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Wow!

Mechanical power of synchron motor

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On page 362 of wildi there is the following equation for mechanical power per phase in watts.  

P= (E_0 *E/ Xs)* sin (torque angle)

I’ve run across a similar equation when practice problems ask for max power , the angle is 90 degrees so

P= (E_0 *E/ Xs)

How do we know when to use this equation instead of just dividing the rated KVA by power factor and efficiency to get Watts?

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8 hours ago, Wow! said:

On page 362 of wildi there is the following equation for mechanical power per phase in watts.  

P= (E_0 *E/ Xs)* sin (torque angle)

I’ve run across a similar equation when practice problems ask for max power , the angle is 90 degrees so

P= (E_0 *E/ Xs)

The power output in watts of a synchronous machine depends on the difference in angle between the internal (or "induced") voltage and the terminal voltage. If you are given enough information in the problem to suss out the angle difference between the two (typically represented by the greek letter delta), and are asked to solve for the active power delivered or the maximum amount of active power that the machine can deliver, then use that formula. 

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4 hours ago, Zach Stone, P.E. said:

The power output in watts of a synchronous machine depends on the difference in angle between the internal (or "induced") voltage and the terminal voltage. If you are given enough information in the problem to suss out the angle difference between the two (typically represented by the greek letter delta), and are asked to solve for the active power delivered or the maximum amount of active power that the machine can deliver, then use that formula. 

Thank you.  Maybe I have been confusing Pout and P_in.  

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