Careful with the sizing of the fan and using natural gas. That's a 452 CFM (only) fan - enoug for electric but not really for natural gas / propane.
Due to the extra heat and poisonous gases that must be removed, the ventilation requirements are considerably higher with a gas brewery as compared to electric.
John Blichmann wrote an article for the November 2012 issue of BYO magazine that summarized ventilation requirements as follows:
Electric based brewery: Divide the element size (in watts) by 17.6 to obtain the required CFM (cubic feet per minute). (Example: I use a 5500W element in our boil kettle. 5500 / 17.6 = 312 CFM).
Gas based brewery: Divide the burner’s BTU/hour rating by 30. (Because of the inefficiencies, a 80,000 BTU burner produces approximately the same amount of heat in the kettle as a 5500W element. 80,000 / 30 = 2666 CFM. You would therefore require a fan that can move 2666 CFM in order to ventilate a gas setup properly if you are running a total of 80,000 BTU in burners at once.)
Kal
I highly suspect that you do not need 2666 CFM to properly ventilate 80,000 BTUs of burner power.
Example #1) My house is heated by a 100,000 BTU natural gas boiler. Natural gas and propane are different fuels, but they aren't that different. Natural gas (CH4) has more H and less C than propane, but that is where it ends.
The boiler in my house is direct vent. That means its combustion is entirely sealed from the air in the rest of the house. All the intake air and exhaust products are piped in and out through 4" PVC ducting. You can even use 3" PVC ducting if the runs are short enough, ie under about 30 feet.
Here is the thing about this boiler. The fan that pulls the air in and pushes the exhaust products out is a tiny 4" squirrel cage. That is all the air that is required to burn 100,000 BTU per hour cleanly. This boiler is certified at 92% efficient. It was tested for a proper air fuel mixture with a mixture analyzer when it was installed. With that air flow there was no CO in the exhaust products, ie its getting enough air.
Example #2) My house came with a big 5 burner natural gas cook top. I guesstimate the total heat output to be over 50,000 BTU. It had a hood and fan over it. The hood was NOT connected to the cooktop in any way, ie it didn't automatically run when you fired up the burner. The fan in it was a low performance 8 inch centrifugal on a variable speed circuit. Often the fan was not running at all when the cook top was being used, often for hours on end.
Example #3) My RV has a 4 burner propane oven/stove combo. Total heat output is probably 25,000 BTUs. Nothing vents to the outside of the trailer. There is a fume hood, but it only runs when the trailer is plugged into 120VAC. There is a CO detector on the wall adjacent to the oven. Trailers are much smaller inside than houses.
I fully agree that one need to have adequate ventilation when burning natgas and propane indoors. People are killed doing this every year. However, if one truly needed 2700 CFM for that 80,000 BTU burner, people would have been killed or as least displayed some form of poisoning in each of these situations.
The actual amount of ventilation needed is much lower than that.
I could calculate how much air it takes to burn propane at a given rate, but its late and I have better things to do.
FWIW,
- we removed the the natgas cooktop and replaced it with an induction model.
- my home brewery in the basement is powered by electricity, not gas, even though I initially designed and ordered parts for a natgas brewstand.
- this web page compares the efficiencies of heating water by various methods.
http://physics.ucsd.edu/do-the-math/2012/05/burning-desire-for-efficiency/
Pot over a flame = 16 to 25%. For reference, 5.5 Kwhr = 18,766 BTU. 18,766 / 80,000 = 23.4% <--- if 5.5 Kw element was as fast as a pot over a flame, this is what the relative efficiencies are.