Today, I'm going to help you get the most out of your loop view. To find the loop view, you're going to head into the Details panel. And right here, you'll see the loop. Let's basically think of it like a diamond loop. And what it allows you to do is see your picture while you're adding detail or sharpening or noise reduction, that kind of stuff. What you might not know is if you were to click on the loop view, it will actually zoom in and then from here, you can move the loop around. Something you might also not know is that you can control how far you zoom in simply by right-clicking on the loop view. You can either go from a one-to-one ratio or you can double that and go to a two-to-one ratio. Excellent buying power. Hopefully you enjoyed today's Friday Quickie. We'll see you back here on Monday for another full tutorial. Have a great weekend, everybody.


Hello, my name is Michael Gustav, I've been in the creative arts for about eight years. I decided to do London this year for the first time with the famous solution, Taiwan Strong. I decided to do a fish and the fish depicting Taiwan and the total depicting solution.


Hello and welcome to lecture number 22 of this lecture series on Introduction to Aerospace Propulsion. Over the last several lectures, we have got introduced to several aspects of thermodynamics, thermodynamic principles and also laws of thermodynamics and how to use them in applications like in thermodynamic cycle analysis and so on. And what we are going to discuss today is a little bit different from what we have been discussing so far in the sense that this particular topic that we are going to take up for discussion today and also in the next lecture combines some of the thermodynamic principles with fluid mechanics in some sense. And so, what we are going to discuss about is on compressible flows and what are the different properties of compressible flows and also we shall look at compressible flow through some simple cross sectional geometries like convergent nozzles and converging divergent nozzles and so on. So, this is a slightly different topic from what we have discussed so far, but obviously this is going to also use many of the thermodynamic principles that we have been discussing in the last few lectures. So, what we shall discuss in today's lecture are the following. We shall be talking, we will begin our talk today with a discussion on what do we mean by one dimensional compressible flows. We shall talk about stagnation properties and then we shall talk about speed of sound and Mach number, how do you define Mach number and then we shall look at one dimensional isentropic flow and variation of fluid velocity with flow area will derive an equation which defines or governs the variation of fluid velocity with area. We will also be discussing towards the end of the lecture on isentropic flow through nozzles. We will discuss two types of nozzles, converging nozzles and converging diverging nozzles. So, these are some of the topics that we shall be discussing in today's lecture and so you might have got a feeling right now that this is going to be a discussion primarily on compressible flows and also on their significance in terms of analysis of compressible flows. Now, during our thermodynamic analysis that we have discussed and also we have solved several problems using the thermodynamic principles. We have always made an inherent assumption that at a given state of a system the density is a constant and so there is an inherent assumption that the flow is incompressible in the sense that we do not really consider variations of density which is not necessarily true in most of the in well in many of the engineering applications. In most of the day to day applications that we are familiar with the velocities of the fluid are very low and therefore, the inaccuracy that we achieve because of assuming incompressible flow is not really high. Whereas, if fluid velocities are very high then it is not really possible for us to make this assumption that the flow is incompressible. The flow no longer remains incompressible and therefore, we should be we should be taking into account the compressibility effects and that is some one of the aspects or that is one of the reasons why we are taking up this topic for discussion today that there are lot of engineering applications where the fluid velocities can be significantly high and therefore, we cannot really assume that the flow is incompressible changes in density are incompressible or changes in kinetic energy are incompressible for that matter. So, what we shall discuss to begin with is that in the basically the significance of the compressible flows and why do we need to discuss them basically because there are some fluid applications there are certain implications where it is important for us to understand or analyze the system in the form of considering the variations in density and also taking into account the variations or effect of kinetic energy though potential energy may still be negligible kinetic energy cannot really neglected. So, the flows that basically involve significant density variations are known as compressible flows though most of the analysis we have considered so far neglected density variations we shall also take a look at some applications where this effect can no longer be neglected. Basically, we shall be again making an assumption here that even the even when we consider the flow to be compressible we shall be analyzing it in a one dimensional sense for an ideal gas with constant specific heat. So, we are going to assume that the specific heat is a constant and the gas is ideal and therefore, we can assume the ideal gas behavior and so on. And where do we see such applications basically in devices that involve flow of gases at very high velocities like in nozzles and so on. And so, it is important that there are we have an understanding of how to analyze such systems which involve very high velocities therefore, changes in kinetic energies can no longer be neglected. Now, so in order that we analyze a system where kinetic energy cannot be neglected we need to define what is known as stagnation property of a system or stagnation property different types of stagnation properties. One of the stagnation properties to begin with we shall understand is the stagnation enthalpy. We already defined what is enthalpy we know now that enthalpy basically represents the total energy of a fluid in the absence of potential and kinetic energy. This comes from the first law of thermodynamics and when we take up this property of enthalpy at high speed flows potential energy being negligible may still be valid, but not kinetic energy. So, we have to take into account the changes in kinetic energy. So, what we do is we basically combine enthalpy with kinetic energy that is we add up enthalpy and part of the kinetic energy and we define a new property known as the stagnation enthalpy. And why is it known as stagnation enthalpy will be clear when we take up when we discuss further. So, this sum of the static enthalpy so called the static enthalpy which I will now be referring to as static enthalpy and the kinetic energy term put together is known as the stagnation enthalpy. So, if you look at the definition of stagnation enthalpy which I had defined stagnation enthalpy which is usually denoted by h subscript 0. 0 is applicable for most of the stagnation properties which we are going to define like temperature pressure and so on. So, stagnation enthalpy h naught is equal to h which is now the static enthalpy plus v square by 2 which is the kinetic energy. And you might notice that these are all per unit mass and therefore, it is specific enthalpy specific stagnation enthalpy is the sum of the specific static enthalpy plus v square by 2. So, the first term on the right hand side is the static enthalpy which does not have a subscript 0 and the second term is the kinetic energy term. And on the left hand side we have the stagnation enthalpy. So, this is how you would define a stagnation enthalpy which is primarily the sum of enthalpy as we had defined earlier which we are now calling as the static enthalpy. So, whenever we take up different properties at high speed cases that is incompressible flows we also define these properties as stagnation as well as static parameters. And so static parameter will become stagnation parameter if the velocities are 0 that is if kinetic energy is 0 then static and the stagnation properties are the same. So, we will now use this principle to define other properties in terms of stagnation parameters like stagnation pressure and stagnation temperature, stagnation density and so on. Now, if you consider a steady flow of through a duct let us say through some diffuser or a nozzle where there is no shaft work there is no heat transfer etcetera. Steady flow energy equation for this is something we have already defined that is h 1 plus v 1 square by 2 is equal to h 2 plus v 2 square by 2. So, the left hand side as we have now defined is this stagnation enthalpy at state 1 which is h 0 1 and on the right hand side we have stagnation enthalpy at state 2 that is h 0 2. So, this means that in the absence of any heat and work interactions if there are no heat transfers or there are no work interactions this stagnation enthalpy remains a constant during a steady flow process. So, this is a very significant property that we need to keep in mind that if there are no heat interactions or there are no work interactions or in a steady flow process for example, then the stagnation enthalpy of such a process remains the same it does not change if there are no heat transfers into the system or from the system stagnation enthalpy does not change. And that is something we have to keep in mind because many of the systems that we are going to consider will have this property to be used that is there is no heat and work interactions taking place like in nozzles and diffusers. And therefore, stagnation enthalpy has to be a constant, but this is not applicable for systems like turbines or compressors because there is a work interaction taking place in the case of compressor work done on the system in the case of turbine work done by the system. So, stagnation enthalpy obviously will not be a constant in such cases it is only applicable for those cases where there are no work or heat interactions. Now, we have now seen that if you look at this duct example that I was mentioning where we had h 1 plus v 1 square by 2 is equal to h 2 plus v 2 square by 2. Let us say that at state 2 the velocity is now equal to 0 that is by some means we bring the velocity at state 2 to be 0. Now, what we have is h 1 plus v 1 square by 2 is equal to h 2 which is also equal to h 0 2 because as we have discussed if velocity is 0, static and stagnation parameters are the same which means that at state 2 the static enthalpy and stagnation enthalpy are the same. So, what this means is that stagnation enthalpy also represents the case where the fluid velocity is isentropically or adiabatically brought to 0 brought to rest. So, at state 2 if you were to bring the fluid to rest we have h 1 plus v 1 square by 2 is equal to h 2 which is also equal to h 0 2 which means that stagnation enthalpy represents enthalpy of a fluid when it is brought to rest adiabatically that is because in this case there was no heat interaction or work interaction across the system boundaries. So, if you bring the fluid to rest adiabatically and in which case stagnation enthalpy basically represents the fluid enthalpy of a fluid when it is brought to rest adiabatically. So, what happens during this stagnation process basically and this is why it is basically called stagnation enthalpy that is the term stagnation comes because we are assuming that the fluid is adiabatically coming to rest that is how the stagnate the term stagnation enthalpy has arise. So, during a stagnation process kinetic energy of a fluid is converted to enthalpy that is internal energy and flow energy which results in increase in the