 Next parameter, mainly for power plants, this is specific fuel consumption. This is for power plants and tells us that if this is our power plant, it produces W dot net. Since it is taking fuel, you will take an m dot air and you will also provide sufficient amount of fuel. I should say sufficient amount of fuel to produce Q dot 1 and appropriate amount of air to provide complete and proper combustion. Then specific fuel consumption is the amount of fuel needed for unit power output. Units of this will typically be Kg per kilo watt or mega watt or you can have it is Kg per kilo joule or Kg per kilo watt hour. Then depending on the plant, there are some other measures. For example, one question was asked, let me go to the next thing. I am not consuming any papers here. What was that five? This is appropriate only for certain kind of machines, particularly machines of the machines which are single equipment and which are the reciprocating type, mean effective pressure. The way the specific fuel consumption is for power plants, there is no fuel consumed in the refrigeration plant. Mean effective pressure can be used for power plants, can be used for refrigeration plants, but it is for machines which use a cylinder piston scheme. Now let us first look at the power plant, a power producing device that is an engine. Now what happens? Here we get into some geometric details. Let us say that you have a machine which works in cycle by moving a piston back and forth and this displacement or movement of the piston is known as a stroke and this volume which is displaced by the piston is known as the stroke volume. Let us say it is an engine and let us have a simple model in which we say that during one stroke a uniform pressure Pm acts on this piston. Then what will be the work done per stroke? Work done per stroke will be Pm multiplied by Vs where Vs is the stroke volume. So this is work done or I should say net work done per stroke and we say that if there are n dot cycle is the cyclic rate that is number of such strokes in which work done per unit time then the power output blue dot net is Pm into Vs into number of cycles per unit time or the cyclic rate. This equation defines Pm the mean effective pressure. I can work it around and write it as Pm equals W dot net divided by stroke volume into n dot. Now see this represents the speed of the engine. The stroke volume represents the size of the engine and this represents the output of the engine. So although Pm has units of pressure notice that it represents the power output per unit stroke volume that for a given size and per unit speed and naturally an engine naturally remember that the speed of the engine will usually be restricted because of some mechanical considerations lubrication dynamics of the cylinder, piston, slider, crank, crankshaft all those mechanical devices they will dictate also the rate at which you can properly open and close valves. So n dot cycle is within a narrow range whereas size is something which we can design and if you want a higher output we can have a higher size. But if a machine has a higher Pm that means for a given Vs it can produce more power output. So here Pm is the higher the better the Pm represents the compactness of a machine. A machine which produces the same output at the same cyclic speed if it has a higher Pm it will have a lower Vs that means it will be a more compact machine. And here I am talking of output if you talk of consumption then the same thing is true of refrigeration plants. Now here I have to use some technical terms whenever you implement a cycle unfortunately I will be talking about power cycle but again as I am saying again and again you can use this for refrigeration cycles also. When you work an engine there is a thermodynamic cycle being executed and using the laws of thermodynamics we get a W dot net. This we can say is thermodynamic power as dictated by the laws of thermodynamics. But this is the power produced by our system which is the working fluid. The working fluid delivers this much net power to the mechanism of the engine. But the mechanism has to overcome friction may be the mechanism has to take care of some devices to make the engine run safely, make the engine run smoothly and take care of some lubrication and some control purposes. So there will be some power consumption let me say by local I mean pertaining to the engine itself and what we will get is the net power to the outside word I will say this is the delivered power. So whenever we talk of an engine when we buy an engine we are interested in the net or delivered power. This thermodynamic power quite often it said to be the power developed by the engine this is the power delivered by the engine and particularly when it comes to the power plant and particularly when it comes to IC engine this power is known as break power and this power is known as indicated power why indicated that we will talk about later and why break because traditionally a break type dynamometer was used to measure this and indicated power because traditionally a piece of equipment known as an indicator is used to determine this power. Now because break power and indicated power will generally be different because there is some local power consumption this may be conscious consumption for running some auxiliaries or control circuits or some safety circuits but it could also be power which needs to be consumed to overcome friction lubrication and such stuff. And now you will notice that because we have a separate BP and a separate IP whenever this W dot net turns up in your mean effective pressure definition or the specific fuel consumption definition or the specific output definition or even the efficiency definition. We have to be careful in classifying it for example efficiency if it is based on break power say W dot net thermodynamics W dot net delivered this will be known as break efficiency because this is based on sorry this will be known as indicated efficiency because