How many BTU's is your Hurricane Natural Gas Burner .......burning

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HomeBrewFoSho

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So I am converting what are actually Banjo burners to run on Natural Gas. I have everything and did my first burn this afternoon. I couldn't see anything . The flame was invisible! hahaha but seriously. So I know I have exactly 7"wc coming out of the elbow my burner screws into which is .252 psi. I measured it with a medical grade digital gauge so I know it's correct =). Anyways, I decided to start at 1/8" drill bit which should give me 56,000 BTU's. Is this enough? Should I be going with more BTU's? And for the real question for anyone who may have done this , do you know exactly how many BTU's your running? I know I could try , try , and try again until I'm happy with it but I figured I would ask. They're kinda a pain to take the burners on and off , and on and off, etc...thanks for the help! :rockin:
 
The limit on drill size for the burner is when the air shutter is fully opened, it would be better to invest in a couple numbered bits and slowly increase the gas jet opening in steps followed by testing. When you adjust shutter on the burner you are looking for yellow tips on the flame, not whole flame yellow. When drilling gas jet opening leave some air shutter adjustment available to tune between cool winter and warm summer conditions.
 
Send me your gauge and I'll tell you!

The only way you can know when you've maximized the orifice size is after you go too far. ;) If you really want to maximize it, be prepared that you might be buying a new orifice.

Williams Brewing sells the NG conversion valve in case you end up deciding to go that route. I bought it for my Hurricane (because I wanted to keep the LP one anyway for portability) and it works quite good.

http://www.williamsbrewing.com/HURRICANE_CONVERSION_VALVE_P2214.cfm
 
The limit on drill size for the burner is when the air shutter is fully opened, it would be better to invest in a couple numbered bits and slowly increase the gas jet opening in steps followed by testing. When you adjust shutter on the burner you are looking for yellow tips on the flame, not whole flame yellow. When drilling gas jet opening leave some air shutter adjustment available to tune between cool winter and warm summer conditions.

Yeah, it's not rocket science I just didn't want to have to keep pulling my burners on and off. My setup is not "assembly/disassembly" friendly but I didn't care that much because I knew once I get it, they'll never come off again.

Send me your gauge and I'll tell you!

The only way you can know when you've maximized the orifice size is after you go too far. ;) If you really want to maximize it, be prepared that you might be buying a new orifice.

Williams Brewing says the NG conversion valve in case you end up deciding to go that route. I bought it for my Hurricane (because I wanted to keep the LP one anyway for portabiity) and it works quite good.

http://www.williamsbrewing.com/HURRICANE_CONVERSION_VALVE_P2214.cfm

If you're burners come off easily just see what size drill bit fits in your conversion valve! Thanks :mug::D
 
Sure, I can do that (assuming I remember). If I forget, shoot me a PM.

Thanks! I'm figuring if we can get a fairly specific BTU that these Hurricanes run the best at on Natural Gas it will make it REALLY easy for anyone else that wants to convert these from propane. Hell , you can make a inH2O gauge out of $2 worth of tubing. Anyone could then read their exact pressure and then drill out the brass accordingly refering to one of those charts. DONE
 
Just thought I would let you guys know some recent test results. I just filled a keggle half way (7 gallons?) and went from 54 degrees to hard boil in almost exactly one hour. I'm running at 56000 btu's and it seems to be a happy medium. I am going to drill extra holes around the bottom back skirt to let out the extra gases which I think will make it burn even better. I was also impressed with how low I could get the burner to go and how clean it burns. More results to come later!
 
That is 10,200 BTUs of heat that you are getting into your pot. Where are you getting the 56,000 BTU number?
 
That is 10,200 BTUs of heat that you are getting into your pot. Where are you getting the 56,000 BTU number?

What? Where are you getting 10k btu's ? I know I have 7"h20 pressure and with the size hole in my brass orifice I should have 56,000 btu's. I'm not sure what you're saying.
 
What? Where are you getting 10k btu's ? I know I have 7"h20 pressure and with the size hole in my brass orifice I should have 56,000 btu's. I'm not sure what you're saying.

Oh, you are calculating the output using pressure and orifice, cool. I was calculating how many BTUs you were actually utilizing. 54F to a boil in 60 minutes in 7 gallons is about 10,200 BTUs, or a 3000W electric heating element, or....
 
That is 10,200 BTUs of heat that you are getting into your pot. Where are you getting the 56,000 BTU number?

Wouldn't all this be affected by ambient temps, elevation, and kettle material? Above and beyond that, the distance between pot and burner would be a factor of heat loss that could also affect the difference between 10,200 BTU's to convert x amount of water at y starting temp to boiling.

