  Kensuke ուḥ Allāh ṁtādānta ṁ relaxā dhana. ṁ ៲ нарāṥa Ṭhure ṁ your honoree hello and welcome to this session namonなので Gene Una Ṭh Rok.... ṇ Ṭhī Hī Ṭhī Hī Ṭhī Hī Ṭhī Hī Hī Ṭhī Hī Ṭhī Hī Ṭhī Hī Hī Ṭhī Hī Ṭhī Hī ammunition offshore. 3 . 1 3 4 5 6 11 11 14 15 26 30 30 31 . . . . . . . . this is the rally line yes this is the rally line i like to repeat it again because here lies the concept other parts are algebraic manipulations which we have already done now the intersection of fanno and rally lines determine the two that is the upstream and downstream the two state points before and after the shock as you know again the fanno line is the locus of state points having the same stagnation enthalpy that means the same stagnation enthalpy or stagnation temperature the energy remaining same it is plotted in T s plane so the same stagnation temperature at all points and the same mass flow rate so the only way of moving along the fanno line is by friction that means with the same stagnation enthalpy and the same mass flow rate the fanno line can be we can traverse along the fanno line with friction and you see the we can flow we can follow the fanno line along the fanno line in such a way that entropy should increase this is because of the second law of thermodynamics for an adiabatic process entropy should increase so therefore the fanno line can be approached in this part either this way or in this part either this way and we have also recognized that this part corresponds to mach number greater than one that is a supersonic flow and this part corresponds to mach number less than one sonic flow this is purely recapitulation and this is the sonic condition mach number one so therefore we have identified that for any flow where the total energy remains constant that means an adiabatic flow the effect of friction in the supersonic region is to change the flow from supersonic to sonic similarly the effect of friction in an adiabatic flow in the subsonic range is to change the subsonic flow to the sonic flow so this is the fanno line characteristics similarly another line which is we have drawn known as rally line this line is the locus of state points where the impulse function remains same that means impulse function remains same that means no friction that is the absence of friction we know the impulse function remains same the same impulse function and the same mass flow rate the locus of state points are known as rally line so rally line in general indicates the heat transfer that means heat added or heat extracted from the system so therefore one can follow the rally line in both the directions that means we can go along this direction and we can also go or come along this direction so this direction while we will go in this direction implies heating of the flow where the entropy is increased while we can go along this direction of the car which implies cooling that the entropy of the working fluid decreases similar to the fanno line this portion of the rally line car represents the sonic flow a subsonic flow that is Mach number less than 1 while this portion of the car represents supersonic flow that is Mach number greater than 1 so therefore one should know that in this case the heating that means increase in entropy means we can flow in this direction or we can go in this direction which means simply that a reversible heating in a supersonic flow changes the supersonic flow to the sonic condition similarly a reversible cooling in supersonic flow changes the supersonic flow to more supersonic region reverse is true for subsonic flow that is a heating in subsonic flow in a reversible manner why I am telling reversible manner this is in absence of friction otherwise the quality of impulse function which is a condition in drawing this line will not be maintained that means a reversible heating in subsonic flow will change the flow to sonic condition while the reversible cooling in the subsonic region will make the flow towards more subsonic region now the shock represents the intersection of these two points with the fanno line the intersection of the fanno line and rally line why because we know that across the shock there is no energy addition so therefore the two points should fall in the fanno line and should correspond to the same stagnation temperature similarly the shock is so thin that we can neglect the effect of friction across the shock so that these two points should correspond to the same impulse function so that these two points must satisfy the rally line flow or the rally line condition so therefore they are the intersection points in rally and fanno line and from the second law of thermodynamics again we have found that if the two points are the end points of a shock wave then the shock wave upstream of shock wave should be this point because the downstream point should be such that it should increase the entropy since the two points corresponds to the condition of no heat addition so therefore entropy should increase so therefore from here we also see that shock always occurs in a supersonic flow and ends up to a subsonic flow with the subsequent increase in entropy so this has to be very much clear so therefore we see another thing from where we can incur that entropy change in the shock process is positive that if we go along the rally line to reach the point x and y not via shock this is