[Stoves] Request for help on TLUD operating data

Crispin Pemberton-Pigott crispinpigott at outlook.com
Sun Aug 17 02:24:47 CDT 2014


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 $B-t(BO2.









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
$B-t(BO2 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.







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 $B!F(B3$B!G(B 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 $B-t(BO2 number. That jump is preceded by (at minute 88) a
sudden increase in gas production, something noted by many TLUD builders.
That jump is caused by the fact there is no damp fuel under the bottom layer
to it flashes dry. In TLUD$B!G(Bs 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







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:







The black line is real time PM2.5[mg/m3], the yellow line is $B-t(BPM2.5[g] and
the purple line is $B-t(BFuel 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$B!G(B
t). This type of HTP analysis provides the information needed to solve
performance problems (in case that wasn$B!G(Bt clear already!)



Regards
Crispin

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