Quote:
Originally Posted by mitiempo
The sizing calc I use gives 4 gauge for 100 amps with a wire run of 6' (total of both wires) with 2% voltage drop.

But you can't use 100A as the input to the voltage drop calculator. I have been arguing this for days and am just about to give up but will try one more time. The peak current from the battery greatly exceeds 100A. Several people have ridiculed me for my explanation but it is nevertheless a fact that the peak current demand an inverter puts on the battery is 50100% more than the DC demand. Everyone "thinks" that inverters draw a pure DC current from the battery. But they don't. They have to draw large current surges 120 times per second. That current surge is needed by the inverter on the positive and negative halfcycles of the AC current. The inverter draws the current surges always in the same direction and just reverses them at the output of the inverter to make pure AC current.
Here is how you can estimate the worst case peak 12V battery current demand:
peak current = AC power out * 2 / 12 / efficiency
So for 1250W
water heater load it is:
1250 * 2 / 12 / 0.8 = 260A.
But the DC load is only:
1250 / 12 / 0.8 = 130A
Some good inverters have filters inside that reduce the peak demand a little bit. But the best ones only reduce it by about 25%. And it adds a lot of cost to reduce the peak current more than that. The larger the AC current demand the larger the peak current that must come from the battery. At the risk of everyone laughing at me again here it is:
The AC current output by the inverter has to come from somewhere. The largest energy storage device is the battery. Ergo, the battery has to deliver most of this peak AC current.
If you don't believe it put an AC clampon RMS current meter on the battery wire while the inverter is running maximum load. There will be AC current there and it is not due to a confused meter. But most meters won't show the peak current. For that you have to use a special wideband current transformer and an oscilloscope which almost nobody has.
Someone will ask why is the worst case peak current twice the DC current? The answer is simple. AC voltage and current numbers are given typically in RMS because it makes power calculations easy. But the peak voltage and current are both 1.414 times the RMS value. So take a
water heater element. We give it 120V RMS and 10A RMS for 1200W. But the peak AC voltage is 120*1.414 and the peak current is 10*1.414 and therefore the peak power is 120*10*1.414*1.414=1200*2=2400W. So 2.4kW is the peak power being delivered to that water heat element when the AC voltage waveform is at its positive or negative peak. That peak happens twice every AC cycle so for 60Hz it happens 120 times per second. Now where do you suppose that 2.4kW is coming from 120 times every second? Most all of it comes from the battery. There is no battery inside the inverter. There are some small capacitors but they don't have near enough
storage capacity to supply that 2400W for any length of time. They quickly discharge and the battery has to make up the difference. Thus 120 times every second a big surge of current has to flow out of the battery. Not coincidentally 120 times a second the current flowing out of the battery goes nearly to zero when the AC voltage waveform is crossing zero.