Theoretical Motor vs. Actual Motor

[EITengineering]

Hi Everybody!

I have a contradiction between theoretical induction motors and actual induction motors I can't seem to wrap my head around.

In theory, motor efficiency is characterized by 1-s where s is the motor slip, i.e. Pout = Pag(1-s), where Pag is the motor air gap.
This implies that maximum efficiency occurs when the motor speed is maximum and s is a very low number.

In reality, I understand motor efficiency is maximum when the motor is running a full load when the speed is not at its fastest. Motor speed is faster when there is less load, yet motor efficiency sharply decreases at smaller loads.

Can anyone explain this?

 

[waross]

Try "In theory, motor efficiency SPEED is characterized by 1-s where s is the motor slip," And 1 is synchronous speed.
"In theory, motor efficiency is characterized by 1-s where s is the motor slip," This statement has no basis in reality.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Pout = Pag(1-s), where Pag is the motor air gap.

Correct.

This implies that maximum efficiency occurs when the motor speed is maximum and s is a very low number.

You are correct that the particular ratio Pout/Pag = (1-s) is a maximum when s is a very low number. Pout differs from Pag primarily due to rotor I^2*R rotor losses, and these losses are in fact lower fraction of Pag as slip decreases. For this reason many high-efficiency motors have lower full-load slip than older motors (this is necessary to minimize rotor losses... it also generally corresponds to lower rotor resistance).

If rotor I^2*R losses were the only ones to consider in an induction motor, then the efficiency would act as you describe. However there are other losses to consider. Efficiency is ratio Pout/Pin (not Pag/Pin). To see how effiency behaves we would also need to consider also effects of other losses included in Pin: stator I^2^R losses and stator core losses and rotor core losses and stray losses.

In reality, I understand motor efficiency is maximum when the motor is running a full load

Not necessarily. Generally somewhere betwen 50% and 100% load. See data sheet for a particular motor if you want an idea of where peak efficiency occurs

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(2B)+(2B)' ?

 

[LionelHutz]

(1-s) simply is not directly proportional to the efficiency. It's one of the terms involved in calculating the motor losses.

Efficiency = Power out / Power in
Efficiency = Power out / (Power out + Power losses)

As the motor is loaded, the power output increases much more than motor power losses increase. So, even though the slip increases and the losses increase, the efficiency still improves.

As a quick check, the motor speed is the highest at no load yet the motor efficiency is also 0% at no load meaning your theory on the efficiency being directly proportional to speed must be wrong.

 

[7anoter4]

What electripete and LionelHutz said is seems to me to be as follows:
If the rotor is locked an induction motor acts as a transformer behaves and the power transmitted to rotor will be
the input power minus the copper stator losses and steel losses[for both: Stator and Rotor].
Pag[internal power]=Pinput[from the electrical supply source]-StatorLosscopper-Losssteel
This power [Pag] will cover the rotor losses and [if it is wound rotor ]the losses in the starting outside resistances.
The magnetic field is "rotating".This field interacts with the induced current in rotor and produces a torque.
Actually, the combined magnetic flux of 3 phases winding system symmetrically disposed on stator periphery will act as if there is a
pole rotor. In the same time it could be demonstrated the relationship between Pag and Torque and synchronous velocity
Pag=k*T*rpmsynch k=constant depending upon unit system.
rpmsynch=frq*60/p p=no.of pole pairs
When the rotor starts to rotate Pag=phno*Rr*Irot^2+Pmec
phno=rotor number of phases
Pmec is the active power =Pout+mechLosses.
mechLosses=ventilationlosses+frictionlosses
The Torque is the same T=Pag/rpmsynch=Pmec/actual rpm
since (rpmsynch-actual rpm)/rpmsynch=s[slip] then Pmec=Pag*(1-s)
Then Pag-Pmec=phno*Rr*Irot^2=Pag*s [rotorcopperloss]
Now, efficiency=Pout/Pinput=(Pinput-StatorLosscopper-Losssteel-RotorCopperLoss-ventilationlosses+frictionlosses)/Pinput
Pinput=Pag+StatorLosscopper+Losssteel
Pout=Pag*(1-s)-ventilationlosses-frictionlosses
efficiency=Pout/Pinput=(Pag*(1-s)-ventilationlosses-frictionlosses)/(Pag+StatorLosscopper+Losssteel)
If ventilationlosses+frictionlosses depends on 1- s, StatorLosscopper+Losssteel does not.
In no-load regime Pag*(1-s)=ventilationlosses+frictionlosses and efficiency will be 0
As 1-s is close to 1, if Pag will increase then efficiency will rise. But since StatorLosscopper rises with Pag^2 from a certain Pag
efficiency will stop to rise and at the end will start to decrease.