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)

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, the 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 to the fact that internal combustion engines have different torque-speed relationships than electric motors do. Examining the different torque characteristics will provide the understanding to make a choice based on solid reasoning — rather than putting faith in a rule of thumb.

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 x EM)

T is torque, lb-ft, and
D is displacement, in.3/revolution

Pump displacement is provided in manufacturer's literature. Continuing with the example introduced at left, if the pump has a displacement of 1.75 in.3/rev., required torque is calculated as follows:

T = (2000 × 1.75) ÷ (75.36 x 0.80)
T = 58 lb-ft

Torque can also be calculated using the familiar horsepower equation:

hp = T × (n ÷ 5250)

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 differences in power characteristics between an electric motor and internal-combustion engine, we'll first examine characteristics of a standard 3-phase electric motor. Figure 2 shows the torque-speed relationship of a 20 hp, 1800 rpm, NEMA Design B motor. Upon receiving power, the motor develops an initial, locked-rotor 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 know as the running torque, which becomes the full-load 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. The difference would be that 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.

Note that Figure 2 shows a temporary large torque excess that can provide additional muscle to drive the hydraulic pump through momentary load increases. These types of electric motors also can be run indefinitely at their rated hp 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 the pump at 1800 rpm, the electric motor has more than enough torque in reserve to drive the pump.