Low Voltage Distribution at Home

Bruthal

Sir Liamalot wrote: »

As a rule of thumb AC priority is heat dissipation whereas DC priority is voltage drop. Making DC conductors a lot more substantial. I use 100A cable to deliver 10ADC 7 meters away @ 12V.

Voltage drop and cable heat dissipation are directly linked. The heat dissipated in the cable is I squared x r. That's the same as V x I where V is the cable volt drop.

100 amp cable is not needed for 10 amps anyway.

V is much lower with batteries, making I much larger. When you have less V a small loss is a large % of power.
Heating cable is a waste of energy.

Bruthal

Sir Liamalot wrote: »

V is much lower with batteries, making I much larger. When you have less V a small loss is a large % of power.

If you have 10 amps flowing in the cable, the losses in that cable, and therefore the heat dissipated by it, will be the same whether the voltage is 12 volts or a million volts.

Bruthal

Sir Liamalot wrote: »

V is much lower with batteries, making I much larger. When you have less V a small loss is a large % of power.
Heating cable is a waste of energy.

Yea you are talking about loss to the load now. Percentage is higher for a lower voltage wattage alright, but it would seem overkill to have 100 amp cable for a 10 amp load all the same.

Probably roughly 1 watt using 25 square, and 2 watts using 10 square for your circuit. Not a big difference.

As I understand it the larger the load the greater the loss. R= V/I
Resistance increases with temperature.

10A is nominal Max. I usually only use 4A most of the time . 20A MCB.
Negligible is entirely subjective. I increased my run-time by 20% pulling out the 1.5mm cable I had previously been using and replacing it with the 5mm. Also now I know that my battery terminal voltage is the same as my outlet voltage so I can run voltage sensitive devices for longer and know the thresholds. On a separate 10A line my absorption fridge now runs at full capacity because the resister that was the old cable was also swapped for 5mm and isn't limiting the supply anymore.
I haven't done a whole lot of maths on it tbh I just feed the variables into a cable voltage loss chart and spec it to <1% where possible.
For a 12v battery the difference between 12.8v and 12.2v is 100% of the usable capacity. This is hugely important when routing chargers as the charger needs to see the battery and not the end of the wires.

There's all sorts of ways around this copper isn't the only solution. More batteries in series, inverters, AC generators (not just the diesel ones) etc.
It's a trade-off of efficiency, expense, reliability and practicality.

Bruthal

Sir Liamalot wrote: »

I increased my run-time by 20% pulling out the 1.5mm cable I had previously been using and replacing it with the 5mm.

You increased battery runtime by reducing circuit resistance? I might be reading what you mean incorrectly.

Bruthal

Bruthal wrote: »

Yea you are talking about loss to the load now.

Ah yes now I get you...I always was. You can fully load mains cable, you need to cater larger cable for your acceptable loss with low voltage.

Bruthal

Bruthal wrote: »

You increased battery runtime by reducing circuit resistance? I might be reading what you mean incorrectly.

Indeed, although not entirely as simple as that. I cut the laptop off at 12.0v laden as this is the lower regulator threshold, with 0.2v drop on the line that meant I wasn't able to use all the power in the batteries. The rest is what was being used to heat the cables which was still a significant proportion although within the cable tolerances.

Bruthal

Sir Liamalot wrote: »

Ah yes now I get you...I always was. You can fully load mains cable, you need to cater larger cable for your acceptable loss with low voltage.

Yea but losses in the cable reduce the overall load on the supply. So if you power something with a 12 volt battery, and increase the supply cable to only slightly reduce the losses, you slightly load the battery higher, so reduce runtime.

brightspark

Here is a calculator you might find useful

http://www.bulkwire.com/wireresistance.asp

It uses AWG sizes, but does give the mm equivalent size

When you mention 1.5mm and 5mm are you talking about the CSA or the diameter?

Bruthal

Sir Liamalot wrote: »

Indeed, although not entirely as simple as that. I cut the laptop off at 12.0v laden as this is the lower regulator threshold, with 0.2v drop on the line that meant I wasn't able to use all the power in the batteries. The rest is what was being used to heat the cables which was still a significant proportion although within the cable tolerances.

0.2v drop in what conditions? What amps and cable run?

Seems a strange drop for the loading of a laptop.

It's only 1.5% loss!
7m run, 1.5mm cable.
Somewhere between 12.0v and 12.8v and 20watts or 90watts at room temperature, depending on what the batteries and the laptop were doing.

brightspark

brightspark wrote: »

When you mention 1.5mm and 5mm are you talking about the CSA or the diameter?

