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Case 68: Pneumatic testing dangers

05.01.2012  |  Sofronas, T.,  Consulting Engineer, Houston, Texas

Use caution on what defines a true ‘safe distance’

Keywords: [explosions] [hydrotesting] [welds] [welding] [tanks] [safety] [pnuematic]

I’ve written previously on hydro and pneumatic testing and how it’s an engineer’s responsibility to question dangerous conditions.1,2 Unfortunately, I still see failures and have been asked what is a safe distance to be at when performing a pneumatic test.3,4

When defining “safe distances” from a pneumatic explosion, a major concern is defining the numerous flying fragments. To understand the energy involved in pneumatic testing, the pressurized air will be treated as a compressed spring. The energy released will be used to propel a fragment horizontally. Fig. 1 shows a fictitious gas spring in a vessel with a spring constant, k, in lb/in., and compressed d, in. The diameter of the fragment is D, and the pressure is p, lb/in.2 In this example, d could be the length of a pipe or the diameter of a vessel, d. The pneumatic spring constant is:

k = p 3 (π/4) D2 / d (lb/in.)

The potential energy (PE) in a spring is:

PE = ½ k X d2 = ½ p X (π/4) D2 X d in.-lb

The kinetic energy (KE) of the fragment W is:

KE = ½ (W/g ) X V2

Equate KE = PE and solve for fragment velocity, V:

V = 17.4D (p X / W )12 in./sec

  Fig. 1.  A compressed air model.

This velocity can now be used in a simple trajectory calculation to estimate the distance that the fragment will travel horizontally. R is the horizontal range that W travels at initial velocity, V, from a height, h, ft:

R = (V/24) X ( 2 X h/32.2 )12 ft

Fig. 2 represents the fragment’s flight path with averaged velocity. Fragments of different sizes will have varying ranges. Stating a safe distance isn’t possible unless you know the size. For example, with a 200-lb/ in.2 pneumatic explosion of a pipe with D = 10 in., h = 10 ft and d = 50 in, a fragment, W = 0.1 lb, will have R = 1,800 ft. However, with a fragment of W = 25 lb and R = 114 ft, there are many possibilities.

  Fig. 2.  Fragment range and flight path of debris.

Detailed trajectory calculations considering departure angles and air friction still require that these variables are defined. It’s always prudent to look for alternatives when pneumatic testing large volumes.

Options such as localized hydrotesting with small volumes involve sections of pipe instead of the whole pipeline. Nondestructive testing of welds and critical areas is another option. Consider reviewing the historical hydrotesting of a modified vessel and only testing the portions that were modified can be used. This may require temporary weld on caps. Strive to recognize codes if you must pneumatically test. Remember: There may be no reasonable “safe distance” or “exclusion zone.” The safe distance specified could reference a blast pressure wave, not flying fragments.

A ruptured 25-lb pipe-end fragment with an air volume of 140 ft3 at 200 lb/in.2 is like a locomotive engine traveling at 15 mph or one pound of TNT. All possess about 1.5 million ft-lb of KE. It is something to seriously consider when discussing pneumatic testing. HP

1 Sofronas, T., “Case 34: Hydrotesting or pneumatic testing,” Hydrocarbon Processing, September 2006.
2 Sofronas, T., “Case 65: Taking risks and making high-level presentations,” Hydrocarbon Processing, November 2011.
3 “Pneumatic Test Explosion in Shanghai LNG Terminal,” Chemical & Process Technology, March 2009.
4 “Pneumatic Test Failure in Mississippi Pipeline Project,” Chemical & Process Technology, July 2009.

The author 

Dr. Tony Sofronas, P.E., was the worldwide lead mechanical engineer for ExxonMobil before his retirement. He is now the owner of Engineered Products, which provides consulting and engineering seminars. He can be reached through the website http://mechanicalengineeringhelp.com by clicking on the comments/question tab. 

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T. Sofronas

Thank you for your comments. The equation was correct in the magazine but in error here. rho 3 should be p and d should be delta in some of the equations. Must have been a problem going electronic. 24 is correct as 1/12*1/2.

vincenzo puccia

probably two printing error: the 3 number in the pneumatic spring constant and the 24 under V (1 ft=12 in, not 24)

Thank You

franco balestri

In the final formula for R, velocity V should be divided by 12 instead of 24?

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