[joe-nwt]

The resistance of the 600kcmil cable is 1908 milliohms, and the 750kcmil cable is 1526.4 milliohms.

Yeah, where did you get these values?

 

[Mike in Canada]

Flesh that out a little for me, seems like something doesn't add up...

Okay, an online calculator says that 30m of 600kcmil aluminum has a resistance of 1908 milliohms.
The same length of 750kcmil aluminum has a resistance of 1526 milliohms. The delta on that is 381.6 milliohms of 'extra' resistance on the 600kcmil wire.

Another online calculator gives cable heat losses.

Cable Power Loss Calculator, Formula, Calculation | Electrical4u

Cable power loss P(loss-kW) in kilo watts is equal to 1000 times of length of the conductor l(M), square of current I(A) in amps, resistivity of the conductor ρ(Ω-m) in ohm-meter divided by the cross section A(sqmm) of the conductor in (sqmm)

www.electrical4u.net

I found an error in what I did. I used the kcmil values for square millimeters, and that isn't accurate.
The new values are 0.8015 kw for 750kcmil and 1.10018 kw for 600kcmil.
The delta is very close to 0.3kw per cable on the whole span. So in the case of my cables (6 of them) that is 1.8kw of loss when drawing 1200 amps on the full cable-set.

If I'm not doing this right, then feel free to correct me, because I now have much MORE heat than I did before. We're at 300W per cable, or 1800 for the set. If I was pulling 1200A all day 300 days per year, that would be about 13 megawatts of power lost just on the delta. That can't be right, can it?

 

[splatz]

If I'm not doing this right, then feel free to correct me, because I now have much MORE heat than I did before. We're at 300W per cable, or 1800 for the set. If I was pulling 1200A all day 300 days per year, that would be about 13 megawatts of power lost just on the delta. That can't be right, can it?

I missed part, how many conductors per phase do you have?

 

[Mike in Canada]

I missed part, how many conductors per phase do you have?

Two conductors per phase. 1200A, so 600A per cable (aluminum single conductor in open air, properly spaced on ladder tray).
Obviously the heat number I came up with is a worst case scenario, but it's still something to consider... providing I am (or the calculator is) doing the math correctly.
Cables do get warm. That has to be power. If a cable is 30m/100' long and it's 'warm' then it's above body heat, so it's, say 45C/110F. If every inch of that cable is that warm then there has to be some watts involved. 300W for one of these cables is only 10W per metre/3 feet. 10 watts seems... reasonable. 300W x 6 cables is 1800W or 1.8KW. 1.8KW x 24 hours x 300 work days in a year (we're 24/5, sometimes 24/6 or 24/7 depending on what's going on). Anyway, 1.8 x 24 x 300 is just shy of 13,000 kilowatts or 13 megawatts. Suddenly this is a big number. If this is correct (and it didn't seem unreasonable before the math was done) then the added heat load over the lifetime of the cable... I mean.. 15 years? Almost 200 megawatts. That might go a ways to pay for the thicker cable.

If I'm doing this wrong then please tell me where I screwed it up.

 

[splatz]

I am just tinkering here but I'm trying to think of the easiest way to figure this out. It's a series-parallel circuit. You have two wires in parallel, in series with the load, in series with two wires in parallel. The current is 1200 amps, so it's going to be 600 amps on each of the parallel conductors. The heating on the wire is based on the voltage drop on the wire, right? So I used the Southwire voltage drop calculator and got this

If the voltage at the end has dropped 0.77%, that means it's

600*(1-0.0077) = 595.38 volts

So the voltage drop on the wire pair on each side is going to be

(600 - 595.38 ) / 2 = 2.31 volts

So the power in each wire is

P = VI =2.31 * 600 = 1386 watts

What do you think, does that make sense?

It should jive with the resistance of the wire, that's a separate calculation but it should yield the same result.

(Edited because I did this for 1200V the first time...)

 

[joe-nwt]

Ignore me all you want but the resistance of 600kcmil wire is 1.908milliohms, not 1908milliohms.

 

[Mike in Canada]

I am just tinkering here but I'm trying to think of the easiest way to figure this out. It's a series-parallel circuit.

I can see hot it could look like a series-parallel circuit, but if all of the resistances are the same, then does it really matter? If the wires are all the same length then they should be all loaded the same, so the load on one wire should be the same as the loads on all of the other wires.

View attachment 183236

Okay, I see where you are going with this.

If the voltage at the end has dropped 0.77%, that means it's

600*(1-0.0077) = 595.38 volts

So the voltage drop on the wire pair on each side is going to be

(600 - 595.38 ) / 2 = 2.31 volts

So the power in each wire is

P = VI =2.31 * 600 = 1386 watts

I agree with the math, and it seems quite reasonable, but you got WAAAAY higher numbers than I did.
With those number that would be 1.386*6*24*300= almost 60 megawatts per year. Yikes!

Oh, hang on.... You did the actual heat loss, not the delta. If you do the same calculation for 750kcmil and subtract the 750kcmil value from the 600kcmil value then you should end up with around 300 watts per cable. At least, that's what I got.

 

[splatz]

If the voltage at the end has dropped 0.67%, that means it's

600*(1-0.0067) = 595.98 volts

So the voltage drop on the wire pair on each side is going to be

(600 - 595.98 ) / 2 = 2.01 volts

So the power in each wire is

P = VI =2.01 * 600 = 1206 watts

 

[splatz]

So by these calculations, the difference is (1386-1206)*6*24*300 = 7,776,000 watt-hours = 7.776 Mwh / year

Units are important - megawatt-hours per year is a lot less intimidating than megawatts - but at these ampacities, based on lets say $0.33CAD per kwh, that's 7776*$0.33 = $2,566.08 annual savings on the power bill. So I think I'd agree that it looks like the payoff on 750kcmil versus 600kcmil is pretty quick at these amounts of power, even if you have to upsize conduits splitters etc. to accommodate.

 

[joe-nwt]

So by these calculations, the difference is (1386-1206)*6*24*300 = 7,776,000 watt-hours = 7.776 Mwh / year

Units are important - megawatt-hours per year is a lot less intimidating than megawatts - but at these ampacities, based on lets say $0.33CAD per kwh, that's 7776*$0.33 = $2,566.08 annual savings on the power bill. So I think I'd agree that it looks like the payoff on 750kcmil versus 600kcmil is pretty quick at these amounts of power, even if you have to upsize conduits splitters etc. to accommodate.

How much heat do you think you are getting from this part? Part of the voltage drop, yes.

 

[243648]

Mike, Joe says that 600kcmil wire is 1908microfarads.

 

[Mike in Canada]

Ignore me all you want but the resistance of 600kcmil wire is 1.908milliohms, not 1908milliohms.

View attachment 183238

Good catch, but I didn't use the resistance to get my final value.

 

[joe-nwt]

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