 Good morning. You know you will recall that so far what we have done is to determine when some energy is deposited at some particular point let us say E0 joules. What is the magnitude of the over pressure which is formed in a blast wave which is generated by this particular energy release. Therefore, what do we do? We said well a blast wave gets generated and it keeps travelling and we considered this the particular case of a spherical blast wave issuing from some energy release over here. We said well the ambient pressure could be rho 0. Let us make a plot over here. The ambient pressure could be let us say P0 the pressure over here versus the distance it travels from the center. The ambient pressure is P0 and supposing the blast wave comes here there is a rise in pressure. The rise in pressure is Ps and the over pressure or the excess pressure behind the wave to the ambient pressure is Ps minus P0 and that we non-dimensionalized with respect to the P0 and this we called as the non-dimensional over pressure. We determined this value as a function of distance and how did we do that? See mind you when some energy gets released some blast wave gets generated. If more energy gets released well the same strength of the blast wave is formed at a larger distance. Therefore, we scale the energy release in terms of explosion length namely a characteristic length r0 and we express the distance rs may be over here the blast wave let us say comes over a distance rs. We non-dimensionalized rs in terms of rs divided by r0 that means we have the scale distance which we called as shock scaling and we were able to determine what did we determine? We said as rs by r0 increases we have Ps minus P0 divided by P0 which keeps falling down which rapidly changes. Well we did this for a strong blast or a strong blast wave. What do we mean by a strong blast wave? In the field near to this that means in the region very near to the source wherein the Mach number of the shock wave which is formed is quite large may be typically greater than around 4. Because when it was greater than around 4 what happened? We had expressions like we had the density behind the blast wave or behind the shock wave divided by r0 was given by gamma plus 1 divided by gamma minus 1 because the term like we had in the denominator over here 2 over ms square since ms is a large number this tended to get cancelled and we got the density similarly we had P0 or the pressure behind the blast wave or let us say the Ps divided by r0 into rs dot square was equal to 2 over gamma plus 1 and the term which had gamma minus 1 gamma plus 1 into gamma into 1 over ms square got deleted got diminished because it is quite a small number therefore it was given by this and similarly u over rs dot was reverse of this. Therefore they were independent of the Mach number and being independent of the Mach number we were able to say at distances close to the source around let us say rs by r0 between let us say 0.3 to 0.4 for which we evaluated the Mach number we said well I can calculate the value of P that is the over pressure that is the pressure behind the blast to the ambient minus the ambient pressure divided by this and this is what we called as the dimensionless over pressure. Let us write the expression what we got we got P minus P0 that means the pressure behind the wave let us denote it by P s or simply P minus P0 divided by P0 we got it equal to 1 over 2 pi into the integral that is the integral we said that at the front well these values do not change with Mach number we said well the slopes behind it are about the same we evaluated the integral which we said was proportional to the energy deposited divided by the kinetic energy of the medium provided it travels at the total as the shock velocity we got this as 0.423 into gamma into we had 1 over rs by r0 cubed and since it is minus P s by P0 minus 1 well this is minus 1 over here. Similarly we had the expression for M s square that is the Mach number of this particular blast wave which propagates out as equal to we had something like 4 pi into i into gamma into 1 over rs by r0 cubed here it should have been gamma plus 1 over rs here it is gamma and this is the way it decays out. Now when we evaluated these values we got for a value this particular value came out to be something like 0.156 because pi is 3.142 i was 0.423 gamma was 1.4 this became a value something like 0.156 divided by rs by r0 cubed minus 1 over here. Now when we got these over pressures what did we find well in the region well let us say that the blast wave starts here it propagates out we got the value of P s that is the behind the shock front we got the P s divided by P0 divided by P0 we got for an rs over let us say 0.2 this we say is rs over r0 for point 2 we got a value around 19 maybe for point 3 we got a value around 5 or 6 or around 8 or so for a value around 4 we got a value around 4 or 5 that means it kept on decreasing and in these cases we said well the Mach number in this case was still high the Mach number in this case was very high the Mach number in this case was around 5 or 6 no around 6 this was around 4.5 and therefore we could get the value