SimPhase Induction Motor Calculations (IMC) program has 4 options. It does the calculations for one three-phase induction motor at a time.

This post is about option 4.

Basically, **option 4 uses the knowledge of some running conditions at one steady-state operating point and it can then calculate the motor running conditions for various other combinations of terminal voltage, frequency and speed control settings**.

Let’s do three simulations with option 4.

The same 25 HP, 460 Volts, 1500 RPM (50Hz) three-phase induction motor is used in all 3 simulations.

Here is the mechanical load for all 3 simulations : The measured loading is 80% of motor rated power when the terminal voltage equals 100% of its rated value. Measurements were made with the motor having no speed control. Also the load torque is assumed to be proportional to the square of the speed (for example the load could be a water pump).

Simulation 1 : Let’s reproduce the conditions when the 80% mechanical loading was measured. The terminal voltage equals 100% of its rated value. The motor has no speed control. Because at this operating point we have no measurement of the speed, both the speed and the slip are unknown before the calculations.

Image 1 shows the running conditions calculated by option 4.

Ideally the power output would equal 80% of motor rated power (by definition it equals the loading). The program calculated a 80.16% power output. Calculated speed is 1461.6 rpm .

Simulation 2 : Same motor and same mechanical load. No speed control. What are the conditions of the motor if the terminal voltage amplitude falls to 90 % of its rated value ?

Image 2 has the results.

Calculated efficiency equals 86.47% (it was 87.84% in simulation 1). Calculated speed equals 1451.9 rpm.

Simulation 3 : Same motor and same mechanical load. But now we add a variable frequency drive (VFD) speed control. What are the steady-state running conditions if the motor speed is set equal to 1000 rpm ?

Image 3 has the calculated conditions.