Motor Starting Active Power?

[Electrichawke]

Hi all,

I have been yearning to ask this question to our experts here for my enlightenment.

When an induction motor whose shaft is connected to a pump is started, a starting current is generated to develop the locked rotor torque required to move the rotor shaft at rest.

The current would be high which is synonimous to transformer energized with its secondary short-circuited and its purely reactive.

Is there any active power (watts) generated during the initial application of voltage (locked rotor torque)? The power factor During starting is low (0.15 to 0.25) based on a starting current say 6.5 x FLA.

 

[Electrichawke]

I have seen folks calculate a “starting kVA” based on LRC x motor rated voltage x sqrt(3) and then apply the Starting PF = 0.15.

This will result to kW and kVars at starting but Im not convinced of the watts generated as the motor shaft just begun to turn.

 

[Compositepro]

As soon as there is any motion in the shaft there is real power that is created. Mechanical power is torque times rpm.

 

[crshears]

"Highly reactive" I would answer Yes to; "purely reactive" not so much. As the OP points out, the power factor of a multiphase induction motor at rest / upon start-up may [will] be very poor, but truly "purely reactive" would imply no real power consumed / converted into rotary motion, hence zero torque, hence motor would not be self-starting . . . but they are self-starting.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[Compositepro]

I2R losses will be most of the real power consumed until the motor speeds up.

 

[dpc]

If the shaft is turning, kW are being used. Also resistive power loss in the motor windings.

 

[che12345]

Dear Mr. Electrichawke (Electrical)(OP)14 May 22 12:34
".... I have been yearning to ask this question ....#1. When an induction motor whose shaft is connected to a pump is started, a starting current is generated to develop the locked rotor torque required to move the rotor shaft at rest.... #2. The current would be high .....Is there any active power (watts) generated during the initial application of voltage (locked rotor torque)? The power factor During....."
#0. The word " generated " [by the load? } is misleading. I think you mean power "supplied " to the load.
#1. At the instant when the power is applied to a locked rotor motor, there is active power (watts) consumed by i[sup]2[/sup] R loses in the rotor and the stator. As the rotor is NOT rotating, there is No mechanical output.
#2. When the rotor starts to move, there is active power (watts) consumed by [ stator, rotor and the mechanical load]. At this instant, the active power (watts) can be calculated by the starting power factor (generally <0.4).
Che Kuan Yau (Singapore)

 

[itsmoked]

Great gobs of active power must be present as demonstrated by these products!

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Gr8blu]

electrichawke How much "foam" gets you the proverbial buzz? And how much "liquid"?
The answer is the same - foam is reactive power, liquid is active power.
When the switch is closed, the winding gets energized. Enough energy (in active ampere) and you create a usable magnet. Until that happens, no relative motion between rotor and stator winding. Once you're sufficiently magnetized, a bit more active ampere gets you turning. Once you're turning, the need for "gobs" of ampere drops and things get a lot easier - quickly - as evidenced by the steady increase in observed power factor.

Is there REACTIVE power at start? YES. Is there ACTIVE power at start? YES. Does the proportion change THROUGH start? YES.

Converting energy to motion for more than half a century

 

[electricpete]

Where does the real power flowing into the motor go?

At the moment power is applied and before the rotor is moving, the real power is going primarily to I^2*R losses and the mechanical power is zero (since speed = 0).

As speed increases (let's say from zero speed up until at least half speed) the current magnitude continuously decreases, the I^2*R losses continuously decrease and the mechanical power (product of torque times speed) continuously increases. A portion of the mechanical power during acceleration will go into accelerating the train's rotating inertia (it increases the kinetic energy of the inertia) and a portion will go to match the friction torque and load torque as a function of speed (those mechanical torques transfer energy into things like friction heating and maybe fluid energy).

> The answer is the same - foam is reactive power, liquid is active power.

Ah, that brings back memories. And makes me thirsty!

=====================================
(2B)+(2B)' ?

 

[LionelHutz]

Considering the windings have resistance, there is real power or watts being used by the motor the instant current begins to flow even if the shaft hasn't yet started to rotate.

I think that the question being asked is based on output power being proportional to speed with the thought being that no real power is consumed by the motor until the shaft is rotating. Possibly also that there should be almost no real power being consumed by the motor when it just begins rotating at very, very low rpm. This simplification doesn't work, the motor has internal losses that consume real power.