small electrically

[mir]

Hi.
a curiosity: I have a generator at 750 kw and a voltage at 20000 v

then

i = w/v = 750000/20000 = 37.5 a
r = v/i = 533.3 ohm

Are these accounts correct?

What kind of cable would I need?

Thank you.
♪

 

[Er Presidente]

Hi.
a curiosity: I have a generator at 750 kw and a voltage at 20000 v

then

i = w/v = 750000/20000 = 37.5 a
r = v/i = 533.3 ohm

Are these accounts correct?

What kind of cable would I need?

Thank you.
♪

depends on the type of cable.
go to the site of the cable bats, there should be a downloadable program for calculation.

Hi.

 

[mir]

I tried the software but it's complicated for me... but the accounts I did above are correct?

Thank you.
♪

 

[mir]

Can someone tell me if the accounts are correct?

 

[Fulvio Romano]

the accounts are wrong, but just as a principle.

first of all the calculation of the current. just if we talk about continuous current, if we talk about alternating current you forgot the rate of reactive power that is still transferred to cable.

then, a 750kw generator means that 750kw is consumption? the maximum load? the nominal load?

the calculation of the resistance then... what resistance is it? dimensionally we find ourselves, but the electronics teaches us that the resistance of a conductor is equal to a fall of potential (in v) divided the current that generated it (in a). If you take the nominal voltage of a generator and divide it for its maximum current you will calculate the resistance that would have the secondary circuit if the generator was short-circuit.

...and I strongly advise you to shorten a 750kw generator!

 

[Fulvio Romano]

Finally, since it seems to me that the electronics is not your daily bread, I am shocking you to deal analytically with the size of the conductor. you have all the rights not to know how to calculate, and simulation programs help little if you do not know what you are doing.

take the technical specifications of the generator, see the section of the cables out and use those. in this way oversized, but at least not wrong.

 

[mir]

You're definitely right, but I was simplifying the account because I thought the most cost was due to cables:

low voltage with 4 cables (triphase + neutral) at 40.75 €/m for 50 meters x 3 => € 24450,00

medium voltage with 4 cables (triphase + neutral) at 33.38 €/m for 50 meters x 1 => € 24450,00

nb: indicative cable prices; in the lower I have "x 3" because I have to carry 3 times the most neutral triphase

 

[maca75]

Are you sure?
40.75€*50*3 does not make 6112.5€
and
33.38€*50 does not make 1669€?
Hi.

 

[mir]

40.75 x 4 fili x 50 x 3

33.38 x 4 fili x 50 x 1

 

[zeigs]

normally for generators the power data is given in kva and not kw just to avoid having to calculate the rectifier. kva and kw correspond in case of load purely resistive, that I know, bulbs.
then the 20kv to what do they refer to? for a three-phase alternator is indicated the chained voltage, and whose the correct formula for the calculation of the current is p [VA]=1.73*v [V]*i [A]the calculation of the resistance you have made makes no sense in this case: it is the resistance that should have a hypothetical resistive load that subjected to the voltage of 20kv dissipates 750kw. A light bulb lady!
to me come about 23a, to spanne the cable is from 6 mm2, but for 20kv has a Mr isolation (in this case the cost does not as much as the insulating). and it is an empirical calculation valid for routes of the order of tens of meters. for longer routes it is better than looking at us someone practical :biggrin: