We are talking about combustion chambers. In the last class, we had a look at various kinds of fundamental combustion chambers, their geometries and a look at how they function. The mechanism of combustion as we know, chemically it is basically a process of oxidation. In case of jet engines, the important issue at hand is how much of heat is released in the combustion chamber as an input to the engine in terms of energy input and that energy input is to be harnessed for doing work in the form of thrust. Now, this energy input as we have discussed in the last class is essentially in the form of heat release by opening up of the chemical bonds, which of course means that there is a lot of chemical reaction that is going on and invariably there are fields of chemical kinetics that do come into the picture to decide how the combustion is actually going to be taking place and how the heat release is actually going to be done. However, as I mentioned the chemical kinetics and the reaction kinetics are fields, which are completely separate fields of expertise and we will not get into those areas, which are pretty comprehensive by themselves. We will stick to the basic physics of combustion and we will look at how the heat release is harnessed into a uniform heat supply into the actual flow turbine that normally is positioned after the combustion chamber. Now, this heat release that is done involves a lot of steps. In the last class, we had a look at some of these steps. Today, we will look at how these various steps and the combustion mechanism that we talked about is put together in a certain number of parameters, combustion parameters that help us quantify what is going on inside the combustion chamber. This quantification is very important because then step by step, we have an idea how the heat release or the energy release can be quantified and what are the various penalties that one may need to pay to get this energy input into the engine. So, in today's class, we will look at a few parameters, which help us quantify some of the mechanism that are going on inside the combustion chamber and some of the things that we discussed in the last class. So, in today's class, we look at combustion parameters, the various parameters that help us quantify what is going on inside a combustion chamber. Now, the first parameter we may like to look at is pressure loss. Now, you see when the flow comes from the compressor, it has been adequately compressed and the amount of compression it has gone through has been decided by the engine designer as per cycle design and this compression that has already taken place is vitally important for the efficiency of the engine and for good performance of this engine. Now, during the process of combustion, there is every possibility that some of this pressure would actually be lost. It is necessary then that the combustion chamber designer takes every care to ensure that the input pressure into the combustion chamber is not substantially lost in the process of combustion and this needs to be taken care of during the process of combustion design or combustion chamber, geometry, configuration and the overall combustion mechanism design. Let us take a look at how this pressure loss is described and in some sense, defined in a very simple manner. Now, the combustion pressure loss that normally we would be talking about can be said to be composed of two different parts. One is the simple fluid mechanic pressure loss or aerodynamic. As the air is going into the combustion chamber, it is invariably subject to a lot of skin friction, lot of mixing that we have discussed about, that is vitally necessary for combustion and turbulence which is also vitally necessary for good combustion. So, these are basic fluid mechanic or aerodynamic pressure losses that take place any fluid flow and in any kind of geometry. The second pressure loss or second kind of pressure loss is related to the rise of temperature due to combustion. Now, the later one actually often is related or described as fundamental pressure loss because it is related to the combustion itself and this is due to the fact that as the temperature rises, the local density actually decreases and the decrease in density pushes the flow faster and as a result there is a local increase in the velocity and this velocity increase locally creates a higher pressure loss because pressure loss is proportional to the velocity squared. So, typically every geometry would have a pressure loss coefficient, so to say and this pressure loss coefficient is then multiplied with the velocity squared to get the pressure loss. Now, this is what is captured in the definition given below that if the flow is flowing from station 1 to station 2 across the combustion combustor, the pressure loss coefficient may be given as p 0 2 minus p 0 1 divided by half rho c 1 square, where c 1 of course is the velocity with which the flow is going into the combustion chamber and rho is the density of the air. Now, this density of the air is what we were talking about just a while back that locally in some part of the combustor, locally this density may actually go down and as a result of going down the local velocity c may go up from inlet velocity c 1 to higher velocities and as a result locally the pressure loss could actually go up. Now, this is a phenomenon which again can be measured, so both the pressure losses can indeed be measured. If we apply the simple incompressible flow theory and we assume to begin with at least that the flow is flowing in a constant area duct that means the combustion chamber can has a constant cross sectional area from inlet to outlet. If we make that simple assumption the total pressure coefficient that we get can be written down in terms of the density ratio and then the temperature ratio and of course the total temperature ratio. Now, we are using incompressible flow theory and assume that the area across the combustion chamber has been held constant. Now, this ratio of total temperatures T 0 1 by T 0 2 is normally of the order of 2 to 3 that means that is the amount of temperature by which the temperature is cooled down from the initial temperature of the combustor to the delivery temperature at the turbine inlet phase or the compressor delivery phase. So, temperature is often brought down by a factor of 2 to 3 and result the pressure loss coefficient would be of the order of 1 to 2 correspondingly. Now, what happens is due to if you look at it the other way some other way due to the strong vortex formation and cross flow that is deliberately induced into the combustion chamber as we discussed in the last class to aid the process of combustion. These essentially entail certain amount of pressure loss and this pressure loss is often of the order of 25 percent of the inlet dynamic head that is a ball park figure, but that is a kind of that means 25 percent of the inlet dynamic head could be lost in various kinds of losses which are absolutely necessary some of it to aid the process of evaporation and mixing and then you have a pressure loss that is inevitable and as a result the uniform exit temperature and low pressure loss are contradictory that means if you want uniform pressure loss at the combustion chamber delivery and you want a reasonably lower temperature so that it is comfortable for the turbines then that is contradictory to the low pressure loss requirement because to achieve those two things you would invariably lose a lot of pressure. Now, this is something which the combustion chamber designer would have to grapple with at the time of designing the combustion chamber. In addition to these you remember we have a restriction of space that is size and volume that is available for combustion chamber so all of it together there is certain contradictory pulls through which the combustion chamber would need to be configured. If we look at the total pressure coefficient we have defined before that is delta p 0 across the combustion chamber divided by half rho c 1 square now this can be written down as we mentioned in terms of two kinds of pressure losses. The first one we said was essentially related to the basic fluid mechanics and the second one is related to the temperature change across the combustion chamber. Now, if you do that the first one often signified by an empirical parameter k 1 and the second one signified by a empirical parameter k 2 they are found for the combustion chamber through various test rig simulations. The first one is through a cold run and the second one is through a hot run.  The second one is directly related to the combustion phenomenon and hence if there is a temperature rise there is going to be certain amount of pressure loss and that is called as we said combustion related pressure loss. So, the first one is fluid mechanic pressure loss second one is combustion related pressure loss and for a given combustion chamber geometry once a certain geometry has been decided upon this can be taken to a test rig and a certain runs or test rig performances can be measured and these two parameters can be quantified and then they can be put in the equation as a given here and can be used for across a large number of mass flows and pressure ratios and fuel flows. So, once we get this equation partly from the rig testing and partly from the design parameters we have a first cut handy equation to give us a reasonable idea about what the pressure losses are going to be across the combustion chamber and this is of course to ensure that you do not lose a lot of pressure that has been developed through the compression process. Now the typical values of design operating point for various kind of combustion chambers that is the canular, the can type and the annular combustion chambers are typically of the order of 35, 25 and 18 respectively these are again as I said ballpark figures the total pressure loss taken as a total pressure that is delivered to the combustion chamber from the compressor could be of the order of 4 to 7 percent of the total pressure and if the combustion chamber designers job to ensure that this pressure loss is kept to the minimum as minimum as possible. So, this is one of the parameters that combustion chamber designer would have to grapple with during the process of design it involves simulation in test rig it could involve simulation in computation computational simulation in both again cold and hot and certain amount of design variables that would have to be probably iterated for and it depends on what kind of combustion chamber you have whether you have annular or whether you have a can type or a canular. So, these are the various issues that would come into picture to keep the total pressure loss across the combustion chamber to the minimum. The next parameter that we would probably like to look at is the combustion efficiency now the total pressure loss that we talked about really speaking has nothing to do with the combustion efficiency as opposed to all the other component that you may have done before where the pressure losses directly impacted the efficiency of those components notably the compressors