PE Problem #51

8/29/2017

An oil field site has a potential for detonation of an equivalent of 180 lb of TNT at its main plant. Determine most nearly the minimum distance where an adjacent oil storage tank would rupture at this energy.
a. 50 ft.
b. 80 ft.
c. 100 ft.
d. 150 ft.

Solution | Posted 08/30/17

8 Comments

8/29/2017 09:45:17 am

I think the relevant section is FPH 2-95 and 2-96... but I'm having a hard time making sense of the explanations and the units are goofy... Scaled distance of ft/kg^1/3 in Table 2.8.1...? Mixing I-P and SI units?

These little things make me worried for the exam because I can get hung up for a long time trying to make sense of errors in the references.

8/29/2017 09:48:57 am

Assuming there is an error, that is. I may not be able to grasp the explanation as well.

Joe Meyer link

8/29/2017 10:09:11 am

Thank you CJ, you're my hero.

I specifically included this problem in the daily problems and not in the book or the weekly exams, as the handbook does not clearly identify how to work the problem and my hope was to see if others felt exactly as I had. You nailed it.

I'm assuming you're seeing that the book discusses a multiplier of "2" for low pressures as well as combined SI and IP with the vague example problem, correct?

The reality of the actual exam is that if a problem is found to have an error, or be based on text that has errors, that problem is then thrown out and not scored on the exam. The exam should not be dependent on areas within the volumes that have errors, and the post-exam evaluation process is supposed to clean those problems up.

Jay

8/29/2017 10:04:11 am

Jay

8/29/2017 10:27:46 am

This is how I understand...

180 lbs of TNT = 82.134 kg.

Table 2.8.1 shows:
20-16 Oil Tank Storage ruptured

Scaled distance of 20 feet per every cube root of one kilogram amount of energy required.

82.134^0.333 = 4.34

4.34 x 20 = 86.6 feet distance maximum.

Does it make sense? Any comments?

Randy

8/29/2017 11:09:21 am

This is how I would do it.

The text is not clear about when to apply the 2x factor for reflected blast waves.

"The ratio of reflected blast wave pressure to incident pressure is approximately 2 for values of P_s0 less than about 2 psig."

What if you have a reflection and P_s0 is greater than 2, as with this case? Is your ratio still two? Does it not apply to greater overpressures?

Ali Khan

8/31/2017 06:21:51 am

I agree with Meyer

Chris D

8/29/2017 01:28:12 pm

B

I used FPH 2-96 table 2.8.1 and calculated a Z in ft/(kg^1/3) for 50' and 80' (180/2.2 to get kg) and got 11.51 and 18.42 respectively. The tank rupture is given as between 16-20(why do they show as 20-16?) which puts me close to the 80' mark.

I'll admit, I went all over looking into SFPE Table 71.2 and trying the R=(Po/E)^1/3 formula in different configurations.

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PE Problem #51

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