 Good morning. In today's class, we will consider the explosions involving storage vessels which contain cryogenic fluids. We will also consider something on rupture of pressure vessels after we deal with explosions involving the cryogenic fluids. But to come back to the particular point of let us say explosions involving cryogenic fluids let us say, we must first define what a cryogenic fluid or a liquid is. Any liquid which has its boiling temperature under normal atmospheric conditions less than minus 150 degree centigrade that is equal to 123 Kelvin is called as a cryogenic liquid or it is also known as a cryogen. You know we find these cryogens are important not only as a fuel like hydrogen is a fuel may be LNG is a fuel. It is also used for several other applications and the industry these cryogens are used and we would like to understand what are the types of explosions which take place when they are kept in a storage vessel. Like for instance, let us say I want to keep a cryogen in a storage vessel. Apparently you know this is the cryogen and we say cryogen is a low temperature fluid that means its boiling point under atmospheric condition is less than 123 Kelvin. Well the vessel has to be insulated very well insulated. Let us say it is insulated using let us say a double wall construction with vacuum or let us say with several other may be foam may be you have multi-layer insulation and all that such that you like to prevent the heat from the ambient. See the ambient is something like let us say at 35 degree centigrade the fluid here is a temperature less than we say minus 150 degree centigrade. Therefore there is always a tendency for heat to flow in and therefore you give sort of a good insulation multi-layer insulation vacuum insulation may be some powdered insulation may be some puff insulation and so on. And therefore let us say this is it the vessel is very well insulated. Now even if you give the best possible insulation there is always some heat in leak which occurs. There is always some heating which is going to take place. You know may be it is small but still it is there because of the large temperature differences. And what does this heating do? It causes may be you know when I heat this particular surface over here little bit of heating may be the fluid becomes less dense over here because whenever I heat a fluid it tends to become less dense and the fluid sort of rises up. Here it rises up and therefore you have something like the fluid which gets heated in the wall region tends to rise up because it is lighter in density compared to the original fluid and it sort of accumulates at the surface. Therefore you have the convective currents over here which go and cause the current to go like this and you have the migration of the heated fluid which is less dense and you ultimately end up with something like a lighter fluid on top may be lighter fluid on top and may be the heavier fluid at the particular bottom. Actually the lighter fluid is at a slightly higher temperature and with respect to time what is going to happen is you have something like lighter fluid at the top the heavier fluid at the bottom and this process of having because of the heat in leak which takes place and in any vessel however well you insulate there will always be some level of some small heat in leak it heats the liquid and you have this process which is known as stratification. In other words the lighter fluid comes to the top it is little warmer and therefore if I were to now plot the temperature gradient if I were to let us say plot the temperature gradient over here what I observe is well at the top I have temperatures which are slightly higher temperature which is slightly higher may be on this scale I plot the temperature and then I have the temperature lower that means I have a warmer fluid on top resting on a relatively lower temperature cryogen you know this is not only true for cryogen it is only also true you know at home if I have let us say a bucket of water and I put something like a heater coil in this particular water I supply I heat the water what is it we observe you know we find that when the water is heated on top of the water you have hot water at the top for quite something in fact you will find boiling of water takes place but the temperature of water in depth is going to be smaller this is also the same thing like stratification may be the less dense water when it is heated water when heated look gets lower in density it rises to the top and you have the warm layer at top which is called as stratification. Therefore we find well you have stratification warm layer and therefore what is going to happen the gas that is the vapor of this particular cryogen which is on top is also warm at this particular temperature it tends to be in equilibrium and if the more and more warmer fluid comes over here well gases generated and therefore in all cryogenic appliances you give something like a vent wall through which if the pressure increases well the vapor gets lost gets released through this particular vent and when it gets released the pressure decreases well the temperature decreases and you can also control the temperature because once the vapor is released the pressure