Equivalent Energy 7

[JamesBarlow]

I know this is a strange question, but I am trying to impress some old college friends with what I do.

What I want to know is, given a vessel with certain pressure and volume, how would you translate that into stored energy.

To get right to the point, I want to tell them that if this vessel were to explode it would have the force of XX sticks of dynomite.

This goes without saying, but I will say it just so everyone is sure. I design all vessels to the appropriate codes and with hefty safty margins to ensure that this will never happen. But if it did.....

 

[reichertc]

I have seen this calculation before for evaluating a safe clear distance to be maintaned around a pneumatic test. Unfortunately, I appear to have misplaced the paper. In the meantime, check out this link on stored energy in a scuba tank...I haven't had time to check it, but it is a good reminder of how to work through the problem.

http://www.combro.co.uk/nigelh/diver/tank.html

 

[JamesBarlow]

reichertc;

Exactly what I was looking for.

Thanks for the help.

 

[butelja]

For your reference, 1 lb TNT is roughly equivalent to 2,000 BTU's (1.56*10E6 ft*lbf) of energy.

 

[reichertc]

Hi there. I managed to dig up one of the formulas that we use for pneumatic testing.

E=P1*V*(ln(P1/P2))

Where E=Stored Energy in kilojoules
P1=Pressure of tank or piping in kPa (absolute)
V=volume of system in cubic meters
P2=Atmospheric pressure in kPa (absolute)

We then divide the stored energy by 5788 kJ which is the amount of energy in one pound of TNT and evacuate the area accordingly.

 

[JamesBarlow]

My next question is how do you figure the evacuation area based on your pound of TNT.

I don't have any experience with explosives so I am still have trouble conceptualizing what these equations mean.

 

[butelja]

That is a very complex thing to analyze. The blast scaling laws are presented in detail in the U.S. Army Technical Manual TM5-1300 Structures to Resist the Effects of Accidental Explosions. This is (or was?) a freely distributed manual pre-9/11, but I'm not sure anymore. There are other open publications relating to protective design for chemical plants and refineries that also cover this area. The Shock & Vibration Handbook by Harris & Crede has some information of blast effects on humans. I'm not about to present the correlation between explosive quantity and blast overpressure over the internet.

Some rough damage information with regard to overpressure is as follows:

0.2 PSI - Breaks Windows
0.5-1.0 PSI - Shatters windows with body penetrating velocity.
1.0-2.0 PSI - Destroys typical wood frame structure
2.0-3.0 PSI - Blows in brick facing of steel frame building

5 PSI - Eardrum rupture
8 PSI - Fatal head injury (due to being propelled into stationary objects)
10 PSI - Serious lung damage
11-15 PSI - Fatal bodily injury (lung rupture or internal organ displacement)

 

[reichertc]

The evacuation areas that we used in conjunction with the formula I suggested are as follows:

<=13lb TNT - 100ft
>13lb,<=26lb - 200ft
>26lb,<=32lb - 300ft

Above 32lb TNT I think that we would probably not allow a pnuematic test.

I don't know what the basis for these numbers are, or any of the safety factors involved, so if you plan on using these for anything to do with safety, I would highly recommend you do some further research.

 

[vanstoja]

There's a one page article in the Sept. 6. 1973 issue of Machine Design by Lloyd M. Polentz titled &quot;The Peril in Pressurized Liquids&quot; which gives equations for fluid energy and pressure vessel strain energy. They are
E_L=P^2*V/2*B and E_S=P*V*s_H*(1.25-nu)/E where

P=pressure, psi v=volume, in^3 B=Bulk Modulus, psi
s_H=hoop stress, psi E=Modulus of Elasticity, psi,
nu=Poisson's ratio.

There's a worked example for 10,000 psi of 10 ft^3 fluid with a Bulk Modulus of 300,000. s_H is 80,000 and nu is 0.27. Liquid energy is about 240,000 ft-lb and strain energy is 37,700 for a total of 277,700 ft-lb. vanstoja

 

[kerntek]

now all you need is the blast over pressure as a function of range and energy of explosion.

there was a neat little d-o-d disc calculator used by civil defence back in cold war days for the calculation of blast pressures and casualty rates from nuclear weapons explosions. The scale was in tons of tnt, but the scaling laws had in fact been derived from measurements of small amounts of explosive.

If you are really keen then let me know and I will try and dig out the details of this calculator for you.

 

[reichertc]

Just found some more good information on this stuff. Check out these websites:

http://www.albertaboilers.com/newsletter/v07i2/v7-iss2-NET.pdf

The referenced web pages are:

http://www-training.llnl.gov/wbt/hc/5060/LiquidGas.html

http://microgravity.grc.nasa.gov/drop2/pdf/restricted_dist.pdf

Check them out.

 

[butelja]

REICHERTC,

I see that your formula for stored energy assumes isothermal expansion. As the potential expansion is very rapid (explosive), wouldn't the isentropic expansion formula be more realistic, and conservative? I can't see how the gas expansion could be isothermal.

 

[reichertc]

butelja,

Good point. The first equation is from an owners standard and the second one is referenced by Alberta's regulatory authority. In other words cheaper vs. safer. Having had no experience with failed pneumatic tests (fortunately), I cannot say which equation would be more accurate. I tend to agree with your thoughts on the subject. I would think that the layout of the system would have a large influence on what actually occurs during a failure.

For example, two identical systems consist of an air receiver and 600 feet of 2&quot; pipe. The first system while being tested fails at the air receiver. The second system fails at the most remote section of pipe. My gut feeling is that the first situation would present a higher hazard than the second. Perhaps this is the kind of situation anticipated by the owner who developed the first formula.

I did run a couple of test cases between the two methods of calculation. Of real interest is not really the stored energy, but rather the minimum safe distance that each calculation technique requires. I have found situations where the first calc is &quot;safer&quot;, and vice versa. (Hardly surprising considering the owners calc is a step function!) I suspect there are conditions where the second calc is substantially &quot;safer&quot;.

Of course, for this owner, we will be using the calc that prescribes the largest minimum distance for any pneumatic testing we do.

 

[reichertc]

butelja,

Once again this issue has reared it's ugly head. I am putting together a procedure for calculating minimum safe distance from pnuematic tests, and since it's going to have my name on it, I'd like as much backup as possible for it. I managed to find some information on blast pressures and fragmentation damage, and would like to get my hands on the technical manual you mentioned to confirm my numbers.

Any idea where I can find a copy of that??

Thanks.

 

[deanc]

Thanks to all for this info.! Very hard to find.

deanc

 

[abbver98]

Vessel stores two kind of energy :
1- fluid energy (generally dominant)
It is always difficult to calculate the energy actually used at rupture since we don't know the fluid relaxation condition. When explosion happens, the fluid relaxation for perfect gas is as :
W= (P1V1 / (g-1)) ( (P2/P1)^(g-1/g) -1)
P1/2 initial/final pressure, V volume, g : gas constant
We can determine the energy variation for steam or superheated water when relaxing to atmospheric pressure by using the entropy diagram.
Ex : for vessel with 1 m^3 , diameter of 1m, under 10 bar, thk 5 mm, the energy is :
-compressed air : 2 300 000 J
-saturated steam : 2 000 000 J

2-wall deformation (spring effect)
low comparing to fluid energy but not negligible for big vessels under high pressure.
For simple cylindrical (for which we neglect the head effect = only uniform traction) the energy is :
W= (V/2)(P^2/e) (D/E)
v volume p pressure e thk D O.D. (or mean) E elasticity modulus
Ex : for vessel with 1 m^3 , diameter of 1m, under 10 bar, thk 5 mm, the energy is 500J.