Ladies and gentlemen, we are now back at the live stream and we are so fortunate to be able to interview Philips Röder. So I have some questions that we have received from the economist from the Eastern European office that we're going to ask Philip and see what his responses are. So Philip, how will the European market be affected if Britain leaves EU, if we have this Brexit as it's called? It's going to be a disaster for two reasons. I think it's worse for the UK to actually exit the Common Market in a sense. But what's really scary is that it might lead to disintegration of the entire European concept and nations like Denmark, smaller countries like Denmark or Bulgaria, that are actually really dependent on a single well-functioning market and if that disintegrates, it's a disaster for businesses across Europe. And why is it that the British people cannot see this? Why is it that they believe that Brexit is the right thing for them, do you think? I think it's a big mismatch and the Brexit debate is really that the UK as an economy understands that it matters to be part of a larger market. But they're also offended and scared of some of the other trends like migration, like tax evasion, other issues where it feels it could be comfortable to regain control of certain policy areas. But having said that, what you sort of overlook in the British debate is and the UK debate is that there's a high price to pay. That is future growth in the UK, which will slump. We're going to see a huge devaluation. We're going to see severe consequences for the UK economy. So how do you believe that the EU and the UK now should resolve this? How should we solve this issue? Well, first of all, it remains a horror movie until the vote is cast. In the case of basically a stay vote, this will still be damaged to Europe because it sort of has emphasized the ideas that you could exit. In the case of a no vote, we're going to see that major devaluation in the UK. We're going to see shop waves of economic turmoil across Europe. And worst case, in a country like Denmark, we might start a debate should we also opt for an exit? Because it's so easy to build anti-European arguments nowadays and it's so tough to argue correct the economics and business case for integrated Europe. Yeah. So one of the reasons why they want to leave the EU is because they believe that they can also have this free trade agreement with the United States. And currently, EU and United States are also trying to find a free trade agreement, the TTIP. So how do you think that this is going to fix into the problem? Honestly, I think that's a bit of a wishful thinking because if you think about it, the biggest challenge that you have in terms of integrating the UK economy is not free trade and goods. It's free trade and services. And it's unsinkable that the UK should be able to negotiate a proper service free trade, let's say with Europe, let alone with the US. And the structure of the UK economy is very service heavy. Just think of the financial district in the city of London. So basically why would anyone in Frankfurt accommodate special regulatory requirements from the UK? Obviously, if you would be a Frankfurt-based financial institution, you would do everything possible to avoid having the UK competition on that market. So to remain or gain access on services just like financial services, it's impossible in Europe and it's going to be even more difficult. Another critique of the European Union from Britain is that they simply put too much regulation on the standards of products, for instance, whereas in the US they have much lower standards. So how far do you believe that the Europeans should go in regulating standards for products in the European Union? I think the way forward in terms of regulation is a mutual recognition. So whenever US authorities say this car is safe for road transportation in Europe, we should say the same. And mind you, in one area there is this mutual recognition and that is in air traffic. As a matter of fact, an airplane that is allowed to land and start in the UK, sorry, in the US is automatically allowed to start on land and be operated in Europe. So if we can agree on mutual recognition in air traffic, why can we not agree on mutual recognition on the way seat belts are fastened within cars in the way that food products should be dealt with or let alone how insurance and financial systems should be integrated. So I think the biggest dilemma here is that there are special interests and fighting down special interests is much tougher than cutting tariffs. Thank you, Philipp. And that was all from the live stream here in the break. We'll see you when the participants arrive and when we'll have the panel debate. Thank you from here.


