DorsetStrider
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| posted on 6/11/08 at 09:42 PM |
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Maths? question
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.
Who the f**K tightened this up!
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martyn_16v
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| posted on 6/11/08 at 09:45 PM |
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Depends on the starting and final pressures I think. Need more info 
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rost
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| posted on 6/11/08 at 09:46 PM |
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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
Nevermind, Didnt account for release pressure.
[Edited on 6/11/08 by rost]
Charlie don't surf!
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DorsetStrider
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| posted on 6/11/08 at 09:49 PM |
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Starting pressure 137bar
Who the f**K tightened this up!
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DorsetStrider
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| posted on 6/11/08 at 09:51 PM |
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Am trying to work out how long
THIS gas cylinder would last for tig welding.
Who the f**K tightened this up!
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rost
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| posted on 6/11/08 at 09:52 PM |
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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.
Charlie don't surf!
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JoelP
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| posted on 6/11/08 at 09:58 PM |
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multiply 2.5 minutes by 137, assuming atmospheric pressure is one bar and teh 18 cubic feet are at atmospheric pressure.
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rost
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| posted on 6/11/08 at 10:24 PM |
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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.
Charlie don't surf!
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JoelP
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| posted on 6/11/08 at 10:26 PM |
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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.
you were right really, i just expanded on your bit when he mentioned it was a gas tank releasing to atmosphere! 
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geoff shep
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| posted on 6/11/08 at 10:27 PM |
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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]
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IDONTBELEIVEIT
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| posted on 6/11/08 at 10:27 PM |
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12 hours constant weld time......@8cfm's gestimatation
Are We There Yet, Are We There Yet!!!!
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DorsetStrider
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| posted on 6/11/08 at 10:39 PM |
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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]
D'OH!
I meant 18 cubic feet an HOUR! not a minute.
Who the f**K tightened this up!
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Liam
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| posted on 6/11/08 at 11:46 PM |
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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
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maartenromijn
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| posted on 7/11/08 at 08:02 AM |
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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.
BLOG: http://thunderroad-super7.blogspot.com/
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matt_claydon
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| posted on 7/11/08 at 08:28 AM |
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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.
No. The bottle is not 1.29 m^3, that would be ridiculous! The bottle is 9.4 litres, which at the given pressure holds 1.29m^3 of (atmospheric) gas.
The answer is 2.5 hours.
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maartenromijn
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| posted on 7/11/08 at 08:29 AM |
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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]
BLOG: http://thunderroad-super7.blogspot.com/
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