the line through which we reach across the that is the shock process this is a dotted line that is the through shock from x we reach y but the x and y can be reached either via rally line or via fanno line if we want to reach via rally line first we will have to heat the supersonic flow and then we reach the sonic region then we will have to cool the sonic flow and come to the subsonic region and this heating and cooling will be such that this stagnation enthalpy or temperature will remain same that means the amount of heat added in this process should be balanced by the amount of heat extracted in this process and since you see the heat addition takes place at a lower temperature while the heat extraction takes place at a higher temperature the entropy of this system will increase while it will move from this point to this point along the rally line you have understood so even along the rally line if we follow we can tell that the entropy of y will be more than the entropy of x or even in the simple t s diagram we see that the entropy has increased so therefore it is very important concept so one should know that the shock takes place when the flow is supersonic and ultimately the flow becomes subsonic after the shock that is the y now last class we also discuss the relationship of the flow properties after the shock in terms of the flow properties before the shock that means we known values of upstream sections of the shock how one can determine the downstream values so this is the Mach number at downstream of the shock as a function of the Mach number upstream of the shock and from this equation we can see that when m x is greater than one m e y will be less than one that means the after the shock flow will be subsonic similarly we have deduced the ratio of pressure after the shock and before the shock in terms of the Mach number at the downstream of the shock well and this can be found that when m x is greater than one p y by p x is greater than one similarly we have found t y by t x as a function of the Mach number at the upstream of the shock that before the shock again by exploiting the equation of state we can find out the density ratio that is the density after the shock to the density before the shock as a function of the Mach number before the shock but now I will tell you one important thing that if we use this equation of state but do not replace p y by p x in terms of Mach number but if we only replace t y by t x in terms of the Mach number upstream of shock we arrive a relationship between density ratio and pressure ratio which is very important so it is simply algebraic manipulations by which we get the density ratios after and before the shock in terms of the pressure ratios or equating the pressure ratio finally we can tell the pressure ratio in terms of the density ratio that means this is simply the manipulations of this equations that means here using the equation of state earlier we derive rho y by rho x as a function of Mach number before the shock m x but without doing that if I only substitute this t y by t x to express rho y by rho x as a function of p y by p x we get this one then from this if we equate for p y by p x we get a relationship which describes the ratio of pressure after and before the shock as the ratio of the corresponding density that is after and before the shock this relationship is very important and known as ranking hook near relation now if an interesting thing you will see that if I we draw a variation for p y by p x with rho y by rho x for a diatomic gas if we take 1.4 for example air usually air is the working fluid under all practical circumstances so air behaves as a diatomic gasses where the ratio of the specific heats is 1.4 now you see if you put this value this value becomes that gamma 1.4 this becomes 6 so therefore we see the inspection of this equation tells us that when rho y by rho x approaches 6 p y by p x becomes infinity that means if we start at 1 for example when rho y by rho x 1 p y by p x this is a trivial solution we already recognized earlier the two state points being same so the curve for p y by p x with rho y by rho x like this that means this is asymptotic to a density ratio of 6. Let us do it in in a log log graph if you do like that you will see that it is better to show 10 then we can say 100 then we can 1000 or 10,000 like that that means even if the pressure ratio tends to infinity for a diatomic gas the density ratio becomes 6 that means the downstream density can only approach to a value of 6 times the upstream density even if the downstream pressure is many many times more than the upstream pressure rho x this is a very important relation moreover you know that if you draw in the same graph the relationship between pressure and density for an isentropic flow that means here again I write p by rho to the power gamma is equal to constant that means if I drawn the same figure the relationship between p and rho for isentropic flow then the equation will be like this this is the p by this is small p by rho to the power gamma is equal to constant. So, this is the equation from here one interesting fact is depicted that in the this range when the pressure ratio is very small for the shock the isentropic curve which is a straight line in a logarithmic plot log log plot it almost coincides with the shock curve this is known as normal