Answer 1:
Yes in principle …using photovoltaic
energy
There is no reason why electrolysis
could not be used as you suggest. The real issue
becomes the efficiency of this process…
how large of a solar collector is needed to
generate the power to make significant quantities
of oxygen and hydrogen? This would require
some extensive chemical engineering calculations.
However, to break the O-H bond of water, its
rather expensive energetically. Why do
electrolysis when you could just effect the phase
change. That is use
distillation?
Boiling sea water requires less energy than
splitting the O-H bond
Another possibility would be to use
photovoltaic energy to distill sea water. That is,
boil sea water and capture and condense the
fresh water. Then the issue becomes getting
the fresh water to where you want it.
You could do this on land with solar
collectors. Pump sea water to a facility and then
use solar power to boil the sea water and condense
the fresh water. Again the issue is the efficiency
and power available given a size of the solar
energy collection apparatus… once you have the
power from that, then its easy to determine how
much sea water can be boiled per unit time to make
the fresh water that is desired. Is this
rate of pure water generation enough to make it
competitive with other methods of fresh water
production?
If you want 1000 kg (one ton) of fresh water
per minute --> that is one ton per minute or 16.7
kg/s. So you need to condense 16.7 kg of boiling
sea water per second. That is, you need to add
sufficient energy to 16.7 kg of sea water each
second to get the sea water to a flash.
Since the enthalpy of vaporization of sea
water is roughly 40,000 Joules / 0.018 kg =
2.2 MJ/kg, that is 2.2 Megajoules per kg. This
is the amount of energy needed to boil 1 kg of sea
water(approximately)
One would also have to add in the heat required
to go from 20 C to 100 C… that is another 50%.
So allowing for that ONE NEEDS roughly 3
MJ/kg per second from the power source to make a
ton of fresh water per minute.
So you now need to see what size photovolatic
system is needed to generate roughly 3 MJ per
second, or 3 MW (3 Mega Watts).
You need to go to the literature and read
about photovoltaic systems… you need one that
can deliver > 3MW
To allow for heat loss and other
irreversibilities you probably need a PhVolt.
generator of say 10 MW.
So, study this and see what size collection
areas are needed given solar insolation in
Southern California. Then send me your musing at
spera@geol.ucsb.edu
The solar energy flux at surface is about 4
J/cm2 min. In a non focusing collector
the temperature can reach 90C. If we operate a
heat engine using the collector as the heat source
and a low temperature reservoir at 25 C,
calculate the area of the collector needed if
the heat engine is to produce
1 horse power ( 740 Watts= 740 J/s). Assume
the engine operates at maximum thermodynamic
efficiency.
ANSWER 6.2 meters
See if you can understand and solve this
problem and report back.
This problem requires consideration of the
second law of thermodynamics. When
operating an engine between a high T reservoir and
low T reservoir the useful work is some fraction
of the heat used to accomplish the work.
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