Simple 3+ Question - Approx Current per Leg

[Beggar]

I'm just looking to bound a problem:

Presuming a 3Ø motor with a full-load draw of 300 amps, what's the approx current draw per conductor feeding the motor?

In the deep recesses of my memory (I took electrical machinery about 20 years ago) I seem to recall that it would be sqrt(2)/2 * FLA. Am I in the ballpark here?

Just for context: I'm working with some incompetent people in China who were supposed to set up a distribution system to provide the required power to my motor starter. They appear concerned that they've undersized the system based only on my single-machine, no-load tests.

My job was to spec the power; theirs was to provide it. I need to decide if they've provided sufficient capacity.

--------------------
Bring back the HP-15

http://www.hp15c.org

--------------------

 

[Beggar]

Oh, and for clarity:

What I'm trying to decide is whether to call a halt to the whole thing, accuse them of incompetence, pull my guys home, and make a big, expensive (politically and financially) stink about it.

I just have to be sure before I go public.

--------------------
Bring back the HP-15

http://www.hp15c.org

--------------------

 

[alehman]

If the motor nameplate says 300 amps, that's the maximum load on each phase. That assumes it is operating at it's rated voltage and rated output.

 

[Skogsgurra]

As alehman says. The nameplate current is the current drawn at full load. If your Chinese friends supply you with a voltage via copper wires that are 240 - 300 mm2 (diameter around 20 mm) and fuses that are somewhere in the 500 - 630 A range (assumimg DOL), then you are fine.

It is even possible to go down to around 150 mm2 if the wires are laid so that they have a good cooling. It is also a question about ambient temperature. If you notice that your Chinese friends are aware of all these things, then you can relax and let them have it their way.

An aside: You do not have to do those engineering school calculations you are talking about. The formula you use is used to calculate RMS from amplitude and that is not needed since it is always the RMS that is given on nameplates and anywhere in daily life. That school thing is only used - in schools.

Gunnar Englund

http://www.gke.org

 

[ScottyUK]

You need to consider volt-drop in the feeder during starting when the current will be 6-8x the 300A nominal. If this cable is of any length then you may find that the cable rating is of secondary consideration. Total volt-drop at the distribution board should not be greater than 10% of nominal, allowing for all drops right through the distribution system.


----------------------------------

One day my ship will come in.
But with my luck, I'll be at the airport!

 

[Beggar]

Thanks, everybody.

Here's an update: The actual FLA is not 300 amps, it's 540 amps!

My Chinese friends are supplying a 1,000 amp breaker with 240mm^2 cable in an enclosed tray.

They, in fact, seem clueless about issues such as ambient temperature and cooling in free air vs a bundle in a tray.

They seem to realize that there's a problem because when we started "playing" with our system (meaning no load), they gathered around the main cabinet and started chattering. Pretty soon an ammeter showed up and the questions started following: "How much power does your system draw?" Our answer: "Just what we told you it does back at the beginning of the project."

Where we are now: I'm hopping a plane to Hong Kong tomorrow (oh yay, more weeks away from home) where I've mandated that they find a LICENSED engineer or commercial electrician with whom I may consult (in English) about their design.

Any of you ever worked in China? It's a freaking nightmare to get anything done, let alone done well.

When we went through a similar project in Europe, we just gave 'em our power requirements and that was that. Oh, to work with competent people again...

--------------------
Bring back the HP-15

http://www.hp15c.org

--------------------

 

[Beggar]

Oh yeah, one question:

To bound the problem, I'm using an ampacity of 700cmil/amp and a 30% derating for the cable being bundled in a tray.

Do those sound like reasonable rules-of-thumb?

--------------------
Bring back the HP-15

http://www.hp15c.org

--------------------

 

[Skogsgurra]

Did you tell them 300 or 540 amps when you started the project?

Gunnar Englund

http://www.gke.org

 

[Beggar]

540.

I just used 300 as an example when I posted here because I wasn't looking at the spec.

What I actually did was give them a facility drawing that included the locations of each of my enclosures that listed the current draw at each one. I then said to presume that each enclosure was pulling FLA simultaneaously with each of the others (which is unlikely, but conservative).

--------------------
Bring back the HP-15

http://www.hp15c.org

--------------------

 

[TMD]

Beggar: In answer to your working in China question, I've been there and many of my colleagues have. It's a nightmare. Patience, persistance and keeping the problem in focus seem to be the keys. Good luck. Oh, and don't expect the Chinese to have any equipment that we (North Americans) take for granted. But you probably already know that.

 

[alehman]

There's not a fixed cmil/amp number. As the cable gets larger, cmil/amp increases rapidly. NEC table 310-16 is the standard guide in the U.S. I realize the NEC doesn't necessarily apply in China, so if there are applicable local rules, those must be followed. Best bet would be to find a local engineer.

For wireways, derating is not required if fewer than 30 conductors are installed and the cross-sectional fill is less than 20%.

Ratings for cable trays depends on the fill and type of tray but in many cases is some precentage of the free-air rating in NEC table 310-17.

http://www.nfpa.org/freecodes/free_access_document.asp?id=7005SB

cmil = mm^2 * 1974

Good luck!