Fragmentation and Safety Distances

Transcription

1 Fragmentation and Safety Distances Examples MNGN 444 Spring 2016

2 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. 2

3 Gurney Constants 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 3

4 Mott Distribution Factor Mott Coefficient for Mild Steel Cylinders High Explosive Mott Coefficient TNT Composition B Pentolite RDX Source: Explosives Engineering Cooper 4

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6 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

7 Solution 1. Mass of Steel: GURNEY M = Vol ρ steel M = π 4 Do2 Di 2 L ρ steel M = π ,850 = 2.93 kg 2. Mass of Explosives: M = Vol ρ CompB M = π 4 Di2 L ρ CompB M = π ,650 = 2.62 kg 3. Fragments initial velocity: V 0 = 2E M C /2 V 0 = 2, = 2, m/s 7

8 Solution MOTT m = M k ln M 0 2M k 2 4. Mott distribution factor: M k = B t 5 3 d t d M k = = = lb 1/2 5. Mass of heaviest fragment: m = M k ln M 0 2M k 6. Fragments cross sectional area: 2 m = ln = lb (0.66 g) d = 3 m d = = 0.2 in (5 mm) A = π d2 4 A = π 52 = 20 mm 2 4 8

9 Solution DRAG 7. Impact Velocity at 50 meters V s = 2, e = 750 m/s NOTE: Hypervelocity impact would take place within the 5 meters range. 9

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11 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 (70 kg) 11

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. 12

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: Z = R WT a P a 1/3 R = Z WT a P a 1 3 R = Comp B in TNT m Surface Burst Double Yield 13

14 Solution: Secondary Blast Damage 1. From the previous example, we predicted a maximum fragment of 1.47e-3 lb: Maximum Impact Velocity = 260 ft/s (80 m/s) 2. Considering the drag attenuation, the minimum distance at which a worst case primary fragment stops being hazardous is: V s = V 0 e A m γ 0 C D R R = m ln V 0 R = , ln Aγ 0 C D V s = 153 m 14

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: Positive Impulse i s t d P s 2 = = 3.2 psi sec Positive Scaled Impulse i s = i s 1/3 M = 3.2 body 165 1/3 = 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. 15

16 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 i s t d P s 2 = = 2.63 psi sec Positive Scaled Impulse i s = i s 1/3 M = 3.2 body 165 1/3 = 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. 16

17 Solution: The following table shows the minimum safety distance for each of the three different blast injuries: Blast Damage Type Minimum Safety Distance Primary Overpressure 17 m Secondary Fragments 153 m Tertiary Decelerating Impact 19 m Taking the most restrictive value, the minimum safety distance would be: 153 m 17