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Calculating cable power losses

Sverige

A Brit in Sweden
Joined
Oct 8, 2020
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59.5N, 15.5E
I watched an interesting video yesterday, which @gnubie posted In this thread. Today I got to work calculating the power losses in my cabling from panel to charge controller and actually ended up installing a second cable in parallel as a result.

My current set up is a 160W poly panel which will charge my 12V SLA car battery at around 10A in full sun. Soon I’ll be adding another identical panel in parallel (due to partial shading, this should work better for me than series connection). So my 10A will soon become 20A, and what a difference that makes to the cable losses (I2R!)

Total cable run from panel location to charge controller is 16 metres. Until today, I had a 4 square millimetre OFC copper cable installed but have doubled that with today’s efforts to 8 square millimetres, by adding a second cable. Using the resistivity figure of 4.61 mOhm per metre from the video gnubie posted, my calcs are as follows:

One panel, 10A (current config)
Power lost in 16m 4mm2 cable at 10A= 100x4.61/1000=0.46W/m x16=7.36W
Power lost in 16m 8mm2 cable at 10A= 100x2.305/1000= 0.23W/m x 16=3.7W

Two panels, 20A (forthcoming upgrade)
Power lost in 16m 4mm2 cable at 20A= 400x4.61/1000= 1.84W/m x16=29.5W
Power lost in 16m 8mm2 cable at 20A= 400x2.305/1000= 0.92W/m x16=14.7W

So with today’s single panel, installing a second cable to halve the circuit resistance has saved me around 3.7W. Taking into account the higher current I’ll be running next week when I add a second panel, I’ve saved 14.7W. Was it worth the cost of the cable? I don’t know. How do you put a value on wasted watts in off grid solar? In the video, the mains tariff is used to calculate the cost of the power losses over the coming years. How would you make a meaningful value comparison for solar?

I was also wondering, how many watts per metre of cable losses are needed before the cable starts to get warm?
 
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I just use this:


Do you have the panel connected directly to the battery, or do you have a charge controller? I certainly hope the latter.

1602694224888.png

5% is acceptable per NEC.

How to you relate that to cost?

3.7W loss.

270 hours of charging at 3.7W is a kWh. Around here, a kWh is about $0.12, so for every 270 hours of charging, I would save $0.12.
 
I just use this:


Do you have the panel connected directly to the battery, or do you have a charge controller? I certainly hope the latter.

View attachment 25006

5% is acceptable per NEC.

How to you relate that to cost?

3.7W loss.

270 hours of charging at 3.7W is a kWh. Around here, a kWh is about $0.12, so for every 270 hours of charging, I would save $0.12.
Yes I use a charge controller.

The cable power loss calculator you linked to is interesting, but produces very different figures to the ones I calculated. Unless anyone can point out a mistake in my sums, I’m going to conclude the online calculator is inaccurate, especially when it produces numbers like this for the case I worked out as 14.7W power loss above:

9643B2F4-8808-4CBF-BD99-881BF4550ABC.jpeg
2.64V drop across a cable carrying 20A will be 52.8W dissipation (P=IxV). Three and a half times more than my sums indicate. I do hope I’m right and they are wrong, otherwise I’m going to have trace heating installed all along my attic space! I suppose the mice will enjoy a comfy winter if that’s the case.

EDIT: It seems 8 AWG is a better approximation to my 8mm squared copper cable. The online calculator still produces numbers significantly higher than my calculated result though, so something is still out of whack. Quite possibly my sums?!
 
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Is your 16m of wire in the opening post taking into account that there are 2 conductors (the + and the - wires) for a total wire length of 32m?
 
Is your 16m of wire in the opening post taking into account that there are 2 conductors (the + and the - wires) for a total wire length of 32m?
No, it is a 16m long two conductor cable, so you’re right, the circuit is 32m long in total I suppose. Maybe that accounts for the difference?

At what level of dissipation will a cable start to get warm and how high before it‘s dangerously hot I wonder?
 
Yes, that would be it as you are doubling your resistance. The temperature mostly comes down to the insulation rating and the cable environment. If the cable is in open air it can get rid of heat much more effectively than if it's in conduit / inside a wall / buried in wall insulation etc. There's a lot of maths involved but there's going to be tables somewhere already set up that tell you the expected temperature rise for a given piece of wire with X watts applied to it. Once you know the rise you add in the ambient and pick an insulation based on that. A lot of insulation is rated to ~75c.

Here's a calculator, don't know how good it is.
 
A wire has a LOT of surface area, so even a notable wattage that would burn your hand in a small package (A 60W light bulb) will be barely warm in open air over 16m of wire with a LOT of surface area to cross sectional area.

Please confirm you're running the wires carrying the "12V" from the charge controller to the SLA over 16m.

I can promise you that the voltage drop calculator is accurate.
 
No, the 16m cable is from the solar panel to the charge controller. From the CC to the battery is less than a metre.
 
Is this an MPPT controller? If so, what is the Vmp of the panel and the Imp of the panel?

Assuming it's around 18V, THAT's the number you should be putting into the drop calculator. That 50% increase in voltage and small decrease in current will have a favorable effect

If it's a PWM, then yeah, battery voltage and Vmp.

If it's a MPPT, you should add your second panel in SERIES.

Assuming 18V and 9A from the panel to the MPPT controller for your original wire:

1602743815888.png
New wire:

1602743874424.png

Second panel in series instead of parallel:

1602743972987.png
 
Thanks, it’s a PWM controller and my panels need to be wired in parallel due to partial shading. If I series connect them I’ll inhibit my production far more than the extra cable losses which come with the higher currents resulting from parallel connection.


With parallel connection I can offset my panel mounting positions so the shade only hits one at a time, and I have decent production for the maximum number of hours.
 
Gotcha. Then battery voltage is the correct input.... however, now you see yet another advantage to MPPT... :)
 
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