So good morning and welcome to the fifth day. Now let us start our presentation. In my whole scheme of thermodynamics till today I have just there are two topics for which I have slides. One was the property relations yesterday that I did because otherwise I will have to spend a lot of time writing all those partial derivatives and Jacobian symbols. And the other one is open thermodynamic systems. There are lots of figures and sketches to be drawn. Now we will look at first illustrations of open thermodynamic systems. Then we will take up a specific case for study and derivation. We do not want to be too mathematical. So we will not consider a very general case. We will consider a specific case which can very easily be generalized. We will see how that generalization is done. We will then apply this to typical engineering systems of our interest. And finally we will take up exercises that the scheme. Now all of us know what are the illustrations of open thermodynamic systems. Open systems are systems through which or into which some material can flow and out of which some material can flow out. It need not be that something has to flow in and something has to flow out. But at least one of these two processes is necessary for us to consider a system to be an open system. An open system need not mean that only a fluid flows in and flows out. It could be even a solid material flowing in and flowing out. So you can consider a large oven through which on a conveyor belt some product moves in, gets baked, cooked or heated or treated inside and moves out at the other end. You can neglect the fluid flow if you feel like but the solid flow is the major flow and that has to be considered. But for us mechanical engineers, particularly thermal engineers, the illustrations begin with turbines, compressors, pumps including fans, then boilers, condensers, heat exchangers. So you can see that the first one are power consumption devices like compressor, pumps, power delivery devices like turbines, fluid movement devices like fans, significant heat transfer devices like boilers, condensers and heat exchangers. There are some simple situations which are also ducts, fluid flowing through a duct of various length and various types. Those are also open systems. Nowadays we are interested in environmental control, physical, psychological, thermal. So we consider our rooms and buildings also as thermodynamic open systems and analyze them. Engineering devices like cars are very complex open systems and finally we ourselves, a human being is perhaps the most complex open system which we may consider. In this course we will not consider and there is no end to this list and last few days I have been bugged by various terms which I have not used but which you find in a large number of textbooks. So I should warn you that an open system is also known as a control volume. That way nobody took an objection when I say closed system or simply a system. Cloth system is quite often known as a control mass. Now that is a proper nomenclature because a closed system does not allow any flow of mass across its boundaries. So the mass inside the system remains inside the system. It is not that the amount of mass remains the same. The identity of the mass, the same molecules remain inside the system. Of course during a chemical reaction they may change their form but at the heart whatever is inside remains inside although it may change its form. The control volume nomenclature for some reason is not as proper as the control mass nomenclature for closed system because if you say control mass means the mass is fixed and identified but a control volume does not mean that the volume is fixed. The open system, any open system can have a change in its volume during a process. So control volume may be considered as volume in which we are interested should not mean that it is a fixed volume whose identity does not or geometric location in space does not change with time. That is not true. It may change. Let us look at some schematics of open system. The inflows and outflows could be through ducts or force through openings. The flows could also be continuously distributed along the boundary. This is what we have done in fluid mechanics. For example when we derive the continuity equation we take a small control volume delta x by delta y by delta z and say fluid may flow in through any surface in any direction. We take care of that when we derive the equation. In the first case where the inflows and outflows could be through ducts and ports you can show or sketch the open system like this. Notice that if you neglect most of the arrows I have a bubble which similar to the one which represents the closed system. We have heat flow in and a heat flow out. Since open systems are generally considered to be something in which flow takes place through the boundary we consider the transient equation. So instead of Q we write the rate of heat transfer Q dot. Instead of W we write the rate of work done W dot. And if you have various ports for example here I have shown two inlet ports through port 1 M dot I 1 is the flow in through port 2 M dot I 2 is the flow in. Similarly there are three outlet ports shown. The outlet port 1, 2 and 3 with flow rates M dot E 1, M dot E 2 and M dot E 3. But if flow rates are in the general case you want the flow to come in and go out at any place then the open system or the control volume could be part of a flow field. And in which case the inflow and outflow may be continuously distributed over the boundary. I have shown a reasonably benign flow field but where the stream lines are more or less parallel to each other. So not a very complex one but if you have a complex flow field you will have churning, swirling even across the boundaries making it a really complex situation to analyze. Now a simple schematic open system which we are going to use will be something which has one inlet and one outlet. So just for the ease of drawing I have kept a more or less rectangular control volume but it could be anything, any shape. And of course since I am using a plane of the paper or plane of the screen I will do sketch everything in two