[Stoves] Re Alexis Belonio article

[Crispin Pemberton-Pigott]

Dear Dr Tom

>The Water gas shift reaction is CO + H2O <==> CO2 + H2.

>At temperatures above 700 C, the reverse reaction makes CO.  
>At temperatures below 700 C H2 begins to be favored.  

Now we are getting somewhere, or at least I am.  If the water gas shift can occur at such low temperatures, it is definitely taking place in a steam injected fire.  However, is there a point at which the excess water has no function?  Of course.

The natural outcome of your chain above is that the H2 is joined to another O, or rather 2H2 joins an O2 to make two water molecules.  And quickly too.

Beyond what level is the water causing us harm?  I think that is the essential question.  There is a rumour that 15% moisture wood has the lowest emissions.  If that is the case, is it because it creates the best (device-specific) environment in which to react all the CO?  I can't see that this would be difficult to work out.

The lengthy list of reaction products (thanks Louis) might be left un-created if the stirring of the flame and the presence of hot steam turns most things into CO2 over a short, oxygenated, distance.

I have been thinking about the loss profile and it seems that the heat to pot losses can be salvaged a bit if the flame is definintely finished by the time the gases come to the pot.  If the excess air can also be managed in spite of the steam, the loss from evaporation can be tolerated for a net gain in stove efficiency.  But this would only be possible for a stove that had a high excess air ratio.

If the steam is directed more to stirring that to injecting air, the excess air ratio might be held in check at perhaps 250%.  

Take the example of 4500 cc delivered over 90 minutes.  Let us assume the operating pressure is 2 bars.  It is going to absorb more heat than at atmospheric pressure.  The reason for this is that although the specific enthalpy of steam decreases with pressure (less heat required to turn it into steam) the specific heat of steam increases and I wat some superheat.

At 2 Bars
50 gm of water per minute = 20,930 Joules to come to 120 from 20 deg
To evaporate it = 110,080 Joules
To superheat to 500 C = 40,299 Joules
Total = 171,309 Joules per minute

Of this, at 1 bar (depressurized) the heat lost at 200 degrees exit temperature is:

Cooling of depressurized steam to 100 C = 10,150 Joules
Condensation at 1 bar = 112,900 Joules
Cooling back to 20 deg 16,744 Joules
Total steam heat available that was lost = 139,794 Joules (82%).

Net heat from steam given to the pot = 31,515 Joules (18%).

Firepower given to be 5500 watts (calculated from fuel burned?)
= 330,000 Joules per minute
Heat absorbed into the pot = 16% = 52,800 Joules per minute

This doesn't look good as steam losses are 2.65 times the heat getting into the pot.

Conclusion:
If this stove were run at 2 bars, the system efficiency would be increased by at least 2-fold if 1/2 the steam were piped into the pot of water (bubbling inside it) and the other half injected into the fire.

If the heat that is generating this steam is being collected from an area of the stove that was previously uninsulated (like a metal combustion chamber that is not preheating air) the calculation looks different because the heat is 'free', or at least some of it is.

If cutting the volume of steam used increases the heat transfer efficiency, it points to poor design on the top end (i.e. reduce the gas path size and use a skirt).

I suspect, Alexis, that your stove has high excess air, too much steam going into it and a poor top end design.

Can you reduce the steam volume used (which will decrease the air flow) and channel the gases past the pot more efficiently?

Best regards
Crispin