Figure 1. The torque-speed curve of an ac electric motor reveals that much more torque can be generated at low speed than is needed to drive a pump at full-load speed.
Correctly sizing an electric motor for a hydraulic power unit is a straightforward procedure. And if load pressure and flow remain fairly constant, determining the power requirement is relatively simple by using the familiar equation:
hp = (Q × P ) ÷ (1714 × EM)
where: Q is flow, gpm (and accounts for the pump’s volumetric efficiency),
P is system pressure at full load, psi, and
EM is the pump’s mechanical efficiency.
For example, assume an application requires a flow of 13.7 gpm at a maximum pressure of 2000 psi, and with a pump efficiency of 0.80. From the equation above:
hp = (13.7 × 2000) ÷ (1714 × 0.80) = 20 hp.
It may seem that an internal-combustion engine as the prime mover would have the same power rating as an electric motor. However, a general rule of thumb is to specify an internal-combustion engine with a power rating 2½ times that of an equivalent electric motor. This is due primarily to the fact that internal combustion engines have different torque-speed relationships than electric motors do, Figure 1. Because the overwhelming majority of HPUs for industrial applications use electric motors as the prime mover, that will be the focus of this article. For in-depth information on specifying internal combustion engines for HPUs, refer to the article beginning on page 32 of the July 2006 issue, or click here.
Pump torque requirements
Power, of course, is the combination of torque and rotational speed. A pump’s torque requirement is the main factor that determines whether a motor or engine is suitable for an application. Speed is less critical, because if a pump runs slowly, it will still pump fluid. However, if the prime mover does not develop enough torque to drive the pump, the pump will not produce any output flow.
To determine the torque required by a hydraulic pump, use the following equation:
T = ( p × D) ÷ (6.28 × 12 × EM)
where: T is torque, lb-ft, and
D is displacement, in.3/revolution
Pump displacement is provided in manufacturer’s literature. Continuing with the example just introduced, if the pump has a displacement of 1.75 in.3/rev., required torque is calculated as follows:
T = (2000 × 1.75) ÷ (75.36 × 0.80)
T = 58 lb-ft
Torque can also be calculated using the familiar horsepower equation:
hp = (T × n) ÷ 5250
where: n is shaft speed, rpm.
Substituting values from the example:
20 = (T × 1800) ÷ 5250
T = 58 lb-ft.
Electric motor torque signature
To understand the performance of an electric motor, we’ll first examine characteristics of a standard 3-phase electric motor. Figure 1 shows the torque-speed relationship of a 20 hp, 1800 rpm, NEMA Design B motor. Upon receiving power, the motor develops an initial, lockedrotor torque, and the rotor turns. As the rotor accelerates, torque decreases slightly, then begins to increase as the rotor accelerates beyond about 400 rpm.
This dip in the torque curve generally is referred to as the pull-up torque. Torque eventually reaches a maximum value at around 1500 rpm, which is the motor’s break-down torque. As rotor speed increases beyond this point, torque applied to the rotor decreases sharply. This is known as the running torque, which becomes the fullload torque when the motor is running at its rated full-load speed — usually 1725 or 1750 rpm.
The torque-speed curve for a 3600-rpm motor would look almost identical to that of the 1800-rpm motor. However, speed values would be doubled and torque values would be halved.
Common practice is to ensure that torque required from the motor will always be less than breakdown torque. Applying torque equal to or greater than breakdown torque will cause the motor’s speed to drop suddenly and severely, which will tend to stall the motor and most likely burn it out. If the motor is already running, it is possible to momentarily load a motor to near its breakdown torque. But for simplicity of discussion, assume the electric motor is selected based on full-load torque.
Figure 1 also shows a temporary large torque excess that can provide additional power to drive a hydraulic pump through momentary load increases. These types of electric motors also can be run indefinitely at their rated power plus an additional percentage based on their service factor — generally 1.15 to 1.25 (at altitudes to 3300 ft).
Catalog ratings for electric motors list their usable power at a rated speed. If the load increases, motor speed will decrease and torque will increase to a value higher than full-load torque (but less than break-down torque). So when operating at 1800 rpm, the electric motor has more than enough torque in reserve to drive the pump.
Those accustomed to specifying electric motors for hydraulic power units may be tempted to size a pump to be driven at 1800 rpm, then specify an oversized motor that can develop enough torque to drive the pump at this speed. This technique will produce a reliable power unit, but one that is relatively heavy, bulky, inefficient, and noisy.
Instead of following this procedure, any of several options should be considered. One option would be to drive the pump at a speed higher than 1800 rpm. Driving the pump at a higher speed decreases its required displacement, thereby reducing its size, weight, and torque requirement.
As far as pump performance, many designs exhibit higher mechanical and volumetric efficiencies when operated at speeds greater than 1800 rpm. On the other hand, operating a pump at a speed higher than what it was designed for will reduce its service life. Therefore, choose a pump speed that offers the best combination of pump performance and motor efficiency.
Internal combustion engines do not exhibit the torque reserve of electric motors — especially when accelerating from rest. Therefore, the pump should be unloaded whenever the power unit is started. This can be done hydraulically or mechanically, through a centrifugal clutch or other type of drive element.
Finally, pump size — and, therefore, size of the motor — often can be reduced by incorporating accumulators into the hydraulic system. If the hydraulic system operates in cycles where full flow is needed only for brief periods, an accumulator can store hydraulic power during periods of low flow demand and release this energy when full flow is needed.
This material was excerpted, in part, from the Fluid Power Handbook & Directory from information originally provided by Ronald R. Gould. P.E.