CHEMICAL ENGINEERING DESIGN & SAFETY

Transcription

1 CHEMICAL ENGINEERING DESIGN & SAFETY CHE 4253 Prof. Miguel Bagajewicz Explosions Explained

2 EXPLOSION A rapid and uniform expansion Expanding Shock wave Initial System Boundary

3 EXPLOSION The expansion ends when the Pressure of the Final System is P=1atm. Expanding Shock wave Initial System Boundary

4 EXPLOSION The limit for the expansion velocity is the speed of sound, achieved only when the initial pressure at detonation is sufficiently high. Expanding Shock wave Initial System Boundary

5 EXPLOSION THERMODYNAMICS Expanding Shock wave Initial System Boundary

6 EXPLOSION The limit for the expansion velocity is the speed of sound, achieved only when the initial pressure at detonation is sufficiently high. Expanding Shock wave Initial System Boundary

7 EXPLOSION THERMODYNAMICS First Law

8 EXPLOSION THERMODYNAMICS Second Law

9 EXPLOSION THERMODYNAMICS Closed System: M final =M initial Expanding Shock wave Initial System Boundary

10 EXPLOSION THERMODYNAMICS First Law Second Law However, the system is adiabatic!!! Q=0

11 EXPLOSION THERMODYNAMICS First Law Closed and Adiabatic system Second Law Explosion is uniform, so the only entropy generation takes place at the boundary, which can be neglected!!!

12 EXPLOSION THERMODYNAMICS First Law Closed and Adiabatic system Second Law 0 The speed of the system as well as its potential energy does not change, Then 0

13 EXPLOSION THERMODYNAMICS First Law Closed and Adiabatic system Second Law 0 We now integrate from t i =0 to t f

14 EXPLOSION THERMODYNAMICS First Law Closed and Adiabatic system Second Law 0 t ˆ(, ) ˆ f M U Pf Tf U( Pi, Ti ) = W = PdV t= 0 Sˆ( P, T ) Sˆ( P, T ) = 0 f f i i

15 EXPLOSION THERMODYNAMICS First Law SUMMARIZING M Uˆ( P, ) ˆ f Patm Tf U( Pi, Ti ) = = W Second Law Sˆ( P = P, T ) Sˆ( P, T ) = 0 f atm f i i

16 EXPLOSION THERMODYNAMICS OBJECTIVE: OBTAIN W!!! This means that one has to compute W = M Uˆ( P, ) ˆ f = Patm Tf U ( Pi, Ti ) = = N U ( Pf = Patm, Tf ) U ( Pi, Ti ) If the initial pressure and temperature is know, we can compute Uˆ ( Pi, Ti ). Now to compute Uˆ ( Pf = Patm, Tf ), one needs the second Law Sˆ( P = P, T ) Sˆ( P, T ) = 0 f atm f i i

17 TANK EXPLOSION- No reaction A 0.1 m 3 tank contains steam at 600 o C and 1 MPA and it bursts. OBTAIN W!!! We need Uˆ ( Pi, Ti ) and M. At 1 Mpa and 600 o C (steam table) Vˆ( P = m 3 /kg. Then = 0.1 m 3 / m 3 /kg=0.249 kg. i, Ti ) Also Uˆ ( Pi, Ti ) = kj/kg. Now we use Sˆ( P ˆ f = Patm, Tf ) = S( Pi, Ti ). Sˆ( Pi, Ti ) =8.029 kj/(kg K). Now using this entropy value at P=1 atm, to find T =248 o f C Uˆ ( P = P, T ) = 2731 kj/kg f atm f M W = M Uˆ( P, ) ˆ f = Patm Tf U( Pi, Ti ) = -141 kj (31 g of TNT)

18 TANK EXPLOSION- No reaction OBJECTIVE: OBTAIN W!!! Gases To use ideal gases we recall that for Then can be obtained using P f =1 atm making the equality * R/ Cp zero. Thus Then T f Sˆ( P = P, T ) Sˆ( P, T ) = 0 f atm f i i ˆ ˆ * Tf Pf = Patm S( Pf = Patm, Tf ) S( Pi, Ti ) = Cp ln R ln Ti Pi T f P = P f atm = Ti Pi * W = N U ( Pf = Patm, Tf ) U ( Pi, Ti ) = NCv ( Tf Ti ) PV i S Where N =. RT i