of P s by P0 well we could calculate these values but what is it we are talking of we are talking of values which are still high let us calculate the value of pressure behind the blast wave well we have something like P s minus P0 divided by P0 at the scale distance of point 2 equal to 19 and we therefore in the blast wave what happens from the ambient pressure it jumps over here therefore what is this magnitude P s minus P0 P s minus P0 is equal to 19 times the ambient pressure well the ambient pressure is 100 kilo Pascal or point point 1 mega Pascal and that is equal to something like 1.9 mega Pascal that is 19 times atmospheric pressure that means a very very strong or a very high pressure ratio is there across even if you consider these values it is quite high but in practice what happens is in practice what happens is let us say we have a glass in a building that means let us let us talk of this well we have let us say a house you have a window you have a window pane the window pane where if supposing a blast comes and hits the window pane the window pane ruptures when the pressure that is the rise in pressure minus the ambient pressure is of the order of 1 kilo Pascal itself the glass pane ruptures therefore we are talking of P s minus P0 mind you this is the shock pressure minus the ambient pressure we are talking of P s minus P0 is equal to 1 ambient pressure we said is point 1 mega Pascal 100 kilo Pascal that is divided by 100 is 0.01 well you know we are talking of high values when we are talking of these equations which we derived we are in in practice we are interested in numbers which are quite small small over pressures can do significant amount of damage and when I try to extrapolate over this when I try to say well I use the same formula and go forward well I cannot solve for these conditions let us say I am I am interested in solving let us say at r is over r naught of let us say 0.6 if I say 0.6 and calculate the Mach number well the Mach number will come out to be even less than 1 which is just not possible because the validity of the equations is only in the strong blast region why was it in the strong blast region we told ourselves well the assumptions were well you are you are the energy which is distributed by the blast wave from the explosion namely e is very much greater than the ambient energy and this is valid that means if I if we write you will recall again we had P s minus P0 divided by P0 into r s by r naught well it kept falling well the validity as per this is only 0.46 but if you look at the strong blast assumption namely m s greater than 0.4 it is only in the region between let us say 0.35 or so 35 to 0.4 that is 0.35 to 4 therefore only in this region whatever we derived was useful I cannot use it for weak pressures like I am interested in the value of maybe a glass rupturing a window pane or a glass breaking which the value is going to be 0.01 which is very much in excess and we also told ourselves I cannot really use this formula when the ambient energy is of the same order as the energy contained or the internal energy contained within the blast wave 0.1 and in these cases the value of the shock from Mach number is low what happens let us go back to this particular equation what I wrote over here let us say P s by rho naught r s dot square let us let us just write it if I write P s divided by rho naught into r s dot square then becomes 2 over gamma plus 1 minus what is it we get let us go back and see what did we get gamma minus 1 divided by gamma plus 1 into gamma into 1 over m square if the at large distances at large scale distances well m s has come down therefore if m s is small this term dominates or this term also plays a role and therefore I will get P s being affected by this when m s is very large well this is negligible and it is just a constant whereas if m s is small well this also begins to play a role and I cannot use the scaling law for which we assume that P s by rho naught r s square is equal to this and we said well irrespective of Mach number I get the value of f which was equal to P at any distance away from the wave divided by rho naught into r s dot square was given in terms of r by r s we had this particular curve and therefore it becomes essential to take care of the m s values similarly I can argue in terms of rho by r naught but stopping over here and proceeding we tell ourselves well not only in the case of glass shattering let us take a take an example when let us say I am standing near near an explosion and let us say my ear sort of is I hear a strong shock wave my my eardrum gets ruptured when the value of P s minus P naught is of the order of 30 kPa I will look at this problem a little later because I am just looking at the total pressure which ruptures the ear we will let us say it is still 30 kPa but we will take a look at reflected pressures after a class or two and you know if we look at this particular value I am talking of P s minus P naught divided by P naught ambient pressure is 100 kilopascal this becomes 0.3 even in the case of maybe our eardrum getting ruptured due to a blast wave I am talking of much lower values and therefore it