 In the last class we have seen how to get the expression for non-dimensional thrust for a turbojet when we had the condition that the flow is optimally expanded through the nozzle optimally expanded through the nozzle is a condition that is most times not satisfied the other condition that is the flow is choked as it leaves the nozzle is the most probable one in case of turbojets okay turbojets as I said in the previous classes other than what was used on concord all other turbojet engines are primarily convergent nozzle they do not use a convergent divergent nozzle this is because the exit pressure after the turbine is very low so you do not have a scope for using a convergent divergent nozzle okay. So let us look at the case where the flow is choked why the exit of the convergent nozzle what does this mean the exit of the convergent nozzle the flow is choked what does it imply to our thrust equation if you remember our thrust equation was f is equal to m.a x right this is our thrust equation if the flow at the exit of the nozzle is choked then what it means is this part is not going to 0 in all the previous expressions that we had derived we had assumed that the exit of the nozzle the exit pressure is equal to the ambient pressure that is the condition for optimally expanded flow now we are saying that the flow is only choked and p7 need not be equal to p0. So here p7 is not equal to p0 so therefore it will produce a pressure thrust also typically pressure thrust is around 20 to 25% of the overall thrust so this is not a small portion so in this class let us look at how to do this analysis when we have this condition before we go there what are you mean by flow in the nozzle is choked when do we say that when do we say that the flow through the nozzle is choked p7 is greater than p1 p7 is equal to 1.2 p0 is equal to 1 is this condition choked then what do we mean when we say that the flow is choked flow through the nozzle is choked yes so if you have a nozzle a convergent nozzle at the exit of it if the Mach number is 1 then we say the flow is choked the reason for that is the reason why we say the flow is choked is what happens when you have Mach number 1 here whatever happens in this side okay you cannot transmit any disturbance or any information upstream because the flow has already reached the speed of sound disturbances propagate upstream at the speed of sound right so if the flow has already reached speed of sound at the exit they can no longer the flow cannot feel anything in the downstream conditions so if m is equal to 1 is reached then the flow becomes independent of downstream conditions and is only a function of upstream conditions right so which is why some people when they plot the mass flow rate with back pressure they say it reaches a maximum that is a little confusing you just need to say that it becomes independent of the back pressure okay typically if you take any nozzle when the flow is not choked the mass flow rate is determined by both the pressures only upon the nozzle being choked then it becomes independent of the downstream condition and becomes only a function of the upstream condition okay so now let us look at the case on our hand where in we have to do the analysis for the flow when the conversion nozzle is choked and this is our thrust equation and we cannot neglect the pressure thrust part okay so we can again use our we can say that f is still very much less than 1 because most times the main gas turbine engine operates with somewhere around 0.02 or 0.04 okay so clearly ratio 0.02 or 0.04 so f is very much less than 1 so this goes to 0 so we can rewrite our expression as f is equal to m.a x v0 x v7 x v0-1 plus I will take out p0 here so I will get p0 a7 now the this portion we can simplify as we have done earlier but in addition we have another component here okay so what is it that we do v0 is equal to a0 m0 right so if you put that and m0 is f is equal to m.a 0 m0 and v7 by v0 I can express it as a7 using a familiar stuff that is we know that we assume that ? and r do not change across the gas turbine engine so you get a7 by a0 as an expression in terms of temperatures so let us do that I can rewrite this as f by m.a a0 is equal to m0 x under root t7 by t0 x m7 by m0-1 plus p0 a7 divided by m.a a0 now what do we do further we both do is the condition that we know we know that m7 is equal to 1 okay at the exit of the nozzle the flow is choked that is Mach number is 1 okay so we can substitute that here and rewrite our expression as so m7 goes to 1 here I can bring in this m0 so I will get only under root t7 by t0- this will become m0 okay so this is the expression that we have now we need to find ratio of t7 by t0 and also p7 by p0 in the earlier case p7 by p0 we used to take it as equal to 1 and find using that find the ratio of Mach numbers now that is not equal to 1 so you need to find two ratios ratios of temperatures and pressures now coming to t7 by t0 we have done this exercise earlier now that just that the nozzle exit is choked do you think this ratio is going to change we need to do a fresh derivation of this ratio or you think it is going to change what is the conditions going to change let us see how it is going to change okay see t7 by tt7 into tt7 by tt6 into tt6 by tt5 tt5 by tt4 okay now what is this this is the only place probably it can change all the rest of the terms are similar to the previous conditions right only thing that will change is probably here right so here we know this is the ratio of we can express this is a ratio of static to stagnation so we can express it in terms of Mach number m7 square then this is one this is