 Good morning. I think let us first recap what we did in the last three classes. We defined an explosion and we said that an explosion is one in which a blast wave gets generated due to some energy deposition at a source. We learned how to get the magnitude or the velocity at which the blast wave propagates due to energy deposition at the source using dimensionless analysis. Thereafter, we were interested in finding out what is the magnitude of the pressure rise across the lead blast wave and also about the wind or the impulse or the change of momentum which is associated due to the movement of the matter which is processed by the blast wave. For that, we wanted to take some examples such that we are very clear about the effects of the explosion. But before we did that part on different types of explosions, we also told ourselves when some energy is released say E0 joules instantaneously at some particular source, it is the blast wave which redistributes the energy and the energy is contained within the lead shock wave and the source. Let us say that the lead shock wave is spherical, we said it could be planar, it could be cylindrical. The energy distributed or deposited at the source is redistributed by the blast wave within the lead blast wave itself and the blast wave keeps propagating outward. In the last class, what we did was we looked at the different types of explosions such that we could relate it and before we go any further on predicting the over pressure or the pressure rise, the impulses behind it. It is necessary to have an idea of what the different types of explosions were and we were able to catalogue the different explosions into something like eight categories. Before these eight categories, we talked in terms of naturally occurring explosions, we talked in terms of accidental explosions. We also talked in terms of intentional explosions and also we know explosions used for constructive purposes and different applications. But for accidental explosions and also some of the intentional explosions, we were able to catalogue the explosions into eight different categories. These eight categories were, maybe let me start with one. We started with something like a condensed phase explosion. By condensed, we mean heavier substances, liquid and solids. We looked at some of the examples which released energy and caused explosions using condensed phase. The examples where the Texas city disaster in which lot of just exploded, we looked at different examples. We looked at explosions occurring at due to cracker industry handling energetic materials and so on. These we found was essentially due to the rapid chemical reactions. We called it as maybe of thermal origin. Maybe it could be of different origins. Maybe when we get into some details, we will examine this. The second one what we said was confined fuel air explosions or confined gaseous explosions. Let me say fuel air explosions. And what happened in the confined fuel? We had some chamber room such as a kitchen or we had an aircraft tank in which we got a combustible mixture. It ignited and just went as a bang and destroyed the place. It was essentially confined. That means in a particular confinement, you have energy release taking place. It could be anything. In the industry like in a workshop, you have a compressor which energizes or which pressurizes a gas bottle. And if by chance the gas bottle bursts, well I have an explosion. I could have the simple example of a balloon which is bursting, which is a confined explosion involving cold gas, cold gas. It could be fuel and air which causes the problem. Third, we said it could also be unconfined. In case of unconfined we looked at spillage taking place or gas getting spilled such as the one we saw at the in Siberia in the Ural mountains. Wherein there was this spillage. The spillage goes along the ground. Two trains pass in the opposite direction. They mix it with air. You have the spark and it explodes. These are unconfined. The fourth one what we considered was believe, BLEVE boiling liquid expanding vapor explosions. What happens in this case? I think I forgot to put this while summarizing the talk. I think I omitted this. But what happened in this case was, well we have something like a particular tank which contains some liquid fuel or some liquid. It could be volatile or could be heavy itself. Because of some fire in some other compartment or other wagons or other places, I have the flame which gets directed over here. The hot flame gets directed over here. The surface gets heated up and when it gets heated up it boils the liquid over here, generates a lot of vapor. The casing also gets weakened because it gets heated and then it bursts. When it bursts, well I have an explosion. I have a huge fireball and an explosion. We saw the example of the crescent city at Illinois. We also told ourselves that in petrochemical industries these do occur frequently such as offshore platforms. We said in Jaipur we had the Indian oil corporation having a fire in Vishakhapatnam. We had such tragedies taking place that is believe explosion. We talked in terms of physical explosions. Physical in the sense we are not really talking of an explosive medium but something like water which is suddenly put in a hot environment and it flashes into vapor and when it flashes into vapor you have a huge superheated steam which releases energy and you have an explosion. We took the example of naturally occurring explosion like at the Krakatov volcano wherein you had immense or very large explosion or when you had a pail of water at flint florn in Ottawa wherein you had such physical explosions. We also told ourselves well even the dust such as what the foodstuffs we eat may be powder it, may be corn flour, wheat flour, may be sugar powder could be explosives when mixed with there especially in a confined geometry. We talked in terms of the explosion taking place at Turin in Italy wherein an young boy he takes corn flour in one hand mixes it he has a candle on the other hand it just bursts into flame and an explosion. The seventh one was wherein we said well you could have the atmosphere player role in mixing the gases we also said atmospheric dispersion. This atmospheric dispersion could come either the