 Good morning. I welcome you all to this session. Today we will be discussing choking in a converging nozzle. Last class we have recognized the effects of variation of area in an isentropic flow and it is the consequence of such effects of area variation in an isentropic flow is the very interesting phenomena in a compressible flow is the choking in a converging nozzle. So, if we see the mass flow rate now the mass flow rate at any section in a compressible flow could be written as rho into a into v where rho is the density a is the area and v is the flow velocity. Now, with little algebraic manipulation this can be written like this the mass flow rate per unit area is equal to rho times v which can be written as p by r t considering throughout this course we will consider the flowing medium or the fluid flowing is a perfect gas. So, which obeys this law of equation of state or the law of state p is equal to rho r t. So, I replace rho as p by r t and v as you know can be replaced as a into mach number where a is the speed of sound. So, this can be written as little manipulation p by p 0 taking p 0 there by r t and we know a the speed of sound can be written as gamma r t where gamma is the ratio of specific heats and mach. So, therefore, m dot by a can be written as a continuation of this is p by p 0 now this r r goes like this. So, we can write gamma by r under root this root r root then p 0 by we can write a root over t 0. So, this root t and t makes root t 0. So, that we can write root over t t that means we bring one root over t 0 forcefully denominator and numerator. So, sorry this is root over t root over t 0 by t into m e this can be written as this gamma r p 0 t 0 are constant. So, that we can write this as root over gamma by r well p 0 by root over t 0 gamma by r p 0 root over t 0 then p by p 0 this is root over into root over t 0 by t into m e. Now, we have already recognized that this stagnation properties p 0 and t 0 bears the relationship with the local properties like this p 0 by p is 1 plus gamma minus 1 by 2 m a square to the power gamma by gamma minus 1. This comes basically from t 0 by t which was first deduced 1 plus gamma minus m a square now if I substitute this p by p 0 and t 0 by t quantities in terms of the mach number what we get we get m dot by a is root over gamma by r into p 0 by root over t 0 by 0 into what we get root over p 0 m a and this p by p 0 we get that this is I am sorry this is p 0 by p and this is t 0 by t all right. So, p by p 0 that means 1 plus gamma minus 1 by 2 m a square. So, it will be minus gamma by gamma minus 1 and this will be 1 plus gamma minus 1 by 2 all right m a square to the power half now this 2 if you combine you get half minus gamma by gamma minus 1 is equal to 2 gamma minus 1 well. So, gamma minus 1 minus 2 gamma. So, this becomes minus gamma plus 1 divided by gamma minus 1. So, I just work it out. So, therefore, we can write m dot by a becomes root over gamma by r p 0 by root over t 0 all right into m a this we can take this is a minus sign. So, that we can write 1 plus gamma minus 1 by 2 m a square raise to the power gamma plus 1 by gamma minus 1. So, ultimately this is a very important expression for the mass flow rate per unit area in a compressible flow through any variable area duct that mass flow rate per unit area you see gamma by r p 0 by root over t 0 m a 1 plus gamma minus 1 by 2 m a square raise to the power gamma plus 1 by gamma minus 1 all right. Now, this expression this expression if you plot graphically or if you inspect mathematically possesses a maximum value now what is that maximum value first of all recognize which is the variable. So, if we consider m dot by a as the dependent variable you see p 0 t 0 are the constants these are the stagnation properties gamma is the fluid property r is also a fluid property that is the characteristic gas constant. So, here mach number is the only independent variable. So, that we can tell that m dot by a is a dependent variable is a function of independent variable mach number and the functional relationship is such that mach number appears in this way the numerator and denominator because of which this function that means m dot by a possesses a maximum for a value of for a particular value of m a. So, our next intention is to find out that maximum value or m dot by a or the value of the argument m a at which this m dot by a possesses the maximum. So, a very simple task is that make this 0 the differentiation of this dependent variable with the independent variable d m dot a d m a to make it 0 first of all we should find out what is d m dot by d m a d m dot a by d m a that means we will have to differentiate this with respect to m a. Let us do this we will see that gamma by r p 0 root over t 0. So, first we take this as the first function whose. So, therefore, please exponent of there there will be 2 gamma minus 1. So, there will be 2 gamma minus 1. So, there will be 2. So, it is all right is it all right now. So, 1 by 1 plus gamma minus 1 by 2 m a square all right. So, gamma plus 1 by 2 gamma minus 1 then the next term will be please tell me minus. So, this may be common does not matter again I can write root over gamma by r p 0 by please slowly we can do t 0 then m a remains as it is m a. So, now what we can do the differentiation of this that means we make a bracket here. So, differentiation of this is this will be minus that means minus gamma plus 1 divided by 2 gamma you also check whether any mistake is done by me or not minus. So, this that means this quantity 1 plus gamma minus 1 by 2 minus 2 m a square x to the power n n x to the power n minus 1 same formula that means minus gamma plus 1 by 2 gamma minus 1 minus 1 and this is multiplied with 2 s very good. So, gamma minus 1 by 2 into 2 m a all