Synchronous Machine

Thread Starter

sitting_duck

Joined Apr 18, 2010
14
Hi,

I'm a mechanical engineering student and I have an exam soon on Electrical Energy Systems. I am pretty rubbish at all things electrical and would really appriciate some help.

I am trying to solve this question on synchronous machines.

The problem

Attached jpg.


My attempt;

I know that the rotor speed is found by;

S=120(frequency)/(number of poles)

If I use the rated frequency I get 1000rpm for my answer.

I am also aware that;

slip speed=(sync speed)-(rotor speed)


How do I determine the synchronous speed from this?
 

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strantor

Joined Oct 3, 2010
5,333
Hi,

I'm a mechanical engineering student and I have an exam soon on Electrical Energy Systems. I am pretty rubbish at all things electrical and would really appriciate some help.

I am trying to solve this question on synchronous machines.

The problem

Attached jpg.


My attempt;

I know that the rotor speed is found by;

S=120(frequency)/(number of poles)


If I use the rated frequency I get 1000rpm for my answer.

I am also aware that;

slip speed=(sync speed)-(rotor speed)


How do I determine the synchronous speed from this?
wrong. that's how you figure out synchronous speed. that's your answer, synchronous speed = 1000rpm
 

t_n_k

Joined Mar 6, 2009
5,455
Hi sitting_duck,

Keep in mind the fact that synchronous machines don't have any slip. They are generally 'locked' to an infinite bus frequency corresponding to a synchronous mechanical speed. Slip is an important property in induction machines which generally run below (& occasionally above) synchronous speed.
 

Thread Starter

sitting_duck

Joined Apr 18, 2010
14
Thank's a million.

How do I determine Eao then?

I know that

Pbus=(EV/Z)*sin(delta)

Qbus=(EV/Z)*(cos(delta))-(v^2/z)


I presume that the last part of the question is important here, That an infinite bus is assumed at zero angle and rated voltage. But I'm not sure how to use this information.
 

t_n_k

Joined Mar 6, 2009
5,455
Presumably the equivalent circuit shown is a per-phase representation of the three phase generator.

For part (b)

At 220V line-line bus voltage the per-phase bus voltage would be 220/√3=127V

The total KVA rating is 4kVA so the per-phase value is 4/3=1.333kVA.

If the machine is supplying rated kVA then the line current would be 1.333kVA/127=10.5A

The per-phase open circuit voltage at rated load and unity power factor would be 127+10.5*j5 = 137.4V at 22.46° power angle. The generator open circuit line-line voltage would then be 137.4√3=238V.

For part (c)

Start with the per-phase induced emf as 40*If=160V. Given this open circuit voltage and the power angle of 10° find the difference voltage between the open circuit value and the load infinite per-phase bus voltage of 127V at 0°. Divide the difference by j5Ω to find the 'injected' line current [including the phase angle]. Then deduce the per-phase power flow using the bus voltage and the 'injected' line current, having regard to the injected current phase angle [i.e. the power factor].
 
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