Hiya all,
Got a question for all the mathmaticians out there...
If a gas cylinder holds 1.29m3 of gas... how long will the gas last if used at a rate of 18 cubic foot per minute?
P.S. I don't know the answer that's why I'm asking the question.
Depends on the starting and final pressures I think. Need more info
At a constant flow:
code:
1 ft³ = 0.028316 m³
18 ft³/min = 0.509688 m³/min
1.29m³/0.509m³/min=2.53 min
Starting pressure 137bar
Am trying to work out how long THIS gas cylinder would last for tig welding.
The pressure doesn't matter IMHO, since the useage is measured in volume. If it'd be in speed you'll need the pressure and outlet
diameter.
Volume divided by volume/time results in time.
multiply 2.5 minutes by 137, assuming atmospheric pressure is one bar and teh 18 cubic feet are at atmospheric pressure.
quote:
Originally posted by JoelP
multiply 2.5 minutes by 137, assuming atmospheric pressure is one bar and teh 18 cubic feet are at atmospheric pressure.
quote:
Originally posted by rost
quote:
Originally posted by JoelP
multiply 2.5 minutes by 137, assuming atmospheric pressure is one bar and teh 18 cubic feet are at atmospheric pressure.
Ah! You're totally right, I didn't account for the release pressure being lower than the tank pressure.
The 1.29 cubic metres must be the volume at atmospheric pressure or else it would be a bloody big cylinder. Do you really mean 14 cubic ft per
minute or per hour. In which case rost's sum would be pretty close at 2.5 hours.
I think 14 cu ft per minute, through 1/4" jet is about 100mph but I may have my sums wrong!
[Edited on 6/11/08 by geoff shep]
12 hours constant weld time......@8cfm's gestimatation
quote:
Originally posted by geoff shep
The 1.29 cubic metres must be the volume at atmospheric pressure or else it would be a bloody big cylinder. Do you really mean 14 cubic ft per minute or per hour. In which case rost's sum would be pretty close at 2.5 hours.
I think 14 cu ft per minute, through 1/4" jet is about 100mph but I may have my sums wrong!
[Edited on 6/11/08 by geoff shep]
Oh dear oh dear guys!
The 'content' of a gas cylinder is the volume of the gas at atmospheric pressure. So if you're welding at 18 CF/hour it'll last
about 2.5 hours. A bit less really cos of course you wont empty it fully.
Liam
Bernoulli says:
p * v = c(onstant)
137 b * 1,29 m^3 = 176,73
Now, the garage where you will do your welding has athmospheric pressure: 1 bar.
At athmospheric pressure:
1 b * x m^3 = 176,73
x = 176,73 m^3
At athmospheric pressure you have 176,73 m^3 of gas.
18 cbf = 0,51 m^3
release the gas at a rate of 0,51 m^3 / hour:
176,73 / 0,51 = 345,1 hours of welding!!
You can weld for two weeks in a row with this bottle! I think it is more likely that you use 18 cbf / minute instead of 18 cbf / hour.
That makes: 345,1 / 60 = 5,75 hours of continuous welding.
quote:
Originally posted by maartenromijn
Bernoulli says:
p * v = c(onstant)
137 b * 1,29 m^3 = 176,73
Now, the garage where you will do your welding has athmospheric pressure: 1 bar.
At athmospheric pressure:
1 b * x m^3 = 176,73
x = 176,73 m^3
At athmospheric pressure you have 176,73 m^3 of gas.
18 cbf = 0,51 m^3
release the gas at a rate of 0,51 m^3 / hour:
176,73 / 0,51 = 345,1 hours of welding!!
You can weld for two weeks in a row with this bottle! I think it is more likely that you use 18 cbf / minute instead of 18 cbf / hour.
That makes: 345,1 / 60 = 5,75 hours of continuous welding.
BTW: This bottle can never contain 1.29m^3 compressed gas!
heigt 0,8 m
diameter 0,15 m, radius 0,075 m
PI * 0,075^2 * 0,8 = 0,014 m^3 !!!
It might be 'Great news !!!' as mentioned in the ad, but I think this guy don't know what he is talking about
Edit: Ahh, matt_claydon, our posts just crossed. Yes, you must be right.
[Edited on 7/11/08 by maartenromijn]