[Stoves] TLUD Test with Switchgrass Pellets

[rongretlarson at comcast.net]

Crispin (cc stoves) 





I like the computations you have provided, but think there are additional efficiencies needing computation. Yours below gives zero benefit to the (very sizeable) charcoal production.- which obviously will entail some howls from those (like me) interested in Biochar. 




First, I think this sort of efficiency measurement should be done without a pot lid. The 22% number which is now measuring something about the tightness of the lid (and not related to the stove) should jump to above 25%, I guess. 




I will send an Excel spread sheet to Erin and Tom on this – so others can play with it. (I am sending it to Crispin separately and can do same for anyone writing me.) That SS shows that about 50% of the initial energy remains in the produced char, so I say that makes the overall stove efficiency, making no other changes in Crispin's work, about 75% (or 25% "loss" – not 75% loss). That is, I claim that just adding the boiling and charcoal-making efficiencies together is valid. The missing 25% is energy that was not captured usefully. 





But, even without adding the charcoal – making efficiency, one can approximately double (as opposed to the above-stated tripling) Crispin's calculated 22% “boil-away” efficiencies by dividing the useful output by the net energy consumed in the boiling task, not by the total amount available . To repeat - I calculate that about half the initial available switchgrass energy is still available in the produced char, and that this MUST be accounted for when comparing a pyrolysis stove with a combusting stove. This comes closer to the efficiency on the water boiling side, but still does not fully capture the value of the char. The lost energy is about half of the non-char energy (not 3/4 of the input energy - as implied with a 22 or 25% efficiency computation). 





To restate, with simple numbers that might result with a re-test without a pot lid, that hypothetically finds that 25% of the energy was productively used in boiling the water and also found that half the energy remained in the char: 




Boiling: eff1 = ¼; (Crispin-computation yesterday, approximately) 







Char-making eff2 = ½ (This value given in many char papers – and I found – almost exactly - in this case) 




My preference: eff3 = eff1 +eff2 = eff1+ 2*eff1 = 3* eff1 = ¾ (I believe that having the same denominator when calculating efficiencies is essential. Of the ¾ useful output 2/3 is in the char and 1/3 is in the boiled water; one quarter of the input energy was not captured.) 




Combined (EPA-like) eff4 = eff1/(1-eff2) = eff1/(1-1/2) = 2 eff1 = ½ (I think this undervalues the char – and has no theoretical basis when computing efficiencies. It does a better job of bringing in the char, but not the correct calculation. This says that about half of the available energy passing the pot is captured. It does not give enough credit to the char. 




The above formula (incorrectly used by almost everyone for stoves) for eff4 was developed at a time when char was not desired and takes advantage of the fact that for small x: 




(1+x) approximately equals 1/(1-x); Example 1.02 approximately equals 1/(0.98) = 1.0204 




Obviously one should not use this formula when x is a big number like 0 .5. 




So - to repeat, Crispin's 22% has been calculated correctly - but it drastically undervalues this particular stove (and all char-making pyrolysis stoves). The standard efficiency formula used by EPA does better, but still undervalues char-making. 




I think more than one efficiency is needed to describe what is happening. 

Ron (no inserts below) 
----- Original Message -----
From: "Crispin Pemberton-Pigott" <crispinpigott at gmail.com> 
To: "Stoves" <stoves at lists.bioenergylists.org> 
Sent: Wednesday, November 16, 2011 7:34:07 PM 
Subject: [Stoves] TLUD Test with Switchgrass Pellets 




Dear Roger S and Switchgrass Fans 



Roger this will be of interest because the fuel came from you! 



In preparation for the demo to the Chinese stove delegation I ran a single test with 8.85mm diameter switchgrass pellets in a Grasifier TLUD stove. 



The stove weighs 645 grams. The fuel added was 1050 grams. 



After it was running well (perhaps 10 minutes because it lit slowly) I put on a donut-shaped steel plate with 4 bolts that served as pot rests. It boiled 1.7 litres of water quickly but I was not tracking the heating rate. I was looking at burn quality and the char produced. Boiling times another time. 



There was too much primary air after a while though it was good to have during ignition. The pyrolysation rate was too high after a while so I closed all the 6mm holes (there were 5) but one. It ‘cooked’ the pot for about two hours. The char remaining was in the form of pellets which remained quite strong. A picture of the char is attached. It evaporated nearly all the water by the time the flame went out. Some 340 g of water remained in the pot (with lid). 



The char was spread on a concrete floor where it smoked for a few minutes. Nothing done except spreading it out to cool. Afterwards it was collected and weighed. 



Fuel (air dry, est 10% moisture): 1050 g 

Dry fuel mass 945 g 

Char yield: 365 g 

Char yield as % of initial fuel mass: 34.8% 



The ash content is about 10% for switchgrass, right? If so: 



Ash content of the initial fuel: 94.5 g 

Ash content of the char: 94.5 g 

Dry Ash-free content of the initial fuel: 850.5 g 

Dry Ash-free content of the char: 270.7 

Carbon content of the char (est 88% of non-ash content): 238 g 



Estimated heat content 16 MJ/kg, x 1.05 kg = 16.8 MJ 

Water heated 1.7 litres = 0.676 MJ 

Water evaporated = 1360 g = 3.070 MJ 

Total heat applied = 3.746 MJ 

Average thermal efficiency from for the entire burn = 22.3%. 



Final note: the char will not burn in this device. 



Regards 

Crispin 


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