 good morning to all of you and welcome to this session of course now last class we are discussing that what will happen if the fluid flows through a convex corner or convex smooth surface let us see that what happens now if we consider similarly a type of oblique shock and you just draw the diagram that v1 is there now it will be deflected as per the geometry this surface is a convex surface this will be the v2 this line is parallel to this and these are the normal and the tangential velocities normal and the tangential velocity now according to our earlier considerations along the shock the property does not change no net force acting along this direction so vs1 is vs2 now for the same vs1 and vs2 the geometry shows clearly vn2 is more than vn1 see that mathematically also you can see that if you consider this angle as beta which was the oblique which was the oblique shock wave angle then the delta is this one so this angle is beta plus delta so this is the angle here beta plus delta because this line is the parallel to the shock so this angle is beta plus delta so vn1 is then what will be that vs1 tan beta where vn2 is vs2 tan beta plus delta and vs2 and vs1 are same so therefore from geometry and from trigonometric relations we see that vn2 that means vn2 is greater than vn1 so this cannot be in a shock so shock is not possible so therefore shock is not possible shock is not possible and therefore this case there will be an expansion wave or expansion flow with expansion waves this we will discuss so before coming to the discussion on expansion waves what we will do first we will make some mathematical analysis let us first do a mathematical analysis like this it is that it like this a plate type of thing let us make a slight change in the orientation theta d theta what happens to the flow condition let us consider first oblique this shock may be oblique shock or a mac wave oblique shock wave just we are now making a generalized thing that there will be a wave which may be mac wave which may be shock wave for that now what we are doing we are making a infinite small turning of the flow that means flow is turned by an infinite so that means the flow is v1 which has the component vs1 and vn1 is turned through an infinite small amount d theta and this value is v1 plus let us consider dvsmall change okay and this is the normal component which is vn1 plus dvn1 and this is the vs1 plus dvsmall change now let us first find out with this small change in this direction what will be the changes in the flow properties now for that we make a representations with infinite small changes in the quantity that velocity quantity similarly the pressure p1 let the pressure p just to represent by p p plus dp so temperature let us consider t here so this will be t plus dt and the density is rho which will be rho plus d rho that means all infinite small change now let us find out whether this changes are positive or negative and here this change on theta in this direction we will consider as a positive change in theta which I will explain earlier how it is known as positive and all this thing at first we consider this type of infinite small change in this initially it was flat so this make a change in the shape that means this is a corner type of thing so where there is a wave and this wave changes the flow turns the flow according to this by an angle d theta that means if you just this is make this is the original direction this is this angle is d theta this angle with the original direction of the velocity now if this be the thing let us find out first the continuity equation so continuity equation we can write that rho vn1 is equal to rho plus d rho into vn1 now we will not write any one we will make vn because here we are already making with the infinite small change so therefore we will consider all as vn vn plus dvn so one I will not use though I have written it in the diagram please change it vn dvn and neglecting the higher order term the product of the higher order term for example d rho dvn we get this equation rho dvn is it plus n that is v dvn that is vn d rho is 0 this we can get by simply telling rho vn is constant making a logarithmic differentiation or simply differentiation rho dvn plus v n d rho is 0 so this is one equation from the differential form from the continuity very simple then if you write the momentum equation with respect to an infinite small control volume here then the momentum equation we said p momentum equation means the momentum theorem in the direction this along this normal direction that means the net rate of momentum a flux normal momentum a flux is equal to the net force in the normal direction that means here area cancels out for both the term both the sides here also it will do like that p minus sorry p plus dp this side is p plus dp this side is p net force is equal to rho vn well into vn plus dvn plus dvn plus dvn plus dvn plus dvn plus dvn minus vn well so therefore this cancels so therefore we get dp is equal to rho vn dvn dp what we get here from here minus dp is equal to rho vn rho vn dvn sorry vn dvn rho vn dvn clear rho vn dvn now if I here what we do if I substitute the dvn from here that means this dvn if I write in terms of minus dp the dvn is minus dp rho by vn then what we get if we substitute dvn here dvn is minus dp by rho vn minus dp by rho vn so vn vn square so you will get a very good relationship that vn square is dp by d rho dp by d rho or vn is