decreases the temperature decreases this layer is in equilibrium with the vapor here and therefore you have some equilibrium type of thing but all said and done you have to release the vapor in order to keep the pressure in a cryogenic storage vessel constant. Therefore you have the venting of gases before coming to a particular problem let us take a look at three or four cryogens which are used and then discuss what are the problems is there any particular problem associated with storing a cryogen in a particular storage vessel in a well insulated storage vessel but as we keep saying whatever be the type of insulation there is always some heat in leak taking place. Therefore let us put a few cryogens well we can talk of liquid helium well liquid helium as a normal boiling temperature at atmospheric pressure of something like 4 Kelvin you could have liquid hydrogen the normal boiling temperature at atmosphere pressure is 20 Kelvin I could also think of liquid nitrogen which has the normal boiling temperature at atmospheric pressure is 77 Kelvin I talk of liquid oxygen which is at 90 Kelvin the boiling temperature and the last one I talk in terms of liquid methane and the temperature boiling temperature is 111 Kelvin at the ambient conditions well these are the five typical cryogens which are cryogens because their operating temperature is less than minus 150 degrees centigrade or 123 Kelvin. Now if we store some of these cryogens in this particular vessel and let us say if it is not totally pure what do you mean by not being pure you know this is an essential part of it like for instance let us say I store LNG you know if I talk in terms of liquid natural gas and we have been considering this we considered the LNG explosion at Cleveland we said well we talked in terms of natural gas explosion at Ural we talked in terms of the liquefied one which entered the gutters and caused the manholes to sort of rocket up or skyrocket therefore you know when we talk in terms of LNG and this LNG is an important fuel nowadays it is transported all over the world in tankers not only that it is also transported in pipelines and therefore and it is also stored it is also stored in tanks like for instance of Cochin harbour we get LNG we then store it in in huge tanks and these tanks are again as I say well insulated with a vent valve such that if you have vapor which is being formed you occasionally vent it out such that you have this and what happens let us consider this LNG LNG is from the gas fields it is got as a vapor and it contains not only pure it is not pure methane it consists 90% of methane it also contains some other substances and the other substances which contains are in addition to methane the formula should have been CH4 it also contains some ethane C2H6 it contains some propane C3H8 and also little bit of let us say may be petrol like for instance gasoline well if I look at the boiling point temperatures you know well liquid natural gas is stored let us say as a liquid it is stored as a liquid we say well it also contains some impurities and these impurities are all fuels mind you it contains some liquid ethane it contains some liquid propane because the temperature is low the temperature is of the order of something like 111 Kelvin which I can say is may be 273 minus 111 which is equal to something like minus 162 degree centigrade because it is below 0 degree centigrade you know at this temperature may be the ethane which is contained over here is also a liquid propane is a liquid but the boiling point of let us say methane is we say is minus 162 degree centigrade of ethane the boiling point is something like minus 85 degree centigrade for propane the boiling point is minus 42.4 degree centigrade and for gasoline it is something like plus 69 degree centigrade therefore when I consider liquid natural gas which is stored in large quantities in in all over the world you know you you have this LNG is not pure methane at this particular boiling point temperature at atmospheric pressure but it also has other traces of let us say ethane propane and gasoline and therefore what is likely let us consider the problem just I keep the LNG in a particular vessel and since it contains some of these heavier constituents mind you C2H6 is heavier than methane propane is even heavier molecular masses higher well gasoline is something like may be C20H something therefore it is even higher therefore when I store these things in a vessel now I will remove the outer layer I just say well I store the LNG over here what is going to happen the heavier constituents such as say ethane propane and gasoline are all at the bottom because it is heavier therefore may be I say gasoline is over here propane over here and therefore the heavier quantities are at the bottom and the lighter methane which constitutes maybe more than 94% is what is at the top of course you have a vent valve over here and if some some because of the we talked in terms of stratification we said well the thing arises and I have a stratified layer on top this is my stratified layer on top okay now what is going to happen it is not that heating leak occurs only from the side walls but heating leak will occur also from the bottom and therefore