So the last topic that we are going to take up in this course is detonations we will spend some time talking about detonation waves the structure of how we have proceeded is as follows we went through the Rankine-Hugonew relations at some stage where we pointed out that the Rankine-Hugonew curve breaks up into the upper branch and the lower branch and the lower branch corresponds to the deflagration of waves and the upper branch corresponds to the detonation branch and with that we then we then said we will now focus on deflagrations right. Now the major problem that we had with the Rankine-Hugonew relations was that we were assuming that there is a inherent flame speed associated with it with which we could construct the Rayleigh line because the Rayleigh line the slope of the Rayleigh line is equal to minus m dot squared where m dot is the mass flux of the flow that is passing through the wave and that relates to the flame speed basically. Therefore by constructing a Rayleigh line we presume that we know what is its slope and that means we know what is the wave speed but in reality that is that turned out to be the Eigen value in the case of deflagrations and we had to get into the structure of the deflagration wave in order to try to find out the flame speed there and so a similar situation remains for detonations as well so we will have to think about that at the moment so but basically I wanted to the bottom line I mean or the commonality or the underlying thing about these things is we are primarily looking at premixed reactants right so that was the framework in which we did this so from doing laminar deflagrations which is essentially plain premixed flames we then moved on to diffusion flames and so on and not strictly in the framework of the Rankine-Hugonew relations but then we now going through all those things we now come back to the detonation part of it and then we ask the first question what is the detonation wave speed right in doing this we recognize that we noticed at the time that in general even if you had what is called as strong detonations they approach the Schaff and Juge detonation unless the system is over driven and weak detonations seldom occurred because it required highly reactive mixtures for the reactions to occur in such a high rates that you have the products that the reactions that are following the wave should happen at such high speed that when the wave propagates at supersonic speeds you now have the products that are following at supersonic speeds as well although not as fast as the waves themselves this is the wave itself so we are not necessarily concerned with weak detonations and we notice that strong detonations tend to become Schaff and Juge detonations therefore it it makes sense for us to confine our attention to the wave speed of the Schaff and Juge detonation wave so that is what we will do now in doing this unlike in the case of deflagrations where we notice that the wave speed is actually an eigenvalue of a eigenvalue problem looking at the structure of the wave where we have to do the mass and energy balances across the or through the wave rather here we notice that the deflagration wave sorry the detonation wave is traveling at supersonic speeds so it quite does not really know what is ahead of it the information does not propagate upstream therefore in a it should be possible for us to actually try to get the wave speed without having to get into the structure so in the case of deflagrations we had this heat conduction that was happening upstream from the reaction zone that was heating up the reactants whereas in the case of detonation wave the detonation wave keeps going and it does not really know what is ahead therefore we cannot really hope to actually look at the structure in order to resolve the wave speed so there must be a trick that is involved somewhere here and I will highlight this trick as we go along but we will pretend that for the moment that we do not have to necessarily go through the structure we will just keep going looking at the kind of Rankine analysis similar to Rankine-Hugonio analysis and then see where we can try to exploit the nature of detonation to look for the wave speed so for the Chapman-Juge wave or particularly Chapman-Juge detonation in this context the product speed is basically the speed of sound so unlike for example in a normal shock where the downstream velocity is always subsonic when the upstream velocity is supersonic in the case of a Chapman-Juge wave the downstream velocity is always sonic locally at the speed of sound so the speed of sound behind the detonation wave is we use the same notation as what we had before a infinity equals square root partial derivative of pressure the downstream pressure with respect to the downstream density at constant entropy constant downstream entropy now here u infinity equals a infinity at the upper Cj point in the Hugonio curve so we have we have m dot equals rho 0 u0 equals rho infinity u infinity which is now equal to rho infinity a infinity so from here u0 is equal to 1 over rho 0 rho infinity a infinity now now a infinity is equals square root of gamma infinity or infinity T infinity so u0 equal to rho infinity over rho 0 square root of gamma infinity or infinity T infinity right so let us call this relationship number one so here what the problem is u0 is actually what we are looking for this very similar to what we did for the deflagrations okay that is related to rho 0 also very similar to deflagrations because there also we were actually trying to look for a relationship for m dot versus the rest and m dot is rho 0 u0 but here what we what we now find see this is the problem what we find is the wave speed which is essentially in a flame fixed coordinate system or a wave fixed coordinate system it is essentially u0 depends on the downstream quantities right rho infinity gamma infinity or infinity and T infinity and that is the problem so we now have to try to find out if we can evaluate those quantities right so so note that the wave speed which is essentially u0 depends on depends on the burn gas properties as a matter of fact the we also know that for example in the lower CJ point as well u infinity