GEK Wiki / Modelling Gasifier Mass Balance
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Modelling Gasifier Mass Balance

Page history last edited by jim mason 13 years ago


(Follow the discussion of this page on the GEK forum)





1kg of biomass requires 1.6kg of air for gasification.  The output wood gas is 2.6kg.

1.6kg of air = 1.328m3

2.6kg of wood gas = 2.738m3

Air Input to Gas Output ratio = 0.485:1  (or 1:2.062)

So as a rule of thumb, the wood gas output volume is 2 times the air input volume (at constant temp)


Tar gasses produced in pyrolysis are 2.2 times more than what we can combust directly and reduce, given the amount of char available.




The Details


Molecular Masses

- Biomass: C H(1.4) 0(0.6) = 23

- Air: N2(78%) O2(21%) CO2(.03%) = 29

- CO = 28 (note: almost the same as air, which makes it extra-dangerous; it is perfectly happy to float around and kill you, rather than sinking to the floor like dense propane(=44), or racing the the ceiling like light hydrogen (=2))

- Steam or water vapour = 18

- canonical woodgas = 28 (much the same as air; SWAGing 20% CO(=28), 10% H2 (=2), 15% CO2(=44), 55% N2(=28))

(daniel, i think this is a bit  pessimistic.  woodgas is better approximated to easy to work with figures as 20% CO, 20%H2, 3%CH4, 7%CO2, and 50%N2.  this is a bit high for hydrogen(really 16-18%), and a bit low for nitrogen.  the co2 should be down in this realm, but is often higher on gasifiers that take off the syngas at high temp before reduction is completed.)


Air= 1.205kg/m3

1 m3 = 1.205 kg

1 kg = 0.8299 m3 


Wood Gas = 0.950kg/m3  (assuming 20% CO, 20% H2, 60% N2)

1m3 = .950kg

1kg = 1.053m3


(At 20C and 1atm.  See Densities of Common Gasses for details.)



Stoichiometry for Combustion and  Gasification


For complete combustion, 1.0kg of bone-dry (0% moisture content) wood needs about 6.5kg of air.  This is referred to as the stoichiometric ratio.  If you've ever done any work with car engines, you might recognize the stoichiometric ratio for gasoline (14.7:1), which is a bit higher.  In a practical combustion device it is usual to provide a certain amount of "excess air", in order to assure complete combustion even with imperfect mixing; this is for both overall efficiciency as well as to keep CO levels down.  A typical wood burner might use 150% of the air theoretically needed; this would be referred to as "50% excess air" (and would be 9.75kg air per kg wood).


For gasification reactions, the usual practice is to give the "equivalence ratio", which is the fraction of the stoichiometric air that you are supplying.  With dry fuels, best results are normally achieved at ERs of about 0.25, with a "typical" range of perhaps 0.20 to 0.33.  ERs lower than this are indicative of mostlly pyrolysis (a lower temperature, rich, energetic gas, but tarry),  ERs higher than this produce hotter but weaker (less energetic) gas.


Factors that will cause you to need an ER higher than the ideal 0.25:


  • gasifier is thermally inefficient (heat loss etc)
  • fuel moisture is high (need to burn some of your fuel, to power the water evaporation)



In gasification we are basically splitting the combustion reaction into two steps.  The combustion reaction is simple:  1.0 kg wood + 6.5 kg air ==> 7.5kg of flue gas (N2, H2O, CO2)


If we run a gasification reaction at a nominal 0.25 ER, we are combining (0.25*6.5) =1.6kg of air with 1.0 kg of wood:


1.0 kg wood + 1.6 kg air ==> 2.6 kg woodgas (CO2, N2, H2, CO, CH4)


To burn this, we add the other 4.9kg of air to this woodgas:


2.6kg woodgas + 4.9kg air ==> 7.5kg flue gas


Note that the overall reaction is of course exactly the same as combustion.


Another interesting thing to look at is the relative quantites of woodgas and air.  4.9 kg of air per 2.6kg of woodgas means that the stoichiometric ratio for woodgas is 1.88:1 (kg air per kg woodgas).  Also, since the density of woodgas is  (ballpark) roughly the same as air, this means that we need roughly 2 volumes of air per volume of woodgas.


If our gasifier operates at an equivalence ratio other than 0.25, you can re-run the above calcs.  The overall amount of air used will be the same (6.5kg of air per kg of wood), but it is divided between the gasification step and the woodgas combustion step.  A little thought will show that a higher ER (more air used in the gasification step) means less air will be used in the combustion step, and further reflection will indicate that this means that less energy will be liberated in the woodgas combustion step.  I.e., higher ERs necessarily produce lower-BTU gas.



Internal Tar-Burning: Stoichiometry, and Energy Budget


In an internal recirculating tar-burning gasifier, we seek to pass all volatiles through a flame (partial oxidation), and then apply that hot gas to a bed of inert char.  The question is, to what degree can we burn the tars?


