Good morning. In the last class, we were looking at the formation of a spherical blast wave from an explosion which released let us say E0 joules of energy. In other words, we are looking at a spherical blast wave which is formed and this spherical blast wave keeps propagating outward. We told ourselves that this lead blast wave at time T is at a distance rs from the source of energy release. We looked at the energy balance and what did the blast wave do? You form a blast wave. Let us take a look at a blast wave. You form a blast wave and it keeps propagating out. At different times, it keeps moving out and therefore this energy which is deposited is dispersed within the region of the blast wave as the blast wave propagates out. We looked at the conservation of energy namely E0 joules of energy which is dissipated in the medium and which creates the spherical blast wave. What did we do in the last class? Let us quickly go through the point because we will continue on the energy release and how the blast wave changes. Therefore, for that I again look at the energy that means we dissipate some energy. The blast wave is at a distance rs from the source. This is the lead shock wave. We look at the energy which is balanced or we look at the energy which is available at some radius r and of width that means we look at a spherical shell of thickness let us say dr at a distance r. Now, the blast wave is at a distance rs from the source. Therefore, maybe this is my distance. This is my distance let us say rs over here. The initial pressure jumps up over here and then maybe the thing expands out. We looked at the expansion in the context of the mass conservation and we found that well most of the mass tends to be concentrated at the front and thereafter there is not much mass left because all the density is over here. Therefore, there is something like a gradient over here. We were interested to write the energy which is contained in this small spherical shell over here element of the shell over here. That means the blast wave is here, the energy gets dissipated. I am just interested in the energy here and we got the expression d is equal to the kinetic energy and what was the kinetic energy? The volume of this element is the surface area is 4 pi r square into dr 4 pi r square into dr which is the volume into the kinetic energy per unit volume is rho u square divided by 2 plus you have the internal energy and initially let us say the medium outside has an internal energy E naught. We have E minus E naught into rho is the change in the internal energy here. We considered that the internal energy change due to the blast wave processing is very much greater than the initial internal energy and this drops out. We wrote E is equal to C vt, C v we said C p minus C v is r and therefore we could write E as equal to C vt, C v could be written as as equal to r divided that is the specific gas constant divided by gamma minus 1 into t rt is equal to p by rho and therefore this equation became d is equal to 4 pi r square into dr into I have rho and what is this rho? Let us be very careful it is at distance r from the source of explosion u at r square divided by 2 plus I have now p at distance r because rt is equal to p by rho p and p gets cancelled divided by gamma minus 1 is the energy which is available in this spherical shell. Now I integrate the total energy from here to here that means I integrate out from 0 to rs and that will be equal to this energy and therefore the expression which we got was E 0 that is the energy in joules which is dissipated to form the blast wave is equal to I now take 4 pi which is a constant outside and now I take integral 0 to rs of I have rho r into the value of u at r square divided by 2 plus I have p r divided by gamma minus 1 into I have r square into dr this is the expression which we got. Now when we look at the distribution we say from starting from rs it keeps decreasing over here or it changes over here. I can as well say that my density at r divided by the initial density I could write it as equal to psi of r by rs because why do I have to write this. Initially I have a jump and then it keeps coming down therefore we wrote this as rho r divided by rho s into rho s by rho 0 rho s by rho 0 is known and therefore the density distribution as a function of the initial density could be written in this particular form. Similarly, I could also write the other expressions where we could write the value of u dot by rs that means let me let me as put down the value of r specifically the velocity behind the shock at the radius r divided by rs dot is equal to let us say phi into r by rs. Similarly, I have p at r divided by rho 0 into rs dot square is equal to f into r by rs all function of the distance from the shock to maybe I am interested at this particular r I am looking at the value of r by rs. Therefore I would like to solve this equation subject to some distributions like this. In the last class when we did the mass conservation we took a power law profile and maybe we will consider some of these things again for energy balance. But as of now I would like to substitute these expressions into this particular energy balance equation and try to solve for it. But while doing so I can also represent this in a