shock curve normal shock curve or Rankine-Hugniott curve Rankine-Hugniott curve. So, the part of the normal shock curve or Rankine-Hugniott curve in the very low range of pressure ratio almost coincides with the curve of isentropic process this leads to the conclusion that if the ratio of pressure is very small in a shock then the relationship between pressure and density can be found out from isentropic relations that means the process of shock can be considered as an isentropic one. So, from this the definition of a weak shock comes a weak shock a weak shock is defined to be a shock where the pressure change change of pressure is very small as compared to the initial pressure. So, define whether the shock is weak or strong a parameter known as the strength of the shock p I define that a strength strength of the shock is defined strength of shock is defined as the difference between the downstream and upstream pressure that means sorry it is small better it is capital that means this is the increase in pressure divided by initial pressure. So, this is defined as the shock strength. So, when the strength of the shock is very small then the shock is known as weak shock. So, for a weak shock the isentropic flow relations for p and rho coincides with the normal shock curve now we see from this equation if we just manipulate from this equation p y by p x is 2 gamma by gamma plus 1 now if both sides I subtract one. So, this side it will be the strength of the shock because strength of the shock is p y by p x minus. So, therefore I can write from this equation by subtracting one from left hand and right hand side that p becomes equal to 2 gamma if you do it it will be coming like this m a x square minus 1. So, an interesting feature comes out that when p is very small m a x square minus 1 is also very small. So, the strength of the shock is very much related to the initial Mach number of the flow now we define a shock as the shock of vanishing strength shock of vanishing strength shock of vanishing shock of vanishing strength ok shock of vanishing strength as the shock where p tends to 0. So, for shock of vanishing strength p tends to 0 m a x tends to 1 from which we conclude that shock of vanishing strength or infinite small strength the m a x tends to 1. That means the Mach number tends to 1 the flow condition is sonic which concludes that a shock of vanishing strength propagates in a medium with the velocity of sound with respect to the medium. That means the shock of vanishing strength occurs when the flow condition reaches the acoustic speed that is the sonic condition earlier we recognize that a small pressure pulse moves in a compressible medium with a velocity equal to the velocity of sound with respect to the medium at that condition. So, therefore we can conclude that a small pressure pulse or a wave with small pressure disturbance is equivalent or a special case of a normal shock with vanishing strength because the normal shock of vanishing strength occurs when the flow condition is sonic that means other we can tell a normal shock a normal moving shock propagates in the medium through the medium with a velocity equal to the sonic velocity of the acoustic speed with respect to the medium. So, this is the thing about the weak shock we can also find out the relationship between density ratio for a weak shock if we express rho y by rho x if we see that this equation when p y by p x is very small then we can infer that rho y by rho x also very small that means the change in the density similarly the change in the temperature will also be very small when for example m a x equal to 1 here t y becomes exactly equal to t x. So, for a weak very very weak shock that is shock of infinite small strength t y will be almost equal to t x similarly rho y will be almost equal to rho x and p y will be almost equal to p x and this sonic condition will be reached all right and the similarly the entropy change will be very small if you look to the relationship for the entropy change you will see the entropy change will be very very small it will be almost 0 and that is the case where we have recognized that Rankine-Hugniard curve that is the curve for the normal shock and the isentropic relation p and rho coincides almost that means we can designate a normal shock of vanishing strength or infinite small strength as an isentropic process similar to that happens in case of a small pressure pulse or a small pressure wave in a compressible medium. I think that this is the course for you in the compressible flow now we will solve some simple and but interesting problems let us see that this problem you see that what is this problem you can take down note down this problem example one so how you can utilize straight forward the formula of compressible flow the straight forward applications example one air flows isentropically through a converging nozzle discharges to the atmosphere discharges to the atmosphere so a converging nozzle discharges to the atmosphere at any section where the absolute pressure is 179 kilo Pascal the temperature is given by 39 degree Celsius well and the air velocity is 177 meter per second well determine the pressure temperature and air velocity at nozzle throat ok so again I read the problem air flows isentropically through a converging nozzle discharges to the atmosphere at any section where the