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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.
Gear pumps are the most common type of positive displacement pump, ideal for transferring high viscosity fluids such as automotive oils, plastics, paint, adhesives, and soaps. As with any pump, proper operation and regular maintenance are vital to reducing costly pump repairs and maximizing pump efficiency. Below are helpful tips for operating your gear pump and ensuring it achieves a long operating life through regular maintenance. As with any pump, proper operation and regular maintenance are vital to reducing costly pump repairs and maximizing pump efficiency.
Gear pumps are the most common type of positive displacement pump, ideal for transferring high viscosity fluids such as automotive oils, plastics, paint, adhesives, and soaps. As with any pump, proper operation and regular maintenance are vital to reducing costly pump repairs and maximizing pump efficiency. Below are helpful tips for operating your gear pump and ensuring it achieves a long operating life through regular maintenance. As with any pump, proper operation and regular maintenance are vital to reducing costly pump repairs and maximizing pump efficiency.
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.
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.
Rotary dryers, kilns, mills and reactors turn on tires and trunnions. Each tire is mounted to the rotating shell and revolves on two roller trunnions. If the alignment between the tire and trunnions is lost due to wear, poor repair, installation error or impact, the equipment will need to be tracked.Â
Rotary dryers, kilns, mills and reactors turn on tires and trunnions. Each tire is mounted to the rotating shell and revolves on two roller trunnions. If the alignment between the tire and trunnions is lost due to wear, poor repair, installation error or impact, the equipment will need to be tracked.Â
The previous article in this series, "Understanding Shaft Alignment: Thermal Growth," explained thermal growth and its affect on proper equipment alignment. A practical example involves a recent project at a wastewater treatment plant in Cleveland that needed realistic cold alignment targets for a 3600 rpm compressor to reach their accurate alignment targets.
The previous article in this series, "Understanding Shaft Alignment: Thermal Growth," explained thermal growth and its affect on proper equipment alignment. A practical example involves a recent project at a wastewater treatment plant in Cleveland that needed realistic cold alignment targets for a 3600 rpm compressor to reach their accurate alignment targets.
A motor brush is not a brush at all in the traditional sense. It is actually a carbon or graphite cube, commonly held in place by a spring, that acts as a conductor between the electrified stationary and rotating parts of a motor. The whole brush assembly is made up of a carbon block, one or more shunts, a spring and a holder. The carbon blocks are easily replaceable and are therefore intended as a wear part to prevent damage to more costly motor components.
A motor brush is not a brush at all in the traditional sense. It is actually a carbon or graphite cube, commonly held in place by a spring, that acts as a conductor between the electrified stationary and rotating parts of a motor. The whole brush assembly is made up of a carbon block, one or more shunts, a spring and a holder. The carbon blocks are easily replaceable and are therefore intended as a wear part to prevent damage to more costly motor components.
Selecting the right pump can pose a significant challenge due to the extensive array of options available for different applications. Each type of pump comes with distinct advantages and disadvantages that necessitate careful evaluation. In this article, we provide a comprehensive overview of prevalent pumping technologies and delve into the limitations associated with pump selection per applications.
Selecting the right pump can pose a significant challenge due to the extensive array of options available for different applications. Each type of pump comes with distinct advantages and disadvantages that necessitate careful evaluation. In this article, we provide a comprehensive overview of prevalent pumping technologies and delve into the limitations associated with pump selection per applications.