becomes necessary that I evaluate the properties far from the site of the explosion that means in the near field I have been able to derive these expressions but I need to get the values in the far field also in the far field the value of r s by r naught will increase and therefore these values the both the over pressure dimensionless over pressure and the Mach number will come down. Therefore how do I do it I have to take into account the value of m square but let us go back and take a look at what is the expression I had for the Mach number we had m square is equal to 1 over 4 pi i gamma into r s this are all constants pi gamma is 1.4 i we said 0.423 I can write m square is equal to a constant divided by r s by r naught cube in the near field wherein I have a strong blast that means the value of the Mach number is high but in the far field what is going to happen m square is going to start affecting my blast and therefore people have been working at it the problem becomes much more complicated and what do you do you do either a numerical solution or you do you can you can weakly bring this in how do I do this let us say I write eta is equal to 1 over m square when m s is a very large number eta becomes a small number and therefore what I write is I write r s from this I can write r s by r naught cube for a strong blast I can write it as equal to let us say a eta that means I just write r s by is equal to c by m s square when the value of the Mach number begins to play a role well the values are not that small over here I can write it as b eta square plus c eta cube maybe I have a series expansion in terms of m s square and such type of analysis has been done taking into account the value of the rise in pressure as a function of Mach number rise in density as a function of Mach number at the blast wave front and when this is done the expression which was got was something like let me put that down it was equal to r s by r naught cube is equal to we had 2.362 eta plus c by m s square plus 4.531 into eta square plus 6.45 into eta cube this was done by by professors Bach and Lee at Megal University and you could have something like the like the dependence of the scale distance as a function of Mach number typically for Mach numbers as low as let us say 1.5 to 2 it sort of matched well and therefore if such asymptotic series could be used what I do is I want to find out the value of m s square at the value of r s by r naught I can solve this equation at given value of r s by r naught and I can calculate for a given value of r s by r naught I can calculate the value of m s square and once m s square is known I can calculate the value of p s by p naught is equal to I use this particular expression again which we derived for a constant shock strength namely we had I had just written the value somewhere over here yes p s by r naught r s dot square is equal to 2 over gamma plus 1 minus gamma minus 1 divided by gamma plus 1 gamma into 1 over m s square since m s square is is somewhat now quite small this this factor plays a role and therefore I can convert this into p naught as equal to 2 over gamma plus 1 and since I have to put it in terms of p naught well I have the gamma and sound speed coming over here 2 gamma m s square minus I get the value of minus gamma minus 1 divided by gamma plus 1. Therefore for a particular value of r s by r naught using this asymptotic type of series I can calculate the value of m s square and using m s square I can calculate this value and get the extension of my curve from a point somewhere over here till the Mach number till much larger value of r s by r naught well we must also remember that when we were doing all this scaling we were also talking in terms of let us say this is the time axis this is the distance from the point of explosion r s we said well initially it is a strong blast that means Mach number is high in the near field then as we proceed it becomes it weakens ultimately it becomes an acoustic wave. In the limit at which it becomes an acoustic wave that means we are talking of the Mach number around 1 we use the acoustic theory and using the acoustic theory I can again calculate the value of p s by p naught and therefore get the value of p s minus p naught by p naught and one such acoustic theory which we will not develop here but we will just presume is what is given as p s minus p naught divided by p naught let us write it out in the limit that the blast wave is approaching a value of around 1 is given as a gamma divided by r s by r naught into b plus I have long r s by r naught to the power half where the value of a is typically around 0.23 to 0.246 while the value of b is equal to 0.36 to 0.45 that means in the far field where in the blast wave decays to an acoustic wave well the over pressure is derived from the acoustic theory and now putting the far field and the near field together that means I have r by r naught that is the distance r s by r naught the value of p s minus p naught divided by p naught what is it we say well then the near field we derived the expressions as we move away the Mach number effects begin to go and the shock further decays over here and in the