### Question:

Several of our utility customers who have heavy commercial/industrial refrigeration loads are located in outlying areas where only single-phase power is available. How do we calculate the value of upgrading to three-phase power?

### Answer:

The following example describes how to calculate the costs/benefits of upgrading to three-phase power. Your customers may also want to consider alternatives such as phase converters and the written-pole motor. The Rural Electricity Resource Council suggests some Three-Phase Power Options for Rural Loads.

#### Grocery Store Refrigeration Energy Savings

A rural grocery is served by single-phase power. The grocery has:

- 6 Amps (A) of motor load (based on full-load amp ratings at 230 Volts) associated with deli-coolers
- 16 Amps (also at 230 V) from two walk-in coolers
- 48 Amps (at 115 V) from four reach-in coolers

The owner wants to install new coolers that operate with three-phase service. The new three-phase line will run for 700 feet at a cost of approximately $7 per foot. The new cooler will have three 2-horsepower compressor drive motors and three evaporator fan motors with a full-load amp value of 8 Amps each.

#### What is the energy savings and simple payback on investment?

First, we'll calculate the initial power draw assuming all motors are fully loaded. Note that fractional hp shaded-pole motors (the cheapest and lowest efficiency motors available) have a full-load efficiency of 64% with a full-load power factor of 63%. The energy baseline for the single-phase system is:

230 V service kW is: 22 A x 230 V x 0.63/1000 = 3.18 kW

115 V service kW is: 48 A x 115 V x 0.63/1000 = 3.47 kW

Total = 6.65 kW

For the new system, we'll assume that the three-phase 2-hp motors are energy-efficient models. For example, Baldor motors (2-hp, 1800 RPM, Totally enclosed) have a full-load efficiency of 86.5%, with a full-load Amp rating of 5 A (at 230 V) and a full-load power factor of 83%.

The kW demand for the three 2-hp compressor drive motors (at full-load) is:

Full-load kW = 6 hp x 1.0 (load factor) x (0.746 kW/hp) x / 0.865 = 5.17 kW

It will be assumed that the evaporator fan motors are fractional hp and inefficient. Their contribution is:

Evaporator kW draw = 24 A x 115 V x 1.732 x 0.63/1000 = 3.01 kW

The total kW draw for the new system is estimated at 8.18 kW. This is more than the old system, but is to be expected as the cooling capacity was significantly increased.

Energy savings will be estimated by assuming that the kW required by the original system is reduced due to the acquisition of energy-efficient three-phase motors. The savings are then:

Savings = 6.65 kW x (1/0.64 - 1/0.865)

The kW reduction is 2.70 kW

The annual value of these savings is (assuming a utility rate of 6 cents/kWh, a monthly demand charge of $15/kW, and 5000 hours per year of operation):

2.7 kW x 5000 hrs x $0.06 = $811

2.7 kW x $15/kW-mo x 12 mo/yr = $486

Total = $1,297/year

The simple payback on the $4,900 distribution line investment is $4900/$1297 = 3.8 years.

Topic: Utility Companies--Transmission/Distribution

Topic: Electrical Systems--General

Sector: Industrial

Sector: Agricultural

Sector: Utility