Ducatiboy stu
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Yes, that's what my equation does.
Enthalpy of vaporisation of water is around 2.26 MJ / kg, so the energy required for boiloff =
2.26 MJ/ kg x (Vstart - Vend) (Eq 1)
Heat capacity of wort is around 4 kJ / kg / oC, (slightly less than water at 4.18 due to the sugar content), so the energy required to heat wort =
4 kJ / kg / oC x Vstart x Trise (Eq 2)
Heat of combustion of propane is around 50 MJ / kg,so the energy from burning gas =
50 MJ / kg x mass of gas burnt. (Eq 3)
Set T rise to 35 degrees, Eq 2 becomes
E = .014 MJ / kg x Vstart (Eq 2a)
Add eqns 1 and 2a to get
E = 2.4 MJ / kg x Vstart - 2.26 MJ / kg x Vend
Efficiency is thus (2.4 x Vstart - 2.26 x Vend) / 50 x mass of gas burnt*.
Multiply both sides by 100 / 50 to get
Efficiency (in percent) = (4.8 x Vstart - 4.52 x Vend) / mass of gas
which is where we started.
I presented the bare finished product above because experience says 99% of the audience hasn't followed this.
* The units are consistent so they drop out of the quotient. The vapour is substantially water so 1 litre lost = 1 kg. The heat capacity of wort is dependent on sugar content, as is the density but they move in opposite directions so assuming 1litre = 1 kg introduces a trivial error.
Thank god for people like you
I was really **** at math in school