slightly different way let me try to tell what I want to do. All what we are telling is we have rs over here at rs I have a step change taking place in maybe the density maybe the particulate velocity maybe the pressure and thereafter it decays down it decays from rs to 0. I am interested in a distance r over here I can as well change the coordinate system such that I am always saying that well the shock is moving I am always having rs as a variable but I am considering at a particular r as the value of r. Therefore I can consider let us say zeta is equal to r by rs and I could say well I am considering the case wherein the lead shock is at 1 I am considering the properties let us say now my coordinate shifts to zeta I am considering the value of zeta over here which is less than 1 I am looking at the property at this particular value of zeta. Therefore if zeta is equal to r by rs well can I change these coordinates then automatically r square becomes equal to rs square into d zeta and well d zeta or dr let us let us substitute the value of dr over here dr from this expression becomes equal to equal to rs for a particular value of rs I have rs into r square is equal to rs square into zeta I am sorry r square is equal to zeta square into rs dr is equal to rs into d zeta and the limits of integration instead of being rs over here it becomes rs by rs which is 1 and therefore the above equation for the energy balance will now be given by E0 is equal to 4 pi into 0 to 0 into now rho r rho r can be written as equal to zeta rho 0 rho r is equal to rho 0 into psi of zeta into u r u r is equal to 5 that means rs dot and what is it we have u r square therefore rs dot square into phi zeta square divided by 2 plus I have the value of pressure r pressure r is equal to rho 0 rs dot square into 5 zeta therefore I have rho 0 into rs dot square into phi f of zeta divided by gamma minus 1 and this value is multiplied by r square r square is equal to rs square into zeta square and dr is equal to rs into d zeta and therefore if I were to simplify this equation I get E0 that is the energy release is equal to now I take the terms outside well rho 0 is the ambient the shock is moving or the blast wave at that point is moving with a velocity of rs dot square therefore I get 4 pi into rho 0 into rs dot square this is the expression for this taking out then I get the value 0 to 1 and what is it I get I get the value of psi at zeta divided by the value of velocity that is the velocity at that particular point is phi divided by phi square psi phi square divided by 2 plus I get f of zeta to the power gamma minus 1 into what is left over here I have rs and rs cube well at a particular radius I am interested therefore rs cube can be taken outside and I have zeta square into d zeta this is the expression I get and therefore now I bring the terms which are outside the integral sign over here I get E0 by 4 pi into rho 0 rs cube into rs dot square is equal to this particular integral 0 to 1 of let me take let me recognize that psi is a function of zeta phi is a function of zeta f is a function of zeta and therefore I can write it as psi phi square divided by 2 keeping in mind that these are all functions of zeta such that I do not need to carry these symbols as I am doing plus f over gamma minus 1 into zeta square into d zeta. Now you know this expression tells us that the energy deposited in the medium of density rho 0 is there this energy if I look at the denominator I have 4 pi rs cube 4 upon 3 pi rs cube this is the volume this is the mass this is something like the kinetic energy term and therefore this expression is an indication of how much of the energy or how much of this energy deposited gets into the kinetic energy of the medium provided that the entire medium let us go back to this if the entire shock medium were to travel at rs dot square that is the net kinetic energy then this expression tells us that what part of the energy travels at the with the kinetic energy such that the velocity is rs dot square and this is what this expression tells and therefore this integral denotes part of the energy which is getting converted into kinetic energy provided all the particles are travelling at the blast wave speed namely rs dot square you know if I look at this expression a little more closely you know yesterday we looked at this point also we said well zeta is equal to 1 at zeta is equal to 1 if I am interested in rho by rho 0 which is equal to your psi and psi we are plotting between 0 to this that is zeta equal to 0 to zeta is equal to 1 we found well the slope is like this similarly if I if I were to take the value of let us say 5 as a function of zeta and you know at zeta is equal to 1 whereas there is a jump in velocity or there is a change in velocity and there is a slope in velocity and similarly for pressure there is a jump in pressure at zeta is equal to 1 and this jump there is a slope that means between 1 and 0 there is a slope over here and therefore you know we expect that to be something like a power law and this is what we said porcel used zeta used these equations again we said well William Ray also used these equations to solve for this and you know there are different ways of solving we could get these things analytically for instance