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Pipe burst pressure

I'm trying to calculate pipe size for a thin copper tube used to feed a 10,000 psi pressure guage. The original was involved in an intense fire and is totally blocked by carbonised hydraulic oil. It's o'd is

1/8" and it's i/d isn't easy to measure due to the blockage. Wall thickness looks to be about 25 thou.

As I understand it the burst pressure of a pipe is given by:

P=(2t x S)/O

Where

P=Burst Pressure in PSI t= wall thickness in inches S is the tensile strength of the material in PSI O is the outer diameter of the pipe

I understand the tensile strength of copper to be about 70Mpa or

10,000 psi

So P = (2x0.025 x 10,0000)/0.125

So P = 4,000 psi

The pump puts out at least 8,000 psi and quite possibly the full

10,000 psi so what's going on - where am I going wrong as the pipe hadn't burst!

Also does anyone know a source of about a metre of thin (ie 1/8" o/d) but thick walled copper pipe?

AWEM

Reply to
Andrew Mawson
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I suspect the formula you used assumes a wall thickness that is thin compared to the OD. I derived the formula some time ago but cannot find the calculation now.

This link

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Gives more encouraging results for a diameter of 0.125 and wall thickness of

0.05 - pretty thick wall but possible?

Regards Roger Woollett

Reply to
Roger Woollett

This is a thin wall approximation for the hoop-stress. Easily derived by considering cutting the pipe through a plane containing the centre line of the pipe.

Force on either side of cut = (pressure) * cross sectional area

So for length L of pipe, cross sectional area = L d

Stress = Force/Pipe thickness = P L d / (2 * L * t) = Pd/(2t)

[And of course hoop stress = 2 * longitudinal stress, which as an aside explains the behaviour of saussages when being cooked!]

For a "thick" cylinder you need to integrate to get Lame's equation and

max circumferential stress = p( (OD)^2 + (ID)^2 ) / ((OD)^2 - (ID)^2)

max shear stress = p (OD)^2 / ( (OD)^2 - (ID)^2 )

Since ID = OD - 2t we get for circumferential stress

p_max = 10000 * ( (OD)^2 - (OD-2t)^2 ) /( OD^2 + (OD-2t)^2 ) = 10000 * ( 0.125^2 - (0.125-0.050)^2 )/ (0.125^2 + (0.125-0.050)^2 ) = 4,700 psi

[Which is still too low....]

However it is clear for p_max of 10,000psi you would end up with t=OD/2 which is commonly sold as copper rod :-)

Does the material need to be copper?

Alan

Reply to
Alan Bain

intense

derived by

(ID)^2)

(0.125-0.050)^2 )

Strictly no - the original is copper or a copper based alloy, and for easy of bending the run it would be preferable, but if you can suggest any other suitable material I am all ears!

AWEM

Reply to
Andrew Mawson

The minimum tensile strength of copper alloy as used in brake pipe is at least 45,000 psi (BS figure). Sch 40/80 pipe is 40,000 psi minimum.

-- Peter Fairbrother

Reply to
Peter Fairbrother
1/8=94 od x 22swg copper pipe with a burst pressure of 18000psi and the calculation is here
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John

Reply to
John

Car brake pipe, in cunifer (IIRC Spelling) or plain copper. Local motor factors should be able to supply it. I might have a length in the garage if you cant get any.

Dave

Reply to
dave sanderson

Car brake pipe, in cunifer (IIRC Spelling) or plain copper. Local motor factors should be able to supply it. I might have a length in the garage if you cant get any.

Dave

Thanks Dave but I doubt it will be as thin as 1/8" o/d

AWEM

Reply to
Andrew Mawson

Car brake pipe is 3/16" or the metric equivalent.

John

Reply to
John
1/8? od x 22swg copper pipe with a burst pressure of 18000psi and the calculation is here
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John

John many thanks - order placed with Abacus.

Andrew

Reply to
Andrew Mawson

Having just looked (at the fittings anyway, goblins have hidden the pipe) it is indeed 3/16"

Dave

Reply to
dave sanderson

Andrew (and everyone else),

I know you've found some suitable pipe now but I though I could add a little to the preceeding discussion. If the wall thickness of a pipe is more than about 1/20th of the diameter then the pipe is considered to be 'thick walled' and the equations discussed previously are generally deemed to be invalid as they assume a constant stress distribution through the wall thickness. For the thick walled case the equations by Lame (pronounced Larmay I think) are used. They are trickier to use as they are a pair of simultaneous equations.

Further info can be had here:

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and an online calculator here:
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regards

Toby

Reply to
Toby

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