Relationship PF with Maximum Power Output in Diesel Generator

Well my friend,

KVA^2=KW^2 + KVAR^2, I believe you aware of this formula.

Here KVAR is called reactive load which results due to resultant of capacitive as well inductive load.

I believe, your load is comprises of very high inductive load, which results in such problem. If you have installed 2-3 induction motor as your load, then I will suggest to use capacitor bank in order to improve the Power factor and efficiency.

Another important observation in your problem is that drop in frequency, which also indicates something towards loading problem of geneartor.

The problem is not with the Alternator, but with the engine Horse Power. The engine HP is arrived at, taking into account an assumed power factor of 0.8 lag. Your alternator can very safely deliver 18kW load at UPF, provided the engine also can take the load. Here in your case, upon loading beyond 12kW, the engine is dying down and hence your frequency and voltage fall.

A Good Answer, Electricalexpert65 is very right.

The problem is not with the Stamford alternator, but in connection with the engine.

Let's make a simple calculation.

For 22.5 KVA alternator, you need 22.5 KW(e) for (PF=1) of electrical power,

and you need 18 KW(e) for (PF=0.8) of nett electrical power output.

Alternators are always have electrical and mechanical losses. (Heat, Friction, Hysterresist, cooling fan etc.)

Assuming the efficiency of the alternator is 0.8

22.5 Kw/0.8 = 28 KW(m) of mechanical power is needed to drive the alternator shaft for the case of (PF=1)

or 18 Kw /0.8 = 22.5 KW(m) of nett mechanic power for (Pf=0.8) is needed.

Naturally this mechanic power is coming from the engine shaft.

Engines are also having losses. (Coupling losses, Vibration losses, etc.)

Now assume the efficiency of the engine at shaft output is (0.8)

Then you need a higher engine output at the shaft, such as

28Kw/0.8 = 35 KW(m) of nett engine mechanic power output. (For Pf=1) or

22.5 / 0.8 = 28.5 KW(m) of nett engine power output (For Pf=0.8).

Above calculations are showing, he needs at least a 30 Kw of engine power at rated RPM.

In any other case, the engine speed will droop proportional to load increase

Kindest Regards.

Dear suvekkmr

Thanks for reply, but I think this problem acquires because of my engine doesn't have enough power so in this load frequency decreased.

Dear electricalexpert65

Thank you for your help.

I think so.

But Im sorry maximum load was 15kw in P.F=1 before the frequency of phase has decreased.(not 12kw)

Samrand,

You say that the motor -generator could not assure the frequency & voltage when loaded up to 12 kW but it worked OK when loading it up to 22.2 kVA it worked.

This is weird because the real/active component of the load (in this case 18 kW) exceeds the 12 kW in case of pf=1.

How did you measure the 12 kW respectively 18 kW/13 kVAr ? Usually via a Watt-meter or, in case of pf=1, via an A-meter & V-meter, but in this case it seems that the values do not match.