Yeah, this forum does not lend itself well to the mathematics behind the theory, I will just give you the numbers and logic of the paradigm.
Given a 20 inch diameter pipe of schedule 20, the internal pressure is 14.7 psi, taken to represent atmospheric conditions. You could take this pipe and submerge it to a depth "h" in water of specific weight 62.5 lbf/ft^3 or:
h = P/(rho g)
h = (14.7 psig/62.5 lbf/ft^3)X(12 in/ft)^3 = 406.4 in.
At this depth, the wall is in neither compression or tension since internal pressure exactly balances external pressure subjected by the water. I would call this the neutral axis, wall stresses are exactly zero.
Continue to submerge the pipe. As the water gradient increases with depth, there shall be a point below this neutral axis to which pressure will crush the pipe. By Thick Wall Pressure Vessel Theory, an element of wall does not care if external pressure is crushing it or internal pressure is blowing it outwards. We can treat internal and external pressure by the same equation, I have chosen the Von Mises-Hencky Equation since it relates hoop, radial and longitudinal stresses as a vector gradient for a triaxial state of stress of an element representative of the wall under the loaded condition.
Grinding through the mathematics, the stress gradient applied to Thick Wall Pressure Vessel for hoop, radial and longitudinal stresses becomes:
S = sqrt(3) P [R^2 / (R^2 - 1)] for R = D/d
D = 20.0 in (OD), d = D-2t = 19.25 in (ID)
"t" is the wall thickness, commecial pipe schedule gives 0.387 inch for your 20.0 OD schedule 20 pipeline.
I set the stress equal to material yield, PVC rigid plastic listed at 5218 psi and a Young's Modulus of 2.8 GPa (metric value in Canada). By giggling around the terms in the Von Mises-Hencky Equation, you would find the required pressure to bring the wall to a yield point, hence the onset of failure to be:
R = D/d = 20.0 in / 19.25 in = 1.03896
P=[5218 psi/srt(3)][(1.03896^2 - 1)/1.03896^2] = 221.7 psi
This corresponds to a depth of h' BELOW the previously established neutral point.
h' = P'/(rho g) for P' = 221.7 psi
h' = (221.7 psig/62.5 lbf/ft^3)X(12 in/ft)^3 = 6130 in.
Therefore, total distance as measured from surface, hence the depth you seek is:
H = h + h' = 406.4 + 6130 in = 6536 in = 544.7 ft
This computation will be influenced by the physical values to the PVC material, dependent upon grade. For purposes of illustration and finding this an curiosity, I used generalized values from a DuPont listing. Be careful to check an MTR from your supplier and make the appropriate changes.
Hope this clarifies your apprehensions. This is a very cute problem.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
