vfd motor-torque & power characteristics

What you were told is right.

Using VSD (Variable Speed Drive) with induction motor, you can get as maximum constant torque as motor rated-torque for any delivered power up to the motor rated-power.

that is OK under the condition of: Power_output = Torque x Speed

The max. power (rated-power) should be respected, i.e. if your load torque = rated torque of the motor, then the load speed should not exceed the rated speed. Such a case is called Constant-Torque-Mode-of-Operation.

If you need to run a mechanical load over the rated speed, then you have to reduce the load torque to maintain the power at its max. value (rated power). Such a case is called Constant-Power-Mode-of-Operation.

Hence, the drive rated-power, as well as the motor rated-power, should match the driven load-power at its (load) maximum speed.

Find the load power at 1750 rpm and choose you motor-drive set above this value.

Good Luck... Samak

Mayby, there is an other solution...

If you have a pump and you let it run faster, you will get more pressure and more volume per time unit out of your pump. Ofcourse you will need more input power at the pump shaft, because you get more water at the output side of the pump.

So at the pump shaft you will need a higher torque at the new higher rotation speed.

In case of an centrifugal pump, the relation old pump shaft torque to new pump shaft torque is equal to the thirth power of the new speed to the old speed.

This means that if you increase the speed with from 1000 RPM to 1500 RPM you will need 1.5 x 1.5 x 1.5 equals 3.375 times the power.

If you double the speed, you need 8 times as much power than original.

On the other side you have you motor.

Your motor is designed for a certain output power at a given speed. From this information you can calculate you output motor torque because output power = speed x torque. This is the maximal torque you can obtain without burning your motor. For this torque you need a certain electrical voltage at a certain frequency.

If you can keep the relation voltage / frequency constant, you can have the motor torque also at higher speeds. And you can have more power out of your motor.

Let me explain with a practical example.

If you have a frequency drive with an input of 400 V than the output voltage will also be 400V. If you program the frequency drive in such a way that the 400V is obtained at 87 Hz, than you get 230 Volt out of the frequency drive at 50 Hz. Because the relation voltage / frequency is kept constant in the drive.

Lets now look at the motor. If you have a motor 230/400V and a supply of 400V 50Hz, than your motor is connected in star (Y).

Now we go the change this in to triangle of delta (D). This means that the motor only can get a supply of 230V at 50Hz without burning.

This 230V at 50 Hz you get out of your frequency drive. If you increase the output frequency, than the output voltage will also increased and you motor coils will be magnetised in the correct way. This will go on until the output voltage of the frequency drive has reached 400V.

Conclusion, you get the nominal motor torque untill the frequency is 87 Hz, and the speed at 87 Hz is 1.74 times higher. So the output motor power will be 1.74 times higher (power = speed x torque). Take care that the motor current, also will be the current in the delta connection. So you need a frequency drive that is also rated 1.74 times higher.

Finally, if you can change you pullys on the motor and pump, of if you can install a stronger motor in the same shaft size, than it is possibel to get more pump output with the use of a frequency drive.

good luck

leurs.electro@telenet.be

"If you double the speed, you need 8 times as much power than original."

I always thought that power was a function of the 'square' x² of the speed (not x³) I thought in the context of air flow the formula contained a square- I am probably wrong

presure is lineair equal to the motorspeed

the quantity per time unit is equal to the square (x²) of the speed

the needed power is equal to the cubic (x³) of the speed