equal to dp by d rho clear just dvn is substituted here minus dp by rho vn now since this change is infinite small we can calculate the process to be free from dissipation and adiabatic so that this can be considered as an isentropic condition that means this can be written as and it can be shown that this way the process is isentropic also del p del rho at constant so therefore this expression is the expression for sound velocity at this state so therefore in this case the normal velocity is sound which means that a infinite small turning due to this infinite small disturbance type of thing infinite small turning infinite small angle of turning d theta makes a wave which is nothing but the mac wave vn is equal to a that means this can be shown this way now that this is simply a mac wave that means if this be the v1 then this is the v1 then the normal component of velocity this is the sorry this is the I am sorry this be the v1 this will look nice this normal component of velocity vn1 simply you write v0 and vn is equal to a and therefore this angle is the mac angle this angle this angle that means this angle means which angle that means this angle well this angle if this is alpha so alpha is what tan alpha is tan alpha is tan is this divided by this so sin alpha not tan alpha I am sorry sin alpha is I am sorry sin alpha is this divided by this that means a by v that is is equal to 1 by mac number mac m so therefore this is the this angle alpha is 1 by m sin inverse that means alpha is sin inverse 1 by m that means this small divergence we will call we will find we will ultimately make a mac wave we will create a mac wave and this will be inclined with this wave by an angle which is the mac angle now next is we do it well after that we write the equality of this equality of this let me do it here otherwise you will be in little trouble that this vs1 is equal to vs2 and vs1 is what now let us consider v and this is v plus dv so therefore if we write this we can write vs1 is vs2 so that means v cos alpha now I define this as alpha cos alpha v cos alpha is equal to v cos alpha this is alpha cos alpha v cos alpha is equal to v cos alpha is equal to v plus dv that means v plus dv and this angle then changes cos alpha plus cos of not plus minus alpha sorry cos alpha minus d theta this angle changes because this is alpha that means this is this is the alpha and this has been changed by d theta so therefore this angle is alpha minus d theta earlier we did it so therefore we can write that equality of this v plus dv cos alpha minus d theta is equal to v cos of alpha well again I draw that figure for you because I understand that you are in trouble sorry this is small angle theta I am just d theta I am just exaggerated it let us consider this that this is the sorry this is the v1 v1 and this is the okay this is the v1 and this is the alpha 1 so therefore this is the alpha 1 okay so sorry I have done some mistake sorry rather I have done some mistake that is why there is a problem I am sorry let we have this thing okay this is d theta so let us make because this is a positive change so therefore we make a thing like this okay as I did earlier so this is this so now this is very clear so this angle is okay this angle is alpha this angle I made a alpha this is the wave and this angle is this angle alpha so therefore this will be this v1 this is vn1 so this this is vn1 and simply v so this will be v cos alpha then what will happen here this will change here so this will be going to this direction this is v plus dv well so this has also a this is vn2 this is vn1 this is vn2 this is again I am writing so this is what is that this is v plus dv now this angle this angle this angle is this is alpha minus d theta clear this angle because this is alpha and this is d theta so therefore this angle is alpha minus d theta which is equal to this angle the same thing now I have to be little fast that now if I write v plus dv equality of the tangential component or parallel component cos of alpha minus d theta is v cos alpha now we have to be little bit fast now let us do it v plus dv now cos alpha sorry cos alpha minus d theta is cos alpha cos d theta minus sin alpha sin d theta is equal to v cos alpha now make the left hand side now v cos alpha cos d theta now for small d theta cos d theta one so v cos alpha now if I multiply with this v sin alpha sin d theta is d theta is d theta v sin alpha d theta plus dv cos alpha because cos d theta is one minus dv sin alpha d theta so dv d theta product of two small term sin d theta is very small sin d theta means d theta dv this is neglected so neglecting the higher order term the product of these two we get what v sin alpha minus v sin alpha d theta plus dv cos alpha is equal to 0 this is a very interesting result we get from which we can write dv by v is equal to minus tan alpha into d theta minus tan alpha into d theta dv by v now tan alpha can be written as minus d theta root over m square minus 1 why because we know that alpha is sin inverse 1 by m for such small turning the small angle d theta so that sin alpha is simple trigonometry tan alpha is 1 minus root over m square minus 1 dv by v sin alpha by cos alpha that is tan alpha dv by v tan alpha d theta which one that will be this is minus this is minus this is plus this minus dv by this oh then that will be plus cos a cos b to plus oh sorry I am sorry you are very correct cos a cos b will vary this is the final result that