See, it might be that at the burner, there's 56,000 BTU / hour, but kettle spacing, surrounding temps, winds, and elevation of the entire rig would impact the efficiency of getting those BTU's into the kettle.
 
How are you calculating that? This burner is clearly putting out more then 10k btu's.

He's calculating the math based on the water volume, time, and temp change only. If you had a perfect 100% transfer of energy, then you could convert that mass of water at that temp to boiling using only 10k BTU's. Since it's not perfect - EVER - it's going to take more BTU's to do it. Those factors that lead to the difference are what my past post was about. Spacing, Combustion efficiency, elevation, humidity, wind, brew kettle material, and maybe even surface area would all reduce that efficiency.
 
Gotcha. The burner is about 4-5 inches from a jeggle. It was 70 degrees today and it was in my garage. Besides drilling the holes in the jeggle skirt I don't see how this setup would be any less efficient than any other burner setup. Unless all gas/propane burners are that inefficient.
 
Most gas burners are about 20% eff. at delivering their heat to a brew pot.

I was just wondering how you were calculating the BTUs, and you explained that. You are measuring gross output, I am measuring the net.
 
Some interesting info:
  • Burners are actually rated by "BTU/hr". "BTU" doesn't technically make sense since the longer you run the burner, the more BTUs expended.
  • 1 pound of propane contains 21,591 BTU of energy.
  • 1 gallon of beer weighs 8.34 pounds.
  • (Pounds of beer ) * (Temp change in F) = number of BTUs.

To determine the BTU ( per hour ) of your burner:

  1. weigh your tank (W1)
  2. Run your burner full tilt for exactly 1 hour. Doesn't matter what you are using it for. Maybe heat up some water or something.
  3. Weigh the tank again. (W2)

Burner Output (BTU/hr): (W1 - W2) * 21,591


To determine how many BTUs went into the beer

  1. Determine weight of beer (see interesting info above) (W1)
  2. Record the starting temp of your water. (T1)
  3. Bring your water up to 200 (roughly) degrees. Best not to boil.
  4. Record the final temp of water. (T2) Make sure to stir briefly before measuring to ensure homogenous mix.

Utilized BTUs: W1 * (T2 - T1)


Heating Efficiency
This is the efficiency of your burner / boilpot combination.

Efficiency = Utilized BTUs / Burner Output * 100%​


Note: If using Natural Gas, replace the 21,591 with 20,161.

It's possible your burner is 56000 BTU. Flame burners throw most of their energy into the air and not into the pot.

Electric, BTW, is 100% effiicient: it all goes in the pot.
 
What is the formula to calculate BTU's from a certain temp to boil?


Weight of a gallon x how many gallons you have to heat x the temp difference to 212.

I then do that equation again but instead of temp dif do 970.

So temp dif of 54 to 212 is....158
weight of 7 gallons is............58.45

8.35 x 7 x 158 = 9,235.1 BTUs needed to get up to 212*
8.35 x 7 x 970 = 56,696.5 BTUs needed to boil
Now add the two.....9,235.1 + 56,696.5 = 65,931.6 BTUs.

The 970 BTUs per pound is needed to kick things into a boil.
Some like Kladue use a higher number for the weight of water. I believe 8.45 ish???
 
Some interesting info:
  • Burners are actually rated by BTU/hr. BTU's doesn't technically make sense since the longer you run the burner, the more BTUs expended.
  • 1 pound of propane contains 21,591 BTU of energy.
  • 1 gallon of beer weighs 8.34 pounds.
  • (Pounds of beer ) * (Temp change in F) = number of BTUs.

To determine the BTU ( per hour ) of your burner:

  1. weigh your tank (W1)
  2. Run your burner full tilt for exactly 1 hour. Doesn't matter what you are using it for. Maybe heat up some water or something.
  3. Weigh the tank again. (W2)

Burner Output (BTU/hr): (W1 - W2) * 21,591

This will only tell you the full tilt number. Max fire is not required to maintain a boil. This test will need to be done again at the maintenance fire also. Or any other rate of fire the user wishes to know about.
 
8.35 x 7 x 970 = 56,696.5 BTUs needed to boil

This is the heat of vaporization. You should only multiply this times the number of gallons that actually boiled off.

For example, let's say you heat 7 gallons to a boil. And you end up with 6 gallons because 1 gallon boiled off. You'd add (1 * 8.34 * 970) to the (7 * 8.34 * 158).
 
This is the heat of vaporization. You should only multiply this times the number of gallons that actually boiled off.

For example, let's say you heat 7 gallons to a boil. And you end up with 6 gallons because 1 gallon boiled off. You'd add (1 * 8.34 * 970) to the (7 * 8.34 * 158).


7 gallons from 54* to 212* requires 9224.04 BTUs. Using your 8.34lbs/gallon

So your saying that the 56,000 BTU/hr burner is going to bring the 7 gallons to a boil in just 6 mins?