fluid temperature and pressure that is if you were to bring a fluid to rest adiabatically it has to reach I mean that energy as part of conservation of energy principle that energy has to get transformed into some other form and what happens is basically that the fluid energy increases in the form of flow energy and so on. So, that ultimately leads to an increase in temperature and pressure of the fluid. So, based on this we shall now define what are known as stagnation pressure and stagnation temperature. So, on the left hand side if you recall we had h 1 plus v 1 square by 2 which was equal to h 2 and internal equal to h 0 2. So, in general we can write h naught that is stagnation enthalpy is equal to h plus v square by 2 which means that for an ideal gas which is where we began our discussion that we are going to assume that the gas is going to be ideal with constant specific heats. So, with constant specific heats for an ideal gas enthalpy is simply equal to c p times t therefore, we have c p t naught is equal to c p t plus v square by 2 or t naught is equal to t plus v square by 2 c p that is left hand side we have a temperature with a subscript 0 that is known as the stagnation temperature. Stagnation temperature is equal to the sum of the static temperature plus one term which we are going to define as the dynamic temperature because that changes with fluid velocity and that is why it is called dynamic temperature. So, if you write the equation for enthalpy in the form of a product of specific ratio of sorry specific heat at constant pressure to the temperature then we have c p t well c p t naught is equal to c p t plus v square by 2 or t naught is equal to t plus v square by 2 c p. So, here t naught is defined as this stagnation temperature and this represents the temperature an ideal gas will attain if it is brought to rest adiabatically that is similar to what we did for enthalpy if you bring a gas to rest adiabatically the temperature that the gas attains at the end of this process that is when it becomes when it comes to rest adiabatically is known as the stagnation temperature. And the second term that is v square by 2 c p corresponds to the temperature during such a process and because it is it can change with velocity it is called dynamic temperature. So, you can immediately see that for non-zero velocities stagnation temperature will always be greater than the static temperature which is something we have defined earlier as well that is as you bring a fluid to rest the energy gets transformed into internal energy and so on which ultimately leads to an increase in its stagnation well increase in its temperature and pressure which we have now defined as stagnation temperature and pressure. And this has to be higher than the static temperature because it has an additional energy term which is the dynamic temperature term. Similarly, we can also define pressure that is stagnation pressure which is basically the sum of the static pressure plus the dynamic pressure. So, stagnation pressure well pressure is equal to p naught is equal to p plus half rho v square I guess you might have learned this in fluid mechanics and this was basically as part of the Bernoulli equation. So, we can immediately see that there is a direct correlation between what you get in thermodynamics with what you feel or what you get in fluid mechanics. So, stagnation pressure is equal to static pressure plus this dynamic pressure and so what we are trying to say here is that there are parameters which we need to take into account when the fluid velocities cannot be neglected. So, kinetic energy needs to be accounted for in calculating parameters. So, similarly the way we have defined for temperature we can also define stagnation pressure. So, the pressure that a fluid will attain when it is brought to rest isentropically is known as the stagnation pressure. And so what we are saying here is that all these stagnation parameters whether it is temperature or pressure or enthalpy will have a value greater than that of the corresponding static parameters for all nonzero velocities. Now, from the ideal gas well we have already assumed that all these gases that we are discussing about will be for ideal gas. So, from the isentropic relations that we had discussed we can see that the ratio of we can will basically relate the stagnation pressure to the stagnation temperature in terms of the pressure and temperature ratios. So, if you were to relate the pressure ratios to the temperature ratios we have p naught by p is equal to t naught by t raise to gamma by gamma minus 1 where gamma is the ratio of specific heats which is equal to C p by C v. Similarly, we can define it for density which is rho naught by rho is equal to t naught by t raise to 1 wide gamma minus 1 where again gamma is equal to the well it is basically the ratio of specific heats. So, these are isentropic relations which we will be using very frequently in our thermodynamic analysis of different cycles and some of them which we have already discussed was not in the form of the ratio of stagnation and temperatures and pressures, but in the form of static pressures and temperature ratios. But in some of the later analysis that we will be doing for aircraft engines where the kinetic energy terms cannot be really neglected these equations will be used very frequently. Now, what we shall do now is to see what happens as if there is a change in let us say the stagnation pressure. So, stagnation pressure well we already seen that stagnation enthalpy does not change as long as there are no heat transfer or work transfer across the system boundaries which means that since stagnation temperature is directly related to stagnation enthalpy this also is applicable to the stagnation temperature and therefore, stagnation temperature also does not change as long as there are no heat and work interactions across the system boundaries. But is it applicable for pressure because pressure is a parameter for ideal gases it is not directly related to the enthalpy it is a temperature which is a direct function where which happens to be related to enthalpy. So, what about pressure? So, to understand that what we will do is to look at a process where we could have a loss in total pressure and we will see what correspondingly happens to the enthalpy. So, what we have done here is that on an enthalpy entropy scale we have h and s scale here and this is let us say a compression process where there is an increase in pressure or even if there is no compression let us look at a fluid which is just a having a certain velocity which means that it will have stagnation parameters which will have to be accounted for which is basically the sum of the static parameters plus the dynamic term. So, the actual state of the fluid is represented here at state one which is shown there and so this is the initial state where it has a certain enthalpy h correspondingly there is an entropy as well and this is on a constant pressure line. So, this pressure line that is seen here is p it is a constant pressure line. Now, what are the corresponding parameters for this particular fluid if you were to look at this stagnation parameters. So, we know that stagnation enthalpy h naught is equal to h plus v square by 2 and so we have h naught which is equal to h plus this dynamic term that is v square by 2. So, this is the isentropic stagnation state that is if the process were to be isentropic then we get the straight line because entropy is a constant. So, it is a straight line it is an isentropic stagnation state where the enthalpy at the end of this process is equal to h naught and the corresponding stagnation pressure is equal to p 0 or p naught, but if the process is not isentropic that is if we look at a process which has frictional losses and some such irreversibilities it cannot be any more isentropic. So, if you look at an actual stagnation state there has to be an increase in entropy. So, there will be a certain positive slope for that particular process, but there are no heat or work interactions which means that the enthalpy should not change because there are no heat or work interactions which are taking place and therefore, enthalpy cannot change in such a process, but because of irreversibilities it is non isentropic and there will also be some pressure loss because of irreversibilities. So, if there is a pressure loss then we have a certain slope for the process and so we have on the same enthalpy line because enthalpy does not change stagnation enthalpy line extended the process ultimately meets the stagnation enthalpy line and the corresponding pressure is the actual total pressure. So, p naught actual need not be equal to p naught isentropic because there could be a pressure loss total pressure loss taking place because of let us say friction, friction can cause decrease in velocities which means that at the end of the process you have a lower velocity and therefore, correspondingly a lower total pressure. Therefore, p naught actual can be less than the p naught ideal or p naught isentropic. So, it is not necessary that the total pressure remains the same at the end of such a process, but what should remain the same is the enthalpy, enthalpy does not change because there are no heat or work interactions taking place. So, this is a very important aspect that we need to keep in mind that in a process where there are no heat or work interactions the stagnation enthalpy cannot change. Therefore, stagnation temperature also does not change, but what is possible is that there could be a difference between the stagnation pressure ideal or isentropic to the stagnation pressure actual which could be because of frictional losses which could lead to non isentropic processes and there could be p naught actual which is less than p naught ideal and in the actual in the ideal case where the if you assume all irreversibility is to be 0, then the p naught actual will be equal to p naught ideal because there are no more pressure losses. So, this is a very important aspect that you definitely need to keep in mind because we will keep using this aspect in many of the analysis that we are going to do in some of the later lectures when we analyze ideal cycles and real cycles of gas turbine engines and so what we have discussed now are on stagnation parameters, stagnation enthalpy, stagnation pressure, stagnation temperature density and so on and the bottom line is that in the absence of heat and work interactions stagnation enthalpy and therefore, stagnation temperature cannot change, but what is possible is that you may have a change in stagnation pressure due to some irreversibilities and non isentropicity of the process. So, what we shall discuss now is a different aspect related to our compressible flows, we will be defining what is known as the Mach number. So, before we define that we need to define modismian by speed of sound. So, speed of sound is basically by definition the speed at which an infinitesimally small pressure wave travels through a medium because sound is a pressure wave which is basically a small pressure wave. So, speed at which the infinitesimally small pressure wave travels through a medium is basically the speed of sound. Now, if you assume an ideal gas speed of sound which is usually denoted by symbol c can be shown to be equal to c is equal to square root of gamma r t where gamma is the ratio of specific heats r is the gas constant for the medium and t is the temperature static temperature. So, c is basically a function direct function of temperature speed of sound is a direct function of temperature. And of course, it also depends upon the ratio of specific heats and the gas constant, but for a particular medium both of these are constant. And so of course, as long as we assume that ratio of specific heats does not change with temperature. So, we can see that speed of sound is a direct function of the static temperature. Now, based on this we are going to define a non-dimensional parameter which is basically taking the