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Fragmentation and Safety Distances
Examples
MNGN 444 Spring 2016
Recommended Literature
1. Explosives Engineering - Paul W. Cooper.
2. Manual for the Prediction of Blast and Fragment Loadings in
Structures โ USDOE.
3. Terminal Ballistics - Rosenberg, Deker.
2
Gurney Constants
3
High Explosive Gurney Constant (m/s)
TNT 2,315
ANFO 2,769
Composition B 2,843
Pentolite 2,970
Composition C4 2,801
RDX 3,205
PETN 3,425
HMX 3,198
Source: Explosives Engineering โ Cooper
Mott Distribution Factor
4
High Explosive Mott Coefficient
TNT 0.0779
Composition B 0.0554
Pentolite 0.0808
RDX 0.0531
Mott Coefficient for Mild Steel Cylinders
Source: Explosives Engineering โ Cooper
5
Calculate the impact velocity at 50 meters of a โworst caseโ fragment generated by a cylindrical Composition B charge encased in an steel pipe. Assume standard fragment shape and ambient air at 20หC. L = 250 mm
Din = 90 mm Dout = 100 mm ฯsteel = 7.85g/cc ฯCompB = 1.65 g/cc
6
7
Solution 1. Mass of Steel:
๐ = ๐๐๐ โ ๐๐ ๐ก๐๐๐ โ ๐ =๐
4๐ท๐2 โ ๐ท๐2 ๐ฟ โ ๐๐ ๐ก๐๐๐
๐ =๐
40.1002 โ 0.0902 โ 0.250 โ 7,850 = 2.93 ๐๐
2. Mass of Explosives:
๐ = ๐๐๐ โ ๐๐ถ๐๐๐๐ต โ ๐ =๐
4๐ท๐2 ๐ฟ โ ๐๐ถ๐๐๐๐ต
๐ =๐
40.0902 โ 0.250 โ 1,650 = 2.62 ๐๐
3. Fragments initial velocity:
๐0 = 2๐ธ๐
๐ถ+1
2
โ1/2
โ ๐ฝ๐ = 2,843 โ2.93
2.62+1
2
โ12
= ๐, ๐๐๐. ๐๐ ๐/๐
GURNEY
8
Solution
4. Mott distribution factor:
๐๐ = ๐ต โ ๐ก53 โ ๐
13 โ 1 +
๐ก
๐โ ๐๐ = 0.0554 โ 0.2
53 โ 3.55
13 โ 1 +
0.2
3.55=
= 6.11 โ 10โ3 ๐๐1/2
5. Mass of heaviest fragment:
๐ = ๐๐ ln๐0
2๐๐
2โ ๐ = 6.11 โ 10โ3 ln
6.46
2โ6.11โ10โ3
2= ๐. ๐๐ โ ๐๐โ๐ ๐๐
6. Fragments cross sectional area:
๐ =๐
0.186
3โ ๐ =
1.47โ10โ3
0.186
3= 0.2 ๐๐
MOTT ๐ = ๐๐ ln๐0
2๐๐
2
(0.66 g)
(5 mm)
๐จ =๐ โ ๐2
4โ ๐ด =
๐ โ 52
4= ๐๐ ๐๐๐
9
Solution
7. Impact Velocity at 50 meters
๐ฝ๐ = 2,234.86 โ ๐โ
20โ10โ6
0.66โ10โ3โ1.2โ0.6โ50
= ๐๐๐ ๐/๐
NOTE: Hypervelocity impact would take place within the 5 meters range.
DRAG
10
You have been tasked to execute the disposal of a cased composition B charge such as the one before. Because you must avoid any risk to yourself and the possible public in the area, you have to estimate a minimum safety perimeter against the hazards that this operation implies. Calculate the minimum safety perimeter for air blast and fragmentation produced by a cylindrical Composition B charge encased in an steel pipe and resting on the ground. Assume standard fragments shape, elevation under 1,500 meters, and ambient air at 20หC. L = 250 mm
Din = 90 mm Dout = 100 mm ฯsteel = 7.85g/cc ฯCompB = 1.65 g/cc
11
(70 kg)
12
Solution: Primary Blast Damage 1. Assuming conservative impulse values, we take the following threshold values
for lung and ear damage:
Lung Damage = 10 psi Ear Damage = 2 psi
We also assume that the people that must be outside the safety perimeter is not wearing any ear protection so the limit peak overpressure will be 2 psi.