we solve the conservation equations in a blast sort of numerically or we use the perturbation techniques to solve for it or we use the similarity solutions we will look at similarity solutions a little later you know we can always get these slopes and therefore if we are talking in terms of strong shocks well the jump conditions at zeta is equal to 1 do not really change with the Mach number and therefore we would expect this particular integral to be near a constant namely if I denote this integral by let us say I should be near about a constant and if you really calculate these things using different numerical methods or as I said may be a perturbation method or let us say the similarity solutions I get I is equal to around 0.421 for the case of the spherical wave which we are considering therefore we tell ourselves well this integral is expected to be a constant and therefore I can write this value let us write this value I get E0 by 4 pi into rho 0 into rs cube into rs dot square is equal to a constant this constant is equal to I which we just now said is equal to this particular integral and this integral or which is a constant tells you what is a fraction of the energy which is deposited gets into the kinetic energy of the medium and this kinetic energy of the medium we evaluated at the shock velocity or the blast velocity at that point in time. Therefore if this is a constant can I use this equation to interpret something well for a given energy release E0 in a given medium of density rho 0 that means let us put down for a given value of E0 joules in a given medium of density rho 0 kilogram per meter cube what is it I get you know E0 I have specified rho 0 I have specified therefore is a constant therefore immediately I tell myself well according to this equation rs cube into rs dot square is equal to a constant what is it I get I get rs cube into rs square is a constant or rather from this equation I get rs dot is equal to a constant let us say this constant is b into rs to the power minus 3 by 2 well if you recall in the second lecture what we had we had used the dimensionless method of deriving the decay of a blast wave we had got the identical expression. Let us take a look at this again we have we have derived rs cube into rs dot square is a constant let us differentiate this if I differentiate this by parts I get rs dot square into differential here that is 3 rs square into rs dot plus I now keep the first value r constant rs cube into I get rs 2 dot into I have 2 into rs dot is equal to 0 constant is equal to 0 let me again repeat rs dot square into differential of the first term 3 rs dot square into into 3 into rs square into rs dot plus I have over here rs cube which I am which I am this is the first term and differential of the second term is rs 2 dot into 2 rs dot and what is it I get immediately I tell myself well in these two terms if I were to take one on the left hand side I have rs 2 dot into rs cube into 2 over here into rs dot is equal to minus 3 on the right hand side I take it on the right hand side I have rs dot cube into rs square and if I were to now take out the common terms over here well rs square cancels here gives me rs and I have rs dot over here gives me rs dot square and therefore I get the value of rs 2 dot into rs divided by rs dot square is equal to minus 3 by 2 which was again the result which we had got from the dimensionless analysis in other words there is a decay that is the shock is decelerating because of the negative sign we got the shock velocity in terms of this and the same results which we got by dimensionless analysis or dimensional analysis we are able to get through the energy conservation. This is point one of the energy equation this must be clear we have solved the energy equation and we get back the condition well in a blast wave what is it we are telling ourselves in a blast wave well you have the streak diagram temperature versus rs you form strong wave and it keeps decaying or rather in a frame of reference of rs dot versus rs you start with a wave and it keeps as the distance increases the velocity at the front decreases well this is the same signature for ms dot also well this comes from the energy equation let us try to see whether I can use the energy equations to better advantage to get some more characteristics of these blast waves mind you we must keep something in mind when we did all this we presumed over here that the initial energy of the medium is small in other words we told ourselves well the blast wave must be strong enough such that the value of E0 can be neglected second is we also use these profiles which was strictly true only for the case when Mach number of the blast wave was quite high that means we are talking of strong blast wave and therefore the conditions what we have derived here are suited only for this initial region or for the initial region over here having said that let us go into some more details some more characteristics of the blast wave and try to solve some some some more param for some more parameters let us pick up on this equation itself let us pick up on this equation namely over here E0 by 4 pi let me write it over here we have E0 that is the energy release in a medium rho 0 is equal to we we we do not need to consider