ratio of the velocity of the fluid or the object to the speed of sound and that is basically known as the Mach number. So, Mach number is basically the ratio of the actual velocity of an object or fluid to the speed of sound. And so in some cases the fluid may be moving and the object is stationary which is what we would do in a wind tunnel testing for example, the fluid is at a certain speed the object is stationary. Whereas, on the other hand an aircraft an actual aircraft moves at a certain speed in a medium where the air is relatively at 0 velocity. So, ratio of that speed to the speed of sound is known as the Mach number. So, Mach number is defined as V by C where V is the velocity of the object or fluid and C is the speed of sound which is equal to square root of gamma R T. So, Mach number is basically a function of the ambient temperature as we have seen which means that it is possible that an object which is moving at the same velocity in two different mediums of two different temperatures will have different Mach numbers. Because even though their velocities are same they are in different mediums which have different temperatures which means that the speed of sound will be different for different mediums depending upon their temperature. And therefore, it is perfectly possible that Mach number of two objects which are moving at the same velocity but in different mediums which have different temperatures the Mach number can certainly be different. And so it is not necessary that if the velocity is the same Mach number has to be the same. Now, if depending upon Mach number if we have a case where Mach number is equal to 1 then such flows are called are basically known as sonic flows. If Mach number is greater than 1 the flow is known as a supersonic flow. If the Mach number is less than 1 it is called a subsonic flow and if Mach number is much greater than 1 usually greater than 5 we refer to such flows as hypersonic flows. And Mach number approximately equal to 1 it is around 1 then we call such flows as transonic flows. So, these are different terms we use for flows depending upon their Mach numbers. So, Mach number less than 1 is subsonic greater than 1 is supersonic and greater than 5 is usually referred to as hypersonic and so on. And so what we will do now is to look at the variation of fluid velocity with the area and we will derive an expression which relates the area different ratios and fluid velocity with the Mach number. So, which means that for different Mach numbers we can see how area and velocities are related. So, for deriving an expression what we will do is basically consider a mass balance for a steady flow process. So, we know that mass flow rate for such processes are equal to the product of density area and velocity. So, rho a v which will be a constant for a steady flow process. So, if we differentiate this equation and divide this by the resultant mass flow rate we can write the above equation rewrite the above equation as d rho by rho plus d a by a plus d v by v is equal to 0. Now, from our steady flow energy equation which we had derived in earlier lectures if you assume work done heat transfer kinetic and potential energy to be more or less 0 then from the steady flow energy equation we get h plus v square by 2 is equal to 0 or d h plus v d v is equal to 0 that is if we differentiate this equation energy equation we get d h plus v d v is equal to 0. So, this is basically coming from the steady flow energy equation where we assume work done heat transfer and potential energy to be 0 and from the T D S equation that was the second Gibbs equation T D S equal to d h minus v d p. Now, for isentropic flows T D S will be equal to 0 and therefore, d h is equal to v d p which is d p by rho because specific volume is the inverse of density and therefore, d h is equal to d p by rho.  So, if we combine this equation with the mass balance equation we get and simplify basically we get d a by a is equal to d p by rho multiplied by 1 by v square minus d rho by d p. Now, it is also known that the ratio d rho by d p for constant entropy is equal to 1 by c square.  So, m square is coming because we will get a ratio of v square by c square which is basically equal to Mach number square.  So, if we rearrange that equation which is basically from d p by rho is equal to minus v d v and.  So, this equation has a lot of significance in the sense that this equation governs the shape of a nozzle or a diffuser in subsonic or supersonic isentropic flows. Now, in this equation area and velocity obviously are positive quantities and if that is. So, Mach number obviously can. So, depending upon the Mach number whether it is greater than 1 or less than 1 area velocity changes can be inferred that is for subsonic flows where Mach number is less than 1 we have d a by d v less than 0. For supersonic flows where Mach number is greater than 1 the rate of change of area with velocity is greater than 0 and for sonic flows where Mach number is equal to 1 then we have d a by d v is equal to 0. So, we will we will try to understand what is the implication of the rate of change of area with velocity depending upon the Mach number. So, as Mach number changes there are changes in velocities with reference to changes in areas. So, let us look at what happens as you change the Mach number and what happens to velocity as you change areas. So, from these equation it basically follows that to accelerate a fluid in subsonic flows you need a converging area that is there has to be a decrease in area because for Mach numbers less than 1 d a by d v is less than 0 which means that if you have to accelerate a flow we have to have a corresponding decrease in area in subsonic flows. Whereas, in supersonic flows you will need an increase in area to accelerate a fluid a diverging nozzle is required at supersonic velocities and what we will also see a little later is that the highest velocity that you can achieve in a converging nozzle is the sonic velocity that is the maximum velocity that you would be able to achieve in a converging passage in subsonic flows will be basically that you would get only a sonic velocity at the end of the converging nozzle. To accelerate a fluid to supersonic velocities you will need a diverging section or diverging area or increase in area after the flow reaches a sonic velocity at the minimum area which is basically known as the throat. So, after the throat there needs to be a diverging section to accelerate a fluid to supersonic velocities and so what I was trying to say is that in subsonic flows if you look at Mach numbers less than 1 as you start accelerating the fluid the maximum velocity or Mach number that you can get at the end of the acceleration is Mach number equal to 1 which is basically known as a sonic velocity. So, a flow which has certain total pressure and temperature at the inlet of the nozzle. So, a nozzle section is shown here will accelerate the maximum that it can accelerate is Mach number equal to 1. So, what if you reduce the area below this that is at the end of this nozzle let us attach another nozzle hoping that this will accelerate it to supersonic speeds, but that is not going to happen what will basically happen is that the section where you get sonic velocity will shift and the section which had sonic velocity earlier will now have a subsonic Mach number. And so the sonic velocity will occur only at the exit of this convergent section instead of the exit of the original nozzle and basically the mass flow rate will now reduce because you are trying to force a certain amount of mass flow through a lower passage area and so you would get a decrease in mass flow. So, we will see later that as Mach number reaches 1 the mass flow is at its maximum and that is known as the choking of a nozzle that is if you were to force fluid through a nozzle till the point that at the exit of the nozzle Mach number is equal to 1 then the mass flow rate that can be passed through such a nozzle has reached its maximum level that is known as choking of the flow which means that if you add another convergent section to that you are not going to increase the Mach number anymore it will in fact lead to reduction in the mass flow rate because your choking area is different now. And so it is not possible for us to achieve supersonic Mach numbers in a convergent section. So in order to get a supersonic Mach number we need to have a divergent section at the exit of the convergent section and such nozzles are known as converging diverging nozzles. We will do some analysis of the variation in pressure across a convergent diverging nozzle with change in back pressure we will do that little later. Now so in summary from the area velocity Mach number relation what we have are the following. So if you look at the first case that is shown here we have a nozzle which has an inlet Mach number which is subsonic. So along the length of the nozzle the pressure and temperature the static pressure and temperature would decrease the velocity and Mach number increases. So this is basically a subsonic nozzle. So area reduces Mach number increases in subsonic flow reverse of that happens at if the area is increasing Mach number is less than one at the inlet the static pressure and temperature increases along the length of the nozzle velocity and Mach number decreases. So this is basically a subsonic diffuser. If you look at the supersonic version of this it is the exact opposite. So if Mach number at the inlet is greater than one if we have to have an increase in Mach number along the length of the nozzle area has to increase. So pressure and temperature decreases velocity and Mach number increases and this is known as a supersonic nozzle and if you look at a supersonic diffuser it is basically decreasing area. So pressure and temperature will increase and Mach number decreases. So a supersonic diffuser at least theoretically is basically a subsonic nozzle. So supersonic diffuser will act as a subs a nozzle in subsonic flow and a subsonic diffuser will act as a supersonic nozzle in at Mach numbers greater than one which is what should happen theoretically. So this what we have discussed now is an outcome of the area velocity Mach number relation which we had derived and so we know how the area will well how the area has to change given a certain Mach number so that you get an increase or decrease in velocity accordingly. Now what we will do is basically relate the stagnation properties that is stagnation pressure temperature and density to the corresponding static parameters like static temperature pressure and density through the Mach number. We have already seen the isentropic relations which relate ratio of temperatures and pressures through the ratio of specific heats that is T naught by T is equal to P naught by P raise to gamma minus one by gamma and so on. So now we will relate the stagnation parameters with their corresponding static parameters through the Mach numbers. Now to do that we have already seen this equation which was relating temperature stagnation temperature to static temperature and the velocity ratios. So T naught is equal to T plus V square by 2 C p therefore, T naught by T is equal to 1 plus V square by 2 C p T. Now since we know that C p is gamma r by gamma minus 1 and C square is gamma r T and also Mach number is V by C. So if you make the substitutions in the dynamic temperature term we get V square by 2 C p T is equal to V square by 2 into gamma r by gamma minus 1 into T which basically is gamma minus 1 by 2 into V square by C square which is gamma minus 1 by 2 M square. So if you substitute this in the first equation we get T naught by T is 1 plus gamma minus 1 by 2 M square. So this relates stagnation temperature to static temperature through the Mach number. Now this can also be extended to the corresponding pressure and density from the isentropic relations. We get P naught by P is 1 plus gamma minus 1 by 2 M square raise to gamma by gamma minus 1 and stagnation density ratios rho naught by rho is equal to 1 