and the turbines in combustion chamber the pressure losses do not impact the combustion efficiency and when we are talking about combustion efficiency we are talking about the efficiency by which the combustion process that means the heat release process is indeed carried out the combustion efficiency can be defined in terms of in two different ways really one in terms of the fuel air ratio for a given temperature rise across the combustion chamber or we can write the total pressure total temperature rise across the combustion chamber for a given fuel air ratio. So, the ratio of these two the ideal and the real is often defined as combustion chamber now as far as one can see in a test rig it is easier to measure the temperature rise across the combustion chamber on the other hand for the operator it is easier for him to measure the fuel flow and the fuel air ratio as a result of which both the definitions are often used alternately to obtain the combustion efficiency. Now, typically the combustion efficiencies are very high indeed of the order of 98 to 99 percent under sea level operating conditions assuming that everything has been very well taken care of and as I mentioned earlier you could have very high combustion efficiency and you could still have very high pressure losses or you could have very low pressure losses and you can still have somewhat lower combustion efficiency. So, the two are not quite connected to each other now at high altitude what happens is that the operating pressure absolute value of the operating pressure falls and as a result of fall of the incoming pressure the combustion process may get affected. Now, combustion as we have discussed before is a very fast burning process and the fastness of the burning process is essentially decided by the high pressure that has been put in the air flow that is coming into the combustion chamber and that high pressure decides the fastness of the burning process. The high pressure essentially decides the potential energy of the air or the working medium and this then takes it very close to the critical threshold energy level at which very fast burning process is possible. If the incoming potential energy in terms of high pressure is not really that high then the fastness of the burning process would be affected and as a result the combustion efficiency would be affected. So, the necessary condition for good combustion is that you have reasonably high pressure that of course, signifies the necessity of having a compressor in front of the combustion chamber. So, incoming pressure and as a secondary requirement incoming temperature if they fall the combustion may become less efficient and this is something which is pretty much acceptable because at very high altitudes your absolute values of pressures are likely to be lower. However, at that high altitude your thrust requirement for the from the engine is quite often lower. So, slightly lower burning efficiency may just be acceptable because your fuel flow or fuel consumption is anyway absolute values of fuel consumption are lower. So, under certain situations slightly lower combustion efficiency may just be acceptable. Now, whereas under certain other operating conditions the high combustion efficiency is a desirable phenomenon and over entire flight schedule of an engine it is necessary that combustion efficiency remains as high as possible to ensure that overall fuel consumption is somewhat on the lower side taken over the entire flight schedule. If we now look at some other parameter that is related to the combustion and that in fact indeed impacts on the combustion efficiency in some way and that parameter is the combustion intensity. Now, combustion intensity is what of course would decide how much heat is released into the combustion chamber and that is into the engine. So, the energy input into the engine is decided by the combustion intensity. Now, combustion intensity is a parameter that can be simply defined as heat release rate divided by the combustion volume into the pressure. Now, you see here the inlet pressure is important and higher the inlet pressure we saw higher is the fastness of the burning process. On the other hand the combustion intensity if it is to be high it has to have very high heat release rate if the pressure is also very high. So, we have to heat release rate is decided by number of operating conditions and the basic chemical composition of the fuel. So, and then we have the combustion volume which is decided by the engine designer. So, in case of aircraft engine the volume available would invariably very low and as a result of which you need to have very high combustion intensity. Now, this availability of low volume in a typical aircraft engine has number of repercussions. The combustion mechanism that we have discussed in the last class tells us that there are steps of combustion that need to be done in a certain period of time and you need a certain space or volume to do all that within the given time. On the other hand as we see here now for combustion intensity to be on the higher side the combustion volume would need to be as low as possible which corresponds to the requirement of aircraft engine. So, there are again contradictory pulls over here to decide what the combustion intensity could possibly be in actual aircraft engines. The heat release rate Q of course is decided as I mentioned by the fuel flow or the fuel itself and hence you need to choose a fuel which has a high value of Q in principally built into its chemical composition. The typical combustion intensity of the is of the order of 2 to 5 into 10 to the power 4 for most of the aircraft engines in terms of kilowatts per meter cube per kilopascals and we shall see that some of these numbers would probably vary a little during the actual operation of the engine. Now, if you look at the dimensions of the combustion chamber the length and the diameter of the combustion chamber are often chosen on the basis of combustion intensity and the volume flow rate. Now, the volume flow rate into the combustion chamber is decided by the engine air flow rate and the engine air flow rate is decided by various parameters like what is the thrust compartment so on and so forth. So, the combustion chamber we are talking about may be an annular combustor or it could be a can type combustor where the air is split into so many smaller combustion chambers and inside of each of them a certain length and diameter are required to be decided upon on the basis of the combustion intensity and the volume flow rate. Now, combustion intensity is chosen on the basis of engine power that is needed. So, the power or the thrust power that the engine has to produce can be deciding factor on what the combustion intensity is going to be. The air velocity in the combustion chamber as it is coming from the compressor and that gives a certain idea about the length that is required and we have talked about the combustion mechanism probably we will have a look at it all over again today again a little later and that tells us that the air velocity that is operative inside the combustion chamber does decide what the resident time is of the working medium air inside the combustion chamber to absorb the heat release and then carry it with it into the turbine. So, this ability of the absorption of the heat release into the working medium air inside the combustion chamber is something that has to be decided by the combustion designer. He has to give the combustion a certain length during which the air flows in and then this flow of air absorbs the heat that is released by the burning of the fuel and then it carries this heat in the form of high energy along with the high pressure into the turbine and certain amount of time and space and of course, volume essentially would have to be given for this working medium air to do this job to absorb the energy input and this is necessary to be done in a manner in a controlled manner and that is why we call combustion a very controlled method of fast burning. The entire process of combustion needs to be controlled all the time during the process of combustion there must be a tacit control over the process of combustion without that control we may have either in extreme case an explosion or the other extreme is the combustion process would get extinguished. So, combustion is a control mechanism of fast energy input into the engine and that control has to be built into the combustion design. It is necessary that combustion chamber geometry, the length, the volume, the shape, the size everything is decided at the time of the design because you do not have a variable geometry combustion chamber as yet. So, the combustion chamber geometry is a fixed geometry there is no variable geometry combustion chamber. It has to be built into the design of the combustion chamber geometry and whatever control that can be actually done during the process of combustion operation would have to be taken into account during the process of design of the combustion chamber. The other things that we need to look at are the weight and overall diameter of the engine. Now, you may be having an aircraft engine that is necessary to be a small engine for example, if you have a military engine you need to make sure that the overall diameter is as small as possible. So, that it fits inside the body of the aircraft and then of course, it is necessary that its weight is as small as possible. So, that you have what the engine designers always call the thrust weight ratio. So, if you want to have high thrust weight ratio and a low diameter to fit into the aircraft you do not have a lot of space available for the combustion chamber and as a result of which quite often many of the designers go for annular combustion chamber which is the easier geometry to adopt really speaking. So, the overall layout of the engine would invariably dictate the combustion chamber type whether it is annular or cannular or can type which is normally not used anymore these days. So, essentially what type of combustion chamber design the size the shape all that is also decided by the overall layout of the aircraft engine because it has to go on board an aircraft it has to fit into the aircraft. If you have land based gas turbine engine some of these things are hugely relaxed because the space is not of great importance a little more space is not going to hurt anybody. But in aircraft engine space is premium and there is very strong restriction on the amount of space and the weight that can be given to the combustion chamber. The weight we are talking about weight one of the reasons is the combustion chambers are wrapped around the basic shaft of the engine. Now, the longer the combustion chamber is remember the basic shaft of the engine also becomes longer the engine itself becomes longer and that increases the weight of the overall engine. So, the combustion chamber shape and size is vitally important specially for aircraft engine applications. Now, let us look at the other parameter that is very important for the combustion process and this is the reaction rate which is