 So, in the last class we started discussing chemical equilibrium and we have shown that at equilibrium delta G is equal to 0, which is the change in Gibbs free energy is 0 for a chemical equilibrium process. And from there we have shown that delta G naught is equal to R T L L K P, where delta G naught is the change in Gibbs free energy at standard state. And R this R is the universal gas constant, temperature in Kelvin and K P is the equilibrium constant. This is defined for standard pressure. So, P naught is 1 atmosphere. Now, we have also shown that the Gibbs free energy for standard state can be obtained by knowing the molar Gibbs free energy of formation of different species. We have also shown that for the formation reactions, since we can estimate how much is the, since we know the Gibbs free energy of formation, we can estimate the K P for formation reactions. Then for a complex reaction and not a formation reaction, we have shown in the last class that by combining different reactions, we can get the expression for K P for little more complex reaction by combining certain few formation reactions. Now, let us continue from there. We have been discussing this K P in the last class. Now, let us look at various forms of this constant. Let us consider this chemical reaction, the generic chemical reaction that we have been discussing so far. This is our generic chemical reaction. What we are now going to talk about is various forms of equilibrium constant for this chemical reaction, generic chemical reaction at equilibrium. Let us consider at equilibrium. So, first of all K P, we have defined K P. We know that K P is a function of temperature and we have defined it as pi i equal to 1 to n that is product of partial pressure at equilibrium of different species rest to repower the difference in stoichiometric coefficient. This we have done in the previous class. Now, we can define the equilibrium constant in terms of moles also, K n. So, K n is defined as pi i equal to 1 to n. The number of moles of particular species i at equilibrium rest to repower again the difference in stoichiometric coefficient where in this reaction, sorry in this equation n i represents the actual number of moles of species i which is typically given as the partial pressure multiplied by volume divided by R T. So, this is equal to P i by P by totals number of moles multiplied by total numbers of moles where N T is the total number of moles in the mixture. So, N T is the total number of moles in the mixture. N T is nothing but the summation of number of moles of all the species i. Similarly, we can define K C that is the equilibrium constant based on concentration. So, that is defined as pi i equal to 1 to n C i e to the power the difference in the stoichiometric coefficient where the C i e represent the concentration of species i at equilibrium. So, that is the equilibrium constant based on concentration. Similarly, we can define based on mole fraction K x. So, on mole fraction based on mole fraction we can get again the concentration of species i as a mole fraction of species i at equilibrium rest to repower the difference in stoichiometric coefficient. Here once again kappa i is the mole fraction of i which is nothing but equal to P i by P that is N i by N T. P i is the partial pressure at equilibrium of species i, P is the total pressure of the system, N i is number of moles of a particular species i, N T is total number of moles. So, this is how it is defined for based on mole fraction. Now, similarly we can also define this property equilibrium constant based on mass fraction y. So, based on mass fraction I can define it as very similar definition where y i is the mass fraction of species i. So, y i is mass fraction. So, that is equal to molecular weight of species i times partial pressure times rho T R. So, that is nothing but molecular weight times concentration divided by density or the density of species i divided by total mass average density that is the definition of mass fraction. So, these are defined properties. Now, I would like to point out one more thing the relationship between all this different representation of equilibrium constant. For that let us first define the change in number of moles in a chemical reaction. So, change in number of moles in a chemical reaction given by delta N is equal to the sum of the number of moles in the product side minus sum of moles number of moles in the reactant side. So, that is the total change in the number of moles. Then it can be shown that K p is equal to K c R T to the power delta N. This is also equal to K n R T by V to the power delta N. Also equal to K kappa P to the power delta N. It is also equal to K y R T rho to the power delta N. Also equal to K kappa P to the power delta N. It is also equal to K y R T rho to the power delta N times pi i equal to 1 to N omega i to the power this. So, this are the different equilibrium constant based on different properties and this is the relationship among them. Now, depending on what is known to us we can estimate the other. So, this is something that is a very important information which is typically used to estimate one of this equilibrium constant or other knowing the other properties. Now, before we proceed further with the chemical equilibrium let us solve a problem to essentially make this concept clearer. Let us consider that we have a mixture of 