plus gamma minus 1 by 2 M square raise to 1 by gamma minus 1. So these are property relations which relate the stagnation parameters to the static parameters through Mach number. Now if Mach number is equal to 1 we have seen at the end of nozzle if Mach number is equal to 1 then if we equate M is equal to 1 in those equations then the properties that we get are known as critical properties and they are denoted by a star superscript star. So if we equate M equal to 1 in them we get T star by T naught is equal to 2 by gamma plus 1. Similarly P star by P naught is equal to 2 by gamma plus 1 raise to gamma by gamma minus 1 rho star by rho naught is equal to 2 by gamma plus 1 raise to 1 by gamma minus 1. So these are equations which relate the critical properties to the corresponding stagnation properties and you can see that it depends only on the ratio of specific heats. So this was about property relations for ideal gases inherently assuming that there is an isentropic flow. Now what we will do next is to analyze isentropic flow through nozzles and we will take two different types of nozzles a converging nozzle and a converging diverging nozzle. So a converging nozzle in a subsonic flow will have a decreasing area as we have seen. So let us look at what happens as you keep decreasing the area that or if you keep the area fixed as you decrease the exit pressure that is known as the back pressure how does it affect the flow parameters. So a converging isentropic flow through a converging nozzle will basically have involve a decreasing area along the flow direction. What we shall do is to consider the effect of back pressure on the mass flow rate and pressure distribution along the nozzle. We will assume that the flow enters the nozzle from a reservoir where the velocities can be assumed to be 0 and so this stagnation temperature and pressure will remain unchanged through the nozzle. So we will not assume any losses taking place in the nozzle. So that pressure is constant and obviously since there is no heat or work interactions stagnation temperature also remains a constant. So if that were the case this is the nozzle that we are talking about which has flow entering through a reservoir which is at pressure P naught temperature T naught exit pressure is P e and back pressure is the term which we can change depending on the back pressure the pressure across the nozzle also will change. So in the first case first scenario we have back pressure is equal to the reservoir pressure there is no flow taking place. So the variation of pressure ratio is a constant. So the pressure would vary in the format shown here. Now as we reduce the back pressure but back pressure is still greater than the critical pressure. Now there is a decrease in the pressure ratio P by P naught and so this is how the variation would be. So from state 1 to state 2 we have a case where back pressure is still greater than the critical pressure. Now if you reduce it further at state 3 where back pressure is equal to the critical pressure the pressure ratio continues to reduce and reaches state 3 let us say and if you were to decrease back pressure below this critical value what basically happens is that there is no change happening it will continue to have the pressure ratio which will continue to drop till the extent that if you reduce the back pressure even further the Mach number at the exit of the nozzle does not change it will remain the same. And so the point when we have the ratio or back pressure is equal to the critical pressure is known as the choked condition or the choked flow and the pressure ratio at that point is P star by P naught we have seen this depends only on the ratio of specific heats. And so for any other pressures which if we continue to drop the pressure back pressure lower than that we will continuously see a decrease in the pressure ratios till the point if we continue to reduce back pressure to 0 the Mach number at the exit is still going to be Mach number is equal to 1 which because basically because it is a choked flow there is no more change happening there. So if you look at Mach number and the pressure ratio plots so we started off at Mach number state 1 so if you look at the mass flow rate it was equal to 0 because the pressure ratio the back pressure was equal to stagnation pressure at the reservoir there was no mass flow mass flow was equal to 0. As you start reducing the back pressure there is an increase in Mach number till a point when it reaches state 3 which was the critical pressure ratio that is P star when Mach number reaches maximum mass flow that is its maximum value. If you reduce back pressure even below that Mach number does not change we get the same mass flow rate and which is basically the Mach number is equal to 1 at the throat mass flow rate remains the same throughout for state 4 and 5 as well from state 2 3 4 and 5 the mass flow rate is the same and if you look at the exit to stagnation pressure ratio after state 3 it remains the same basically because it is the choked condition and it has not changed after the flow reaches the critical state. So this is how the variation of a flow would be through converging diverging nozzle and from the above what we basically see is that when exit pressure is equal to back pressure well exit pressure P will be equal to back pressure for back pressure greater than or equal to the critical pressure and exit pressure will be equal to critical pressure for P be less than P star. So for all back pressures lower than the critical pressure the exit pressure will be equal to critical pressure the Mach number is unity and mass flow rate is maximum which is basically the choked flow a back pressure lower than the critical pressure cannot be sensed in the nozzle upstream flow and it basically does not affect the mass flow rate. So back pressure this is basically relating the back pressure and the exit pressure to the critical pressure as you change the back pressure values. So let us now look at the same scenario for a converging diverging nozzle because we have seen that the maximum Mach number that you can achieve in a converging nozzle is unity for achieving supersonic Mach number you need a diverging section after the throat.  So for back pressures which are different from what it should be you may not really achieve a supersonic flow. So let us analyze a flow a supersonic flow as we change the back pressure. So this is a converging diverging nozzle there is a converging section a throat where the area is minimum and a diverging section P naught is the pressure in the reservoir and P e is the exit pressure P b is the back pressure. So let us say P b is pressure which is currently higher than the stagnation pressure P naught as you reduce it to P a there is a flow which takes place and as you increase the P b what basically or as you reduce back pressure how it affects the flow as you reduce back pressure which is still greater than the critical pressure then we get initially an increase the static pressure reduces velocity will increase and then in the diverging section the static pressure will increase. Now as you reach the critical pressure which is P star which happens when P is equal to P c or back pressure is equal to P c the flow reaches its minimum static pressure here which is equal to P star and you get Mach number is equal to 1, but it does not achieve a supersonic flow subsequently it becomes subsonic and the static pressure continues to rise. So if you reduce the back pressure even further what happens is after the throat the static pressure begins to continues to drop up to a point where there is a sudden increase in static pressure which is basically due to the occurrence of a shock. So at the in the diverging section of the nozzle there is a shock which causes after which the flow becomes subsonic we will discuss more about shocks in the next lecture and then exit of the shock the flow becomes subsonic and it take static pressure again rises like in a subsonic flow. So you have still not gotten a subsonic supersonic flow at the exit of the nozzle it is still subsonic flow because of the presence of a shock. If you reduce the flow back pressure further we get a supersonic flow because the shock which was there in the diverging section will continuously move outward as you reduce back pressure eventually the shock will be pushed out of the diffuser and the flow will become supersonic at the exit of the nozzle. So if you look at the Mach number plot for state A there was no Mach number because there was no flow for state B the Mach number increases and then it decreases in the diverging section state C it reaches Mach number equal to 1 because that is the critical condition and then of course it does not become supersonic it again decelerates and becomes subsonic. So we have subsonic flow all the way here at state D the flow is supersonic after the throat continues so till this point after which there is a shock and because of the presence of a shock the flow becomes subsonic and we still have a subsonic flow at the nozzle exit. If the back pressure is lower than what is happening at state D then the shock continuously moves towards the exit and for all other states which is E and F and G we will have continuously have a supersonic flow all the way to the exit of the nozzle. So we have now achieved a supersonic flow all the way up to the exit of the nozzle and this was possible only because the back pressure was adjusted to values which were lower than what we have seen here lower than pressure at D that is P D and P E and for pressures lower than that we get a supersonic flow continuously all the way up to the exit of the nozzle. And so in the Mach number plot you can see that it goes up to Mach 1 at the throat after which if the back pressure is not low enough it can again become subsonic or if it is low up to a point you may get a shock in the normal shock in the divergent section which means that you get a supersonic flow up to the shock and downstream of the shock it again becomes subsonic and if the back pressures are lower enough then we get supersonic flow all the way up to the exit of the nozzle. So this is the variation of supersonic flow or this is how you would achieve a supersonic flow by changing the exit back pressure and it is not just enough that you put a divergent section at the end of the from the throat and still get a supersonic flow that will happen only if the exit back pressure is low enough. So let me wind up today's lecture where we have discussed at least had a very quick introduction to some of these topics which are related to compressible flows and we had some discussion on one dimensional compressible flows on stagnation properties on the speed of sound and Mach number. And we have seen variation we have derived an equation which relates the fluid velocity and Mach number with the flow area and we have also seen isentropic flow through nozzles two types of nozzles converging nozzles converging diverging nozzles because our interaction for this was a very brief one and there are separate courses which are offered on compressible flows which are known as gas dynamics and so on. And so this is just to give you an idea of what is the importance or significance of compressible flows and why you need to understand stagnation properties and take due care in calculating properties which involve significant kinetic energies. So in the next lecture we will continue our discussion on compressible flows. We will discuss little bit more on shock waves and expansion. We will discuss about normal shocks oblique shocks and Prandtl mayor expansion waves. We will be discussing about two duct flow cases one is duct flow with heat transfer and negligible friction which is basically known as the Rayleigh flow. We will also discuss about duct flow with friction but without heat transfer known as the Fano flow. So these are some of the topics which are in continuation with our discussion on compressible flow. We will take up for discussion during our next lecture.