the integral let let me put it in terms of the fraction what we had got we got E0 divided by 4 pi rho 0 r s cube into r s dot square is equal to I now I want to slightly transform this equation into something which is more usable which will help me to scale different types of blast waves and let us therefore divide the numerator and denominator by P0 that is E0 by P0 divided by 4 pi rho 0 I also divide the denominator by P0 into I get r s cube into r s dot square is equal to I over here I have just copied this equation here I have E0 4 pi rho 0 r s cube r s dot square is equal to I I get this particular expression now let me do something let me multiply the denominator by gamma and also divide by gamma such that I still retain the same if now I look at this particular equation I have gamma P0 by rho 0 is equal to A0 square that is the sound speed in the free stream medium that is ahead of the blast wave and I also know well r s dot square divided by A0 square is equal to the mark number of the shock mark number ms is equal to r s by A0 and therefore substituting this expression in this particular one and what is it we will get we will get the value of E0 by P0 divided by 4 pi gamma into ms square is equal to I which is a constant we say is equal to around when the for the spherical case with which we are right now dealing it is equal to something like 0.423. Therefore now let us take a look at this particular expression whether it will help us to simplify things well in this equation 4 pi gamma into ms square I think I have still to write the value of r s cube is equal to I now if I were to look at the parameters now I say well E0 is equal to energy release in joules what is a joule joule is a Newton meter if I look at the value of pressure pressure is Pascal which is Newton per meter square and if I look at the dimensions of E0 by P0 I find that the dimension is equal to Newton meter by Newton per meter which is equal to meter cube Newton and Newton gets cancelled it becomes meter cube or rather if now I say I am looking at the value of E0 by P0 to the power one third I get the unit as equal to meter. Therefore now I tell myself well E0 by P0 is equal to meter cube or E0 by P0 to the power one by three is a meter and therefore I can talk in terms of E0 by P0 as a length scale which is associated that is a length scale associated when an energy is impulsively released in a medium whose density is P0 that means I say length scale of an explosion. I define length scale of an explosion as equal to E0 by P0 and it is tells me if an energy E0 joules is deposited in a medium of pressure P0 I get a length scale and this length scale of an explosion I denote by R0 so much meters. In other words when I have an energy release E0 joules in a medium of P0 the ratio of E0 P0 is equal to the explosion length square and therefore I use this explosion length in this particular expression and what is it I get? Let us put that down. I get now the value of over here I get R0 explosion length square divided by 4 pi gamma into the value of M square into RS cube is equal to I or rather from this can I get the value of let us say I take the R0 I bring it here I get explicitly the value of M square that is the Mach number of the blast wave what it is going to come? It is going to come as equal to 1 over 4 pi into gamma into I into I get RS by R0 cube this is the value of the Mach number of the blast wave. Now what does this expression tell us? All what we are telling is instead of considering the distance if I consider the distance which is divided by the equivalent explosion length that means I have the distance which is scaled then I get a value of Mach number square which is given by this. If I look at this expression well gamma is 3.14 for if I am sorry pi is 3.142 gamma for particular air is 1.4 I we said is a constant around 0.423 then what we find M square is a function of only RS by R0 cube. In other words what has happened? When we reduce the energy release in terms of an equivalent length scale namely the explosion length what is it we have done? Let us take a look at the figure again and try to interpret this result. We go back to the streak diagram with which we are all familiar by now. We say this is the distance travelled by the lead blast wave. Initially if some energy is deposited in a medium I get this to be my blast wave. Let us say that the energy release is E01 joules I form a strong shock wave with decays. If now I deposit much higher value of energy what is going to happen? Well the shock is going to be get started strongly and decays after a longer time. This is for the case of E02 joules in which E02 joules is greater than E01 joules. If now I deposit even a much larger value of energy what is it I get? Well since the energy release is higher it travels a further distance this is E03 joules and therefore depending on the energy release I get for the same time maybe I get a blast which is further away or rather if I were to put this particular figure in terms of let us say the Mach number or the rs dot as a function of rs what is it I get? You recall we looked at the decay initially I get for E01 I get this type of result maybe for E02 maybe I get some result like this we use the blue color maybe I get E02 like this