dimensions. So the control volume is the CV and the extent of our control volume will be from the inlet I and exit E. I have shown just one inlet and one exit and we have decided that in the inlet duct somewhere at a convenient place we decide on a plane and we say that is the inlet port. And in the exit duct we decide on a plane which is which we call the exit port. Again for convenience we have as I will say soon we will consider the areas or the surfaces to be such that the local velocity is normal to these surfaces. Of course we can cut the plane like this but then we are unnecessarily complicating the situation. Because if we have to do that we will do it our expressions will just become that much complex. Now the situation is the state of the control volume that what is inside the control volume will be represented by the state variables volume, mass, entropy and energy. And we will consider these to be functions of time. So things are changing with time. So we will have V as a function of t, M as a function of t, E as a function of t and S as a function of t. Now let me be clear that we are overloading the symbol V here. V will be used for volume as well as for velocity. But there is unlikely to be any confusion because the status of volume is minor. The status of velocity that the inlet and exit ports is a higher status. Not only that the velocity that the inlet and exit ports will have subscripts like I and E whereas the volume of the control volume at most will have a subscript C V. We will also assume that the fluid at the inlet and the fluid at the exit are in local equilibrium. Now this is a new term and the idea is if we want the whole control volume to be perfectly in equilibrium that means we will have unique set of properties and that means nothing will be changing from inlet to exit. Now that is not a very interesting situation for us to consider. So what we say is let things change but if I take a sample of a fluid at the inlet and the sample of a fluid at the exit I will find that the state can be uniquely determined or almost uniquely determined for our purpose. And hence we say that the state of the fluid at the inlet and state of the fluid at the exit are assumed to be in local equilibrium. Then we consider that the situation in inlet and exit is one dimensional with everything uniform across the cross section. By everything I mean not only the state but also the velocity. This allows us to write algebraic equations and does not force us to use integrals when we consider flow of mass or flow of energy across the inlet surface or the exit surface. So this is a simplification and we know that if this assumption is not valid how to extend this to a non one dimensional situation. Because of this assumption we can say that at the inlet port the state is area A i density rho i. Now although I have written specific values I have not used the word specific everywhere but specific volume V i specific energy E i velocity V i and of course we have assumed V i is normal to A i. Again V i normal to A i is an assumption which is not a very rigid assumption. If it is not normal to A i we will have to use the component of V i whenever necessary component of V i normal to A i. So we will have to add appropriate sin theta or cos theta terms and similarly at the exit state we will have area A i density rho i volume V i energy E i velocity V i etcetera as needed and of course the assumption is the exit velocity is normal to the area A i. Now continuing with the situation we will say the rate of heat transfer to the control volume from its surroundings is q dot t the same sin convention as earlier. The heat absorbed by our system is positive. So our nomenclature for q dot t will be the rate of heat transfer to the control volume from its surrounding and we will call this q dot could be a function of time t. About the work transfer we have to be careful because the rate at which work is done by the control volume on the surroundings will be noted as W dot s and it could be a function of time and this s subscripts comes from a traditional term shaft work or shaft power although today we know that this power need not necessarily be absorbed or delivered through a shaft but the old nomenclature shaft power remains and although we do not use the nomenclature anymore the subscript s continues to be used. W dot s includes all components of work except that required for making the fluid flow into and out of the control volume. So some of you have been asking questions about flow work. Flow work is not to take care of it separately. W dot s although it comes out of a shaft connecting rod mechanism but it is essentially expansion and compression work. Similarly we could have stirrer work, electrical work any other component of work is accepted. Now our job is to derive equations of thermodynamics we are not going to use any new principle and all that we have to do is convert the equations which we have derived using the concepts of thermodynamics for closed system to open system transform those to a form which is applicable to open system. So we have to consider an equivalent closed system for analysis and for that first thing we will define are inlet and exit plugs and the idea is at the exit plane let the exit plane be at this location at time t at shown since fluid moves out and we have a uniform velocity normal to the plane of exit. Let us say after some time interval delta t that is at t plus delta t the exit plane so at t plus delta t the particles of fluid which were at this location move some distance and are at this location. Whatever is the fluid between the original location of the exit plane and the new location of exit plane I call the exit plug. Similarly looking at the inlet here we will say that let at inlet the fluid at time t the fluid at just before the inlet be at t at time t plus delta t let it be at the inlet plug. So that means during the period t to t plus delta t this much fluid moves out of the control volume that is in the exit plug and during the same period of time this fluid in the inlet plugs moves into the control volume. Now we need to have nomenclature for these positions. So notice that at time t the exit plane is at e f which is the exit port whatever fluid was at e f at time t plus delta t moved to e prime f prime the plane e prime f prime. Similarly at time t the fluid which was at the plane b c or c b moved at t plus delta t to c prime b prime. And one should notice that as the plane moves there is no fluid flow across that plane.  