This is the Big Picture, an official television report of the United States Army produced for the armed forces and the American people. Now to show you part of the Big Picture, here is Master Sergeant Stuart Quain. This is an age of streamlining. Everything from automobiles to guided missiles today reflects the dynamic look of the future. Your Army is no exception. Its modern fighting division is lean and functional, simple in organization and efficient in performance, keeping pace with the many marvelous inventions at its disposal. The Pentomic Division, so named because it consists of five battle groups, is a semi-independent, self-contained fighting unit designed to operate effectively on an atomic battleground should the need ever arise. Today's Pentomic Division is characterized by the ability of fewer men to deliver a more devastating concentration of firepower on any would-be aggressor. It is an ultra-modern idea, but the story behind it is as old as the history of man.  He increased his knowledge, constantly improved the surroundings in which he found himself, and learned over the centuries how to survive hostile nature as well as hostile fellow human beings. The forward-looking and ingenious ones managed to stay alive only through superior military organization, tactics, and weapons. The long range and rapid fire of bow and arrow artillery deepened the battlefield and ultimately led to larger and more powerful weapons. The catapult could punch an opening in a castle wall. The crossbow, with its deadly accuracy and tremendous impact, was thought to be a weapon so terrible that it would end war. But heavier armor was developed for protection, and the short range of this weapon was soon regarded as a primary disadvantage. The English longbow used against armored French knights at Crécy and Agincourt represented the pinnacle of firepower until the age of explosives. Here indeed was a new type of environment. It developed slowly. The early guns were less accurate, had a slower rate of fire, and a shorter range than the longbow which they replaced. But guns were being enlarged and consistently improved. They increased in destructive power and in bulk. As better metals and machines were developed, the tools of war became more plentiful and efficient. Not only tools of war, but the course of human history itself was changed by a violent explosion in the New Mexico desert. Age we live in was born. Who has the United States Army charged with supervising the initial development of this mightiest of all explosives, adapted it to the Army arsenal, and reorganized to defend against it? Why not call on the Army for an answer? Well America is planning to survive, I can assure you. Our Army has been reorganized in complete recognition and acceptance of the unparalleled changes in our environment. The knowledge that atomic weapons are in our hands and in the hands of our potential enemies greatly influences our thinking. It has meant a reevaluation of our approach to modern war, the war of the future. To be sure we have also reorganized to wage modern non-atomic war, so-called conventional war. Conventional weapons and equipment of the Field Army have been and are being made. Lighter, stronger, longer lasting, and with an increased radius of action. So we do stand ready to fight a modern war with improved conventional weapons. But because of the continuing threat of atomic weapons and missiles, with their far greater range and vastly broadened target area of destruction, the Army has been recast to fight and win on the atomic battlefield also, wherever it proves to be. This film is a combined progress report and forecast to tell you something about the Army's doing and will continue doing to adjust to the impact of the changing scene. To meet and master the realities of man's environment in this modern age. The Army is prepared today as always to fight a conventional non-atomic battle. It is also ready to employ nuclear weapons should it ever be necessary. One low yield atomic warhead will release as much destructive energy as 21,600 105mm howitzer high explosive shells. The old Army Combat Division with its triangular three regimen structure has been recast in pentagonal form in order to better adapt it to any situation. Its five mobile battle groups operating semi-independently provide more firepower with less manpower than ever before. In any future combat there would be no static front line and the depth of division sectors would be much greater in order to reduce to a minimum a division's vulnerability to atomic weapons. The pentomic unit embodies the concept of lean active mobile task forces capable of dispersing widely on any battlefield. With his troops so dispersed how could a commander bring the full strength of his pentomic division into offensive action against an enemy? There are three basic elements in the answer to that question. Wide-ranging communications. Wide-ranging communications increases the commander's control even with wide dispersion. A network of radial relay circuits tie in combat support and administrative units with a communication center in such a way that even with an atomic blast the system can continue to function over a variety of other traffic patterns. The equipment which makes up this communication system must be sturdy and powerful but also light enough for easy transportation. Helicopods into which the equipment for an entire command post can be assembled can be deposited wherever they are needed and can be quickly put into action as a communications nerve center. Long-ranging firepower of the second element in the pentomic concept is comprised of many new weapons including missiles. Many of these missiles such as the honest John Rocket can deliver either a high explosive or an atomic warhead. Missiles with atomic capability are assigned to missile commands. Units set up to provide atomic fire support to allied or U.S. ground forces. Honest John is self-propelled on its launching platform so when it's air-transported it's ready to go. A smaller version, the little John, supports ground troops in any terrain at short notice. The mighty Redstone Chief Operational Missile for the Heavy Missile Command is ready to support a pentomic division from far behind the battleground. The Redstone was the forerunner of the Army's Jupiter Sea which launched Explorer, the first satellite sent into space by the United States. The smaller missiles can be moved with the battle groups in the field. Even the corporal, a medium ballistic missile with excellent range and a devastating impact, is highly mobile. This weapon has both atomic and non-atomic capability. It is estimated that four corporal battalions firing atomic missiles would exceed the power of all American artillery in World War II. La Crosse is a highly accurate surface-to-surface missile for close support of ground troops. Its pinpoint delivery makes it an ideal weapon for destroying enemy bunkers or fortifications. It can be moved speedily into position and be ready for action within minutes. Nike Ajax, the Army's ground-to-air defense missile, is capable of destroying fast-moving enemy aircraft. Nike Hercules is a faster, higher-flying brother of Ajax and is capable of carrying an atomic payload into large formations of high-altitude bombers. The Hercules attains a speed of more than 2,000 miles an hour and is guided by radar on its deadly journey through the sky. Latest in the Nike family, not yet operational, is the Zeus, called an anti-missile missile, built to shoot down enemy ballistic missiles. While Nike Ajax and Hercules are designed for the upper reaches, Hawke is used to handle low-flying aircraft. Hawke is a homing-type missile. Once on its own, it seeks its prey independently and pounces on it with sudden surprise. Dart is the soldier's missile, an amazingly destructive weapon that weighs less than 100 pounds and can be carried with infantry tools. Swift and lethal can knock out any modern tank. A David among Goliaths, weapons of the Pentamic Army are radically new. Some, like the M60 machine gun, are simply newly designed models of weapons used by the Army for many years. The M60 fires the standard NATO 7.62-millimeter cartridge and replaces the older machine guns, both the light and heavy. Its light weight means that it can easily move forward with the assault troops all the way to the objective. M14 Springfield is a lighter, more efficient rifle for the modern infantrymen. It is capable of either fully automatic or semi-automatic fire. It also fires the standard NATO cartridge, which can penetrate a steel helmet at 1,200 yards. It will replace three standard weapons, the M3 submachine gun, the 30-caliber carbine, and the M1 Garand rifle. Weighing less than nine pounds, this weapon was only recently accepted for general use by the Army after passing rigorous tests. Far-Ranging Mobility is the third element of the Pentamic Unit, linked with wide-ranging communications and long-ranging firepower. Helicopters are basic to this concept of mobility. Skimming over any kind of terrain, they can dart in quickly and unload infantry units ready for battle, providing a dimension to two-povement never before possible. Weapons and supplies can be carried short distances and delivered to troops in the combat area, saving precious time, which might be lost by ground transportation. Enough ammunition to keep light artillery batteries operating for hours can be delivered in minutes by a flight of these whirly birds. They are vital to the fast-moving, hard-hitting Pentamic Army. The helicopter's ability to make near-vertical takeoffs and landings in any type of terrain renders it ideally suited to perform reconnaissance. It is also used to evacuate the wounded in order that they may receive medical attention as soon as possible. Many