How fast should 1 gallon of water boil with a 230k BTU 23 jet burner?

[-CHRIS-]

I am not a home brewer yet, but based on everything I am buying for making maple syrup I soon will be. You guys seem to be the expert in home-built burners so I figured I would post here. I just built the following natural gas burner and I wanted to see if I did it correctly. I used the generic 23 jet 230k BTU burner that is sold on Amazon.com The tips of the burner are 6" below the bottom of the pot. I boiled 1 gallon of water today and it took 9.35 minutes. If I do the math, it seems too slow.

ambient temp 14 degrees F
starting temp 48 degrees F
Ending temp 210 degrees F
1 gal of water is 8.35 pounds
1 BTU = 1 degree F temp raise in 1 gal of water in 1 hour
9.35 minutes to boil the water
so...
(210-48)x8.35 x (60/9.35) = 8756 effective BTU transferred from the burner???

I realize this setup is not 100% efficient, but 4% efficiency seems insane unless my math is wrong.

I am wondering if the burner needs to be mounted higher or lower and/or should I place a ring around the pot to keep the heat in. Interestingly enough, the metal sides only hit around 250 degrees so I am not losing a ton of heat to the sides.

Thanks in advance!

Chris

 

[lousbrews]

In general combustion is a very unfavorable process for heat transfer. As far as moving the burner you only want the bluest part of the flame in contact with the stove as that is the hottest part of flame thus ensuring the most heat transfer.

As far as the math I didn't check.

 

[brewman !]

I am not a home brewer yet, but based on everything I am buying for making maple syrup I soon will be. You guys seem to be the expert in home-built burners so I figured I would post here. I just built the following natural gas burner and I wanted to see if I did it correctly. I used the generic 23 jet 230k BTU burner that is sold on Amazon.com The tips of the burner are 6" below the bottom of the pot. I boiled 1 gallon of water today and it took 9.35 minutes. If I do the math, it seems too slow.

ambient temp 14 degrees F
starting temp 48 degrees F
Ending temp 210 degrees F
1 gal of water is 8.35 pounds
1 BTU = 1 degree F temp raise in 1 gal of water in 1 hour
9.35 minutes to boil the water
so...
(210-48)x8.35 x (60/9.35) = 8756 effective BTU transferred from the burner???

Nope. The BTUs into the water are (210-48)F x 1BTU/FLb x 8.35 Lbs = 1352.7 BTU. The time component has nothing to do with it.

When you do this calculation, you are assuming that none of the energy was put into the phase change component (water to water vapour), in spite of seeing actual steam. In reality you might have brought your water to 90% of its phase change energy. The phase change energy (enthalpy) for water is about 980 BTU per pound.

What you really need to do is use a thermometer to record a temperatures lower than the boiling point instead of relying on the water reaching a boil as your determination of temperature.

In the mean time, the energy into the system is 230K x 9.35/60 = 35.84K BTU, assuming that you were running the burner at full blast and the propane tank was supplying sufficient pressure and the propane combusted properly.

I am not an expert in these matters.