Positive displacement type, membrane filter press feeding pump {the capacity (Rpm) decreases when the discharge pressure rises}
Step 1 : Load Characteristic
When selecting frequency drive, the motor load type must be considered. We must analyze for our application the torque and power demand in function of speed.
Referring to the pump catalogue we find correct pumping curves:
- We determine viscosity and density of the slurry ( our case : η =115 cP, ρ = 0.89 kg/dm³ )
- We determine working range for the pump:
- pressure(s) in function of rotation speed
- capacity in function of rotation speed
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We verify power and torque demand for the pump in the expected working range of speed / capacity / pressure.
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We fill up the empty filter with high capacity, once the pressure increases we decrease capacity (our regulator PID). Then we locate the working points on the graph.

Note:
is the power demand for working point.
is the torque demand for working point.
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By linking the working points, we obtain the expected power demand curve and torque demand curve.
Step 2 : Motor plate data
To obtain requested range of speeds for our pump, we reduce the motor rotation speed through reducer with reduction I =10 . We will loose some power due to efficiency of motoreducer (about η = 0.98 for our case ), but available torque will increase 10 times, while the speed will be divided by 10.
From the motor “name plate” we verify all information necessary for the drive choice. For missing data one must refer to the manufacturer catalogue.

Example of Three Phase Induction Motor Plate
Step 3 : Choice of Frequency Drive
The “motor + frequency drive” must give sufficient torque for whole range of speeds where we want to operate. We keep in mind the proper cooling of the motor at low speed.
Finally we interpose the obtained torque for our application and available torque that we can obtain from our motoreducer driven by frequency drive. We easily notice that operation “our” motor in hypersynchrone zone ( above 50 Hz ) still covers our torque demand. In other words: if we operate motor between 25-100 Hz (750 – 3000 Rpm), we have neither problem with power, nor torque.

To size necessary electrical power Pel of “our” drive:
P shaft (Power on the shaft)
η (Motor efficiency)
cosφ ( Power factor ) |
= 15 kW
= 88%
= 0.9 |
As result we choose motor 1500 Rpm, 15 kW and our speed operation range would be 750 – 3000 Rpm. The necessary drive power : 19 kVA |
Profits for “our” case :
- We choose motor 15 kW and not of 22 kW that may result from first approach.
- Motor cost – lower
- Electrical control, cables – smaller
- We need 19 kVA drive not 28 kVA
- Drive cost – lower
- Panel size, control, cables – smaller
- Power supply estimation – lower
- Motor reducer – smaller
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