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First, here are a few definitions to establish a common framework for this discussion.

It takes 1 BTU/hr to raise 1 lb of water 1 deg F.

1 liter of water weights roughly 2 lbs, so it takes 2 BTU/hour to raise 1 litre 1 deg F.
Water at home temperature is 60 F, it will boils at 212 F. There is 152 F difference.

1 hour = 60 minutes = 3600 seconds.


So, to raise 1 litres of water from 60 F to boiling at 212 F would require 1*152*2=304 BTU/hour.

This analysis does not take into account any thermal loss (cooling) because the loss would vary based on outside temperature, material of the vessel, surface area of the opening, etc. Regardless if you use electricity or propane, insulating your hot liquor tank and brew kettle will help you achieve your target temperature more quickly and maintain that temperature with less energy.

Generally speaking propane burners are rated on BTU/hour or simple BTUs. They range from 35.000-200.000 BTU/hour. This is the amount of energy produced by burning the propane, but not necessarily transferred to heat the water. So, it is not a perfectly efficient transfer to your HLT or boiling kettle. There will be further loss from the vessel itself since it is not a perfect conductor. It is difficult to determine the actual amount of energy produced by the burning propane that actually makes its way into heating the water. Therefore, I will make an assumption that only 50% of the energy produced by a propane burner actually heats the water. A quick Google search turned up the King Kooker turkey fryer produces 54K BTUs/hr. Adjusting for efficiency, the propane burner provides 27,000 BTU/hr (from 54,000*50 % per hour). Based on this value, the propane burner could bring 2 lbs of 60 F water to boil at 212 F in (304/27,000*3600 seconds) = 40.53 seconds. Obviously, a more powerful burner will heat water more quickly, but with greater fuel consumption.

The source for this information is given as a Related Link. One should recheck the accuracy of the math used above for adequate result.

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