is equal to a infinity right and m dot being equal to rho 0 u0 equal to rho infinity u infinity is valid regardless of where we are on the Hugonio curve right so if you now plug in u infinity equal to a infinity that is valid for both upper CJ and lower CJ points so strictly speaking this could be a CJ detonation wave or CJ deflagration wave at this moment right so we haven't quite strictly speaking exploited the fact that we are particularly interested in the detonation wave at this stage so keep that in mind and look for where we are doing this so what we want to now try to do is to explore or chase the valuation of the burn gas properties so let us let us try to do that one itself implies that rho 0 squared u0 squared equals m dot squared equals gamma infinity p infinity rho infinity you can actually get this quite easily now in the Rayleigh line we have p infinity minus p0 divided by 1 over rho infinity minus 1 over rho 0 equals minus m dot squared and then we plug this in here right so from this we can get 1 over rho 0 minus 1 over rho infinity we swap this so that we can get rid of the negative sign in the m dot squared so that would be p infinity minus p0 divided by gamma infinity p infinity rho infinity that sets coming from here right now let us call this to and this would be pretty interesting and important at some stage in the future from now on what we are going to do something that could be a little bit boring alright but we will come back to this and say well we can forget about everything that I said which is kind of complicated and boring but we can just go back and look at this and then see if we can exploit something here so let me just go through whatever needs to be gone through and then we will come back to this so here what we want to do is multiply above by p0 plus p infinity so we have p0 plus p infinity times 1 over rho 0 minus 1 over rho infinity equals we now get p infinity squared minus p0 squared divided by gamma infinity p infinity rho infinity so why did we do this this is where we are now going to make the point that we are looking at a detonation wave we have already made the point that we are looking at a Chapman Juge wave right and we pretend that we do doing detonation but not quite yet until now right so what we are what we want to do is to notice that across the detonation wave the pressure increases manifold right so many times and therefore p infinity is larger than p0 quite larger than p0 actually right now what we don't want to do at this stage at this step is to say that p infinity is much larger than p0 so I used the words carefully there I said quite larger than not really much larger than so if you now say quite larger than right then we can we can then easily say that p infinity squared is much larger than p0 okay p0 squared so what we are going to right now do is to say suppose p infinity squared is much larger than p0 squared this is much less of an approximation when compared to directly supposing that p infinity itself is much larger than p0 okay so this is this is alright this is this is okay but what we are going to do what we what we will find is this is not going to really help us directly it is going to lead us to go into some circles and then we will come up with the iterative scheme of solving this right so that is the boring part that I was just talking about and then we go through all that and then I am going to come back and say wait for an engineering purpose can I relax and say I do not want to actually multiply by p0 plus p infinity here and then get this p0 squared to be considered to be much larger than p0 squared directly into can I say if p0 it is p infinity itself is much greater than p0 okay instead of looking at the squares I will do that quite some time later okay but this is the point where we are beginning to talk about detonation so it is essentially the idea and this allows us to get around the structure of the wave okay by doing by recognizing this we do not have to bother about getting the structure to unlike unlike we did in the deflagration but there the physics demanded it okay so the the upstream conduction demands that we consider the structure of the wave here we did not have to do that instead we actually short circuited by going through this approximation so now let us suppose we will now continue to assume that p infinity squared is much greater than p0 squared and that is going to imply that p0 divided by rho 0- p0 divided by rho infinity plus p infinity divided by rho 0- p infinity divided by rho infinity that is opening up these parentheses here that is equal to p infinity divided by gamma infinity rho infinity the pins we have got rid of the p0 squared so we remind with p infinity squared one of the p infinity p infinity is there gets cancelled with the one at the bottom so you get only this at the bright end side and so you use the equation of state r0 t0- rho 0 divided by rho infinity r0 t0 plus rho infinity by rho 0 r infinity t infinity- r infinity t infinity that means wherever we have a p infinity by rho infinity we use an r infinity t infinity p0 rho 0 p0 over rho 0 we use r0 t0 but when you have this mixed kind of quotients we now take the ratios of densities and then use the equation of state wherever applicable and that is going to mean that on the right hand side we also write r infinity t infinity divided by gamma infinity so multiply this is the kind of boring part I was talking about so it is may basically some algebra