Sverige
A Brit in Sweden
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?
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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