induction motormotortorque
In induction motor, Torque is proportional to square of applied stator voltage (I agree that) but some source says torque is proportional to current. For me something proportional to square of volt should be proportinal to square of current as well. Is that right?
This source may be meaning under constant voltage torque is proportional to current? May it be?
Relating a motor's current to a fan output airflow requires you to know details about the motor, the fan, the mechanical system surrounding the fan, and the ambient atmosphere.
The motor current depends on the motor speed, as shown by a speed vs. current curve.
The motor speed depends on the motor's speed-torque curve vs. the fan's speed-torque curve.
The fan's speed-torque curve depends on the inlet air density.
The speed of the air exiting the fan presumably depends on mechanical details such as - the pressure drop in the fan's duct work, the open/closed position of air dampers, and so on.
It should be evident there is no 'simple' relation between fan motor current and fan output airflow!
I will add two useful pieces of information.
The "Fans and Blowers Energy Efficiency Reference Guide" (published under the branding of BC Hydro, Manitoba Hydro, and others) is an excellent introduction to fan and blower design. This is required reading for all electrical engineers dealing with fans and blowers.
Contrary to your statement that fans have constant torque: Some loads have a constant torque, but fans aren't one of them. From a book by Schneider Electric we have the following table:
Here is an example of a motor's torque-speed curve and speed vs. current curve.
Here is an example of a centrifugal fan's load-torque curve.
Here is an example of a centrifugal fan's power vs. airflow curve (note - power = torque × speed.) Note that the relationship is non-linear.
In AC induction motors peak torque, that which is used to accelerate it from a stop or to recover from a step change in load, is related to the square of the voltage change. So if the voltage is 80% of normal, the peak torque capability of the motor will be only 64% of normal; at 70% voltage, you are down to less than half of the available peak torque. That means the motor is more easily able to stall and even if that doesn't happen, the slip of the motor will increase so if it pulls the same current, it will do less work, meaning the efficiency becomes poor and more of the energy consumed just becomes heat in the motor. Or if doing the same work, it will pull more current and current comes with heat. the latter scenario, more current, leads to hopefully tripping an overload protective device but the lower efficiency may increase the motor heating effects even though the current is low enough to not affect the protective device.
If the load on the motor has decreased as the voltage dropped, or the motor was originally grossly over sized, that may not be a problem. But generally a motor is selected to provide just enough power as was necessary.
Best Answer
Very likely true when the rotor is locked, mostly true when slip is a good fraction of 1, and pretty much utterly false when the motor is running close to its synchronous speed. By their nature, induction motors have zero torque exactly at their synchronous speed, no matter how much power they may be burning up for other reasons.
Are you sure that this unnamed source is talking about induction motors? For a 100% efficient motor the output power will equal the input power; at a fairly constant speed, a fairly efficient motor will have an input power roughly equal to the output power -- if you assume constant excitation voltage this would translate to a torque that's proportional to the proportion of current that's in phase with the voltage.