You’d be surprised how much warming occurs from the time cold solvent is introduced to the time the last bit of solvent leaves the material column, even without a soak period, especially with un-jacketed columns.
From a purely theoretical perspective I can think of three mechanisms that lead to substantial heat gain—consider that the stainless spool is the thermodynamic boundary:
- Heat gain to solvent due to equilibration of column to solvent temp (biggest driver)
- Heat gain due to latency from filling up and draining column (medium driver)
- Enthalpy of solution for dissolution of cannabinoids, etc in solvent (almost negligible)
Assumed Variables:
Column OD = 6" =0.152 m
Column Length = 48" =1.219 m
Column Thickness = 6mm =0.006m
Density of Stainless Steel = 8000 kg/m3
Heat Capacity of Stainless Steel (c_ss) = 0.5 kJ/kg*K
Heat Capacity of Butane (c_but) = 2.3 kJ/kg*K
Heat Capacity of Biomass (c_bio) = 1.5kJ/kg*K
Volume of Stainless Steel Column = V=piL(r_o^2-r_i^2) = 0.00336 m^3
Mass of Stainless Steel Column (m_ss) = 27kg
Mass Butane (m_but) = 12kg
Mass Biomass (m_bio) = 5kg
Solvent/Biomass = -40C
Ambient/Stainless Temp = 20C
Part 1
C_ss = c_ss * m_ss
C_bb = (c_bio * m_bio) + (c_but * m_but)
T_f = ( C_ss*T*_ss,0 + C_bb**T_bb,0 ) / (C_ss + C_bb)
T_f = -21.2C (47% increase in temp)
Part 2 (lol)
assuming 5 mins to inject and transfer to collection
Convective Heat Gain = Qcon = hAdT
(dT in this case is just the temp difference between the ambient surroundings and the solvent/biomass—since the temp of the solvent is differential due to what was shown in part one, just assume an average for a more accurate calc—so instead of a 60K difference, it’s closer to 50.5K)
h = 5 - 10 W/m^2*K
A = pi*ODL + 2pir^2 = 0.6205m^2
Qdot_con = 298W
Qcon = 298W*5min = 89.4kJ
Radiative Heat Gain…I’m tired of typing shit out, it’s getting ridiculous, just trust me, is 14.1kJ
Qtot = Qcon + Qrad = 89.4 + 14.1 = 103.5kJ
Change in Temp from Conv+Rad = 103.5kJ/C_ss + C_bb = 2.1C
T final final =- 21.2 + 2.1 = -19.1C
This is all theoretical, but you can assume that it’s a bit more considering that frost increases heat flux over time, the differential effects intensify the heat gain
IME I’ve injected -40C solvent and gotten close to 0C by the time it was in collection
$0.02
P.S. excuse any math errors, the formulas should be right though