# AC motor load torque calculation

by RomanU
 P: 6 Hello! Not sure whether it's a right place to post. My question is from a field of electric drives. I can't figure out how to choose an induction motor for a certain load. Load is a simple cylindric flywheel fitted directly on a rotor. Flywheel has a known mass and dimensions. I know that a motor would accelerate untill it's torque will be equal to the load torque. So to select an appropriate motor I need to know it's torque rating(torque at ). The question is how to calculate the load torque of a flywheel? Excuse me for my English. Regards Roman
 P: 2,378 Well the torque will be whatever aerodynamic drag you have on the cylinder and the bearing loss. But, if you have an AC induction motor it will NOT accelerate until the load torque = motor torque. AC induction motors have their speed regulated by line frequency. Once this machine is up to speed is there any other load besides the bearings and air friction? What are the dimensions of the cylinder?
 P: 6 I think a motor will accelerate until it's torque equals a load torque. And the speed therefore depends on the load (assuming that frequency and voltage remains the same). Below is an illustration. Several motor torque-speed characteristics are shown there. This cylindrical flywheel I described is a "constant torque" load i.e. torque doesn't depend on the speed rotation (black line in the picture). The only thing I can't understand is how to calculate this load torque. Averagesupernova: Cylinder may have any diameter, height and mass. I need a general formula to calculate a torque in order to select an appropriate motors.
P: 4,667

## AC motor load torque calculation

You are correct in stating that a motor speeds up until its torque equals the load torque. But no torque is required to maintain a constant RPM on a flywheel (excepting for friction and windage). Flywheels are used to supply high instantaneous torque to a load, while the motor is used to replace the flywheel energy lost. A typical application of a motor flywheel combination is for an air compressor, or a motor flywheel alternator system, when the alternator output is pulsed. In these cases, the motor needs enough torque to speed up the flywheel from the minimum to maximum RPM between compression strokes.
 P: 2,378 You both are INCORRECT in stating that a motor (at least AC induction motors, and AC is in the title of this thread) will speed up until load matches motor torque. An AC induction motor will not contiuously speed up with no load. As long as the load torque is within the motors specs, it's speed will be regulated within a few RPM of nameplate spec'd speed regardless of speed. Universal series wound motors such as a portable electric drill motor are a different story. They require some sort of load to prevent them from running wild. - So Romanu, you tell US what type of motor you want to use.
 P: 6 I was talking about AC induction motor, asynchronous motor in other words. Averagesupernova, I cant agree with you. AC motor speed depends on: voltage, frequency, and a load. Changing any of them leads to a change in speed. As far as I know without a load a motor will speed up untill it reaches the noload point (closer to the horizontal axis on the graphic). Motor won't continiously speed up because it's rotor itself is a load. I'll state a problem in other words: How to find out mass and dimensions of the flywheel (a cylinder) so that a motor will be running at nominal speed? I know how to do it with other types of loads: hoisting, belt conveyor, but not with a simple cylinder on the shaft.
 P: 2,378 Now are you talking about a change in speed of a few RPM or what? I will admit that much. But a 20% change in RPM from load to no load is NOT an induction motor or is a damn poor one. - I think you better have a read here: http://en.wikipedia.org/wiki/Induction_motor
 P: 4,513 The greatest torque on the motor will be in acting against the angular inertia of the load. This is greatest at omega_0. You may want a motor with a significant start-up phase.
 P: 2,378 Since we don't even know how large this 'flywheel' is, it's hard to say what type of motor to use. I have several bench grinders with a grinding stone on each end. I'm pretty sure one of them is 1/3 HP 3500 RPM motor. I suspect split phase, not capacitor start.
 P: 619 RomanU, you asked: How to find out mass and dimensions of the flywheel (a cylinder) so that a motor will be running at nominal speed? I know how to do it with other types of loads: hoisting, belt conveyor, but not with a simple cylinder on the shaft. The point is, there is no special how to at all. A flywheel is no load in the sense of taking power away from the motor (it stores energy, so it does not remove energy from the motor), and thus requires no energy at all, other than the very small amount consumed in its bearings. About the only concern is to avoid putting too large of a flywheel on the motor, one that will prevent the motor from coming up to speed quickly during startup. If startup is dragged out, the high currents associated with start up can cause overheating in the motor. During startup, the motor is supplying power to the flywheel bring it up to speed. It is only after the motor and flyhweel are up to steady operating speed that the flyhweel takes no operating power from the motor. With a large flywheel, a small flywheel, a fat flywheel, or a skinny flywheel, or no flywheel, the induction motor is going to run very close to rated speed when there is no load on the motor.
 P: 6 Ok. Thanks a lot, Dr.D. Now everything is clear to me with a flywheel i.e. it's no load. Averagesupernova, thanks for the wikipedia link. But where is it stated that RPM doesn't depend on a load torque? I know the Equation of motion: dw/dt = (Te-Tl)/J I might be wrong, but doesn't it mean that acceleration stops at the moment when Te=Tl (motor torque = load torque)? As you guys clarified a flywheel is almost a zero load (low torque), that's why motor runs at speed close to synchronous. But if there is load torque RPMs change. Am I right?
 P: 28 With an AC induction motor, when the load increases, torque produced by the motor will increase to maintain RPM. When the load decreases, the torque produced by the motor will also decrease so the RPM is somewhat constant. This occurs from the amount of "magnetic slip" of the rotor and the driving phase or phases. The rotor must lag the current by some amount for the motor to run. The further the rotor slips from that range, the higher the induced current and magnetic field. This property keeps RPM somewhat constant with varying torque to match the load, only fast load variations cause entire phases to slip, the flywheel is there to store energy during those periods.
 P: 6 bipolar, Thanks, I was wrong than.
 P: 28 It isn't very intuitive, especially when working with different types of motors, where the formula you posted for speed is correct. I scratched my head the first time I saw the internals of one. I took it as handwavium proof that Physics works at the time.
 P: 4,513 Some of these posts might sound a bit misleading. There is a certainly energy, converted from the line, through the motor, to kinetic energy stored in the the flywheel. The rotor itself will also have stored kintetic energy. Are you using single phase or three phase power?
 P: 6 3 phase supply.
 P: 4,513 Somehow I lost track of this thread. The quantity you are looking for is called moment of inertia. It serves the same function as mass in the equation F=ma Momentent of inertia, $$I = \int r^2 dm$$ F=ma is replaced with $$T=I\alpha$$ T is the torque and $\alpha$ is the angular acceleration. $\alpha = d\omega / dt$. $\omega$ is the angular velocity.
P: 619
 Quote by Phrak The greatest torque on the motor will be in acting against the angular inertia of the load. This is greatest at omega_0. You may want a motor with a significant start-up phase.
Phrak, I think this is in error, at least to some degree.

The greatest motor torque will exist when working against the greatest combination of inertial load (I*alpha) plus load torque. I think you assumed that the motor is starting with no load, and that may, or may not, be a good assumption.

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