# Calculate ideal unloaded rotor speed

I have the following problem: MY ATTEMPT

Okay, I've found two formulas that will definitely help me solve this problem:

The first is $$\S=\frac{N_s-N}{N_s} \$$, where $$\S\$$ is the slip, $$\N_s\$$ is the synchronous speed, and $$\ N\$$ is the rotor-speed.

The second formula is $$\N_s=\frac{120f}{P}\$$, where $$\ f\$$ is the frequency and $$\P\$$ is the number of poles. Combining these two formulas and isolating $$\N\$$ we get:

$$\N=\frac{-120Sf}{P}+\frac{120f}{P}=\frac{-120 \cdot 0.05\cdot350 \text{Hz}}{8 \:\text{poles}}+\frac{120\cdot 350\text{Hz}}{8\:\text{poles}}=4987.50 \: \text{RPM}\$$

So a rotor speed of 4987.50 RPM. But is this what is meant when they ask for the "ideal unloaded rotor speed" or have I calculated something different?

I hope someone can clarify this for me.

• Do you know the definition of "ideal" in this context or in engineering in general? Do you know why induction motors have slip?
– user80875
May 19, 2020 at 14:01
• No not at all too be honest. My instructor has barely touched on motors and that subject, so I am not too confident in this part of electrical engineering. According to Andy aka's it seems like the nominal slip tells how much lower the rotor speed is compared to the synchronous speed. But I am not sure WHY induction motors have slip in the first place.
– Carl
May 19, 2020 at 14:11
• Carl, you have thought enough about the question to doubt that you understand it. I believe that is the intent of the question. It is not a question that should have been asked of someone with your experience. I have explained in my answer.
– user80875
May 19, 2020 at 14:35

Slip in an induction motor is what makes the motor capable of producing torque. Nominal slip is the slip that the motor has when is is producing the nominal, AKA rated or nameplate torque.

In the normal load range, the slip of an induction motor is a fairly linear function of torque.

In most engineering contexts, "ideal" means neglecting all losses and characteristics that make things complicated.

However neglecting those things does not mean that they can always be neglected or forgotten. An unloaded motor is one that has nothing connected to the motor shaft. That does not mean that no torque is being produced, there are still mechanical losses consisting of bearing friction and aerodynamic drag inside the motor and due to the cooling fan that may be on the end of the shaft opposite the drive end. Those losses are neglected fo an "ideal" unloaded motor. That means that the speed of an ideal unloaded induction motor is the synchronous speed. Mention of the slip could be considered something thrown in to make this a trick "question." Asked of someone who should have learned a lot about motors including their losses, it is a question that is designed to make the student think carefully about what is being asked.

• I would express the answer as below but arbitrarily close to synch speed. May 19, 2020 at 14:33
• Yes, that addresses the unasked question: "how does an ideal motor actually reach synchronous speed"? Even the Steinmetz equivalent kind of falls apart for an "ideal" motor.
– user80875
May 19, 2020 at 14:41 The key words here are "ideal" and "unloaded"

I have not regarded the word "nominal" to mean it has a 5% slip at full load because it would be impossible to accurately predict what slip it has when unloaded. I have used this as the definition of the word nominal: - To my way of thinking this means that the "nominal slip of 5%" means at virtually no-load (barring the windage and friction of the rotor). Having said that, many induction motors will have a full load speed at around 90% to 95% slip; such as is depicted in this picture: - Nevertheless, I have used the word "nominal" to mean very lightly loaded.

A nominal slip of 5% (rotor unloaded as per the pasted text in the question) means that it is rotating at 5% lower than synchronous speed. Synchronous speed for a motor with 8 pole pairs operating from a supply of 350 Hz is 5250 rpm (Calculator here) and so 95% of this is 4987.5 rpm.

But is this what is meant when they ask for the "ideal unloaded rotor speed" or have I calculated something different?

I think it is correct.

But I am not sure WHY induction motors have slip in the first place.

If the rotor was travelling at synchronous speed, it would be travelling at the same rotating speed as the revolving magnetic field hence, to the rotor, there is no $$\\frac{d\Phi}{dt}\$$ to induce a voltage into the shorted rotor and therefore, there can be no magnetism produced by the rotor hence it can't spin at synchronous speed.

It has to spin slightly slower than sync speed in order for a voltage to be induced in the rotor so that rotor current can flow in the squirrel cage and, produce a counteracting magnetic field (that allows a torque to develop that produces motion).

– user80875
May 19, 2020 at 14:05
• @CharlesCowie what do you think the real question might be? May 19, 2020 at 14:09
• @Andyaka I would interpret "nominal" slip to mean slip at nominal full load. May 19, 2020 at 14:23
• @BrianDrummond the question says "unloaded" in the pasted text. May 19, 2020 at 14:26
• @Brian Drummond; I would say that the question is intended to make the student think carefully about what is really being asked.
– user80875
May 19, 2020 at 14:31

"So a rotor speed of 4987.50 RPM."

That's what I got. Unless your course has given you a definition of nominal slip that doesn't use the "ideal unloaded rotor speed" you've finished.

No slip = no relative motion = no induced currents = no torque. I interpret "ideal unloaded rotor speed" as synchronous speed since that is the ideal speed when zero torque, internal or external.

But here I would be inclined towards the 5% slip and assume by "nominal" they mean the slip when no external load applied since it still takes force to turn an unloaded rotor and therefore requires slip. 5% slip seems rather high though for no load.