fuel which spills over and it vaporizes it moves by the atmosphere it could have that or else may be even the toxic gases like we had at the Bhopal gas tragedy wherein you had people dying because in a cold day when there is temperature inversion you have the gases which cannot really escape into the atmosphere it causes a problem. We also talked of the great smog of London and the last one which was number 8 was nuclear explosions. In nuclear explosions we talked of loss of coolant analysis loss of coolant type of explosions we call loss of coolant type of accidents wherein may be because of the hot metal coming in contact with water which is used for cooling it generates hydrogen and this hydrogen mixing with air causes an explosion. We also said well fission and fusion are cases wherein intentionally these are exploded but we will not consider fission and fusion type of accidents in the course which we are doing and these were the type of eight different categories of explosions. This gave us a picture of what happens in an explosion. We also know that a blast wave is generated in an explosion and now our next aim is well can we sort of predict the pressure rise across a blast wave point one and also what is the impulse or the blast wind or the wind which is associated which causes momentum and disrupts the things from the scene of the explosion and this is what I want to do today. Therefore let us put the schematic again what we want to do in this particular talk today. In other words we told ourselves well I have on a streak diagram in which the y-axis is the time the x-axis is the distance. We said well and when I release energy spontaneously at a point a strong shock gets generated it keeps weakening it becomes an acoustic wave in the far field this is acoustic here it is a strong shock as it were and what was the condition in a shock there is an abrupt rising pressure well this is what happens a shock attenuating and we discuss this in terms of lightning if you are very near to the source of lightning what happens well very near to it I hear a very loud bang if I am far away I hear a rumble like an acoustic signature. Therefore see we are interested in finding out what is the over pressure generated at different points. Now to be able to do that we immediately tell ourselves well the problem is little complex because the lead shock I have an explosion taking place let us say some energy getting released here the front the wave front is continually decaying in other words if ever to plot the velocity which we did do as a function of distance we said r s dot goes as a constant into r s to the power minus 3 by 2 the velocity gets decaying the blast wave keeps decaying as the distance proceeds point one we also know that if I were to look at this figure a little more closely let us try to follow the path of the particles which are processed by the shock to do that I take let us say at this when the shock wave comes at this particular point maybe it picks up a particle here let us say that the identity of this particle is let us say at a distance a from the origin over here I call it as the distance x is equal to r s a well the particle enters the shock over here it rises in pressure then it expands out and therefore this is the type of trajectory which this particle entering at a would have in other words I am defining this particular particle therefore I am looking at the Lagrangian point of view namely following the particle as it moves behind the shock wave it is ahead of the shock wave it is stationary well a particle which is away from it at B is processed by the wave when it has decayed out in strength therefore the over pressure here is low it also follows the particle it also follows after the shock it has a particular trajectory and so on a particle let us say this is at r s B I have another particle at C let us say r s distance r s C from the source over here it again goes behind therefore when I look at the properties of this particle a it is also influenced by the blast maybe at this particular time the blast is here it is influenced by the blast at this particular time the decay of the properties therefore the problem becomes very complex it is not only it is decaying but the properties of the medium enclosed by this are also governed by subsequent motion of the blast wave and therefore with the problem becomes highly what we said earlier was highly transient and not only transient terribly unsteady and terribly non-linear things are decaying things are affecting one is affecting the other highly non-linear and predictions become a little more difficult but we should be able to predict it we should be able to write the governing equations and predict it but before I get into some detailed predictions can we make the problem a little simpler understand the problem a little better and how do we do that instead of looking at the decaying shock wave can I sort of presume or look at the properties let us say on the streak diagram t versus x now I say well if the shock wave to move at constant velocity can I predict the properties behind it and therefore let us therefore start with this and then put the decaying wave a little later and then look at the decaying effect therefore let us do the problem for a wave traveling at constant velocity namely the shock wave traveling at constant speed predict the properties behind it and then extrapolate it to the case of a blast wave therefore in today's class I will specifically look at maybe the properties behind a shock wave moving at constant speed therefore let us take a look what is it I want to do let us be very clear about the problem let us say I have something like this a pipe or something let us consider it is a unit area cross sectional area is a unit over here 1 meter square such that I do not need to carry my area of cross section into it I presume well I have a shock wave which is traveling at constant velocity r s dot when I say r s dot I mean the velocity of the shock wave and it is equal to d r s by dt over here which is r s dot let us also presume that this shock is moving into a medium and this medium has properties let us put it in a slightly different color the properties of this medium are let us say initial pressure