that we go through rho 0 divided by r infinity t infinity and that is going to get you a quadratic you might recall doing something like this with the Hugonio curve and the Rankine the on the Rayleigh line put together earlier on to get the upstream Mach numbers but here our goal is slightly different so we get rho infinity over rho 0 the whole square minus 1 over gamma infinity plus 1 minus r0 t0 divided by r infinity t infinity times rho infinity over rho 0 minus r0 t0 divided by r infinity t infinity equal to 0 let us call this 3 now this is a quadratic in rho infinity over rho 0 right now but you see what about of the coefficients the coefficients involves gamma infinity r0 r t0 r infinity t infinity right so those are the coefficients there so this solving this will give rho infinity over rho 0 in terms of gamma infinity r0 t0 r infinity t infinity so course we are still chasing that means we wanted to go back here and then say can I get rho infinity well you can but it is going to be in terms of gamma infinity r infinity t infinity we need to know those things as well right so that is okay so we are not introducing anything new we are you we are trying to count only the old things so that is all right so we can easily now so also use p infinity p infinity is related to rho infinity by rho 0 and t infinity by essentially the ratio of equations of equations of state at the product and reactant conditions so that is p infinity equal to rho 0 divided by rho infinity r0 t0 divided by r infinity t infinity times p0 so let us call this for the reason why we we number these equations is we will now go through an iterative procedure iterative iterative solution procedure the first thing we do is assume a p infinity we do not know what p infinity is but let us assume it and then I make a mistake right p equals rho or t so we need to have p infinity equals is that right I want to make sure I do not make that mistake but I did so assume t infinity 3 now next thing we do is calculate the equilibrium composition yi infinity based on p infinity and t infinity that is assumed that is you once you get to step 3 you have a p infinity and t infinity to work with which you now use the equilibrium composition there and then so once you do this you now use the energy balance and check whether this is going to resolve things so check for a infinity from the Higonia relation in fact we had the mass balance here and the Rayleigh line is a combination of statements of mass conservation and momentum conservation that is what we have been working with we have no used energy conservation yet so we will try to now try to use that as a check right so we will we will try to use the Higonia relation we had H infinity – H 0 equals one half 1 over infinity plus 1 over rho 0 times p infinity – p 0 and which means the sensible enthalpy part of it HS infinity – HS 0 so what we do is we now split this in a sensible enthalpy and heat of formation so the differences in the heats of formation of the products and reactants altogether is the heat release right so that is Q so that would be Q plus one half 1 over rho infinity plus 1 over rho 0 times p infinity – p 0 here then we want to use the internal energy that is but HS is sensible enthalpy is equal to internal energy plus p over rho therefore HS infinity – HS 0 equal to a infinity – E 0 plus p infinity over rho infinity – p 0 over rho 0 okay let us now substitute this up there substituted this in the Higonia then we get a infinity – E 0 equal to one half 1 over rho infinity plus 1 over rho 0 times p infinity – p 0 – p infinity divided by rho infinity plus p 0 divided by rho 0 plus Q so simplify this you have some opportunity to cancel things therefore we get E infinity – E 0 equals Q plus one half p infinity plus p 0 times 1 over rho 0 – 1 over rho infinity so here we can now go back and use this relationship number 2 and what you get you get a p infinity plus p 0 over here that is what we try to multiply but then if you now use 1 over rho infinity no not – 1 over rho infinity as this you naturally get a product of p infinity plus p 0 times p infinity – p 0 which will lead to p infinity square – p 0 square right so using 2 a infinity – E 0 equals one half p infinity square – p 0 square divided by gamma infinity p infinity rho infinity plus Q here again you now try to approximate saying p infinity square is much greater than p 0 square and therefore that is going to lead us to a infinity – E 0 that is equal to one half p infinity divided by gamma infinity rho infinity plus Q plus Q and p infinity over rho infinity is R infinity T infinity therefore this is R infinity T infinity divided by 2 gamma infinity plus Q which is E infinity equals E 0 plus or infinity T infinity divided by gamma infinity plus Q so you call this 5 then we have assumed a T infinity so the equilibrium composition is also now going to give you a a gamma infinity and or infinity so now you have for the assumed value of p infinity and T infinity you get gamma infinity and or infinity with which you can check right so check if the above equation is satisfied right so once you know that this is satisfied you are okay with the choice of T infinity that you made in step 2 all right of course you still have the outer loop of having assumed a P infinity so once you do this solve for so this is this is step 3 that we have just finished step 4 would be solved for rho infinity over rho 0 using the quadratic equation that is number 3 right and step 5 