given in terms of a number of parameters. Now, pressure to the power n, n depends on the number of molecules involved in the process of combustion and it is normally 1.8 for hydrocarbon fuels which are normally used in aircraft combustion. E is the activation of energy and f t relates to the various forms of molecular energy that is involved in the process of combustion that is the rotational, translational and the vibration energy which are released when the heat is released and the molecular energy that it acquires from the heat release manifest itself in the form of these three forms the rotational, translational and vibrational energy which is then released to the air that is mixed with the air that is passing through and it acquires all this energy into itself. Now, that all that together that is there is a molecular kinetics there is a there is a reaction kinetics which is all together decide what the reaction rate of the combustion process is going to be. Now, reaction rate is important because that decides what the reaction time is going to be and that reaction time is given in terms of again total pressure and the value of n that we have talked about and that would decide what the resident time of the gas inside the burner is going to be. Now, the resident time of the gas inside the burner can be given in terms of the length of the combustion chamber divided by the velocity through the average velocity through the combustion chamber as we have discussed the velocity may vary a little locally inside the combustion chamber and the length of the combustion chamber can be decided of course, by the density of the combustion chamber the area across the combustion chamber which to begin with we can assume to be constant and then all that decided by the mass flow of the gas through the combustion chamber. So, all that together decides what the resident time is going to be now you see the reaction rate the reaction time and the resident time are connected to each other. The reaction rate as we just saw is decided by the molecular kinetics of the reaction kinetics phenomenon the reaction time is decided primarily or directly proportional to the pressure that is available and hence we have argued that it is necessary that you allow air to go inside the combustion chamber with as high pressure as possible to impact on the reaction type which as we just mentioned is related to the fastness of the burning process and then the resident time which is again related and that is decided as we now see related to the average combustion chamber density the flow in the inside the combustion chamber the area the length of the combustion chamber and then the mass flow through the combustion chamber. Now, these are the parameters that actually impact on the combustion mechanism which we discussed in the last class. This is what the combustion mechanism is all about and the things that we talked about the physics is a combustion that we talked about we are bringing in a lot of vortices we are creating artificial vortex system we are deliberately slowing down the flow all that is done essentially to impact on these parameters. These parameters are impacted by the physics of what we discussed in the last class may be we will have a look at it again today all over again to see that these parameters which we are talking about and as I mentioned we are not going to talk about a molecular kinetics, but they have impact and they can be somewhat controlled in addition to the molecular kinetics through certain physical manipulation of what is happening inside the combustion chamber and if we have a good control over what is happening we have a control over the reaction time we have some control over the resident time and that resident time permits the entire combustion process to be completed the heat release to be completed and then the entire heat release to be passed on to the passing air. Air has to absorb the heat and then go on from the compressor to the turbine. Now, this is important and that is why it is necessary that you do all that within the given resident time and hence the combustion chamber designer should have reasonable notion of what is the resident time available for the given volume and the given length of the combustion chamber. If we now look at how the length of the combustion chamber can be decided from the earlier parameter we can say that the length of the combustion chamber is directly related to the gas that is flowing the resident time the density and the area and hence we can say that some of these parameters the density can be related through the thermodynamics to the pressure to the power 1 by gamma. Now, what happens is the burning size of a burning length can be decided with the help of essentially the pressure and the temperature and here what is written as the length is proportional to the pressure to the power minus r divided by root over total temperature T 0 C c of the combustion chamber or at the combustion zone. Now, where r is 1.51 for hydrocarbon fuels that are used for combustion in aircraft engines and as we have stated before it is n is equal to 1.8 and r is equal to 0.714 for other kind of fuel where n is equal to 1. Now, the increasing compression ratio the combustion chamber size and length may be reduced. Now, this is something which what we are discussing that if you have high compression ratio the combustion process can be faster and the heat release rate can be higher and then the combustion chamber size and the length can be reduced. And as a result of this we can say that for large engines the each can of the if you consider each can as a combustion chamber the each can chamber size may not be large if the operating pressures are kept reasonably high. So, the question is what is the pressure operating pressure that you can adopt for the particular engine and then what is the mass flow through the engine that you have with you and these are the parameter that will impact on what is the combustion chamber length and the size and of course, the shape of the combustion chamber. So, these are the things that decide what all the geometry of the combustion chamber is likely to be. Now, if we go back and take a quick look at how the combustion phenomenon is actually tried to be brought under control or kept under control and as I mentioned this is normally a fixed geometry combustion chamber you do not have a variable geometry combustion chamber at least not as yet. And as a result of which once a geometry is created you are fixed and again as I may have mentioned before each combustion chamber for every engine of aircraft use is an unique combustion chamber. You normally cannot use one combustion chamber used in one engine for another engine completely new combustion chamber would need to be designed for any new engine that is created. Now, let us take a look at this combustion phenomenon the flow as I mentioned is coming from the combustion from the compressor at very high pressure and as we saw in the last class about 20 percent or so comes inside the main combustion zone and it comes through what we call swirler and these swirlers give a huge swirling motion to the flow and then we have this some kind of a bluff body over here which creates more of swirling flow more of local vortices and these vortices essentially create a huge recirculation around this zone as we discussed in the last class and this recirculation zone essentially aids the process of evaporation and mixing. This is pre mixing before the combustion we have two rounds of mixing one before the combustion one after the combustion the first mixing is before the combustion this is the evaporated fuel from liquid to gas would need to be mixed with air. So, you have a gaseous mixture of air and fuel before the combustion and this needs to be uniform across the combustion cross section. So, in the combustion zone it is necessary you create a uniform mixture gaseous mixture of air and fuel and this needs to be artificially done through the process of recirculation zone that is created over here and as I mentioned this is a fixed geometry. So, you have to design a priory and ensure that it happens all the time during all the process of combustion during various operations of the engine. Now, if we do that if we do that we have a uniform gaseous mixture of air and fuel and that goes into the primary zone and this primary zone is what we call the combustion zone and in this zone you have this curved surface over here which is called the flame front and as we discussed in the last class this flame front is where the flame is burning and the heat is being released. So, in the primary zone from here to here over this zone you have a uniform mixture of air and fuel that is burning that is moving through the zone through the flame front and the heat is being released. At the end of the primary zone one has to ensure by design that the entire heat has been released by chemical reaction from the fuel into the air and fuel mixture and this released energy now needs to be uniformly distributed across the entire combustion chamber. So, we have the secondary air coming in through these vent holes from the sides and now this is the 75 percent of the air that was bypassed 75 to 80 percent of the air that was bypassed and now they are coming in through these vent holes to create the secondary zone and this is necessary essentially to ensure that you have uniform distribution of energy or heat across the entire combustion chamber volume. So, the uniformity of distribution of heat is extremely important in certain operating conditions it may happen that a very small remnant of the burning process or combustion may still be carried on in the secondary zone. The tertiary zone is extremely important to bring down the temperature to something that is comfortable to the turbine. So, second while the secondary zone is important for creating a uniform temperature distribution the tertiary zone is important for creating a lower temperature zone that is comfortable to the turbine and ensure you still have a uniform distribution of temperature pressure and finally, delivery to the turbine with uniform velocity distribution. So, the combustion chamber design and this what we are seeing here is a can type combustion is essentially to ensure that you carry out all these steps step by step in a step by step sequential manner. So, that you have a uniform temperature distribution at very high temperature and retain very high pressure as delivered from the compressor and deliver it to the turbine. So, there are number of steps as we can see and as we have discussed before all these steps would need to be completed in a controlled manner and somewhere within that you also have the molecular kinetics or the reaction kinetics which happen essentially in the primary zone during the process of heat release. So, all these things need to be done in a controlled manner and during the process of combustion as I mentioned it is a fixed geometry combustion the only control you have is the fuel flow rate. So, there is no other control and hence the combustion chamber design must