maybe when I talk in terms of E03 I get a result like this that means I get at same distance when energy release is higher I expect a stronger shock and this is what it tells and therefore what is it we are telling? Well rs dot depends on this rs dot divided by A0 is ms over here it has this dependence and what happens instead of using rs if I use the value of rs by r0 what is it I get? Now I am able to convert this figure into a figure which says if I use the scale distance rs by r0 and I am looking at ms square from this particular expression all what it tells is well I have a single curve for the value of E01 E02 E03 irrespective of the energy release which gets apportioned here I get a single curve and therefore I am in a better position to solve the equations and this way of characterizing the distance traveled by the shock in terms of the explosion length is what is known as shock scaling that means we are able to scale the Mach number of the shock in terms of the scale distance and I do not really need to consider the energy gets embedded here and therefore I do not need this multiple energy curves to do that I can solve the equations straight forward that means let us think in terms of a problem all what we say is the following we deposit some energy in the medium let us say E0 joules if the pressure is P0 I get the explosion length r0 is equal to E0 by P0 for the spherical as 1 by 3 and once I know this well I use this particular expression r0 cube by 4 pi ms square is equal to 1 over 4 pi gamma i into rs at the particular distance rs since r0 is known I can calculate the value of my ms square. Therefore let us take a look at this expression again we plot some figures over here I show this on this figure maybe I look at ms square I substitute the value of these things and what is it I get I get the value ms square is equal to 0.1343 divided by rs by r0 cube. Now if I were to plot this what is it I get I get may be the value of ms square ms let us say I take the under root and I plot it as a function of rs by r0 and what is it I get when rs by r0 is 0.2 the value of my mark number is 4.1 from the above equation I when the value of rs by r0 is 0.3 I get the value as 2.2 when the value is around 0.4 and how do I get it I just substitute the value here I divide this this gives me the value when it is 0.4 the value is 1.47 1.45 I am sorry and when it is 0.5 the mark number is 1.037 in other words 4.1 2.2 1.45 1.037 and therefore this is my mark number distribution and what is it I must infer from this well I made a strong blast assumption but I keep getting numbers and strong blast is normally we found when we were looking at the density ratios yesterday we found that well it must be we get the constant values of rho by rho 0 p by rho 0 rs dot square and u by rs when the mark number is typically greater than around 4 and therefore the validity of this particular results are only in this region. In this region the mark number influences let us say the any the jump conditions influences the profile and in this case the predictions may not really be that good. Therefore what is it we have done using energy balance we are able to get the lead shock mark number and if I get the value of ms square I will be in a position to get my pressure also that means I have when the shock is propagating away at the lead shock well I have the pressure initial pressure is p0 I have a jump in pressure to a value ps and then it decays further I can find out what is the jump in pressure and if I know the jump in pressure I can always calculate the jump in pressure is ps the initial pressure is p0 and ps my p0 is the increase in pressure across a lead shock wave and this is something which is known as over pressure. Let us try to get an expression for the over pressure also we do it it is quite simple we have done this already in a sense in that we know the value of the shock pressure that means we say p by rho 0 into rs dot square we say it is equal to 2 into gamma plus 1 for a for the particular case of a strong blast wave that means in this particular region well I can simplify this and get it as p by rho 0 divide by p0 in the numerator divide by p0 in the denominator you have rs dot square well let me also divide the numerator and denominator of this particular expression in the denominator by gamma again and now I have gamma p0 by rho 0 is equal to ms square and therefore I can write the value of p by p0 we use the same color shock p by p0 divided by gamma into ms square is equal to 2 over gamma plus 1 or rather I get the value of p by p0 is equal to I get 2 into gamma plus 1 into gamma into ms square and now if you were to look back at the expression for p divided by p0 in the third and fourth classes we had derived the expression p by p0 is equal to 2 gamma by gamma plus 1 ms square minus gamma minus 1 by gamma plus 1 since this is a we are talking in terms of large mark numbers well gamma minus 1 by gamma plus 1 is negligible and this is the pressure. Therefore let us calculate the pressure behind the shock to be able to distinguish between p and r I can write this as ps that is behind the particular lead blast wave I am interested in the value of ps over here and therefore let us derive it we all what we do is we substitute the value