[Stoves] H2O produce burning ethanol

[Crispin Pemberton-Pigott]

Dear Philip

>...Or are you using the word "exclude" in the sense of "reduced by"? 
>In which case, yes, you are correct.

That is indeed what I meant,

>That is close enough to your 1.32kJ to suggest, again, we are
>talking the same basic language, and including the heat from 
>condensing the water produced in combustion.

Yes.

>So the heat in to the pot should be the LHV, NOT the HHV, period.  

Agreed.  What I wanted to see was a worked calculation to see if it includes
the heat 'lost' by not cooling the moisture and gases from (say) 100 to 20,
the initial ambient temperature use to calculate the HHV.

I offered the three 'reduced by' numbers I can think of:

HHV - condensing heat - hot gases - hot water, or
29.67 - 2.65 - 0.070 - 0.393 = 26.557 kJ/g of Ethanol or 20.953 kJ/ml

Including the latter two increases the 'discount' by 17.5%.

For natural gas water production would be quite a bit higher so not
including the cooling of the water would make a bigger difference.

I have read and read about how LHV's are calculated and no one ever seems to
include the heat of the water between 20 and 100 degrees.  Perhaps it is
just easier that way.

Dr Reed, Dr Bond, Dr Verhaart... anyone? Can you please comment on the issue
of whether the heat offered to the pot should be based on a 100 or 150
degree rejection temperature?

Thanks
Crispin