PE Problem #92

10/11/2019

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 10/14/19

7 Comments

Jacob

10/11/2019 10:44:37 am

C)?

MDS

10/11/2019 11:01:20 am

C. ~100

Assuming a ground level explosion.

Overpressure/Scaled distance from FPH Table 2.8.1

Used with Figure 2.8.4 or
Used with the energy scaled distance equation on FPH 2-95

Alex

10/22/2019 10:09:03 pm

MDS I believe you are correct. Some how the solution changed from Oil Storage Tank(20 ft/kg^1/3) to Concrete Block Failure(28 ft/kg^1/3). (20 ft/kg^1/3) (2*82 kg)^(1/3)=109 ft

Richard Koenig

10/11/2019 11:23:12 am

b.80 feet -
1. convert 180 lb to kg = 180*0.4536= 81.65 kg
2. from Table 2.8.1 - see "Oil storage tank Ruptured. - 20-16 ft/kg
3. 20 ft/(kg^1/3) x (81.65 kg)^(1/3) = 86.7 feet,

note: raise only the kg to the third power.

See FPH 2-96 Table 2.8.1 and example problems above.

Shawn B.

10/11/2019 06:07:11 pm

b. 80 feet

Based on FPH Table 2.8.1:
Avg. ~18 ft/kg^1/3 for Oil storage tank rupture

Also assuming ground level explosions

10/11/2019 11:55:54 pm

b. 80 FT

Amr

10/12/2019 05:59:53 pm

b. 80 ft - FPH Table 2.8.1

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