of ms square as we got by solving the energy equation either this or let us put the ms square as we got here we got ms square is equal to 1 over 4 pi gamma I into rs by r0 cube so that we can determine the over pressure with particular distance. Let us do that now we therefore write from the expression over there ps by p0 is equal to we get is equal to 2 gamma by gamma plus 1 into ms square but ms square is equal to 1 over 4 pi gamma I therefore this I can write is equal to 2 gamma over gamma plus 1 into I get now 4 pi gamma I into rs by r0 cube or rather in this I find gamma and gamma gets cancelled this cancels to giving 2 I get the value as equal to 1 over 2 into gamma plus 1 2 into gamma plus 1 into I into 1 over rs by r0 cube in other words I get the value of ps by p0 and what is the over pressure over pressure at the shock front is equal to I have a jump in pressure minus the ambient pressure therefore the dimensionless over pressure is equal to ps minus p0 divided by p0 that means you have ps by p0 minus 1 therefore the value comes out to be 1 over 2 I into gamma plus 1 into 1 over rs by r0 cube this is the value of the over pressure and mind you now we know the value of over pressure or dimensionless over pressure as a function of the shock distance or distance divided by the explosion length over here. Let us take a look at this particular expression and see whether we can interpret this how the over pressure behaves with distance we will substitute the values and what is it we get let us get the values we get ps minus p0 divided by p0 I now have I is equal to 0.423 the gamma plus 1 is 2.4 I substitute the value it gives me 0.156 divided by rs by r0 cube over here this is the value of ps by p0 over here I subtract 1 therefore I should have subtracted 1 over here I get this is the value and now if I plot this expression that means I plot the over pressure ps minus p0 divided by p0 as a function of rs by r0 what is it I get let us again do this similar to what I have done over here therefore Mach number I get at a value of rs by r0 is equal to 0.2 the value of the over pressure that is the dimensionless over pressure gives out to be 19 at the value of 0.3 rs by r0 is equal to 0.3 the value comes out to be 3.8 at the value of 0.4 the value comes out to be 1.44 and let us take one last value 0.5 at which the value comes out to be something like 0.25 in other words if I join the values what I get is well the curve looks something like this and what is it we find well at 0.2 we found out that the Mach number was 4.1 well strong blast assumption is valid 0.3 the Mach number was 2.2 for which the over pressure you know in other words you know when the Mach number has decayed to these type of values well I cannot really expect the over pressures to be correctly predicted because we found well the strong shock wave assumption is not valid I cannot write the density ratio is gamma plus 1 gamma minus 1 or 2 over gamma plus 1 is equal to the shock pressure divided by rho 0 rs square and all that and therefore maybe in this region in the region of the strong shock wave that is in the near field to the explosion I would expect this expression to be really valid whereas in the far field this expression may not be totally valid in other words I have been able to get but mind you you know in a particular region I can still predict my over pressures and therefore using the energy equation I am able to get the over pressure values well this we have done quite a bit now what are the let us quickly summarize what little we have understood for the blast wave using the mass balance and energy balance what did we tell ourselves well from mass balance what did we get we found the following namely when we have a strong blast wave that is in the area near to the source of the explosion what is it we got we found that maybe I have an explosion over here I form a blast wave over here near to this well most of the mass the blast wave that is the blast wave as it moves it sweeps collects all the material and locates all the material in the zone of the lead shock wave and this is what I have lot of mass here travels with a high velocity and this is what creates my compression or the or the damage using the energy balance what did we get we have the over pressure shock pressure minus the initial pressure divided by p0 is a function we get of 1 over rs by r0 cube 1 over r0 cube over here minus 1 and therefore I am able to predict for a given distance what is the type of pressure I have over here similarly it is possible for me to predict the velocities but let us quickly discuss what we have achieved so far in terms of some scaling that means shock scaling in which I express the distance in terms of r0 has really been useful and I am able to get the over pressure from a blast wave and over pressure is important because that is what creates a lot of damage for me because the high pressure compresses and damages a building knocks down people and all that therefore we use shock scaling over here can I think of anything further if I look at this particular expression what is it I get can I talk in terms of other forms of scaling can I say how does a blast wave scale for over pressures in other words let us ask our question scaling for over pressure