it is based on IP whereas the delivered power is BP this will be known as the break efficiency. Similarly, when it is specific fuel consumption this will be known as indicated specific fuel consumption this will be known as the break specific fuel consumption indicating it is specific fuel consumption based on the break power. Similarly, when it comes to the mean effective pressure technically quite often known as MEP this will be indicated mean effective pressure and this will be break mean effective and so on. Now we come to again the implementation we say that look the processes which we could implement where easily easy implementation was possible for these kinds of processes the constant volume process in a closed system adiabatic process expansion compression in a closed or open system or isobaric heating cooling in an open system and we looked at the parameters like efficiency work ratio mean effective pressure and so on. So, let us look at the let us evaluate our traditional Carnot cycle using a gas as a working fluid and we have sketched the T s diagram let me sketch the PV diagram unfortunately no Carnot diagram on the PV plane is ever drawn to scale and that is why even if you try to draw it crudely you will be able to you will be showing it as a crooked non-linear or curvy linear parallelogram working between two temperature levels T 1 and T 2 and two isentropic at s 1 and s 2. The problem that arises is although efficiency of this is maximum for a given T 1 and T 2 when it comes to other parameters you will find and I recommend that you sketch the Carnot cycle using air as the working fluid and sketch it to scale take T 2 to be say 300 Kelvin near our room temperature. Take T 1 at say 900 Kelvin not very high but good enough we expect an efficiency of 67 percent 66.66 percent with this. But now when you actually sketch it recommendation sketch this to scale now remember that assume that your engine is of a certain size and of a certain volume and so T 2 and V 2 gives you P 2 this is P minimum. Let us say the P minimum is 1 atmosphere and that will give you the amount of mass in the Carnot engine. Now we have decided that T 1 is 900 K we decide the maximum pressure that we can have because the cylinder will have to be made strong enough to handle that pressure. Let us say the higher pressure is 50 bar then you know the temperature you know the pressure you know the mass you can determine what this volume is. So you can plot the all the things you will notice that the Carnot cycle is a curvilinear parallelogram but it does not look anywhere like this. It looks so thin that I cannot properly sketch it on this diagram I have plotted it and if I find that plot I will put it upon Moodle and you will be surprised to know that the work ratio for a Carnot cycle is almost equal to 1 it is of the order of 0.95, 96, 98 of that order. That means a significant part of the work produced during the expansion process is governed is consumed back in the compressor process. It is a very thin parallelogram and because of that we can show that the PM is very very low and because the PM is very low you will find that the Carnot cycle with a gas as a working fluid is never implemented in practice. Carnot cycle as defined in textbooks is never implemented in practice but then what is implemented? Modifications of Carnot cycle are implemented. Now remember the attraction of a Carnot cycle is that it is a 2T cycle heat absorption at one temperature, heat rejection at one temperature but the Carnot cycle if you say is a TSTS cycle one process in which temperature is maintained constant second process in which entropy is constant, third process in which temperature is constant again at some other value, fourth process in which entropy is constant at some other value. For it to be a 2T engine it is necessary for us not to change this isothermal heat absorption and heat rejection processes but we can change this isentropic processes and when you change these isentropic processes two things happen. First thing you are modifying the cycle but remember that if you want the modified cycle to have a Carnot efficiency of 1 minus T2 by T1 two things are necessary. First it should be a 2T cycle that we are not changing because we have one constant temperature process and another constant temperature process but when you replace these two constant entropy processes by say constant volume or constant pressure processes you will notice that a reversible constant volume process and a reversible constant temperature process will not be an adiabatic process that means the working fluid will need to be supplied heat or from the working fluid we will need to absorb heat extract heat as we execute those processes. So it does not remain a 2T cycle if you directly do it but we overcome that hindrance by using an internal heat exchanger. I will show you the scheme. If you replace this S by two constant pressure processes you end up with a cycle and of course with an internal heat exchanger we end up with a cycle which is known as the sterling cycle and if you replace these by its two constant volume processes and internal heat exchange is managed by heat exchanger you end up with what is known as the Ericsson cycle. Let me go to the next page and sketch a big TS diagram actually on this smooth tablet it is difficult for me to sketch straight lines. Let us say we are sketching horizontal lines seem okay vertical lines are bad. Let us say this is T1 this is T2 the S is long just because I want to show the three cycles next to each other. The Carnot cycle will be like this heat absorption at T1 heat rejection at T2 and these are isentropic so these are adiabatic process. Now when you take the Ericsson cycle you have heat absorption at T1 no doubt you have heat rejection at Q2 no doubt but you need to absorb heat during this constant volume constant pressure process because it is reversible constant pressure process. Similarly you need to reject heat during this constant