#### What is in this article?:

- Hydraulic power units
- Gas and diesel engine power
- Torque behavior of engines

A hydraulic power unit driven by an electric motor must be sized differently from one driven by an internal combustion engine — due to differences in their torque-speed curves.

When specifying components for a hydraulic power unit, the prime mover is sized based on torque, speed, and power requirements of the hydraulic pump. This is fairly straightforward for electric motors bec ause they generally have a starting torque that far exceeds running torque. Often, though, designers specify motors sized larger than necessary. This results in wasted energy because the motor operates at less than maximum efficiency.

Diesel and gasoline engines are another matter. They have a much flatter torque-speed curve, so they deliver roughly the same torque at high speed as they do at low speed. This means an internal compustion engine may develop high enough torque to drive a loaded pump, but not enough to accelerate it to operating speed. Consequently, with all other factors being equal, a power unit requiring an electric motor of a given power rating usually requires a gasoline or diesel engine with a power rating more than double that of the electric motor.

### Selecting the optimum motor size

The cost of electricity to operate an electric motor over its entire life span generally is many times that of the cost of the motor itself. Therefore, sizing the motor correctly for a hydraulic power unit can save a sizable amount of money over the life of the machine. If system pressure and flow are constant, motor sizing simply involves the standard equation:

*hp* = (*Q *×*P*) ÷ (1714×*E _{M}* ),

where:

*hp*is horsepower,

*Q*is flow in gpm,

*P*is pressure in psi, and

*E*is the pump's mechanical efficiency.

_{M}However, if the application requires different pressures during different parts of the operating cycle, you often can calculate root mean square (RMS) horsepower and select a smaller, less-expensive motor. Along with the calculation of rms horsepower, Figure 1, the maximum torque required at the highest pressure level of the application also must be found. Actually the two calculations are quite simple.

For example, such an application might use a 6-gpm, 3450-rpm gear pump to power a cylinder linkage that operates for an 85-sec cycle, Figure 2. The system requires 3000 psi for the first 10 sec, 2200 psi for the next 30 sec, 1500 psi for the next 10 sec, and 2400 psi for the next 10 sec. The pump then coasts at 500 psi for 20 sec, followed by 15 sec with the motor off.

It's tempting to use the standard formula, plug in the highest-pressure segment of the cycle, and then calculate:

*hp* = (6 × 3000) ÷ (1714 × 0.9)

= 11.7 hp for 10 sec.

To provide this power, some designers would choose a 10-hp motor; others would be ultra-conservative and use a 15-hp motor; a few might take a chance with 7½ hp. These motors in open drip-proof C-face models with feet would carry a relative price of about $570, $800, and $400 respectively, so there could be savings of $170 to $400 per power unit by choosing the 7½-hp motor — if it will do the job.

To determine this, first calculate the horsepower for each pressure segment of the cycle:

*hp _{1}* = (6 × 2200) ÷ (1714 × 0.9)

= 8.5 hp for 30 sec.

*hp*= (6 × 1500) ÷ (1714 × 0.9)

_{2}= 5.8 hp for 10 sec.

*hp*= (6 × 500) ÷ (1714 × 0.9)

_{3}= 1.9 hp for 30 sec.

The RMS horsepower is calculated by taking the square root of the sum of these horsepowers squared, multiplied by the time interval at that horsepower, and divided by the sum of the times plus the term (*t _{off}* ÷

*F*), as indicated in Figure 1.

Substituting the example values into the boxed equation and solving reveals that *hp _{rms}* = 7.2. Thus, a 7½-hp motor can be used from the standpoint of horsepower alone. However the second item, maximum torque, still must be checked before reaching a final decision. The maximum torque required to drive this particular pump will be found at the highest pressure - because the gear pump's output flow is constant. Use this equation:

*T* = *DP* ÷ (12)(6.28) *E _{M}*, where

*T*is torque in ft-lb, and

*D*is displacement in in.

^{3}

For this example,

*D*= (6 × 231) ÷ (3450)

= 0.402 in.

^{3}

Then

*T*= (0.402 × 3000) ÷ (12 × 6.28 × 0.9)

= 17.8 ft-lb.

Because electric motors running at 3450 rpm generate 1.5 ft-lb/hp, the 17.8 ft-lb of torque requires 11.9 hp (17.8÷1.5) at 3000 psi. This matches closely enough for the example application. (At other standard motor speeds: 1725 rpm generates 3 ft-lb per hp; 1150 rpm, 4.5 ft-lb per hp; 850 rpm, 6 ft-lb per hp.)

Now the second criteria can be checked against what the suggested motor can deliver in torque. What is the pull-up torque of the 7½-hp motor selected? Because the torque is least as the motor accelerates from 0 to 3450 rpm, it must be above 11.9 ft-lb with an acceptable safety margin. Note that a motor running 10% low on voltage will produce only 81% of rated pull-up torque: in other words, (208÷230)^{2} = 0.81. Reviewing motor manufacturers' performance curves will show several available 7½-hp models with higher pull-up torque. Any of these motors could be a good choice for this application.

Both motor criteria now have been verified. The RMS horsepower is equal to or less than the rated motor's horsepower. The motor's pull-up torque is greater than the maximum required.