when heating leak occurs from the bottom you already have a dense denser relatively denser may be ethane propane gasoline at the bottom and therefore whenever some little bit denser liquid is heated by by heating leak from the bottom well its density decreases but when its decrease when its density decreases for these heavier substances its density could still be less than the liquid methane on top therefore even though it's slightly warmer it tends to still kick be at the bottom because even though it is heated more than the liquid over here more than the liquid methane over here its density is still not less than methane for it to migrate up and therefore it continues to get heated till a particular time when its density let's say liquid density of the corresponding heavier constituents becomes less than that of the liquid methane and therefore you have higher energy or higher temperature over here because it's more than this and this layer of heated sort of heavier constituents of gases is known as thermal overfill therefore now we know when it gets heated because of the further heat in leak well it rises to the top and sort of you know when it rises to the top it it it mixes with the stratified layer and this this what we call as something like a rollover that means the heated thermal overfill comes and heats the top layer which is already a little warmer but mind you now the temperature difference here is quite high it comes to the top mixes with the stratified layer and you get a discrete increase in temperature of the surface layer and mind you the gas over here is at the original stratified layer temperature over here and all of a sudden this increase in temperature is not is such that the vapor pressure of this is not in equilibrium with the pressure here suddenly there's a spurt in formation of a gas something like a flash vaporization and this flash vaporization as we saw in the last class it generates so much of gas that it is not sufficient that it's not the vent hole is not sufficient to vent it away with the result the pressure increases rapidly and you have the burst of the storage vessel therefore whenever we talk of explosions involving storage vessels holding cryogens we talk of thermal overfill that means the initially that means the the the cryogen is not a pure cryogen maybe it contains some substances some some cryogens which are or some other substances which are relatively denser and they can be get heated to a more they can be get heated to a larger temperature before that density matches with the layer above it and all of a sudden when and or when it gets heated to a level where in its density becomes equal or less than this well it rises up we call this as thermal overfill it mixes with the stratified layer on top namely a rollover process in which thereafter this flash vaporizes because it is at a higher temperature and the bursting of pressure vessel occurs therefore whenever we handle cryogens which are somewhat impure you have mixture of some other substances which are there such explosions are likely but in the case of LNG well it it is obtained as a mixture of some substances it is not pure methane by itself and therefore LNG has to be the storage vessel has to be designed with care such that the thermal overfill and rollover will not cause an explosion. Having said that it is also important that we talked of LNG on and off during the earlier classes we find that well you know we have tankers in the ship which are designed to take this low temperature fluid across the different oceans we have the LNG over here and you know in if by chance let us say a ship containing the LNG leaks out supposing there is a leak on the surface of water in in a sea then you have a LNG which is being formed and we said this is very dangerous why did we say it is dangerous because if LNG leaks over let us say the surface of the earth because the LNG is very cold it immediately cools the earth and also maybe it tries to it flashes into vapor because its boiling temperature is very low but when it flashes into vapor it absorbs the latent heat from the liquid over here and therefore the rate of evaporation comes down with time whereas if the spill occurs over water in a ocean because of the turbulence of the ocean and also the convection movement of water below the convection of water which takes place you know the spill continues to evaporate you have a very copious amount of LNG and therefore LNG requires some caution when we transport it by sea or even for that matter LNG being a cryogen it has to be handled with care and there are problems such as as we say explosion in pressure vessels holding LNG and it has to be adequately designed to hold the additional pressure. Having said that you know we were looking at the explosions involving cryogening fluids there is another instance or another phenomenon which can again cause the explosions in in the storage vessels. Let us take a look we said earlier that you know you have the storage vessel you have the cryogen which is kept in a storage vessel now let us say I have a storage vessel like this I have a vent valve because whatever said and done you cannot you cannot avoid the stratification you have a stratified warm