right this can be written as if you take 2 m a minus gamma by r. So, this is equal to 0 d m a by this now first of all I write the expression of this gamma by r p 0 by root over t 0 and if I take this as the common. So, 1 by 1 plus gamma minus 1 by 2 m a square all right gamma plus 1 divided by 2 m a square 2 gamma minus 1 then what will be there this will be 1 please 1 minus this will be what gamma plus 1 by 2 gamma minus 1. So, I what will be this gamma minus 1 and gamma minus 1 will be this 2 and this 2 will cancel. So, therefore, 1 minus m a square what will be the result m a square this will be gamma plus 1 all right please any problem you tell me gamma plus which one will be plus please tell me 1 this expression this will be plus why this will be minus. So, you check it this will be plus. So, 1 minus that will be m a square gamma plus 1 now when I divide this this will be cancelled that means simply this will come down 1 plus gamma minus 1 by 2 m a square. So, is it all right 2 will be there sorry 2 will be there 1 plus gamma minus 1 by 2 ok. All right this will be the expression now for maximum for maximum value of d m dot a by d m a for maximum value of m dot by a for maximum value of m dot by a this has to be 0. So, definitely this has got a maximum because the second derivative is the minimum well. So, therefore, the maximum value that we can check we are not going to check it here. So, if you make it 0 then we will get 1 minus m a square gamma plus 1 all right divided by 2 into 1 plus gamma this quantity 0 which gives m a is equal to 1 which gives m a is equal to 1 ultimately if you take this 2 plus gamma minus 1 m a square minus m a square gamma plus 1 which will give simply this is a very very interesting conclusion. That means the mass flow rate per unit area at any cross section reaches its maximum value when m a the Mach number at this section reaches 1 what is the physical significance of it. Now, we will come to the physical significance of it now. Now, let us consider flow through a nozzle or any divergent it may be a convergent divergent duct. Now, we know that at a given set for a given steady condition the mass flow rate is constant through the duct mass flow rate is constant through the duct. So, the mass flow rate per unit area is usually maximum at the minimum section minimum section is here and this is the discharge section for a converging nozzle and this is the throat for a converging diverging duct. So, the Mach number 1 means that if we go on increasing the flow velocity or the subsequent Mach number we will see the value of this m dot by a the maximum value of m dot by a at any section. That means, for example, in a converging duct this is at the discharge plane this reaches maximum if we go on varying flow velocity by changing the pressure at the downstream for example, then we will see this m dot by a will reaches maximum when this section will attain a velocity corresponding to Mach number is equal to 1 this is the physical significance that mass flow rate per unit area becomes maximum when the Mach number at that section becomes 1. Now, let us see that this will be explained more clearly if we observe this. Now, let us consider this thing first of all you must know that certain terminologies that for example, there is a convergent duct first of all we will discuss the convergent duct. Now, the pressure of the ambience where the convergent duct discharges the fluid discharges is discharged from the convergent duct is known as back pressure of the mass that means, the downstream pressure of the mass and the inlet plane pressure is the inlet pressure p i which usually we refer as stagnation pressure if it is coming from a big reservoir with this nomenclature. Let us now consider a situation like this let us now consider a situation like this let us now consider a situation like this let us consider a converging nozzle is discharging fluid from a reservoir this is the direction of flow. So, this is a reservoir this is a reservoir where the fluid is substantially at rest and the properties are corresponding to rho 0 t 0 p 0 t 0 rho 0 the stagnation property. Now, therefore, p 0 is the pressure that is the inlet pressure to the nozzle the stagnation pressure now you see this the nozzle is discharging into this chamber which is connected like this with a valve to atmospheric pressure what is the purpose of it physically what you want to mean that this the if this be the back pressure that is the pressure of the chamber where the fluid is discharged from the convergent nozzle convergent nozzle is isentropic that means, is adiabatic properly insulated this our purpose is to vary this back pressure of the nozzle for a fixed value of this stagnation pressure that is the inlet pressure to the nozzle. So, this can be done practically like this that if we open the valve first if the valve is closed this pressure is equal to the pressure p 0. So, when you gradually open the valve this pressure is going to be reduced and the flow in the nozzle will be increased. So, if we widely open the valve this pressure will be ultimately p atmospheric that atmospheric pressure atmospheric pressure is the minimum pressure which we can achieve, but this pressure can be made higher than the atmospheric pressure at different values by closing the valve gradually or other ways starting from the fully closed position opening the valve gradually we can reduce the pressure from a very high value which may be equal to p 0 under the fully closed position to a value of the pre atmospheric pressure. That means, this is the way in practice we can reduce or we can change the back pressure of the nozzle. Now, what we will see let us draw two figures one is with the length of the nozzle let us consider this is the exit plane