would be fine p infinity from 4 that is here right then 6 check if this is the same as assumed right so whenever we say check we would like to do to we would like to think about two things one it is always but within a certain tolerance okay so you are not going to exactly match maybe you can match but to the first decimal place second decimal place third decimal place or as a percentage whatever it is you want to you want to hold hold to a tolerance and if it is satisfied then you proceed if not then you have to repeat the inner loop right so in this case for example you now assume T infinity and then you go through this check and if you are satisfied within the tolerance you proceed if you are not satisfied within the tolerance right then you assume a different T infinity in step to go through the check for the assume P infinity and then once you have converged on a T infinity then you proceed to step 4 step 5 and then check so when you check of course it is within a tolerance and within the tolerance if you have not satisfied then you have to go back and change your P infinity goes through the whole thing again right so so you do you do this of course it is a bit of a tedious process so once you do this so finally step 7 would be once P infinity row infinity T infinity and or infinity or finalized use one to find you not that is almost like the post processing right now of course we find that there is kind of like a loop within a loop and it is going to take a while for you to converge so instead of going through this the next best thing that you can do as I pointed out earlier is do not push yourself to assume P infinity squared is much greater than P naught squared directly say can I can I can I can I adopt P infinity itself as much greater than P naught right so a good approximation is to directly use P infinity itself as much greater than P naught in 2 right that is what I that is what I said we will go back to so that implies 1 over rho 0 minus 1 over rho infinity equal to we earlier had P infinity squared minus P naught squared divided by gamma P infinity rho infinity we threw away P naught squared in preference to P infinity squared cancel one of those P infinities at the top with one at the bottom and so on but now we have a P infinity minus P naught divided by gamma P infinity rho infinity P naught is thrown away in comparison with P infinity the P infinity directly gets cancelled with the one at the bottom so you are left with only 1 over gamma infinity rho infinity right so this means that rho infinity over rho naught is simply equal to gamma infinity plus 1 divided by gamma infinity HS infinity minus HS naught which is not equal to Q plus one half P infinity minus P naught times 1 over rho naught plus 1 over rho infinity in fact we have we have just written it here still around so here what we want to do is to this something that we did during the ranking you go in your days that is we want to write this as gamma infinity over gamma infinity minus 1 P infinity over rho infinity minus gamma naught divided by gamma naught minus 1 P naught over rho naught the way we get this is we now say HS infinity is CP infinity T infinity CP infinity is gamma infinity gamma infinity or infinity divided by gamma infinity minus 1 and then you have a T infinity or infinity T infinity is P infinity divided by rho infinity similarly for the not conditions so if you now do this and then the next step you do is you notice that you are not going to have a big difference in gamma infinity and gamma naught that between the two okay to think about this gamma is basically ratio of specific heats ratio of specific heats depends on whether the gases are monatomic or diatomic or polyatomic whether it is linear polyatomic or non-linear polyatomic and so on now it is monatomic you are going to have something like sorry if it is diatomic you are going to have something like 1.4 if it is monatomic it is going to be like 1.667 and so on and if it is polyatomic you may have something less than 1.4 right 1.3 so all these things are around of the order of 1 if you now think about order of sub magnitude it is all of the order of 1 and mostly for mixtures of gases in both cases it is going to be in general a polyatomic system and so we expect that you do not even have this variation right so we do not expect a big difference between gamma infinity divided by gamma infinity minus 1 and gamma naught divided by gamma naught minus 1 so for an order of magnitude point of view we notice that just like how we now threw away P not itself as opposed to P infinity right we now say let us get rid of this entire term all right so if you now do and similarly we can do do it here as well in preference to P infinity so wherever we find P infinity minus P naught kind of thing you now get rid of this so with this we now say gamma infinity divided by gamma infinity minus 1 times P infinity minus 1 half P infinity times rho infinity divided by do not plus 1 equals rho infinity Q some more simplification well basically you can pull out P infinity all right and then plug your rho infinity over rho naught as gamma infinity plus 1 divided by gamma infinity here and pull this P infinity out you have something in terms of gamma infinity here there is also going to be in terms of gamma infinity so you will now have one big function that is a that is gamma infinity that is in terms of gamma infinity so quickly going through that divided by gamma infinity minus 1 minus half