Doh! My bad. I'll be the first to admit I made a math error. My calculator was set to a 40 gallon pot size and I didn't change that value. So re-checking my calculations based on the formula at the Physics Classroom, the real math is...
So 48 minutes based on 100% thermal transfer and no losses.
So now the results makes more sense to me. At 120 minutes he had a 40% efficiency. That would make more sense given the external heat pads, the absorption of the pot, and the losses to the air.
That's something worth noting on my Excel calculator. The fields in yellow require manual entry. So if you use the unit convertors at the bottom, make sure you type those values into the relevant fields. When I changed the volume value from 40 gallons to 5 gallons, I didn't do that.
-Tim
- Calculate the delta between start and goal temp
- Start temp = 20*C
- Goal temp = 100*C
- Delta = 80*C
- Calculate the volume in Grams
- Grams = Gallons * a conversion factor of 3785.4118
- Total Grams for 5 gallons = 18927.059
- Multiply Delta * Volume * Energy Constant (4.186 Joules)
- Total energy required to boil = 6338293.518 Joules
- Divide Total Joules Required by Energy Applied (2200 Watts)
- 6338293.518 / 2200 = 2881.042508 seconds
- Total seconds / 60 = 48.01737514 minutes
So 48 minutes based on 100% thermal transfer and no losses.
So now the results makes more sense to me. At 120 minutes he had a 40% efficiency. That would make more sense given the external heat pads, the absorption of the pot, and the losses to the air.
That's something worth noting on my Excel calculator. The fields in yellow require manual entry. So if you use the unit convertors at the bottom, make sure you type those values into the relevant fields. When I changed the volume value from 40 gallons to 5 gallons, I didn't do that.
-Tim
The post above shows a 70 C change in 60 minutes for 25l of water. This is a total energy of 4200 J/litre/C x 25 litres x 70C = 7.35 MJ supplied to the wort in 3600 seconds, for an average power of 2041 W supplied to the wort. With an average power supplied of 2200W, this is an efficiency of 93%. This seems very good, but not implausible, and certainly not unphysical. I'd like to be sure that the volumes and temperatures were properly calibrated, have an accurate specific heat and know that the power draw was constant before being more precise than estimating that at 90-95% efficiency.
I don't know where this figure of 38 MJ required to boil 5 gallons comes from. It's wrong by a factor of 4 or so.