Yet, when you do the math my way we get exactly the required amount needed to boil in the one hour that is being claimed by the poster of this thread. As things get closer to a boil less BTUs are needed and with a constant input of fire things will get more hot at a faster rate. You are right with the 8,000 ish BTUs lost to evap. But more is needed then the 10,000BTUs for a temp change.
 
I thought that was the original question. My mistake if not.

Yes, you right. I was just stating that this is not a full spectrum math figure. Rate of consumption needs to be done at different firing rates if one cares too know. Back pressure on a gas line makes a difference in actual flow rates.
 
7 gallons from 54* to 212* requires 9224.04 BTUs. Using your 8.34lbs/gallon

So your saying that the 56,000 BTU/hr burner is going to bring the 7 gallons to a boil in just 6 mins?

Most of those BTUs are not going into the boil. They're going around the pot and off into space. Thus, the poor efficiency (20%) of a flame.

If you apply that efficiency to 56,000, you end up with about a 10,000 BTU into the pot, per hour. So I guess what I am saying is that it will take about 1 hr to get the pot boiling.
 
Here is a perfect example. This place claims that 54,000 BTUs/hr will bring 7.5 gallons to a boil in just 9.8 mins. Unless I am mistaken, this would mean that there is only 6% efficiency going on here. They state to be safe to double the heating time. So we are now at 20 minuets to a boil. So this would mean that his burner is only about 25,000 BTUs. I am just learning this stuff so I could very well be wrong.

I copied this from here,
http://www.lalagniappe.com/mall/lobbycookerfaq.htm



Q. How many BTUs do I need?

A BTU (British Thermal Unit) is the amount of energy required to raise 1 lb of water 1 degree Fahrenheit. You will usually see propane burners rated in BTU's, but what they mean is BTU's/hr.

Here is a simple way to see how many BTUs you need for your pot size. Water density is 8.3 lb/gal. To raise 1 gallon of water (1 x 8.3 = 8.3 lbs) from 70 to 212 deg F in 1 hour you will need 8.3 x 142 = 1,178.6 BTUs.

Using this BTU requirement for each gallon of water you can figure out how many BTUs would be required to boil your pot of water in one hour. For example a 30 quart pot (7.5 gallons) (full) would require 7.5 X 1,178.6 BTUs = 8,839.5 BTUs to bring the pot to a boil in one hour, assuming 100% efficiency. Of course 100% efficiency isn't realistic. Assuming 100% efficiency a 54,000 BTU/hr cast burner should bring that pot to a boil in 8,839.5 / 54,000 BTUs = .16369 hours or 9.8 minutes. Anyone that has ever tried to bring a full 30 quart pot to a rolling boil knows that it doesn't happen that quickly.

How much heat is lost before it even gets to the pot is hard to calculate. Outside temperature and wind each can negatively impact the efficiency of the heat transfer from the flames to the pot. A safe estimate would be 50% efficiency, so doubling the heating time would probably be realistic.

Evaporation takes away heat and to hold a rolling boil will require additional heat besides that required to raise it to a boil. Without going into a technical explanation just take my word that boiling away 1 gallon of water per hour will require approximately 8,000 BTUs/hr.

So, a 54,000 BTU burner should comfortably boil a 30 quart pot in 20 minutes or less and comfortably hold that pot at a rolling boil. However, if you move to a 60 quart pot, the heating time doubles and now you are sitting around for 40 minutes or more waiting for the pot to boil. Somewhere between a 30 quart pot and a 60 quart pot you probably need to move to a jet burner which produces 110,000 BTUs and therefore cuts the heating time approximately in 1/2.

Be skeptical of BTU/hr output claims.
 
Most of those BTUs are not going into the boil. They're going around the pot and off into space. Thus, the poor efficiency (20%) of a flame.

If you apply that efficiency to 56,000, you end up with about a 10,000 BTU into the pot, per hour. So I guess what I am saying is that it will take about 1 hr to get the pot boiling.


So your saying a 3000 watt electric element would have done the same thing as his burner?
 
So your saying a 3000 watt electric element would have done the same thing as his burner?

That is exactly right

So, a 9000W kettle would do it in 1/3 the time, or about 19 minutes.

A 3000W heating element will heat as fast as a 55K BTU burner
 
Here is a perfect example. This place claims that 54,000 BTUs/hr will bring 7.5 gallons to a boil in just 9.8 mins. Unless I am mistaken, this would mean that there is only 6% efficiency going on here. They state to be safe to double the heating time. So we are now at 20 minuets to a boil. So this would mean that his burner is only about 25,000 BTUs. I am just learning this stuff so I could very well be wrong.