I know yes cos a minus b is cos a cos b sorry plus sin a sin b the minus will come if it is sin a minus b or cos a plus b correct very correct so there will be minus d theta by root over I am happy is very good dv by v is equal to minus d theta by m square now we go to the well energy equation now we go to the energy equation I am in little we know that energy equation we wrote that 2 gamma that is h 1 c p t 1 that is written as gamma by gamma minus 1 into p by rho plus v square why 2 because v square by 2 I am just I am writing this thing it is v square by 2 gamma by gamma minus 1 is what c p into r c p is r gamma gamma minus 1 that means it is c p by r and p is equal to rho r t so therefore it comes from what it comes from c p into t but 2 c p t this is c p into t gamma r by gamma minus 1 r t is p by rho and v square by 2 so I am multiplying 2 on both the sides that means this is the static enthalpy plus the kinetic energy we discussed is earlier that is why I am writing straight because here I do not want to repeat it again now for an infinite small change the quantities are shown like this then rho plus d rho yes plus p now if you do it now you again you see that 2 gamma by gamma minus 1 into p by rho plus v square is equal to this side 2 by gamma minus 1 now you take p by rho common that means you take p from the denominator and rho from the numerator if you take p from the denominator this will be p plus d p to the p 1 plus d p by p to the power 1 that is 1 plus d p by p that means 1 plus d p by p so therefore 1 plus d p by p I am writing in one shot and rho if you take 1 plus d rho by rho to the power minus 1 and if you make the binomial theorem and second order terms you neglect then it becomes is equal to 1 minus d rho by rho only here this will be multiplied so that finally we can write 1 plus d p by p which is there already p common d p so denominator contribution is that 1 plus d rho by rho to the power minus 1 will be multiplied for which only one term will come 1 plus d p by p minus d rho by rho all right sorry this is in the first bracket I get plus v square plus 2 v d v I just again neglect the higher order term so v square v square will be cancelled so this thing will be cancelled 2 gamma by gamma minus 1 p by rho 1 will not be there so therefore 2 gamma by gamma minus 1 p by rho d p by p minus d rho by rho will become equals to 0 so ultimately will become equals to 2 v d v so ultimately we can make this one like this 2 gamma by I can write it p by rho into d p by p minus d rho by rho into d p by p minus d rho by rho d rho by rho is minus 2 v d v if there is any mistake you tell me this is ok now what we can do we use a formula that d p by d rho we know that d p by d rho is equal to for an isentropic process because already we have taken the for the process the entropy is constant and that becomes equal to a square so we will use that if you use this that and what you do this gamma p by rho already you use as s square gamma p by rho so therefore you can write 2 gamma gamma p by rho s square so just see that what I am doing this I am using and d p by p I am taking common then what is there so d p by p that is p by rho d rho by d p and if we use this equation this will be ultimately 1 by 1 minus gamma is equal to minus 2 v d v what I am doing gamma p by rho is s square d p by p I am taking common that means this will be d rho by d p into p by rho and d rho by d p into p by rho is 1 by 1 minus gamma so if you equate this 2 then you will find this then gamma minus 1 in gamma minus 1 will cancel so therefore from here we get straight away d m d p by p we get straight away that d p by p d p by p is what d p by p now gamma minus 1 and gamma minus 1 will change so therefore gamma only gamma that means 2 2 calcins minus v d v by gamma that gamma will not be there so gamma minus 1 so that gamma so only gamma v d v by no gamma will be there 1 minus 1 by gamma 2 a square rho there gamma will not come gamma p by rho is a square so this gamma will not come so therefore gamma v d v divided by a square that means d p by p is equal to minus gamma v square if I multiplied gamma by a square v gamma m square sorry this 1 d v by v gamma minus sorry this is gamma m square because I multiplied with v v square by a square is m square now we know the value of d v by v so d v by v we have already found out just I see d v by v is equal to minus d theta by root over m square minus 1 so therefore we get d p by p is equal to d v by v m square d v by v is again I see minus d theta root over m square so minus minus can sense so it will be gamma m square root over m square minus 1 into d theta so now I just I will just multiply gamma make it like this d v by v d v by v well so this is d v by v is this one and I get d p by p is this one from the energy equation now after that I will do for another same thing I will do for another same thing is the mach number now mach number you can define in terms of its square mach number square is v square by a square because a unnecessarily contains under root of that that is why we are now if we consider a is gamma a square is gamma p by rho then this will be gamma p because a square I am writing gamma p by rho that we know already we wrote so therefore if we make a differentiation logarithmic differentiation then 2 m d m by m is equal to 2 v 2 m no 2 d m by m this will be 2 v 2 d v by v 2 d v by v I am taking the logarithmic 2 d m by m ok 2 m that means 2 l n m d m by