is p0 let it be atmosphere in which case the pressure is equal to 100 kPa or 10 to the power 5Pa let it be at initial temperature T0 let the property let me let it have no air let it be stagnant it has no velocity let it have a density let us say rho 0 and then behind the shock because it is getting processed the pressure could be let us say p over here the temperature could be T the density could be rho and because the shock is moving it also drags the particle let the velocity behind it be let us say u over here velocity of the particle. Now when the shock is moving at constant velocity at least I neglect the effect of dk I want to know what are the what is the value of p rho t and u for this constant velocity shock and this is what I want to do today. Therefore what I do is I have to write the equations of motion relating sort of the portion over here which is the free unaffected zone into which the shock is traveling this is the zone in which the pressure rise has taken place and I want to be able to predict p t and rho over here you know to be able to do a problem with wave moving is difficult and what we do is maybe we keep the frame of reference of the shock to be stationary in other words I sit on the shock wave such that the my plane of reference is the shock itself and when I look at this what is going to happen well the medium is going to move towards it at velocity r s dot well the pressure of the medium is p0 t rho 0 and t0 and what happens to the process gases well it is at pressure p it is at temperature t it is at density rho and it is now moving since I am having the shock stationary in other words earlier the shock was moving the velocity behind is equal to r s dot minus u over here and this is the frame of reference I do well I consider unit surface area such that I do not need to bring area into account and for this particular problem in the frame of reference of the shock I would like to write the continuity equation the momentum equation and energy equation and try to solve for the shock properties that is the properties I want to determine p rho temperature and the value of the velocity behind the shock well let us do it let us now write the equations down see so far we have not said what is the medium the medium could be a solid substance could be a liquid substance could be a gaseous substance could be anything and therefore let us let us write the continuity equation well continuity equation says that the mass is conserved rather the mass which is entering the shock we are studying in the plane of reference of the shock rho 0 into r s dot into unit surface area is equal to rho into r s dot minus u over here what is happening well the the mass flux which is entering is equal to rho 0 into r s dot mass flux which is leaving is rho into r s dot minus u which is the continuity equation call it as continuity equation equation 1 what is the momentum equation well rate of change of momentum is equal to the impressed force you are considering unit surface area therefore force is pressure and the pressure is equal to pressure increase is equal to p minus p 0 where does it come from it comes from the mass flux that is rho into rho 0 into r s dot which is the mass flux it could have been I could have even written this over here into the change in velocity change in velocity is equal to r s dot minus I have r s dot minus u which is the velocity behind and therefore this gives me the value as rho dot r s dot into r s dot minus this this becomes plus here because this is the velocity behind is equal to u is equal to p minus p 0 which is the momentum equation well I have to write the energy equation but before writing the energy equation let us see whether we can make out something from these two equations before I write the energy equations and I have a I have some reason for doing this so far we have not said what is the type of substance the properties of the substance we said it could be a solid it could be a liquid it could be a gas into which the shock wave is moving and if I want to solve these equations well I could write r s dot minus u let us go back to this figure you have the velocity behind the shock as r s dot minus u let us call r s dot minus u as u 1 you know it simplifies instead of carrying this term I just put it as u 1 in which case what I get here is I get this is equal to rho into u 1 over here you know I have the momentum equation here which is equation 2 over here and now I write r s dot minus u is equal to u 1 therefore I can write this as equal to rho rho naught into r s dot into r s dot minus u 1 over here and therefore I could again write this as equal to rho naught into r s dot square into 1 minus u 1 divided by r s dot and if we write the continuity relation wherein I say r s dot minus u as u 1 I get the value of u 1 over r s dot is equal to rho naught by rho and this I can write as rho naught into r s dot square into 1 minus u 1 is equal to rho naught by rho into r s dot that means it is equal to u 1 divided by r s dot is equal to rho naught by rho and if I were to take rho naught outside this becomes rho naught square into r s dot square into since I have taken rho naught outside this becomes 1 over rho naught it becomes 1 over rho naught minus 1 over rho over here and this becomes the value of p minus p naught or rather what is it I get I get the equation as the value I get as rho naught square that is the initial density square into r s dot square is equal to p minus p naught into divided by 1 over rho naught minus 1 over rho and this is what I get by solving the continuity equation 1 with the momentum equation 2. Now see I have been able to get some relation between pressure and density in terms of the shock speed over here but I want explicitly the relation between p rho with p 0 rho 0 let us try to get rid of this term and get the general types of solutions for which I look at the energy equation. Now to be able to solve the energy equation well I have to go back to my scheme again well I have the gas coming at a velocity r s dot it is leaving with the velocity now I say u 1 because u 1 was equal to r s dot minus u we have the properties here p rho t we have p 0 rho 0 t 0 over here I want to write the energy equation and let us further assume