URGENT - Eletrical board consumption 1

[pintodacosta]

Hi all,

I have an issue that i have doubts, and that i describe above:

I have a 3phase (400V)feeder capacity 800A in the substation, from this feeder exit the 3 phases (R,S,T), each phase has 2 parallel conductors of 240mm2.

Question1:

I measure the currents in each cable of each phase, and i got the following results:

Phase R:
Cable1: 54A
Cable2: 54A
Phase S:
Cable1: 55A
Cable2: 55A
Phase T
Cable1: 54A
Cable2: 54A

My doubt is regarding the total current that the feeder is supplying to the eletrical board to where the 3phases are connected.
How do i calculate it?

Question 2:

The 3 phases mentioned above arrive to a distribution board (board A). No neutral available.

I want to feed from this board A, 3 more boards (230V), this means i need a neutral cable. The feeder in the substation has the possibility of connecting a neutral cable. Is possible to get the neutral from my feeder located in the substation, make him arrive to the neutral busbar in my board A, and from there power the 3 monophase boards that i need to get power to??
Is there the need of making any protection to the neutral cable??

Please advice.

Tx in advance regards

 

[Parchie]

1) You have 110 A per phase, give or take (you have 2 cables sharing the load; 2 x 55 = 110A). Your 800A capacity is still lightly loaded.
2) Yes. You can supply single-phase loads at 230V using the neutral to pair with any of the phase wires. I don't have my book here but we usually size the neutral the same as the phase conductors in case you can't balance your single-phase loads. If you can, there are 4-pole breakers you can buy, the fourth pole intended for the neutral conductor.
Hope that helps.

 

[xnuke]

Why not just put in a isolation transformer downstream of Board A and ground the neutral of the transformer, then power your 230 V single-phase panels from it?

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[pintodacosta]

Hi,

1st tx all for the answers.

when i have a 3phase feeder, the current given by him or absorved by the load, is always equal to current in 1 phase? or the total current absorved is the sum of the current in the 3phases.

rgds

 

[Skogsgurra]

Current is current. Is current. So, there is no such thing as "sum of currents" - at least not in a three phase system.

If you are thinking about Total Power, then it is another mug of beer. Which one is it that you really worry about?

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[pintodacosta]

Skogsgurra,

The doubt is the following (sorry is a basic doubt, but i was never able to clarify it.

I have measure the current in phase R, of a 3phase system, 2 cable runs per phase, and the result was 50 Amps in each cable of phase R.

identycal currents were measured in the phases S and T.

So, the current in phase R is 50A + 50A (2 cable runs in parallel) = 100A

The total current absorved by my load is 100A or is 100 A * 3 = 300A (100A from phase R, 100A from phase S, 100A from phase T??

Rgds

 

[waross]

Current is current.
In a three phase system, KVA = Volts x Amps x √3 / 1000
KVA ignores the displacement between the Current phase angle (A) and the Voltage (V) (Expressed as the cosine of the displacement angle).
KVA can not be directly summed with KVA when the displacement angles are not equal.

In a three phase system, kW = Volts x Amps X √3 x (cos of the displacement angle) / 1000
As you can see, the cosine of the displacement angle is considered in this equation. As a result, kW may be summed directly.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[pintodacosta]

Bill,

i remain with the same doubt....

according the current measured for phase R, what is the current absorved by the load?

 

[xnuke]

The line current is 110 A in each of the three phases, as indicated above (or 100 A, given your later post). We do not sum line currents at the load. Why? Because it simply doesn't matter. The line current is used to size the conductors that carry the current to the load, and that's the only current that matters when viewed from outside the load. We don't concern ourselves with the phase currents inside the load, because for the same power and voltage ratings, the line currents will be the same regardless of whether the load is connected in delta or wye. Therefore, we simply don't worry about "current absorbed by the load." Technically, in a balanced system, the three current phasors sum to zero.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Skogsgurra]

Your last question. The current "absorved" (we usually use the word absorbed, or even better, consumed by the load is, according to your first data, 108, 110 and 108 A for phase L1, L2 and L3 (or A, B and C) or even R, S and T (where is that terminology used, except for motors?).

I cannot understand what your problem is. The conductors can carry the current, no probs at all. Even if they are Al instead of Cu, you didn't state which. But, no worries.

Why is it so urgent to have this non-question answered?

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[waross]

Do you feel as if we are fanning the flames of a blazing marsh-mallow?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[itsmoked]

URGENTLY

Keith Cress
kcress -

http://www.flaminsystems.com

 

[pintodacosta]

Hi,
my confusion is this:

The feeder that i have has a 800A free to deliver. Actually there are some loads connected, I measured the line currents, and at the moment there are aproximately 100A per phase. In therms of feeder occupation or usage, Iam using 100A (thismeans i have 700 A free), or i am using 3 x100A (this means i have 500A free).

It was mentioned that was urgent, due to the date that i must provide an answer.

Rgds

 

[davidbeach]

Time to dig your Power System Analysis textbook out of the back of the closet and review the basics of 3-phase power system.

 

[xnuke]

Your question has been answered several times above. Since you have to keep asking, it's clear you simply didn't understand the answer, and are not qualified to work with power systems. You should find a job where you won't be asked to do so.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[stevenal]

100 A: Yes!
3 X100: Meaningless! No!

 

[lukin1977]

Feeder Capacity of 800A means 800 A per line NOT the sum of the 3 lines.
So yes, you can load each of the lines with 700 A more

 

[pintodacosta]

Tx all for the clarifications.