Part 23 (2/2)

[Illustration: FIG 54--MORTAR FOR MEASURING THE BALLISTIC POWER OF EXPLOSIVES _A_, Shot; _B_, Steel Disc; _C_, Section of Mortar (Cast Iron); _D_, Wooden Plug holding Explosive (_E_); _F_, Fuse]

Mr T Johnson made so 29 Ibs, and he adopted the plan of r results:--

Range in Feet

Blasting gelatine (90 per cent nitro-glycerine and nitro-cellulose) 392 Ammonite (60 per cent Anite (60 per cent nitro-gelatine and gun-cotton) 306 Roburite (AmNO_{3} and chloro-nitro-benzol) 294 No 1 dynaelatine) 264 Stonite (68 per cent nitro-gelatine and 32 per cent wood-un-cotton and nitrates) 223 Carbonite (25 per cent nitro-gelatine, 40 per cent wood-meal, and 30 per cent nitrates) 198 Securite (KNO_{3} and nitro-benzol) 183 Gunpowder 143

~Calculation of the Voluas evolved in an explosive reaction may be calculated, but only when they are si made at 0 and 760 mm Let it be required, for exaralycerine The explosive reaction of nitro-glycerine may be represented by the equation

C_{3}H_{5}O_{3}(NO_{2})_{3} = 3CO_{2} + 2-1/2H_{2}O + 1-1/2N_{2} + 1/4O_{2} By weight 227 = 132 + 45 + 42 + 8 By voluhts of the several products of the above reactions are calculated by ht of 1 litre of hydrogen at 0 C and 760 rrm

” H_{2}O = 9 x ” = 08064 ”

” N_{2} = 14 x ” = 12544 ”

” O_{2} = 16 x ” = 14336 ”

The voluases at 0 and 760 ramme as the unit of mass, is found to be 2232 litres Thus:--

Volume of 44 of CO_{2}, at 0 and 760 mm = 44/19712 = 2232 litres

18 ” H_{2}O ” ” = 18/08044 = 2232 ”

28 ” N_{2} ” ” = 28/12544 = 2232 ”

32 ” O_{2} ” ” = 32/14366 = 2232 ”

Therefore

132 grms of CO_{2} at 0 C and 760 mm = 2232 x 3 = 6696 litres

45 ” H_{2}O ” ” = 2232 x 2-1/2 = 5580 ”

42 ” N_{2} ” ” = 2232 x 1-1/2 = 3348 ”

8 ” O_{2} ” ” = 2232 x 1/4 = 558 ”

____________

16182 ”

Therefore 1 gralycerine when exploded, produces 16182 litres of gas at 0 C and 760 as at the temperature of explosion, we simply apply the law of Charles[A] Thus--

V : V' :: T : T' or V' = VT'/T

in which V represents the original voluinal temperature on the absolute scale

T' ” new temperature of the same scale In the present case T' = 6001

Therefore substituting, we have