Forums » Electrical Engineering

Cable sizing

• • 1 posts
February 6, 2019 12:24 PM PST

Hello,

I have a problem with the cable sizing.
And I have:

Three-phase
20m cable length
Max. 100 amperes of current
cos of 1
Copper spez.resistance (y = 56)
and a maximum voltage drop of 400 V, deltaU (3%) = 12V

Now I have the formula for voltage drop in a three-phase network to A (conductor cross-section)

so:

((Root 3) * length * current * cos) / (Y * deltaU) = A (cross section)

((Root 3) * 20m * 100A * cos1) / (56 * 12) = 5.12 mm²

So I would have to use at least 5.12 mm² or 6 mm² in those conditions.

Something is looking wrong. What did I do wrong here? Is something wrong with the calculation?

regards

Raj

• • 141 posts
February 6, 2019 1:42 PM PST

I used an online calculator which is not as accurate.  I substituted #10 wire for 6 mm2 & 65 feet for 20 meters and it had a 2.8% drop.

The problem with the calculator is when there is no drop to worry about you will get a conductor smaller than the load.  When this happens I just use whatever conductor is appropriate for 100 amps- Obviously, 6 mm2 is not large enough so you probably need about 35 mm2

This post was edited by Dennis Alwon at February 6, 2019 1:44 PM PST
• February 6, 2019 1:58 PM PST

Slow poster;  Following Dennis' post

Disclaimer:  I am not familiar with metric sizing conventions for copper conductors  So hopefully someone will jump in and educate me.

Just looking at the physics:

Took me a while to figure out the Copper spez.resistance (y = 56) is the conductivity.  The number I saw was 58.14 x 10E6 - close enough, so that works.

So, yes, 6mm2, 20 meters, 100A will give a 9.6V drop phase to phase.

However, I wouldn't expect that to be true for very long.  6mm2 copper at 100A is going to be red hot very quickly.  Doesn't one normally spec 35mm2 wire for loads in the 100A range?

What is the context?

carl

This post was edited by Deleted Member at February 6, 2019 2:00 PM PST
• • 141 posts
February 6, 2019 3:06 PM PST

Carl, I believe the issue is when using the VD formula when there is no vd appears to give a value much smaller than what is required.  I have done this many times.

The exact problem above using an online calculate comes up with 10 awg for 100 amps.  10 awg is about 6 mm2 so obviously the formula is not correct in some instance.  Whenever a value is calculated lower than allowed by code I just use the code compliant conductor size.  Does that make sense?

• February 6, 2019 4:48 PM PST
Dennis Alwon said:

The exact problem above using an online calculate comes up with 10 awg for 100 amps.  10 awg is about 6 mm2 so obviously the formula is not correct in some instance.  Whenever a value is calculated lower than allowed by code I just use the code compliant conductor size.  Does that make sense?

Yes.

The formula is good - took me a while to figure that out.

I tend to not use online calculators - I can't verify the algorithm.  So, just looking at the physics and verifying by a different route

I looked up the resistance of 6mm2 per km.  R(per km) = 2.80 ohms/km

Vd (phase to phase) = R(per km) X (20m/1000m) x 100A x SQRT(3) = 9.7V at 20C

Within seconds that piece of wire is really hot - maybe not red-hot, but certainly past insulation failure.

From what I am seeing, normal conductor sizing for 100A is 35mm2.  If that is true, what is the context for considering using 6mm2?

Just wondering

carl

• • 129 posts
February 6, 2019 11:16 PM PST

I agree with Dennis and Carl. please forget that formula • • 62 posts
February 7, 2019 12:20 AM PST
krishn Raj said:

Hello,

I have a problem with the cable sizing.
And I have:

Three-phase
20m cable length
Max. 100 amperes of current
cos of 1
Copper spez.resistance (y = 56)
and a maximum voltage drop of 400 V, deltaU (3%) = 12V

Now I have the formula for voltage drop in a three-phase network to A (conductor cross-section)

so:

((Root 3) * length * current * cos) / (Y * deltaU) = A (cross section)

((Root 3) * 20m * 100A * cos1) / (56 * 12) = 5.12 mm²

So I would have to use at least 5.12 mm² or 6 mm² in those conditions.

Something is looking wrong. What did I do wrong here? Is something wrong with the calculation?

regards

Raj

Yes, it is not correct. See this... 