layer at the top and this keeps on the amount of stratified layer is shown shaded over here this is the stratified layer and therefore the evaporation takes place and therefore you have to essentially vent of the gases you know when you went off supposing there is a sudden vent of gases and if you went out the gases what is happening there balance gas because of the expansion that means the pressure gets reduced there is expansion the gas over here cools down when the gas cools down well you know it is almost at the saturation temperature saturation temperature corresponding to this and if it cools down well it is below the boiling point of the vapor at the particular condition of pressure over here and therefore we expect when it cools down it should sort of get back from the vapor it must become a liquid but the the process of condensation that means the vapor becoming a liquid takes some time and very often with sudden venting what happens is you form something like a meta stable vapor which is a vapor at a temperature less than the boiling point corresponding to this particular pressure in other words you know it is tending to condense to the liquid but it is still a vapor and it is there as it is and therefore when when you have regions cold regions over here that is the walls which are cold and therefore all of a sudden this meta stable vapor cannot exist as a vapor it is waiting to condense and therefore all of a sudden you have condensation taking place and therefore you have this condensation taking place of the vapor may be near the surface and any condensation to condense you know you have to reject heat and therefore the condensation has some heat that means and this heat which gets rejected can drive a shock wave and such a shock wave driven by condensation is known as a condensation shock. We can readily show this on the our Eugenio namely I have pressure versus 1 over rho the initial conditions corresponding to the cool down conditions can be shown by the initial conditions P0 and rho 0 and what is it or let us say 1 over rho 0 because my scale is rho 0 here I have P0 into 1 over rho 0 and since some heat is associated with condensation I have something equivalent to a condensation process over here which is let us say meta stable and when it condenses I have heat which is getting generated I can always form a shock with some particular velocity and this velocity the Mach number of the condensation shock is in general small it is of the order of 1.4 in fact these are encountered in any process in which you have a rapid cooling of the vapor like in a wind tunnel you have this problem in you could have it in the case of these storage vessels but anyway you know you have a shock and these shock will increase the temperature will also increase the turbulence will lead to more flash vaporization and this could also be a cause for explosions of storage vessels therefore we can summarize by saying well when we consider explosions of maybe cryogenic liquids in storage vessels well the thermal overfill and rollover associated with different constituents in the cryogen and secondly condensation shock could contribute to the explosion well this is about the explosions of cryogenic storage vessels let us move to the next topic which is let us say something similar we say well why cryogenic fluids you know any pressure vessel can can explode like for instance I have let us say a compressor the compressor is used to charge a storage vessel of air let us say I keep on supplying the pressure at higher and higher pressure well by chance if my regulation system is such that I supply I pressurize it to a very high value well this storage vessel could just explode could just rupture therefore we would like to find out what is the energy release during rupture of a closed vessel now to be able to do that let us again formulate it it is quite simple we have done something similar earlier let us consider a storage vessel of let us say volume v let us say it ruptures at a pressure p0 and let us say the volume is also let us say v0 of this particular pressure vessel let us say that the thermodynamic properties of the gas are such that it has a specific heat ratio gamma the specific heat ratio gamma is equal to cp by cv we have been looking at it for the different gases and we will find well gamma also plays a major role let this pressure vessel which explodes at a pressure p0 let the ambient pressure be p ambient let the ambient temperature be also t ambient let the temperature here be let say t0 corresponding to pressure p0 over here you know I want to find out what is the energy released in this particular rupture and how do I say well the energy released in this rupture let us say e is equal to the initial internal energy which is contained by the gas over here mind you it is a constant volume explosion therefore I say well the initial value of the internal energy over here minus the final value of the internal energy of the same volume here and what is this equal to this is equal to per unit mass I have small u that is so much kilo joules per kilogram into the initial mass of gas let us say is m0 over here m0 into ui minus when the explosion has taken place in the same volume I have