[Gasification] back to hard "tinkering"

[John Bertl]


Peter / Belize
wrote:
>A 1000 gallon rain vat costs $500 US -- That is a round cylinder 5 ft 
>diameter -- 6.5 feet long -- You end up with 13 feet of trough when split.
>total surface area facing the sun would be: 104 ft sq.

I am not sure how you did your math as I cannot repeat it.
I calculate the area of the two half cylinders facing the sun
as being 65 square feet.  (5ft wide x 6.5ft long x 2 cylinders)
65 sqft is the area perpendicular to the sun.

If I calculate the area of the surface of a cylinder, which I
think you might have done, I get 102 sqft not 104
(3.14159 x 5ft x 6.5ft)

Additionally I would like to add that a half cylinder will not
focus the sun’s rays on a collector because it has a common
radius.  The only shape that will focus the rays of the sun is
a hyperbola.  A skilled person could crudely warp the walls
of a half cylinder into a hyperbola shape by cutting off the
ends and attaching hyperbola shaped ends.

This is not to take away anything you have said as you are
basically correct.  The work is defined as the enclosed area
of a Rankine Cycle T-S  diagram.

See the diagram here

http://en.wikipedia.org/wiki/Rankine_cycle


There are four processes in the Rankine cycle, each changing
the state of the working fluid. These states are identified by
number in the diagram above.

    * Process 4-1: First, the working fluid is pumped from low
        to high pressure by a pump. Pumping requires a power input
    * Process 1-2: The high pressure liquid enters a boiler where it
       is heated at constant pressure by an external heat source to
       become a saturated vapor.
    * Process 2-3: The superheated vapor expands through a turbine
       to generate power output. This decreases the temperature and
        pressure of the vapor.
    * Process 3-4: The vapor then enters a condenser where it is
        cooled to become a saturated liquid. This liquid then re-enters
        the pump and the cycle repeats.


The theoretical maximum efficiency of a heat engine equals the
difference in temperature between the hot and cold reservoir divided
by the absolute temperature of the hot reservoir. To find the absolute
temperature in kelvins, add 273 degrees to the Celsius temperature.
Looking at this formula an interesting fact becomes apparent.
Lowering the temperature of the cold reservoir will have more effect on
the ceiling efficiency of a heat engine than raising the temperature of
the hot reservoir by the same amount. In the real world, this may be
difficult to achieve since the cold reservoir is often an existing ambient
temperature.

See the T-S diagram here

http://en.wikipedia.org/wiki/Carnot_cycle

In the diagram above, the work is the white area W.
The higher we raise temperature Th, the more work.
The lower we lower temp Tc, the more work.
Anything that increases the area increases the work.

Peter / Belize
wrote:
>The system you describe probably has over all
>efficiencies of a fraction of one percent -- ergo -- they do not publish 
>that minor detail --
>eh??

This is very true because the efficiency is
calculated as the the area under the
curve from the high temperature
down to absolute zero being 100%
Therefore, a temperature difference
of a few degrees near ambent
temperature is almost zero efficiency

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