Similarly the inlet plug means that as if you have a piston here just left of the c b plane and from time t to time t plus delta t you have moved it from c b to c prime b prime. So now this gives us an idea of the equivalent closed system. This equivalent closed system occupies the space a b c d e f a at time t. So it occupies the place a b c d e f back to a that is at time t it includes the inlet plug and the control volume but does not include the exit plug. Similarly at time plus at t plus delta t it will go from a to b prime c prime then d and then e prime f prime and then back to a. So at time t plus delta t it will include the control volume and the exit plug but it will not include the inlet plug. So as I have already said it occupies the space a b c d e f a at time t and a b prime c prime d e prime f prime a at time t plus delta t. And as we have said and notice no mass flows across the boundaries of this system during this period. This period is the time period from t to t plus delta t. So this is the closed system and hence we now apply the conservation of mass to this system. For closed systems we did not have to specifically apply but now since we have to transfer or transform our relations for closed system to those for our open system we need to consider specifically the conservation of mass equation. Then we apply the first law and then finally we apply the second law. The initial state of the system it shown here by means of a gray overlay. I am not sure how many of you are able to see it but in front of my screen at least the grayness starts from c b and ends at e f occupying the whole of the control volume and the inlet plug. So our closed system at time t is this from c b to e f. And at time t it will be at time t plus delta t. You can see the inlet plug has gone out of our closed system. The exit plug has been included in our closed system. So all that has happened is some fluid has moved into our closed system and out of our closed system. So this is at time t this is at time t plus delta t. With the transmission I am not sure whether the grayness is seen but I will upload the slides and then by appropriate settings of your screen you should be able to see this. Now again some nomenclature. We talk about the processes and interactions. First the process for our control volume. We have said that the control volume is represented by mass, energy, volume and entropy. Let us say during the period t to t plus delta t mass should go from m at t to m at t plus delta t energy from e t to e of t plus delta t volume from v t to v t plus delta t and entropy from s t to s t plus delta t. During this period the rate at which heat transfer takes place is q dot. So the amount of heat absorbed is q dot delta t. We have to be careful about the amount of work done. One component of work is w dot s into delta t but there will be some work interaction associated with pushing out the fluid through the exit plug that we call we and some work associated with pushing or pulling in, sucking in the fluid through the inlet plug that we call w i. Now let us apply conservation of mass. This is essentially algebra. Since our system, equivalent system is closed, mass of the system at t plus delta t is mass of the system at t. The mass of the system at t plus at t is mass of the control volume at t plus the mass in the inlet plug. Remember that at t our control mass or closed system consists of the control volume and the inlet plug. So mass of the system at t is mass of the control volume at t plus mass of the inlet plug. Similarly mass of the system at t plus t plus delta t is mass of the control volume at t plus delta t plus mass of the exit plug because at t plus delta t our closed system occupies the control volume and the exit plug. We expand the masses in the plugs using the local density, the area of flow, the velocity of flow and the time period. So, here you will notice that the two assumptions are used. First we write this as an algebraic equation and not an integration over area. So, this assumes uniform property. Second thing, we have simply ai into vi for the volumetric flow rate. This indicates that the assumption that vi is normal to ai is used. If it were not normal to ai, we will have a component indicator like sin theta or cos theta. So, the mass in the inlet plug B C C prime B prime is rho i ai vi delta t. Similarly, the mass in the exit plug mass in E f f prime E prime will be rho e A e V e delta t. We know that mass of the system at t plus delta t is mass of the system at t. Substitute for various terms and you get this equation. Transpose and divide throughout by delta t. The expression changes like this. Take the limit as delta t tends to 0 and you have our mass conservation equation. V m C v by dt is rho i ai vi minus rho e A e V e. Now, it is very convenient to use the nomenclature for conservation of mass. Rho i ai vi is known as the rate of inflow of mass into the control volume. Typical unit kg per second. Similarly, rate of outflow of mass m dot e is rho i A e V e. So, we have our conservation of mass equation in this compact form. B m C v by dt is m dot i minus m dot e. That is rate at which the mass of the control volume changes with time equals the rate at which mass flows in minus the rate at which mass flows out. This is the basic form of conservation of mass. By basic I mean based on these assumptions that we have one inlet, one exit and local equilibrium and uniform velocity and uniform state and velocity normal to area. Now, we have to apply first law. We start by applying first law for the equivalent closed system. And of course, I have written it in slightly different way, but this is essentially the equation which I have been insisting that you write q equals delta e plus w. I have just transferred terms from one side