Airblast from Underwater Explosions - dm2_080077

Custom Search

Solution:
(1) For (Pso)max = 15 psi, W/V = 0.017 from Figure 14.
Therefore, V = 125/0.017 = 7,350 ft3
(2) For a cube, L = V1/3 = (7,350)1/3 = 19.4 ft
Therefore, length, height, and width of cubicle should be
19.4 feet
(3) For a given distance, R, of 200 feet, the scaled distance
R/W1/3 = 2001(125)1/3 = 40 ft/1b1/3
(4) From Figure 13, Pso = 1.0 psi behind the back wall,
1.5 psi behind the sidewalls and 1.8 psi out the open
front.  (Pso)max = 15 psi, which is greater than all
the other pressures listed above.  Therefore, the
pressures are correct.
c. Airblast from Underwater Explosions.
Problem:
Determine the overpressure at a position 10 feet above the
surface and 60 feet from Surface Zero (y = 10 and R = 60)
produced by 1,000 pounds of TNT detonated 25 feet below the
surface.
Given:
(1) Type and weight of explosive.
(2) Location of explosive.
Solution:
(1) Calculate the scaled charge depth, [lambda]d.
(2) Calculate the scaled horizontal distance, [lambda]x.
(3) Calculate the scaled vertical distance, [lambda]y.
(4) From the appropriate figure, read off the overpressure.
Include accuracy of +/- 30 percent.
Calculation:
Given:
(1) Charge weight, W = 1,000 lb of TNT.
(2) d = 25 ft, R = 60 ft, and y = 10 ft
Solution:
(1) [lambda]d = d/W1/3 = 25/1,0001/3 = 2.5 ft/lb1/3
(2) [lambda]x = R/W1/3 = 60/1,0001/3 = 6 ft/lb1/3
2.08-61