very distant field wherein you have the wave becoming almost an acoustic wave well it is like this and we get the composite over the entire thing this is the near field this is the field wherein m s effects begin to shape up the effect of internal energy of the free medium begins to shape up and this is where you have the acoustic effect. Therefore this is for right up to 1.01 1.02 I can predict using such equations and all these were put together by Professor Strahlov and the type of figure which we got was something similar to what I show in the slide which is shown on the which is shown here what is it we see well you calculate and this is shown by the dotted line over here and what is it we see this is the strong blast region initially I am just going to take a look at the dashed line over here we say in the strong blast region I have this particular curve coming over here and maybe this is followed by this effect of the Mach number decaying the energy release and then subsequently being followed by the acoustics over here this is the type of figures which are normally available but you know in this figure I also show some experimental points I show well nuclear explosions if they take place they seem to tally quite well with the with what we have been doing so far and now I need to qualify this and how do I qualify it let me get back to the board and see whether we can understand this in in some better way all what I tell now is well I have been able to get this figure which I say is for an explosion and what did we consider we consider when some energy is released at a point let us say our energy is released we told ourselves well as spherical wave propagates and spherical blast wave propagates out we did not really consider the dimensions of this particular point which is getting liberated at which the energy is gets liberated in other words we never even made a mention well the volume at which the energy gets liberated is so many cubic centimeters or cubic meters and so on we just assume that a wave gets that that means we are we are not even giving any credence to the volume we just said well it could be a point release at this particular point we are releasing energy is zero joules but in practice we all know well you know if I have something like a sphere of a given volume I keep on pressurizing it this has a particular volume and I keep on pressurizing it at very high pressure well the case burst and I have an explosion taking place a blast wave propagates out you know this could be like this it could be a cylindrical it could be a square or it could be a vessel which is rectangular which I pressurize and therefore in this case I get sphere in this case in the neighborhood I will get something in the near field which is a representative of this it could be planar waves which are getting generated if I consider may be a cylindrical vessel well I could also get waves which are cylindrical and it just moves forward from this as a cylindrical geometry we did talk about some of these things earlier why we talked in terms of the explosion at the Boston Marathon earlier this year what was there in that what happened there was this particular pressure cooker in the pressure cooker some energetic material was kept over here and it exploded and therefore you had a given volume and therefore the explosion would the blast wave in the near field would be something similar to this and later on we may be we must see what is the shape of the thing but the point I am trying to make is well I have a finite volume in the finite volume if an energy is released what is the effect on the on the value of the over pressures and for that we try to again tell ourselves well we simplify the problem we tell ourselves well I am interested in say a spherical volume a sphere let us say the radius of the sphere is RE and the energy gets liberated in this particular volume this is where now I say the energy is getting liberated and the energy and this volume now ruptures and I have something like a spherical blast wave which moves out from this volume it moves out over here and how do I now model this and now I have a finite volume can I say something about in the near field far field how the explosion will behave therefore let us again put it down we say well the energy is released is zero joules is liberated in the volume what is the volume I say a sphere of radius RE therefore I have four upon three five into RE cubed is the volume well how do I liberated maybe I put some combustible it increases in pressure and temperature and therefore the when the internal energy per unit mass is let's say E it explodes therefore I am saying well the density at the point of explosion is row at that point the energy is E over here and what can I now write some equation for this you know E energy is equal to we say per unit mass is equal to it is at constant volume you have CV into T is the energy per unit per unit mass over here well we already know that CV can be written in terms of gamma and specific gas constant that is CV is equal to R over gamma minus one therefore I can write the energy per unit mass of the thing which burst over here is