1 mole of nitrogen and half mole of oxygen and they are heated at 4000 degree Kelvin. So, because of and 1 atmosphere pressure and 1 atmosphere pressure. So, because of that there is a resulting reaction and we get the final equilibrium mixture consist of nitrogen, oxygen and some nitrogen, nitric oxide NO NOX. We can also assume the initial temperature of both nitrogen and oxygen T i is 298.15 Kelvin that is essentially the standard temperature. So, nitrogen and oxygen are supplied at a standard temperature and then they were heated steadily. Now, the question is here how much heat will be required to take this to 4000 degree Kelvin temperature. So, how much delta q is equal to what how much heat is required. So, first for this problem let us look at what is known to us. First we want to estimate this A B and C how much how much is the total what is the total so that the final composition. In order to do that first we do that atomic species balance. So, we have two species nitrogen and oxygen. In the reactant side we had we have two atoms of nitrogen. So, this is equal to then 2 A plus C in the product and the oxygen in the reactant side we have one atom of oxygen this is equal to then 2 B plus C. So, if you simplify this we get A is equal to 1 minus C by 2 B equal to half minus C by 2. So, then at equilibrium this is what is our final composition. So, if you can estimate C we know how much is the final product composition. So, my final equilibrium reaction then is this looks like 1 minus C by 2 times N 2 plus half minus C by 2 times O 2 plus C N O. Now, what is given by this? Given to us that the initial temperature is T i the final temperature of the mixture is 4000 Kelvin. So, here what we are solving is that what we are saying that initially the mixtures were at this temperature first they were heated to this temperature then reaction occurs. It is equivalent to saying that the reaction occurs at this temperature and then the heat takes this to this temperature. So, both of them are equivalent that we have already discussed. Now, if we had a stoichiometric reaction. So, this reaction as you can see clearly is not stoichiometric, but for a stoichiometric reaction it will be half N 2 plus half O 2 giving N O. So, if you have to produce 1 mole of N O we need half N 2 and half O that is my stoichiometric reaction and the value of K p for this reaction can be obtained from thermo chemical tables this is equal to 0.03. Now, as we can see here from our discussion we had 2 species we had 3 unknowns. So, you are not able to solve this. Now, we are be using that information the K p information to solve this. Now, this K p is applicable at equilibrium and this stoichiometric reaction is the equilibrium stoichiometric reaction. So, for this reaction K p is equal to P N O by P N 2 to the power half P O 2 to the power half. Now, what is the partial pressure of N O in this mixture? It is the number of moles times the total pressure divided by total number of moles. So, it is a mole fraction times the pressure. So, first of all P N O is kappa N O times the total pressure and the mole fraction is the number of moles of N O in the final mixture which is C divided by total number of moles which is let us say at the for the time being is N T we will estimate that also times pressure. Similarly, the mole fraction of N 2 in the final mixture is A by N T times P and A according to this estimate is 1 minus C by 2 times N 2 by A by N T times P. Similarly, mole fraction of O 2 is B by N T times P. So, this is equal to half minus C by 2 by N T times P. Now, these are our actual mole fractions. So, we can estimate the partial pressure for every molecule here now using this because the partial pressure of N 2 will be nothing but mole fraction multiplied by total pressure sorry this P will not come here sorry because we are talking about just the mole fraction. So, now we come back to this equation. We write the partial pressure for all of them power half half minus C by 2 by N T times P to the power half. So, you can see here now we have one equation which relates which is a function of only C. So, if I now expand this first of all this N T will cancel off pressure will cancel off. So, we have a simplified expression which gives us the value of C in terms of K P. So, by simplifying this we get K P is equal to C upon 1 minus C by 2 to the power half times half minus C by 2 to the power half. After we expand this putting K P equal to 0.3 you get this. Now, let us solve for C is a quadratic equation. So, it can be very easily solved this gives C is equal to 0.1825 or minus 0.212516. Now, if I look at this two numbers the negative value is physically not possible because there has to be some amount of nitrogen oxide present nitric oxide present. So, therefore, this is the correct value once we have C then we can estimate A and B also. So, my final mixture is final mixture which is A N 2 plus B O 2 plus C N O is equal to 0.9087 N 2 plus 0.40875 O 2 plus 0.8125 N O. This is how we get the final composition. Once we have the final composition now we know how much is our final temperature we can get how much heat is required. So, let us look at that first of all delta H of formation for N O is equal to plus 21 per 5 8 kilo calorie. Now, this is a constant pressure process. So, delta Q is equal to delta H. So, essentially we have to find out the total enthalpy