Diameter

brightspark

Sir Liamalot wrote: »

Diameter

You do realise then that the 5mm diameter is 11 times the size of the 1.5mm cable

Bruthal

Yea 1.5 diameter is 11 times smaller than 5 diameter. 25/2.25.

Photo shows that as well.

brightspark

Sir Liamalot wrote: »

I find that using 2 conductors often has a contact resistance imbalance and afterwards doesn't fully equalise the loading. In terms of running two 1.5mm cores where what's needed is a 3mm core.

If you are working with diameters then then 2 by 1.5mm cables would have a CSA of about 3.53mm sq

The 3mm has a CSA of 7.07mm sq... Twice the size




(Pi approximated at 22/7, no Pi on the calculator I used, but it's not important anyway)

Bruthal

Bruthal wrote: »

Yea but losses in the cable reduce the overall load on the supply. So if you power something with a 12 volt battery, and increase the supply cable to only slightly reduce the losses, you slightly load the battery higher, so reduce runtime.

The load being an op-amp doesn't care unless it's a really big resistor it just uses more current. If you increase the supply voltage by reducing the resistance you use less amps for the same watts.
Peukert compensation of batteries means that the less the instantaneous current draw on a battery the more the battery gives you over time. It's about internal resistance and chemistry.

brightspark

brightspark wrote: »

If you are working with diameters then then 2 by 1.5mm cables would have a CSA of about 3.53mm sq

The 3mm has a CSA of 7.07mm sq... Twice the size

I don't.
I work with acceptable losses and use single conductors.
That was an off the cuff example. Please excuse my lack of pi.

Bruthal

Sir Liamalot wrote: »

The load being an op-amp doesn't care unless it's a really big resistor it just uses more current. If you increase the supply voltage by reducing the resistance you use less amps for the same watts.

If its a switching regulator yea. If its a fixed load, then increasing the voltage increases the load wattage.

The watts dont just stay the same and current goes down, for a fixed load I mean, as opposed to a switching regulator.

Id still say 100 amp cable is overkill for 10 amps, it was a fair jump in cable size you went for. Although 100 amps may be a bit high for 16 square cable, so 100 amp cable may be a slight overstatement you made. Or maybe not.

I do use Pi when ordering pizza, two small are often a better deal than a large :pac:

Bruthal

Sir Liamalot wrote: »

I don't.
I work with acceptable losses and use single conductors.
That was an off the cuff example. Please excuse my lack of pi.

The PI doesnt matter for comparisons since its a constant in all actual size calculations.

brightspark

Sir Liamalot wrote: »

I don't.
I work with acceptable losses and use single conductors.
That was an off the cuff example. Please excuse my lack of pi.

Pi isn't relevant apart from determining the cross sectional area. I only mentioned it in case someone pointed to an error in my calculations.



it's the radius squared that creates the difference in voltage drop/current carrying capacity

you would need 4 x 1.5mm diameter cables to have the same volt drop as a 3mm diameter cable, not 2.

If you double the diameter, you square the area, and as was said earlier it's the C.S.A. that matters.




Most electricians here, especially those who mainly work with mains voltages would be more used to seeing C.S.A. cable sizes.

Bruthal

Bruthal wrote: »

If its a switching regulator yea. If its a fixed load, then increasing the voltage increases the load wattage.

The watts dont just stay the same and current goes down, for a fixed load I mean, as opposed to a switching regulator.

Indeed; resistive fridge got more effective. Regulators got more efficient.

To achieve meter accuracy to socket I wanted < 0.5% drop. It's only 14meters of cable lads hardly gonna break the bank. I've mentioned it's not a practical solution to upscale.

Fair enough I'll use CSA in future. I find diameter easier to eyeball.

brightspark

Sir Liamalot wrote: »

I do use Pi when ordering pizza, two small are often a better deal than a large :pac:

Are you sure?

http://lifehacker.com/an-engineer-explains-why-you-should-always-order-the-la-1532897984

A Supermacs leaflet I have here has a large pizza 14 inch @ €16, and the small 10 inch @ €11.

The 14 inch has almost double (49/25) the area of the small 10 inch at less than 1.5 times the price

The two small ones would cost you €22, and you only get 2% more pizza, than if you had paid just €16 for the large one.


(Apologies if I've gone too far off topic)

It did when I was in college and ordered a lot of pizzas. :rolleyes:
Depends on the menu, etc.