again one after burner first one is nozzle this is after burner this is tt flow through turbine then tb flow through combustor or burner then tc flow through compressor and this is flow through intake which is again one and this is ?0 okay now here we know that m7 must be equal to 1 so we can get an expression for t7 by t0 as tt okay now we also know what is tb if you look into your notes the previous class we derived this what is tb ?b by tc ?0 okay and if you plug that in you will get t7 by t0 is equal to tt instead of tb I have ?b by tc ?0 into tc ?0 okay so this cancels off and I am left with ?b okay so this is the expression for t7 by t0 for the condition when the nozzle is choked now what happens to the compressor turbine power balance does it change because of this or whatever we had derived remains the same see what is happening is something downstream of the turbine right so whatever we had derived is upstream of this so this should not change so what we had derived for the compressor turbine power balance does not change so we had derived that from compressor turbine power balance we had derived that tt must be equal to 1- ?0 by ?b into tc-1 this remains the same okay now we have to find an expression for tt here so let us do that I can rewrite this as tt is equal to ?b- using this expression into this equation I will get t7 by t0 is equal to 2 into ?b so this and this cancels off I am finally left with 2 into ?b- ?c ?0 okay so we have been able to get the temperature ratio here that we were looking for right now what are the other things that we need to derive here we have got this part temperature ratio done we still have to find the pressure ratio and we have to derive an expression for this quantity here okay so let us do that now again cascading pressures we are looking for an expression for p7 by p0 we can write it as p7 by pt7 into pt7 by pt6 pt6 by pt5 now when we do this in the previous case we had assumed this I mean we had assumed the flow to be optimally expanded so this became one okay now it is a not equal to 1 so we will get the first term is ratio of static to stagnation conditions so I can express it in terms of Mach numbers okay the next one indicates flow through nozzle we have assumed all efficiencies we are doing this analysis assuming all efficiencies to be one so this is one again the next is flow through the after burner we have not switched it on so this is again one and what is pt5 by pt4 this is flow through turbine so this is this is pressure ratio so this is pt into pt4 by pt3 is flow through combustor combustor the pressure is the same because we are assuming an ideal cycle here so this is one and this is flow through compressor so this is pC and this ratio is for flow through diffuser or intake this is one because we are assuming efficiencies to be unity and the last term is ?0 to the power of ? by ?-1 okay so I can write p7 by p0 is equal to this is one so I will get 2 by ? plus 1 into ? t to the power of ? by ?-1 into ?c to the power of ? by and we do know that ?t and ?c are related through compressor turbine power balance and if we were to take that into account we can show that the ratio p7 by p0 would be taking into account compressor turbine power balance we saw that the compressor turbine power balance was unaffected and we could get this expression for ? t now using this in that expression there we can get okay I have only substituted for ? t there ?c ?0 divided by ? so this is the expression that we have now we have been able to get expression for two quantities the last quantity that we need an expression for is we need an expression for okay so how do we go about doing this how do we get this expression we just like to get an expression for F you said we said compressor turbine power balance and then to get an expression for ? t in terms of ?c sorry in the last case energy balance across the combustor what was what we used to get an expression for F to get an expression for ? t we said compressor turbine power balance and we got the expression similarly is there something that we can do here to get this expression yeah what we need to do is we need to look at mass flow rate through the nozzle we will see how we can use that to get this expression mass flow rate through the nozzle what is the expression that we know we know that m.a x 1 plus F that is mass of air plus mass of fuel burnt this must be equal to ?7 v7 x a7 right now again we can make this approximation that F is very much less than 1 so therefore this goes to 0 so I get m.a is equal to ?7 v7 a7 okay now what is ?7 in terms of pressure m.a is equal to using the equation of state I can write p7 by R T7 for ?7 x v7 I can rewrite it as what mark number x a7 right so x a7 so I get what we are looking for is p0 x a7 by m.a a0 okay so what do we need to do here this we know is this quantity is what is this because this is flow is so m7 is equal to 1 so I am left with what is a7 a7 is nothing but ?R T7 okay so if I substitute for a7 here and rewrite m.a I will get p7 under root ? divided by this would not be there R T7 a7 okay now what do I need to do I need to multiply by if I multiply by a0 on both sides I get one part that is m.a a0 okay so let us do that I get m.a a0 is equal to again a0 is nothing but right so I will use that on the right hand side so I get p7 x ?R T0 okay so R and R cancels off here so I am left with p7 a7 x ? right now what do we need to do again so if I I want m.a by a0 by p7 p0 by a7 right so I can do this I can write this as m.a a0 divided by p0 a7 must be equal to p7 by p0 x ? x under root T7 T0 by T7 okay so we have been able to reduce m.a a0 x 2 known quantities