absolute pressure is 179 kilo Pascal's the temperature is given by 39 degree Celsius at a particular section ok and the air velocity is 177 meter per second at that section determine the pressure temperature and air velocity at nozzle throat it is a very simple problem and straight forward application let us find out that this is the problem a converging nozzle is there so this is the discharge plane now one thing is that it is discharging into atmosphere that means p b is p atmosphere now we do not have to be biased either way for example when you see that ok they has the problem has told determine the pressure at the throat converging nozzle throat means is discharge area so definitely pressure is not atmospheric pressure it is a choking condition otherwise the problem could not have been given do not be tempted like that it means so happen the choking may not be there in that case the pressure at the discharge may be p atmosphere so first of all we will have to check whether the pressure at the throat or the exit section atmosphere or not how do you check that now this nozzle may have a stagnation situation which can be simulated as a big reservoir that this converging nozzle may be related to a or may be connected to a big reservoir that means with respect to the flow through this nozzle is an isentropic flow there is a stagnation situation well there is a stagnation situation so corresponding to this stagnation situation the properties are p 0 t 0 rho 0 we know all those things now here we have been told that at any section at an arbitrary section the pressure let this section is 1 let this section is throat t so section 1 we have been given p 1 is 179 kilo Pascal t 1 is given 273.515 plus 39 that Kelvin and v 1 is given 177 meter per second ok so now we know that at any section the flow properties are related to the stagnation properties like this p 0 by p 1 is equal to 1 plus gamma minus 1 by 2 the local mach number whole square m a square rather to the power gamma by gamma minus simply we should start from t 0 t 0 by t 1 1 plus gamma minus 1 by so from here using these equations now we can find out this stagnation pressure how we can find out we know v we know t 1 so we can find out m a 1 how because a at this section a 1 is root over gamma r t 1 with this t 1 we can find out a 1 ok now before that we can check for this condition p 0 by p 1 that I will do afterwards now let us consider this a 1 is root over gamma r t 1 so we can find out p 0 so the value of p 0 comes out to be and also the value of t 0 ok the mach number comes out to be I am telling you this a comes out to be the problem answer is that 354 meter per second if you substitute this and with this a if you calculate if you calculate the mach number the mach number will be well the mach number will come out to be 0.5 0.5 yes mach number will be straight away 0.5 if you put this mach number and solve this with the value of gamma taken as 1.4 whenever the working fluid will be here we can take the value of gamma considering the air as a diatomic gas 1.4 so you get the value of p 0 very simple problem p 0 is 212 you can check it if you perform this calculation and t 0 well p 0 t 0 of course you can calculate it is not given here so t 0 also you can calculate by substituting the mach number 0.5 here with the gamma 1.4 p 0 comes out to be this once the p 0 is coming out we can check whether the nozzle is choked or not that means we can find out the corresponding pressure at the throat if it is choked now at the condition of the choked that m a is equal to 1. So let us find out what is the critical pressure p 0 by p star that means that if we put m a is equal to 1 this will be gamma plus 1 by 2 whole to the power gamma by gamma minus 1 which means that we can write this thing p star is equal to p 0 2 by gamma plus 1 by gamma by gamma minus 1 and this quantity with k gamma is equal to 1.4 is 0.528 this quantity is 0.528 of p 0 with gamma is 1.4. So if you perform the calculations we will see that p star is more than that that means p is 0.528 now p 0. So therefore we see that p 0 if you consider p 0 is the atmospheric pressure. So therefore we see that p star is 0. Sorry p 0 we have found already 1 what is p 0 202 what is the value 212. So what is the value of p star 111 so therefore automatically this is more than the atmospheric pressure p atmospheric is 101 kilo Pascal. So therefore nozzle is choked so that means the condition at the downstream will correspond to the choking condition. So therefore p star will be the p exit. So if I denote the throat as t here you have denoted t so p exit or the p throat. So throat condition means the exit condition will be p star here the nozzle is choked the nozzle is choked. Now when the nozzle is choked straight forward I can write t 0 by t star is what is that 2 by gamma plus 1 that means substituting the value of t 0 we can find out the value of t star and the value of v at this that means t throat is equal to t star. So value of the throat is equal to v star which is equal to a star which is equal to gamma r t star. So this value becomes equal to well 111.936 this is 112 kilo Pascal. So only this problem refers to the pressure at the throat but you can find out what is the value have you calculated. So you can calculate from it the value of the velocity at the throat it will be simply gamma r t star