Hybrid Car – More Fun with Less Gas

Anybody know where to get a small steam turbine? - Page 5

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Posted by Curbie on March 25, 2009, 8:36 pm
 
wrote:


As I understand it, heat comes from the sun in the form of solar
radiation, is then converted to heat (somehow) and the heat is
delivered to the Striling and only then can you play with Carnot. You
seem to be skipping solar radiation to heat conversion which will eat
a huge chunk of energy? More on this later...

I'm all for the continue option, you can't accomplish anything without
trying. I think A little more work on heat volume and temperatures are
needed.

This is where I'm getting gummed up with metrics, I don't understand
how 1kW of solar radiation is directly equitable to Carnot?

This is where I think there are some flaws. This is critical to the
functioning of the machine and should be calculated.

Here's the "meat & potatoes": the surface AREA of your receiver is
104 In.^2 =(2 * PI()) * (0.375 / 2) * 88 * percentage of surface
absorbing concentrated light.
NOT the 6.25 In,^2 you need for a "Concentration Ratio" that will
produce 1440F

If your already getting the results you need for screw the math, but
if your not, I recommend reading the parts of the Newton manuscript
that will help you convert solar light into heat (below).

Good Luck.

Curbie

http://etd.lib.fsu.edu/theses_1/available/etd-01052007-181917/unrestricted/Newton_Manuscript.pdf


Posted by Morris Dovey on March 26, 2009, 2:15 am
 
Curbie wrote:
 > Morris Dovey wrote:


Yes - losses will be substantial. I expect that reflection will be the
primary loss mechanism, with re-radiation right behind that. I've been
thinking of using coarse abrasives to scratch the surface to increase
the opportunity for multiple reflections - and I've been thinking about
a couple of ways to reduce the re-radiation. I've also thought about
using an evacuated glass envelope around the collection tube, but don't
have the budget to try that on the first model.


Carnot established an upper limit to the efficiency of any Carnot cycle
engine, based only on the temperature extremes. For an ideal engine with
a given heat input, that efficiency determines the mechanical energy output.

In my example, a 1kW (heat equivalent) input to an ideal engine with a
Carnot efficiency of 0.746 would produce a mechanical energy output of 746W.


It is, which is why I walked through the method of calculation a second
time. :)

However, the engine /won't/ be ideal - and one of the objectives of this
entire project is to discover /how far/ it is from ideal, the specifics
of /how/ it's not ideal, and to puzzle out what can be done to make it
/closer/ to ideal. I don't expect instant gratification...


Nope. I've already /measured/ that half of that trough width can produce
725F. Doubling the trough width while maintaining the same "bright line"
width will deliver twice the energy to the same area - so I expect a
temperature at least close to double the 725F, or 1450F.

You can even see a photo of the very first measurement attempts at

    http://www.iedu.com/DeSoto/Projects/Stirling/Heat.html

(about 2/3 down the page)

The interesting question, as you mentioned, is: "how well does any of
this do for heating the air inside the tube?"


I've been doing a passable job of converting radiant to heat - if I do
say so myself :) (sample at the link in my sig) - but I'm planning to
spend part of the week-end going over the materials you provided (thanks
again!)

The math provides the hooks on which I hope to hang understanding.
Whenever the math and the results don't agree, that's a signal that
there's something I need to understand better. Let's keep the math. :)


Thanks! A lot of my good luck so far has flowed from conversations like
this one.

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html

Posted by Tim Jackson on March 26, 2009, 8:10 am
 Morris Dovey wrote:

That is flawed physics.  First you need to be thinking about absolute
temperature not relative, and second, in radiation the temperature goes
as the fourth root of energy.


Tim Jackson

Posted by Morris Dovey on March 26, 2009, 1:41 pm
 Tim Jackson wrote:

Might work for you, but I'm working with old-fashioned temperature which
represents energy density per unit mass.

I think of and talk about temperature in units with which the person I'm
talking is familiar - and yes, convert (whatever) to Kelvins for
calculations.

Just out of curiosity, do /you/ have a thermometer graduated in Kelvins?

:)

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

Posted by Tim Jackson on March 26, 2009, 3:20 pm
 Morris Dovey wrote:

No I don;t have a thermometer calibrated in kelvins, nor one that will
measure 1450F.  But nonetheless when you are talking about radiation you
have to convert to absolute temperature to do any useful calculations.

I'm not criticising your units, but I'm giving you a C- for your physics.

It works like this.

The maximum temperature you can achieve in your target is that
temperature where the self-radiated power is equal to the incident
power.  This is calculated by Stefan's Law.
(http://en.wikipedia.org/wiki/Stefan%27s_law )
which says that the emitted power is proportional to the fourth power of
the temperature.

So if you double the incident power you will increase the absolute
temperature by about 20%

If you started with 725F, that's 658K, doubling the power will get you
783K, which I make to be 950F.

This of course is the 'ideal' situation, and does not account for any
power you may be removing to drive an engine, this of course has to be
subtracted from the incident power before doing the temperature calculation.

So for example if you were capturing a kilowatt of sunlight and getting
725F at your target without load, then it was radiating a kilowatt back
into space.  If you doubled the power by using a bigger mirror around
the same target then you could draw off one kilowatt while maintaining
725F. Two kilowatts come in, one gets radiated, just the same, the other
goes into your engine.

Tim

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