Dear Crispin,
Crispin Pemberton-Pigott wrote:
Dear Frank
I have completed the exercise of deciding what to calculate, what to plot
and especially how to limit the number of numbers floated.
For the PM, the measurement is taken by the Dusttrak DRX in milligrams per
cubic metre for the sample it has measured. Suppose it is 2.5. Of course
that sample is diluted by the condensing particle diluter, so the 2.5 has to
be multiplied by the dilution factor. Suppose it is 11. Then the DRX reading
of 2.5 is multiplied by 11 giving 27.5 mg/m^3.
The turbidity reading is somehow converted to weight? And do you have an
idea of the range of particle diameters of the PM? What I am thinking is
if the particles you are looking at is greater than 0.7 um they can be
collected on a Whatman GF/F filter and carbon determined in a Leco
(large boat) C/N analyzer. We would then know the real energy loss in
the PM.
Next the volume of gases produced by burning the mass of fuel that
disappeared during the time period during which the DRX took measurements is
calculated. This means the stoichiometric combustion gas volume. It varies
with the fuel composition so it is necessary to know what the fuel contains
or else take a pretty good guess. Let's say it is 8 kg of gases per kg of
fuel missing from the fire (remember the ash does not burn off).
This because the carbon in the wood converts to CO2 during combustion
using both the 50% oxygen in the wood and from the 20% in the outside
air and hydrogen to H2O increasing the weight approx. 8 times (?)
This 8 kg is multiplied by the excess air in the stack (it is always in a
stack). Suppose it is 250%. Lambda is excess air plus 100% so Lambda is 3.5.
Where does this + 100% come from?
The mass of stack gases is then 8 x 3.5 = 28 kg with the small caution that
excess air and stack gases are not exactly the same density (but very
close). Let's say it is 28 kg of gases per kg of missing fuel. That is
28/1.18 = 23.73 cubic metres, which is the same as 23.73 litres of gas for
each gram burned.
Going slow here..... : )
The 1.18 value is converting grams gas mixture to ~mole of the mix?
so 28/1.18 = ~22.4 liters? You are using 23.73 liters that is a better
estimate?
This the same as 16 g oxygen = 22.4 liters?
In this way I obtain the volume of gases passing in the time period that the
PM measurement was taken. As the mass burned is measured over the same time
period, they correlate.
weight of gas = (wt. organic matter + wt PM)
If 5 g of fuel is burned, that is 5 x 23.73 = 118.65 litres of gases with a
PM concentration of 27.5/m^3. So 27.5/1000*118.65 = 3.2 mg of particles
during that 10 seconds.
This is repeated for each data set recorded during the test, typically 1800.
At the end of the fire, there is a little bit of PM measured by the DRX but
the mass burned is SO small that the PM mass per 10 seconds is negligible.
At some point it reduces to be less than the resolution of the scale the
stove sits on. There is no point continuing the test after that. Even if it
were to be left for an additional hour, it makes nearly no difference to the
PM total because the burn rate is so low.
The total emission of PM is the sum of the masses calculated each 10
seconds. Dividing by the number of MegaJoules released from the fuel (as
received) gives a figure for PM2.5 emitted /MJ. I have attached an example
of the output. The chart contains a plot of the PM values calculated per 10
seconds (black) as well as a cumulative total (yellow). The total mass
burned (missing mass, not mass of whole coal) is also plotted so you can see
the burn rate changing. The attached chart is the baseline stove in
Ulaanbaatar.
This PM figure is 'portable' as it can be carried anywhere to make direct
comparisons between different stoves burning different fuels.
The only thing remaining is to calculate the thermal efficiency as well. The
PM mass will have to increase because of the extra fuel burned to deliver
the needed number of 'net MJ' delivered into the room.
Whats important here is to keep the ratio of PM/MJ at its best (lowest)
value while 'working'. The time needed for the stove to burn to do the
job is the thermal efficiency (as I think about it).
This calculation will
be made using the O2 level in the stack, the stack temperature at the point
where it leaves the living space, and two constants calculated so as to be
able to use the Siegert method for determining the efficiency. This method
is used internally by the Testo gas analyser. There are two methods of doing
it but that is another conversation.
I guess these stack measurements and calculations need be done (during
working) if we want to compare TE for stoves in a catalog to be able to
pick one. We still need to (1) state the fuel type/quality being used,
perhaps (2) have a fan of a certain power to blow air over a stove
(standardize for heating a room) and (3) water boiling or (4) something
frying (to simulate cooking). All keeping the PM/MJ ratio within a range
we claim the stove will do. I suggest we should work the stove for hours
when collecting data for comparison. Keep adding fuel, keep the fan on,
keep adding water to boil and keep the frying pan loaded for a ~4 hour
time. This reduces the error that happens when starting the stove and
determining the amount of ash produced from the fuel at the end. I think
this error is too large a percentage in the WBT due to the time the
stove is operated is so short.
You are using coal from one coal mine? without more drying? You have it
easy. : )
Thanks
Frank
What do you and (everyone else) think so far?
Regards
Crispin
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--
Frank Shields
Soil Control Lab
42 Hangar way
Watsonville, CA 95076
(831) 724-5422 tel
(831) 724-3188 fax
[email protected]
www.compostlab.com
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