# 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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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 http://www.engineersedge.com/pipe_bust_calc.htm
Gives more encouraging results for a diameter of 0.125 and wall thickness of 0.05 - pretty thick wall but possible?
Regards Roger Woollett

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
wrote:

feed
intense
is
derived by

line
aside
(ID)^2)
(0.125-0.050)^2 )

o/d)
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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Andrew Mawson wrote:

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
1/8� od x 22swg copper pipe with a burst pressure of 18000psi and the calculation is here http://www.abacus-tubes.co.uk/pages/view_products.php?category=soft_copper_tube_in_coil
John
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
1/8� od x 22swg copper pipe with a burst pressure of 18000psi and the calculation is here http://www.abacus-tubes.co.uk/pages/view_products.php?category=soft_copper_tube_in_coil
John
John many thanks - order placed with Abacus.
Andrew
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Jan 10, 2:36�pm, "Andrew Mawson"

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: http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Machine%20design1/pdf/module-9%20lesson-2.pdf
and an online calculator here: http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html
regards
Toby
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Jan 8, 9:54�pm, "Andrew Mawson"

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
wrote:

feed
o/d)
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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Car brake pipe is 3/16" or the metric equivalent.
John
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Having just looked (at the fittings anyway, goblins have hidden the pipe) it is indeed 3/16"
Dave
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

## Site Timeline

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.