necessarily have the geometric control at the time of design itself. Just to complete the picture over here the can type that we are seeing is quite often also referred to as a flame tube inside which the flame is localized and securely maintained and held during the entire operation of the engine it is necessary that at no point of the operation of the engine the flame is extinguished and hence this is often called a flame tube to ensure that inside it the flame is continuously burning and the combustion is continuously taking place. The ignited plug that you see here is used normally only once during the process of initiation of the combustion the injector that you see here and we will be discussing the injection process later on may be in the next lecture actually ensures that you have sufficient injection of fuel to for the corresponding operating point of the engine and that is done by the engine control system. So, these are the various operating conditions of the combustion chamber that ensure that you have a continuous combustion you have a control combustion and you have complete combustion given the shape the size and the space that is allotted to the combustion chamber. So, we can sum up by saying that the development of a practical combustion system it requires good mixing the primary mixing or the premixing before the combustion requires a good circulation zone and this is created by placing a baffle or a bluff body in the flow. The swirling vane is attached to the front baffle so that swirling flow is set up you have to create that swirling flow under all operating conditions of the engine and that brings down the actual velocity from 150 which is what is delivered from the compressor to about 40 to 50 meters per second and this is what we discussed a little in the last lecture also. And the baffle and the flame tube and the swirler are combined to give the primary combustion zone or the primary zone inside which the combustion is securely held and continued through the entire process of operation of the engine. So, this is absolutely important to ensure that combustion is continuous complete and efficient during the entire process of operation of the engine. The development of a practical combustion system also ensures that the baffle that you have creates a low forward velocity the spiral flow that it creates helps what we had called in the last class the process of atomization that is the fuel when it is injected immediately breaks up in very small droplets as we saw in the last class and this creation of the small droplets is extremely vital to the process of evaporation and mixing and this process as we mentioned earlier is called atomization and this atomizing is an important phenomenon for the process of efficient burning or efficient combustion. The secondary mixing which takes place after the combustion is important for uniform distribution of the combustion products and uniform distribution of the heat that is released from the combustion and then the flame tube is to ensure that this takes place in a secured manner and the vent holes that are available around the inner liner or the inner flame tube or the inner can ensures that the flow that comes in from the air that comes in creates a secondary zone which ensures you have uniform temperature profile at the end of the secondary zone through a process of good mixing inside the flame tube whereas, in the primary zone the geometry is fixed the dilution of the combustion product is important in the secondary zone. The two stages of mixing one in the primary zone one in the secondary zone it is important that in the secondary zone for aircraft gas turbine you may have a tertiary zone and this is important for the life of the turbine. The life of the turbine is essentially dependent on the combustion delivery and this delivery is essentially created by the mixing process which may require two stages that is the tertiary zone may be required to ensure that the temperature that is going from the combustion chamber to the turbine is something comfortable to the turbine. The outer casing which normally delivers 75 percent of the air and envelops the entire all the cans that are inside essentially delivers the air the cold air so to say into the secondary and the tertiary zone after the combustion has been completed. So, the 75 percent of the air as we talked about do not take part in the combustion process and this inner can or the flame tube has the small vent holes and these vent holes are designer holes they have to be designed. So, typical geometry of the vent holes their positions their sizes of the holes are to be designed into the flame hole the flame tube and they then create a proper mixing process inside those mixing zone. So, creation of those vent holes is extremely important and it is the combustion chamber designer who creates those designed holes. The annular combustion chamber which we have talked about are expected to be used more and more in the gastrobines in the future because they are easier geometry and they are used in smaller small gastrobines even today in large numbers. The issue there is you do not have the localized combustion or secured combustion zone or secured primary zone all the combustions are taking place in an open annular space. So, the security that is provided by the can is not available anymore and as a result it may impact on the combustion efficiency a little and the combustion intensity a little and hence the annular combustion chamber may have somewhat lower efficiency and may have a somewhat lower combustion intensity. It however requires lesser of the cooling air for the total combustion process and hence it is lighter it does not have all the weights of all those cans and hence it is lighter and it essentially easily qualifies for application in military aircraft engines. Essentially an annular combustion chamber is nothing but very similar to the reheat chamber which people have been using for a long time. The temperature and velocity profiles are not as uniform as in the can type or the annular type because you do not have that kind of control as it is used by using the flame tubes or the can types. So, annular combustion chamber may not deliver a very uniform temperature velocity profile on to the turbines. Now, these are some of the things that the combustion chamber designer would have to take into account before deciding what kind of combustion chamber he wants to use and what kind of combustion chamber he would like to have for his kind of gas turbine engine. We have talked about combustion mechanism, we have talked about the combustion parameters that impact the combustion mechanism and how by design some of these parameters and some of these mechanisms can be kept under control on a continuous basis throughout the operation of the engine. In the next class we will be talking about certain issues that are related to the operation of the combustion. We will be talking about combustion stability and correspondingly combustion instability and we will be talking about the fuel injection system how the fuel is injected into the combustion chamber and the injectors themselves that is what we will do in the next class.


Welcome back to this course on nanostructured materials. We are in the module 4 and we are discussing optical properties of nanomaterials. This is the lecture number 10 of module 4 and we have 12 lectures in total in this module. This is going to be the second lecture on optical properties of nanostructured materials. So, in the previous lecture on optical properties, we discussed some of the basics of what happens when a particle size decreases, what happens to its optical properties, what happens to its color and why those changes in color take place. Specifically, we discussed about the nature of bonds and the nature of bands and the different discrete levels, which can be incorporated within the energy gap by impurities, which can cause different low energy absorptions, which can cause two photon absorption and emission. Then, the change in the shape of the density of states versus energy plots, how you can continuously change the density of states and energy in a bulk solid, which you cannot in say a nano well or a nano wire or a nano dot that is a quantum dot. So, what happens when you change a bulk particle to a nano sized particle and how it changes with the dimensionality, whether you lower the dimension to nano size along only one dimension or two dimension or all three dimensions will give rise to different changes in the density of states versus energy. That will bring rise to changes in the absorption of energy and the excitation of an electron from a valence band to a conduction band and what are the different types of properties, which you can expect in a semiconductor like cadmium selenide, which we saw how it gave different types of colors in colloidal solutions of cadmium selenide having the same material, but having different sized particles. That was one example, then we looked at different other types of optical transitions, the fluorescence and phosphorescence in these systems and what happens when the size decreases to the nano dimension. So, let us continue on that and on the second lecture. So, in semiconductors you have a conduction band and you have a valence band as discussed earlier and you have an energy gap and when you make a small sized semiconductor particles, then you have more discretized energy levels instead of a band, where you have a continuum of available energy states. Now, you have discrete energy states of course, the band gap is there and the band gap is enhanced this is much larger than the band gap in a bulk semiconductor. So, the band gap also increases along with discretization of the energy states. So, this effect is also called quantum confinement because we are confining the electrons and because the energy gap is becoming higher and electron cannot be excited easily unless you give higher energy. So, increase in the band gap and confinement appears for particles with around 10,000 atoms and that may be the size of like 0.5 nanometer particle when compared to a bulk particle which is around 1 micron. Now, when you have an electron which moves from the ground state which is the valence band to the excited state which is the conduction band. So, you have a hole left behind and that hole is positively charged because it is the absence of the electron which went into the conduction band. Now, this pair of electron and hole together because in this semiconductor now you have an electron in the conduction band and a hole in the valence band together these two particles will be called an electron hole pair and it is also called as an exciton. So, what is an exciton? An exciton is a electron and hole pair and that has been created because the particle has absorbed energy which has been given and that energy is larger than the band gap and so an electron is excited and it leaves behind a hole in the valence band and together the electron and hole pair is called an exciton. So, this is the basics of absorption of radiation creating excitons. Now, what