m is equal to 2 d v by v plus d rho by rho minus d p by v this will be 2 v 2 d v by v I am taking this is ok 2 d m by m is equal to 2 d v by v d v now you just substitute these values but before that you have to know d rho by rho d rho by rho is another one which you have to find out now before that we find out what is d rho by rho how you will calculate d rho by rho it is very simple we have considered a isentropic flow p rho to the power gamma square so take a logarithmic differentiation so d p by p is the plus or there we can take that side minus gamma d rho by rho minus d p by p plus gamma d rho by rho is 0 minus so d rho by rho is minus 1 by gamma d p by p and that it is alright d p by p is plus gamma d rho by rho is 0 ok p rho to the power gamma is constant for the isentropic flow for the isentropic flow p rho to the power gamma is equal to constant we are considering the flow where the entropy does not change that is the isentropic flow so therefore d p by p plus gamma d rho by rho is 0 minus gamma d rho by rho so minus 1 by gamma no no p p into rho to the power gamma is constant is the p rho to the power minus gamma exactly you are correct p rho to the power minus gamma p by rho to the power gamma is constant I am sorry so therefore it will be then plus plus p rho to the power minus gamma I am sorry p by rho to the power gamma sorry sorry p by rho to the power gamma is constant p by rho to the power gamma is constant. So, d p by p is equal to gamma d rho by rho. So, d p by d rho by rho is 1 by gamma d p by p. What is d p by p? d p by p 1 by gamma m square that means that gamma will not be there. So, therefore very good d rho by rho will be m square by root over m square minus 1 very good I am sorry p v to the power gamma is constant, but it is p by rho to the power gamma is constant I am extremely sorry this will be p into rho to the power minus gamma constant very correct p by rho to the power gamma constant. So, therefore there will be a minus sign m square root over m square. So, this is d rho by rho now if you substitute now d v by v d v by v d m by m now d v by v where we have got again I am telling that d v by v in term that minus d theta by root over m square minus 1. So, d p by p is gamma m square by root over m square minus 1 d theta and d rho by rho if you substitute all those things finally you can write d m by m becomes equals to if you do it you will see d m by 1 you can write like this one line minus 2 plus m square for d rho by all m square minus 1 will get common minus gamma m square because d theta by root over m square minus 1 is common to all. So, if you clear this thing you will get this is a very standard expression 1 plus gamma minus 1 by 2 m square this is a very standard expression minus d theta by this is minus d theta I am writing minus here m square minus 1. So, this will come like that finally you will get this now if you compare all these things again I am showing it d v by v is minus d theta by root over m square minus 1 d p by p is gamma m square by root over m square not rather I write it again that will be better for you. So, now I write everything d v by v is equal to minus d theta by root over m square minus 1 now d p by p I write which we deduced d p by p is gamma m square by root over m square minus 1 d theta d rho by rho is 1 by gamma. So, m square by root over m square minus 1 d theta good and d v by v d m by m sorry then d m by m is equal to 1 plus gamma minus 1 by 2 m square this into minus d theta. So, this is plus d theta this is plus d theta. So, minus d theta I write here which I write earlier root over m square minus 1 now you see the beauty is that we made a change like this this is the concept now where the way was like this which changes change the thing like this this is d theta now this d theta is positive in the this is considered to be positive in this direction this changes is positive this is because this makes a change in a way that it makes this velocity that there is a reduction in the tangential in the normal component of velocity or we find out that p 2 is more than p 1 that here we will see that if we consider this d theta as a positive one then we see that the positive d theta. That means in this direction a change will make the velocity d v less than 0 if here if d theta is greater than 0 then what happens d v less than 0 d theta greater than 0 d v d p d p greater than 0 d theta greater than 0 d p d theta greater than 0 d rho greater than 0 and d theta greater than 0 d m less than 0 this is typically the shock wave that means where we can positive change in the angle here the velocity decreases pressure increases rho increases and magnetic but the other way if d theta is less than 0 then d v will be greater than 0 and d p will be less than 0 d rho will be less than 0 and d m will be greater than 0 that means the flow will accelerate with a change in a reduction in pressure and density and with a flow will accelerate that means with a increase in velocity with the increase in Mach number also with a reduction in pressure and the reduction in density and this is known as expansion waves or expansion flow that I will explain now I will show you is there anything is okay now what happens I will just tell you now if you consider a change like this now any finite change a now you think a change that is a finite change in the angle that means a concave corner or a concave wedge type of thing concave a surface like this that is