change in this process which is equal to how do we do it? We assume that first the reaction had occurred at standard state and then the products are heated to the final temperature. So, total delta H is equal to first of all heat of total heat of formation which is equal to nothing but this we have seen in the previous class rather two class back I think. So, this is the total heat involved in the formation reactions. This is total heat of formation of all the product species total heat of formation of all the reacting species. So, that is the heat involved during the reaction at standard state and then the products are heated delta H P from this standard state of 298 Kelvin to 4000 Kelvin which is my final temperature. Now, if I look at this in the product side we have that composition. So, we have 1825 times delta H F of N naught. So, delta H F of N naught plus the N O 909087 times delta H F naught of N 2 plus 0.40875 delta H F naught of O 2. So, this is the total change in the formation enthalpy and then the heat required to heat it up. So, that will be equal to now this amount of nitrogen is heated from 298 to 4000. So, that is first of all let us look at nitrogen. So, this amount of nitrogen is there initially in the final which is heated now for nitrogen then oxygen is heated again from the initial temperature of 298 Kelvin to the final temperature and N O is heated again from 298 to final temperature. Now, what is the heat of formation of nitrogen? It is 0 because nitrogen appears as nitrogen molecule here. So, this is 0 heat of formation of oxygen is also 0 and these values are we can get from thermo chemical tables. The enthalpy for nitrogen for oxygen for N O at different temperatures are tabulated in thermo chemical tables. So, we can get that once we do that if I estimate the final heat is 5.53 kilo calorie. So, 5.53 kilo calorie of heat must be supplied in order to take to carry out this reaction. This is an endothermic reaction. So, much of heat is required. So, that explains how we estimate. So, in this problem we showed first that how do we estimate the final composition and then instead of solving for adiabatic flame temperature we said that the flame temperature is known how much heat is required. So, this is how we do it. I would like to point out one more thing here that for the estimation of K p we talked about Gibbs free energy. Many times Gibbs free energy is not available to us, but typically the enthalpy variations changes are given. So, how do we get the Gibbs free energy or rather the K p if enthalpy change is given. So, for that we use Clausius Clapeyron equation. Let us see how we get that. We have already shown that ln K p is equal to minus delta G naught by R T and we have shown that Gibbs free energy is equal to H minus T s. Therefore, at standard pressure this was defined as standard pressure H naught minus T s. So, therefore, the change in Gibbs free energy is equal to delta H naught minus T delta s naught and therefore, we can write delta G naught by T equal to delta H naught by T minus delta s naught. Now, let us look back again at this equation and let us differentiate it with respect to temperature. So, if we do d d T of 1 K p this is equal to nothing but 1 by R minus 1 by R d d T of delta G naught by T that comes from by differentiating this. So, we define first of all from here we have shown that d d T of ln K p is equal to this. Now, let us look at first this term d d T of d ln T. So, first let us look at d d T of delta G naught by T delta G naught by T if according to this is equal to delta H naught by T minus T delta s naught minus delta s naught. So, this will be equal to minus delta H naught by T square plus 1 by T d d T of delta H naught minus T d d T of delta s naught. So, because H naught is also a function of temperature. So, we can write this like this. Now, let us look at the Gibbs equation. It says that d H equal to T d s plus V d p. Since we are considering a constant pressure process therefore, d p is 0. So, this term is 0. So, d H equal to T d s. Therefore, we can write this as d of delta H naught is equal to T d of delta s. Therefore, d d T of delta H naught equal to T times d d T of delta S naught. Here we are now we are imposing the condition that the enthalpy and the entropy here at essentially the process is at standard pressure. Now, if I look at this expression and this expression we have this minus this appearing here. Therefore, this term is 0. So, therefore, this term goes to 0. So, what we are left with is only this portion of this expression. Therefore, once we put it here this term d d T of G naught by T delta G naught by T is equal to nothing but minus delta H naught by T square. So, if I put it here this comes to be equal to delta H naught by R T square. So, here now we have got an expression for K p in terms of delta H naught which is something typically a known property universal gas constant and temperature. So, this equation is called Clasius-Caperron equation which gives us the value of K p if delta H naught is known. So, now so far we have been discussing typically simpler reactions one step reaction, but in our rockets we have complex reactions involving may be multiple reactions. What we know is initially the reactants that we are supplying we may not even know what is the final products. If we know the final products let us say we know the final products