[webinar] Embracing Digital Transformation in Maintenance & Plant Operations | March 13 at 10AM EST – Register Now
A Guide to Matching Electric Motors with Hydraulic Power Units
A Guide to Matching Electric Motors with Hydraulic Power Units
Ronald R. Gould, CFPE
Sizing electric motors correctly for hydraulic power units can save a sizable amount of money over the life of the equipment. If system pressure and flow are constant, motor sizing simply boils down to the standard equation:
hp = QP / 1714EM
where: hp is horsepower Q is flow in gpm P is pressure in psi, and EM is pump’s mechanical efficiency.
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, 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. 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 seconds. 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 horsepower, 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 7.5 hp. These motors in open drip-proof C-face models with feet would list at about $570, $800, and $400 respectively, so there could be savings of $170 to $400 per power unit by choosing the 7.5-hp motor – if it will do the job.
To determine this, first calculate the horsepower for each pressure segment of the cycle:
hp1 = 6(2200) / (1714)(0.9) = 8.5 hp for 30 sec. hp2 = 6(1500) / (1714)(0.9) = 5.8 hp for 10 sec. hp3 = 6(500) / (1714)(0.9) = 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 (toff /F), as indicated in the box at the bottom of this page.
Substituting the example values into the boxed equation and solving reveals that hprms equals 7.2. Thus, a 7.5-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)EM
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 per hp, the 17.8 ft-lb of torque requires (17.8 / 1.5) or 11.9 hp at 3000 psi. This checks out 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, 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 7.5-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: i.e., (208/230)2 = 0.81. Reviewing motor manufacturers’ performance curves will show several available 7.5-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 to match the motor with the appropriate hydraulic power unit. 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.
It is very difficult to calculate and predict the speed of an air cylinder; there are too many variables which influence speed and on which it is difficult to gather accurate data. We must rely mainly on good design practices and use the benefit of past experience. It is better to have the cylinder speed too fast to begin with because it can always be reduced with a speed control valve, but if it is too slow to begin with, it is more difficult to increase its speed later.
It is very difficult to calculate and predict the speed of an air cylinder; there are too many variables which influence speed and on which it is difficult to gather accurate data. We must rely mainly on good design practices and use the benefit of past experience. It is better to have the cylinder speed too fast to begin with because it can always be reduced with a speed control valve, but if it is too slow to begin with, it is more difficult to increase its speed later.
On the design of a new machine which is to be run with a hydraulic motor, a determination of required speed and horsepower must be made so a model with suitable ratings can be selected. This article describes several methods of making such a determination. Designers who are experienced only in selecting electric motor drives need to be careful in designing hydraulic drives because of important differences between these two motors.
On the design of a new machine which is to be run with a hydraulic motor, a determination of required speed and horsepower must be made so a model with suitable ratings can be selected. This article describes several methods of making such a determination. Designers who are experienced only in selecting electric motor drives need to be careful in designing hydraulic drives because of important differences between these two motors.
Don’t run centrifugal pumps on the end of the curve. Pumps are designed and selected to operate near their highest efficiency point. If they operate at the right-hand end of the centrifugal pump curve the likelihood of cavitation increases.
Don’t run centrifugal pumps on the end of the curve. Pumps are designed and selected to operate near their highest efficiency point. If they operate at the right-hand end of the centrifugal pump curve the likelihood of cavitation increases.
Regardless of the approach taken to measure true dynamic movement, coupled machines need to be set to cold alignment targets that will reflect the actual changes in the shaft alignment. This will lead to lower vibration levels, increased Mean Time Between Failures, decreased maintenance expenditures and increases in production. Much like the philosophical change from aligning shafts with dial indicators to aligning shafts with laser based systems, these types of measurements will take some time to be generally accepted and routinely practiced.
Regardless of the approach taken to measure true dynamic movement, coupled machines need to be set to cold alignment targets that will reflect the actual changes in the shaft alignment. This will lead to lower vibration levels, increased Mean Time Between Failures, decreased maintenance expenditures and increases in production. Much like the philosophical change from aligning shafts with dial indicators to aligning shafts with laser based systems, these types of measurements will take some time to be generally accepted and routinely practiced.
Such a question often comes up when it is time to recondition a mechanical seal: We use several different brands of mechanical seals in my plant. When a pump is taken out of service, we usually send the seal back to the original seal manufacturer for reconditioning. One of my suppliers has offered to recondition all of my seals, regardless of brand. Is this a good idea? Most process plant maintenance departments return their used mechanical seals to the original seal manufacturer (or Seal OEM) for reconditioning.
Such a question often comes up when it is time to recondition a mechanical seal: We use several different brands of mechanical seals in my plant. When a pump is taken out of service, we usually send the seal back to the original seal manufacturer for reconditioning. One of my suppliers has offered to recondition all of my seals, regardless of brand. Is this a good idea? Most process plant maintenance departments return their used mechanical seals to the original seal manufacturer (or Seal OEM) for reconditioning.
It's important to find out if material flows are present or not throughout a bulk-processing facility; material cost savings and increased plant efficiency can offset an investment in monitoring. Most bulk solids processors can do this using low-cost acoustic emission-monitoring technology.
It's important to find out if material flows are present or not throughout a bulk-processing facility; material cost savings and increased plant efficiency can offset an investment in monitoring. Most bulk solids processors can do this using low-cost acoustic emission-monitoring technology.
