Would this be right?

[chee_dee]

Hi All, Newbie here.

Love the site. My first Post.

Would I be right in doin da following?

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)

50 I(line) = 50/1.732 I(phase)
I(phase) = 28.9 Amps

Can I now put this on a three phase supply meaning Each Line wud draw 28.9x2 amps? thus equaling 57.8Amps 3? Supply?

Meaning that a 57.8 A 3 ? Supply would be sufficient for feeding 6 50 A (L-N) Units.

Thanks in advance

 

[hardworkingstiff]

Hi All, Newbie here.

Love the site. My first Post.

Would I be right in doin da following?

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)

50 I(line) = 50/1.732 I(phase)
I(phase) = 28.9 Amps

Can I now put this on a three phase supply meaning Each Line wud draw 28.9x2 amps? thus equaling 57.8Amps 3? Supply?

Meaning that a 57.8 A 3 ? Supply would be sufficient for feeding 6 50 A (L-N) Units.

Thanks in advance

That would be incorrect. Pretty simply, you will connect 2 loads per phase. 2x50=100 amps per phase.

You could also look at it this way.

50(amps) x 230(volts) x 6 = 69KVA.

69000/(400*1.732) = 99.6 amps per phase.

 

[chee_dee]

That would be incorrect. Pretty simply, you will connect 2 loads per phase. 2x50=100 amps per phase.

You could also look at it this way.

50(amps) x 230(volts) x 6 = 69KVA.

69000/(400*1.732) = 99.6 amps per phase.

Okay, why do *sqrt(3) with da phase voltage?

if u *sqrt(3) with the Line Voltage(230) then you get the phase voltage(400). Why do u multiply it again? Or am I missing something??

Thanks for the reply

 

[chee_dee]

Maybe I explained my question wrong in the first post.. Actually what I want to know is about the relation between Phase Current(phase-phase) and Line Current(Phase-Neutral)

Say u have 3 Units to supply with 230v 50A Single Phase(Europian, between Line and neutral)

Can u supply these 3 units with a 400v 50A 3 phase supply or is it not that simple?

 

[bob]

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)
50 I(line) = 50/1.732 I(phase)
I(phase) = 28.9 Amps
Can I now put this on a three phase supply meaning Each Line wud draw 28.9x2 amps? thus equaling 57.8Amps 3? Supply?
Meaning that a 57.8 A 3 ? Supply would be sufficient for feeding 6 50 A (L-N) Units.

Thanks in advance

Your calculation "50 I(line) = 50/1.732 I(phase). I(phase) = 28.9 Amps" only applies to a 3 phase delta installation. If you are supplying "230 volts 50amp
single phase", Your calculation is not correct. On a delta system supplying
230/120 volt single phase, you can not use the 3rd phase. It sounds like you
want 120/240 volts single phase from the utility.

 

[petersonra]

Hi All, Newbie here.

Love the site. My first Post.

Would I be right in doin da following?

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)

50 I(line) = 50/1.732 I(phase)
I(phase) = 28.9 Amps

Can I now put this on a three phase supply meaning Each Line wud draw 28.9x2 amps? thus equaling 57.8Amps 3? Supply?

Meaning that a 57.8 A 3 ? Supply would be sufficient for feeding 6 50 A (L-N) Units.

Thanks in advance

Presumably you would be able to connect them line to line on a 230V 3 phase system, but you would use 50A per unit, so you would be at 100A per phase for 6 units.

if you connected them L-N on a 230V 3 phase system you would only have 133 volts on the unit, not 230V.

 

[charlie b]

I am writing up a separate reply. But let me just invite attention to the OP's home country. I think we are dealing with a system that has 400 volts, phase to phase, and 230 volts, phase to neutral.

 

[charlie b]

Can u supply these 3 units with a 400v 50A 3 phase supply or is it not that simple?

It is nothing like that simple.

For starters, let me suggest that you now and forever banish from your thoughts any notion of ?current per phase.? It is nonsense. To be specific, when you mention having current of 28.2 amps on one line, and then said that you need to double it (to 57.8 amps) to get the total current, that is nonsense. That is like saying that if you drive your car from home to the grocery store, and then drive it back, that equals a total of two cars. Not so at all; it is the same car going out, and the same car coming back.

Current is ?charge in motion.? One amp represents a certain amount of charge moving past a given point within a given amount of time. It is like standing on a bridge and counting how many cars pass under you in one second?s time. Current flows from the source to the load on one wire, and then flows back to the source along another wire. The total amount of current leaving (along however many wires are connected to the source) must necessarily and must always equal the total amount of current returning (once again, along however many wires are connected). So there is no meaning to the addition of ?current leaving? plus ?current returning? equaling ?total current.?

The only time there is any meaning to a comparison of ?line current? and ?phase current? is in an analysis of what is happening internal to a transformer, as compared to what is happening external to that transformer. Depending on whether the transformer is connected as a WYE or a DELTA, the current in one of its windings may be the same as the current leaving on one of the lines, or it may differ by the factor of the square root of three.

Lou?s calculation, in post #2, is correct. It is also the only ?safe? way to deal with a combination of single phase and three phase loads. You first calculation the power (KVA) of each load. Then you add them to get total KVA. Finally you calculate current by taking total KVA, dividing by the phase-to-phase voltage, and dividing again by the square root of three.

 

[LarryFine]

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)

I think we are dealing with a system that has 400 volts, phase to phase, and 230 volts, phase to neutral.

Pretty simply, you will connect 2 loads per phase. 2x50=100 amps per phase.

. . . you would use 50A per unit, so you would be at 100A per phase for 6 units.

All of this combines to be the correct answer.

 

[bob]

We have to supply 6 units with 230v 50 A Single phase(Line-Neutral)
50 I(line) = 50/1.732 I(phase)
I(phase) = 28.9 Amps
Can I now put this on a three phase supply meaning Each Line wud draw 28.9x2 amps? thus equaling 57.8Amps 3? Supply?
Meaning that a 57.8 A 3 ? Supply would be sufficient for feeding 6 50 A (L-N) Units.

Thanks in advance

If the voltage is 230/400 then it appears that the system is a wye system.
Note the OP says

6 units with 230v 50 A Single phase(Line-Neutral).

If it is a wye his caculations

50 I(line) = 50/1.732 I(phase) I(phase) = 28.9 Amps

is still incorrect. Iline = Iphase if it is a wye.

 

[chee_dee]

Thanks for the replies guys.. think I got it now. ja it's Wye.

(1) Calculate individual Single Phase kVA............. kVA=P x I
(2) Add them up and, because the supply's 3phase, Use that kVA to determine the size of the 3 phase supply...............kVA=1.732 x V x I.

That's it....

Now if there were 7 units ( not able to balance on the 3 phases). Do I just devide them up on the phases and then use the Current value of the Highest Leg to determine the 3phase kVA..

And also how balanced must a 3phase system, be percentage wise, to call it balanced??

Thanks again all for the replies.

 

[hardworkingstiff]

Now if there were 7 units ( not able to balance on the 3 phases). Do I just devide them up on the phases and then use the Current value of the Highest Leg to determine the 3phase kVA..

That's what I would do.