[Stoves] Understanding TLUDs, MPF and more. (was Re: Bangladesh TLUD )

Crispin Pemberton-Pigott crispinpigott at outlook.com
Mon Dec 11 19:29:40 MST 2017


Dear Andrew



TLUD makers may want to save this post in order to explain why biomass can, under certain conditions, be converted to char without the addition of air. If you cannot see the whole spreadsheet paste-in, or if it is folded, widen the window or zoom out.



CPP>>There is enough oxygen in biomass to almost completely combust the hydrogen content.



AJH>Doh not that hoary canard again, all the hydrogen in wood is already bonded to an oxygen or carbon atom, oxygen has to be added to extract heat from the wood. The same is true for other partially oxidised hydrocarbons like methanol, acetone or ether.



Yeah we were here before, and I save you the embarrassment of pointing out your error. Still, you insist, so here is something from a your proper chemist.



Biomass can be characterised using a standard formula using C:H:O:S:N:Ash in a fixed ratio. The energy derivable from it can be approximated using the following formula:



HHV (in kJ/g) = + 0.3491C + 1.41 H  - 0.1034 O + 0.1005 S - 0.0151 N - 0.0211 A

(Ref. Dr Tom Reed - the proper chemist. I only managed A Levels chemistry.)



The minus sign in front of the O is the energy that must be invested in the reaction to separate it from the cellulose/lignin/etc in the biomass (on average).



Note that the signs for Carbon, Hydrogen and Sulphur are positive. Note also that Oxygen's negative is balanced almost exactly by Sulphur's net positive contribution. Nitrogen is a little net negative, as is ash.



There is very little sulphur in biomass, but in spite of that, there is almost enough energy in that tiny bit to pull out all the Oxygen from the biomass and make it available to the Hydrogen and Carbon. In fact a bit of H has to be burned by the O to release the last bit of O. That means should the sulphur be oxidised using the little O it needs, enough energy is released to make all the rest of the O available to burn other things. (In practise some of the S would remain bonded to the C.)



So let's burn some biomass (on paper) using Dr Reed's formula. I will calculate directly the LHV on the far right to save time.


Sample Hardwood and Softwood

Moisture

H2O

C

H

O

S

N

Ash

Total

LHV

Eucalyptus Camaldulensis

10.0%

10.0%

44.18%

5.28%

39.57%

0.01%

0.27%

0.68%

100.00%

16.17

With 0% fuel moisture

0.00%

49.09%

5.87%

43.97%

0.01%

0.30%

0.76%

100.00%

18.26




So, there is lots of energy available if we burn the wood.



Suppose we remove or not burn and of the carbon, leaving only the other elements. That will give us the energy available from the non-carbon fraction, virtually all of which is released during pyrolysis. In this next case I am going to double the mass of fuel so the non-carbon fraction is 1 kg. I am doing that because the carbon I will assume to be inert, not absent. I want to see how much energy is available from the combustion of 1 kg of "non-carbon" which is approximately 2 kg of dry wood. This is the reaction that takes place in a 'perfect' char making pyrolysis front wherein 100% of the carbon is retained in the form of char.


Sample Hardwood and Softwood

Moisture

H2O

C

H

O

S

N

Ash

Total

LHV

Eucalyptus Camaldulensis

10.0%

10.0%

0.00%

10.38%

77.73%

0.02%

0.53%

1.34%

100.00%

1.72

With 0% fuel moisture

0.00%

0.00%

11.53%

86.37%

0.02%

0.59%

1.49%

100.00%

2.20




With no moisture at all there is 2.2 MJ/kg of net energy available to drive pyrolysis. With 10% moisture in the wood, there is still (net) 1.72 MJ/kg available to drive a pyrolysis reaction using only the oxygen in the fuel and burning none of the carbon at all.



The output is NO, SO2, a little H2S, and H2O. If we were to burn the hydrogen only, there is enough oxygen available to burn 93.6% of it in this species. It is the heat from this process that delivers the net 2.2 MJ available from pyrolysis. There is an H:O ratio below which there is not enough heat generated to drive the pyrolysis without adding air. That ratio does not occur in biomass because there is always a little excess hydrogen available.



If one was using 5% moisture wood pellets the net reaction energy available would be 1.96 MJ/kg LHV. This is easily enough to drive a pyrolysis reaction in an insulated space indefinitely without air provided there was a continuous supply of biomass fuel available. It might not work well with a descending front. We can check.  It will definitely work with a horizontal or rising front (BLDD and sideways). In the case of a forest fire, the process is largely sideways and can continue for days underground, fueled by the dry roots. Underground coal seam fires use a different chemical pathway, although it is also an MPF. A few are known to have burned for centuries with no air.



For those interested to know at what moisture level (WWB) the energy from pyrolysing wood is net zero (assuming you could get it started) the answer is:


Sample Hardwood and Softwood

Moisture

H2O

C

H

O

S

N

Ash

Total

LHV

Eucalyptus Camaldulensis

46.0%

46.0%

0.00%

6.23%

46.64%

0.01%

0.32%

0.81%

100.00%

0.00

With 0% fuel moisture

0.00%

0.00%

11.53%

86.37%

0.02%

0.59%

1.49%

100.00%

2.20




At this level of moisture, it would take all the available energy from reactions to completely dry the fuel and leave 100% of the carbon behind.



>I won't trouble to correct the other spin in your post other than to agree with Paul that we all understand the a descending or migratory pyrolysis front  became a term used to describe Top Lit UpDraughtand isn't nor was ever used in other contexts.



I will respond to this later. I am checking with a real gasification scientist first.



Regards

Crispin





Andrew
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