that in addition to properties being p 0 rho 0 and t 0 and let the initial enthalpy of the gas here be h naught let the enthalpy of the gas in this region be h in addition to pressure p t and rho over here therefore what is it now I consider this shock which is steady as a control volume it is a steady shock moving at constant velocity and therefore when the shock is stationary gas is coming at constant velocity leaving at constant velocity u 1 and the equation for this control volume is well I have h 0 plus the kinetic energy per unit mass this is enthalpy so much joules per kilogram per unit mass into r s dot divided by 2 is equal to h plus I have u 1 square divided by 2 u 1 is equal to r s dot minus u is the value of u 1 this is the energy equation and now if I want to solve this I have to express enthalpy in terms of these particular properties and depending on the substance the enthalpy could be a function if it is a non-ideal gas well h could be a function of temperature and pressure if it is a solid or liquid substances well it is very cohesive the enthalpy is more complicated function of pressure temperature and density and therefore at this particular point we would like to introduce some simplification for the substance which is involved. Let us take the case of the substance being let us say a gas therefore now even though the momentum equation what we derived here is irrespective of the substances which are associated in which the shock is travelling when I look at the energy equation it becomes necessary for me to specify whether the substance is going to be a solid liquid or gas you know we know for substances such as solids and liquids the enthalpy is much more complicated and the relation between pressure temperature and density is through the equations of state are much more complex therefore I take a gas and in this particular gas I say let us also further assume that this gas is ideal what do we mean by a gas is ideal we say well for an ideal gas it is a pure substance enthalpy is only a function of temperature internal energy is only a function of temperature and therefore h is a function of temperature alone for an ideal gas and the moment we say h is a function of temperature internal energy is a function of temperature we also know that for an ideal gas the equation of state becomes p v is equal to m r t mass of the gas having a volume v therefore we start with an ideal gas and we also say in addition to being ideal let the gas be perfect what do we mean what is the difference between a perfect gas and an ideal gas for a perfect gas well d h by d t which is the specific heat at constant pressure well c p is equal to del h by del t at constant pressure and c v is equal to del u that is the internal energy by d t by del t at constant volume or we do not even need to do this because h is only a function of temperature therefore I can as well write it as equal to d h by d t is only a function c v is equal to d specific internal energy by temperature and therefore for a perfect gas I can simplify this equation and write it in the following way let us write the energy equation now we say that the gases are perfect gas and therefore I write h 0 h minus h 0 is equal to I bring h 0 on this side is equal to r s dot square divided by 2 minus u 1 square divided by 2 and h minus h 0 I can write as c p h is equal to c p t and therefore it is equal to t minus t 0 is equal to half into r s dot square minus u 1 square over here. Now c p can I express it in terms of the gas constant well we know c p minus c v is r and also we know that the ratio of specific heat c p by c v is equal to the specific heat ratio gamma and therefore I can simplify with this by saying c p into 1 minus 1 over gamma is equal to the specific gas constant r well if we are talking of specific gas constant which is unit is joule per kilogram kelvin and therefore c p is equal to gamma r divided by gamma minus 1 gamma minus 1 by gamma gamma r by gamma minus 1. Therefore the energy equation when we put some restrictions and tell ourselves we are only interested at present in a perfect gas over here into which a shock is moving what is it we start getting we get the equation as gamma r by gamma minus 1 into t minus t 0 is equal to I have half into r s dot square minus u 1 square over here. Now we look at this a little more closely let us first take a look at the left hand side well for the left hand side I have gamma by gamma divided by gamma minus 1 into r t minus r t 0 well p v is equal to r t m r t or we have p v let me use this part of the board p v is equal to m r t or p by rho is equal to v by m is rho m over v is rho p by rho is equal to r t and therefore this I can write as equal to gamma by gamma minus 1 this is the shock gas I have the pressure and density given by this this is the initial gas r t 0 is p 0 by rho 0 and this becomes my left hand side on the right hand side if I were to do a similar simplification what I get is half into r s dot square into 1 minus u 1 square divided by this should have been r s square r s dot square we have r s dot square here it should have been r s dot square 1 minus u by r s dot square and therefore this becomes half into r s dot square this is what I have taken outside therefore r s dot square into 1 minus u 1 by r s from the continuity equation namely u 1 by r s dot is equal to rho dot by rho and therefore we get the expression as coming over here is equal to rho dot square divided by rho square let us be clear this is equal to rho into u 1 is equal to rho not into r s dot therefore u 1 by r s dot is equal to rho 0 by r s dot so that we get the value of this over here and therefore now if I take rho dot square outside I get half into rho dot square into r s dot square into 1 over rho dot square minus 1 over rho square and now we have really got an expression for rho dot square into r s dot square by solving the momentum equation and this is where it is we have rho dot square into r s dot square is p minus p 0 into 1 over rho 0 minus 1 over rho and therefore this I can now write as half into p minus p 0 into divided by 1 over rho 0 minus 1 over rho this was the expression we got for rho dot and this is multiplied by 1 over rho 0 square minus 1 over rho square and this we find