to the other. Now, let us see what the terms turn out to be. Delta e is e of the system at t plus delta t minus e of the system at t. Now, the system at t is control volume at t plus the inlet plug. So, e of the system is expanded as e of the control volume at t plus the energy associated with the inlet plug. Similarly, at t plus delta t, the system occupies control volume and the exit plug. So, e of the system at t plus delta t is written as e of the control volume at t plus delta t plus energy of the exit plug. Now, energy of the inlet plug is the mass inside the inlet plug multiplied by its specific energy. So, what you have seen here in brackets is nothing but mass in the inlet plug rho i A i V i delta t multiplied by specific energy at inlet and mass in the exit plug is also written in the same fashion. The sorry energy of the exit plug is also written in the same fashion. The mass in the exit plug multiplied by specific energy at exit. So, using this expanded forms, the expression for data e which is the left hand side of this equation turns out to be delta e equals e C v t plus delta t minus e C v at t plus rho e A e V e delta t E e minus rho i V i A i delta t into E i. The next thing is about Q, Q is simple. This is the shorter form and then Q is the Q dot delta t. We do not have any complications here and this expression about that is delta e written in terms of m dot e E e delta t and m dot i E i delta t. So, notice from this equation that if we were not to have any mass inflow and mass outflow delta e would have been simply these two terms and which is what we expect if it were a closed system. So, now on this slide a bit cluttered slide we have started using the first law for the system and we have expanded expressions for delta e and for Q. Let us see what we do with W. We expand W as the work required to push the fluid out plus the work required to pull the fluid in plus any other component of work. Any other component of work is W dot s delta t and these two works the so called flow work terms work required to push out the fluid through the exit plus work required to pull in the fluid through the exit are written as W e and W i. Now go back to our figure and see how we can expand W e and W i. I have the terms here. Let us look at it like this. We consider the exit plug to be created by movement of a piston from the inlet plane of the exit the initial plane of the exit at t to the final plane of the exit at t plus delta t. So, the movement of that piston will be V e into delta t. So, the volume occupied will be A e V e delta t and the pressure into the change in volume will be P e A e into V e delta t. Another way of looking at it is P e into A e is the force acting on the piston and V e into delta t is the displacement of that force. So, finally we end up with the same expression. And we notice that if I multiply this by rho e and divide this by rho e then I will be able to write this in the form more compact form P e V e m dot e delta t. In a similar fashion we can treat the work required to pull the fluid in at the inlet. The expression will be very similar to W e with two differences. One the subscript e will be replaced by the subscript i and two there will be a negative sign because the pressure of the fluid is outwards whereas the fluid is being pushed in. So, this negative sign indicates that generally this will be a negative number. So, for the flow work term we have W e plus W i equals m dot e P e V e delta t minus m dot i P i V i delta t. It should be remembered that this P e V e comes out of this algebraic manipulation. Now, let us substitute all this in the first law. If you substitute this in our basic statement of the first law delta e is Q minus W. Here in this equation we have the left hand side which is the expanded form of delta e. The first term here is Q and the remaining terms here represent minus W. Now, if we transpose term we will get the following expression. The terms are transposed and combined because we see that on the left hand side we have an m dot e. On the right hand side also we have an m dot e. On the left hand side we have an m dot i and on the right hand side also we have an m dot i. So, terms containing m dot are all pushed to the right hand side and we get the modified form algebraically modified form of the first law in which we have change in energy of the control volume on the left. In here this is the change in energy of the equivalent closed system, but here the left hand side is change in energy of our control volume and that is expressed in terms of a Q dot term, a W dot s term and a term pertaining to m dot i and a term pertaining to m dot e. And the term pertaining to m dot i contains E i which comes out of the change in energy and P i V i which comes out of the flow work.  Now if we go back here all that we have to do is divide throughout by delta t and take the limit as delta t tends to 0 that is our standard trick from calculus.  There is a name for this term, but we will not mention that just now, but even at this stage one should remember that this is the so called shaft work it does not include the flow work required to push fluid out and pull fluid in. The flow work is hidden here in P i V i terms and P e V e term and this E i and E e along with m dot are actually part of the change in energy of the equivalent system terms although we have pushed them to the right side and combined them with the flow work term. Now generally we apply open system equation to systems with which we are familiar and our systems are systems through which a simple fluid flows through the system. For such a fluid our experience is that the specific energy at inlet and exit is made up of typically three components the internal energy the mechanical kinetic energy and the gravitational potential energy. This is not the full expansion this is a restricted expansion the general case is still this. So tomorrow if you have an if you are considering a magneto hydrodynamic situation then you will have to have an electrical energy and a magnetic energy and other complicated terms along with u i and v i and g z. But in our case if we assume a fluid which is non