a concave change then this can be considered as sum of many infinite small changes and what happens a number of waves will be generated where the Mach will be changes and these waves will be ultimately converging and we will give you a finally converging wave that is the converging wave that is the shock wave that is the oblique shock oblique shock wave this nature will be like this but if we have a thing like this then what will happen this will start somewhere here that may be for example here not it may not be very sharp then what will happen this type of thing will give a diverging type of wave and this wave is known as these are known as these are the this is just for drawing I am showing this is the diverging wave this is known as expansion wave expansion wave this is expansion wave this will never convert this will never convert expansion wave and within this the flow takes place like this ultimately goes on diver like this and throughout this expansion wave region the flow is remains isentropic the entropy change will be very less and this flow is known as Prandtl Meyer flow Prandtl Meyer flow this flow is known as E me Prandtl Meyer flow E me why Prandtl Meyer flow this flow is known as Prandtl Meyer flow this is an now for example if this is also like this but if this is a sharp one and this is a sharp one this I have shown also then it will give a converged oblique shock but this will give a this type of fan type of thing so that is why this is known as this is known as expansion fan expansion fan so this is the thing that means if this is a sharp corner like this is a convex all these oblique shock waves march one another and gives rise to a single oblique shock wave across which there will be an increase in pressure increase in density and a decrease in velocity and Mach number and just the opposite will be there in the case of a concave surface like this when the flow takes place there will be a series of expansion waves and which will be diverging in nature and create a fan like thing and if we think of a sharp corner then they will make a fan like that this will be attached here and throughout and the same thing will happen I will explain afterwards in case of a flow fluid at some pressure P higher than P atmospheric pressure if this valve is opened or the diaphragm is raptured then the flow will commence then the flow will commence and the expansion will go like that and you will see that there will be a zone of expansion wave that will be created here so that I will explain afterwards so this is the picture by which an expansion wave is created now what we will do we will now calculate this d theta Mach number relationships what is that now what is most important is that d m by m d theta let me write that d m by m is 1 plus gamma minus 1 by 2 m square into minus d theta divided by root over m square minus 1 now for an expansion wave I now do not give this minus d theta sign rather rather first of all let me give this minus d theta sign then I will tell you now if we integrate this with respect to m that means that is m square root over m square minus 1 by m root over m square minus 1 by m into 1 plus gamma minus 1 by 2 m square d m if we integrate this this is a 1 plus gamma minus 1 m square whole to the power minus 1 yes whole to the power minus 1 d m whole to the power minus 1 d m rather I must write in a different fashion root integration root over m square minus 1 divided by your correct gamma minus 1 by 2 m square into d m by m I think this is ok now if I calculate from 1 theta 1 to theta 2 and we can calculate from m 1 we can integrate from m 1 to m 2 so we can get a relationship that what should for a given change in the angle what should be the change in the mach number so we can make an integration now better giving any limits because we do not know the limits we better integrate it simply by considering that it indefinite integral we do not take first of all this minus sign because it is the case where we only consider the concave surface so we take theta is equal to integration of this root over simple integration of this root over 1 plus gamma minus 1 by 2 m square d m by m now if you make this different indefinite integral you have to impose some constant some constant because the integration constant will be there you have to impose some condition or the constant instead of impose some condition to eliminate the constant of integration then we consider at m is equal to 1 theta is equal to 0 arbitrary that means we consider that when the flow is sonic that m is equal to 1 then theta is equal to 0 but this thing I will explain after what that when m is equal to 1 what could have been the value of theta that I will tell after what that means now we put a condition that at m is equal to 1 when the m is equal to 1 then theta is equal to 0 if we put a condition and integrate this then you will get a relationship that I will now write to you that integration I am not doing that can be done by making integration by parts and just I am giving you the result now theta is root over gamma plus 1 by gamma minus 1 tan inverse root over gamma minus 1 by gamma plus 1 m square minus 1 minus tan inverse root over m square minus 1 now this is the expression where theta is 0 arbitrarily when m is equal to 1 this I will be discussing after what when specially I will solve the problem this arbitrary nays will be understood that how it is taken care of let us first for the sake of