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 – USDOE.

3. Terminal Ballistics - Rosenberg, Deker.

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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 – 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 – Cooper

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Calculate the impact velocity at 50 meters of a “worst case” fragment generated by a cylindrical Composition B charge encased in an steel pipe. Assume standard fragment shape and ambient air at 20˚C. L = 250 mm

Din = 90 mm Dout = 100 mm ρsteel = 7.85g/cc ρCompB = 1.65 g/cc

6

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Solution 1. Mass of Steel:

𝑀 = 𝑉𝑜𝑙 ∙ 𝜌𝑠𝑡𝑒𝑒𝑙 → 𝑀 =𝜋

4𝐷𝑜2 − 𝐷𝑖2 𝐿 ∙ 𝜌𝑠𝑡𝑒𝑒𝑙

𝑀 =𝜋

40.1002 − 0.0902 ∙ 0.250 ∙ 7,850 = 2.93 𝑘𝑔

2. Mass of Explosives:

𝑀 = 𝑉𝑜𝑙 ∙ 𝜌𝐶𝑜𝑚𝑝𝐵 → 𝑀 =𝜋

4𝐷𝑖2 𝐿 ∙ 𝜌𝐶𝑜𝑚𝑝𝐵

𝑀 =𝜋

40.0902 ∙ 0.250 ∙ 1,650 = 2.62 𝑘𝑔

3. Fragments initial velocity:

𝑉0 = 2𝐸𝑀

𝐶+1

2

−1/2

→ 𝑽𝟎 = 2,843 ∙2.93

2.62+1

2

−12

= 𝟐, 𝟐𝟑𝟒. 𝟖𝟑 𝒎/𝒔

GURNEY

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Solution

4. Mott distribution factor:

𝑀𝑘 = 𝐵 ∙ 𝑡53 ∙ 𝑑

13 ∙ 1 +

𝑡

𝑑→ 𝑀𝑘 = 0.0554 ∙ 0.2

53 ∙ 3.55

13 ∙ 1 +

0.2

3.55=

= 6.11 ∙ 10−3 𝑙𝑏1/2

5. Mass of heaviest fragment:

𝑚 = 𝑀𝑘 ln𝑀0

2𝑀𝑘

2→ 𝒎 = 6.11 ∙ 10−3 ln

6.46

2∙6.11∙10−3

2= 𝟏. 𝟒𝟕 ∙ 𝟏𝟎−𝟑 𝒍𝒃

6. Fragments cross sectional area:

𝑑 =𝑚

0.186

3→ 𝑑 =

1.47∙10−3

0.186

3= 0.2 𝑖𝑛

MOTT 𝑚 = 𝑀𝑘 ln𝑀0

2𝑀𝑘

2

(0.66 g)

(5 mm)

𝑨 =𝜋 ∙ 𝑑2

4→ 𝐴 =

𝜋 ∙ 52

4= 𝟐𝟎 𝒎𝒎𝟐

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Solution

7. Impact Velocity at 50 meters

𝑽𝒔 = 2,234.86 ∙ 𝑒−

20∙10−6

0.66∙10−3∙1.2∙0.6∙50

= 𝟕𝟓𝟎 𝒎/𝒔

NOTE: Hypervelocity impact would take place within the 5 meters range.

DRAG

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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 20˚C. L = 250 mm

Din = 90 mm Dout = 100 mm ρsteel = 7.85g/cc ρCompB = 1.65 g/cc

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(70 kg)

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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.

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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:

𝒁 =𝑹

𝑾𝑻𝒂𝑷𝒂

𝟏/𝟑→ 𝑅 = 𝑍 ∙

𝑊𝑇𝑎𝑃𝑎

13

→ 𝑹 = 1.3 ∙2.62 ∙ 1.33 ∙ 2 ∙ 293

1.013

13

≈ 𝟏𝟕 𝒎

Comp B in TNT Surface Burst Double Yield

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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 “worst case” primary fragment stops being hazardous is:

Maximum Impact Velocity = 260 ft/s

𝑽𝒔 = 𝑽𝟎 ∙ 𝒆−

𝑨

𝒎∙𝜸𝟎∙𝑪𝑫∙𝑹 → 𝑅 =

𝑚

𝐴𝛾0𝐶𝐷∙ ln

𝑉0

𝑉𝑠→ 𝑹 =

0.66∙10−3

20∙10−6∙1.2∙0.6 ln

2,234.86

80= 𝟏𝟓𝟑 𝒎

(80 m/s)

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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:

𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 → 𝑖𝑠 ≈𝑡𝑑 ∙ 𝑃𝑠2

=3.2 ∙ 2

2= 3.2 𝑝𝑠𝑖 ∙ 𝑠𝑒𝑐

𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑆𝑐𝑎𝑙𝑒𝑑 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 → 𝑖𝑠 =𝑖𝑠

𝑀𝑏𝑜𝑑𝑦1/3

=3.2

1651/3= 0.58 𝑝𝑠𝑖 ∙ 𝑠𝑒𝑐/𝑙𝑏1/3

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.

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Solution: Tertiary Blast Damage 2. Starting with a new limit for the incident overpressure of 1.5 psi (“guessing”),

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

𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 → 𝑖𝑠 ≈𝑡𝑑 ∙ 𝑃𝑠2

=3.5 ∙ 1.5

2= 2.63 𝑝𝑠𝑖 ∙ 𝑠𝑒𝑐

𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑆𝑐𝑎𝑙𝑒𝑑 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 → 𝑖𝑠 =𝑖𝑠

𝑀𝑏𝑜𝑑𝑦1/3

=3.2

1651/3= 0.48 𝑝𝑠𝑖 ∙ 𝑠𝑒𝑐/𝑙𝑏1/3

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.

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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