magnetic non dielectric non charge then it is easy to say or a good approximation to say that e i will be u i plus v i square by 2 plus g z i and e e will be u e plus v square by 2 plus g z e. Now combine this with p i v i and p e v respectively and now we see that we have a combination u i plus p i v i here and the u e plus p e v here which we have defined as the enthalpy at inlet and enthalpy at exit respectively. This common combination of u i and p i v i or u and p v which you see here as well as here was the motivation or provocation for us to define this combined property h i. And if you are ambitious you can now say that look quite often we are going to have h i and v i square by 2 together. So in compressible fluid flow both these terms will be significant and we will call it the stagnation enthalpy and coming here this is some sort of a extended enthalpy. So you can say this is the total enthalpy including all components of a i as well as p i v i. Whereas if you consider only the internal energy component of e i and combine that with p i v i you get h i and h i plus v i square by 2 you get the stagnation enthalpy. This is nomenclature but this is the traditional expansion and in terms of this expansion we get our so called default or standard form a reasonably general form of first law of thermodynamics for an open system with one inlet and one exit. So now we have two major equations derived one was the conservation of mass second one was second one now is the conservation of energy. And once you remember that this is a reasonably general form of the first law for open systems. I will not show you the details but at this stage it is proper for us to do some generalization. For example if we have more than one inlet or let us consider first a D generalization let us say that we have a situation where there is no inlet only exit a cooking gas cylinder when it is used in the kitchen all that you do is extract the gas out of it in which case all that we will do is drop this term. We fill in air in a car tire we assume there are no burst so there are no leaks a situation in which mass flows in but nothing flows out in that case we will neglect this term. And suppose we have a situation where we have more than one inlet for example some tank in which various streams are being pumped in for reaction or for mixing very common in chemical industry very common in food processing and pharmaceutical industry. In which case all that we will do is we will sum this up the m dot i will be summed up over various i's i1 i2 i3 so we will have to have a summation term. Similarly if there are more than one exits there will be a summation term here over the appropriate number of exits. Again if the flow is not uniform across a port well we will have to go back to the m dot suppose m dot i term and integrate this as rho i, v i, d a i everything will change will be a function of the location on the inlet area similarly on the exit area and that way we can even write a very general form of the first law with integration over a control surface. The surface which encloses the control volume sometimes known as the control surface. Now let us see where we feed next. Now let us go to some further items applications to typical systems and also the second law. First we will do second law for open systems then we will do special cases of the laws for open systems then we will apply them to typical engineering systems and then we will be ready for solving exercises. But before we do that a quick review we have spent about one hour so we will now have some symbolism and a quick review. What we have done so far we have looked at a typical open system with one inlet and one exit and instead of that big complicated diagram with inlet ports and exit ports we need now to use only a simpler schematic for showing an open system and the schematic could be as simple as this. Just a bubble now for ease of sketching on a computer I have used a rectangle with rounded edges but the shape and size is left to you. But actually in practice we have certain standard symbols for pumps, heat exchangers, turbines, compressors, valves so we tend to use those symbols but thermodynamically it is nothing but a block and we will show the inlet not by any port or duct just by an arrow showing this is the inlet another arrow showing this is the exit and we will have an arrow showing the q dot interaction and we will have an arrow showing the w dot s interaction. Quite often we will simply be dropping this s knowing fully well that w dot is the total work content work interaction except the flow work and the flow work term will always be hidden in the enthalpy at inlet and enthalpy at exit. We will not be specifically indicating that flow work is being taken care of it will be taken care of automatically. The first task will be to derive an expression for the second law for open system. We can derive an expression for the second law of open system using that technique which was developed so far go back to our system diagram with inlet plug exit plug apply the second law and come to a conclusion but we will not do that we will do something different we will use analogy. In this analogy will also help us appreciate the similarities and differences between the laws for closed system and laws for open systems. First let us look at conservation of mass. For a closed system we have d of dm system by dt is 0 the mass of the system does not change. In fact our statement is much stronger than this says the this only says that the numerical value of the mass of the system does not change with time but our conservation of mass is much stronger than that it says whatever is the mass inside the system and is identified as inside the system remains inside the system. For a control volume we have rate of change of control volume is m.i minus m.e so notice that on the left hand side when we replace control volume system by control volume at the right hand side we have the term containing inlet mass and exit mass inlet mass flow and exit mass flow. In a similar fashion when it comes to first law the so called conservation of energy for a closed system we have if we differentiate our equation with time d e system by dt is q dot minus w dot. For a control volume we have d e cv by dt is q dot minus w dot plus m.i ei minus m.i ee but here you see a formula where w dot is split into w dot s and the flow work component. So that is why I have not written this in terms of enthalpies I have kept it in terms of energies. Now imagine in a similar fashion what will you do for second law I will leave it to you to do at an exercise I have the slides ready with me and I will join you in a few minutes by taking a small break.