pressure process because it is a reversible isobaric process and for this you will have to absorb heat at temperatures other than T1. So what is done is we say that look whenever the working fluid reduces its temperature below T1 that heat is being extracted and it is stored somewhere and is made available to the same working fluid when it starts heating up. So we set up an internal heat exchanger so called a regenerator to do this and that way we change our Carnot cycle to the Ericsson cycle. Similarly if instead of two constant pressure processes you take two constant volume processes the constant volume process lines would be slightly steeper although they do not seem to be very steep here than the constant pressure processes on the TAs diagram and again you have Q1 Q2 and an internal heat exchanger known as a regenerator to see to it that even the sterling cycle remains a 2T heat engine. So with this heat exchanger with the ideal internal heat exchanger each one of this is a 2T heat engine so the efficiency of each one of them will be 1 minus T2 by T1. Now with this modification we still retain the 2T characteristic of the modified Carnot cycle but we need an ideal heat exchanger internal to it than ideal regenerator. We will maintain the Carnot efficiency but now the question is do we really improve the work ratio? It turns out that the work ratio is improved but not very tremendously and the problem of implementing with a gas an isothermal process where you have to do work transfer as well as heat transfer together still remains and because of this of these three cycles you will find that if at all any cycle is implemented and that is the sterling cycle. The sterling cycle is not really implemented as an engine although attempts are being made and there are some working engines working on the sterling cycle but for refrigerators particularly for cryogenic refrigerators the sterling cycle is considered a very good option. So the basic sterling cycle or its modifications are the devices by which we do all our very low temperature refrigeration known as cryogenic refrigeration. But remember that whether you take the Carnot cycle whether you take the Eriksen cycle or whether you take the sterling cycle we end up with the situation that our heat absorption has to be under isothermal conditions heat rejection also has to be under isothermal conditions and hence these become difficult to implement it with an ideal gas or even any gas as a working fluid. And I will first take a few questions and then go over to our cycles discussions. I have hands raised from VNIT, Nagpur then KJ Somaeya Mumbai. VNIT, Nagpur do you have any questions? Over to you. Hello, why the whole lake do not get freeze till up close depth? Why the only upper layer of lake gets frozen? Okay I am taking over I understood the question. Your question was why is it that the lake, lake gets frozen only on the surface and not all over. The reason there are many reasons for this each contributing to some extent. First thing remember that the inside the earth it is always warm. If you go down a few meters the temperature is the mean annual average temperature which depending on the location. For example, you take a place like Delhi minimum ambient temperature of the order of may be 5 degree C maximum ambient temperature of the order of 45 degree C. So, it is reasonable to assume that if you go down something like 45 meters the mean temperature in the earth will be of the order of 20 degree C. Consequently the lake any lake is usually exposed to reasonable warmth at its depth. And when it freezes it freezes because the ambient air temperature goes down and that is it loses heat from the surface. And once a thin layer of ice forms remember ice and water may have almost the same order of magnitude of thermal conductivity. But through ice the heat transfer is only by conduction whereas if it were fluid by natural convection or may be by surface wave some sort of if not forced induced convection it will be well mixed. All these things combined to give you a layer of ice below which water will remain. And that is and but the layer of ice can be pretty thick in fact sometimes vehicles can be driven over it. So, may be that is the reason lakes get frozen only on the surface and not at the depths over to you. Over and out. K. J. Somaia over to you. Yes sir. Good afternoon sir. My question is the dual cycle is a modified form of this diesel cycle whether it is used for operational cycle for the diesel engine over to you sir. We will be discussing this during our lecture in the afternoon over to you. Okay sir thank you sir. Ph. G. Coimbatore over to you. Good afternoon sir. This is a question regarding the figure you drew in the morning's class. You stated the relation between engine refrigerator and heat pump by drawing the temperature limits sir. You had a three temperature limits one is environment temperature another one is lower than the environment another one higher than the environment. And considering the case of engine it is rejecting heat to the T environment but you showed that it is slightly higher than T environment. Can you please explain thank you over. Let me explain I think this is the figure that you are talking about I showed a this is the temperature axis and I sketch this as the environment temperature and remember that an engine rejects heat to the ambient a refrigerator also rejects heat to the ambient but a heat pump absorbs heat from the ambient. And remember that a heat transfer will not take place unless there is a small temperature difference in the ambient direction. So if you have an engine it takes in heat from the T source when it rejects there has to be a small delta T. And hence the heat rejection temperature to a small extent will have to be higher than that of the ambient. Similarly when you have a refrigerator taking in heat from some cold space at T naught it rejects it to the ambient. So I should really show it like this a delta T