can I can I really solve for this what do you mean by scaling for over pressure supposing I have an explosion in which let us say that the explosion length is r0 that means I deposit some energy e0 and I get the explosion length as r0 because energy release in a medium p0 is able to define it I am able to get may be at some distance r is away I am able to now predict what is my over pressure namely ps minus p0 by p0 minus 1 this is the value of the over pressure dimensionless over pressure. Now instead of depositing e0 by p0 which is r0 cube supposing the energy deposited is such that I now have an energy deposited which is equal to lambda times lambda could be a multiplication factor that means I deposit a value of explosion length equivalent explosion length to be lambda r0 the same over pressure as per this expression will come out to be at a distance of lambda rs away will give me the same value why is it because lambda rs by lambda r0 comes out to be identical even though the energy is released is higher well the scale of over pressure over here and scale of over pressure is same and this type of scaling laws in which I change my energy level in terms of r0 to lambda r0 is known as Kranz Hopkinson scaling law for over pressures. What is its state? It states if an observer is situated at a distance lambda rs away from an explosion whose explosion length is lambda r0 he will feel the same over pressure as an observer who is stationed at a distance rs away from an explosion whose explosion length is r0 as simple as that this is Kranz Hopkinson scaling law and it is used left and right in the explosion industry because we would like to find out what is the energy released from an explosion how it affects the pressure at some distance away and Kranz Hopkinson scaling is widely used. Let us try to interpret this in a slightly different form. We draw the last figure which I draw now is the figure of over pressure. I tell myself well I have ps minus p0 divided by p0 which is the dimensionless over pressure as a function of rs by r0 we got this particular curve we told our curve is valid only in the near region of the blast in which the blast wave is strong and now once I know the energy released from the explosion I know the ambient pressure I can calculate r0 I want to calculate the over pressure at a given distance if the distance falls within this which is let us say around 0.25 or so I can predict my over pressure and we are able to do some problems now. But if we look at certain books they do not use the soccer law in terms of explosion length. They rather plot instead of writing r0 they plot the scale as equal to energy release to the power 1 by 3 because r0 is equal to e0 by p0 to the power 1 by 3 p0 for most blasts in air is around 1 atmosphere pressure that is 10 to the power 5 Pascal and therefore they express it in this form and they get the value of ps minus p0 by p0 and therefore they give a curve like this. But instead of looking at energy since we are looking at energy release from an explosion and this energy release from an explosion is directly proportional to the mass of explosive used very often and why instead of even using energy over here they put mass of the explosive used to the power 1 by 3 and this is how many books like Baker's book write in terms of rs by mass of the explosive or mass of the particular explosive used we will take a look at this a little later and this is how figures are drawn. Well, this is Kranz-Hopkinson scaling law we are able to get the over pressures but something which we must keep in mind. See I keep qualifying each time we can predict in the near field of the explosives where in the blast wave is high. Why? Because we showed earlier that the value of rho by rho 0 at the shock front ps by rho 0 rs dot square and the value of u divided by rs dot are all about the same in this particular region because the jump conditions did not really depend on the Mach number square. Whereas in this region it is a function of Mach number square therefore the integral which we used here in the expression as equal to 0.423 may really not be valid and therefore it is only in this region which we can use. Let us try to get some more information of that and for that let us look at the assumptions what we made and with that maybe we will be a little more wiser. What is the assumption which we made while deriving this over pressure relation? Well we assume the value of the initial internal energy of the medium is very much less than the internal energy of the medium which gets enhanced due to the blast wave rather you will recall we had a spherical blast wave. We talked in terms of a small segment or a small annular area spherical shell over here in which we said it is equal to E minus E0 is equal to E that means we neglected the value of E0. That means what is it we have neglected? We have neglected the internal energy of the medium here. Can we find out under what conditions for all what value of r is by r0? Does the internal energy of the medium become significant? Let us say in this particular figure if I find that r is by r0 for r is by r0 greater than this E0 becomes significant. Well I can immediately say in this region well the initial internal energy of the medium has to be considered and