The Prime Minister Croxy visits us as the leader of one of America's closest and most stalwart allies. Italy in recent years has emerged as one of the principal nations of the Western Alliance. Prime Minister Croxy, by assuming leadership of the Italian people, has quickly established himself as a major figure in global politics. And we're impressed by the dynamic and statesman-like leadership provided by the new Italian government. We're particularly pleased by this opportunity today to deepen our consultations and broaden our cooperation with our trusted Italian friends. Italy is playing a crucial role in the NATO effort to correct the imbalance in nuclear forces in Europe. In our discussion today, Prime Minister Croxy and I reviewed the INF negotiations between the United States and the Soviet Union and reaffirmed our goal an equitable and verifiable agreement. If Soviet entrants in transigence continues, we're committed to NATO deployment of U.S. missiles. We remain hopeful that an understanding will be reached and we'll continue negotiations even after initial deployments. Indeed, for two years, the United States has negotiated in good faith while the Soviets have continued to deploy theater nuclear weapons. The Prime Minister and I also reviewed developments in other arms control fields. The United States and Italy share a common approach to the forthcoming conference on disarmament in Europe where we will be attempting to reduce the risk of surprise attack in Europe by improving observation capabilities and by providing a better understanding of military forces and activities. We're striving for reductions of conventional forces and to establish parity in ground forces in Central Europe. I expressed to the Prime Minister our deep appreciation for the positive contributions Italy is making to Mediterranean security and to the search for peace in the Middle East. Italy was among the first nations to offer its forces for service in the Sinai to help implement the historic Camp David Accords. Italian and American soldiers now are bearing the difficult burdens of peacekeeping in Lebanon. We are cooperating there not only in keeping the peace, but in making peace as well. The Prime Minister and I reviewed our common efforts to promote genuine political reconciliation within Lebanon and to work toward the restoration of Lebanese sovereignty throughout that troubled nation. I cannot overemphasize the importance of the Italian diligence in this crucial effort. We examined the world economy, discussed our respective efforts to deal with the economic challenges we face. I expressed our understanding of the difficult economic problems Italy presently confronts and our support for the tough but essential measures the Italian government is taking to restore balanced and non-inflationary growth. I expressed the hope that our own strong recovery would help Italy lift itself out of its present recession. I have been especially impressed by Prime Minister Croxy's commitment to attack organized crime and narcotics trafficking. We share a crucial interest in efforts to suppress the international traffic in narcotics. According to the Prime Minister and I have agreed to establish a joint Italian-American working group on organized crime and narcotics trafficking. Prime Minister Croxy comes to our shores as an especially welcome guest. In international affairs where the element of personal trust is so essential, such meetings as this are invaluable. I'm delighted to say that the Prime Minister and I have laid the basis today for continuing the open, cooperative, and mutually helpful relationship which has been traditional between our two governments and their leaders. Prime Minister Croxy will visit us today in the quality of the leaders of one of the most narrow and resumé of the United States, Italy, in recent years has emerged as one of the main nations of the Western Alliance. Prime Minister Croxy, assuming the direction of the guide of the Italian people, has soon affirmed which important figure in the world policy. We are really struck by the dynamicity and the quality of the statistics that characterize the leaders of the new Italian government and we are particularly pleased with this opportunity to deepen our consultations and expand our collaboration with our trusted Italian friends. Italy is a role of crucial importance in the NATO effort to re-equilibrate the relationship of nuclear forces in Europe. In our meeting today, Prime Minister Croxy and I went through in a magazine the Treaty on Missiles in Europe, which is in progress between the United States and the Soviet Union. We have reaffirmed our objective, that is, the achievement of an equitable and verifiable agreement. If the Soviet transigence will continue, we are committed to the NATO settlement of the United States Missiles. We continue in the hope that an intention will be added and we will continue in the Treaty even after the initial signings. Indeed, for two years the United States has continued to negotiate in good faith, just while the Soviets have continued to place their nuclear weapons in the theater. Prime Minister, we have also examined the developments in other sectors of the limitation of the weapons. The United States and Italy share a common approach to the next conference on disarmament in Europe, in which we will try to reduce at least the risk of any possible attack of surprise in Europe by operating better observation systems and promoting a better understanding of the military forces and their activities. We strive to obtain further reductions of conventional forces and to establish a