we do not know what are the reactions occurring. So, now let us try to get the reactions which are occurring in a practical system where we have more complex chemical equilibrium. So, the next thing that we are going to discuss is complex chemical equilibrium and how to get that condition. So, next is complex chemical equilibrium. Now, we have seen some example in the previous class that at one of the examples where the reaction was occurring at high temperature we had four species and nine unknowns. So, atom balance was not able to provide us enough equations to solve the problem. Even in the previous case we used one K p equation, but when we are going to more complex systems if we have S species and from C elements then from atom balance for C elements we get C equations. So, we get C equations from atom balance which we have already discussed many times. Now therefore, this S minus C is our additional required equations which we need to close this problem. So, this term S minus C is called the degree of reaction freedom. A reaction having only one degree of reaction freedom is called a simple reaction like the reaction we took in the example. But if the degree of reaction freedom is greater than 1 then that reaction is said to be complex. So, if S minus C is greater than 1 that is called a complex reaction. Now, we want to work with a complex reaction. We have seen earlier that for any reaction no matter whether simple or complex we have seen that at equilibrium d G is equal to 0. And what is d G if we have S species what is total d G for that total change in free energy this is nothing, but sum over this S species the gives free energy for independent species individual species times the change in number of moles of the species. So, that is the total d G and this is equal to 0 at equilibrium we will use this relationship to obtain the equilibrium relations. Let us start with an example that will make it easier to explain. The example we have already seen similar example before that we let us say we have some methane and some oxygen as reactant and the products what we have is some unburned methane some oxygen some water vapor some carbon monoxide some carbon dioxide these are the products. Now, this is typically what happens we have carbon monoxide carbon dioxide water vapor unburned hydrocarbon oxygen everything is present at equilibrium. We want to find out then what are the reactions equilibrium reactions that are occurring what are our elements in this case the elements are carbon hydrogen and oxygen. So, we will get three equations from there then in the final product. So, our chemical reaction is something I do not I am not mentioning how much it is there the final composition will have something like this. So, in the final composition we have N C H 4 moles of methane N O 2 moles of oxygen N H 2 O moles of water vapor N C O moles of carbon monoxide and N C O 2 moles of carbon dioxide. Now, we want to find out relationships to get this only then our final composition will be known. So, in order to do that let us first to get the equations that we are looking for first the three elements are this. Let us say that depending on the initial composition we have N H number of hydrogen. So, then N H is equal to in the in this product side we will have 4 N C H 4 plus 2 N H 2 O right because 1 C H 4 molecule will give 4 hydrogen atoms 1 H 2 O molecule will give 2 hydrogen atoms. So, if N H is the initial number of hydrogen atoms this is will be the product side. Similarly, we can get N O number of atoms of oxygen. So, that will be from atom balance 2 N C O 2 plus 2 N O 2 plus N H 2 O plus N C O. Similarly, we have N C is the number of atoms of carbon. So, that is equal to N C H 4 plus N C O 2 plus N C O. So, now our atom balance is done. We get only 3 reactions and we have 5 unknowns. So, we need 2 more equations somehow. So, let us look at the changes. First of all what is the change in number of atoms of hydrogen D H N N H is 0 right because that is going to be conserved. So, this is equal to 0 and that then is equal to nothing, but change in this plus change in this. So, this is equal to this therefore, from this 3 equations similarly, we get D N O equal to 0 and D N C equal to 0. So, from this 3 we get 3 equations which are D N C H 4 plus D N C O plus D N C O 2 plus D N C O 2 is equal to 0 2 D N O 2 plus 2 D N C O 2 plus D N H 2 O plus D N C O equal to 0 4 D N C H 4 plus 2 D N H 2 O equal to 0. So, this represent 2 D N C H 4 plus 2 D N C this represents D N O and this represent D N H. So, the change in number of atoms we get 3 equations here. Now, as I have just mentioned that we have 5 species in the product and 3 elements. So, the degree of reaction freedom in this case is 2. So, we need to have 2 more equation. Now, what we do is since we have 5 variables, what we will do is first let us choose 2 of these variables as independent variables and express the other variables as function of these 2. So, what will be the most logical choice? Let us say we choose methane because that is our field. So, we choose one of them methane as one of the independent variable and the other independent variable let us say is carbon monoxide because typically in stoichiometric reaction we should get carbon dioxide. Since we are getting carbon monoxide that is some intermediate reaction occurring. So, we choose that as one of the reactant