equal to R into the temperature at which it burst into gamma minus one PV is equal to MRT or P by row is equal to RT this becomes P by row into one over gamma minus one we have been doing this such type of calculations earlier and therefore if I were to write the value of the energy which gets released let's let's put that down we get therefore E naught is equal to 4 upon 3 into pi into RE cube into row into P by row into one over gamma minus one or rather row and row get cancelled and this is the pressure of the explosion let me call it as PE to distinguish it from the ambient pressure P naught and therefore I get the value of E zero is equal to 4 upon 3 into pi RE cube into P divided by gamma minus one let us divide both the sides by the ambient pressure and if I do that I get E zero by P zero is equal to I get 4 upon 3 pi into RE cube divided by gamma minus one into the value of PE by P naught or now but we know that E zero by P zero is what we called as the explosion length cube we define E zero by P zero to the power one by three was equal to R naught therefore this becomes your R naught cube and this becomes your R naught cube and therefore what is it I get I get the value of RE by R naught cube is equal to I take this on this side and I bring the other terms on the left hand side I get 3 into gamma minus one divided by 4 pi 3 gamma minus one divided by 4 pi into P naught divided by PE in other words we get an expression for the size of your vessel that is the radius of the vessel divided by R naught is given by this particular expression and taking the value of gamma is equal to 1.4 pi is 3.14 I get the magnitude of this and if I solve for this what is it I get let me get you the precise values the values are the 3 into 0.4 divided by 4 into 3.142 this becomes equal to 9.55 into 10 to the power minus 5 10 to the power minus 5 of P zero by PE or rather what is it I am looking at I am looking at this value therefore I can write the value of RE let me use the other part of the board or rather I get the value of RE by R naught is equal to I get 9.55 into 10 to the power minus 5 to the power 1 by 3 into I have to get the value now I get P naught by PE exactly what I have got over there is what I have written over here and this comes out to be equal to 0.046 into well this must also be one third because RE by R naught was this into I have P naught by PE to the power 1 by 3 in the limit now I tell myself in the limit where PE is a very large number maybe you know the the explosive is such that I have intense pressure pressure tends to infinity when PE tends to infinity well the value of RE by R zero is there that means RE by R zero when I have a high pressure explosion tends to zero well I am getting the point point explosion like what we have been considering we do not even need to consider the effect of RE when the pressure is very high but when the pressure rises let us say instead of saying very high let us say when the pressure a vessel could explode when the pressure is around let us say 100 MPa if PE is equal to 100 MPa that means equal to 1000 atmospheres the value of P naught by PE is equal to 0.1 is the ambient pressure 0.1 MPa divided by 100 that is equal to divided by 100 or rather when I take the cube root of this particular expression therefore I get the value of RE by R naught is equal to I get no I should not have written this the value here is equal to 9.55 into 10 to the power minus 5 to the power 1 by 3 as it were now when I when I put this value over here and I substitute the value of 9.55 into 10 to the power minus 5 to the cube root into 1 over 1000 to the power cube root I get a value around 0.046 in other words even if the pressure is something like 100 MPa instead of being infinity I get RE is a small number and therefore what is it I find I find well RE is very much less than R naught when the pressure is high and therefore we can consider these things where RE to be negligible when it is very much less than R naught and therefore whatever we derived for a for a for a blast wave progressing from a point source is valid as long as the pressure of your explosion is quite high. Let us put the whole thing together now that we have done this problem and ask ourselves some particular questions the questions we would like to ask under what conditions can I use what are the predictions which are done for point source we tell ourselves when RE is less than R naught well it seems to be valid because RE is something like 0.04 times when the pressure is high when it is less than this we also know that region of interest is Rs by R naught strong blast is something like less than 0.3 well RE by R naught was something like 0.04 whereas Rs by R naught is still less than R naught but still Rs is much greater than RE therefore we find that when RE is very much less than Rs at which we are interested and this Rs is again less than the value of R naught well for this we can say well the point assumption is valid and we can go ahead and predict over here but when we go back and look at the figures what I just showed on the slide we what do we find we find when we have PS minus P naught divided by P naught and