final mass over here and let us say that the internal energy is small uf so much kilo joules per kilogram therefore it is equal to mf minus uf over here therefore I want to solve this I am interested in finding out how much energy is released during the particular rupture and therefore let us put things together I have e is equal to m0 you know I can write internal energy we have seen internal energy is equal to cv t that means t u that is the initial temperature over here minus mi into cv into the ambient temperature because once rupture has taken it is the ambient gas which is occupying occupying this volume well the final pressure is equal to the ambient pressure because nothing is there over here we also have well cp minus cv is a specific gas constant or cv I take it in terms of gamma over here is equal to r over gamma minus one specific gas constant in terms of kilo joules per kilogram kelvin and therefore I can write this as equal to m0 into r t0 minus I have mi into I have r into t ambient divided by gamma minus one all what I have said is I have substituted cv is equal to r gamma minus one and I take gamma minus one common in that particular denominator and here the assumption made is gamma for the gases over here and gamma for the gases here is same which need not be so a high temperature gas has a lower value of gamma as we shall shortly see but when I look at this particular expression from gas equation I have pv is equal to mrt specific gas constant and therefore I can write this mrt0 is equal to the initial value of pressure let us say p0 is what is said is the pressure over here into v that is the initial volume the volume does not change we said that the volume is v0 minus the final volume is again v0 mrt0 corresponds to the ambient pressure over here therefore this becomes p0 v0 minus p ambient v ambient divided by gamma minus one or rather the energy released if I take v0 common because the same volume is there I get v0 into p0 minus p ambient divided by gamma minus one which is the energy which is released during the rupture of a pressure vessel. If I say well this is there I find you know I find that well you have gamma and you know for air gamma is 1.4 you find that small changes in gamma is going to affect the energy release therefore let us take a look at gamma what are the factors which influence the gamma because you find well it is not only the volume and the initial pressure but it is also the the value of gamma which is influencing the energy release and therefore let us let us see what what gamma represents in fact if you go back and see the Chapman Juge velocity for a detonation we also got it as equal to 2 gamma square minus 1 into q there also it was a strong function of gamma and we found well gamma changes between the unreacted gas reacted gas we took it to be common but here also it is sensitive to gamma therefore I think it is necessary to be a little more clear about what are the factors which affect gamma well let us say if I have something like a mono atomic gas something like helium let us say it is a simple gas you know it is just a gas which has let us say mono atomic it has only three degrees of freedom it can move in three directions therefore the degree of freedom it has is three if I take a diatomic gas like air like I have nitrogen oxygen and which constitutes air I say it is diatomic well it has it is it is double therefore it has three translational degrees of freedom it can also rotate in two directions therefore the degrees of freedom are five therefore if I keep on increasing the the complexity of a gas like for instance I take something like let us say di trichloro some are refrigerant let us say trichloro trifloro let us say methane which is used as a refrigerant you know if I consider this you know it is much more complex and the number of degrees of freedom are quite large in this case it is it is it is of the order of something like maybe 10 15 or so and therefore you know when I heat a gas again you know what I do is I increase the degree of vibration levels of the gas goes up and therefore its degrees of freedom that means I convert some of the energy into the vibrational modes of a gas and if it is vibrational modes of the gas the degrees of freedom goes up if it is rotational modes the degrees of freedom goes up and it is possible to connect the gamma to the degree of freedom as say 1 over 2 by the degree of freedom this is this comes from the kinetic theory of gases and therefore you find that for monatomic gases such as helium which is a degree of freedom of 3 it is equal to 1 2 by 3 is the value of gamma which is equal to 1.67 for a diatomic gas it is 1 2 by 5 which is equal to 1.4 and for a gas like trichloromethane or dichlorodifloromethane it is of the order that is gamma is of the order of around 1.1 and therefore for a heated gas in which it is very hot maybe the gamma will be very low and when we talk of a low value of gamma apparently I am in my denominator I am having gamma minus 1 therefore the denominator becomes a small number and therefore the energy release goes up. Therefore let us do a small problem to find out the sensitivity of the energy release in the explosion of a pressure vessel with respect to