therefore my predictions are valid only in this particular band or in this particular band of r is by r0. Let us do that problem therefore we tell ourselves well the shock is over here the source is here this is the shock which is r is distance away and what is it we are telling ourselves well the energy release of the by the source we say is again E0 joules and if the initial internal energy of the medium has to contribute what is it we are saying E0 joules must be of the same order as the internal energy E0 of the medium but what is the internal energy of the medium is per unit mass therefore we tell ourselves when E0 is equal to 4 upon 3 that is the volume occupied by the particular volume within the lead shock wave 4 upon 3 pi into rs cube into the density which is the mass which is available here is equal to into E0 that means the internal energy of the medium here is of the same order as energy release well I should be able to get this particular point and beyond this what is happening is the initial internal energy affects my shock or affects my over pressure affects my Mach number of the shock and therefore I can look at this particular expression and get a feel for what is the value of r is by r0 from this expression which will decide whether the the decide the region in which my predictions are quite reasonable let us do this to be able to do this I I I just simplify this expression I get the value of E0 divided by 4 upon 3 pi into rs cube divided by rho 0 is equal to E0 and what is E0 internal energy per unit mass which is equal to CVT that is specific heat at constant volume into temperature because we know that d by by d of temperature is equal to CV CVT is the value and CV we have seen earlier can be written as again we do that CP minus CV is equal to specific gas constant CV is equal to therefore r divided by gamma minus 1 therefore I I get the value of CVT as equal to RT by gamma minus 1 or this is equal to P0 and the ambient temperature is T0 this is equal to P0 by rho 0 and therefore this part CVT I can replace by the value of P0 by r0 is RT0 into 1 over gamma minus 1 I carry this forward here and therefore CVT can be replaced by or E0 can be replaced by the expression 1 over gamma minus 1 into P0 by rho 0 and if I were to solve this particular expression over here what is it I get I get E0 I bring P0 over here in the numerator divided by what is it I get well rho 0 rho 0 gets cancelled over here I I get 4 upon 3 pi into rs cube is all what is left over here is equal to 1 over gamma minus 1 have a portion the terms E0 by P0 is what we call as cube of the explosion length E0 by P0 to the power 1 by 3 energy release in the medium 1 by 3 is equal to r0 and therefore what is it I get I simplify this and I get the value of rs by r0 cube bring it on the right hand side over here must be equal to I get the value of gamma minus 1 divided by 3 comes on top 3 into gamma minus 1 divided by 4 pi 3 into gamma minus 1 divided by 4 pi is equal to this particular value and what is the value now I am able to find out the value of rs by r0 at which the value is such that if rs by r0 is greater than this value this value let me get the precise number for you the value of rs by r0 is equal to put the value 3 into 0.4 divided by 4 into 3.1 into 4.2 which is equal to 0.095 or rather the value of rs by r0 taking the cube root comes out to be 0.46 and therefore what is it we tell ourselves if the value of rs by r0 is greater than this particular value well my predictions are all wrong because the initial internal energy begins to play a role and predictions will be much lower than this but we also found when it is 0.4 the mark number of the shock has already decreased to a value around 2 or something and it is still in the weak blast region the predictions what we have done so far are therefore valid only for values around let us say less than around 0.25 or 0.3. This is where I conclude today and to be able to put things in perspective what is it we have done we have looked at the energy conserved by the lead shock wave we have been able to get an expression for PS that is the over pressure behind the over pressure of the blast wave and so in addition to this we talked in terms of Kranz-Hopkinson scaling law we talked in terms of Sack scaling law which is able to give us this expression we also did in the last class the mass balance which told that whenever I have a strong blast region well the mass gets concentrated at the front and this mass concentration is something like a hammer or something like a snow plow which moves my material at high velocity and does the damage. Therefore in the next class what we look at is we are looking at only the strong blast region we would like to take a look at the weak blast region such that we can have some expressions for this but in the weak blast region it is somewhat more difficult because as we tell ourselves the initial conditions are dependent on Mach number therefore we look at the weak region and get the overall that means in both the strong blast region and the far field region in which the blast wave has decayed we will try to get the over pressures and in addition to this we will also take a look at impulses and with that we would be done with the blast waves well thank you.