we have Rs by R naught over here well the point explosion gave this we also had some some deviation over here for finite volume that means when some explosive is burst over here that means these are the effects due to the finite volume effects that means the pressures what we get may not be as high as we have assumed in this particular analysis. Having done that let us take a look at what is the effect of geometry like for instance I have a cylindrical vessel which explodes or let us say a rectangular geometry which creates a plane wave in the near front you know what is going to happen well in this case I am going to get a cylindrical wave which propagates out from this it is going to go maybe at larger times well I am going to get another cylinder over here it is going to propagate out but we also know because the pressure behind the blast front is quite high and the density is high the temperatures here are very high therefore the sound speed is very high and therefore the high sound speed will smear out these corners and therefore what is happening is these corners will get smeared out and in the as the blast wave propagates out in the near field maybe the geometry of the of the source will have an effect but thereafter the effects will get nullified because and ultimately in the far field I will get a spherical wave. Having said that let us go back and again address the slide what we had we said well we were able to predict the over pressure as a function of the ambient pressure dimensionless over pressure as a function of the scale distance distance scale by r0 we were able to get the point source when we have the effect of the volume coming here well there are some deviations these are the experimental points but other than that maybe as we go ahead the all the points match with this and this is what we use to determine the over pressure from the explosions sorry that means we have been able to get the effect of over pressure as a function of distance and let's do one or two small problems such that we are very sure about what we have learned so far let me do two model problems with that maybe we will be able to see how to predict the over pressures in an explosion let me consider one one one problem like this let us assume I release an energy at a point or at some small source as equal to let us say 70 megajoules of energy is what I release and the ambient pressure at which I release this energy is equal to 0.095 mega Pascal in other words we are talking of a situation wherein some energy is released over here all of a sudden instantaneously the energy release is 70 megajoules the ambient pressure at which this energy released is 0.095 mega Pascal 0.095 is very near to 0.1 mega Pascal and this might be at a slightly increased altitude and now I am interested in finding out what is the magnitude of the over pressure maybe I deposit this energy at a distance let's say 1 meter away from the explosion site I want to evaluate the value of the over pressure what is over pressure the pressure behind the wave divided minus the ambient pressure I want at 1 meter I also want to find out what is the value at let's say a distance of let's say 20 meters away and at a distance of 100 meters away that means I am interested in finding out the over pressure at a distance of 1 meter away from the explosion site at a distance of 20 meters away and at a distance of 100 meters away therefore this is the problem given and how do I solve it maybe I will make use of both the charts and whatever we learned so far we will try to do this problem and with that maybe we will know how to do such problems in practice well the first thing we say is well I convert this energy into something like a explosion length and explosion length R0 is equal to E0 divided by P0 ambient pressures to the power 1 by 3 in this particular case the energy released is equal to 70 mega Pascal that is 70 mega joules 10 to the power 6 joules the value of pressure is equal to 0.095 into 10 to the power 6 Pascal units please be careful about units this is joules therefore Pascal therefore you have joules is equal to Newton meter divided by Pascal is Newton by meter square therefore the numerator should be in joules P0 must be in Newton per meter square of Pascal you have 1 by 3 and this gives me the value of meter therefore I get the value of R0 this I simplify let me get the value is equal to 9.03 meters let me erase this out now I am interested what will be the over pressure at a distance 1 meter away therefore RS is equal to 1 meter at RS is equal to 1 meter the value of RS by R0 is equal to 1 divided by 9.03 which is around 9 that is 0.11 is the dimensionless value of the scale distance well you know we have been telling ourselves well at distances less than around 0.3 whatever we derived is valid and therefore I can write PS minus P0 divided by P0 was equal to the equation we had reduced to the form let us first write the equation 1 over 2 pi i into gamma minus 1 into RS by R0 sorry should have been in the denominator RS by R0 cube minus 1 which we said again was equal to 0.156 divided by RS by R0 cube minus 1 and therefore I put in the value of RS by R0 is