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?