exit plane of the nozzle e and let the pressure at the exit plane be denoted as p e now we draw space is not much. So, we draw two figures one let we consider the pressure variation along the length and velocity along the length of the nozzle. So, now we will see physically what happens when the valve is completely closed there is no flow throughout the fluid will fill and it is a static fluid filled. So, p 0 p b p e all same that means we can draw a line like this where this is p 0 is equal to p e is equal to p b that means the stagnation pressure pressure at the exit plane and the back pressure all are same. Now, what happens in practice that if we gradually open the valve p b is getting reduced that means we gradually open the valve and set a value of p b which is lower than p 0 that means we set a value of p b here this is the first one p b one then what will happen the flow will take place through the nozzle small flow will take place and pressure will be gradually decreasing to this and the exit pressure in the nozzle will match to the back pressure that means the exit pressure in this case p b one is equal to p b or other way you should write p is equal to p b one here the p e will equal to p b one and what will happen the velocity also will increase will from a 0 value to a maximum value at the exit plane if we further open the valve so that we reduce the pressure to a another lower value p b two for example p b two that means second case then also the nozzle flow will increase because the pressure difference is more so there will be a continuous expansion and the exit pressure will match to the back pressure of the nozzle that means the nozzle plane the exit plane pressure the pressure at the exit plane of the nozzle will be equal to the back pressure that is it so velocity also curve will show like this there will be more velocity at any point in the nozzle the velocity will be more so this is the discharge plane so now if we still open the valve again more so that p b is further reduced then it will so happen that at some condition let us consider that p b three when the flow rate will continue to increase and pressure will continuously decrease in such a way that means this is one two so this three condition represents a flow velocity here v at the discharge plane which is equal to the velocity of the sound or rather we can write Mach number is equal to one that means if I draw the graph of Mach number it will be from zero to some value zero to some value below one so this will correspond to a value zero to one there is no space otherwise I could have shown this curve so it will be a similar curve like that that means if we go on reducing the pressure we will see that a pressure will be reached here which will correspond to a flow velocity which is equal to the Mach number one it is obvious because we have already recognized the sonic properties that p by p star if you remember that becomes equal to what is the value gamma plus one by two to the power gamma by gamma minus one or we can write p zero by p star that is p star by p zero p zero is two by gamma plus one to the power gamma by gamma minus one that means corresponding to a stagnation pressure p zero there is a value of p star where that means there is a value which is p star where the flow velocity will be equal to the velocity of the sound that is v r t star that means this is equal to this is known as v star and v star is equal to this is equal to a star that means the Mach one is reached and in that case the back pressure p b three in this case will be equal to the exit plane pressure and that will correspond to our p star that is the condition at the corresponding to m e is equal to one now what will happen physically we can still lower the pressure this pressure may be still higher than atmospheric pressure. So, physically we can still go on lowering the pressure or even if it is atmospheric pressure we can attach it to a vacuum pump. So, physically we can still lower the pressure from this p star which corresponds to a sonic condition here what will happen if we lower this pressure here that is a case four p b four which is lower than the p star then what will happen further expansion is not possible that means the nozzle cannot increase the velocity more than Mach one that means if we express still if we read by reduction of this pressure flow velocity will still increase that means the velocity will reach more than the sound velocity that means this part the latter part of the convergent duct will act as a supersonic flow that will not be done because the area reduction is not accompanied with acceleration in supersonic flow. So, therefore the further acceleration is not possible that means the velocity here will still remain as a that means the Mach one will be still remaining and the corresponding pressure p star will remain like that that means by a further reduction in the back pressure the exit plane pressure conditions for example the pressure in the exit plane will not reduce further it is not only the pressure, but the velocity corresponding flow properties will remain same as that corresponding to Mach one case that means in that case the curve will remain same that means even if we make another lower back pressure the curve will remain same the pressure curve the velocity curve will remain same. So, what will happen in practice the fluid will suffer a sudden pressure jump from the exit plane to this region in the surrounding ambient this will be caused by a sharp discontinuity in the pressure field which is a three dimensional phenomena and through a series of shock waves. So, there is oscillation and finally it attains a steady back pressure value similarly with this back pressure. So, therefore when the back pressure is reduced