equal to 0.11 this gives me 0.156 divided by 0.11 cube minus 1 and this gives me a value equal to 117.2 minus 1 which is equal to 116.2 is the value of PS by P0 by P0 but my P0 is equal to 0.095 mega Pascal therefore the value of PS minus P0 is equal to I get 0.095 into 116.2 so much mega Pascal and this works out to be 11.02 mega Pascal this is from the strong blast assumption if I want to use the figure I go back to the slide I am interested in the value of 0.11 and if I take the value of 0.11 maybe I take RS by R0 0.11 over here well I get back something like 100 and if I look at this yeah it is something like 116, 117 is what I get and this is how I can use this figure or I can use since I am in the strong blast region I can as well use the equation which we derived in the earlier classes well this is how you evaluate at the distance of 1 meter away when I look at the distance of 20 meters away let us again do this problem well at 20 meters I am writing the value of RS by R0 is equal to 20 divided by 9.03 which is your R0 meter by meter this is equal to 2.2. Now you know 2.2 we said well the strong blast assumption is valid only when RS by R0 is less than around 0.4 or so because after that the Mach number drastically decreases I cannot use these equations which I derived I have to necessarily use the chart or do some more calculations to therefore let us use the chart which is available in the slide for this particular purpose therefore I take a look at 2.2 RS is equal to 2.2 well this is 1 2.2 is over here when I take the value well it comes to something like 10 to the power 0 something like 1.5 therefore I get the value of from the chart I get the value of PS minus P0 by P0 at the particular value RS is equal to 1.5 and therefore I get PS minus P0 is equal to 1.5 into 0.095 mega Pascal which is equal to 0.143 MPa which if it is expressed in kilo Pascal it is equal to 143 kilo Pascal this is the over pressure let us do the last part wherein I am also interested in the value of over pressure at 100 meters well at 100 meters what happens the value of RS by R0 is equal to 100 divided by 9.03 which is equal to 11 and at the value of 11 I again read the value and if I look back at the distance of this is 10 over here I am looking at 11 the value if I were to put down what is the value I get at 11 well I am somewhere over here something like I am having 10 I am having 11 over here if I get the value it is somewhere over here therefore the value I think will come out to be something like 0.11 let us put down the value 11 difficult to do it if I have if I were to use a scale I will say 0.1011 or 0.01 is the value of PS minus P0 divided by P0 and therefore in a similar way I get the value of PS minus P0 is equal to this is from the chart I get the value as equal to 0.095 into 0.01 which is equal to 0.00905 MPa that means we are talking of something like 0.95 kPa or something like 1 kPa which is what is the magnitude required to break a glass window pane or a glass window pane let us say therefore we are able to find out the over pressures using the chart and also using the strong blast assumption and we do such problems. I take the case of trying to find out if the over pressure in other words if the over pressure at a given distance from an explosion say at a distance of 50 meters from the site of an explosion is let us say 30 kPa then and the ambient pressure is equal to let us say 100 kPa that is 0.1 MPa that means if the measured over pressure at a distance of 50 meters from the site of an explosion is 30 kPa and the ambient pressure is 0.15 0.1 MPa find the energy release in the explosion. How do we do this problem? Well in this particular problem it is given to me that PS-P0 is equal to 30 kPa the value of P0 is given as 100 MPa 0.1 MPa therefore I get PS-P0 is equal to 30 divided by 100 is equal to 0.3 is the value. Now I also know that this value is reached at a distance rs is equal to 50 meters away. Therefore I again I know well this is my over pressure this is my distance and therefore I use the chart again and I want to find out the value of the value of rs by r0. Therefore when PS my P0 is 0.3 I want to find out the value of rs by r0 I go to the slide again I find that the value of rs by r0 when it is 0.3 this is 0.1 0.3 somewhere here somewhere over here you get the value around 1.5 or so this is 0.3 comes out to be something like 1.5 let us say rs by r0 is 1.5 and therefore what is it I get now I get the value of r0 from this expression as equal to 50 divided by 1.5 is it alright let us again do this the value is 0.3 at 0.3 the value of rs by r0 is 1.5 r0 is equal to 50 by 1.5 which is 33.3 meters is the value of r0. If r0 is 3.3 meters well the value of E0 by P0 to the power 1 by 3 is equal to 33.3 and therefore I know P0 is equal to 10 to the power 5 Pascal that is 0.1 MPa I can find out the energy release and this is how we go about doing the problem. Therefore what is it we have done so far you know by now we must be clear about how to do these problems involving over pressures in a blast wave and what is it we have done we started with a strong blast assumption we just said well it could be extended