On 08/09/2010 02:08 AM, Ken Gotberg wrote:
Hi David
Boy, I don't have a clue about what you
are talking. A cubic centimeter (cc) also called a milliliter (ml)
takes one calorie to raise its temperature one degree centigrade also
called one degree Celsius. To raise one cc one hundred degrees
Celsius takes one hundred calories and to raise one thousand cc or ml
i.e. one liter takes one hundred times one thousand and is equal to
one hundred thousand calories. Since one calorie is equal to 4.184
Joules, it takes 418,400 Joules (0.4184 MJ) to raise one liter of
water one hundred degrees Celsius.
You are correct, I slipped a decimal. Mea culpa.
These are your assumptions about the
temperature increase. If we add this to the 2.2 or 2.26 mega Joules
we get 2.26 + 0.4184 MJ or 2.6784 MJ. If you put this on an exam,
depending on what was being taught, you would probably fail because
1. you are assuming that everything is linear which it isn't and 2.
there are far too many digits in the answer when we aren't even sure
about the 2.2 MJ per liter.
Okay, and the 1° C/1 gm/1 cal holds ONLY for water, not water mixed with
CaCl2, nor does the boiling point stay constant with changing
concentrations. Yes, the heat required to change water from fluid to
gas at 100° C should be an approximation close to 2.2 MJ/L, but the
temperature may need to be much higher, or the CaCl2 may have a
different specific heat, let's use 2-3 MJ/L as an engineering approximation.
At 2 MJ/L, you will need 8 MJ/person/day, at 3 MJ/L, 12 MJ/person/day.
Is this correct? That looks like 3 KWH/person/day, assuming 100%
efficiency, just to make water vapor. Then there is the cooling (since
humans do not do well with 100° C live steam), which requires energy.
For a family of 4 at 10 KWH/day/person means (with a solar efficiency of
under 25%) 40 KWH/day/person, 160 KWH/day incident radiation. I bet
your proposed system is less than 25% efficient and 160 KWH/day
represents a large area.
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