[Stoves] Saving the WBT

Fri Aug 16 16:52:27 CDT 2013

Ron,

I look forward to hearing Jim's, or anyone else's approach to the difficult
problem of accounting for the energy. Whoever comes out with a method there
will be another right around the corner. This is non-ending so there is no
need to wait. My suggested approach is not a comparison - just a different
way of looking at it. Hopefully one that will work without all the errors
regarding calculating the remaining chars. 

I am thinking of a new approach where we do not need to handle char at all.
I noticed when using the GEK and Tom Reeds TLUD that when fresh biomass ran
out the secondary flame went out, or very poor flame. Just add more biomass
and you are in business. Hot coals several inches below the pot did a poor
job of heating the pot - so why even consider them? Its only the fresh tars
that heat the pot and all that other energy just heats the stove body.
Important to heat the stove body and aid in breaking the bonds to release
lumps of tars and complex organics free to head to the secondary.  But IF
(Big IF) they do not significantly heat the pot we can rule them out it
saves that problem of all the difficult calculating.  If you were to fill a
rocket with char and blast air on the char would you get a secondary flame?
The stove body would get red hot but the pot only a few inches away would
heat up slowly without the flames licking the bottom. Lots of useless heat. 

The question is can we take a block of wood and determine the weight
fraction that will contribute to the secondary? And the fraction that sits
and combusts in the stove body? I think the pipe will do that. 

Something different to talk about.

Thanks Ron for the reply.

Regards

Frank

Frank Shields

Control Laboratories; Inc.

42 Hangar Way

Watsonville, CA  95076

(831) 724-5422 tel

(831) 724-3188 fax

frank at biocharlab.com

www.controllabs.com

From: Stoves [mailto:stoves-bounces at lists.bioenergylists.org] On Behalf Of
Ronal W. Larson
Sent: Friday, August 16, 2013 1:27 PM
To: Discussion of biomass cooking stoves
Subject: Re: [Stoves] Saving the WBT

Frank and list:

    Abut 1.25 hours after yours was a message fem Jim Jeffords on a webinar.
He is obviously going to talk on Tuesday about how to handle remaining char
when calculating efficiency.  It was not obvious to me how you are handling
char in your example.  Could you give an example where the remaining char by
weight was 25% of the input biomass weight.  Maybe 30% into the cook pot.
Or any numbers you want.  Then we can hopefully compare your approach with
Jim's

Ron

On Aug 16, 2013, at 12:12 PM, Frank Shields <frank at compostlab.com> wrote:

Greetings Stovers,

All this talk about the ocean water got me thinking about the Water Boiling
Test.

I would like to suggest a new way of testing and reporting results:

1)  Procedure

2)  Justification

3)  Calculations

Procedure: We take some oven dried wood and place in a pipe. Add both end
caps, loosen one, weigh and place in an oven at ~450c. Then cool and weigh.
The loss in weight is the volatile fraction of the fuel. This is the
fraction that provides the energy to boil water. We determine the energy of
this fraction and that is the energy of the fuel. Keeping track of the fuel
weight we use we determine the total usable volatile-energy.

We put the pot of water on the stove, measure the temperature of the water,
start the fire and monitor the water temperature. We keep the fire going
until the water is at 'simmer' then keep steady for 30 min. Adding no more
fuel we then we manipulate the fire to keep the secondary burn going as long
as possible. Soon as the secondary burn goes out we pull the pot off and
measure the area under the temperature plot for energy that went into the
pot. Energy in the pot / volatile-energy X 100 is the efficiency(?).

Justification: When you are boiling water it is only good as long as the
secondary burn is going. When that goes out, even with a glowing stove
below, the water heating process slows way down because, as I learned in
Stove Camp, we need the heat forced hitting the bottom of the pot to stick
to it, go through the pot and heat the water.

Biomass fuel has two types of energy; 1) the tars (C-H-O) that create the
secondary burn and 2) the chars (C-C) that only heat the stove body.
Important for the chars to heat the stove body but there is more than enough
with a good insulated stove and all that extra heat is wasted - not used to
heat the pot. When biomass is heated between 300c to 450c tars of massive
C-H-O structures go to the secondary burn and ALL C > CO2, and all H > H2O
releasing massive energy just at the pot bottom. The C-C bonds (chars) left
need to go C (solid) - CO volatile) forms releasing energy only in the stove
body. The CO (volatile) goes to the secondary burn (adding to the energy of
the tars) to go CO > CO2 releasing a relatively small amount of energy.
Under the best of conditions all the C goes to CO (not CO2) in the stove
body but this is such a small amount of energy compared to the tars
providing ALL their energy to heat the pot I suggest we can ignore (or
estimate) the CO > CO2 added energy.

This being the case if we use only the volatile fraction as the total energy
then once the secondary burn stops all the rest of the material in the stove
body can be ignored.

Calculations: A block of 100 g dried wood contains 44g C, 50g O and 6g H.
Let's say 22g C goes to the secondary as tars to heat the pot and 22g C left
behind to heat the stove. This can be determined (if needed) in the lab
measuring C and H in the biomass and C and H in the char left. The weight
loss in the pipe contains 28.2 % carbon and 7.7 % hydrogen for the starting
energy value figuring all H and all O are included in the tar fraction.

Now we need to use Bond Energy (I need help) to determine the energy value
we give for all the tar carbon going all the way to CO2 and the hydrogen
going all the way to H2O. We sum the Bond Energies in the tars as the Total
Energy of the fuel. Add to it (ignore or estimate) the Bond Energy of the CO
to CO2 in the chars.

Bond Energies:

C - - O = 360 kj/mol

H - - O = 366 kj/mol

What is C>CO2 and H>H2O?

I realize if one has H2 and O2 that nothing happens until you provide enough
energy (light a match) to break the H-H and O-O bonds to re-create H2O in an
explosion. In this stove case there might be enough of the extra heat in the
stove body to break apart the tars into C and H and O so we can just
calculate them going completely to their end components.  

Regards

Frank

Frank Shields

Control Laboratories; Inc.

42 Hangar Way

Watsonville, CA  95076

(831) 724-5422 tel

(831) 724-3188 fax

frank at compostlab.com

www.compostlab.com <http://www.compostlab.com/> 

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