9 W(1) Energy loss = 263.9 W ÷ 1,000 x 6,000 hr/yr = 1,583 kWh/yr For Motor No. 2: 3-Phase Watts loss = (22.5)² x (0.167) x 3 = 253.6 W Energy loss = 253.6 ÷ 1,000 x 6,000 = 1,522 kWh/yr Savings in the wire, due to the more efficient motor: 1,583 - 1,522 = 61 kWh/yr (1)Note that the I²R losses in each phase are multiplied by 3 to determine the losses in all three phases. (Be careful not to use √3, a common mistake.)
at \$0.09 per kWhat \$0.07 per kWh
Value of Additional Savings in Wire Losses \$5.49 \$4.27

Admittedly, these additional savings look small, but they’re 1.8% of the savings of the more efficient motor, and they don’t cost one cent extra. They come from the reduced I2R in the wire, they add to the justification of the more efficient motor, and they are a free tag-along.

In a similar manner, the more efficient motor will also have an effect on the energy losses in the transformer supplying this system. For this example, we will start out with a standard-efficiency transformer, and compute the effect of motor selection on the transformer loss. In other words, we will estimate the losses that will not occur because less power is being drawn through the transformer. (Later, we will calculate the savings from a high-efficiency transformer.) The model used here has a stated efficiency of 98.47%.

The reduced loss in the transformer due to the motor selection can be approximated from the formula:

 Reduced loss = [(reduced motor loss) + (reduced wire loss)] x (transformer inefficiency) Reduced Loss(kW) = [(15.51 kW - 14.94 kW) + (0.264 kW - 0.254 kW)] x (1 - 0.9847) = 0.0089kW Multiplying by 6,000 hours of operation, reduced loss = 53.4 kWh/yr

We can now see that the total annual savings from selection of the high efficiency motor is: 3,420 kWh (\$307.80) in the motor itself, plus 61 kWh (\$5.49) in the No. 8 wire, plus 53 kWh (\$4.77) in the transformer, totaling 3,534 kWh (\$318.06 at 9¢ kWh). All this as a result of an extra up-front investment of \$183.50 in a premium-efficiency motor. And remember: the \$318.06 annual savings continue, year after year, following the initial short payback period of less than seven months.