[Stoves] Radiation Analysis for Gasifier

Crispin Pemberton-Pigott crispinpigott at gmail.com
Sat Mar 17 16:03:23 CDT 2012


Dear Ron and Marc

 

Before addressing the corrections or the exactitude of the analysis I want to first recall the purpose of the calculation Marc has done. He is looking to see if it is possible for the heat transfer efficiency to have doubled by placing a dome of mesh that got very hot under a pot. The postulations about how it might have accomplished the feat are separate from the question of if it can be done.

 

The figures Marc chose are reasonable, even favourable to the case. In order for the dome to have doubled the cooking efficiency (reducing the time to boil) it would have to be shown that from a low radiant baseline the addition of a high radiant object could account for the change.

 

If the absorbed power with the dome was at a rate of 1.65 kW, as Marc has calculated, then the baseline case is half of that, viz 0.83 kW. While there is certainly a radiative element in the baseline case, we do not know what it is because we do not have a photo of the stove taken in the IR band. Let us suppose it was 20% radiant and 80% convective, giving some credit for hot cases such as water vapour and CO2 being emissive in the IR.

 

Next, assume a baseline thermal efficiency for the whole system of 30%. That means the pot was absorbing 0.825 kW from a 2.75 kW fire if I do the sum correctly to an additional decimal place. 

 

Of that 0.825 kW, 20% is radiant and 80% convective. That means 165 Watts of radiant heat and 660 Watts of convective heat.

 

In order for the net power to double, the ‘wasted’ heat would have to be converted to IR and emitted to the pot. As is shown in the drawing, there is quite a substantial area of the pot+housing that is ‘not pot’. Disregard the low incident angle of the IR, even though an experiment reported here this week showed a significant change in the efficiency in the reception of low angle IR. 

 

If an additional 0.83 kW was to be obtained from the 2.75 kW fire, assuming no change in the fire power or reduction in the excess air, the radiant contribution would have to rise from 165 Watts to 825+165 = 990 Watts, plus retain the 660 Watts of convective heat. The system efficiency will have to rise to 60% from 30% to achieve this. That seems unlikely but let’s not draw conclusions just yet.

 

What Marc is showing is whether this is possible at all. Using the generous temperature of 750 C and a wire mesh of 10% coverage and an unbelievably generous emissivity of ε = 1.0 for the wire the emitted power is 301 Watts, 1/3 of the needed IR power, and not yet discounting for the fact that only about 60% of the radiant energy is hitting the pot. If you consider that the mesh radiates downwards as well as out and away, the % is probably even less.

 

Let’s be generous and add more surface wire: 40% wire coverage. The power emitted jumps to 1205 Watts which is above the needed 990. Add the more realistic emissivity for stainless steel wire of ε = 0.6 and it drops to 723 watts. Then reduce the temperature to a more realistic 650 C and it drops to 477 W. Factor in the losses to the local environment that is ‘not pot’ and it drops to 346 W. As the pickup of heat is not 100% efficient, the IR heat available is even less. My guess is closer to 250 W (about 70%).

 

Even if there was zero heat transfer from IR in the baseline, an increase of 250-350 Watts is not enough to cut the cooking time in half – it is still 3-fold short of making this happen. And that still has to be factored for the mesh area which is probably less than 40%, and we must consider the round wires emitting in all directions.

 

Conclusion: there is absolutely no way for a radiant dome to double the cooking efficiency of this particular stove. Whatever the differences are between the two burners, the improvement in IR is at the most no more than a few % because the radiant heat from the baseline is not zero and a realistic calculation of what it could be gives about 250 Watts absorbed IR energy from the dome, or 16% of the total heat getting into the pot.

 

All of the above does not say that a radiant mesh dome can’t increase the efficiency of the stove. It just shows it can’t double the it. Because there is a real possibility it will help, this spreadsheet can be used to optimise the effect, and to calculate what effect a radiant structure might have. Designers, rejoice.

 

Regards

Crispin

 

Final numbers used:

 


Prepared 3/10/2012 by Marc Pare

	

Reviewed and Revised by Crispin Pemberton-Pigott 2012/3/17


Re-released 2012/3/17

		
			

Dimensions


radius_pot

mm

125


radius_dome

mm

125


gap height

mm

40

			

Area of mesh


percent metal

 

0.3


A

mm^2

49087.38521


A_mesh

mm^2

14726.21556


 

m^2

0.014726216

			

Radiation


stefan boltzmann (σ)

W/(m^2-K^4)

5.67E-08


emissivity (ε)

 

0.6


T_mesh

K

923


T_pot

K

333


 

 

 


q

[W/m^2]

24272.81586


Q

W

357.4467187


 

kW

0.357446719

			

Power to boil water in 1L, 222s scenario


Spec Heat Water

kcal/kg-C

1


Temp Difference

C

74


Density Water

kg/m^3

1000


Volume Water

L

1


 

m^3

0.001


Energy required

kcal

74


 

kJ

310.06


Water evaporated

g

25


Latent heat of Evap

J

                 2,257 


Heat absorbed by pot

J

             366,485 


Time

s

222


Power

W

                 1,651 


 

kW absorbed

                   1.65 

			
			

Percent of heat that might be contributed to cooking by Radiation from a red hot mesh dome under the centre.

 

21.65%

			
			
			

Bonus View Factor Calculation

			

View Factor


r_1

mm

125


r_2

mm

125


a

mm

40


R_1

 

3.125


R_2

 

3.125


X

 

2.1024


F_1-2

 

0.727

			

Percent of heat actually contributed to cooking by Radiation from a red hot mesh dome under the centre.

 

15.74%

					

 

 

From: rongretlarson at comcast.net [mailto:rongretlarson at comcast.net] 



 

Marc, Matt etal

    Two problems I see with your analysis.

   First is minor -  Your equation 1 show a linear variation with temperature, whereas it should show a 4th power.   But you were using the proper fourth power in your Excel spread sheet - so this was just a typo.

   More serious is your assumption that the metal portion of the mesh is 10%.  This is appropriate only for a very few mesh per inch and fine wire.  My guess is that Paul's mesh could be more like 30-40% - which will change your conclusion a great deal.  See pages like:

    http://www.twpinc.com/wire-mesh/TWPCAT_12/p_014X014S0170W48T

   So this is to ask Paul Olivier for a visual check on what he was using in his particular strainer.  A manufacturer and model number would be helpful, if available

   Conversely, I worry about assuming the mesh was as high as 750 degrees - based on the color in Paul's photo.  But I am used (vaguely - long time ago) to looking at solid materials through a peep hole in ceramic kilns.  The openness of the mesh must affect our visual color/temperature calibrations.  Anyone up on that?

   I'd like to know more about the maximum possible kiln power level - by knowing the amount of rice husk consumed per unit time  (same as question asked by Crispin, I think).  From this we can start to compute the convective heat transfer coefficient.  In other words, what part of the output energy was not getting into the cookpot?  I think we can assume a larger portion of the radiative energy was captured than of the convective.

   Also the amount of water evaporated should be easy to measure rather than guesstimate.   I also would feel better running longer and using the weight evaporated for these energy capture-power computations.

Ron

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