PSV requirement for tube rupture - "open" system

[Helmoltz]

Dear all,
I'm facing a though problem, and nobody at work can provide me with a complete answer

System: Shell & Tube heat exchanger, 46 bar g steam in tubes, 1 bar g process gas in the shell. Design pressures are 52 bar g for the tubes and 5 bar g for the shell.
The shell diameter is about 30".
The process side (i.e. shell side) is ultimately open to the ATM without any obstacle (i.e. a 24" outlet line with no valves).

Question: Is a PSV required for tube rupture?

The company I'm working for has a long experience in such plant configuration, and PSV were never added, because if you calculate the relieving flow rate due to tube rupture, such flow can be easily managed by the outlet pipe itself leading to a moderate increase in pressure.

I recently HAZOPed this unit with a very demanding client, and they accepted this answer for the steady state.
However, they want me to evaluate the transient pressure increase due to tube rupture, and in case this is higher than the 5 bar g (design pressure), provide PSV on the shell.

I would sincerely appreciate any suggestion.

 

[don1980]

If the low-pressure side (LPS) has an outlet that is always open, and large enough to prevent overpressure, then a PSV isn't necessary. You already have overpressure protection through the open pipe to atm. Open atm outlets are acceptable for protecting a vessel.

The transient analysis suggested by the client would be potentially justifiable if the shell side (LPS) was filled with liquid. When you have a very high pressure gas/vapor/supercritical fluid on the tubeside, and a shellside (LPS) filled with liquid, then a sudden tube rupture can result in catastrophic failure of the shell, even if the shell side has a large PSV. The shell can "explode" due to the sudden high force applied by tube-side fluid. The explanation is pretty simple when you think about what happens in cases like this, but you have to think about it in terms of what happens on a time scale of milliseconds rather than seconds. When the tube breaks, a sudden shock force is applied to the liquid around that break. Liquid is incompressable, so that force is immediately exerted on the shell containing that liquid. The inertia of the liquid prevents it from moving out of the way quickly enough to create a pocket for the gas to expand into, or for the gas to reach the shell side PSV. In other words, the force is applied so quickly, the shell side might as well be filled with a solid rather than flowable liquid. A shell-side PSV can't protect from this scenario because it can't respond (open) fast enough. The shock force from a tube break can greatly overpressure the shell in a couple of milliseconds, before the inertial of the PSV spring can respond, and before the liquid can accelerate out of the way. A relatively large rupture disk is required in such cases - it's the only device that will respond quickly enough to protect from this scenario..

What I described above is a unique risk that only exists when you have a liquid-filled shell and an extremely high pressure gas on the tubeside. [Different companies have a different criterion for how high the tubeside pressure needs to be before being concerned about this scenario. The general consensus is that it has to be greater than ~ 750-1000 psig.] If you have a vapor in the LPS (shellside), the sequence of events is fundamentally different. Vapor, of course, is compressible so the time-to-overpressure is much higher - it's high enough for a PSV to open defend against the scenario.

 

[moltenmetal]

No PSV is necessary in your case, nor is a higher design pressure required in the absence of a PSV.

If the shell were liquid filled and open to the atmosphere, it could potentially be a different matter as don1980 describes.

 

[Helmoltz]

Thank you for your replies.
I totally agree with the position that only a liquid-filled shell would be susceptible to mechanical damage.

Do you have any idea how to prove that a gas-filled shell will not overpressurize?

The Client's remark is that when a high pressure area is suddenly opened to a low pressure area, a shock wave (with magnitude about 1/2 of the high pressure) travels through the system. I found some info from literature about this phenomenon ("shock tube"), but in my opinion the described system is quite different from mine, since we are talking about a very small volume (one tube) against a "huge" low pressure volume (vessel plus piping). If it were full of liquid, than problems may arise (due to inertia and incompressibility of the medium), but being full of gas I would say that such shock wave will dissipate long before damaging the shell.

I'm not sure about how to approach the calculation of transient pressure increase in order to show that pressure will always be lower than 5 bar g - any suggestion?

Thanks

 

[LittleInch]

You could try a transient pipe or pipeline simulation with your vessel as a 30" pipe with your 24" pipe going to your outlet, plus what ever flow you have going through the shell normally to create your 1 barg pressure then add a very fast opening valve - you should be able to set an instantaneous opening - with your tube size and pressure and see what happens...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[don1980]

The phenomenon I described above doesn't occur if the shell side is gas.

What you have is a plain vanilla relief design. Gas is compressible. When the tube breaks, the pressure rises to set pressure and the PSV opens. Nothing unusual there. The fundamental difference is that with a compressible fluid, PSVs have time to open.