is equal to a square minus b square is equal to a plus b into a minus b and therefore this is equal to a minus b therefore this expression becomes let me write it on the other side of the board here I can now delete the continuity equation I can delete the momentum equation and write the final equation as equal to gamma over gamma minus 1 into let me write it p by rho minus p 0 by rho 0 is equal to half what I have here p minus p 0 into all what I get is I get this to be cancelled I get 1 over rho 0 plus 1 over rho over here and this becomes the equation we obtain by solving the continuity momentum and the energy equation and what does it tell? It tells what are the properties behind the shock p and rho as a function of properties ahead of it p 0 and rho 0 and the only other factor which comes is the specific heat ratio of the gases. Well this equation is what is known as a Hugoniot and since we are using it for a shock it is known as a shock Hugoniot let us see what does it give let us say I am interested in the properties behind the shock which is p I look at the properties ahead of the shock maybe let me sketch it again I have the shock which is stationary I have which we kept stationary the gas is moving with a ahead of it with a velocity r s dot the pressure is p the density is rho the initial pressure is p 0 the density is rho 0 and what does it give? It gives a variation of p as a function of 1 over rho and it depends on the initial condition well p 0 could be somewhere here the 1 over rho this is let us say p 0 this point could be 1 over rho 0 this is the initial point and for this initial point how do the states change and this becomes what we call as the shock Hugoniot. The Hugoniot equation tells what are the properties behind the shock as a function of the properties ahead of the shock that is pressure and density behind as a function of pressure and density ahead of it and this is the initial point let us call it as a and for all these points are the properties corresponding to the shock and see why do you get so much multitude of properties well we have not specified in this equation the velocity at which the shock is moving depending on the velocities I could have a number of properties which are there and this is what we call as a shock Hugoniot. Let us go further see we want specifically the relation between the value of let us say p as a function of rho therefore let us try to plot p by p 0 as a function of 1 by rho divided by rho 0 and let us therefore simplify this Hugoniot shock Hugoniot equation further and let us do it I want to specifically write p by p 0 therefore I simplify the shock Hugoniot equation as gamma divided by gamma minus 1 I take p 0 outside I let me also take rho outside and therefore I get over here p by p 0 I have taken rho outside therefore it becomes rho by rho 0 therefore this will be clear I have taken p 0 outside therefore p comes over here I have taken rho outside rho gets out here rho comes on the top minus rho by rho 0 is equal to on the right hand side I have half into I again take p 0 outside I have therefore p by p 0 minus 1 into p 0 and I also take rho in the denominator let us say p 0 by rho over here and if I take rho over here I get rho by rho 0 minus plus 1 over here therefore you know in this equation see I am wanting a relation between p by p 0 and rho by rho 0 and that is the reason I am doing this I find since I have taken these two outside these two cancel on the left side and the right side and therefore now if I were to write the expression for p by p 0 as a function of rho by rho 0 I take it on the other side I have p by p 0 into gamma by gamma minus 1 on the left hand side now I have p by p 0 with a half and therefore I get here minus half because I am still keeping it on the left hand side I bring it here minus half and minus half of what it becomes 1 plus rho by rho 0 and now I go to the other side I have this value therefore I get gamma by gamma minus 1 into the value of rho by rho 0 and what is it I am left with I have taken this now I have minus half I brought p by p 0 I still have minus half over here into what happens p by p 0 I have taken therefore this becomes p by p 0 into this now I have minus half into rho by rho 0 plus 1 over here this I simplify it I want to write it let us check this again I have p by p 0 into gamma minus 1 and then I brought p by p 0 on this side minus half into rho by rho 0 plus 1 over here then I take rho by rho 0 on this side it becomes gamma by gamma minus 1 minus sign becomes plus rho by rho 0 then I have minus half into rho by rho 0 plus 1 therefore now I want to simplify this term I have p by p 0 into now I write well it becomes minus half gamma by gamma minus 1 minus half which is equal to 2 gamma it becomes gamma plus 1 divided by gamma minus 1 minus I have the value of half that is equal to rho by rho 0 that means I have 2 by gamma minus 1 over here that means 2 gamma by 2 gamma minus 1 that means 2 gamma minus gamma minus 1 which becomes 2 therefore I have 2 over here is equal to the same thing I write on the other side I have half over here and therefore I have rho by rho 0 which is here therefore I write now gamma by gamma minus 1 minus half which is equal to gamma plus 1 divided by gamma minus 1 and the half I have already taken outside therefore I have minus the value of this is rho this becomes minus 1 over here and therefore what is it I get I get the value of p by p 0 is equal to gamma plus 1 divided by gamma minus 1 into this was rho by rho 0 rho by rho 0 into rho by rho 0 minus 1 divided by I have gamma plus 1 divided by gamma plus 1 divided by gamma minus 1 minus rho by rho 0 and this becomes an explicit relation which relates the pressure ratio that is the pressure behind the shock with the pressure ahead of the shock in terms of the density behind the shock divided by density and this equation is known as the Rankine-Hugoniot equation. We could similarly have an equation for instead of I could put rho by rho 0 and find it in terms of pressure and if you do this you will find that the expression comes out to be very similar even though the signs will be slightly different on the right hand side you will get rho by rho 0 is equal to gamma plus 1 divided by gamma minus 1 