Hello everyone. I just now signed up to this site because I wanted to make a propane cannon and I had a few questions.
-It is said when threading a BernzOMatic torch head you use a 1/8" NPT(National Pipe Thread) The torch head is much to big to even begin to thread it with a 1/8". I found this information here: http://www.burntlatke.com/thread.html Yes I took the bent part of the head off.
-What is the purpose of having two ball valves?
-You attach the pressure gague either to the torch head or have a hose leading to it (nothing between) correct? Then you use the ball valves for? I understand why you would need one but not two.
Thanks
First propane combustion.
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SpudStuff
- Sergeant 5
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The torch is the perfect size for 1/8" threads
the pipe between the two valves fills with the propane metering it. so you
close the second valve
Open the first valve
Close the first valve
open the second valve
close the second valve
fire
The pressue guage is between the torch head and the
Shit.
just Click Here
the pipe between the two valves fills with the propane metering it. so you
close the second valve
Open the first valve
Close the first valve
open the second valve
close the second valve
fire
The pressue guage is between the torch head and the
Shit.
just Click Here
Last edited by SpudStuff on Fri Apr 28, 2006 11:51 pm, edited 1 time in total.
what happens when you metre propane, is you turn on the propane with the 2nd ballvalve open when the regulator indicates the right mix (cant be bothered to explain, check wiki.) you close the first ball valve, and open the second one, that is how propane is metred., i think you have to thread it with 1/4,3/8 or 7/16.EDIT: you beet me. 
First you must understand that several spudders have found a 4% mixture to be the best mix, and the rest of us have followed there claim ever since. This is going to be injected with atomsphere (plain old air surronding us). Atomsphere is 14.7psia so that means that the propane your injecting is only 4%. Below I will show the pressure of the propane inside your cannon:
X/ 14.7= 4/ 100
X/ 14.7=.04
X= 14.7*.04
X= .588 PSI or 15.288 PSIA
Since it would be very hard to flip open a ball valve and fill your chamber to .588 PSI, that is why we use a metering system. So instead we fill a small chamber to a higher pressure to increase accuracy when it comes to metering, then it is dumped into chamber and disperses to ~.588 PSI.
Let me walk you threw a simple calculation which actually can be done on online spud gun calculators. Lets say your chamber is 1000mL or 1L, and that you want to fill your metering pipe to ~50 PSI. Now that we have the constants of the mixture being 4%, metering pipe at ~50 PSI, and 1,000mL chamber.
First we need to find the ratio between the chamber and metering systems pressure.
X= 50/ .588
X= ~85
Now we know that it takes one unit of volume in the metering system to equal ~85 units of a chamber. Easier explained as that we need our chamber to be 85 times bigger then the metering system. Now we simply need to find the size of the meter pipe by using 1 to ~85 ratio.
X= 1000mL/ 85
X= ~11.76mL
Now we know in this instance that are metering system will be ~11.76mL big, IF we meter with ~50 PSI, and are chamber is 1000mL.
X/ 14.7= 4/ 100
X/ 14.7=.04
X= 14.7*.04
X= .588 PSI or 15.288 PSIA
Since it would be very hard to flip open a ball valve and fill your chamber to .588 PSI, that is why we use a metering system. So instead we fill a small chamber to a higher pressure to increase accuracy when it comes to metering, then it is dumped into chamber and disperses to ~.588 PSI.
Let me walk you threw a simple calculation which actually can be done on online spud gun calculators. Lets say your chamber is 1000mL or 1L, and that you want to fill your metering pipe to ~50 PSI. Now that we have the constants of the mixture being 4%, metering pipe at ~50 PSI, and 1,000mL chamber.
First we need to find the ratio between the chamber and metering systems pressure.
X= 50/ .588
X= ~85
Now we know that it takes one unit of volume in the metering system to equal ~85 units of a chamber. Easier explained as that we need our chamber to be 85 times bigger then the metering system. Now we simply need to find the size of the meter pipe by using 1 to ~85 ratio.
X= 1000mL/ 85
X= ~11.76mL
Now we know in this instance that are metering system will be ~11.76mL big, IF we meter with ~50 PSI, and are chamber is 1000mL.
Well supposibly propane comes from the tank at ~90 PSI. I know why it is ~90, but will take to much time to explain. Anyways, you dont want to set right at ~90 PSI, or within a few times of using the propane will be below ~90 PSI and mess up your calculations. So instead you meter it to something lower. I for example do ~50 PSI to get more use out of my tanks, this also increase the size of the metering pipe.
As you might have caught on before, ~50 PSI is a variable that can be adjusted you just needed it to be a constant to figure out the math. You can follow my same steps, but change the PSI and chamber volume. 1,000mL is rather small in realitly, I was just using it as a example. You have a IM service? This would be hell a lot easier over a IM chat...
As you might have caught on before, ~50 PSI is a variable that can be adjusted you just needed it to be a constant to figure out the math. You can follow my same steps, but change the PSI and chamber volume. 1,000mL is rather small in realitly, I was just using it as a example. You have a IM service? This would be hell a lot easier over a IM chat...