independent variable. So, then what we can do is from these 3 equations we can write from this equation D N H 2 O is equal to minus 2 D N C H 4 let me call this equation 1. Then from this other equations we get D N O 2 is equal to half D N C O plus 2 D N C H 4 let me call this equation 2 and similarly we get D N C O 2 from this equation is equal to minus D N C H 4 minus D N C O let me call this equation 3. So, essentially what we are doing is we are representing D N H 2 O D N O 2 and D N C O 2 as function of D N C H 4 and D N C O. So, change in moles of methane and carbon monoxide. Now, we have chosen methane and carbon monoxide as independent variables. So, now we have these 3 equations. Next step now is bringing our discussion on free energy. What is the total free energy? The total free energy change total free energy change is D G. So, that free energy total free energy is nothing but free energy change is D G C H 4 species times D N C H 4 plus G C O times D N C O plus G C O 2 times D N C O 2 plus G O 2 times D N C O 2 plus G O 2 times D N O 2 that is it 1 2 3 4 whatever D N H 2 O. So, this is my total change in this free energy all 5 species are considered. So, let me call this equation 4. Now, what we do is from this expression let us eliminate this this and this then it will remain only a function of C H 4 and C O. So, after eliminating this we get D G is equal to G C H 4 plus 2 G O 2 minus 2 G H 2 O minus G C O 2 times D N C H 4 plus half G O 2 plus 2 G C O minus 2 G C O 2 times D N C O. So, that is the total change in this free energy in terms of the free energy of all the species we are considering times the change in number of moles of methane and carbon monoxide. Now, from the equilibrium D G is 0. So, this term must be equal to 0. Now, let us see this is not 0 and this is not 0. Therefore, in order and this is positive this is actually this is non zero this is non zero. Therefore, in order for this to be 0 this must be 0 and this must be 0. So, in order for this to be 0 mathematically what we have is D G D N C H 4 equal to 0 D G D N C O equal to 0 and that means that this term is equal to 0 and this term is equal to 0. So, what we get is G C H 4 plus 2 G O 2 minus 2 G H 2 O minus G C O 2 equal to 0. This then gives me after eliminating these things and the and from this I get G O 2 plus 2 G C O minus 2 G C O 2 equal to 0. I get these two equations. Now, these two equation corresponds to the equilibrium condition for two equilibrium reactions. First of them comes from here. This will be valid for an equilibrium reaction at standard temperature and pressure which will be C H 4 plus 2 O 2 equal to 2 H 2 O plus C O 2. If I have an equilibrium equation at standard temperature and pressure then the total change in gives free energy delta G naught for this is equal to 0. We have already talked about it. Therefore, in order for this to be correct we need to have this condition. Actually, it may not be a standard temperature and pressure. Any reaction can do that because delta G naught delta G for a chemical reaction at equilibrium is 0. Therefore, in order for this to be correct we have this equilibrium reaction. Similarly, for this to be correct we have another equilibrium reaction which is now this reaction C O is something as I said is appearing as one of the intermediate products. So, this is appearing from this reaction. Now, I have these two equilibrium reactions. Once we have these two equilibrium reaction what is the advantage I am gaining now? For this equilibrium reaction I have a K p. For this equilibrium reaction I have another K p. So, and those K p's can be easily obtained. For example, if I look at this reaction then K p for that reaction is equal to P C naught P O 2 to the power half by P C O 2. So, this can be further written as mole fraction of C O times total pressure mole fraction of O 2 times total pressure to the power half divided by mole fraction of C O 2 times total pressure. So, this is equal to K C O K O 2 a kappa O 2 kappa C O 2 times pressure to the power half. So, for the given pressure if you know the value of K p if you know the value of K the total pressure we now get one relationship rating this three. Similarly, we can get for the other reaction K p is equal to P H 2 O square P C O 2 by P C O 2 by P C H 4 P O 2 square in this equation pressure term will not come out. So, finally, this will be equal to this is square this is square. So, now, I get this two equations for anything is missing here this will be half there will be half here. So, I have this two equations for K p's and K p's as I said is available from the standard table we can find out K p. So, once K p's are known or we can estimate K p from the definition that as a Cassius-Caperron equation or from delta G we can estimate K p for this equilibrium reactions. Once K p's are known now I have one and two these two equations relating this five mole fractions and three we had already got before from atom balance. So, now, I have five equations five unknowns which can be then now solved for the final composition. So, that is how now I will get the final composition. So, now, let us look at more complex reaction I will not I am not going to derive it, but I will show what are the reactions equilibrium reactions that will be there. So, we will look at a very complex reaction. Let us say we are looking at combustion of methane in air at very high