the value of gamma. Therefore let us do this problem let us say I have a pressure vessel which holds air whose gamma is 1.4 at a pressure of let us say 10 MPa. Let us say the ambient is ambient pressure is 100 kilo Pascal which is 0.1 MPa. This is in the other case and this vessel let us say this vessel burst at a pressure of 10 MPa. I also consider another example wherein for the same volume let us say the volume is 1 meter cube so that I do not need to carry the volume. I have the same 1 meter cube vessel having a pressure of let us say the same pressure 10 MPa. Ambient pressure is 0.1 MPa as earlier but it contains let us say refrigerant gas and this refrigerant gas has a gamma let us say equal to 1.1. Now I want to find out what is the energy release in the two cases. I just use my equation we have derived this E is equal to 10 MPa that is 10 into 10 to the power 6 Pascal minus 0.1 into 10 to the power 6 Pascal that is P not minus P ambient divided by gamma minus 1 and into the volume which is 1 so much joules is what is generated over here. In this case I have a similar expression I have 10 into 10 to the power 6 Pascal minus 0.1 into 10 to the power 6 Pascal divided by in this case gamma minus 1 is equal to 1.4 minus 1 in this case it is 1.1 minus 1 into volume so much joules which is generated therefore in this case I take 10 to the power 6 outside 10 minus 0.1 is 9.9 divided by 0.4 joules and in this case I have again 10 to the power 6 into 9.9 divided by 0.1 you know if I solve for this I get a value of the energy release being something like 24.75 into 10 to the power 6 I can call it as mega joules in this case it is 99 mega joules and therefore you find yeah you know the energy released when I have something like a complex gas is much very much higher for the same conditions compared to air the question is why should this happen why should an explosion of the gas at the same pressure same initial conditions same volume and same ambient conditions release much more energy the thing is that a more complex gas because of its different modes different degrees of freedom because of its structure can hold much more internal energy that means the initial internal energy of this gas is very much higher than the internal energy of this gas and since we are talking of the drop in internal energy well the complex gas gives me much more energy but why should the internal energy change and not the enthalpy h is equal to u plus pv it is this which decides the internal structure which holds the energy this is just flow work which does not change substantially therefore cv tends to be high for the complex gases and since it has high it holds much more energy this is the logical conclusion having seen the values or having estimated the energy release in a rupture of a pressure vessel let us go ahead and try to ask one more question you know we said energy is liberated in the explosion of a pressure vessel that is the rupture causes energy to be released and it is this energy which creates a blast wave and we will see the effectiveness of the energy release when we talk of TNT equivalence after some three classes however in all let us consider this energy release could be very slow could be very fast whenever we talk of a constant volume explosion well it just goes with a bang that means the energy release is very fast compared to a constant volume explosion if I have something like a constant pressure explosion what do you mean by a constant pressure explosion well let us say well I have a piston which is driving the gas at constant pressure and the volume increases well that could also drive provide some energy with which I could have an explosion taking place and what is the value of the energy release let us say I have a constant pressure therefore I have something like a piston pushing the gases at constant pressure and let us say there is no change in the pressure of the gases let us say that the initial volume is V2 and the final volume is let us say V1 that means I am talking in terms of constant pressure into V2 minus V1 is the work done which is the energy of the a constant pressure process what we are talking of you know what would be if I were to consider this constant volume explosion derived from the thermal energy which is supplied to this particular volume and also a constant pressure explosion by the thermal energy supplied for this particular change is there something like this being more effective this being more effective I want to find out what is the conversion factor for the thermal energy into the the energy that is the mechanical energy obtained in the particular rupture therefore let us let us let us do this particular problem it gives us a feel of how much energy how spontaneous energy is takes place in a constant volume and the effectiveness of the explosion and therefore when I consider a constant pressure explosion well we say energy is equal to P into we say it is equal to the V2 minus V1 and what is the heat which is drives this particular explosion the energy supplied is equal to Cp that is the mass of the gas into Cp into I have T2 minus T1 or rather T1 minus T2 that is the heat that is this the final temperature minus the initial temperature and I can