mathematics consider some arbitrary condition that theta that is the turning is 0 when m is equal to 1 just mathematically I get now here another very interesting observation is that when m tends to infinity then what is the value of this turning this will not infinity because tan inverse infinity is pi by 2 tan inverse theta limit as theta tends to pi by 2 is infinity okay no not sorry sorry sorry sorry limit tan theta limit tan theta tan inverse infinity is equal to pi by 2 I am sorry tan pi by 2 is infinity so therefore tan inverse infinity is pi by 2 that means limiting value that is the limiting value I wanted to write that when theta tends to pi by 2 I wrote correctly limit tan inverse theta theta limit tan inverse theta theta tends to pi by 2 it is alright so this is tan inverse infinity is pi by 2 so therefore theta is this is the trigonometric relationship because tan pi by 2 is what that sin pi by 2 cos pi by 2 cos pi by 2 is sin pi by 2 1 cos pi by 2 it is okay means tan pi by 2 is infinity okay because sin pi by 2 1 cos pi by 2 is 0 so therefore it is infinity so therefore this will be root over gamma plus 1 divided by gamma minus 1 into pi by 2 minus pi by 2 that means is equal to pi by 2 into root over gamma plus 1 by gamma minus 1 and this value for gamma is equal to 1.4 for air this theta max or the limiting value you can take max because m tends to infinity is 130.5 degree okay now what happens is that 130.5 degree that means if the turning angle is more than that what happens I just show you if the turning angle is more than that let us consider a turning like this that this is the turning which is equal to theta max let this is theta max so a flow which is taking place this way is turned like this and they may go to this the pressure may fall to zero and if there is a turning which is like this then there may be a pressure zone which here is in this zone the pressure is vacuum pressure vacuum pressure so this zone is actually the vacuum pressure zone or the zero pressure zone that means if the turning angle is more than this maximum angle but we are not interested in this type of situation now so only this turning angle this concept is there now I will tell you we will solve some problem then only these things will be made more clear otherwise it will be difficult just let me do these things where is the problems yes let us make let us solve some problems now example one air flows at m1 is equal to 1.8 with a pressure of 90 kilopascals and a temperature of 15 degrees Celsius down a white channel the upper wall of this channel turns through an angle 5 degree away from the flow leading to the generation of an expansion wave leading to the generation of an expansion wave find the pressure Mach number and temperature behind this expansion wave find the pressure Mach number and temperature behind this expansion wave now what is this let us see the problem problem is like this air flows at Mach number 0.8 with a pressure of 90 kilopascals and this down a white channel the upper heel of the channel that means this the upper wall of this channel turns through an angle of 5 degree so therefore what happens there is a this type of flow takes place that means the flow is taking place flow changes like this the flow changes its direction like this okay now here the Mach number is 1.8 well this pin is not working well Mach number is 8 T1 is 15 degrees Celsius I think this pin what better P1 is 90 kilopascals okay then what happens that this one now this one M2 P2 and T2 now you see this one will make you clear about that arbitrary constant that from the flow leading to the generation of an expansion wave so first of all what you have to do we have to find out from the isentropic table first of all this flow is an isentropic flow with this Mach number 1.8 what is the value of P0 by P1 from isentropic table the value of P0 by P1 from the isentropic table is 5 point that is already done 5.746 and the value of T0 let this section is 1 this section is 2 so P01 T01 by T1 and that is 1.648 try to understand that now in isentropic table if you remember earlier I told you that isentropic table shows P0 by P1 T0 by T1 rho 0 by rho 1 A star by A all these things I told and there one theta value was there at the end for all isentropic table there is one theta value that theta value is the theta value for the Prandtl Meyer expansion or expansion wave which is also isentropic they follow these isentropic table values so therefore corresponding to this Mach number 1.8 I the P0 by P1 this one I the T0 by T1 I read a value of theta and that theta value is equals to the value 20 theta becomes equal to 20.73 degree now you understand what does it mean that means this Mach number is generated by turning an angle 20.73 degree Celsius where the Mach number was 1 you understand that means from a one Mach number this turning will be then the final turning angle theta this was the theta so theta for this thing theta 2 that the final theta is 20.73 degree plus 5 degree is 25.73 degree against this theta we will find out the P02 by P2 we will find out T0 2 by T2 from the isentropic table and what we will find out we will find out the value of M2 the Mach number everything we will find out if we know this thing so therefore we will find out the Mach number P02 now P02 by P2 now this value will be 25.73 so P02 by P2 in this case is 7.585 T02 by T2 is 1.784 and M2 is 1.98 well that means try to understand what is the