relationship of parity in fact of terrestrial forces in Central Europe. I expressed to President Consiglio our deep recognition for the positive contributions made by Italy in the security of the Mediterranean region and for the search for peace in the Middle East. Italy was one of the first nations to offer its armed forces to serve in the SINAI to help in the effort to act on the historic importance of Camp David's Accords in that region. Therefore, the American and Italian soldiers are currently holding together the difficult honor of maintaining peace. We are collaborating not only in the safeguarding of peace but also in the establishment of peace. The Prime Minister and I have examined our respective common efforts to promote a real political reconciliation between the Gulf of Lebanon and to work for a reprisal of the Lebanese sovereignty throughout that worked nation. I could not exaggerate the importance of the leadership that characterized the Italian part in this crucial effort. We have also passed on to the world economy review and we have dealt with our respective efforts to resolve the economic challenges that we have faced. I have expressed our understanding of the economic problems so difficult that Italy is currently facing and our support to Italy for the so spinous but indispensable measures that the Italian government is taking to renew a balanced growth that is free and not inflationary. I have also expressed our hope that our energy recovery would help Italy to rise up the China and get rid of the current recession in Italy. I was particularly struck by the commitment from President Crags in the campaign against the organized crime and drug trafficking. We have in common a crucial interest in every effort to repress the international drug trafficking. That is why I, President of the Council, have determined an intensified cooperation between our respective collaborators to fight more and more the organized crime and drug trafficking. President Crags arrives at our response as a particularly welcome guest in the international affairs in which the aspect of personal trust is so essential. These meetings are of incalculable value and I am very happy to say that the First Minister and I have set the foundation for the continuation of this open relationship and collaboration of Appoggia Mutuo, which has been a traditional capital between our two governments and our respective leaders. Thank you very much. I am particularly grateful to President Reagan for the warm welcome he has given me during my visit to the United States. I consider this welcome as a tribute first of all to the Italian nation. I personally thank President Reagan for the live attention he has given to the members of his government and he has wanted to show them the problems and points of view that he has raised in the name of the Italian government. I have transmitted to President Reagan the message of friendship that the government and the Italian people revolve around the government and the American people, a friendship that belongs to the history of our democratic tradition and together to the future of our people. During the meetings that have been very cordial, open and I think of great utilitarian Cyprus, we have deepened with a common constructive and numerous issues, essential issues for the organization of peace in Europe and the world, problems and prospects of the international economy and the development of the Italian-American relations, issues of common interest for the industrial cooperation and for a deep fight against the great crime and international trafficking of drugs to be connected. The United States and Italy want firmly the peace they consider with absolute conviction as the most precious good for humanity and they are both determined to resolve through a method of pacification the controversies and international conflicts. In the field of the Atlantic Alliance, our governments will operate for that, at the end of the strengthening of peace, to maintain open and commercial prospects in the field of disarmament and the control of armaments, to guarantee peace and security. On the issue that at the moment dominates the relations between the East and the West relative to the nuclear balance in Europe, we intend to maintain a close consultation and constructive solidarity that is in action and develop a line of conduct that is consistent with the decisions already adopted by our countries. We have followed the way of the negotiation that must remain open in any case, also in the presence of a program of modernization of NATO's missile systems, where this was necessary, from the negotiate and negative line of the Soviet Union and the part of Warsaw, our purpose is to continue to solicit the Soviet availability for non-prejudicial treaties in order to reach a reasonable and controlled agreement that is worth guaranteeing a balance of forces established at the lowest possible level. A policy of peace must act in every area of the world, to end conflicts and reduce the most dangerous points of crisis in full respect of the sovereignty of the States and of the rights of the people. In this logic, Italy intends to ensure its constructive contribution, assuming, as in the case of Sina and Lebanon, direct responsibility in a humanitarian and peaceful mission for a perspective of unity, independence and reconstruction of Lebanon. But a policy of peace must naturally connect with a wider international cooperation. For this reason, we have confronted our points of view on the difficult phase that crosses the global economy in the prospect of favoring a stable and lasting reprisal. Italy and the United States must promote further development of the volume of their exchanges and of their cooperation, which will make even richer the excellent political relations between our countries. For the purpose of a more effective fight against the great organized crime around the international currents of drug trafficking, which challenges and threatens our societies, a special commission will be instituted and the necessary actions will be predisposed because it can carry out its functions with full efficiency. In today's