temperature this is air at very high temperature. So, that is at temperature is so high that we have dissociation also then for that case the composition is going to be like this as you can see they are considering lots of species including the formation of NOx including the formation of NOx we are considering and then we are considering dissociation of NOx we are considering this equations. So, this is my total number of species or the product species. Now, let us say what is the degree of reaction freedom here I have carbon hydrogen nitrogen oxygen. So, I have four atoms and how many unknowns do I have 1 2 3 4 5 6 7 8 9 10 11 12. So, we have four atomic species and 12 unknowns. So, my degree of reaction freedom is 8. So, I need to have eight equilibrium reactions like we have done like here to close the problem and those eight reactions are going to be like this that essentially is obtained following the process I have just discussed. So, this eight equations are 1 2 3 4 5 6 7 and 8. So, as you can see now I have this eight equilibrium equations and this eight equilibrium equations now I can get 8 k p values and now I will have the remaining 8 equations I can get from this k p values I can complete the problem and solve for the entire composition. So, this is typically what is done in solving the complex chemical equilibrium problems. So, this brings us to our discussion on the chemistry part. So, let me just summarize what we have learned and what can we get from here. We have looked at chemical equilibrium, we have defined various properties like k p delta g naught etcetera etcetera and then we have shown that if only we know what are that species going to be in the product side, we can form certain equilibrium reactions. Once we form those equilibrium reactions for those equilibrium reactions k p values can be obtained from the tables then we have the additional equations to close the problem particularly with complex reactions where the degree of reaction freedom is more than 1. So, now one more point I will have to point out here that this k p values as we have said is a function of temperature. So far in this case we are considering that we know the temperature. Now, if I combine this discussion with the discussion of adiabatic flame temperature, what did I said at that time that first we choose the value of adiabatic flame temperature. So, let us say we choose the value of adiabatic flame temperature then we formulate this, we get the k p value for that temperature, we solve for the final composition like we have done also for the problem that we solve today. So, we solve for that we get the final composition. Once we have the final composition we do the energy balance that is how much heat is going to heat up like the problem we solved. Then we see that from the energy balance this mass energy is produced by a chemical reaction and finally, going to the final temperature this much is required do we have the match. If we do not we change the initial case of the final temperature and repeat this process. So, essentially the problem that I have worked out we repeat that again and again for different temperatures till we get the final composition as well as the final all the equilibrium reactions will emerge now. So, everything we will get we get the final composition we get the final temperature. What will that give me once we have this we get the chamber temperature T c naught and remember that pressure is also included here. So, pressure is something that needs to be provided. So, pressure is one of the parameters that will appear we get T c naught and once we have the mixture we can get the molecular weight first of all of the mixture because we know the entire composition we can get the molecular weight of the mixture. Once we have the molecular weight we can get the value of r for the mixture. Similarly, we can get C p for the mixture because that is mass averaged we can get C v for the mixture because that is also mass averaged. Once we have this two we can get gamma for the mixture which is C p by C v. So, now we have gamma we have T c naught pressure is already given. So, now we have all the things required to estimate the thrust and start designing the nozzle also. So, therefore, this is how this is what comes from the combustor combustor design where after the combustion when we specify the fuel and oxidizer we know how much is coming at what pressure at and what temperature we get the final temperature considering adiabatic mixture as well as adiabatic combustion as well as the final composition. Then we now have all the necessary ingredients to carry forward this analysis. So, that is what I wanted to discuss in the combustion part. One point I made here is that we need to know the fuel composition. So, that is what I am going to start and discuss in the next couple of lectures. We will first radicate one lecture to solid propellants and then to liquid propellants. We talk about the fuels we talk about how the solid propellant rockets work. Then we talk about liquid propellants what are the liquid propellant rockets what are the components and how they work and that completes in our discussion on chemical rockets. After that we will dedicate couple of lectures on electric propulsion. So, that will complete the course. So, I will stop here today and next class we will discuss solid propellant rockets. Thank you.