Contrary to popular opinion, bigger is not always better. A case in point is the electric motor. Pump users tend to want a little extra power, “just in case.” That’s why automobile manufacturers can still sell cars with 300 hp engines, when the speed limit may be less than 70 miles per hour. But, like those gas-guzzlers, oversized electric motors cost more to run — sometimes a lot more. In addi- tion to wasting energy, they often trigger expensive demand charges on utility bills due to low power factor.
Fortunately, it’s easy to determine how much power a pump load requires — without expensive equipment or engineering expertise. In collecting the data, just make sure the pump motor is operating at peak continuous load.
The illustration on the next page shows the essentially linear relationship between percent load and current from no-load current to nameplate currentof a pump motor. Notice, though, that zero load does not equal zero current. As- suming it does equal zero will cause mistakes in determining the required power, with the error inversely proportional to the load. That means that the biggest errors will occur when evaluating motors most in need of being matched to the pump load.
Although the percent load a motor carries could be determined from a graph, it is easy (and more accurate) to calculate the actual load using the following equation:
hpr =hpn (1–FLAa –FLAnl)
where hpr is required power in hp,
hpn is nameplate power in hp,
FLAa is actual current draw in amperes, and
FLAnl is current drawn at no load in amperes.
To obtain good input data, run the motor uncoupled at no load and measure the current with a clamp-on ammeter. Don’t take any shortcuts here. The no-load current will be higher if the motor is coupled than if uncoupled. Even though the driven equipment (pump) might not be doing any work, some power is required to overcome friction in driving it. To avoid errors, use the uncoupled current.
Next, document the nameplate current and the current at the motor’s actual load. Because an undersized motor presents other problems, the safest method is to measure the current over the entire cycle of operation. If the load is seasonal, record the current during peak load.
A realistic example
It’s not hard to find oversized motors in industry. In one case, a plant had a 40-hp motor driving a hydraulic pump for a molding ma- chine. The motor had a nameplate current rating of 50 A and drew 15 A when operating uncoupled — slightly more than one-third of full- load current. When driving its peak load, it drew 37 A. A quick calcula- tion reveals that the actual peak load was just over 25 hp:
hpr = 40 [1–(50–37)/(50–15)]
hpr = 25.1 hp
Substituting the 40-hp motor with a 25-hp one decreased full- load current from 37 A to 30 A. The plant had been paying for, and wasting, a lot of electricity.