gamma infinity plus 1 divided by gamma infinity plus 1 equal to rho infinity Q and then you put things together so we get P infinity times gamma infinity plus 1 divided by 2 gamma infinity times gamma infinity minus 1 equals rho infinity Q but then keep in mind that this itself after the simplification still has remnants of rho infinity over rho naught right so and this rho infinity then gets cancelled with that right so then you get P infinity equal to 2 Q 2 Q rho naught times gamma infinity minus 1 now if you did not make a big fuss about having to find gamma infinity of course you can assume equilibrium find the composition and so on but the idea was you do not go through all that right that means you now say let us not worry about the variation of gamma between reactants and products you assume some gamma that is common and unknown then you directly get a value for P infinity just knowing the heat release and the density of the reactants right the interesting thing here is Q is like joules per kg and rho naught is kilograms per meter cube so rho naught Q is like joules per meter cube so this is basically something like a volumetric heat release of the reactants right that means the denser the reactants greater the pressure all right that means denser the reactants it compacts more heat within it per unit volume and that means greater the downstream pressure in fact we have been exploiting the notion that the pressure itself is much greater than the initial pressure and that is really the hallmark of detonation waves so what is actually great about detonation waves is not as much about the kilometers per second kind of velocities but what is what is utilized in a detonation wave in terms of applications is the pressure so you have this huge pressure build up behind the wave and as the wave propagates it now has a rf action that follows it but sucks everything in and then destroys the materials that it passes through or the or the matter that it passes through so here what we are basically seeing is if you were if you we are actually talking about gases all over but if you now think about reactants that were actually in solid form and correspondingly you gave rise to some heat okay because of the gas evaporating and then reacting sorry the solid evaporating and giving rise to the detonation wave in the gas phase all right then the volumetric heat heat of the of the reactants is a lot higher and therefore solid detonation reactants give rise to a much higher increase in pressures right so you get you get this physically meaningful relationship from from making this approximation here and then proceeding instead of getting stuck in this iterative loop so finally so now we want to go back to 1 and say u0 is equal to 1 over rho 0 square root of gamma infinity p infinity rho infinity that can be written as 1 over rho 0 square root of gamma infinity p infinity rho infinity over rho 0 times rho 0 and therefore we can write this as still because the rho 0 there and so we can write this as square root of gamma infinity p infinity divided by rho 0 times gamma infinity plus 1 divided by gamma infinity cancel the gamma infinity equal to gamma infinity plus 1 and so the gamma infinity gets cancelled and p infinity divided by rho 0 is 2q times gamma infinity minus 1 that is equal to square root of 2 gamma infinity squared minus 1 times q so what we can see is note that p infinity goes as q but u0 goes as square root of q so you do not quite get you know the same effect out of the heat release and let us just push this a little bit more and we are pretty much there to see the result here so since p infinity equals 2q rho rho 0 times gamma infinity minus 1 p infinity divided by rho infinity is equal to 2q rho 0 divided by rho infinity times gamma infinity minus 1 then whenever you see a rho infinity over rho 0 plug in a rho infinity gamma infinity plus 1 divided by gamma infinity so you say 2q times gamma infinity gamma infinity minus 1 divided by gamma infinity plus 1 and that gives you q equal to so p infinity over rho infinity to start with is put it this way this is nothing but r infinity t infinity which is r u t infinity divided by w infinity right so from here we can get q trying to put these two together right q is r u t infinity divided by w infinity times 1 half gamma infinity plus 1 divided by gamma infinity times gamma infinity minus 1 so from here what we can see is q goes as t infinity divided by w infinity and then putting these two together right we have u not essentially goes as t not sorry t infinity divided by w infinity to the half right now for those of us who are familiar with rocket propulsion we get a very similar result for the exit velocity of the rocket from a gas dynamic nozzle undergoing supplied by combustion gases at a temperature t infinity and molecular weight of gases being w infinity right and this is what translates to something called as specific impulse now if you have a detonation engine pulse detonation engine the detonation wave keeps on propagating out and the at this at the speed and what we find is that the speed of the wave is also directly proportional to t infinity divided by w infinity the path the same dependence so effectively a pulse detonation rocket engine is not going to have a specific impulse that is way too different from a ordinary chemical rocket so there is no way of beating around nature in this case right so you are pretty much going to get the same result as we can see from this.