I copied this from here,
http://www.lalagniappe.com/mall/lobbycookerfaq.htm

I'm sure the burner is what they say, 54,000 BTU. That just means how fast it can blow propane out it's nozzles.

Their numbers are fine. About 9000 BTU to heat 7.5g from 70 to 212. If you assume that all the burner's expended BTUs are going into the pot, it will take 9000/54000 = .16 hr. or 10 minutes. Thing is monti, most of the flame is going around the pot. I can't tell you how much, and neither can they. Factors like your pot design, burner stand, and wind have a big effect. I'd would say that their 50% is probably very optimistic too. Maybe someone can take the formulas I posted earlier and do some real world tests.

I only use electric so I can't add anything further here.
 
Though the heat transfer with electric is nearly 100% to the liquid, there is still heat lost through the walls of the kettle and into ambient air. I'm looking forward to testing this on my new rig which will heat HLT with either propane or 5500w element.
 
Here you go guys,

Copied from here,
http://www.stanleymotorcarriage.com/Parts/Superheater.htm

The steam that is produced when water is boiled is called saturated steam. Saturated steam is nothing more than steam at the same temperature of the water from which it came. It does not take a lot of heat energy to raise the temperature of water from the freezing point (32° Fahrenheit) to the boiling point (212° Fahrenheit). To accomplish the 180° Fahrenheit temperature change of one pound of water requires about 180 British Thermal Units (BTUs) of heat energy. However to change the water from it's liquid state at 212° F to a gas or vapor (steam) state at 212° F requires an additional 970 BTUs of energy added (latent heat of vaporization) to the water. Simply stated, a total of 1150 BTUs of energy is required to change a pound of water at 32 ° F into saturated steam at 212° F. Thus at atmospheric pressure, like a pot of water on a stove, 84% of the heat consumed by the water in turning it into steam goes solely to convert the liquid water at 212° F into a gas at 212° F
 
Making the assumption that most of the heat is blowing up the sides of the kettle is not too accurate, most of the energy is passing as radiant heat from the flame, with a lower level component of gas heat transfer. While electric systems are nice and clean, the amount of heat delivered is limited by supply system constraints, and overall delivery is usally a fraction of gas burner systems.
 
Making the assumption that most of the heat is blowing up the sides of the kettle is not too accurate, most of the energy is passing as radiant heat from the flame, with a lower level component of gas heat transfer. While electric systems are nice and clean, the amount of heat delivered is limited by supply system constraints, and overall delivery is usally a fraction of gas burner systems.


Kladue,

Am I wrong in the 6% utilization number, based on the 10,000 BTUs needed to boil? Maybe I inverted the numbers? If I am right then it isn't even as good as 6%.
 
While electric systems are nice and clean, the amount of heat delivered is limited by supply system constraints, and overall delivery is usally a fraction of gas burner systems.

Disregarding the math, my experience says you are quite wrong. I can heat/boil 14 gallons in my 5500W electrric keggle much faster than my old burner did with 7g.

BTW, 5500W = 18,700 BTU/hr.

I'd like to know what those supply "system constraints" are. Truly, as I want to know the details here.
 
OK, I just ran a test! That was sorta fun.

Data
  • 10 gallons water in my keggle.
  • Start Temp: 70F
  • End Temp: 209F
  • Elapsed Time: 40 minutes

Total BTU's expended

lbs * delta-T = BTUs
(10*8.34) * (209-70) = 11,593 BTUs

Effective Power Rating of my Electric Keggle:
11,593 BTUs / ( 0.67 hrs ) = 17389 BTU's / hr.

Efficiency of My System
17,389 / 18,700 = 93%

That efficiency is not 100% because of heat loss through walls of keggle, etc. Also, I have no idea what the power looks like and if I am truly getting 5500W.

Here is the graph from my test (from a digital temp meter with logging). Click the graph to get a big one.

 
The supply system constraints are related to average residential service panels of the 150-200 amp range and a supply voltage of 120/240 Volts. Most are loaded to the point that adding more than 50 amp load is not a good idea. With a 50 amp breaker you are limited to 80% continous load for safe operation and that limits sustained KW to 9.6Kw (32,765 Btu's). The heating to boiling process analogy is filling a bucket with a hose, you have to fill the bucket (970 Btu's / LB) before it overflows (Boiling). The larger the heat source( larger the hose) the faster the bucket fills, then to keep bucket overflowing (boiling) takes much less energy as only the water evaporation and heat loss though sides are the loads.
Electic systems are nice and clean and easy to automate but they will take a bit more time to get to boiling than a properly built gas fired or steam heated system. If electric systems is your thing go for it, it is just another way of getting the job done and making beer, I tend to favor the brute force gas/steam approach to brewing.
 
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