we discussed earlier also what happens as the size decreases? As the size decreases the band gap changes. So, we can tune the band gap it is called the tunable band gap in semiconductors and this has been plotted for several semiconductors and for bulk silicon for example, you have a value of 1.14 electron volts and for bulk gallium arsenide you have a value of 1.5 electron volt. Now, these values can change depending on the size of the nanoparticle. So, here you have some quantum dot which has a band gap of around 5 electron volts when it is very small around say 10 angstroms or 0.1 nanometer and when it is like 30 angstroms. So, 30 angstroms 1 angstrom is like 0.1 nanometer. So, 30 angstrom is like 3 nanometers. So, if you do this study then you can see the band gap varying from 5 e v to around 2 e v when you are changing the size of the quantum dot. The diameter of the quantum dot is changing from say 1 nanometer because 10 angstroms is 1 nanometer and this is in angstroms. So, 1 nanometer particle has a band gap of 5 electron volts and 3 nanometer particle has a band gap of around 2. So, you can vary a considerable amount say around 3 electron volts in this particular semiconductor where by just varying the size or the diameter of the particle. Now, this change in the diameter and the band gap reflects it in its colors. So, if you plot the optical density with energy for particles of the same kind. So, you have cadmium selenide particles these are quantum dots and these are having different diameter. So, you have 21 angstrom. So, it has an absorption at around 2.6 electron volts. When you go to higher particle size say 40 angstroms which is 4 nanometers your band gap is now close to 2.01 or 2.02 electron volts. So, from 2.02 electron volts at 4 nanometers which is a larger particle if you decrease the size of the particle to around 21 nanometer the band gap has increased to around 2.56 electron volts and this shift in the band gap can show the variation in colors. So, a large particle will have a smaller band gap and will have a color which is towards the red if it absorbs and this shift to higher energy in as the size of the particle become smaller you are shifting to a higher energy this is typical of size effects quantum size effects. And also it shows you the discrete structure of the spectra in addition you also see that the intensity of the absorption. So, it becomes more intense more sharp and more intense as the size of the particle is decreasing. So, if when it is 21 angstrom is much sharper and when it is 40 angstroms it is little broader. So, not only the wavelength or the energy shifting but the intensity also changes as you change the size of the quantum dots or the size of the semiconductor and this can be cadmium selenide it can be studied in other quantum dots as well. Now, this so far we were looking at semiconductors which have a valence band and which have a conduction band and there is a band gap and you are changing the band gap of the semiconductor by changing the size of the particle. So, that is about semiconductors in metals what happens because in metals you do not have a band gap. So, the color of a metal nanoparticle normal bulk metals are normally black in color or very shiny and but when you make nano sized metal particles colloidal solutions you will see colored solutions even for metal particles. So, but the absorption phenomena is not the same as in semiconducting nanoparticles because metals do not have a band gap whereas, semiconductors have a band gap. So, what is this color in metals due to? So, if you look at metal nanoparticles there is this property of surface plus bond resonance which occurs especially if your nanoparticles are metals. So, in this what happens when if you apply an electromagnetic field the electromagnetic field is like a light is also electromagnetic field. So, whenever you shine light there is electromagnetic field and if that nanometer size particle the size is smaller than the wavelength of the light which you are passing. In that case there can be a oscillation of the conduction electrons because in metals you have conduction electrons the electrons are present in the conduction band. So, these electrons can have coherent oscillation if the size of the particle is much smaller than the wavelength of the light which you are passing. So, if the wavelength of the electromagnetic radiation which we is light here is much smaller than the size of the metal particle then the conduction electrons of the metal particle can oscillate coherently that means in phase with the conduction band electrons and this is basically due to the interaction with the electromagnetic field and this is called surface plasmon resonance plasmon is basically quantized oscillations of conducting electrons of the conduction band. So, all the metals will have conduction electrons and they if they oscillate as light falls on it then you will have surface plasmon resonance. So, this is the schematics where you have this metal particle and this is the electric field and this shows the oscillations and this electron cloud. So, this electron cloud can oscillate along with the electromagnetic radiation which is light if the particle size is much smaller than the wavelength of the incoming light beam. So, a wavelength of the incoming light beam is of this length because you have one crest and one trough makes the wavelength and you see the particle is much smaller than the wavelength of the light coming. Suppose the particle had been this big and this is the wavelength then you will not see this surface plasmon resonance. So, this you can see when the size of the particle is much smaller than the wavelength of the light and then you will see these oscillations of the conduction band electrons and that is called the plasmon. Now, when you have that when you have this kind of particle which you consider as a dipole. So, you can have in a dipole positive and negative arrangement creating a net dipole moment if there is a shift in the electron density then you can write the equation for the dipole moment as a function of the dielectric constants and the applied electric field. So, if the metal cluster is placed in an electric field because electromagnetic radiation or light is associated with the oscillating electric field it also has oscillating magnetic field, but light has a oscillating electric field. So, that means whenever you have light you have an electric field and the metal particles are now in the electric field. So, the negative charges will be displaced from the positive ones in the presence of a electromagnetic field and then you can write this equation where this is a electric polarizability if this whole quantity where epsilon is the dielectric constant of the particle and epsilon m is the dielectric constant of the medium. So, and E naught is the applied electric field from the electromagnetic radiation. So, if this entire term is the electric polarizability alpha then you can write alpha is equal to this and then you see in the denominator if you want to make alpha very large then you need this to be very small. Now, this can be very small or you can make this whole thing go to infinity when this goes to 0. Now, here this will go to 0 when epsilon plus 2 epsilon m becomes equal to 0 and that can happen only when epsilon is equal to negative of 2 twice of the epsilon of the medium. So, that is what is written in a different way that if the magnitude of this term which is in the denominator becomes a minimum then this will become very high. So, you want to know when alpha will be high that is when will the electric polarizability be very high if this is high then this will be high. So, now so there are two things one is the electric the dielectric constant of the particle and the dielectric constant of the medium and these two have to be taken into consideration and depending on their numbers you can find out when this will become a huge value and that is when we call surface plus 1 resonance to occur or S P R to occur when this value will become a very high and that will happen. So, you will have surface plus 1 resonance when epsilon prime becomes equal to negative of 2 epsilon m that is because of this equation that this is in the denominator and you want to make this whole term very large. So, that will happen only when epsilon plus 2 twice epsilon m becomes equal to 0 and this epsilon of the nano material is can be written as the epsilon 1 and epsilon 2 with the epsilon 1 is the real part and epsilon 2 is the imaginary part of the complex dielectric constant of the metal. So, similarly you can write the dielectric constant of the medium in which you have this metal particles embedded. So, what you basically want is this equation that is the real part of the dielectric constant of the metal should be equal to negative of twice of the dielectric constant of the medium and then you will have a resonance of the oscillating electrons. So, these is the oscillating electron cloud. So, it is up there and down there and that will oscillate such that there will be a resonance between the incident light wave or the optical beam and that can happen exactly when this value matches. So, this is the criteria for surface plus 1 resonance to occur and that depends on the dielectric constant of the metal and the dielectric constant of the medium in which the metal has been impregnated or supported and or the medium through which the light is being passed. Now, according to the Drude model we can rewrite the real part of the dielectric constant and the imaginary part of the dielectric constant using these equations where we incorporate the dielectric constant at infinite or very high frequency and we incorporate the plasma frequency omega p and there is a dielectric relaxation frequency omega d and. So, using these equations you know the relation between the real part of the dielectric constant to the plasma frequency. Similarly, the complex part of the plasma of the dielectric constant can be written in terms of the plasma frequency and the relaxation of the damping frequency. Now, you can ultimately find out that the plasma frequency of the metal is related to the density, the number density of the electrons, the velocity of the electrons at the Fermi energy and the mean free path of the conduction electrons. So, the plasma frequency you can kind of understand if you know the mean free path of the conduction electrons and the velocity of the electrons at the Fermi energy which is called V f. If you know that then you can determine the omega p which is the plasma frequency and that is related to these numbers here. So, here this is the various constants are there the electronic charge, the mass and the dielectric constant dielectric permittivity of free space. So, if you know the plasma frequency of course, you know the plasma energy from that one can find out what kind of particles will show what kind of plasma frequency or plasma energy and from that you can find out if the plasma frequency or energy is lying in the visible region or not. If it lies in the visible region then due to this