PE Problem #39

7/25/2018

An oil production facility is concerned about spacing of hazards prone to detonation. What would be the distance from a 181 lb TNT equivalent vessel explosion that failure of concrete block walls would occur?
a. 154 ft.
b. 200 ft.
c. 244 ft.
d. 317 ft.

Solution | Posted 07/26/18

5 Comments

FPE

7/25/2018 09:05:12 am

b. 200 ft.
199.55 ft Ref NFPA FPH 2-96

Colleen

7/25/2018 10:18:36 am

I got A. 154 ft

Per T.2.8.1 in FPH p. 2-96, the standoff distance associated with the consequence of failure of concrete walls is 28-20 ft/kg^(1/3). When I used the greatest distance of 28*(181)^(1/3), I get about 158 ft, which is closest to answer A.

Jimmy

7/25/2018 11:53:56 am

Damn, I learned something: we are given lbs of TNT, not kg. Also, note the example in the FPH says double the blast wave energy when you use Table 2.8.1. So, if by chance you forget to convert to kg, AND forget to multiply by two, you get 158, which happens to be really close to 153.3 (if you do it correctly on your EIGHTH try, um, like I finally realized) which is most nearly 154. What are the odds the real exam will be so forgiving? Pay attention to simple errors that you tend to repeat--I blow units REGULARLY.

Colleen

7/25/2018 12:01:27 pm

Better to mess up now than on the real exam!

One of the AHJs I work with said when he took the exam there were a TON of questions with mismatched units; lesson learned is to definitely have conversions on hand quickly. I'm not positive, but I think there might be something in the back of the SFPE Handbook. Hopefully our fate won't be the same!!

Pari

7/26/2018 12:06:55 am

200

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