what is the value of t star what is the value of t star 2 by gamma plus 1 oh sorry it is sorry it is other way gamma plus 1 by 2 when you make t star it will be t 0 into 2 by gamma plus 1 I am sorry it will be gamma plus 1 have you calculated this ok. If you calculate then you will get the value so t star n. So therefore in this case the most important part is that you have to first check whether the nozzle is choked or not that means we will have to find out the stagnation conditions stagnation pressure stagnation temperature stagnation that means usually the back pressure is given whether the back pressure is lower or more as a higher lower or greater than the critical pressure that is p star corresponding to this stagnation pressure that means we can use this formula. So if we see the back pressure is lower than the critical pressure that means the nozzle will be unable to expand up to the back pressure it will choked that means the finally it will expand up to the critical condition critical pressure but if we see this critical pressure is lower than the back pressure then the nozzle will be able to expand up to the back pressure. That means it is simply expansion up to the back pressure continuous expansion up to the back pressure will be the pressure at the throat ok. Now next problem you please write another problem air flows another simple problem steadily you write this problem air flows steadily and isentropically in a converging diverging nozzle at the throat the air is at 140 kilo Pascal's the throat condition is given absolute that means this is the pressure in absolute and at 60 degree Celsius well at the throat the air flows steadily and isentropically in a converging diverging nozzle this is a converging diverging nozzle at the throat the air is at 140 kilo Pascal's and at 60 degree Celsius the throat cross sectional area is 0.05 meter square at a certain section in the diverging part in the diverging part at a certain section in the diverging part of the nozzle of the nozzle of the nozzle well the pressure is 70 the pressure is 70 kilo Pascal's absolute all pressures are in absolute. Calculate the velocity simple problem calculate the velocity calculate the velocity and area of this section and area of this section. Now I repeat this problem again air flows steadily and isentropically a steady and isentropic flow in a converging diverging nozzle here the nozzle is a converging diverging at the throat air is at 140 kilo Pascal's absolute and 60 degree Celsius the throat cross sectional area is 0.05 meter square at a certain section in the diverging part of the nozzle the pressure is 70 kilo Pascal's absolute calculate the velocity and area of this section. Now the tips for this type of problem is like this when you have a converging diverging nozzle now you may have two situations flow in this direction one is the venturing another is a nozzle though it is told a converging diverging nozzle rather I should converging diverging duct I will make it flows through a converging because the nozzle what may confuse it. So, if I tell a converging diverging duct so then you can take it both as a nozzle or a diffuser. So, what is the check now let us consider this is a throat and some section in the divergent portion is one now throat conditions are given p t is equal to 140 kilo Pascal's. So, t t is equal to 60 degree Celsius that means 273.15 plus 60 that means 330.15 kilo and area is given 80 is 0.05 meter square now since p 1 is given as 70 kilo Pascal's. So, is it a venturing flow or a nozzle flow convergent divergent nozzle nozzle flow yes that first you will have to understand because in the divergent part the pressure is decreased that means until and unless the flow is supersonic and increase in area cannot decrease the pressure which means that this part the Mach number is greater than one that means it is a convergent divergent nozzle which means that this part the Mach number less than one and nozzle is the condition where Mach number is one. So, therefore, we are identify these as a convergent divergent duct since there is an expansion or a decrease in pressure in the divergent part of the duct this is the only tips or the crack of the problem. So, therefore, we can find out the respective p 0 t 0 we can find out the respective p 0 respective t 0 what is t 0 t 0 is equal to t 0 if you t 0 by t t t t is the t star t t is the t star that is t 0 by t star is gamma plus 1 by 2. Similarly, p 0 is p t into gamma plus 1 by 2 to the power. So, taking the value of gamma is equal to 1.4 and p star p t t t is the p star t star because p t is p star because throat corresponds to the sonic condition for a convergent divergent nozzle. So, therefore, if we know the t t and p t we can find out the corresponding stagnation properties for that flow that is p 0 and t 0. So, this p 0 and t 0 values comes like this p 0 is if you calculate p 0 comes out of the 0 to be 265 kilo Pascal's and t 0 comes out to be 400 k. Well, now what you will find out now since you find out the we have to find out what calculate the velocity and area of this section. Now, to have to find out the velocity at that section first of all we will have to find out the mach number at this section if we know the mach number at this section and other properties like p 1 t 1 we can calculate the velocity. So, to calculate the velocity we