Rider509
Well-Known Member
Let's run some rough numbers and use an extract of four grams per ounce as a reasonable starting point. So if I process 16ozs that will yield 64g of extract. 64g * 0.12CO2= 7.68gCO2 or .1745molCO2. Solving for P in the provided equation at 121C assuming 50% headspace (250ml) in the reaction vessel yields... holy hell, Batman!
P = 331.7662065867 pound per square inch.
The reaction vessel is rated to 3MPa or 435psi so I'm golden there, but DAMN!
With an increase in headspace to 400ml:
P = 207.35387911669 pound per square inch.
And that doesn't even take into account the additional pressure of the vaporized EtOH with a molar mass of 46.5g/mol.
I guess I'll be reducing the extract down to a very viscous consistency and possibly be running smaller batches. The first run should be interesting.
P = 331.7662065867 pound per square inch.
The reaction vessel is rated to 3MPa or 435psi so I'm golden there, but DAMN!
With an increase in headspace to 400ml:
P = 207.35387911669 pound per square inch.
And that doesn't even take into account the additional pressure of the vaporized EtOH with a molar mass of 46.5g/mol.
I guess I'll be reducing the extract down to a very viscous consistency and possibly be running smaller batches. The first run should be interesting.