below the critical pressure this pressure is known as the critical pressure that I will come afterwards below the pressure corresponding to the sonic condition at the exit plane the pressure will remain same that means which corresponds to this condition and the velocity will remain same that is equal to root over gamma t star and a sudden pressure jump through a series of shock waves will take place. So, the velocity curve and the pressure curve will remain same let us examine what is the graph of or figure of mass flow rate in this condition in this condition if you see here if this condition if you draw the mass flow rate you will get a picture like that if you draw the mass flow rate as you have already seen that mass flow rate equation per unit area the mass flow rate equations per unit area here you see the mass flow rate equations per unit area P 0 by t 0 root over gamma by r this quantities there usually t 0 p 0 are taking like with mass flow rate per unit area this is the usual convention that t 0 root over t 0 by p 0 is taken and if we take the if we draw the graph or figure for this mass flow rate variation per unit area with the back pressure p b with the back pressure p b well then what happens is that let p b by p 0 when this is one that means here you see that when the valve was completely closed that means this is the line so there was no flow no velocity that means this starts from this point as we go on reducing the p b the mass flow rate per unit area increases and it increases up to the point when p b well p b big b I have given p b reaches p s asteric p b reaches p asteric then the mass flow rate will not change further because we have already seen that mass flow rate per unit area this becomes a function of Mach number where it at least the maximum value when Mach number becomes equal to one that means this is the condition when Mach number one is reached that means during this condition this this region p b is equal to p e but during this condition p b is less than p e and p e becomes equal to p s asteric that means here if you reduce the p e the exit pressure at the nozzle discharge plane that is the pressure at the nozzle discharge plane is also reduced and match to it p b and the mass flow rate increases because these are the situations where you can cause an increase in flow through the nozzle by reducing the pressure so you go on reducing the pressure as it happens for all incompressible flow the flow rate through the nozzle will gradually go on increasing so continuous expansion will take place and a continuous acceleration will take place so in those cases the exit plane pressure will always match the corresponding back pressure but when this pressure reaches the pressure corresponding to the situation when the Mach one is reached here then a further reduction in back pressure that you can do the will not change the exit plane conditions that means the exit plane pressure will remain as that corresponding to the Mach one situation so this is this situation during when the mass flow rate will not change is remain constant so this situation is known as choking choking in nozzle choking in nozzle and the flow is known as choked flow then the flow is known as choked flow flow is known as choked flow so under this condition the flow is choked choked means no further change in the flow can take place or rather further increase in flow can not take place by an reduction in back pressure now let us examine few more things now we know that the relationship between the stagnation properties sorry sonic sonic and stagnation properties if we recall we know that if we start from t star t star to t zero is equal to it will be just the reverse two by gamma plus one is it true similarly p star by p zero will be if we see that earlier we have recognize the sonic properties in relation to stagnation properties gamma by gamma minus so rho star by rho zero is equal to two by gamma plus one by gamma minus so you see now that when it is not the nozzle which is of concern when these properties are fixed that means the stagnation state is fixed our sonic state is also fixed for example for a given stagnation pressure the sonic the pressure for the sonic condition is fixed this condition is known as critical condition critical condition sometimes in respect to nozzle flow this asterisk we can reduce as c that is we write as p c that means asterisk can be the star can be written as c so that you can replace rho star is rho c that means for a given stagnation pressure inlet pressure to the nozzle the critical pressure that is the pressure where the mac one is reached is already fixed provided the gamma is fixed which is a fluid property so it depends entirely similarly for the stagnation temperature is fixed the critical temperature is fixed if the stagnation density is fixed the critical density is fixed that means this critical state is fixed corresponding to a stagnation state similarly if we just have a look the mass flow rate equation m dot by a you see it is a function of mac number so therefore if i put m a is equal to 1 if put m a is equal to 1 and find out from here what is the m dot a maximum please find out the value of m dot a maximum maximum is equal to root over gamma by r p 0 by root over t 0 so simply it will become it is 1 that means 2 plus gamma minus 1 that gamma plus 1 that means it will be 2 plus 1 by gamma gamma plus 1 by 2 to the power gamma plus 1 by 2 gamma minus 1 therefore we see that the maximum mass flow rate that m dot by a per unit area is also fixed by the stagnation properties p 0 t 0 gamma r the fluid properties which i want to mean that this reaching or attainment of the critical condition depends on the stagnation conditions along with the corresponding mass flow rate so therefore