surface plasma resonance you will see a color if it does not lie in the visible region you will not see the color. So, if you look at that the if you calculate the plasma frequency of different metals and with different sizes then you can calculate you can see which one will be red in color which one will be blue in color etcetera. So, in this chart you can see that if the plasma on energy goes from say 3.5 electron volts there to around 0.5 electron volts you scan the region from ultraviolet to the infrared and here different particles with different morphology have been have do show different colors which has been represented here for example, if you have silver nanospheres. So, if you make silver particles of a particular size then you will be having a color which is in around blue in color, but it may also have green color depending on the size of the nanoparticles. So, there is a broad range of plasma on energies for silver nanospheres from starting from around 410, 420 nanometers to around approximately 580 nanometers. So, it is a big energy region that means with silver nanospheres you can get different energies from the green to the violet by changing just the size of the nanospheres of silver. However, if you have gold shows much more variety and much more range. So, if you have gold nanospheres you can vary the plasma on energy and the color associated with that plasma on energy from say 600 nanometers to something around 780, 760 nanometers just by changing the diameter of the gold nanospheres you can get all these colors from the yellow orange to the red using gold nanospheres, but you cannot get blue color with gold nanospheres as shown using surface plasma on. If you have a structure which is not a sphere, but made of gold and it is in the form of nano shells. So, it is like a strip of gold particles in a circular fashion with a certain thickness without any thing filled inside. So, that is like a nano shell or a nano egg if you can make it can vary from 600 nanometers to 9000 nanometers is a very large range. So, changing the shell thickness and changing the diameter of the nano shell you can have all kinds of colors from green, yellow, red, brown and even colorless those particles which have a plasma on energy in the mid infrared say 3000 nanometers. You are already in the infrared very large wavelength very small energy you can get colorless particles of gold. So, you can not only get brown color red color you can get colorless particles of gold as shown in the figure. Then you can get using nano rods you can get from nearly bluish in color to up till brown color with nano triangles gold can be also made in triangular form not only as a disk not only as a sphere not only as a nano rod gold can also be made as a triangle. So, these triangles are very useful because see where they absorb these triangles can absorb in the infrared mid infrared whenever you can have anything absorb in the infrared that means you can heat that material. So, if you have gold nano triangles in your body say you inject gold nano triangle somewhere people have shown how then it will heat up the cells around it. So, this is using near IR or mid IR energy which gets absorbed by this nano triangles of gold it can heat the surroundings by picking up energy. So, range of energies are possible for the same material if you change the dimensions or you change the shape you can get a large range of energies using the SPR band. So, this is a very particular for metal nanoparticles this is not for semiconductor nanoparticles this is for metal nanoparticles for semiconductor nanoparticles of course you can change color, but then you have to worry not about surface plasmon, but you have to worry about the band gap and how what is the band width how discrete are the levels etcetera. So, that is for semiconductor nanoparticles and this is for metal nanoparticles where the color can come from the surface plasmon resonance. Now, this plasma oscillation is basically a fluctuation of the free electrons the density fluctuation of the free electrons or free electrons means those electrons which are in the conduction band. In the conduction band the electrons are not held to the core tightly and so the electrons are supposed to be free to move. So, they may be called as free electrons in metals many times we use the term free electrons basically they are electrons in the conduction band and they can supposedly move around. That means, delocalized over all the different atoms in the cluster or in the nanoparticle. So, if you have a fluctuation density fluctuation of free electrons which are present in a metallic solid then you can have what is called the plasma oscillation. So, the plasmon's oscillate at the plasma frequency which is related to the free electron density and the effective mass. Now, if you confine these plasmon's this was in the bulk that they are oscillating with a plasma frequency when you confine the plasmon's to surfaces which can interact with light. So, if you have a wave which is going through the material and the material can interact with the slide then it can form what is called a propagating surface plasmon polariton. So, that is called an SPP band surface plasmon polariton and that happens when plasmon's are confined to surfaces they are not present in the bulk they are only confined to surfaces and then this kind of surface polaritons can form and that results if you have confinement that means, if you decrease the size of the particle then this surface plasmon polariton modes they get affected. So, earlier if you look in the bulk the plasma frequency is related to the free electron density and the effective mass in this fashion where in the nanoparticles where you observe the surface plasmon polaritons where confinement is there because now you cannot have the plasmon's in anywhere in the bulk they are only in the surface. So, confinement effects are there in this polaritons and this equation of the plasma frequency changes and now you see a factor 1 by 3 has come in the in finding out the frequency of the nanoparticles when it is showing resonant SPP modes or resonant surface plasmon polariton modes. So, different whether the plasmon's are confined or they are in the bulk you will have different plasmon frequencies that is what is being told and hence you will have different colors because the frequency will be related to energy being absorbed and that will be related to the colors which are given out or which you see for the different nanoparticles. Now, Michael Faraday as discussed earlier was the first to report the synthesis of colloidal metal particles and in this case the colloidal metal was gold. So, it was a gold particles which are kept here in this solution and you can see they are stable even today the particles do not settle down it is a clear looking solution is basically made up of gold nanoparticles in this colloidal solution and long back in 1908 me tried to explain the phenomena of the color of these metallic colloids and that he did by solving Maxwell's equation and he predicted that for homogeneous metal particles you will have optical extinction only when the size of the particle here it is written 2 r that means twice the radius as the diameter of the particle is much smaller than the wavelength of the light which you are using to see the particles. So, me predicted that this kind of optical extinction involves scattering as well as absorption. So, there is an incoming beam there will be an intensity of the incoming beam and when it gets scattered or it gets absorbed then that intensity gets lost and that is also you can say is called extinction. So, me tried to explain this extinction based on scattering and absorption and he predicted that this kind of extinction will be valid when the size of the particle is much much smaller than the wavelength of the light and that is what we discussed earlier also that you will have this oscillations of the conduction electrons of the metal particles when the size of the particle is much smaller than the wavelength of the light. So, the wavelength of light is this big and the size of the particle is only this big because the size is much small then you see this SPR band and then you can see colors of metals in colloids as seen by Michael Faraday long back around 1856 and 1857 and that is remarkable Faraday's contribution not only in other fields of electromagnetism and laws of electrolysis, but also in nanoscience where the first metal nanoparticle was synthesized in the laboratory by Michael Faraday as gold nanoparticles. So, this is the solution of gold nanoparticles and the color explained by Mies theory based on extinction due to scattering and absorption for particles which are much smaller than the wavelength of light. Now, when you have two particles two metal particles how will these plasmon interact? So, will there be a shift in the surface plasmon band? So, if we try to understand the interaction between particles and the fields around these particles an isolated sphere or a particle the polarization direction does not matter it does not matter whether vector is pointing this way or that way, but when there are two particles it will matter in which way these dipoles are being shown and how they are present? If they are present in the same in the same direction then the interaction it will be in one way and in this case it will be in a different way. So, this is a longitudinal coupling of the two metal particles to the oscillations of the conduction band electrons in the two particles and this coupling results in a shift of the SPR band to lower frequency. So, the restoring force in the longitudinal coupling is actually reduced when this one couples to this one. So, that is because they produce a resultant which is opposing the original plasmon vector. So, the restoring force is reduced by coupling to the neighbor and the SPR the band shifts to lower frequency where the force is reduced you shift to lower frequency and if it has a transverse component. So, there are these two particles metal particles and these are the vectors and if they have a transverse interaction transverse coupling then the restoring force is increased by coupling. So, the resonance will shift to higher frequencies. So, there are two possibilities of coupling of two metal nanoparticles the coupling with can be longitudinal and then longitudinal means the two dipoles are kind of arranged in the same direction and they have a restoring force which reduces tries to reduce the coupling. So, basically the resonance the SPR band will shift to lower frequency or lower energy whereas, in the transverse case the SPR band will shift to higher frequency. This is with the interaction of particles because in many systems nanoparticles will not be alone they will have neighbors and so particle particle interaction will be there between these metal nanoparticles and one has to understand what happens to the plasmon's dipolaritons when the coupling takes place. So, the size dependence that is say gold particles we say bulk gold is yellow in color and when we decrease the size of the particles