Power Factor Question

[scott88]

Hi,

I have been given a question that I dont fully understand or better put couldnt explain my answer 100% confidently , can anyone confirm and expand on their answer please;

The effect of a sudden increase in load on an electrical generator not controlled by an automatic voltage regulator will be to;
A. Reduce the volts, lower the frequency, raise the power factor
B. Reduce the volts, lower the frequency, lower the power factor
C. Increase the volts, lower the frequency, raise the power factor
D. Increase the volts, raise the frequency, raise the power factor

I believe the answer to be "B" but have also been told "A" by a colleague which created some doubt. Its the power factor part that gets me confused, I was working on the theoretical understanding and assumption that the generator was running at unity power factor 1, and a sudden increase in load would result in an immediate surge of current, a reduction in voltage, reduction in the frequency and that the power factor would fall to 0.8/0.9 lagging, depending on the load. If the unit had an AVR, the excitation would increase to match the voltage and the prime mover would increase, applying mroe torque to match the load.

Thanks,
S

 

[waross]

The answer may depend on the power factor of the suddenly applied load, but if we assume that the generator is running at unity power factor it gets real easy real quick. From unity PF the only way is down.
I'll have to think about the answer when the generator is not running at unity.
Generally when the voltage of a generator is lowered, it will produce less VARs but the changing frequency will also have an effect on the PF of the loads.
I have to think about this, possibly for several days.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[wroggent]

V_t=generator terminal voltage
E_g=excitation voltage
d=power angle
X_g=positive sequence synchronous reactance

Generator current:
I=(E_ge^jd-V_t)/(jX_g)

Apparent Power:
S=P+jQ=V_tI^*=V_t(E_ge^-jd-V_t)/(-jX_g)

Thus

Real Power:
P=V_tE_gsin(d)/X_g

Reactive Power:
Q=V_t/X_g*(E_gcos(d)-V_t)


I would suggest pretending the bus that the generator terminals are connect to be infinite. If you know the power factor before the load is added you can know what it will be after you add a known load. Determine the current to the infinite bus:
I=(P/(V_bus(p.f.))<-cos^-1(p.f.)
(minus sign before cosine is for lagging, no minus sign for leading p.f.)
for before and after the load is added. Then, the excitation voltage is E'<d=V_bus+jX_gI
Assuming the mechanical input to the generator is constant: if d is larger after the load has been added, the generator accelerated and the frequency is higher, and the opposite is true as well. You can also see the effect on the generator voltage.

In summary, I suppose your question could be restated. In the interpretation I've presented, adding a load is the same as changing the power factor. You must know how the power factor changes to know how the voltage and frequency will change.

Also, you may want to look up the swing equation.

Also, I haven't had any coffee yet this morning so I may be wildly incorrect.

 

[ScottyUK]

I think wroggent might need that coffee! The infinite bus model can't be applied here because both voltage and frequency appear to be variable in all possible answers.

The effect of adding load to the machine depends on the nature of both the load being added and of any existing load on the machine. The question does not contain sufficient information to answer definitively. Find a new lecturer.

 

[LionelHutz]

Yup, not enough info to answer.