right we already know the expression for p0 by p7 by p0 and we also know expression for T7 by T0 so we have been able to reduce it to this form now let us go back and substitute it and see what we can get if you remember our expression for our expression for f by m.a a0 we had got was under root T7 by T0 – m0 this part was taken care of and p0 a7 by m.a a0 x what was the p7 by p0 – 1 so we have got expression for this as well as this so when we substitute we get this part remains as is T7 by p0 – m0 plus p7 by p7 a p0 a7 by m.a a0 is 1 by ? x okay now we can simplify here and rewrite this expression as okay now we know expression for T0 T7 by T0 we also know expression for p7 by p0 so we will substitute that and see what is the final form that we can get it is a really big expression okay this is the final expression that we get now if we were to substitute here what we did earlier that is does this produce static thrust or not so let us do that for static thrust m0 is equal to 0 and ?0 is equal to 1 so if you substitute that we will get this is the expression that one can get and further simplification is possible on this I will leave it as an exercise to you okay you can take out ? plus 1 and further simplify it this part also you can take out okay so we have been able to derive this expression for non-dimensional thrust now this is a fairly complicated expression compared to that we derived for optimally expanded flow through the nozzle and much more complicated than the expression for ramjet okay now the next part that we need to address is what ISP okay so ISP part what we will do is ISP we know the expression for ISP by a0 as right and we have spent a considerable amount of effort trying to derive this expression we already got this expression now what we need is 1 by f now in the previous case when the flow was optimally expanded through the nozzle we had derived this expression for 1 by f now what we need to look at is what is the difference between the expression that we derived there and will there be a difference here now again if you look at the flow process what is happening in the nozzle is much more downstream then what is happening through the combustor so the expression for 1 by f that we had derived earlier is the same so I can use 1 by f that we had derived earlier as Q by CP t0 into this expression remains the same so I know that 1 by f is known f by m.a is known so if you substitute this expression there sorry this expression you will get ISP okay fine now having derived these expressions let us look at what is it that we can understand from these expressions remember we did the same exercise we found out what is the Mach number at which the non-dimensional thrust would be a maxima depending on what values of ? B and other things for the ramjet let us do the similar exercise for turbojet okay now in a turbojet we will again look at the condition where the flow through the nozzle is optimally expanded simply because it is something that is easier to do in the classroom doing something on this is a little more complicated so I will look at what we can find out using the expression that we derived for optimally expanded flow okay so if you look at the expression for optimally expanded flow through nozzle f by m.a a0 was m0 into ? B- ? C ? 0- ? 0- okay fine this was the expression that we had if you see here as ? B increases what should happen see as ? B increases what should happen to f by m.a a0 this also increases because you can pull out the ? B terms and you see that you get 1-1 by ? C ? 0 so as ? B increases it is fairly obvious from this expression that as a consequence f by m.a a0 also increases right which means what that if you have a large enough ? B then the size of the engine is going to be smaller and smaller for the same thrust okay fine or if you keep the dimension same right then your thrust is going to increase if you increase ? B okay this is preferred because you will have lesser drag on the engine okay then this is fairly clear then what what we need to look at is what happens to what is there an optimal value for the compression ratio if I fix ? B and ? 0 is there an optimal value for compression ratio okay if I am flying at a cruise Mach number let us say if I am flying at a cruise Mach number then my ? 0 gets fixed remember when in this previous class one of the previous class we had derived an expression wherein we looked at what is the range and how the overall efficiency gets affected by it right in that we had said most of the flight takes place in the cruise range right so if you substitute ? 0 for cruise and you know the ? B value then you can find an optimal expression for ? C let us do that in the next few minutes so we are looking for optimal value of ? C given ? B and ? 0 so how do we go about it again take a derivative with respect to ? C so you get – ? B ? C ? 0 this goes to the power of – 1 2 the derivative of the terms containing ? C that is the first term is 0 then you have a – ? 0 then again you have ? 0 plus because this becomes minus so you have ? B by ? C square ? 0 okay fine derivative with 1 goes to 0 so you get this expression now this goes to the denominator because you have multiplying the I mean you have a power – 1 so what you need to look at is only this part of the expression because for this to be for f by m dot a not to be maxima this should go to 0 this can only go to 0 if the numerator goes to 0 okay so this is the numerator in this so for maxima ? 0 must be equal to 0 which means that ? C must be equal to okay fine sorry yeah so you get this condition so we will stop here and continue in the next class thank you.