the mass flow rate per unit area m dot a and the pressure which will correspond to the critical condition here all will depend pressure or temperature or the density which will correspond to critical condition here and the corresponding mass flow rate per unit area which will be the maximum above which we cannot go even if we reduce the pressure depends all on stagnation condition now the question can come that if the critical property critical pressure has reached that p b has reached to this value p c corresponding to p c where p is equal to p c in that case if we use the same p c and if the nozzle is reduced that means if i reduce the nozzle area nothing will happen the p c will remain same that means the same curve will reveal and the value of m dot by a will remain same as i already wrote here where i have written in value of this will remain same so only thing is that total mass flow rate will change that means mass flow rate per unit area is uniquely fixed is uniquely fixed where when while this stagnation properties are fixed so while you decrease the area the total mass flow rate will change that means it will be reduced if you increase the area it will be increased so therefore the conclusion is that mass flow rate per unit area coming out from a converging model of course at any given stage this is the maximum here at the exit plane and this value will reach a further maximum with the change in the sets that means set means for a given stagnation pressure if you go on changing the back pressure when the mach number one is reached and at that condition the pressure is p star which is known the critical conditions or the p c is given by this expression if we consider gamma is one point four which usually as a thumb rule we use that for a diatomic gas gamma is one point four and if you substitute here this formula always use t zero becomes point eight three three p star usually use point five two eight two and corresponding row star from this formula becomes point six three three point zero point zero point zero point so these are the values that means if I know a particular stagnation state I can tell the critical so therefore for any nozzle so now we can find out that if I have a stagnation pressure I can tell so this must be the back pressure point five two eight two times this stagnation pressure for which I can get the maximum flow that means at when the pressure will reach to a value of point five two eight two times this stagnation pressure the sonic condition at the discharge plane of the nozzle will arrive so a further decreasing back pressure will not increase the flow in the nozzle under this situation the flow through the nozzle is said to be choked that means flow is choked so a further increase in the flow cannot take place a very interesting physics a very interesting physical explanation for this is like that now as we have already recognized earlier that a in a compressible flow a disturbance propagates as a pressure wave upstream with a velocity equal to the velocity of sound relative to the fluid medium now you see when the nozzle is choked now we have recognized the choking in the nozzle that means if the back pressure is reduced to a value when the sonic condition is reached at the exit plane no further change in the back pressure that means if you still reduce the back pressure will not reduce the will not increase the flow rate through the nozzle now if you think in this manner that how the flow rate is changed physically when the back pressure in the nozzle is above the critical pressure reduction in the back pressure how does it change the flow when you reduce a back pressure for example you operate a valve at the downstream where the people does in practice what happens a disturbance is created at the downstream which is transmitted as a signal in the form of a pressure wave go to the upstream and changes the flow so that it sends a new signal with an increased flow through the nozzle so this is precisely the physical picture but when you reach the sonic condition at the throat then a further change in the back pressure which can create a disturbance cannot go upstream as a messenger to change the signal to give another increased flow rate because in that case the absolute velocity of the disturbance becomes zero because its velocity towards the upstream direction relative to the fluid flow is the sound velocity if you recall that so when the fluid is flowing in the sound velocity it has reached at this end so this wave becomes a stand steel wave it cannot penetrate it cannot go because the velocity becomes zero there if you calculate the absolute velocity at the discharge end which is created which will shoot proponent upstream the velocity is zero because this relative velocity is acoustic velocity it is zero it becomes exactly zero so it cannot move upstream to change the condition so therefore the condition in the nozzle remains as it is that means the pressure distribution in the nozzle the velocity distribution in the nozzle all the conditions that all the sections remains unchanged along with the mass flow rate through the nozzle and the mass flow rate under this condition is maximum and the condition at the discharge plane of the nozzle that is the nozzle orifice of the discharge plane whatever you call is the sonic condition this flow is known as choked flow and this situation is known as choking in a converging nozzle the flow is choked all right so this depends only on this stagnation pressure corresponding to a stagnation conditions stagnation pressure temperature and density the critical pressure density and temperature depends and also the mass flow rate per unit area which is the maximum for that situation any question please please any question please ok thank you