Solar Requirements - Confirmation
I'm taking a 3 month mountain biking trip through British Columbia this summer. Most of the time will be spent camping at National/Provincial parks without access to electric. When I really break it d...
Forum Discussion
"Let's say we have a converter / charger designed to deliver 60 amps to the batteries while bulk charging @ 14.4 volts. 60 x 14.4 = 864 watts, in which case a Honda EU1000i which is rated at 900 watts continuous would seem to be suitable. However, let's say that converter's PF is rated @ 0.7, which is pretty common for many converter / chargers. Plug that into this watts > VA calculator and we see that what we actually need that EU1000i genset to produce is 1234 VA.... Problem is it can't, which is why an EU2000i would be the more appropriate choice in this case."
VA applies to the input watts, not to the output watts. Converter efficiency is about 85%, so that 864 output means 1016w input. Now apply the 0.7 PF and get 1452 VA required.
In real life with my Kill-A-Watt my 55 amp converter (PowerMax PM3-55ADJ) powered by my Honda 3000:
124.7v, 11.06a, 980w, PF 0.7, 1383VA, doing DC 56.8a, set to 14.8v with battery passing through 13.7.
There is some confusion whether to use the converter's set voltage or the battery voltage for the output watts. Using converter's- 14.8 x 56.8 = 841w and with 980w input that is 86% efficiency.
While we are at it, the PF corrected 100 amper has this on the Kill-A-Watt
122.7v, 15.57a, 1854w, 0.97PF, 1910VA doing 102a set at 14.8v with battery passing through 13.95. Using 14.8 x 102 = 1510w output. with the 1854w input that is 81% efficiency.
So that is where I suspect PF correction comes at the cost of some efficiency 81 vs 85. But it is a trivial loss in the bigger picture.
If the efficiency were 85 instead of 81, input would be 1776w vs 1854, but with PF 0.7 instead of 0.97, VA would now be 2338 instead of 1910.
VA applies to the input watts, not to the output watts. Converter efficiency is about 85%, so that 864 output means 1016w input. Now apply the 0.7 PF and get 1452 VA required.
In real life with my Kill-A-Watt my 55 amp converter (PowerMax PM3-55ADJ) powered by my Honda 3000:
124.7v, 11.06a, 980w, PF 0.7, 1383VA, doing DC 56.8a, set to 14.8v with battery passing through 13.7.
There is some confusion whether to use the converter's set voltage or the battery voltage for the output watts. Using converter's- 14.8 x 56.8 = 841w and with 980w input that is 86% efficiency.
While we are at it, the PF corrected 100 amper has this on the Kill-A-Watt
122.7v, 15.57a, 1854w, 0.97PF, 1910VA doing 102a set at 14.8v with battery passing through 13.95. Using 14.8 x 102 = 1510w output. with the 1854w input that is 81% efficiency.
So that is where I suspect PF correction comes at the cost of some efficiency 81 vs 85. But it is a trivial loss in the bigger picture.
If the efficiency were 85 instead of 81, input would be 1776w vs 1854, but with PF 0.7 instead of 0.97, VA would now be 2338 instead of 1910.
