Faradays heating estimate for 8Xa and Variable Plate Cell

Calculate Water volume (4 in inside diameter)

4”dia, plate height plus 4” water over plates

V=PI()* radius^2*height = 201 cu in

Calculate plate volume (1/4 in plate backer)

V=height*width*thickness*2

V=3*12*(1/4)*2 = 18 cu in

Water volume = total volume – plate volume

Water volume = 201 – 18 = 183 cu in

Calculate lbs of water

1 gal water = 231 cu in, and 1 gal water = 8.35 lbs

Water volume = 183 / 231 = .79 gals

Water weight = .79 * 8.35 = 6.6 lbs of water in container

Calculate how much power goes toward heating the water in the container

Total input power 80V@ 4A = 320W

48700cc/hr = 48.7L/hr = 0.8117L/min

Watts to make gas = .8117(L/min) * 140 (W/L/min) = 113.6W

Calculate watts to heat water = Total input power – power to make gas

Watts to heat water = 320 – 113.6 = 206.4W

Calculate water temperature rise in 20 minute run

206.4W = 11.748BTU/min

BTUs in 20min = 11.748 * 20 = 235BTUs

1BTU = raise 1lb water 1degF

Water temperature rise in 20 minutes = 235 / 6.6 = 35.6degF

If the temperature of the water started at 70degF and ran for 20 minutes then the ending cell temperature should have been at 105.6degF. Reports indicate the cell was cool to the touch, I calculate it should have been warm to the touch.

I know there is a first hand report that documents a demonstration which says how long the cell ran and didn’t get hot, but I couldn’t find it, so I used 20 minutes.

What do you guys think? Did I make a mistake in my calculations or logic?

Calculate Water volume (4 in inside diameter)

4”dia, plate height plus 4” water over plates

V=PI()* radius^2*height = 201 cu in

Calculate plate volume (1/4 in plate backer)

V=height*width*thickness*2

V=3*12*(1/4)*2 = 18 cu in

Water volume = total volume – plate volume

Water volume = 201 – 18 = 183 cu in

Calculate lbs of water

1 gal water = 231 cu in, and 1 gal water = 8.35 lbs

Water volume = 183 / 231 = .79 gals

Water weight = .79 * 8.35 = 6.6 lbs of water in container

Calculate how much power goes toward heating the water in the container

Total input power 80V@ 4A = 320W

48700cc/hr = 48.7L/hr = 0.8117L/min

Watts to make gas = .8117(L/min) * 140 (W/L/min) = 113.6W

Calculate watts to heat water = Total input power – power to make gas

Watts to heat water = 320 – 113.6 = 206.4W

Calculate water temperature rise in 20 minute run

206.4W = 11.748BTU/min

BTUs in 20min = 11.748 * 20 = 235BTUs

1BTU = raise 1lb water 1degF

Water temperature rise in 20 minutes = 235 / 6.6 = 35.6degF

If the temperature of the water started at 70degF and ran for 20 minutes then the ending cell temperature should have been at 105.6degF. Reports indicate the cell was cool to the touch, I calculate it should have been warm to the touch.

I know there is a first hand report that documents a demonstration which says how long the cell ran and didn’t get hot, but I couldn’t find it, so I used 20 minutes.

What do you guys think? Did I make a mistake in my calculations or logic?