I have a three phase induction motor with name plate details:

  • frequency = 50 Hz
  • PF = 0.86
  • RPM = 1475
  • efficiency = 94.2
  • voltage = 415 V
  • ampere = 77 Amps
  • HP = 60

Now this motor is operated at 40 Hz and would like to calculate the load % of this motor.

So, I followed this fact sheet page 3, and used this formula to calculate load %.

$$\text{load}_\% = \frac{P_i}{P_n} \cdot 100$$

where: \$P_i\$ - input power, \$P_n\$ - rated power

$$P_n = \frac{\text{HP} \cdot 0.7457}{\eta_{fl}}$$

where: \$\eta_{fl}\$ - efficiency at full load

To find out \$P_n\$, I used the rated value of HP = 60 and \$\eta_{fl}\$ = 0.94. As the motor is operated at 40 Hz, which is less than the rated frequency (50 Hz), is it correct to use 50 Hz motor rated readings to calculate load % which is operated at 40 Hz?

Or do I need to make changes in the calculation?

  • 1
    \$\begingroup\$ Can the load current still be 77A, or must it be less? Heating is the main limitation. Small motors are often cooled with a fan on the same shaft, so cooling falls at lower speed, limiting I. Big motors are often cooled by an external fan, and full cooling is maintained down to stand-still. Which is yours? \$\endgroup\$ – Neil_UK Dec 16 '17 at 7:36

When operating an induction motor at a lower frequency, the most important change in performance is the reduction in speed. The synchronous speed is given by RPM = 120 X f / P where f is frequency (Hz) and P is the number of motor poles (an even integer number). The rated speed at 50 Hz for the given motor is 1475 RPM. The synchronous speed is a little higher. The nearest speed that satisfies the synchronous speed formula at 50 Hz is 1500 (1500 = 120 X 50 / 4), so this is a 4-pole motor. Since the rated speed is given as 1475 RPM, we know that the slip at rated load is 1500 - 1475 = 25 RPM. Horsepower = Torque (Lbs - Ft) X RPM / 5252, so the rated torque is 214 Lbs. - Ft. Since the motor's rated power is given in horsepower rather than watts, I am assuming that all of the ratings should follow the USA units of measurement. Units can be converted as necessary.

At 40 Hz, the synchronous speed will be 1200 RPM. The slip at rated load can be held to 25 RPM at reduced frequency by reducing the voltage the proper amount, approximately maintaining a constant V/Hz ratio. That means the rated speed will be 1175 RPM at 40 Hz. The rated load torque can be maintained at the 50 Hz value. Rated power at 40 Hz will then be 216 X 1175 / 5252 = 48.3 Hp.

If the motor voltage can be controlled in a way that will maintain the slip RPM, the motor's rated torque can be maintained all the way to zero speed. In order to operate at that torque continuously, the motor will need to be adequately cooled. At 40 Hz, the motor may not need any special cooling provisions. Below that, adequate cooling will become progressively more difficult. Determining and maintaining the optimum voltage also becomes progressively more difficult as the frequency is reduced.

With the optimum voltage applied, the motor current will be approximately proportional to torque and will remain constant as the speed is reduced with constant load torque.

  • \$\begingroup\$ This in agreement to my answer. Torque is same. Hp is reduced 80% of rated and pF, excitation loses are increased 25% which are normally only 10% of full current loss \$\endgroup\$ – Sunnyskyguy EE75 Dec 16 '17 at 19:40

No load excitation current depends on some friction but mainly inductive reactance. Thus no load current rises by 50/40 or +25% which adds conduction loss and reduces full load torque current allowed. Since Hp =k * Torque * RPM and RPM is controlled at no load by line f , another reduction in Hp to 4/5 or 80% of rated Hp with about a 3% rise in winding temperature

Torque is the same from starting but due lower back EMF from 80% Rated RPM, torque at rated speed increased for the same applied voltage, conduction losses increase about 5/4 or 125% at no load and 105% at full load current. RPM is 4/5 of rated speed and Hp is 4/5 of rating for same current but at 5% high ‘C rise from lower induction impedance losses.

In summary, slightly hotter, 20% slower but a bit more torque avail. due to lower BEMF. Which may cause more bearing wear if unbalanced from vibration displacement at lower f. And efficiency drops up to 5%


Actually, you cannot calculate the load ON a motor, you can only measure it. The load is whatever you connect to it.

You can calculate the load CAPABILITY of a motor, and that would follow what Mr. Cowie has said. More simply, (F2/F1) x P1 = P2, where F = Frequency and P = power. So 40/50 = .8, x 60HP = 48HP. So at 40Hz, you motor is now CAPABLE of 48HP. the magic of the VFD is that this motor is still capable of delivering the same TORQUE at 40Hz as it did at 50Hz, and most machines work on torque, not power. Power in a rotating machine is simply a shorthand expression of torque at a given speed so when speed drops, so does power.

But just because you have lowered the speed does NOT necessarily mean you have lowered the load on the motor. Probably, bot not absolutely. That's technically up to your load profile.


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