to lower Mach numbers and we use the particular curve what we had in which case we had over pressure on the y axis Ps by P0 divided by P0 as a function of rs by r0 we use these curves in the strong blast region in the weakened region to be able to predict the over pressures if over pressures are available I can predict my energy release and this is what we do in blast waves. Having said that let me spend a moment or two in the last part namely about impulses with which we will continue in the next class. We will just introduce the subject and carry forward in the next class and what is the impulse you know when we talk of over pressures what we are basically considering is we are having a blast wave which is created and the effect of blast wave is to create something like a pressure which crushes but we also told behind the blast wave because of the varying pressures you have the wind that means you have something like a blast wind and because of this blast wind only it is called as a blast wave because it creates that wind and what creates the wind well let us assume I have an explosion over here the blast wave is travelling forward maybe I am standing over here and watching a particular point over here at which a wave comes and let us say once the explosion is initiated the wave comes here after a particular time that means I have my time axis over here after a particular time the wave comes here it is the arrival time of the wave and once the wave comes till the wave comes the ambient pressure is just the ambient pressure which is P0 which is equal to let us say 0.1 MPa once the wave comes well the pressure shoots up I have the over pressure comes down to PS and the wave continues to move forward when the wave moves forward at this particular point the gas is expand but it is still processed by this and therefore the gas moves forward the pressure falls and if by chance the pressure falls below the atmospheric well it could come over here and it could go over here we say this is the time at which the pressure reaches the ambient due to the expansion is equal to Ta plus T plus and it comes back over here into Ta plus T plus plus T minus well this is the duration at which I have the wind blowing in this direction or that means I have from the high pressure gases it blows outward over here this is the value of T plus and in this region I have the wind blowing inwards because the pressure is there I have T minus over here in other words I have a region having positive pressure above the ambient I am having negative pressure over here this is in an ideal case of let us say wave which is travelling forward and what happens how do I define the wind condition well the particles are carried forward I have something like a momentum when I have the momentum of the gases you know initial momentum is when it is at rest I have change of momentum and what is rate of change of momentum we had talked of this earlier rate of change of momentum is equal to I get d by dt of the movement momentum change what is available and this is equal to the impressed force because of this pressure if I cut off in terms of unit surface area well for unit surface area I can say well force is equal to pressure by area or rather I say rate of change of momentum or rate of change of momentum per unit area is equal to what is the difference between the excess pressure and the ambient pressure that means it is equal to something like Pt minus P at any time over here minus the value of Pa over here or the this is the rate of change of momentum per unit area therefore the change of momentum and change of momentum is what we call as impulse therefore impulse per unit area is equal to I have to integrate this expression into Pt minus Pa into the value of dt therefore I have something like a positive impulse over here because the Pt is greater than Pa I have a negative impulse because the Pt is less than Pa I can define the wind effects in an explosion as being let us say we have positive impulse I call it as positive impulse per unit area I have negative impulse per unit area and this is given by integral 0 to t of Pt minus the pressure at any time divided by minus the value of the ambient pressure into dt and what is the time from the arrival time that is ta to the value of ta to t plus is the value of positive impulse and what is the negative impulse we have ta plus t plus is the starting over here it finishes off at ta plus t plus plus t minus into the value of P in this case it is this is higher therefore I am just looking at the negative impulse it is P0 minus Pt into the value of dt and therefore I say specific impulse per unit area the unit being force into distance Newton second per meter square is given for positive impulse and for negative impulse by this particular expression in the next class what I do is I will not do the details but since we know the rate at which pressure is changing behind the explosion I know these profiles I will try to find out the values of the positive impulse and negative impulse and then discuss further how the effect of an explosion is to damage objects around it well thank you then we will see in the next class