Consider that wood is roughly 80% volatiles and 20% fixed carbon.  Pure carbon has a stoichiometric ratio of 12.7 (see below for calc), so that 0.2kg of carbon needs 2.54kg of air.  Since 1kg of wood needs 6.5kg of air for full combusion, that means that the 0.8kg volatiles fraction needs the other 6.5kg - 2.54kg = 4kg air (engineer's math here... ;-).


Now let's say we're running our gasifier at an ER of 0.25.  So for each 1kg of wood, we pyrolyze the wood to 0.2kg of char and 0.8kg of tar gas, and partially combust the tar gas with 1.6kg of air, producing 2.4kg of very hot gas.  We pass most of this very hot gas over the char, which will back-react (a la Boudard) until the temp drops to about 800C.  The rest of this gas passes upwards through our fuel bed, running the updraft portion of our gasifier, which provides us with our tar gas to burn.



(daniel.  we need some more work in this part.  this is the core of the question.  we are not completely burning all the tar gas or we would overwhelm the amount of char available to reduce the resulting co2 and h2o,  we can only burn that fraction of tar gas that the resulting products of complete combustion have adequate char to reduce back.  if we exceed this rate/volume, we move into overpulling the reactor, and co2 and h2o exiting the bottom without being reduced. 

the ratio of c to tar gas after pyrolysis by weight is 1:4. 

each c is 12 mass.  each tar molecule is CH2O or 12 + 2 + 16 = 30mass

so if each c is 12 and each tar is 30 by weight, then the molar ratio of C:Tar after pyrolysis is  1/4 = 12/30x or  1:1.6

combusting tar gas is 1:1 tar to oxygen, or  CH2O + O2 = CO2 + H2O.  or 30 + 32 = 18 + 44 = 62

so for every tar molecule we combust with oxygen, we now have two molecules of combustion product. 

so our original C:Tar of 1:1.6 is now C:CombustionProduct 1:2(1.6) or 1:3.2


the reduction reactions are C + CO2 = 2CO  and C + H2O = CO + H2.  thus we need one C for every one combustion product molecule to be reduced (whether h2o or co2) 1C + 1ombustionproduct = 2 syngas out

as the reduction reactions need 1 C for every 1 Combustion Product, but our C:CombustionProduct ratio is actually 1:3.2 we have 2.2x more tar gas than we we can directly combust to create the needed combustion products for reduction.  this then, is the amount of extra tar that we need to crack thermally.  2.2x times what we can combust. 

i am not terribly confident of my figuring here.  can the chemists among us confirm what i did here?)






Let's SWAG what temperature rise this combustion will produce.  (An interesting factoid that I will use here (thanks, Dad!) is that one standard cubic foot of air, oxidized to completion with a typical fuel such as CH2 (oil distillate) or CH2O (biomass) , will liberate roughly 100 BTUs of energy.  In metric terms, that is 3000 kJ per kg of air).  Since we're not burning to completion, I'll SWAG that we will get 2000 kJ per kg of air that we consume (justification: in the reaction chain C -> CO -> CO2, about 1/3 of the energy comes from the first step, and about 2/3 of the energy comes from the second step).


Producer gas (and air too) has a heat capacity that is conveniently close to 1000J/kg/degree-K.  So our 1.6kg of air gives us 3200kJ of energy, which is applied to 2.4kg of gas, which has a heat capacity of 1kJ/kg/deg-K.  Result: a temperature rise of 3200/2.4/1 = 1333K.  So we should expect to see a flame temperature in the 1350C ballpark.  This is an acceptably high temperature to gasify our char.


(and if we preheated the 1.6kG combustion air to 600C, that's good for another 600*1.6/2.4 = 400C temperature rise, which would indicate ~1750C)


Good news:

  • some tars will be destroyed by being oxidized
  • we have an adequately large energy budget to produce high enough temps (1300C+) that we can thermally crack the tars that weren't oxidized
  • we have enough high quality heat (heat in excess of 800C ) to run the reduction reaction.




Calculating the Stoichiometric ratio for Carbon


C  (12g/mol)  + O2 (32 g/mol)  --> CO2  (i.e. we need 32 grams of O2 for every 12 grams of C that we burn).


Air is 21% O2, so after the math settles out (32g-O2 / 12g-C * 1 g-air / 0.21 g-O2), we get:  12.7 kg air per kg of carbon.


Comments (1)

(account deleted) said

at 11:11 am on Mar 26, 2009

Daniel this is really excellent work. Even I can follow along on this good of explanation. I can weld and operate and even observe in-procees and post-procees results but without the math ability I cannot out side my fields of experience predict sucsessful results. Means making a lot of protptypes. Your explaination with just bigger number/littler number ratio understanding is understandable to even me. Now since I've already stuck my foot in my mouth with both JimM. and MaxG. could you please explain the VOLUME relationships between the ungasified solid biomass and the produced gas that must be accommodated in the physical sizing through the reactor and filter train? Thank You.

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