into p by p 0 in this case it will be plus 1 divided by gamma plus 1 divided by gamma minus 1 into plus of p by p 0. This is also Rankine-Hugoniot equation therefore these two equations are known as Rankine-Hugoniot equations and describe what is the value of the pressure ratio in other words all what we are talking is a shock moving in a medium in which the pressure is p 0 the upstream value is rho 0 the undisturbed medium this is the disturbed medium p and rho we are relating the value of p by p 0 p by p 0 and expressing it as a function of rho by rho 0. This is what we call as Rankine-Hugoniot equation and the Rankine-Hugoniot equation is just an extension of the shock-Hugoniot in which explicitly we express the pressure ratio and the density ratio I am sorry I should have been rho by rho 0 just like I write p by p 0 p by p 0 p by p 0 rho by rho 0 over here and this is what is the Rankine-Hugoniot equations. Therefore let us now summarize at this particular point of time before we proceed let us take stock of the problem what we wanted to do and then proceed further therefore what is it we wanted to do and where are we now we started off by saying we wanted the properties behind the shock wave which is moving at constant velocity like we have a pipe the shock is moving at r s dot into a medium whose pressure is p 0 rho 0 we wanted the value of p and rho once I know p and rho if the gas is an ideal gas I know p is equal to rho r t I can also find out the temperature. What did we get we got the Hugoniot that means I write the continuity momentum and energy relation relating this and this I took the condition of the plane of reference as the shock wave I got the Hugoniot equation which I called as the shock Hugoniot because I am specifically looking at this particular wave as a shock wave I got the shock Hugoniot and then I related the value of rho by rho 0 as a function of p by p 0 and also p by p 0 as a function of rho by rho 0 which we called as the Rankine-Hugoniot equations. Now our next step is well I am writing the equation for this discontinuity see mind you but in the first class when we when we looked at the waves we also talked in terms of acoustic waves or rather the sound waves can we can we say in what way this wave which is a shock wave is going to be different from an acoustic wave after all you know for an acoustic wave also you remember we solved it we said it is isentropic we also told ourselves well if it is isentropic I can write as p v to the power gamma is a constant or rather p by rho to the power gamma is a constant and therefore we said and we also derived an expression for sound velocity and we said sound velocity is given by under rho gamma Rt. Can we now relate the pressure and density behind a shock wave to pressure and density behind an acoustic wave what happens if the process is isentropic like in sound wave let us spend a couple of minutes on it before proceeding further with this particular problem. Let us now say yes I have been able to relate the pressure let us say I look at the Rankine-Hugonio equations and what is the Rankine-Hugonio equation tell me well I have p by p0 as a function of 1 divided by rho by rho 0 and what is it I get my initial point is p0 my initial point is rho 0 which in this non dimensional plot becomes 1 it becomes 1 because I am p by p0 this is point 1 this is point 1 this is my initial point over here and the Rankine-Hugonio equation tells me well the properties are all like this these are the different values of p by p0 as a function of rho by rho 0 and if I were to plot the isentropic solutions on this particular plot you know where is the sound wave or if I have something like the process of shock being isentropic will I get the same value or not let us try to do that therefore now I erase this equation well I say well for a sound wave p p by rho to the power gamma is a constant and therefore now I can write p by p0 is equal to rho by rho 0 to the power gamma and now if I were to plot it in this instead of plotting this why not plot it as a function of rho by rho 0 so that you know the thing gets diverted if I plot it as a function of rho by rho 0 what is this figure going to be instead of being like this is going to be like this over here therefore let us plot that because we are looking at a compression process which is a shock and therefore let us now maybe make a plot it as rho by rho 0 as a function of p by p0 and what is it this is my initial point one and if for an isentropic process I have this equation well this is the type of the solution what I get if the process is isentropic if I have a shock and if I were to plot this particular equation which is a Rankine Hugonio equation what is it I will get I will get the solution to be something like this in other words the pressure ratio is higher for the same density ratio and the reason is it is not isentropic there is some dissipation there is some it is no longer adiabatic and that is no longer reversible and therefore I get a higher pressure ratio compared to the isentropic assumption of p by rho to the power gamma is a constant in fact on the state properties I cannot show this by a continuous line because it is it is something which is which is not reversible and therefore normally we show it by a dashed line and we say well the shock properties are denoted by this particular dotted line across a shock therefore we are we are now very clear about Hugonio that is the shock Hugonio we are also clear about the Rankine Hugonio equation which comes wise for solving for the properties behind the shock as a function of properties ahead of the shock but we have also forgotten what we wanted to do we wanted explicitly when the shock is propagating in a particular medium what is the value of pressure what is the value of density and now what we have got is a family we have got may be number of pressure ratios are possible but which is the pressure ratio which is really going to be there when the shock wave is rs dot let us say rs dot is let us say 500 meters per second is it going to be this point is it going to be this point that means into this plot I have to put a velocity if I can put a velocity and say for this velocity it is going to be here for this rs dot it is going to