Two pm alternators in parallel query

  • Thread starter b.shahvir
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In summary: so the total power delivered to the load is not constant when the phase of one generator is changed.
  • #1
b.shahvir
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Hi Guys, :smile:

My query is a bit abstract in nature! But i will be very grateful if someone can provide me with the solution to my query. Here goes;

Consider two Permanent Magnet (PM) Alternators Alt 'A' & Alt 'B' connected in parallel, supplying equal power to a purely 'Resistive' load (unity p.f.). Both the Alternators have equal stator synchronous reactances (stator resistances can be neglected). Please note, they are not connected to the local power grid (infinite bus). Now, if i increase the driving torque of say, Alt 'A', then what changes will occur with respect to the following electrial parameters:-

1) Will magnitude of induced emf 'Ea' of Alt 'A' increase/decrease ? Please explain.

2) Will magnitude of induced emf 'Eb' of Alt 'B' increase/decrease ? Please explain

3) Will magnitude of terminal voltage 'Vt' increase/decrease ? If the magnitude of 'Vt' remains constant or decreases (due to armature reaction), then please explain.

4) Can someone please provide me with a practical phasor diagram for the above query. I am asking for one, since the phasor diagrams in textbooks have lot of assumed conditions & hence do not relate to practical conditions!

Any kind of help will be highly appreciated.


Thanks & Regards,
Shahvir
 
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  • #2
I simulated your problem with LTSpice. I had two 100-volt 60-Hz in-phase voltage generators in parallel, each with a 1 ohm series resistance. The two outputs were combined in parallel, and the summed current fed to a 10-ohm load resistor. Each voltage source put out about 4.5 amps, and the current in the load resistor was 9 amps.

I then delayed the phase of one generator by 45 degrees. Although the current in the load resistor remained at 9 amps, the current out of each voltage source jumped up to about 35 amps, and the two output currents were nearly 180 degrees out of phase. I suggest that if you build this circuit that you fuse each alternator at perhaps 20% above the expected peak current.
 
  • #3
Bob S said:
I simulated your problem with LTSpice. I had two 100-volt 60-Hz in-phase voltage generators in parallel, each with a 1 ohm series resistance. The two outputs were combined in parallel, and the summed current fed to a 10-ohm load resistor. Each voltage source put out about 4.5 amps, and the current in the load resistor was 9 amps.

I then delayed the phase of one generator by 45 degrees. Although the current in the load resistor remained at 9 amps, the current out of each voltage source jumped up to about 35 amps, and the two output currents were nearly 180 degrees out of phase. I suggest that if you build this circuit that you fuse each alternator at perhaps 20% above the expected peak current.


Dear Bob, :smile:

Thanx for your help...but there is a problem here which i am trying to understand since long.:confused: If phase of anyone generator (say Gen A) is increased/decreased, induced EMFs of Gen A and B i.e. Ea and Eb would also increase/decrease respectively. As a result, terminal voltage Vt and bus frequency would also rise/fall accordingly. Then how can output current (at 9 amps) and output power remain constant ? ?

Actually, i had considered similar components i.e. 1 ohm series resistances and 10 ohm load resistance. Ea of Source A was considered a bit higher than Eb of Source B, the circuit considered was DC. I then applied KVL as well as Superposition Theorem to the loop and found that out amps of Source A had increased while that of source B had decreased suitably...but Source B was still taking up load, albeit much lesser than before. Also, Vt and hence load amps had increased accordingly!

Maybe in the above problem, if you consider series impedances (to simulate synchronous reactances of Gen A and B) instead of series resistances but keeping the load purely resistive... then this arrangement might simulate an increase in Vt and cause load current I(load) (and hence output power) to increase accordingly.

Also, I humply request you to provide me suitable phasor diagrams for the same based on the simulations if its not much trouble. :smile:
Plz help me understand this concept as i am pondering on it for quite some time. I will be very much grateful! :frown:

Kind Regards,
Shahvir
 
Last edited:
  • #4
You are correct in that the model I used for the two alternators is incomplete. In actuality, the permanent magnet motor rotors have inertia, and so if the phase of one leads the other at some given time, then by the exchange of out of phase currents, the relative phase of the two alternators will begin to oscillate, something akin to the phase oscillations of a salient pole synchronous motor when the load is changed. Also, in my model, the power loss in the two series resistors (I squared R losses) goes up nearly a factor of 100 (from 4.5 amps to 35 amps) when the two alternators are 45 degrees out of phase. If the mechanical power driving the alternators cannot supply the required extra torque, the system will stall.

I do not have the tools to give you phasors. Perhaps you could run a Spice model with rotor inertia included.
Bob S.
 
  • #5
Bob S said:
You are correct in that the model I used for the two alternators is incomplete. In actuality, the permanent magnet motor rotors have inertia, and so if the phase of one leads the other at some given time, then by the exchange of out of phase currents, the relative phase of the two alternators will begin to oscillate, something akin to the phase oscillations of a salient pole synchronous motor when the load is changed. Also, in my model, the power loss in the two series resistors (I squared R losses) goes up nearly a factor of 100 (from 4.5 amps to 35 amps) when the two alternators are 45 degrees out of phase. If the mechanical power driving the alternators cannot supply the required extra torque, the system will stall.

Dear Bob, :smile:

But could you help me with the doubt i mentioned as regards load current remaining constant? It is somewhat difficult to grasp that in spite of rise in bus terminal voltage Vt, load current I(load) remains constant. Kindly help me out here please! :frown:

Kind Regards,
Shahvir
 
  • #6
Someone please reply to my query :frown:
I'll be very grateful

Kind regards,
Shahvir
 

What is the purpose of connecting two pm alternators in parallel?

The purpose of connecting two pm alternators in parallel is to increase the overall power output. When two alternators are connected in parallel, they work together to supply power to the same load, resulting in a higher power output than a single alternator could provide.

How do you connect two pm alternators in parallel?

To connect two pm alternators in parallel, the positive terminals of both alternators are connected together, as well as the negative terminals. This creates a common connection point for both alternators, allowing them to work together to supply power.

What are the benefits of connecting two pm alternators in parallel?

Connecting two pm alternators in parallel has several benefits. It increases the overall power output, improves system reliability by allowing for backup power in case one alternator fails, and reduces stress on each individual alternator by sharing the load.

What are the potential issues with connecting two pm alternators in parallel?

One potential issue with connecting two pm alternators in parallel is unequal load sharing. If one alternator has a slightly higher output than the other, it may end up carrying more of the load, which can lead to overheating and potential failure. Other potential issues include synchronization issues and increased complexity of the system.

How do you ensure proper load sharing when connecting two pm alternators in parallel?

To ensure proper load sharing, it is important to use alternators with similar power ratings and characteristics. Additionally, proper synchronization and control systems can be implemented to ensure that both alternators are working together efficiently and sharing the load evenly.

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