[Stoves] H2O produce burning ethanol

[Crispin Pemberton-Pigott]

Dear Philip

Thanks a bunch for stating it so clearly - greatly appreciated.  I will try
to do the same for the problem I face in recommending a value for LHV when
testing stoves.  This is directly relevant to the present discussion of the
heat to be subtracted from the charcoal heat offered to a pot (deducting for
the charcoal left in the ash that was not burned).

So this is Stoves 101:

>Ethanol produces water when it burns:
>C2H5OH + 3O2 ----> 2CO2 + 3H20
>46g  +   96g ----> 88g  +  54g

>So 1g ethanol will give rise to 54/46 = 1.174g water
>Thus the LHV of 100% ethanol = HHV - 2.261 x H2O
>=29.67kJ/g - 2.261kJ/g x 1.174g/g =27.02kJ/g

In the USA this figure is often cited:
HHV 29.66
LHV 26.7 MJ/Kg

Close enough? No problem.

You define HHV as:

"It is the heat liberated by the combustion of a fuel at 20 (or 25,
depending on the precise definition) deg C with the all products at 20 (or
25) deg C and any water of combustion in the vapour phase at that
temperature."

This is what I referred to as the cooling of the gases to get the HHV.  No
problem.

The 'standard' definition of the LHV is to exclude the heat released when
condensing the water produce during combustion (only).

For biomass burning LHV is usually given as HHV - 1.32 MJ/Kg.

However as some have pointed out, the LHV is not really the heat offered to
the pot, because all the gases are rejected by the stove at some temperature
above ambient.  Some standardize the gases at 100 C, some 150 C.

The cooling of the gases (not the condensing of H2O) to ambient yields heat
that is not available to the pot and never will be unless it is _just_
starting to heat up such as when the moisture condenses on the pot (as you
noted).  You made reference to the HHV for the ethanol stove pots in the
early boiling phase in your comments a couple of months ago.  Agreed, but
the standard LHV is not the same as the heat offered to the pot.  That is a
problem in my book.

Baldwin, in an attempt to address this, calculated that a rejection
temperature of 100 C was a reasonable.  While your example uses 120, I feel
that whatever it is we should agree on it, and have a stated way of
calculating it.

>So if you burn a gram of ethanol, and you let the products escape at 120deg
>C you will have available the HHV less the heat required to heat the
product
>gases from the standard temperature to 120 deg C.  

And the water is included in that list of 'product gases'.

I am with you on that 100%.  That is what Baldwin was talking about except
that he suggests we use 100 C. The Americans use 150 C in industry. I don't
mind which as long as we have some justification for it.

Here is exactly the sort of calculation we need:

>The heat capacity of CO2 at constant pressure, for instance, is given by
>10.34+.00274T-195500/(T^2)cal/deg.mol, with T in deg K, 0 deg C= 273.1 deg
K.

...and a similar calculation for any other gases.

>You then have to integrate this expression over the 100 deg temperature
rise
>and convert from cal/mol to J/g and you find you drop the available heat by
>about 0.07kJ/g ethanol.  

Yes. Baldwin proposed the same correction proposing HHV-1.39 instead of
HHV-1.32 for wood fuels. He did not include the heat needed to bring the
water from 20 to 100. He did not explain why the heat for the water was
omitted. It is a lot more than the adjustment for gases.

>From a gm of ethanol, there will be 1.174 gm of water, which if rejected at
100 C contains:
1.174 x 4.186 x 80 deg = 0.393 kJ/g

Remember the condensing heat = 2257 x 1.174 = 2.650 kJ

"Stove LHV" should be:

HHV - condensing - hot gases - hot water, or

29.67 - 2.65 - 0.070 - 0.393 = 26.557 kJ/g of Ethanol or 20.953 kJ/ml


Can everyone measure their exit gas temperature and use that to fine tune
the figure?

Suppose we told people to measure the exit gas temperature and make a
calculation based on that.  Won't work.  We can't use the temperature of the
exiting gases directly because it is not only CO2 and steam.  The excess air
will cool the gas mix and unless we know the excess air ratio, a simple
calculation cannot provide a meaningful answer.  Where the exit temperature
might be 200 C, with lots of excess air it is cooled to perhaps 125 and the
thermal efficiency is actually lower, not higher, as a lower temperature
would tend to indicate. That is why a combustion analyser measures the
excess air when determining the system efficiency to get the amount of heat
lost in the (cooler) excess air.

Thus, in the absence of a combustion analyser, the only practical thing we
can do is to give an different LHV that realistically gives the heat offered
to the pot, determined by calculation only and used as a 'Stove LHV'.  If a
stove has a high excess air ratio, it will perform poorly because it is not
taking advantage of heat that really is there but can't be captured and a
thermometer will not tell us what is going on.

Given that stoves do not, on average, reject heat at 100 C, what value
should be used and what is a meaningful heat value for cooking stove fuels?

Thanks
Crispin