Tony Bryer wrote:

Are there any scientists here who can tell us how much heat could
potentially be recovered if you could condense all the water vapour in
the combustion gases. I remember enough school chemistry to know that
it's CH4+ 4O = CO2+ 2H2O - so every 12g of gas burned makes 36g of water

If you take the density in its gaseous state at 1.82 kg/m^3 - your 3:1
water to gas weight ratio above would yield 5.46 L of water vapour.
Which seems a bit much...

Are you sure that 36g is the mass of just the water? If it were the
total mass of the combustion product (i.e. inc CO2) that would make more
sense with a water vapour yield of 3.64 L

(if you take the latenet heat of vapourisation of water as 2.26 MJ/kg
the 5.46L figure would give you a total of 12.3MJ as oposed to 8.23MJ
for the lower figure)

Since the figures are normally quoted at about 36 / 43MJ for the nett /
gross energy yield of natural gas the latter one seems closer.

vapour. And the latent heat of water is easy to find. But I couldn't find
what gas weighs given that it is under pressure when metered.

It is under very little pressure when metered (0.021 Bar) so you can
probably discount that.


--
Cheers,

John.

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