[Stoves] Request for help on TLUD operating data

[Ronal W. Larson]

Crispin:

	1.  Thanks for the several figures and explanations below.  Unfortunately,   I need considerable more to figure these out.  If there is any report on this for any high quality TLUD type stove, I’d love to read that.

	2.  To try to obtain an oxygen balance, you have difficulty because you are not measuring water content in the exhaust gases.  I think my need is much easier and simpler.  You must be measuring the free oxygen in the exhaust stream, if you are deducing the O2 content of NO and NO3.  The free oxygen content (that meant when we talk about roughly 21% oxygen) is the number I am most interested in.  Can your supply the time history of that single quantity for the two stoves you are describing below?  (And any other stoves).  What number are you recommending for O2 in the exhaust stream?

	3.  The abscissas and ordinates on all the figures are very hard to read.  If you could give the min and max values that would help a lot.

	4.  Do I understand that between the red and yellow vertical lines that CO/CO2 is greater than 2%?  How large might that get?

Ron


On Aug 17, 2014, at 1:24 AM, Crispin Pemberton-Pigott <crispinpigott at outlook.com> wrote:

> Dear Ron and Gas Calculators
>  
> 1.      I look forward to your chart.
>  
> I will do this is two stages. First, a chart shows the result of the following calculation:
>  
> Measure as many gases in the stack as you can and calculate the total oxygen present expressed as O2. If it is in the form of NO, the you take NO/2 to give an O2 value. If it is NO3 then you take the NO3 ppm x 1.5 to give O2. And so on. Add up all the O2 in ppm and plot it on the chart.
>  
> In theory the total O2 will be 209,450 ppm because that is the atmospheric concentration. In the plot below, the value is used as a data quality check to see if the total is wildly at variance from this figure. If it is, it represents a hardware problem such as a cell reading very high or low. The plot we are interested in is the Green line, ignore the others. The name of the line is ?O2.
>  
>  
> <image003.jpg>
>  
> You can see that the value gets as low as 195,000 ppm(v).  The suppression of the line from the expected value is because of two things: the burning of Hydrogen to form water vapour which was not in the list of gases measured and the evaporation of water from the fuel itself which dilutes the sample. This dilution is frequently not considered when making mass balance and volume calculations (big error).  The (sort of) spikes downwards are caused by refuelling the stove with moist fuel, in this case placing fuel on top of an existing fire.
>  
> The fire in the example above is not a TLUD and is not operating as a gasifier (though obviously any fire involves making gases and burning them). Note that there is virtually no fuel moisture or hydrogen by the time the fire reaches the char burning stove which is the last hour of the burn. The ?O2 line is nearly on the theoretical level.
>  
> Here is the same chart type showing the same calculation for a TLUD burning 14% moisture fuel in a gasifier that operates as a pyrolyser then a char burner.
>  
> <image005.jpg>
>  
> Notice the shape of the suppression is completely different. In the beginning there is the ignition of the kindling materials which evaporates moisture from the upper layer of fuel before it starts to emit gases.  That is the initial suppression. The situation returns to normal (almost) after 20 minutes as the fire develops because the hot fuel starts to burn at a much lower firepower than during ignition. The difference between 209,450 and 205,000 is the water vapour (which is not measured ) created by burning fuel hydrogen.
>  
> The values plotted are independent of power, remember, it is just the sum of what is found compared with what ought to be there if there was no fire at all.
>  
> Once the fire starts to go really well, it evaporates fuel moisture at an increasing rate all the while burning some hydrogen. The sum of the suppression is the total of the water vapour from H2 combustion plus evaporated fuel moisture.
>  
> The down-peak is the point at which the fuel pyrolysation rate maximises. When it is completely charred, the char burns but as you can see, there is a lot of hydrogen still emerging. The plot for an Oorja stove burning wood pellets is similar but with a much faster upward recovery.  This test was ended when 90% of the initial fuel mass was burned, but if it had continued the line would have been horizontal at just under 209,450 (caused by residual hydrogen making unmeasured water vapour).
>  
> An interesting recent development is that we have a way of getting a reasonable H2O measurement in real time even with high concentrations from very hot stacks so in future I hope to show additional information.
>  
> The salient point of the second chart is that the concentration of water vapour in the char bed of a pyrolyser rises a lot. It would be incorrect to say it is the difference between the theoretical 209,450 and the observed 182,500 represents the value we are discussing (in the char bed). This measurement is made in the stack or hood.  The gases are not yet burned in the char bed therefore has a completely different emissions profile. To find out what is happening inside the char bed the sample has to be drawn from within it, about the middle, perhaps. 
>  
> Assuming, just as an example, that the ratio of the number of moles of gas in the char bed and in the stack was 1:3, then the suppression would be 31% meaning the water vapour concentration in the gas is 310,000 ppm(v).
>  
> The calculation is
>  
> (209,450-182,500)/[(182,500/3)+(209,450-182,500)]                          [2]
>  
> where ‘3’ is the ratio between the number of moles of combustion products and the number of moles of gas emerging from the bed.
>  
> The resistance to flow would be calculated on the basis of the temperature at the time and gas composition. The latter evolves quite a bit.
>  
> One last note, look at the sudden drop in the production rate of water vapour at minute 92 when the bottom of the pile flashes dry, creating an upward jump in the ?O2 number. That jump is preceded by (at minute 88) a sudden increase in gas production, something noted by many TLUD builders. That jump is caused by the fact there is no damp fuel under the bottom layer to it flashes dry. In TLUD’s with poor secondary air management, there is usually a peak in CO production and a concomitant peak in the CO/CO2 ratio at that time.
>  
> Note the black line at 88-92 minutes in Chart 3
>  
> <image008.jpg>
>  
> The green line is excess air (calculated on a chemically balanced basis).
>  
> That flash of gasification of the bottommost fuel later at minute 88 is the immediate cause of the rise in the CO/CO2 ratio. If you can mentally integrate the two lines in Charts 2 and 3 you will appreciate that the best combustion (in this stove) happens when there is still some water vapour being formed. In this stove the secondary air is not particularly well designed. Here is the real time PM and cumulative mass burned chart:
>  
> <image011.jpg>
>  
> The black line is real time PM2.5[mg/m3], the yellow line is ?PM2.5[g] and the purple line is ?Fuel mass burned.
>  
> Notice that the burn rate was low between minute 18 and 35 and it was smoky enough to make about 5g of PM2.5. If the secondary air was better arranged, it would have passed the UB-CAP minimum 90% reduction requirement (it didn’t). This type of HTP analysis provides the information needed to solve performance problems (in case that wasn’t clear already!)
>  
> Regards
> Crispin
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