write well Cp minus Cv is R therefore Cp is equal to gamma R by gamma minus 1 therefore the heat which is supplied is equal to m gamma R T2 minus T1 divided by gamma minus 1 and now I do the same thing mR T2 is equal to the value of pressure being constant P into the final volume let us say V2 minus I have mR T1 the same pressure into V1 divided by gamma minus 1 we have to take gamma into mR T is the P2 V2 I still have to retain this or rather Q is equal to I have gamma divided by gamma minus 1 into I have P V2 minus P V1 divided P V that is about it that means I have gamma by gamma minus 1 into P into I have V2 minus V1 what is the work done work done is equal to P into V2 minus V1 this is P into V2 minus V1 into gamma minus 1 therefore if I take the ratio of the work or the energy which is released in the particular constant pressure process let us say I have E I say constant pressure process divided by Q in a constant pressure process I have P V2 minus P1 P V2 minus V1 cancelling and I get E by Q is equal to gamma minus 1 divided by gamma if I do the same thing for a constant volume explosion we form that the energy released in the constant volume explosion is equal to V into gamma minus 1 into we had P that is the P0 minus P1 or P ambient and we also find that the energy heat which is supplied to do this I can write it as equal to I have M C V into the T2 minus T1 which I can now write as as equal to M R divided by gamma minus 1 into T2 minus T1 again writing M R T2 and M R T1 as equal to P0 V and P1 V I get this equal to V into P0 minus P1 divided by gamma minus 1 and therefore for a constant volume energy release divided by constant volume thermal energy what is required is equal to V same expression therefore it is equal to 1. Therefore we find that the since gamma is of the order of 1.4 the energy release in a constant pressure process is much lower and we know that a constant pressure process is a slow process whereas when I talk in terms of a constant volume explosion it is a sudden process and therefore now I can plot as a function of time the value of let us say energy divided by the heat release I say well for a sudden process I have this equal to 1 and as I move towards a constant pressure process the value I get is gamma minus 1 divided by gamma therefore the suddenness of an explosion is important to be able to convert all the thermal energy into the mechanical energy you know this completes what we said is about the the rupture of pressure vessels about the energy release and we find yes now we can estimate the energy release but there is one one small problem which very often the chemical industries face and let us quickly go through it you know the problem is like this you know in in any chemical plant or any any plant you have a series of reactors you know we are you are trying to form some substance like like for instance let us let us go through one one particular example let us take the example of the explosion which happened at Flixborough in UK where and on June 1 1974 there was a spillage of let us say cyclohexane which mixed with air and and just ruined or made Flixborough into a ghost city you know what happens in a chemical plant you have somewhat like you have reactors several reactors which are essentially storage container storage vessels you have let us say reactor one reactor two over here I have reacted three over here yeah several of these reactors and all these reactors are always connected by pipelines because from one reactor I transported to the second reactor to the third reactor to the fourth reactor and so on now the question is supposing by chance in one of the reactor there is an explosion I do not want my entire plant to get to to to get damaged I would like to still protect my second reactor and third reactor and somehow ensure that the explosion from reactor one does not get into reactor two or rather I am interested in so designing this this pipeline which connects my different reactors should be such that let us say its diameter must be d and let us say its length must be l I am interested in the connection having a particular diameter d and a length l such that explosion from one reactor does not get into this even if this explodes can I still save this is the question therefore we are looking at this particular connecting or let us say the pipeline and we immediately say well I have studied this if d is less than the quenching dimension or rather in the case of detonation if d is less than the critical diameter well my problem is solved film cannot travel but the problem tends to be a little more complex there are additional factors which are involved and that is what I want to highlight now you know let us say well in the reactor one I want to simplify it I want to put the scheme as well I show the reactor through some another geometry let us say I have one reactor in which there is some explosion taking place well it is a donor as far as an explosion is concerned it is connected to a next vessel which receives the explosion I call it as a receptor well I am interested what must be the diameter of this pipeline and what must be the length of this pipeline to make sure that the explosion does not travel from one one particular