Alternator torque when connected to a phase-shifted AC source

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The discussion focuses on modeling the torque experienced by an alternator connected to a phase-shifted AC source. The initial approach involves treating the alternator as an AC power source in series with a resistor, leading to confusion when connecting two alternators with a phase difference. The user grapples with the implications of phase shifts on torque and power, questioning why the power remains consistently signed regardless of the phase difference. Insights reveal that connecting alternators out of phase can result in high torque and potential short circuits, emphasizing the importance of proper phasing. Ultimately, the user finds clarification in a referenced paper that addresses the interactions between the currents and voltages of the alternators.
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An alternator rotates at constant frequency. It is connected to a phase-shifted AC power source at the same frequency. What is the torque experienced by the alternator depending on angle?
Hi,

I am trying to figure out the torque experienced by an alternator to plug into the swing equation.

I am not sure how to exactly model the alternator for that. With my current approach, I only get confused.

My idea is to model the alternator as a AC power source with fixed frequency in serial with a resistor. If I then short-circuit the alternator, the torque experienced is directly proportional to the produced power, right?

So say the frequency f=1Hz, the peak voltage is Umax=1V and the resistor is R=1Ohm.

Then, the voltage of the AC power source over time is Us(t)=Umax*sin(2pi*f*t)=sin(2pi*t) volt.
And the current over time is I(t)=Us(t)/R=sin(2pi*t) ampere.

Therefore the power over time is P(t)=Us(t)*I(t)=sin(2pi*t)*sin(2pi*t).

So far this is correct, or is it? The power is always positive (or always negative, depending on sign).

Now, I am confused what happens if I connect it to another alternator instead of short circuiting. Say an alternator with the same properties, but a phase shifted by angle phi.

What I would expect to happen is that the torques experienced by both alternators, integrated over one full revolution, would be in the direction that reduces phi towards zero, if phi is small to begin with. Say phi is in the interval [-pi/2, pi/2] at the start, then the torques would cause phi to reduce, if the alternators were not fixed to a given frequency.

However, in my model above, if I would connect two alternators in parallel, I get confused.

The alternators are modelled as AC power sources with a resistor in series again. The resistor is connected to the phase of the alternator. By connecting them in parallel, I mean connecting their grounds together and connecting the resistors together, such that we get a loop. I can draw a picture if required, but right now I have no paper.

Now we can again take the voltages over time which are:
U1(t)=Vmax*sin(2pi*f*t)
U2(t)=Vmax*sin(2pi*f*t+phi)

The voltage between the AC power sources would be
U(t)=Vmax*sin(2pi*f*t)-Vmax*sin(2pi*f*t+phi)

And the current
I(t)=U(t)/2R

And power
P(t)=U(t)I(t)

According to Wolfram alpha, when integrating this over one revolution, I get
P(phi)=-2pi(cos(phi) - 1)

Compute 'integrate (sin(x-phi)-sin(x))^2 over x from 0 to 2pi' with the Wolfram|Alpha website (https://www.wolframalpha.com/input/?i=integrate+(sin(x-phi)-sin(x))^2+over+x+from+0+to+2pi) or mobile app (wolframalpha:///?i=integrate+%28sin%28x-phi%29-sin%28x%29%29%5E2+over+x+from+0+to+2pi)

This does not seem to make sense, as independently of the sign of phi, the sign of the power is always the same. I would think that depending on the sign of the phase difference, the power would be positive or negative, such that the phase difference would reduce. However, like this it seems like one alternator would always be accelerated (or always decelerated), independent of the sign of phi.

This does not make sense, as Wikipedia says, that running an alternator in a power network with a slightly leading phase will cause it to produce more current and hence more power, which should create forces that slow it down.

I hope this is not too complex, but can anyone point out where I went wrong?
 
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Is this homework? If so, we can move it to the Engineering homework forum.

I once worked at a place where the operator inadvertently put a 5 megawatt generator online without properly phasing it. A big BOOM, broken parts, and magic smoke came out. It was down for several months.
Somebody else will help you with the math, but the result is a short circuit at the instant of making connection. If the generator is connected 180 degrees out of phase, the result is a short circuit driven by double the voltage of the generator. The torque is very high, so it will not stay out of phase for very long.
 
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jrmichler said:
Is this homework? If so, we can move it to the Engineering homework forum.

I once worked at a place where the operator inadvertently put a 5 megawatt generator online without properly phasing it. A big BOOM, broken parts, and magic smoke came out. It was down for several months.
Somebody else will help you with the math, but the result is a short circuit at the instant of making connection. If the generator is connected 180 degrees out of phase, the result is a short circuit driven by double the voltage of the generator. The torque is very high, so it will not stay out of phase for very long.
Thanks for the reply! This is not homework, I am just doing some self-studies over the holidays. I graduated already :)

Thanks for the insights! So even at 180° phase shift, if we assume no damage to any parts, the alternators would sync? Then my mental model is very much off, because I thought at 180° they would not shift phase at all.
 
I can't update my answer above anymore, so sorry for double posting. I now found a paper that answers my question: https://www.nature.com/articles/nphys2535

The equation for the current sums up the currents through an alternator caused by each alternator, including itself. The voltage is the internal voltage of the alternator. Then the power is just the multiple of them.
 
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