Re: Re: Reinforcement Pad Weep hole

From: <Christopher>
Date: Wed Sep 14 2005 - 16:39:00 EDT

I couldn't stand it, I wasted 20 minutes or so making a brief assessment of what kind of pressures you'd have to build up to bulge a doubler plate. You can make a few assumptions and get a pretty good idea of what needs to happen. Assume--

--the doubler behaves as a flat annular plate

--The doubler OD, Do, is about twice the ID, based on reinforcement
boundary requirements, and that the ID and OD are fully connected to the nozzle wall and the vessel wall respectively. The stress in the doubler (Timoshenko, Strength of Materials) due to pressure, Pi, over the faying surface is S = 0.185Pi(Do/td)^2 with td as the doubler thickness.

--The opening diameter is half the vessel diameter, D, and the doubler
thickness equals the vessel thickness, t. The ratio of vessel diameter to thickness, D/t, is about 2xAllowable stress/internal pressure, which works out to be about 100 for a design pressure of 400 psi and an allowable stress of 20000 psi. For ordinary vessels the ratio ranges from 20 to 100 or more.

--For the opening size specified, Do = D so Do/td = D/t. The pressure,
Pi, required to yield the doubler is Sy/0.185(Do/td)^2 = Sy/0.185(D/t)^2. A vessel carrying an allowable stress of 20 ksi and a yield stress of 40 ksi with D/t = 100 and Do = D (corresponds roughly to an internal pressure of 400 psi) requires roughly 22 psi to bulge the pad. if D/t = 50 the pad requires 86 psi to bulge.

If air were trapped between the wall and the doubler the internal absolute pressure is proportional to the absolute temperature. To go from atmospheric pressure to 22 psi requires a temperature rise to 675F from atmospheric if the volume remains constant. If D/t = 50 the doubler gets thicker and the pressure to bulge is 86 psi. That would take a temperature rise to just over 2600F. Note the trapped volume will increase as the doubler deforms, so the required temperature to achieve a given pressure gets higher. The high temperature has to occur throughout the entire volume of air and be maintained long enough for yielding to occur.

The situation is about the same if the area inside the pad contains steam or any other gas. Water remains liquid at higher temperatures than 212F if the pressure goes up, so any liquid trapped between the requires a higher temperature to become steam at the required pressure.

I conclude that it's not likely that gas can be heated high enough to bulge the doubler under ordinary conditions. I don't think this is a likely cause of the doubler deformation or the cracking that Chakraborty describes. Given that Chakraborty's vessel didn't get the proper Q/A and was welded wet, I daresay the weld broke because it was embrittled. Although the pressure (4 kg/cm^2 = 60 psi) is the right order of magnitude, it's hard to imagine that the volume change to a 6 inch dent would allow such a pressure build-up.

So now I've stuck my neck out, just waiting for someone to point out the hidden flaw in my work...

Christopher Wright P.E. |"They couldn't hit an elephant at <a href="/group/PipingDesign/post?postID=_kr1dToU0SYfPzN20wAxtG5IFWXvfnikPjp3BNzHDYn2VyGfS8N1zKfxzwFjr0CrK-HLpn23cvgh7A">chrisw@skypoint.com</a> | this distance" (last words of Gen.

.......................................| John Sedgwick, Spotsylvania

1864)
<a href="http://www.skypoint.com/~chrisw/">http://www.skypoint.com/~chrisw/</a> Received on Wed Sep 14 16:39:00 2005