MYTH: Clean water! A wonderful dream :))))
If you ignore the method of imploding the Brown’s Gas over the contaminated water (this would cause an explosion) and use instead a separate chamber to implode the Brown’s Gas, then allow the already existing water vapor in the contaminated chamber to be sucked into the Brown’s Gas container, to be condensed – YES it will work! BUT the electricity required to produce enough vacuum by this method would be much better spent driving a conventional vacuum pump. You’d get several times more water without the possibility of explosion.
See For Yourself:
Let’s do some math. We’ll assume we have excluded all air from a container containing contaminated water and that container is at 20°C. Water vapor in the contaminated container would be at a pressure of 0.02 kp/cm2 and would contain 0.01729 kg/m3. One liter of water turned to pure Brown’s Gas would occupy about 3.7 m3 of volume. We introduce the water vapor to the vacuum created by the imploded Brown’s Gas; which at .01 kp/cm2 and 20°C, water vapor density is 0.0072516 kg/m3. Thus, one liter of Brown’s Gas (imploded) would recover a maximum of 0.037 liters of water (less because Brown’s Gas water already occupies volume and as gas from the contaminated tank rushes into the Brown’s Gas tank, Brown’s Gas tank pressure rises).
Each liter of water takes about 10,800,000.00 joules to create 3700 liters of Brown’s Gas, recovering a maximum of 0.037 liters of water. Assuming temperatures remain the same. If we assume this process took one hour, then we use a constant of 3 Kwh to purify 37 grams of water per hour using this Brown’s Gas technique.
If We Use A Simple Vacuum Pump:
We move water vapor from an environment of 0.01 kp/cm2 to 1 kp/cm2 (vacuum to atmospheric pressure). Increased pressure at the same temperature causes the excess moisture to condense out. The moisture holding capacity of vapor per cubic meter is fixed by temperature, not pressure. Thus pumping 3.7 m3 to atmospheric pressure recovers about 0.024 liters of water. Our pump would use about 0.1 Kwh of power.
If We Use A Simple Evaporation System:
Working at atmospheric pressure and heating the contaminated water (with solar is best, but we’ll assume using electric) to 30°C, then we could recover 0.024 liters of water with 0.02 Kwh of power.
Does It Make Sense To You?
Why purify water with Brown’s Gas when a simple evaporation system is simplest and least expensive to operate? The evaporation system is simple and inexpensive to build, is not dangerous and uses only a very tiny fraction of the power used by the Brown’s Gas technique. Of course, evaporation systems using solar power do require some space and are not really portable.
Are You Starting To Get A Picture Here?
Yes, Brown’s Gas can do a lot of things, but can it do those things in a more practical manner than other (already existing) options?
At Eagle-Research we dispel myths, and we also find practical uses, see Fabulous Uses :))