reactor to the other therefore now I tell myself supposing let us say that the diameter of this particular connector is such that it is somewhat same as the surface area of this particular vessel or rather if I save volume and I say mean surface area if the surface area cross-sectional area of this vessel is same as something like let us say this diameter is d is same as let us say surface area corresponding to volume one is same as let us say pi by 4 into d square well I have such a huge wind that may be a flame gets started you know whenever gas is getting generated you know it really cannot build up pressure here because the the vent is so large that it just allows the gas to go through and therefore in this case there is no danger at all but to think in terms of a large vent and why does it happen the vent area is so large that means before pressure with time tends to get build up at the low pressure itself it starts venting the gases and therefore you have something like a gas going over here and therefore it is a unburned gas mixture which goes out even if it is burned the if the diameter is less than quenching thickness well it quenches nothing is going to happen well it's it's safe but in practice it's not possible in practice d the diameter or the surface area is going to be very much smaller than this and therefore since the vent area is much smaller than the original cross sectional area this happens like a constant volume vessel or a constant volume reactor which is exploding and in a constant volume explosion the maximum value of pressure is pm which we have hydrocarbon we said it's between 5.7 to something like 7 atmospheres if it is hydrogen it is of the order of 11 atmospheres and therefore I have a constant pressure explosion which gives me a high value of pm and therefore what is going to happen you have a high value of pm and therefore the gas comes in here you know it comes at high pressure and it and it expands into this donor into this receptor vehicle that means let's say v2 as a hot jet if this hot jet comes into this particular reactor over here and if this hot jet immediately mixes with the cold reactive gas over here or with the with the gases in this particular volume what is going to happen well this jet gets extinguished and therefore it is not possible for you to to initiate combustion over here and since I am that means I am talking of if the mixing of the jet with the gases here is very rapid that means the mixing time let's say tm is a very small number well I cannot ignite this gases because it gets quenched. However if the hot jet comes and the chemical reaction occurs so rapidly that heat is getting generated well this also explodes or rather if the chemical reaction time let's say tc of the gases here is a small number well this this container or reactor explodes therefore we find the ratio of tm by tc if the mixing time is small well I am safe therefore for safety I say well that mixing time divided by the chemical reaction time must be less than a particular value and this also therefore becomes an important parameter it's not only the quenching distance and mind you you know whenever we do these things you know we also talked in terms of quenching distance divided by the flame thickness or critical tube diameter for a detonation let's let's concentrate on flame d cube by tf we said is is of the order of something like 5 or 10 or something of this order we also were able to put the quenching diameter in terms of a peclet number that means the convection heat transfer divided by the conduction heat transfer you know but it's very difficult to to theoretically determine this particular diameter and in practice what is done is we we derive it from experiments and the quen the diameter so derived from experiments is known as maximum experimental safety gap and it is known as MESG it is derived for the different gases and we know what must be the type of diameter what is used it is also necessary that the length must at least be of this diameter such that the heat losses and the active radicals loss to the walls will quench it and in practice we use the maximum experimental safety gap but in addition to the safety gap we must also look at the phenomenon by which things takes place the relative size of this relative to this volume the mixing times and chemical reactions times in order to make sure that explosion in one vessel does not get into the other vessel well this is all about explosions in the cryogenic vessels rupture of pressure vessels and also something on containment of explosions in a particular reactor and not spreading to the other in the next class what we do is we go to a slightly different topics we have been restricting ourselves to gas phase and dust explosions we will take a look at the condensed or the solid explosives in the next class in the next two or three classes we will deal with condensed explosives thank you an announcement please the lectures between numbers 28 to 32 covered the different categories of explosions the references for further reading on these lectures namely those between lectures 28 to 32 and a set of homework problems pertaining to these specific lectures are given in the downloads of this video course