thousands of American lives were saved in Kolia as a result of this kind of merciful helicopter mission. Fixed-wing light aircraft also perform a number of important jobs. One of the most vital of these is observation for artillery flying. These planes are also used for wire laying and supply drops, employing the latest techniques and equipment. The light plane is well suited to help give the Pentamic unit its mobile character. Fixed-wing aircraft, like the helicopter, prove themselves invaluable in securing reconnaissance information. Equipped with television, photographic and other surveillance devices, these aircraft provide the kind of widespread observation of the battle area needed by ground troop commanders. On the basis of information transmitted to trackers on the ground, commanders are aided immeasurably in their fight against what has been in all wars the greatest and most frustrating problem, the gathering of accurate reliable information about opposing forces. This experimental radio-controlled drone carrying an aerial camera is an example of surveillance devices to aid the commander. In the not too distant future, drones will fly with one or more electronic devices such as radar, infrared, motion picture and television cameras. Controlled from a remote ground position, this equipment will be able to gather information long in advance of the most forward reconnaissance elements and with no danger to our personnel. Sky cavalry, a new concept incorporated into the Pentamic division, uses aerial steels to sweep over wide fronts and extended distances in the battle area on the cavalry's traditional mission of reconnaissance. Part of the sky-caps function is to transport organic infantry and reconnaissance platoons to critical areas. Some ground reconnaissance elements are armed against enemy attack with a recoilless rifle mounted on a jeep. These units are linked by radio to aerial reconnaissance. Warning and advising each other, aerial and ground observers combine to operate as a wide-ranging, far-seeing information-gathering team. An ultimate objective of the Pentamic Army is realized in the airborne division, first to be given the pentagonal form. It is completely transportable by air. Formally, supplies, repair equipment and supporting elements were dropped piecemeal or were sent into the division battle area only after ground transportation was possible. Now, the entire division moves through the air. Most of it is air droplet. Much material is air landing, but every man and every piece of equipment necessary to the division's fighting strength on the ground is airborne. The armored unit has always had many of the desired features of the new Pentamic Army. It is a mobile, maneuverable force capable of massive firepower. For still greater mobility, the Army is bringing forth a new light tank. It carries a deadly 76-millimeter gun and can be transported by air. Designated the T92, it is faster, smaller and more agile than its predecessor. A highly maneuverable armored steam, it can turn quickly in its own length. The low silhouette helps protect the vehicle from the menace of enemy anti-tank weapons. It would also serve to lessen the effect of any atomic blast. If it is completed in a critical area, it can add the needed punch to assault troops. The armored personnel carriers of the armored division provide swift ground movement for the modern combat soldier. These carriers are capable of transporting the entire complement of one reinforced battle group with all weapons and essential equipment. Rugged and amphibious, the light armor of these vehicles and their capability of being tightly closed give excellent protection on the atomic or non-atomic battle vehicle. They can move troops into an objective area quickly and without regard for the type of terrain. Moving from one part of the battle area to another, each personnel carrier can deliver a rifle squad equipped for its specific mission and also carry ammunition and a three-day supply of rations. However, he is transported to his destined point of contact with the enemy. By air drop or light aircraft or armor plated carrier, he is still the individual soldier who, just as he has on battle fields of the past, will carry the heaviest share of fighting responsibility on any battle field of the future. The atomic war he may be forced to fight will be a war of dispersion, which he will be able to win with the aid of wide-ranging communications, on-ranging firepower, on-ranging mobility. Let us look now at a combat exercise embodying these three elements of the pentamic concept. Let us imagine that our army has decided on an offensive thrust to seize a key area deep in enemy territory. The plan calls for an all-out airborne drop on the target. The idea is to affect the deepest possible penetration in a hurry and make the target area a base for further offensive action. Air Force planes, especially designed for this kind of operation, discords the soldiers as rapidly as possible. Each man jumps with his own unit and all of the equipment he will need when he lands. Large pieces of equipment are dropped with the men, including a heavy self-propelled gun. The weapon has been thrust and rigged in such a way that it will always land upright and ready for action. Half a dozen immense 100-foot parachutes are required to deliver its bulk safely to Earth. The airborne attackers carry out their operation with split second precision, and each piece of equipment is unlimbered the moment it strikes the ground. Firepower is needed now. Supported by heavy fire, the airborne infantrymen advance. Even dispersed over wide distances, the troops are still connected to their command with the latest of communications equipment, such as the two-way radio set contained in the helmet or the improved sound-powered telephone. Intelligence transmitted from the forward reconnaissance element speeds the infantry attack on the objective. Another unit of armor strikes to link up with the airborne attackers, now creating chaos in the enemy rear. Aerial observers have reported, however, that a bridge has been blown up by the enemy and armor's path. If this were to delay armor's advance, it would seriously hamper the army's effort to set up a base for powerful coordinated firepower. In other times, it would indeed have created a delay, perhaps a fatal one, but the equipment of today's army is equal to the crisis. To permit the tanks to cross over the area blasted by the enemy, a mobile assault bridge moves with the armored column. Developed by the Corps of Engineers, the bridge can handle any vehicle up to 50 tons. It rides on a standard tank chassis and is operated by a two-man crew. Little time is lost as it prepares to span a ravine which is holding up the armored column. Covering fire is furnished the crew by a waiting tank. For a moment, the bridge rears up before extending itself across the castle, like a great cautious monster trying to find a safe footing for the impatient armor. A jeep moves up first and finds one tread solvent enough for its narrow wheelbase. Although the ravine took only minutes to span, armor has been held up long enough. Without the mobile assault bridge, such a delay in the advance might have been measured in hours, critical hours which now have been saved. SkyCav has been covering combat surveillance, exchanging information with our ground forces, making sure no profitable target is offered the enemy. The commander calls for artillery support to knock out enemy weapons which might direct fire on our assaulting units. His request is relayed through the communication system to a corporal missile battalion in the rear. Missile artillery and other weapons combine for mass firepower and help to neutralize enemy positions. Bridges retrieved for instant use elsewhere, now that the armored column no longer needs it at this point. A plastic assault pipeline is used to refuel armor from a distant gasoline source. Immediately needed supplies are ferried in and dropped by cargo helicopters. These same aircraft are utilized on the return trip for the rapid evacuation of wounded. The objective has been seen, but the assault troops do not rest. The troops remount and move in small active units to hold and enlarge the area. Secure in their new position, they move forward once again, unleashing a new attack upon the next objective. We've tried to give you a brief rundown on thinking and planning for the Pentamic Army and show you what we have done and are doing to materialize that thinking in positive action. What we haven't shown you is the vigor with which new weapons, new equipment, new organizations and new ways of doing things are being made ready for phasing into the Army. Some of these are being phased in now. Some are for next month. Some for a half dozen years from now. The flexible structure of the Pentamic Army will accommodate all new tools as they become ready. Ours is a modern Army, for and of the modern American soldier who was born into a land of applied science, who cut his teeth on radio grids and model airplanes, who grew up with mechanical marvels and electronic wonders as robot companions that he learned to adjust, maintain and improve. To this resourceful soldier, we offer the Army of the future today. The enormously flexible structure of the Pentamic Army will accommodate all new weapons, methods and ideas as they are available. Ours is a modern and dynamic Army, as strong as the men who serve in it, as responsive to the challenge of the future, as the country it serves. Now this is Sergeant Stuart Queen, your host for The Big Picture. The Big Picture is an official television report for the armed forces and the American people, produced by the Army Pictorial Center, presented by the United States Army in cooperation with this station.


