Load / Current / Speed Relationship in Motor 1

[noel0589]

Hi all,

I am trying to get an understanding of what goes on in a motor as load is increased. Say we have an elevator; with the weight of the cab alone, it will draw 200 starting amps and run at 120 amps at the lines. As load is increasingly applied, the motor speed will increasingly fall from near synchronous speed. This results in less internal inductive resistance and subsequently a higher current draw. So at full load, the motor will still draw 200 starting amps, but current at full speed with the load will be higher than 120 Amps. The speed of the elevator will be less as load is applied.
Also, the RPM of the rotating magnetic field is always the same. What load does is impede the "natural" movement of the rotor and a motor is designed to handle up to a load that will not impede the speed of the rotor such that the current drawn will not be so high as to burn out the windings. Please correct me if I am wrong. Thank you!!!

 

[Skogsgurra]

It is very difficult to understand what you are trying to say, noel. If you are saying that the speed goes down when the motor is loaded and that the motor current also goes up, then you are correct.

The speed decrease (known as "slip") goes from zero percent at no load to something like 2 - 5 percent (depending on motor size and rotor construction) when fully loaded. So, a 1500 RPM motor at no load and 1440 RPM at full load (if slip is 4 percent at full load) will run at about 1470 at half load.

Hope this helps.

 

[aolalde]

An induction motor has two regions. Starting area; below the breakdown torque and load area; after the breakdown torque.
Each motor has specific design parameters but in general the breakdown torque matches around 90% of the synchronous speed (the speed of the magnetic field, Ns = 120*Freq/Poles).

To produce torque the rotor current must increase ( F = Bg*L* Ir). Bg is the air-gap flux density, L rotor bar length, Ir rotor bar current and F is the force developed in one rotor bar.

The current will be proportional to the rotor induced voltage and bar impedance.
Ir = Vr/Z
The voltage induced must increase to produce the current required to match the load torque (Vr = k*dFlux/dt). For that reason the rotor speed reduces and allows the rotor bars to cut the magnetic flux at a higher rate (increased slip). The torque load could be increased until the maximum torque or breakdown torque is reached, at that condition the rotor stalls. Note that the rotor frequency (Fr) is very low in the load area (Fr = slip* Fl)as compared to the line frequency (Fl). In reality the motor is only allowed to exceed around 10 to 15% the full load current to prevent overheating damage.
The starting condition is another story.

 

[DickDV]

noel, I don't know if you are dealing with IEC or NEMA motors but, either way, get a torque-speed curve for the motor type you are interested in.

That curve reveals more about a motor in simple form that anything else I've seen.

A slip-current or torque-current current is less common but probably available somewhere. I would start with the manufacturer.

 

[noel0589]

aololde, your post is highly knowledgable, so much so that I don't know if what I wrote is wrong or if you are giving me a more in depth analysis of my basic understanding. Please let me know as I try to study your post. Also, please check the other post from before on “Problem / Failure Analysis”. I had a question for you. Thank you!!!

 

[aolalde]

I will try to analyze your post:

”I am trying to get an understanding of what goes on in a motor as load is increased.
Say we have an elevator; with the weight of the cab alone, it will draw 200 starting amps and run at 120 amps at the lines. As load is increasingly applied, the motor speed will increasingly fall from near synchronous speed.” That is correct.

“This results in less internal inductive resistance and subsequently a higher current draw.” You are partially correct in reality it is the motor impedance change. Impedance is the resultant of Resistance and Inductance.

“So at full load, the motor will still draw 200 starting amps, but current at full speed with the load will be higher than 120 Amps.” Correct , the 200 amperes starting current are function of the line voltage.

“The speed of the elevator will be less as load is applied.” Correct the slip increases when load torque is increased to the motor shaft. slip; s = (RPMsyn-RPMshaft)/RPMsyn

”Also, the RPM of the rotating magnetic field is always the same.” Correct, the synchronous speed ( RPMsyn) depends on the line frequency and motor winding number of poles.

“What load does is impede the "natural" movement of the rotor”. Correct

“and a motor is designed to handle up to a load that will not impede the speed of the rotor such that the current drawn will not be so high as to burn out the windings.” Not so correct, electric motors will try to take any load imposed until it stalls and it will burn out unless an overload relay is added.

 

[jraef]

Excellent response. I voted you a star.

"Venditori de oleum-vipera non vigere excordis populi"