Electrical Generator

It's a well-kept secret:

http://cr4.globalspec.com/search/sitesearch?do=show&sort=textmatchrank&srch=Power%20factor&order=asc

The current lowest value occurs at p.f. 1, so the current is higher at p.f. 0,80.

Can you give exact calculated current at both pf?

It responds to calculation. No joy in the links in #1↑, then?

What is the voltage value ?

230 Volt

Divide that into the kVA rating to get current.

Electronick is the right one. Did beat me to the answer. Kudos.

If OP would have asked (for his homework) about the breakdown of the current into the real and circulating parts, a diagram would have be drawn.

Answer:

real, less than 100%

circulating, less than 20%

You work out the details.

You know, I have a fashion for these matter´s! Tank you :)

For a constant power rating current is greater for 0.8 power factor.from the equation

p=VICOSø

...and for a constant current the power delivered to the load is greater at unity power factor.

All the original poster has to do now is select which one is the desired one.

The Genset has a maximum Current rating.

This is given for a p.f. = 0.8 as a standard. Therefore, your 5kVA will give a maximum of 5 x 0.8 = 4kW active power. Since the voltage output is 230V, then the maximum current will be form the formula: P = V x I x p.f. which gives I = P / (V x p.f.)

I = 4000 / (230 x 0.8) = 21.74 A

For p.f. = 1, then the P = 230 x I x 1.0 and P = 5000 x 1.0 = 5000 W

Which gives I = 5000 / (230 x 1.0) = 21.74 A {the same as before}

In conclusion, the important item to consider when talking about p.f. is the amount of active power you get when p.f. varies. Also, the p.f. depends on the loading circuit and not on the genset. If the load has a p.f. of 1, but the genset is rated at o.8 p.f., then you must make sure that you do not overload the prime mover by extracting 5kW instead of 4kW! This brings you back to the formula: P = 230 x I x p.f.

Therefore, I = 4000 /(230 x 1) = 17.39 Amps only if the p.f. = 1 for this genset.

Otherwise, if you draw 21.74 A at p.f. = 1, you will be loading at 5 kW which might be more than the driving motor can handle.

Thank you so much!

Now I dont have doubt!

Thanks once again!

Dear Mr.amitsonpethkar,

The 3 phase Circuit Power is calculated by

Power in KW = 1.732 (Root 3) x V X I X P.F

If PF is 1, the current will be the lowest value.

DHAYANANDHAN.S

Your question lacks the proper framing to rule out many of the well stated answers here, since you did not state the quality of the load or the quantity precisely. You did mention the capacity of the genset. For constant total load, the vector sum of total resistive load and total reactive load must be held a constant. Also, it is a quite reasonable assumption to not allow voltage droop as the load shifts from resistive to reactive. Here we are referring to a single phase genset.

Example #1: pf=1, and cos Θ =1 P=4KW, V=230 V, therefore I=17.391 A

Example #2: pf=0.8, cos Θ =0.8 P=4KW, V=230V, therefore I=21.739 A , so quite clearly in a shift from resistive to reactive power the current demand increases, all else being the same.

The generator will be most efficient at unity power factor. If the power factor is at 0.8, which is very common, the difference can be supplied by capacitors to correct the power factor to 1.0. This is very easy if the load is constant but very difficult if the load is constantly changing. Capacitors cost less than generators and require no fuel. Mathematics aside, power factor is exactly what it says it is, the number multiplied times the generated or supplied power to get the actual power. At 0.8 power factor you get 80% of the power available as usable power, therefore you need 1.25 times the current at 0.8 power factor to the same work.

Yes, our larger substations all incorporate capacitor (banks).