When working with linear functions, we often times have to solve linear equations, and we've seen some examples of that beforehand, like if you have something like y equals 2x minus 1, and you want to figure out when f of x equals 4, you just plug in 4 for the y-port and solve for it. It's not too complicated. Now, there are situations where something like the following might arise. This is sort of like a worst case scenario when it comes to solving linear equations. Take something like 1 half x plus 5 minus 4 equals 1 third 2x minus 1. Now, is solving such a thing just an algebraic chore, or is there actually situations where you might want to do something like this? Well, first of all, I want to mention that although it's not simplified, the left-hand side is a linear function. One of the benefits of not simplifying is we can see some of the transformations that have been done here. If you were to take the standard unmodified, untransformed linear function, you would actually get the identity function like this. That is, you get the graph y equals x. This is sort of like the untransformed linear function. When you look at things like this, what have we done to it? Well, we've shifted it to the right by 5, 1, 2, 3, 4, 5. We've stretched it vertically by a factor of 1 half. Changing the slope is changing the graph here, so it's kind of like now we lower the slope to get something like this. Oh, and then we also moved it down by 5. I guess I moved it to the right, so I should be getting something like this. Then I moved it down by 5 by 4, 1, 2, 3, 4. You get a graph that kind of looks like this. We've changed this line with transformations. Moved it to the right, moved it to the left by 5, compressed it vertically by 2, moved it down. We can do that. Same thing kind of going on over here. There's some type of horizontal stretch, vertical stretch. There's a horizontal shift. There's reasons why we might want to leave things unfactored, right? Because we can see transformations when they're not simplified right there. Another thing I should mention is that why would we want to set these two linear expressions equal to each other? Well, one thing that happens could be something like the following. Say we have two linear equations like this. Their graphs are going to be lines. Where is this common point of intersection? Where is this point, which we can see graphically right here? If we had something like f of x equals 1 half x plus 5 minus 4 and g of x equals 1 third 2x minus 1. If you were trying to figure out graphically where these two lines intersect each other, you're looking for some x-coordinate for which their y-coordinates are the same. Specifically, you'd be looking for when f of x equals g of x. Is there a common x-coordinate for which their y-coordinates f of x and g of x respectively are equal to each other? Well, solving that equation is exactly this equation right here. So when we start wondering where two graphs intersect each other, which is actually a very important question. When we start asking ourselves when two functions intersect each other, whether graphs intersect each other, that starts to mean that we get a whole linear expression on the left and a totally different linear expression on the right. We want to solve that. That comes down to this type of linear equation. So there's some stuff going on there. Whoops, I lost my screen. Here it is.  So let's solve this equation. When it comes to solving linear equations, there's a basic template that one can go through and you can do it step by step. The first thing I do when I work with a linear equation, if I notice any fractions involved in it, like say you have one half and one third, we don't necessarily like doing arithmetic with fractions. You know, fractions are our friends, but it's kind of like that friend that's always borrowing your stuff. It's somewhat demanding. And so the first thing I want to do when it comes to an equation like this is this is sort of an optional step, but I like to clear the denominators. That is, I can actually remove all fractions from consideration. That way I can kind of postpone the difficulty of working with fractions to clear the denominators. What that means is we're going to first identify the denominators in presence. We have a one half, we have a one third. We want to identify the least common denominator. That is the least common multiple of the denominators present. And in this situation, we have two times three, which is equal to six. So what we're going to do is we are going to take our equation, one half x plus five minus four is equal to one third, two x minus one. And we're going to multiply both sides of the equation by the least common denominator. So we're going to multiply both sides of the equation by six, do six right here. Now on the left hand side, because we have actually two terms, we have a sum one half x plus five and a negative four. We're going to distribute the six onto both terms. Make sure you distribute. On the right hand side, we don't have a difference going on. We actually have a product we're not going to distribute. Now some people might think, well, why don't you distribute the six onto both pieces? And I can sort of counter that with the following. If I have something like two times three and I times that by four, do I distribute the four onto both pieces and end up with eight times 24? I'm sorry, eight times 12. And then continue on from there. No, that's not what you do. You simply just do two times three, which is six times four, which is that's where the 24 came from. So we don't multiply each factor in the product by the third factor. We only distribute when we have a plus or a difference, right? So if you have like two plus three times by four, that's the situation when you would distribute here. And so we're going to distribute the six on the left hand side right here. So this would give us a six over two X plus five. We're going to get a four times six. And then the right hand side will look like six over three and then two X minus one. When you distribute, when you multiply your six by a product, you're only going to multiply one of the factors. And the fact you want to multiply by is the fraction. That's what kind of initiated this whole process. Notice that six divided by two is a three. So we get three times X plus five minus 24. And then six divided by three is a two. You get two X minus one. Now, the fact that these factors are themselves sums and differences now means at this stage we can distribute the three. We can distribute the two and we end up with three X plus 15 minus 24. And then the right hand side, we're going to get four X minus two. So what we've done so far is we were able to clear up the denominators and we multiplied everything throughout. So now everything's been expanded out. This kind of leads into the second step when it comes to solving a linear equation. This is the step I refer to as combine like terms. And in fact, if I was going to separate these two steps, I would say that, you know, we'll draw this line to kind of separate here. Everything above this line was just the purpose was to clear the denominators. You'll notice that on this equation that has a star next to it, there's no more fractions. We can get rid of the fractions if we don't want to deal with them by multiplying by the LCD. Now we're in a situation where we want to combine like terms together. We want to add together the common factors of X and then we want to add the constants together. Oftentimes in order to get your variables together like the X is right here, you're going to first have to distribute because these variables are trapped in a cage, the cage of multiplication. The distributive property helps us get them out. So we get to a place like this one right here, 3x plus 15 minus 24. If you see like common terms on the same side of the equation, feel free to combine these together. 15 minus 24 is a negative 9.  That's the first part of combining like terms, but we also have to combine terms that are on opposite sides of the equation, right? We have sort of like this Berlin wall that separates the east and the west side of the equation. We want to reunite the X's. We want to reunite the constants. How do we do that? Well, the idea is what's good for the goose is good for the gander. As long as we do the same thing to both sides of the equation, equality will be preserved. So for example, if I subtract 3x from both sides, what you'll see here is that the 3x minus 3x on the left will cancel out. And on the right, the 4x combines with the negative 3x to give us negative 9 is equal to x plus 2. Was that a plus 2 or a minus 2? It's a minus 2. So we combine together the X's on the right-hand side. Why did I move things from the left to the right? It's mostly because I want to have a positive coefficient. It's just sort of an optimistic thing to do there. And now to solve for X, we have to get rid of the 2, so do its inverse operation plus 2 on the right. We add 2 to the left. And so we end up with X equals negative 7. And that is then our proposed solution to this equation. Now, good practice here would be to check our solution. We want to check to make sure that we have the right answer here. So coming back to the original equation, the original equation has been, you know, that's not filthy by the touch of man right here. We plug back the number into that situation. And so check the left-hand side. If you look at the left-hand side of this thing, you have one-half. X we think is negative 7 plus 5 minus 4 to that. Negative 7 plus 5 would be a negative 2. So we give one-half times negative 2 minus 4 there. One-half times negative 2 is going to be a negative 1 minus 4. That should give us a negative 5. That's what we think the left-hand side should be. We then want to compare that to the right-hand side. The right-hand side was just one-third, 2 times negative 7 minus 1. Negative 7 times 2 is a negative 14. Negative 14 minus 1, that gives us a negative 15. And the negative 15 over 3 gives us a negative 5. And so we can see that the left-hand side and the right-hand side agree with each other. So we did find the solution to this linear equation. When it comes to solving a linear equation, one only expects a single solution. There's only going to be one X intercept of a linear function. And so when you're looking for linear equations, no matter how complicated they're going to be, you only expect there to be one solution. We don't get any other situation when it comes to true linear functions.