19 TANK EXPLOSION- No reaction OBJECTIVE: OBTAIN W!!! Gases Then W = N U P = P T U P T = NCv T T * ( f atm, f ) ( i, i ) ( f i ) * R / Cp PV * Pf = P i S atm W = N U ( Pf Patm, Tf ) U ( Pi, Ti ) Cv Ti T = = i RTi Pi W PV R P P R / Cp i S * f atm = Cv = P i * 1

20 TANK EXPLOSION- No reaction A compressed air tank (V S = m 3 ) at 25 o C and 250 bar (high!!) explodes. OBTAIN W!!! Data Cp of air is 29.3 J/(mol K). For ideal gases Cp-Cv=R * R / Cp PV * Pf = Patm i S W = Cv 1 R P =8,853 kj i This is equivalent to ~2 kg of TNT. Energy of TNT=4600 kj/kg. This is a sizable explosion!!! It could destroy everything within 4 m, produce substantial structural damage within 7 m, minor damage within 35 m, breaking glasses within 80 m and ear drum rupture within 5 m. These numbers are approximate, of course,

21 TANK EXPLOSION- No reaction If the tank (15 kg) is accelerated with that energy, at 45 degrees, how far would it go? We assume all the energy goes to the tank and no friction. E=m v 2 /2 v 0 =sqrt(2e/m)= =sqrt(2* 8,853,000 J/15)=1,086 m/s (sound=343m/s; expect sonic boom!) v x =v 0 cos(45), v y = v 0 sin(45) gt ; d= v x 2t max v 0 sin(45) gt max=0 d=sin(45) cos(45) v 02 /g d= 120 km!!!!

22 TANK EXPLOSION- No reaction If the tank (15 kg) is accelerated with that energy, at 45 degrees, how far would it go? We assume all the energy goes to the tank and we consider friction. v 0 =1086 m/s; Drag force is f= C D ρa v 2 /2 Force = f + g=m a; a x =-k v 2 cos(θ)/m a y =-(k v 2 sin(θ)+g) Too complicated to solve analytically!!! But we can solve a few meters and compare to the case without friction.

23 TANK EXPLOSION- No reaction

24 TANK EXPLOSION- No reaction If the tank (15 kg) is accelerated with that energy, at 45 degrees, how far would it go? We assume all the energy goes to the tank and we consider friction. v 0 =1086 m/s; Drag force is f=c D ρa v 2 / meters instead of 120 Km.!!!!!!

25 TANK EXPLOSION- No reaction If the tank (15 kg) is accelerated with that energy, at 45 degrees, how far would it go? We assume all the energy goes to the tank and we consider hitting a concrete wall first. v 0 =1086 m/s; Momentum conservation mv 0 +Mv M = mv 0 =(m+m)v sys. v sys = v 0 m/(m+m) But v sys ~0. Actually M is the mass of the wall and the earth attached to it. Thus, the energy absorbed by the wall is 8,853 kj. To calculate the damage we need mechanical engineers.

26 TANK EXPLOSION- reactions The initial temperature and pressure are needed. Assume propane in air. The reaction is: C 3 H O N 2 3 CO 2 +4H 2 O N 2 Initially the # of moles is N b =24.8 and the temperature and pressure are T b P b. After the reaction: N i =25.8, Thus P i = P b N i T i /(N b T b )

27 TANK EXPLOSION- reactions The initial temperature and pressure are needed. C 3 H O N 2 3 CO 2 +4H 2 O N 2 We now need the temperature after the explosion, but before the expansion starts. First Law says: 0. Then and assuming ideal gases. Uˆ( P, T ) = Uˆ( P, T ) b b i i ( 3C )( ) V, CO2 4CV, H 2 O 18.83C V, N 2 Ti Tref + + = ( C )( ) V, C3H 8 5CV, O C V, N 2 Tb Tref = + +

28 TANK EXPLOSION- reactions The initial temperature and pressure are needed. C 3 H O N 2 3 CO 2 +4H 2 O N 2 ( 3C )( ) V, CO2 4CV, H 2 O 18.83C V, N 2 Ti Tref ( C )( ) V, C3H 8 5CV, O C V, N 2 Tb Tref and one can obtain T i + + = = + + What if there are solids or liquids or the ideal gas assumption cannot be made? Solids and Liquids: Tb Uˆ ( P ˆ ˆ b, Tb ) H ( Pb, T ) = H ( Pref, Tref ) + C p( T ) dt Tref Non ideal gases: Need Uˆ ( Pb, Tb ) and Uˆ ( P need to be bniti /( NbTb ), Ti ) obtained using the equation of state.