be here then I am in business otherwise I do not know what to do therefore it is necessary for us to look at the Rankine Hugonio equations again or look at the shock Hugonio equation again but we have to define the velocity on this plot and let us see how to define a velocity such that I can say if the velocity is here it is over here the velocity is say 600 meters per second it is over here and therefore the second part is let us take a look at velocities and how to represent it on the plot of p by p0 versus 1 over rho by rho 0. But let us take a look at the momentum equation which we have solved some time ago what did we get we got rho 0 square into rs dot square is equal to p minus p0 divided by 1 over rho 0 minus 1 over rho please check we have we have got this we use the continuity equation and momentum equation to give this form therefore now I say well I have rs dot square is equal to 1 over rho 0 square into I get p minus p0 into now 1 over rho 0 minus 1 over rho I am sorry about this it is the final minus initial please check it again this is the value we have 1 over rho 0 minus 1 over rho therefore now I take 1 over rho 0 outside and now I get 1 minus rho 0 by rho over here now I find well rho 0 and this gets cancelled over here I can also take p0 outside and therefore this becomes p0 by rho 0 into p by p0 minus 1 divided by 1 minus 1 over I have this becomes 1 1 minus rho 0 by rho or rather this becomes 1 minus rho by rho 0 let us be very clear 1 over rho 0 minus 1 over rho I take rho 0 outside therefore if I take rho 0 outside rho 0 comes on top or rather it comes in the bottom over here it is rho 0 and therefore I rate 1 minus 1 over rho minus rho 0 and therefore what is it I get I get the value of rs dot square is equal to p0 by rho 0 into this particular value of p by p0 minus 1 divided by 1 minus 1 over rho by rho 0 over here in the denominator. What does this tell us let us take a look from the plot well again I have to plot the value I have to put the ordinates and the abscissa let us put it down. Let us say that the initial point is here p0 rho 0 this is the point 1 initial point this is 1 into 1 because pressure is p0 by p0 is 1 rho by rho 0 is 1 and let us say that the final point is somewhere over here corresponding to this. What does this tell us it is p by p0 minus 1 that means we are looking at this particular value p by p0 minus 1 we are looking at this particular value divided by 1 minus rho by rho 0 this is the rho 1 minus rho 0 1 minus 1 over rho by rho 0 this is equal to p by p0 minus 1 over here and therefore if I were to join these two points the slope of this particular curve is this particular bracket over here and if it is multiplied by the value of the initial pressure to the initial density well it gives me the value of rs dot square. In other words this line gives me an indication of the initial velocity when I multiply the slope by initial pressure and the divided by initial density and this line is known as the Rayleigh line. It gives me an idea of what the initial velocity is and therefore if I say well from the momentum equation I am able to get the Rayleigh line which is given by this expression and now if I look at the net problem what is my problem my problem was to be able to determine the value of the final pressure and the final density therefore all what I have is I have a shock ugonium initial point this is the shock ugonium which is passing now I have the shock velocity which is going at a given value of rs dot for a given value of rs dot well I have only a particular value of velocity which comes and hits here and therefore this is my value of pressure and this is my value of density therefore by solving the Rayleigh line along with the rancun ugonium equation or the shock ugonium equation both are same I am able to get the final point and I can get the final pressure and the final density because the initial density is known I get the value and I can get the properties of my medium. This is how we calculate the properties behind but I think why not put it in a proper form see the point is can I get an explicit relation also therefore let us quickly repeat what little we have done and set course for how to get an explicit relation see I would like to know well p by p0 as a function of shock velocity as a function of rho 0 as a function of p0 is what this is what I am aiming at I tell myself well from the shock ugonium and the Rayleigh line I am able to get p by p0 and I am able to get p by p0 I am also able to get rho by rho 0 as a function let us say of let us say rs dot rho 0 and p0 this I am able to get but this is graphically using the using the shock ugonium and the Rayleigh line let us try to solve for it explicitly an expression to be able to get this back to be able to get that expression we need to do a little more homework and let us see what little we will have to do in that we tell ourselves well I am able to get the Rankine-Hugonio equation may be from Rankine-Hugonio equation can I put the Rayleigh line into the Rankine-Hugonio equation and solve specifically for the pressure and the density behind the shock to be able to do that it is better to do it in a non-dimensional case. In other words instead of writing that the shock progresses with a velocity rs dot or rather in the plane of frame of reference that means the shock stationary the gas is moving towards it with a velocity rs dot instead of carrying the term rs dot can I non-dimensionalize it and put all the expressions in terms of p by p0 so also have don't carry the units of meter per second or kilometers per hour and all that in the shock speed. Therefore the non-dimensionalization we do is may be with respect in the free medium we have p0 rho0 t0 which gives me a sound speed a0 in the free medium and we already know well a0 is a function of the initial temperature or rather we have derived this expression a0 is equal to gamma specific gas constant into t0 and if I write Mach number as equal to rs dot divided by a0 well it becomes non-dimensional I express it in terms of Mach number but what does the Mach number represent it tells me the ratio of the shock speed or say this is the shock Mach number the velocity of the shock divided by the density of the medium upstream