American Welding Society Forum

Could any of you give me a very rough idea of how long 4.2m^3 of Oxygen would last welding 2mm-10mm steel and cutting 4mm - 25mm steel. I mean rough as in... a few minutes, an hour, a couple of hours (Straight). I read from the back of the Harris booklet one foot is exactly 30.4cm. So in feet the tank would be 4200cm / 30.4 = 138 ft^3. I know that my tote torch uses 1.1 ft^3 in about 5 minutes (With a seriously hot flame on) so i can expect the tank to last about... 138 x (5x60) = 41400 seconds. Or... 41400 / 60 = 690 m or 690 / 60 = 11.5 h. Does that seem right? 11 and a half hours is a long time to me for 4.2m^3 of gas.

Now all I need is for my plans to be cut short by the evil that is CGU House Insurance. : ) I can sense it will be because a torch would just be too useful.

John H. UK

Now all I need is for my plans to be cut short by the evil that is CGU House Insurance. : ) I can sense it will be because a torch would just be too useful.

John H. UK

Dear John,

First of all, you should do the right calculations. It's true that 1 ft equals 30,48 cm, although it's easier to do the calculations either in millimeters or meters. So, 1 ft equals 304,8 mm or 0,3048 m. But never mind, let's go on with 30,4 cm.

Now, you say that "in feet your tank would be 4200 cm/30,4 = 138 cu.ft." What's that? How can you divide cm (4200) per cm/ft (30,4) and obtain the result in cubic feet? There's something wrong in the calculation. What's 4200 cm? The tank diameter? If you want to know the circumference, then you should multiply 4200 per Pi (3,1416). On the other hand, one cubic foot equals 28,3 liters, or 0,0283 cubic meters, if you prefer.

So, do the calculations correctly before going any further.

Giovanni S. Crisi

Sao Paulo - Brazil

First of all, you should do the right calculations. It's true that 1 ft equals 30,48 cm, although it's easier to do the calculations either in millimeters or meters. So, 1 ft equals 304,8 mm or 0,3048 m. But never mind, let's go on with 30,4 cm.

Now, you say that "in feet your tank would be 4200 cm/30,4 = 138 cu.ft." What's that? How can you divide cm (4200) per cm/ft (30,4) and obtain the result in cubic feet? There's something wrong in the calculation. What's 4200 cm? The tank diameter? If you want to know the circumference, then you should multiply 4200 per Pi (3,1416). On the other hand, one cubic foot equals 28,3 liters, or 0,0283 cubic meters, if you prefer.

So, do the calculations correctly before going any further.

Giovanni S. Crisi

Sao Paulo - Brazil

4200cm^3 is the volume of oxygen in the tank, not the physical cylinder volume. It'd have to be a pretty big tank if it was 4.2m in diameter, I did use the ^3 for the 4.2 indicating a cubic number. If one foot equals 30.4cm then to every metre will go 3.28 of them. If we change the metre into a ^3 then there will now be 100 times as much inside it. So 1000 / 30.4 (The distance of 1ft) = 32.89 feet. I have 4.2m^3 (Or 4200cm^3) in the cylinder so that will be 32.28 x 4.2 = 138 feet^3 or there abouts. If you think of a one metre cube on the floor, there will be about three blocks going across one side, down the other, then three going up. So there is 9 on the first layer, 18 by the next and 27 by the last. If i had four of these cubes i would have 27 x 3 = 81 small blocks. 81 isn't too far off 138 considering I have not used the 30.48cm or added on the extra for the .2 of a metre. There is around 2.5cm in one inch and 12 inches in one foot. Which means there are 12 x 2.5 = 30cm about in one foot. So in one cm there is 1 / 30 = 0.03333333 feet. So in 4.2m, or 4200cm, there is about 0.03333333 x 4200. Or 140 feet. I'm talking about actual gas present in the cylinder, not it's size. Think about it this way. If I was to turn my 30.4 into a cubic number it'd be 30.4 x 30.4 = 924.16cm^3 per cubic foot. Which means I'd only get 4200 / 924.16 = 4.53ft^3 for my 4.2m. I have a tiny tote torch that I can hold in my hand that contains only one point one foot ^3. Where as the cylinders are nigh on a metre tall and around 20cm O.D.

I have the sinking feeling I'm doing something very wrong here like in an exam but I have checked with other people who think it's 138ft^3.

All the best,

John H. UK

I have the sinking feeling I'm doing something very wrong here like in an exam but I have checked with other people who think it's 138ft^3.

All the best,

John H. UK

Here are my thougts, 4.2m^3 would equal a volume of 1m x 1m x 4.2m, which woul equal 100cm x 100cm x 420cm. There is 30.48cm in a ft, 100/30.48=3.28, 420/30.48=13.78. Therefore, 100cm x 100cm x 420cm = 3.28ft x 3.28ft x 13.78ft = 148 ft^3.