thing that we are finding out from this Mach number that to get this Mach number through an expansion because this theta is kept against a Mach number from now your concept will be clear from this expression what is that expression I just wrote after integrating this thing immediately I wrote an expression where is that well where is that expression just now theta yes I wrote this expression this expression is valid provided theta is equal to 0 if this condition is satisfied that means when I compute this isentropic table at a given Mach number this theta means from Mach number 1 this Mach number is obtained with this angle of turn so therefore for this Mach number if we have a angle of turn 20.73 through an expansion wave we can generate this Mach number from Mach number 1 so therefore the if we have to find out the property here by turning with respect to an expansion way we have to add 5 degree with this value as if from Mach number 1 this turning is been made so then we relate this to the Mach number 1 condition through the calculation of theta for this Mach number 1 point that is the beauty of this type of problem so we find out this against this turning angle we get p02 by p2 t02 by t12 now p01 and p02 is same because this is the isentropic flow so therefore p02 by p2 is this p2 is p02 by this one and p02 is p01 which is p1 into this one so therefore we can calculate that p02 is p02 divided by 7 point 585 and p02 is p01 and that is is equal to 5.746 into p1 is 90 kilo Pascal divided by 7.585 and that becomes equals to what that becomes equals to well p2 I am giving the value 68.2 kilo Pascal and similar way if you calculate t2 that is we can calculate t2 in the similar way because we know t02 by t2 so t2 is t02 divided by 1.784 and that becomes equal to t02 is t01 and t01 we have calculated 1.648 into t1 that means 1.648 into t1 t1 is 288 divided by 1.784 and that gives you a value which is already calculated and that value is what t2 value is minus 7 degree Celsius so this way you can calculate okay we can solve another problem but I do not think time is left so we cannot solve any problem anymore time is over well another problem just I give you for your exercise a simple wing just you write this problem may be modeled as a 0.25 meter I will give you in wide flat plate set at an angle of 3 degree to an air flow at Mach number 0.25 okay and the pressure in the the pressure in this flow being 60 kilo Pascal assuming that the flow over the wing is two dimensional estimate the lift and drag force per meter span due to the wave formation on the wing so this is a very interesting problem so this problem actually you can solve the problem is like this there is a flat plate it is modeled by a flat plate which is 3 degree like this now when the flow takes place like this what happens there is a this is a shock and oblique shock and they are takes place this expansion wave so float goes by expansion wave in the upper part and the shock wave this is the shock wave shock wave by lower part so there will be an increase in pressure this is one this is two this is three you can find out the pressure here in this upper part from the expansion wave theory you can find out the pressure p2 this is p1 is given t1 is given m1 is given t1 is not given p1 is given however so you can find out the p2 and p3 when you can find out the p2 and p3 then you can make the vertical component as the lift force and the horizontal component as lift force this width is given per meter span in this direction I can tell you the well this result that the lift force will be lift force will be 6.23 kilo Newton per meter of span length and drag force will be very small 0.33 this is only because of the pressure that means you calculate the oblique shock concept as I did with this mach number finding out the shock wave angle find the turning angle is 3 degree and find out this p3 similarly sorry find out the p2 here sorry find out the p2 and find out the p3 by using the expansion wave theory we for this mach number mach number is what mach number is 2.5 you find out what is the value of theta with that add 3 degree and for that turning angle you find out the ratio of pressure make the stagnation to static stagnation pressure same as that so you can find out the static pressure here you can find out this pressure from the oblique shock angle for this m1 and this delta here the deviation is delta then you find out the shock wave angle beta and then modify the mach m1 sin beta find out the shock table normal shock table what has the pressure ratio you find out p2 the problem is simple but very interesting that you can find out the aerofoil wing it happens that one part there is an expansion wave another part there is a shock wave takes place so because of the shock wave this part the pressure is more this pressure will be more p2 will be more and here the p3 will be less so finally if this is the 3 degree so it will be p3 lift will be p3 minus p2 into that is a vertical direction lift and horizontal direction drag is cos of 3 degree and drag will be p3 minus p2 sin of 3 degree so that I give you the hint that this way you can solve this problem okay today after this so therefore with this I close the lecture on this introduction to compressible flow and I will close this course here so I think you just go through all the courses all these materials that have been taught and I wish you all the best to learn the subject okay thank you