will have to know mach number that means m a 1 how to know the mach number we know p 0 we know p 1. So, therefore, we relate this p 0 by p 1 this we use this equation. So, only you will have to see the which equations we will be using m a 1. So, gamma by sorry gamma by gamma minus 1. Means we know this value p 0 is 265 p 0 is 265 and p 1 is given in the problem as 70 and this mach number is found out and we see that this mach number corresponds to a value which is more than 1. Obviously, and it is 1.52 this is by calculations all of you understand we find out mach number when we find out the mach number what we can find out first of all we can find out the t well we can find out similarly the t 0 by t 1. Now, when we know the mach number we can find out. So, we first find out the mach number by equating this p 0 by p 1 we are given value of given the value of p 1 not the value of t 1. So, once we find out mach number. So, put this mach number and get the value of t 1 the temperature at that section which comes to be 274 k. Now, everything is known if I know t 1 I know a 1 as root over gamma r t 1. So, the value of a at that section becomes well the value of a is not worked out in this problem. So, a becomes root over gamma r t 1 and we can find out v 1 as mach number that is 1.52 times the a 1 and that value becomes equal to 504 meter per second. Now, to find the area of cross section velocity and area of cross section. Now, area of cross section here to be found out by equating the mass flow rate under steady condition. So, if I write the continuity equation that is the mass flow rate at steady condition at throat and at this section then we can write that m dot is equal to density at the throat into area at the throat into the velocity is equal to density rho 1 v 1 a 1. So, what has to be found out a 1. So, see that whether everything is known rho throat is rho star that we can find out a throat is a star and v throat is v star. So, a star is given in the problem 0.05 meter square v star is what v star is same a star is root over gamma r t star. So, we know this thing. So, v star we know. So, a star is given in the problem and rho star we can find out similarly from this type of relationship. That means I write it that is rho 0 by rho star is equal to 1 plus gamma minus 1 by 2 into 1 by gamma minus oh you cannot see rho 0 by rho star. That means we can find out the rho star alright. That means the quantities that is density area and velocity required in determining the mass flow rate at the throat we can find out the same mass flow rate if we equate with the quantities at the desired section at the steady state rho 1 v 1 even. So, v 1 we know rho 1 we can find out by using the same equation same type of equation which connects rho 0 by rho 1 with the Mach number. That means it will be this rest to the power 1 by gamma minus 1 rho 0 by rho 1. That means rho 0 by rho 1 for any local property will be gamma minus 1. That means once we know the Mach number at any section we know the relationship of the flow properties like density temperature and pressure with this stagnation properties through this relationship for isentropic flow. So, we know rho 1 v 1 is already known. So, we can find out a 1 the cross sectional area well. So, this cross sectional area after calculation comes to be well the cross sectional area after calculation comes to be 0 point a 1 comes to be here I write a 1 can you see yes comes to be 0 point you can check your calculation 96 meters. So, this two equation two problems sorry two problems highlight the basic understanding of the compressible flow and the straight forward application of the formulae. So, what happens is that when this relates to a problem of convergent nozzle relates to a problem of convergent nozzle you first try to find out this stagnation conditions from the conditions given if stagnation condition and the straight forward given that is all right otherwise you find this stagnation properties and check whether the back pressure which is usually given for a problem is lower or greater than this critical pressure. That means, you find out the critical pressure that is when this sonic condition is reached compare that with the back pressure and determine whether the nozzle is choked or not and accordingly solve the problem. Similarly, when the problem will be posed tactfully through a convergent divergent duct do not hurriedly consider this duct as a nozzle or continuously a diffuser. So, it will act continuously a nozzle a diffuser when one part is subsonic other part is supersonic that means, the supersonic flow will be diffused in this part and again a sonic subsonic flow will be this will acting as a continuous diffuser or a continuous nozzle. So, in that case the conditions will be accordingly given, but it can act as a venturi meter. So, that you first decide that whether it is a nozzle or a diffuser or a nozzle and diffuser mix that means, throughout the flow is subsonic and accordingly you find for example, if it acts as a nozzle continuously that means, this is the sonic condition this part is subsonic this part is supersonic. If you find that this is acting as a continuous diffuser then also this is a sonic in that case it is supersonic and it is subsonic. So, you first decide that from the data given and accordingly you solve the problem. Thank you. Thank you.