148 ft^3 at 14.7psi (atmospheric pressure) would equal 73 ft^3 at 15psi gauge pressure (29.7psi absolute pressure, 14.7 + 15). Your torch uses 1.1 ft^3 in 5 min or 13.2 ft^3 per hour. 73/13.2=5.5 hrs of total Oxygen, with something more like 4 to 4.5 hrs of usable Oxygen.

Does this sound like reasonable estimate?

148 ft^3 at 14.7psi (atmospheric pressure) would equal 73 ft^3 at 15psi gauge pressure (29.7psi absolute pressure, 14.7 + 15). Your torch uses 1.1 ft^3 in 5 min or 13.2 ft^3 per hour. 73/13.2=5.5 hrs of total Oxygen, with something more like 4 to 4.5 hrs of usable Oxygen.

Does this sound like reasonable estimate?

The 4.2m^3 is the volume of O^2 at cylinder pressure already. It's amazing how confusing metric and imperial can make things. The reason this all started is the tote cylinders are imported from the US so all the measures are in imperial on their backs.

'There is 30.48cm in a ft, 100/30.48=3.28, 420/30.48=13.78. Therefore, 100cm x 100cm x 420cm = 3.28ft x 3.28ft x 13.78ft = 148 ft^3.'

This looks right to me ^ because it makes perfect sense.

So if we take it as already at pressure there is 148ft^3. Which is still...

5mins x 60 = 300 seconds of burning per 1.1ft^3 which means in one hour, or 60 x 60 = 3600 seconds, the torch would burn 3600 / 300 = 12 of the mini cylinders. Which is 1.1 x 12 = 13.2ft^3, as you said. Which means 148ft^3 will take... 148 / 13.2 = 11.212121 Hours. Or like you say about 11 - 10.5 hours of usable O^2.

We should have just stuck with the accurate, loads, lots, some gauges!

You don't want to know about the trouble I've had getting this on the house insurance, took about a million and one calls between the insurance agency and the fire inspectors.

Thanks very much for the help!

John H. UK

'There is 30.48cm in a ft, 100/30.48=3.28, 420/30.48=13.78. Therefore, 100cm x 100cm x 420cm = 3.28ft x 3.28ft x 13.78ft = 148 ft^3.'

This looks right to me ^ because it makes perfect sense.

So if we take it as already at pressure there is 148ft^3. Which is still...

5mins x 60 = 300 seconds of burning per 1.1ft^3 which means in one hour, or 60 x 60 = 3600 seconds, the torch would burn 3600 / 300 = 12 of the mini cylinders. Which is 1.1 x 12 = 13.2ft^3, as you said. Which means 148ft^3 will take... 148 / 13.2 = 11.212121 Hours. Or like you say about 11 - 10.5 hours of usable O^2.

We should have just stuck with the accurate, loads, lots, some gauges!

You don't want to know about the trouble I've had getting this on the house insurance, took about a million and one calls between the insurance agency and the fire inspectors.

Thanks very much for the help!

John H. UK

I believe 4.2m^3 is the gas volume at atmospheric pressure (14.7psi). A cylinder 20cm dia X 100cm high would have a volume of about 1.1 ft^3. 20/30.48=.656 ft, 100/30.48=3.28 ft, V=R^2 x pi x length = 1.1 ft^3. Cylinder pressure is approximately 2000 psi.

1.1 ft^3 at 2000 psi would contain the same amount of gas as 74 ft^3 at 29.7psi and 148 ft^3 at 14.7 psi (atmospheric pressure).

1.1 ft^3 at 2000 psi would contain the same amount of gas as 74 ft^3 at 29.7psi and 148 ft^3 at 14.7 psi (atmospheric pressure).

I checked with BOC and they think it may also be at atmospheric pressure but they're not sure. Wouldn't it be much easier if they gave it as per cylinder pressure? They also have a price list that lists the prices of the cylinder then the fill cost and then on the propane it has (Kg). That made me wonder, is that the cost of each kilo you ask to be added or just telling me it's measured in Kg? It can't be for how much you ask for because it seems to change for each cylinder.

I almost bought a Murex kit, purely because it's a BOC franchise, but it only welds 8mm and cuts 20mm. I've order the Harris 843MX which is one of the first torches, bar the Murex DH's, that has a 'grip' rather than a barrel. It's £340 for the outfit but it looks incredibly comfortable to hold so if anyone wants to know what it's like to use let me know. There aren't any pictures of it on their site yet because it's releatively new.

Why doesn't some goggles place start fitting those layered lenses you can get in new shades that darken to UV as well? Auto-Crystals are far too much.

Thanks again,

John

I almost bought a Murex kit, purely because it's a BOC franchise, but it only welds 8mm and cuts 20mm. I've order the Harris 843MX which is one of the first torches, bar the Murex DH's, that has a 'grip' rather than a barrel. It's £340 for the outfit but it looks incredibly comfortable to hold so if anyone wants to know what it's like to use let me know. There aren't any pictures of it on their site yet because it's releatively new.

Why doesn't some goggles place start fitting those layered lenses you can get in new shades that darken to UV as well? Auto-Crystals are far too much.

Thanks again,

John

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