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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.
Torque behavior of engines
A gasoline engine has a dramatically different torque-speed curve, Figure 4, than an electric motor does. This means a gasoline engine exhibits a much less variable torque output throughout its speed range. Depending on their design, diesel engines with the same power ratings may generate slightly higher or lower torque at lower speeds than gasoline engines do, but diesels exhibit a similar torque curve throughout their operating speed range.
Calculations above determined that 58 lb-ft of torque is required to drive the pump at any speed. Referring to Figure 4, the 20-hp gasoline engine develops a maximum torque of only 31 lb-ft — clearly not enough to drive the pump. This is because its 20-hp rating is based on performance at 3,600 rpm. Maximum torque occurs at speeds near 3,000 rpm, but is still well below the 58 lb-ft required by the pump. Even if the engine produced enough torque at this speed, power would still be inadequate due to the lower speed.
This is where the 2½ sizing rule comes from. An HPU requiring a 20-hp electric motor to drive the pump at 1,800 rpm would require a gas or diesel engine rated at about 50 hp. Moreover, these values are based on an engine operating at its maximum torque and power ratings. However, manufacturers recommend that gasoline and diesel engines only operate continuously at about 85% of their maximum rated values to prevent seriously shortening of their service lives. So referring again to Figure 4, a 20-hp gasoline engine would develop just over 26 lb-ft of maximum torque, and only 24 lb-ft at 3,600 rpm.
It is also interesting to compare this performance with fuel consumption. The fuel consumption chart, Figure 5, shows that a 20-hp gasoline engine achieves greatest fuel efficiency at about 2,400 rpm, where it consumes just over 8.2 lb/hr (0.41 lb/hp × 20 hp). At 3,600 rpm, the engine would be considerably less fuel efficient.
Actions to take
By now it should be clear that specifying a gasoline or diesel engine to drive a hydraulic power unit follows a different procedure than that for specifying an electric motor. If you are accustomed to specifying electric motors for hydraulic power units, you may be tempted to size a pump to be driven at 1,800 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 would be to drive the pump at a speed higher than 1,800 rpm. Pump literature for mobile equipment should list ratings at a variety of speeds. If it doesn't, consult the pump manufacturer. Driving the pump at a higher speed decreases its required displacement, thereby reducing its size, weight, and torque requirement. So operating the power unit at higher speed more closely matches engine performance to the application by increasing torque produced by the engine and reducing the torque required by the pump.
More specifically, operating the pump in our example at 2,800 rpm would increase engine torque to more than 30 ft-lb and reduce torque required by the pump to perhaps 38 ft-lb. Although the engine torque still would fall short of that required, it obviously comes much closer to matching pump torque than when operating at 1,800 rpm.
Designers may be tempted to run a gas or diesel engine at or near the speed at which it exhibits optimum fuel efficiency. However, an operating speed where the engine produces maximum torque generally takes priority. This is because if the engine doesn't generate enough torque at its optimum fuel efficiency speed, a larger engine would be required. But a larger engine consumes more fuel, which would defeat the purpose of trying to conserve fuel by operating at a specific speed.
In addition, pumps generally have a speed range at which they are most efficient. So even if an engine operates a few hundred rpm above or below its optimum fuel efficiency speed, torque produced and pump dynamics generally have a more pronounced effect on overall efficiency of the power unit. Therefore, the speed at which the gas or diesel engine operates should take all of these considerations into account.
As far as pump performance, many designs exhibit higher mechanical and volumetric efficiencies when operated at speeds greater than 1,800 rpm. On the other hand, operating a pump at a speed higher than what it was designed for would reduce its service life. Therefore, it is important to choose a pump speed that offers the best combination of pump and engine performance.
Perhaps an even better alternative would be to provide a gearbox or other type of speed reducer between the engine and pump. Although this would add components to the power unit, it would increase torque and reduce speed while allowing both the engine and the pump to operate at their optimum speeds. The additional cost of the speed reducer may be offset by the lower cost of a smaller, lighter, and less-expensive engine.
Because gas and diesel engines do not exhibit the torque reserve of electric motors — especially when accelerating from rest — it is especially important that the pump be unloaded whenever the HPU is started. This can be done hydraulically, or mechanically through a centrifugal clutch or other type of drive element.
Finally, as with HPUs driven by electric motors, pump size — and, therefore, size of the prime mover - 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.
Ronald R. Gould, P.E., is vice president, engineering, Advance Lifts Inc., Lake Charles, IL.
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