The slip is an important value when it comes to the calculations of the steady-state running conditions ( input current, input real power, power factor, efficiency, output torque, etc. ) of an induction motor using its equivalent circuit.

With the speed in per unit, the slip is defined using following equation :

The scientific literature has many articles on the subject of motor calculations using the equivalent circuit but I find that some important issues with the slip value are rarely discussed.

### The basics

The steady-state operating point of an induction motor is the intersection of the torque-speed characteristic of the motor with the torque-speed characteristic of the mechanical load. The steady-state slip of a motor is the slip at this intersection point.

The torque-speed characteristic of an induction motor is a function of the stator voltage.

**Therefore the steady-state slip is a function of the loading and a function of the stator voltage.**

What slip value should we use for the calculations with the equivalent circuit ?

We need an accurate value for the slip because the torque of an induction motor varies a lot for very small variations of the speed (and thereby of the slip) in the operating region (see image 1). The slope in the operating region is even steeper with big induction motors.

Let’s see some options that we have:

### Option 1 : Get a measurement of the motor speed

Using a tachometer.

That allows us to use equation 1 for the slip.

Resulting equivalent circuit calculations are very accurate.

If the stator voltage or the loading changes, we need another measurement of the speed.

The calculated torque gives us an important information about the amplitude of the load torque at that speed.

But we rarely have a measurement of the speed.

### Option 2: How about using an approximative slip value in equivalent circuit calculations ?

If you are using the nameplate full load speed, be advised that NEMA and IEC standards allow a 20 % error on nameplate slip speed. For a 1800 rpm motor with a nameplate full load speed equal to 1750 rpm, it means that the full load speed can be any speed between 1740 and 1760 rpm.

The slip speed is the difference between the synchronous speed and the full load speed.

If the approximative slip is a full load slip, it must be scaled using an estimate of the mechanical loading (if the loading is 80 %, the full load slip must be multiplied by 0.8).

Often the voltage for our calculations is not the rated voltage. **It is very difficult to adjust an approximative slip for voltage deviation**. The adjustment depends on how the load torque varies with the speed. This adds to the inaccuracy of option 2.

Nameplate values are values at rated voltage.

Let’s analyse the accuracy of option 2 with a small example:

Question: Let’s suppose that, for a specific loading at a specific voltage, the exact slip of a running induction motor is 0.0175 (1.75 %) but we don’t know it because we have no measurement of the speed. What happens if we use a slip equal to 0.0100 (1.0 %) in the calculations ?

Answer: Even though we make a very small error (0.75 %) on our estimate of the slip value, we are making an error of about 43.% on the torque (1.0/1.75 = 0.57). We are underestimating the steady-state torque.

For a small 0.75 % error on the slip we ended up with an error of about 43 % on the torque (and a similar error amplitude on the real powers). Not good.

This option is easy but the calculated results can easily become very inaccurate.

Induction motors are nonlinear electrical components. We need to put more effort into the calculations.

Final note about option 2: **A generic slip value taken from the scientific literature should never be used for calculations with the equivalent circuit. This is simply not accurate enough.**

### Option 3: Find the intersection point

As mentioned, the intersection of the two torque-speed characteristics is the steady-state operating point. **The slip at the operating point is the best slip value we can have without a measurement**.

Like option 2, option 3 requires an estimate of the mechanical loading.

While option 2 was very simple, here we need some computing to find the solution. But the results are more accurate than option 2. And we can find a good solution at any voltage.

SimPhase IMC Program uses option 3. It can give accurate results for any voltage deviations (even large ones).

### Conclusion

Induction motors are very nonlinear machines. Simple calculations are often not a good idea.

If we have no measurement of the speed, computing the operating point (using an estimate of the voltage and an estimate of the loading) is the best option. It is a little more complex but it is a lot more accurate.