13
Solution: Primary Blast Damage 3. Determine the scaled distance using the blast overpressure graph:
4. Minimum Safety Distance for ear damage due to blast overpressure:
๐ =๐น
๐พ๐ป๐๐ท๐
๐/๐โ ๐ = ๐ โ
๐๐๐๐๐
13
โ ๐น = 1.3 โ2.62 โ 1.33 โ 2 โ 293
1.013
13
โ ๐๐ ๐
Comp B in TNT Surface Burst Double Yield
14
Solution: Secondary Blast Damage 1. From the previous example, we predicted a maximum fragment of 1.47e-3 lb:
2. Considering the drag attenuation, the minimum distance at which a โworst caseโ primary fragment stops being hazardous is:
Maximum Impact Velocity = 260 ft/s
๐ฝ๐ = ๐ฝ๐ โ ๐โ
๐จ
๐โ๐ธ๐โ๐ช๐ซโ๐น โ ๐ =
๐
๐ด๐พ0๐ถ๐ทโ ln
๐0
๐๐ โ ๐น =
0.66โ10โ3
20โ10โ6โ1.2โ0.6 ln
2,234.86
80= ๐๐๐ ๐
(80 m/s)
15
Solution: Tertiary Blast Damage 1. Using the values of overpressure and positive phase duration predicted for the
ear damage, we proceed to check if there is any risk of tertiary blast damage at 17 meters from the blast:
๐๐๐ ๐๐ก๐๐ฃ๐ ๐ผ๐๐๐ข๐๐ ๐ โ ๐๐ โ๐ก๐ โ ๐๐ 2
=3.2 โ 2
2= 3.2 ๐๐ ๐ โ ๐ ๐๐
๐๐๐ ๐๐ก๐๐ฃ๐ ๐๐๐๐๐๐ ๐ผ๐๐๐ข๐๐ ๐ โ ๐๐ =๐๐
๐๐๐๐๐ฆ1/3
=3.2
1651/3= 0.58 ๐๐ ๐ โ ๐ ๐๐/๐๐1/3
At a distance of 17 m from the blast, an impact at more than 10 fts is expected to occur for a human body of 165 lb. This value is higher than the general tolerance limit for the human body. For this reason, we must find by iteration the safety range for the tertiary blast damage.
16
Solution: Tertiary Blast Damage 2. Starting with a new limit for the incident overpressure of 1.5 psi (โguessingโ),
we obtain an scaled distance Z = 1.5. With this new scaled distance, the expected positive phase duration will be td = 3.5 ms. Next, we repeat the previous calculations
๐๐๐ ๐๐ก๐๐ฃ๐ ๐ผ๐๐๐ข๐๐ ๐ โ ๐๐ โ๐ก๐ โ ๐๐ 2
=3.5 โ 1.5
2= 2.63 ๐๐ ๐ โ ๐ ๐๐
๐๐๐ ๐๐ก๐๐ฃ๐ ๐๐๐๐๐๐ ๐ผ๐๐๐ข๐๐ ๐ โ ๐๐ =๐๐
๐๐๐๐๐ฆ1/3
=3.2
1651/3= 0.48 ๐๐ ๐ โ ๐ ๐๐/๐๐1/3
Finally, using the scaled distance formula and Z = 1.5, we can obtain the range at which an limit impact velocity of 10 fts is expected to occur for a human body of 165 lb. The value obtained is approximately 19 meters.
17
Solution: The following table shows the minimum safety distance for each of the three different blast injuries:
Taking the most restrictive value, the minimum safety distance would be:
Blast Damage Type Minimum Safety Distance
Primary Overpressure 17 m
Secondary Fragments 153 m
Tertiary Decelerating Impact 19 m
153 m