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

wrote:

*>Curbie wrote:*

*>> Ok,*

*>> *

*>> Your Stirling cycle web page looks fine which confuses me all the*

*>> more. Where did you consider gas law in your post? I don't know spit*

*>> about the metric system except that two liters is about the size of a*

*>> coke bottle, they didn't teach it when I went to school in the*

*>> stone-age.*

*>I have the same problem with metric - but this is the first problem of *

*>its kind that I've tackled, so I converted familiar units to metric and *

*>forged ahead in the hope that most of the physics/engineering experts *

*>from whom I might need to beg help would be more comfortable with that *

*>system.*

*>The approach I took in my post was "follow the power". I can guesstimate *

*>the input power, and can identify losses at the focal tube and losses *

*>within the engine because of the Carnot cycle efficiency limitations.*

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 know that there will be other losses since the mirror won't be a *

*>perfect reflector, the focal tube won't be a perfect absorber, air *

*>viscosity increases with temperature to prevent perfect flow, etc ad *

*>nausea. I doubt that I'll ever be able to isolate and individually *

*>quantify all of the loss mechanisms. My choices were to either shrug and *

*>continue or declare myself incompetent and quit. I shrugged.*

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.

*>> After reading, and re-reading your post I'm still at a loss over the*

*>> same two questions I'll re-word slightly:*

*>> 1) Whether your moving a solid or liquid you still have to produce*

*>> enough VOLUME of heat to heat the of gas to the desired pressure, for*

*>> EACH cycle, to do the designed work.*

*>At this stage in /my/ development, there isn't a design output *

*>requirement. I'm coming at this as a learning experience (which allows *

*>me to re-invent wheels as necessary) to discover, given a particular *

*>engine configuration, just how much work the thing can be tweaked to *

*>perform...*

*>I think [1] is true. From the energy perspective, if I increase the *

*>energy level of the gas by 1kW, I should be able to see 1kW worth of *

*>effect - and if I then reduce the energy level by 1kW, I should be able *

*>to see 1kW worth of effect in the other direction. *

This is where I'm getting gummed up with metrics, I don't understand

how 1kW of solar radiation is directly equitable to Carnot?

*>Carnot tells me that *

*>this isn't quite so, and that my effects will be dependent on the *

*>temperature extremes - and provides a formula for how much of my input *

*>power can result in actual output power: (1 - Tc/Th, temps in K).*

*>Since I have (a guesstimate of) the input power*

This is where I think there are some flaws. This is critical to the

functioning of the machine and should be calculated.

*>, and values for the *

*>temperature extremes, I can approximate the maximum (ideal) output power *

*>directly, without needing to involve pressure and volume variables.*

*>What this tells me is that if I can manipulate the Th and Tc values so *

*>as to produce a Carnot cycle efficiency of 74.6% (and if my engine were *

*>"ideal"), an input of 1kW would result in a 746W (1 hp) output.*

*>The big question is: "Just how far from 'ideal' will the engine be?" The *

*>only way I know to get an answer is to build it and see how well it does. :)*

*>> 2) VOLUME of heat from a parabola is a function of the area of the*

*>> collector and TEMPERATURE of heat is a function of the "Concentration*

*>> Ratio".*

*>> *

*>> I got into my solar concentrator spread-sheet and set-up a 2m^2*

*>> concentrator and adjusted the receiver area for a temperature of*

*>> 1440F, I got a geometric "Concentration Ratio" of 127 to 1 and*

*>> requiring a receiver AREA of 6.5 Inches^2 or 403mm^2. Since you are*

*>> talking about a parabolic tough 4x8 feet, I think your talking about a*

*>> 48" long receiver that has a total receiver AREA of 6.5 Inches^2?*

*>The actual width will be close to 88" and the target area will be 3/8", *

*>for a ratio closer to 88/0.375 or 234:1. 3/8" * 48 = 18 in^2. (I plan to *

*>wear welding goggles.)*

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

*>> I based my spread sheet on*

*>>*

*
http://etd.lib.fsu.edu/theses_1/available/etd-01052007-181917/unrestricted/Newton_Manuscript.pdf *
*>> I think submitted by an engineering student as his master's thesis. I*

*>> bring this up only because I know you've done software simulation and*

*>> you might be looking to cross-check for results. I like this Newton*

*>> manuscript because he proved all his calculations with answers which*

*>> is great for the metrically-challenged (me).*

*>One of the nice things about the simulator is that I can enter all my *

*>information in familiar units and do a front-end conversion to metric, *

*>then display results in both units.*

*>I'm very conscious that the simulator (so far) only simulates an ideal *

*>engine. The only thing I hoped for in writing it was a tool that allowed *

*>a "sanity check" on actual results. As I learn more about the real-world *

*>operation of these things, I may be able to make the simulator more *

*>realistic.*

*>> http://eosweb.larc.nasa.gov/sse/RETScreen/ is NASA site where you give*

*>> it your latitude and longitude and they give you a pile of data*

*>> including "Daily solar radiation".*

*>I appreciate the links. I've saved 'em for when I've had more sleep.*

*>> You have heat volume and peak temperature calculations which are over*

*>> my head, but I'm easily confused.*

*>I suspect that you aren't - and I think this stuff does take a bit of *

*>time to internalize. All the calculations I have on the web page were *

*>either lifted directly or derived from a handful of university web sites *

*>and wikipedia articles - and I've had a good bit of help from some of *

*>the folks who hang out on alt.solar.thermal.*

*>If it weren't for the Internet I'd still be saying: "I wonder if..."*

Posted by *Morris Dovey* on March 26, 2009, 2:15 am

Curbie wrote:

> Morris Dovey 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...*

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.

*> 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?*

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.

*> This is where I think there are some flaws. This is critical to the*

*> functioning of the machine and should be calculated.*

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...

*>> for a ratio closer to 88/0.375 or 234:1. 3/8" * 48 = 18 in^2.*

*> 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*

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?"

*> 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).*

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. :)

*> Good Luck.*

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:

*> *

*> 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.*

*> *

*> *

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:

*> Morris Dovey wrote:*

*>>*

*>> 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.*

*> *

*> 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.*

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:

*> Tim Jackson wrote:*

*>> Morris Dovey wrote:*

*>>>*

*>>> 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.*

*>>*

*>> 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.*

*> *

*> 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?*

*> *

*> :)*

*> *

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

>Curbie wrote:>> Ok,>>>> Your Stirling cycle web page looks fine which confuses me all the>> more. Where did you consider gas law in your post? I don't know spit>> about the metric system except that two liters is about the size of a>> coke bottle, they didn't teach it when I went to school in the>> stone-age.>I have the same problem with metric - but this is the first problem of>its kind that I've tackled, so I converted familiar units to metric and>forged ahead in the hope that most of the physics/engineering experts>from whom I might need to beg help would be more comfortable with that>system.>The approach I took in my post was "follow the power". I can guesstimate>the input power, and can identify losses at the focal tube and losses>within the engine because of the Carnot cycle efficiency limitations.