Everybody forced here with being case number 60 of 2022 top stribble threads nine box hobby case here Bigger team style Andrew R. Cardinals Bobby D. Padres Brian B. Brewers Chris D. Braves Chun Wei with the Dodgers Edward P White Sox and Astros Eric R. Angels Mariners Yankees Glensy Royals Greg B. Guardians Harry C with the Red Sox Jake S. Twins John F. Rays John M. Diamondbacks and A's Costa K. Blue Jays Kusei Nationals Matt B. Matt's Matthew K. Giants Nick L. Pirates Rob L. Phillies Ryan H. Cubs Scott F. Tigers Shane W. with the Marlins and the Rangers Steve C. Orioles Trenton with the Rockies Viviana with the Reds. Good luck everybody. Here we go What's up War Eagle? I'm with War Eagle or Tolok alone might be the best player of any sport of all time Yeah, I just don't think Smoltz was as good as Glavin or Maddox as far as like big like overall Like all things considered Right. Am I wrong in saying that? Does anyone disagree with that? I Disagree Yeah, Johan was on his way Joseph. Johan was insane in his twins years. So good. Hope you're all having a great night That Nolan Ryan guy don't doesn't come back that guy. I'll get mad if he sees chat It's a 299 Mets Max Scherzer It's a 150 Vladimir Guerrero Exposed go to the Nationals Triple Jersey Walker Bueller to 36 Dodgers game used no hits Rookie Otto Royals MJ Melendez to 75 on card Melendez Yeah, Bob Gibson. I think yeah Gibson co-facts Randy Johnson Is it biased to say Pedro Martinez? Issue with Pedro is his prime was arguably the best ever But it didn't last more than like, you know what six years or so to 259 Corbyn Burns Brewers to 125 Kershaw Dodgers Redemption Dodgers Freddie Freeman triple threads auto rally card amber parallel Congrats Dodgers. That'll be numbered to 18. This is going to Chunwei C There we go Dodgers and Jersey card Kiroloff twins to 48. Oh, that's awesome smack. I remember that Breeze Hall silver bar happy for your friend. Thank you for if I'm sure you refer them. Thank you I put Clemens up there. I think as far as like as far as mod if we're talking modern era only I think I put Clemens up there The 299 manor Amir as Red Sox are at least in the discussion to 125 Anthony Rizzo Yankees Guardians Jose Ramirez to 18 triple relic game used Go J ran for the Guardians and Jersey Auto Cardinals Jack Flaherty to 75 Can't vote for Pedro Think back here. We got extra relic. It looks like so that's pretty cool extra hit Judge to 299 Yankees and number to 150 Buster Posey Giants Yeah, Clemens deathling belongs in the Hall of Fame triple relic auto Elohim in as 13 of 18 White Sox rookie results Jersey card judge number to 48 Yankees game used That's called Dallas overtime elite got some good prospects and angels Jersey tonight Mike Trout game used So we are sold out on baseball tonight tomorrow. However, Sam has a 20 box Bowman mixer a 20 box Bowman mixer you guys check it out right here Scott years 18 through 22 products from years 18 through 22 Including 19 Bowman draft jumbo two boxes of that 21 draft 22 drafts. There's some awesome stuff in this one 2018 Bowman Chrome, maybe some Otani is in there It's a 199 willy maze for the Astros or to Giants. I don't know I said Astros Giants Ozzie Smith to 25 Cardinals Blue J's Boba Shet triple relic auto to 27 200 hits 30 homers 40 doubles and Jersey card twins Max Kepler to 18. Hey, that's first break mojo smack It's a thing. What is going on with this pack? Invisorated all right, Bobby Whit rookie Aronato to 199 Cardinals in the 75 Mookie Betts Dodgers Nice judge to 27 Yankees triple relic game used lid lifter lid lifter Redemption Phillies triple threads auto single jumbo relic Zack Wheeler. Oh, we're talking about Kyrie now.  I tell you what no one cares more about themselves in the NBA than Kyrie Irving Cares a lot about himself to 199 Torkelson Tigers rookie and a 259 Peter Lonzo Mets Triple relic auto for the Mets Darryl Shawbury David Wright Peter Lonzo to 27 great card for the New York Mets The worst team in basketball right now Jersey for the Tigers Miguel Cabrera to 48. I Believe it's the Hornets Really they were doing decent decently. Let's let's see. Let's see Donnie. Let's see who's the worst worst in the east is the Pistons Then it's the Hornets Rockets are terrible spurs are terrible, too So it's Hornets Pistons rocket spurs He's gonna go he's only gonna go somewhere that trying to win 199 Javier Bayez Tigers to 99 Alex Bregman Astros Astros grand walk-off triple relic Jose Altuve to 36 and Jersey auto Phillies Reese Hoskins to 75 The issue with Kyrie Joseph is if you can get him to buy in like down the stretch and get him to play Like he makes your team a lot better like really no matter who you are at this point That's the issue You just don't know what you're gonna get with Kyrie He could just be like, you know what I'm not gonna play for like three games and just won't do it Yeah, you don't know what you're gonna get Like if he's invested within your team like he was really good earlier this year for the Nets Like they were on fire until he got hurt and then Just too many antics with that guy just antics all the time not to mention the BS that he did earlier this year. I don't want to talk about that Always something with Kyrie man always something with Kyrie. It's a 50 Tom Siever Mets and a 259 hunter green reds Triple relic Walker Bueller to 27 Dodgers low OPS and Jersey Auto Cardinals Jack Flaherty to 99 He's a little like hard. I'd rather have hard on my team than Kyrie. I think Just from like a team perspective To 75 will the Amaze Giants in 259 each your own Mariners There we go. John Smoltz four of nine triple relic auto Braves Good company There is a the G is damaged here. You can see it's part of like it's just like twisted Little die-cut piece. I'll send a box wrapper to the brave spot and Jersey car with a loose thread Trey Mancini Orioles to the 27. Yeah, it just kind of stinks it. He's so arrogant and selfish 299 Nationals rookie CJ Abrams to 150 Javier Baez Tigers Triple Jersey Auto Marlins Jazz Chisholm Jr. to 99 on-card jazz auto. Yeah, I'm sure Katie hates him and Jersey card Braves Ian Anderson at 36 Yeah, it happens it's just a few guys though, it's not even like a widespread thing 299 Bryce Harper Phillies to 125 Lou Gehrig Yankees. I think it was worse a few years ago triple relic red Sog Xander Bogart's three of nine and Bat Relic Auto Cubs Patrick Wisdom five of ten. Yeah, I wonder who will who like who are the potential targets people are saying? Yeah, Cleveland needs to make a trade but they they should not trade for Kyrie I'd be counterpressed. They already have a great point. Are they done of them? That's I mean it would be pointless, but I just think You think a Lakers would Joshua what would the Lakers is interesting because you know, he likes LeBron So like don't mommy play But I don't know Wasn't the whole thing with that. He didn't want to be in LeBron shadow. I Don't know whatever. I don't know man Who knows what he changed his mind every week? You know what I mean? Devers 259 Red Sox into 299 Nolan Aronato Cardinals That was in the moment Triple relic Mike shots 18 angels Donnie's his agent. He's losing the moment guys. All right leave him alone And sunny gray Jersey out of twins to 99. No calves shouldn't want them calves already made one huge trade They don't need to make a huge trade. They need to make a trade at the deadline Yeah, moving Levine makes sense to for the for the bulls Wander there Durand rookie to 299 Red Sox and Torkelson to 125 Tigers future phenom triple Jersey Auto Reds Jonathan India to 75 and Jersey Card Reds Joey Votto to 18 Game used Ryan is Emoting it up right now all the emotes. Hope we see a book in the last few here Or that last box. I guess there's not one in this one Lou Bob the 199 White Sox and a 50 Jacob deGrasse Mets That's right Josh, there's a few more Carlton Fiske a 27 triple Jersey Auto White Sox nice Fisk and Wander Franca rookie Jersey to 36 rays player war. Hopefully my version ships pretty soon here Beats me to 150 Hank Aaron Braves and a 199 Anthony Rizzo Yankees and Triple Jersey Tatis to 36 for the Padres young star and rookie auto twins Royce Lewis 65 of 199 twins on card rookie auto Can't change your wand. I kind of like that. You can't change it. You kind of have to redo it You shouldn't you shouldn't be able to pick your house the sorting hat No, Donnie I guess if you if you choose like character traits, I think they should be able to be like that You know what I mean? Like you choose certain traits and it's like, okay. Well, you're gonna be this If you can if you get the If you get to pick an animal name it we're gonna name Get an owl and I don't know the name Maybe like Maybe a Riddler Expecting a pun Wrecking that's all name. It's a 75 Boba shed blue jays to 259 angels show. Hey, it's honey It's spelled like Wrecking not record Wrecking Triple relic labor Torez Yankees game used around Or no abundant walk-offs.  All right. No huge hits in that case. I was hoping for something bigger. We'll recap here guys We got one more baseball break up next to There's 16 to go on illusions football eight bucks number 16 And then there's 10 teams left on an impeccable football case tonight. All right, relax We had Fernando Tatis Jr. Glaber Torez Fernando Tatis Jr. Wanda Franco Joey Vado Mike Trout Xander Bogarts Ian Anderson Tramansini Walker Bueller Jose Altuve Miguel Cabrera Aaron Judge Max Kepler Mike Trout Judge again, Jose Ramirez Alex Karoloff Walker Bueller Josh Lowe hun January you are the one autos now Josh Lowe hundred you first autos Royce Lewis Jonathan India Carlton Fisk as a white sock sunny gray Patrick Wisdom jazz Chisholm Jr. John Smoltz to nine Jack Flaherty Reese Hoskins match triple strawberry right Alonzo. That's a good one Zach Wheeler redemption for the Phillies Boba Shet the 27 blue Jays Aloy Jimenez to 18 white socks Jack Flaherty redemption Freddie Freeman to 18 Dodgers and MJ Melendez That's gonna do it. Thanks y'all for good on to five star