or rather what does this really denote let's let's let's spend a moment on this I can therefore write it as equal to rs dot square divided by a0 square to the power half. Now I say well rs dot square is an indication of the kinetic energy associated with the shock or associated with the flow of gases over here a0 square we found a0 square is equal to gamma rt or the initial temperature gamma rt0 t0 tells me what is the temperature of gases or what is the level of the molecular energy which is available or something like an internal energy of the gas which is available therefore Mach number is a representation of the kinetic energy of the gases to the molecular or the internal energy of the gases and as the internal energy increases for a given velocity well the shock Mach number comes down it is just the ratio of the kinetic energy to the internal energy and therefore we will use the Mach number instead of the shock velocity to be able to derive an explicit expression. Let's do that let's get started you know for that we again revisit the problem we reformulate the problem let's look take a look at the continuity equation and momentum equation and the energy equation again what is the momentum equation let me start with momentum p minus p0 was equal to we had rho 0 into rs dot which is the mass flux which is coming and there was a velocity change and what was the velocity change initial velocity was rs dot the velocity behind the shock was u1 over here this was my momentum equation and now we would like to express it in terms of let us say we will get rid of rs dot we will get rid of u1 we will put it in terms of Mach number therefore the problem which we consider is well I have now these shock which is traveling at velocity rs dot or or equivalently in the frame of reference of the shock the gas is moving at a velocity rs dot it's moving here with a velocity u1 the properties here are p0 rho 0 and t0 the properties behind are p rho and t over here instead of rs dot I put in terms of ms over here the sound speed in this medium is a0 the sound speed in the medium processed by the shock is at a higher temperature therefore the sound speed is a over here and therefore the Mach number behind of the gases which is leaving the shock wave in the frame of reference of the shock wave is equal to m which is equal to u1 divided by the sound speed of this medium and therefore I would rather translate this momentum equation into an equation which contains the shock Mach number ms over here Mach number of the gases behind the shock and this is what I will do now let's quickly derive it and then leave the final expression for the next class let's just simplify the momentum equation let's see where we could end up with therefore rs dot square over here let's take a look at this rho 0 into rs dot is equal to rho u and therefore I get p plus rho into u1 square is equal to p0 plus rho 0 into rs dot square how did this come rho 0 rs dot is equal to rho u1 and I have u1 it becomes u1 square rs dot is left rs dot and rs dot get cancelled this is my equation and now can I put this in this particular form I take p outside I get p plus I have rho by p into u1 square is equal to I take p0 outside I get 1 plus rho 0 into rs dot square over here and here since I have taken p0 outside it should be rho 0 by p0 now in this particular expression if I were to multiply on this particular fraction by gamma here and gamma over here and also over here gamma and gamma and now I look at this particular expression gamma p by rho here also gamma p0 by rho 0 what did we find earlier we found that when we derived an expression for sound speed a0 square is equal to we said is equal to dp by d rho or delp by del rho and that was equal to gamma p by rho or this was what gave us gamma rt and therefore if sound speed is given by gamma p by rho well gamma p by rho is the sound speed behind the shock gamma p0 by rho 0 is the sound speed ahead of the shock and therefore I can write this expression as equal to p into 1 plus I have u1 square and I have still gamma over here into a square behind the shock which I call as a1 now here because it is gamma p by rho it is behind the shock corresponds to 1 which I could either call as a or a1 distinct from the other side wherein I get p0 into 1 plus I have gamma into rs dot square divided by a0 square and now we know u1 by a is the Mach number behind which is m rs dot this is rs dot divided by a0 is equal to m a shock Mach number and therefore the momentum equation now becomes let us write it I have p into 1 plus gamma m square why m square we said well u1 by a is m therefore u1 square by a1 square is m1 square is equal to p0 into 1 plus gamma into ms square shock Mach number. Therefore we now have a relation between m and ms and our effect is we can write this equation as p by p0 is equal to 1 plus gamma m square divided by 1 plus gamma m square. Therefore what is it we do we will proceed with this in the next class and we will get an expression linking the Mach number behind the shock with the Mach number ahead of the shock and once we do that we will be able to correlate it with the properties. To summarize in this particular class we started with the discontinuity moving we looked at the properties ahead of the shock and behind the shock and we got an expression namely the shock hugonio which linked the properties then we said that the shock moves at a particular velocity which was defined by the Rayleigh line the intersection of the Rayleigh line with the shock hugonio or the Rankine-Hugonio relations gave you the properties namely pressure and density dependence with respect to a head and behind the shock and then we wanted to get an explicit relation and for that we introduced the Mach number ms of the shock and the Mach number behind the shock and we are now trying to solve the relation between m and ms. We got the momentum equation in this form we will again get the value of rho by rho 0 and put it together we will be able to get the Mach number behind the shock as a function of Mach number ahead of the shock. This is what I do in the next class and with that we will be able to get this shock relations and use it for predicting the pressure and impulse for a blast wave. Well thank you then.