all kinds of colors can be seen. Now, all these colors at different size of the gold particles are not coming due to one reason there are two reasons why you see color in gold particles one is that the geometric factor the blue the purple the large size particles the color is explained by me theory where the particle acts as a scatterer. So, light scattering by the metal sphere can be explained is used to explain by me theory which gives rise to the gold blue and purple colors these colors for large particles this is when the metal nanoparticle is larger than approximately 30 nanometers. Then the these particles the reason because of this large particles having colors is that some of the electrons which are oscillating with the field are not in phase. So, some electrons get behind phase and this is called phase retardation or retardation effect and for that the color can change. So, this change in color for the larger nanoparticles of metals is explained by me theory where the light scattering by the metal spheres involves electrons. But all the electrons are not oscillating with the light in phase some electrons are lagging in phase and this phase retardation can cause these colors the subsequent at the small size level when you are below a certain size say 5 nanometers 3 nanometers 4 nanometers at that size the changes in color which is more towards reddish brown to orange and then finally, it becomes color less at very small size those are due to quantum size effects which we had discussed earlier. In quantum size effects we had discussed as you reduce the size of the particles the band gap is increasing the energy states become more discrete and this is called the quantum size effect and the color changes associated with that or the changes in optical properties associated with these small particles say less than 5 nanometer particles is due to quantum confinement and can also be said is due to the quantum size effect. However, the color changes at large sized particles at 30 nanometers 25 nanometers etcetera is explained by me as due to scattering from scattering of light leading to its resonance with the conduction band electrons which is called the SPR. But all the electrons are not resonating in phase some electrons because of the large size of the particle are lagging and hence it causes a shift in the color in the large size particles. Whereas, in the quantum regime when you are talking of very small particles the variations in the color are due to quantum effects are due to changes in the density of states with energy are changes due to the bandwidth or the discretization of the energy levels. So, there are two different area two different phases by which you can explain the color. So, the color the range of size of particles depends on the color depends on the range of size of particles in one range the higher range the color can be explained by the me theory and at lower dimensions the color changes can be explained using the quantum confinement or quantum size effects. So, at very small size the colors are basically reddish brown to orange or color less and those are due to quantum size effects. But at large size this blue violet and red at this size you can explain using light scattering by a sphere by as explained by the me theory. Now, what is the effect of the surrounding medium if you have a metal particle and you have some solvent molecules say water or ethanol or cyclohexane how does the surface plasmon band or the S P R peak which is shown here like a how does this change as a function of the environment. So, here you can see that you have spherical silver particles and these spherical silver particles have an absorption around 440 nanometers. So, this is the absorption of the spherical silver nanoparticles. Now, if you move them from suppose they were in air now if you move the air remove the air and add some other liquid which has a refractive index of say 1.44 then you get the plot B. So, from A which has a maxima at 440 nanometers you now get B which has a maxima around 500 nanometers. So, this is a shift of 60 nanometers by changing the environment from air to some liquid which has a refractive index of around 1.44 and then further change can be observed if you change the refractive index of the medium from 1.44 to 1.48 you will get the next plot and like that there is a continuous shift in the maxima of the absorption of this energy due to the surface plasmon resonance by the change in the environment. So, it is highly sensitive the S P R band is sensitive to the environment and this sensitivity of the S P R band for metal nanoparticles especially of gold silver etcetera can detect adsorbate induced changes in the dielectric constant of the surrounding nano environment. So, the sensitivity of the S P R band can be utilized for a device which can sense what is happening in its environment. If you make any changes in its environment it can immediately detect because it will change its absorption of the surface plasmon bond.  So, it is a sensitive tool and if properly applied you can make it a chemo sensor that is sensing molecules of different types or biosensing which is again sensing molecules of different type, but of biological relevance that means of relevance to life and medicine. So, you can use the S P R band very effectively in biological applications in medical applications for chemo sensing and biosensing. So, this particular example as I mentioned was for silver nanoparticles and the particle size the particle size was fixed what you change is the environment around the particles. And as you change the environment of the particles that is you change the liquid in which the particles are present and change the refractive index of the liquid then the absorption the surface plasmon band changes and you can follow these changes very effectively. Now, one thing is about environment the other is about surface modification what about if you put some molecules or some functional groups on the surface of these gold particles. So, there will be what is called surface modification and the optical properties of metal particles like gold particles or silver particles etcetera are influenced by their environment. And if you change using a controlled manner the surface you can alter their properties in a very controlled manner. So, you can do controlled properties of nanoparticle assemblies by the separation between the particles that is how much distance the two interacting particles should be. If you can control by using some methodology then there will be very interesting approach towards biosensing by controlling the inter particle interactions within these assemblies of metal nanoparticles. Also you can code these metal particles with a uniform shell of inert material like silica or titanium etcetera you can cover them these metal particles with silica titanium etcetera and control the distance between the particles. So, the two metal particles will be separated even if they try to touch each other since there is a outer shell they will be away from each other by at least twice the thickness of the one shell or some of the shell thickness of the shells formed around the two particles. So, you can control interactions by controlling the distance by keeping the particles at different distances chemically or physically or by introducing shells around the particles with particular shell thicknesses which will allow one to keep the particles at specific distances and hence control the surface plasma. Now, further so to that this is an example of silica coated gold nanoparticles. So, you have got gold nanoparticles here and these gold nanoparticles have been covered with silica. So, that is what I was telling that if you consider this particle inside and this particle inside they may have important interactions, but those can be controlled by varying the shell thickness of silica on gold particles. So, different types of thicknesses can be formed and what is given is you can make particles with 10 nanometers with 23 nanometers with 58 nanometers and with 83 nanometers different size of particles and they are gold coated with different shell thicknesses. So, the particle size is not varying the shell thickness is varying and in this case the particle is small of say 15 nanometer and the shell thickness is also of similar order of around 10 nanometers. So, this is most uniform here of course, the particle size the core has increased although the thickness has been kept constant in all of them of 15 nanometers. So, this type of silica coated gold nanoparticles are of importance for many applications and their thickness of the shell can be monitored and controlled and that will control when you try to bring the particles together the metal particles will come to some close distance and that distance can be controlled by controlling the shell thickness that is the amount of silica around the gold particles. Now, what happens to the spectra that is the absorption when you have this kind of shells around the gold particles. So, this is a plot of the absorbance of this gold colloids which have been coated with silica and you see the dark bold line is the surface plasmon band of the gold particles with no silica shell. So, there is this is the surface plasmon band which is coming around 520 nanometers approximately now that has zero shell thickness. So, there is no silica on top of these gold particles but when you add or make a 2 nanometer thick shell around this gold particle you see that the intensity has gone up and there may be a slight shift in the peak of this gold silica core shell nanostructures. When it is 4 nanometer you can clearly see the shift in the surface plasmon as well as the intensity will get affected and this is the pictures of these particles of gold with different silica with a 60 nanometer silica shell and what has been changed is the refractive index of the liquid in which these particles have been dropped. So, you have the gold particles which are covered with 60 nanometer silica shell and you put them in some liquid which has a refractive index of 1.45, 1.42. So, you are varying the refractive index of the liquid of the medium and you see the shift in a color using the same nanoparticle 15 nanometer gold particle and 16 nanometer thick silica shell on top of the gold particles and you are adding the same particles to four different solvents and what is different in these solvents is the refractive index of the solvents and you can see variation of these gold silica nanoparticles in the depending on the environment and this has lot of importance in future applications. So, with that we come to an end to today's lecture and this was the 10th lecture of the module 4 and the second lecture on optical properties and in our next lecture we will complete our study on optical properties of nanostructured materials and that will be our 11th lecture and then the last lecture will be the 12th and last lecture of the module and the last lecture of the course which will be on mechanical properties of nanostructured materials. So, thank you very much and we will meet for the next lecture.


