[s]

There is another way, if you can;t modify in internal connections

you can use a capaitor with the following arrangement

1- the connection should DELTA

2- the delta poles are A,B,C

3- L1 is connected to A, N is connected to B, the capacitor is connected to A and C

 

this should do the trick

Regarding the capacitor rating, frankly i don't know the calculations, maybe Mr.Spir (My mentor) can help with this.

[s]

Ok, 

Firstly , out of personal experience I don't recommend messing around with generators unless you have the experience, patience and time,

Second, how many persons have the access for the generator? is it only you? or there is many people have the access?

 

Your problem is either electrical or mechanical,

If it's mechanical that means the generator temperature is at critical levels and the cooling system needs maintenance ( here PLEASE do not do it yourself unless you are qualified ) 

If it's electrical, that means the temerature sensor is either needs re-adjusment or replacement, also check the control unit settings, there is a big chance (if you are not the only one has the access for generators) that someone messed with control settings,

 

Anyway, refer to the manual and check if your problem is mentioned there,

 

Regards

[s]

if it has terminals of +12 and -12, yes you can

[s]

Consult the manufacturer

[s]

Power factor concept is applicable on loads not sources

[s]

You should consult the manufacturer

[s]

What are the primary and secondary voltages of this transformer?

[s]

Try to analyize the power quality and measure the THD ( total harmonic distortion) 

[s]

What about power factors of those loads?

it's vital to know them, P= 1.73 x V x I x cos(phy)

[s]

If this cooker electricly is just an electrical heater you can try converting the cooker itself into 1 phase

It has three windings, seperate the connections between windings, and energize every winding seperately,

[s]

What do you mean by calculating time?

[s]

You can think about it like this

the coltage is 2 v

3000Ah means that if the current is 3000A the battery will provide electrical energy for 1 hour

so simply the formula is like this : E= 2 x 3000 = 6000 WH = 6 KWH energy required

 

P.S : this calculations are for IDEAL enviroment, but could be applied 99% of times,

[s]

1/Rt=1/R1+1/R2+1/R3+………………….+1/Rn

Rt = Total resistance

n = number of parallel circuits

 

Also the same formula can be used for Z 

[s]

NO CAN DO, since P load > S transformer, end of story

[s]

Hi

One mistake at my post

P is the real power in Watt , S is the apperant power in VA

S=Square root (Psqr + Q sqr)

Just replace the word “REAL” instead of “apperant” in my previous post

[Author]

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