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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.
Failure analysis of a gearbox, fan, bearing and shaft. When equipment breaks unexpectedly it is good practice to investigate why it happened so that the root cause can be rectified and the problem prevented from again happening.
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Diagnosing a bent shaft with vibration analysis is easy to do, especially when you can visualize its movement using 3D animations. Vibration patterns for different faults can look similar to each other. This is why it is always best to start with the machine, consider what faults are most likely and then take data in the proper test points and axes to find the most likely faults. It can get complicated, but 3D animations make it easier to visualize what is going on.
Diagnosing a bent shaft with vibration analysis is easy to do, especially when you can visualize its movement using 3D animations. Vibration patterns for different faults can look similar to each other. This is why it is always best to start with the machine, consider what faults are most likely and then take data in the proper test points and axes to find the most likely faults. It can get complicated, but 3D animations make it easier to visualize what is going on.
Many people operate rotary screw compressors with oil seeping from around the shaft. This housekeeping nuisance and waste of expensive oil can be avoided with a better understanding of the shaft seal. The mechanical seal and lip seal are the most common types of shaft seals used on Rotary Screw compressors. The following provides further information specific to each of these styles.
Many people operate rotary screw compressors with oil seeping from around the shaft. This housekeeping nuisance and waste of expensive oil can be avoided with a better understanding of the shaft seal. The mechanical seal and lip seal are the most common types of shaft seals used on Rotary Screw compressors. The following provides further information specific to each of these styles.
In theory, machine alignment is a straightforward process, but in real world applications, it is often compounded by structural faults such as ‘soft foot’, piping strain, induced frame distortion, excessive bearing clearance or shaft rubs. These pitfalls can turn a simple job into an all day affair - frequently with unsatisfactory results despite a conscientious effort and a considerable investment in manpower and downtime. This paper will examine how to eliminate some of the typical reasons why alignments are unsuccessful.
In theory, machine alignment is a straightforward process, but in real world applications, it is often compounded by structural faults such as ‘soft foot’, piping strain, induced frame distortion, excessive bearing clearance or shaft rubs. These pitfalls can turn a simple job into an all day affair - frequently with unsatisfactory results despite a conscientious effort and a considerable investment in manpower and downtime. This paper will examine how to eliminate some of the typical reasons why alignments are unsuccessful.
It turned out that the photoelectric sensor could only be adjusted to stay on all the time or off all the time when the cans were going by at any higher rate than Jog speed on the Seamer. We adjusted and adjusted the sensors. The sensors were replaced with new sensors. It did no good. The lines were stopped while we traced all the control and power wiring so we could try to determine if there was a problem in the wiring.
It turned out that the photoelectric sensor could only be adjusted to stay on all the time or off all the time when the cans were going by at any higher rate than Jog speed on the Seamer. We adjusted and adjusted the sensors. The sensors were replaced with new sensors. It did no good. The lines were stopped while we traced all the control and power wiring so we could try to determine if there was a problem in the wiring.
I was asked recently to investigate and solve an overheating problem in a mobile hydraulics application. The hydraulic system that was overheating comprised a diesel-hydraulic power unit, which was being used to power a pipe-cutting saw. The saw was designed for sub-sea use and was connected to the hydraulic power unit on the surface via a 710-foot umbilical. The operating requirements for the saw were 24 gpm at 3000 psi. Heating of hydraulic fluid in operation is caused by inefficiencies. Inefficiencies result in losses of input power, which are converted to heat. A hydraulic system's heat load is equal to the total power lost (PL) through inefficiencies.
I was asked recently to investigate and solve an overheating problem in a mobile hydraulics application. The hydraulic system that was overheating comprised a diesel-hydraulic power unit, which was being used to power a pipe-cutting saw. The saw was designed for sub-sea use and was connected to the hydraulic power unit on the surface via a 710-foot umbilical. The operating requirements for the saw were 24 gpm at 3000 psi. Heating of hydraulic fluid in operation is caused by inefficiencies. Inefficiencies result in losses of input power, which are converted to heat. A hydraulic system's heat load is equal to the total power lost (PL) through inefficiencies.