This is stock number 9J106A. We are here at Summit Automotive in Fond du Lac, Wisconsin. You're new and used Jeep and SUV headquarters. Today we are checking out this super clean 2019 Jeep Cherokee Latitude. This vehicle has the 2.4 liter multi-air motor which pumps out 184 horsepower. From this HD video you will be able to tell that this Jeep is extremely clean all the way around. Billet Silver is the color. We shoot all of our videos in 1080p, 60 frames per second, so if you have HD capabilities on your computer tablet or a smartphone device turn them on right now because it is like you're right here looking at the Jeep with me. And definitely your best way to check out the vehicle before seeing it in person. And if you want to check out more pictures in the upper right hand part of your screen is a link to our website. Click that and check out some more photos and descriptions and pricing and everything there. This vehicle comes with 17-inch painted alloy rims and it has Firestone destination 22560R17 tires and they have probably about 70 to 80% of the tread left on them. They went to a more of a Billet Silver painted rim as opposed to like the bright silver with the new body style here. You get the LED running lights. It has LED headlamps and LED fog lamps. Front bumper is in really nice condition. No dents or dings or cracks on that. You get these seven slotted grille of course on the Jeep. The hood, no dents or dings on there. And the passenger side fender is in excellent condition as well. No dents or dings on there. Passengers side rim is in excellent shape as well. No scuffs or scrapes there. And as you go down this side of the Jeep you can see just how clean the body is, how reflective and mirror like that paint is. That's why we take these HD videos so if you are far away or even if you're close by and just can make the trip down but you're still interested in purchasing the vehicle you can see the vehicle, hear the vehicle and have confidence in what you're looking at before you even get here. In fact we offer a Summit Auto Fastpass option where you can apply for financing, get payment quotes and even appraise your trade right on our website. So if you want to do that in the upper right hand part of your screen another link to our website click that and go there. Back rim is in excellent shape. No scuffs or scrapes in the back tires have just as much tread as the front tires. In fact with that Summit Auto Fastpass, actually we got capitalist fuel fill. They went to that for 2019. Summit Auto Fastpass is the easiest way to purchase the vehicle. Do everything right online and then you show up and basically you'll have your paperwork all ready to go. And uh yeah pretty easy. Built silver really good. A lot of metal flake in the paint. You can see the rear bumper was nice and clean. LED tail lamps. This is a manual raised gate. Back storage area is very nice and clean and those seats do fold down for extra storage. Has the tire inflator kit from Mopar because this is a front wheel drive vehicle and the front wheel drives only come with the inflator kits. Has the shocks are working real well and you'd expect them to be on 2019. No dense dings or cracks in that rear bumper. Rear quarter is in nice shape as well and the driver's side quarter is in really nice condition too. Back rim no scuffs or scrapes on there and down the rest of the side of the vehicle just as clean as that passenger side. Has the entering go system. The heated power mirrors. Inside the latitude package gives you the leatherette and cloth interior. There are no rips. There are no tears on the seats. Power driver seat with lumbar. We have factory floor mats throughout on this vehicle and this one does have auto headlamps. Power windows, power locks and power mirrors. As we hop inside the vehicle here you can see that this one has 11,455 miles. You get the digital speedometer, outside temperature and compass display. The leather wrap steering wheel is an excellent condition as well. Cruise controls on the right, Bluetooth and information center controls on the left. On the back of the steering wheel are your audio controls. This one has the 7-inch radio, AM, FM and Sirius XM radio capabilities. This is the 4C radio so you can hook up your Android Auto and Apple CarPlay devices to this vehicle and do that there.  And those are dynamic grid lines so as you turn the steering wheel they turn as well. If you don't like doing your climate control on the radio you can do it down here. It has stability control and your start-stop feature. Just press that button and turn off the start-stop feature. Augs and USB right there has the 9-speed automatic transmission passengers side floor mat and seat are in excellent condition. No rips or tears on that and the headliner is in really nice clean conditional rips or tears up there. It does have map lights up here and we'll take a quick look at the back seats and then check out under the hood. Didn't want to point out it does have that keyless entry so you just have that fob in your pocket how the enter and go system works and if you put your hand in the door handle it'll unlock the doors for you. Back seats are just as clean as the front seats. There are no rips there are no tears back here.  Factory floor mats throughout back here as well and you do get a USB and USB-C plugins back here to on each side and these seats do fold down for extra storage which they go down really nice and flat which is nice if you have to haul a bunch of extra stuff and we'll take a quick look under the hood. Finally child door locks back here. I'd personally like to thank you for checking out the video today and hopefully from this HD video you will have been able to tell just how clean this Jeep is all the way around inside and out. Under the hood we have the 2.4 liter multi-air motor and Jinbei is very clean runs very smooth. This Jeep has been fully safety to and inspected by our service shop has a fresh oil and filter change. All the fluids have been checked and topped off. The vehicle has been gone through mechanically 100% and is 100% ready to go. To see more pictures of this Jeep or one of our other 450 new and used cars, trucks, SUVs, minivans, Wranglers, halftons, three-quarter tons, one-tons, you name it we got to go to our website www.summitauto.com full pictures and descriptions of every single vehicle from two locations all at summitauto.com and if you'd like to check out more HD videos you can go to youtube.com slash summitauto. In fact in the bottom right hand part of your screen is a link to subscribe to our YouTube channel. On the right a link to more used inventory videos like this one and in the bottom if you haven't been to our website a link to this vehicle on our website click those check us out and we really look forward to helping you with this super clean 2019 Jeep Cherokee Latitude. Thanks again.


You may be familiar with the term acid to describe a substance that has a pH lower than 7, turns blue litmus paper red, and turns universal indicator solution, or paper, red, or orange, all depending on its strength. An acid is simply a substance that dissociates to release a hydrogen ion or proton in water. Let's start by thinking about citrus fruits, oranges, lemons, and limes. Many citrus fruits have a very defining sour taste to them. This sour taste is due to the presence of a certain weak acid, citric acid. It is a common misconception that citric acid is vitamin C. In fact, vitamin C is ascorbic acid, and although it is found in citrus fruits, kiwi fruits and red peppers have a much higher ascorbic acid concentration. Another food item that you may be familiar with is vinegar. Vinegar is a very dilute solution of ethanoic acid, also known as acetic acid. Like citrus fruits, vinegar has a very distinctive sour taste. Rhubarb leaves are toxic because of a high concentration of oxalic acid. Oxalic acid, however, is still found in low concentrations in spinach, parsley, and chives. Let's open a cold soft drink can or bottle in our minds. Most likely, you imagine many bubbles rising from the drink. This happens because carbonic acid is formed when carbon dioxide is fully dissolved in soft drinks during the manufacturing process. Here is a challenge for you. You know that carbon dioxide can dissolve in water to give carbonic acid. Can you write down the equation for this reaction? Pause the lesson and continue when ready. The reaction equation is carbon dioxide and water equilibrates to give carbonic acid. Notice the reaction arrow that shows you that this is a reversible reaction. This carbonic acid can dissociate to give hydrogen carbonate and a proton. In fact, an acidic solution is made because of the presence of these protons. The extent of the dissociation of carbonic acid is critical in maintaining the optimal pH. Our blood has a very specific pH 7.41. If the concentration of protons is too high, it will increase the acidity of blood. Even at 7.35, the medical condition is known as acidosis. If the concentration of protons is too low, it will decrease the acidity of blood. Even at 7.45, then the condition is known as alkalosis. If you exercised for an extended period of time, your muscles will begin to cramp. This cramping feeling is due to the presence of lactic acid produced by your muscle cells when there is insufficient oxygen. Milk contains lactose and certain types of bacteria can ferment lactose into lactic acid. The lactic acid then reacts with a certain milk protein called chazen. The combination of lactic acid and curdled chazen gives yogurt and sour cream their texture and taste. So to summarize, there are many kinds of acids, not just hydrochloric acid and sulfuric acid. A lot of these acids can be found in everyday food items, inside our body and in many other real-life applications.