meeting, President Reagan and I have confirmed the sentiments of friendship that bind our countries, reaffirming the commitment of our governments to the defense of peace. Thank you. I am particularly grateful to President Reagan for the warm welcome accorded me during my visit to the United States. I consider this welcome primarily as a homage to the Italian nation. I also thank President Reagan for the special attention that together with members of his government he has dedicated to the problems and to the views which I've had the opportunity to expose to him this morning on behalf of the Italian government. I myself have conveyed to President Reagan the message of friendship from the government of Italy and from the Italian people, to the government and to the American people also, a friendship that belongs to the history of our democratic tradition as well as to the future of our two countries. During our talks which have been friendly and open and I believe reciprocally very useful, we have carefully examined with mutual constructive hopes several matters. These matters are essential to attain peace, both in Europe and in the world, problems and prospects of international economy, the development of Italian-American relations, issues of common interest for industrial cooperation and for an in-depth struggle against crime and the international drug traffic connected to it. Both the United States and Italy are firmly seeking peace which they consider with absolute conviction the most precious gift to mankind and are both committed to solve international problems and disputes through peaceful means. Within the framework of the Atlantic Alliance, our two governments will work toward strengthening peace aiming to maintain open prospects for negotiations in the field of disarmament and arms control so as to guarantee peace through security. On the question of nuclear balance in Europe, which presently is the most important in East-West relations, we intend to maintain the present system of close consultation and of constructive solidarity and to develop a line of conduct coherent with decisions already adopted by our countries. We have followed the path of negotiations which must remain open in any case. Should a launching of the program for the modernization of NATO missile systems occur due to the negative negotiation policy followed by the USSR and by the Warsaw Pact, our intention is to continue to urge Soviet willingness towards serious negotiations without preconditions aimed at achieving a reasonable agreement providing for some means of control and some guarantees toward an equilibrium of the agreed forces at the lowest possible level. A policy of peace which must operate in every region of the world so as to terminate conflicts and to reduce the most dangerous fictions in full respect of the sovereignty of nations and of the rights of peoples. In this spirit, Ely intends to assure its constructive contribution by taking upon itself direct responsibilities in humanitarian and peace missions such as it's the case for the Sinai and Lebanon aimed at achieving unity, independence and the reconstruction of Lebanon. However, a policy of peace must necessarily be linked to a wider international economic cooperation. This is why we have compared our points of views on the difficult period of the world that the world economy is experiencing in view of fostering a stable and lasting recovery. Italy and the United States must promote a further increase in the volume of their trade and cooperation which will further strengthen the excellent political relations already existing between our two countries. Also in view of an ever effective fight against organized crime which is involved in international drug traffic and which is challenging and threatening our societies, a special joint commission will be set up and the necessary steps will be taken so that it may operate in the most effective way. In our meeting today, both President Reagan and myself have reaffirmed the feeling of friendship which binds our two countries. We are affirming the commitment of our governments for the safeguard of peace. Thank you.


Hello and welcome to the session. In this session we discuss the following question that says, find the points on the curve y equal to xq at which the slope of the tangent is equal to the y-coordinate of the point. Let's now proceed with the solution. The given curve is y equal to xq, let this be equation one. We have to find the points on this curve at which the slope of the tangent is equal to the y-coordinate of the point. For this we suppose that the required point be x1, y1, that is x-coordinate is x1 and y-coordinate is y1. Now the point x1, y1 lies on the curve given by equation one. Therefore, this means we have y1 is equal to x1q, let this be equation two. Next differentiating equation one with respect to x we get dy by dx is equal to 3x square. Now dx at the point x1, y1 is equal to 3x1 square. Now this dy by dx at the point x1, y1 is the slope of the tangent at the point x1, y1. The question is given that the slope of the tangent at that point on the curve is equal to the y-coordinate of the point. So according to the question x1 square would be equal to the y-coordinate of the point which is y1, three equation three. Now from equation two we have y1 equal to x1q. So on the equations two and three that is these three equations we have square is equal to x1q which means x1q minus 3x1 square is equal to zero or you can say x1 square into x1 minus 3 the whole is equal to zero which means that either x1 square equal to zero or x1 minus 3 is equal to zero. That is we have x1 equal to zero or x1 equal to three. So we now get two values for x1 either zero or three put in x1 equal to zero in equation two we have y1 equal to zero therefore we say for x1 equal to zero we get y1 equal to zero. Next put in x1 equal to three in equation two we get y1 equal to that is twenty seven therefore we say for x1 equal to three we get y1 equal to twenty seven. Hence we have three twenty seven the required points zero zero and three twenty seven. So this is our final answer and the study solution of this question.