above the ambient. Whereas for a heat pump it rejects heat to the T let us say hot space absorbs work. Since it absorbs heat from the ambient it will have to be at a temperature slightly lower than the ambient. These delta T's which I have shown here should be as small as possible because that is better from a thermodynamic point of view over to you. Thank you sir over. That brings us to the end of this quick interaction after lunch. Let me now go over to my tablet and continue from where I left off. In the morning we looked at various parameters, classification, etc. And then we said that the modification of the Carnot cycle we looked at two modifications the Ericsson type modification and the Sterling type modification. The question was raised as to whether it is right that Ericsson is two isothermals and two constant pressures and Sterling is two isothermals and two isometric or isochoric constant volume lines that is right. One thing you should remember is the following. Carnot is T S T S, heat addition in this T, heat rejection in the other T. If you take Ericsson it is T P T P again heat addition in the two isothermal processes and these two reversible processes whereas to be a heat exchange between these two processes. Similarly, Sterling it is T V T P again heat absorption and heat rejection in the two isothermal processes and these two reversible isometric or constant volume processes have to exchange heat internally with each other using a heat exchanger. That is necessary because ideally all processes in these cycles have to be reversible. Then we notice that the real problem with these cycles is the Q 1 and Q 2 at constant temperature. That is the issue. That is very difficult to implement in practice using a gas whereas heating at constant pressure is easy to do. So perhaps for a gas the easiest cycle is the cycle in which Q 1 and Q 2 would be constant pressure processes because these are easy to implement and the remaining two processes would be isentropic processes. We know that isentropic processes can be implemented reasonably easily in either an open system or a closed system. That brings us to the next cycle and also remember that constant pressure process is easy to implement in an open system whereas constant volume heating and constant volume cooling may be implemented in a closed process. These modifications that heat addition and heat rejection could be at constant volume and constant pressure processes whereas the other two processes could be isentropic. Give us rise to two or more cycles. The most important of them are Brayton, Otto and Dez. The fourth one is sometimes called an Atkinson cycle but it is not really used in practice. It is a good approximation for some type of cycle. So that brings us to the Brayton cycle. The Brayton cycle is a PSPS cycle. So the heat addition would be here, heat rejection would be in the next isobaric process and since the isobaric processes are reasonably easily established under open system conditions, the Brayton system cycle is almost always implemented as a multi equipment cycle and this is the basis for all gas turbines and jet engines of all kinds. The T-S diagram of the cycle looks like this. It works between a higher pressure and the lower pressure limit. Wherever you have, it is always a lower temperature limit and a higher temperature limit. T low, T high and again a P low and a P high. You start with the lower temperature and lower pressure gas usually at ambient conditions and compress it ideally adiabatic reversible, so isentropic to the higher pressure you go from 0.1 to 0.2. Then from 2 to 3 at the higher pressure you have the process isobaric heat addition. So you can say 1 to 2 is this process, 2 to 3 is this process at constant pressure. Then expand it isentropically, reversibly isentropically, so adiabatically you come to state 4 that would be this process and cool it from 4 to 1 that completing the fourth process. The actual implementation needs to be shown because it is a multi equipment cycle needs to be shown in a block diagram. Compressor, heat absorber, cooler 1, 2, 3, 4. Show directions of flow, show compressor, heater, cooler, turbine. You will have a W dot turbine, you will have power absorbed by the compressor, W dot C, Q dot 1, heat absorbed by the heater, Q dot 2, heat rejected by the cooler. And at this stage before talking about the open brick cycle, make them understand that this is not a constant temperature cycle. So although it works between a higher temperature T H and a lower temperature T L, it is not really a 2 T heat engine because during heat absorption the temperature rises from 2 to 3 along this process and during heat rejection it lowers itself from or it gets lowered from 4 to 1. Consequently, we do not expect the Carnot efficiency to be obtained here, but we can do a standard analysis of the Baton cycle. In standard analysis what we do, we assume the working fluid to be an ideal gas with constant C p, C v. All processes are ideal. As prescribed constant volume, constant entropy etcetera, quasi-static and here we do not have to say, but internal combustion if any is replaced by let me say external heating and if necessary we assume that it is a closed cycle. This is not a strong assumption, we do not really need to make it. And the standard analysis all of us are familiar with is found in any textbook and show that the standard analysis of the, call this the standard efficiency of a Brayton cycle will turn out to be 1 divided by 1 over R p raise to gamma minus 1 by gamma and bring to their attention two properties 1 gamma is the ratio sorry not 1.4. It is the ratio of C p by C v that is the characteristic of the fluid and R p is the pressure ratio or the compression ratio of the Brayton cycle and that is p higher divided by 1 over R p lower. And at this stage you can give exercises on determining the specific output and all that. It is necessary at this stage to show them that the actual Brayton cycle quite often known as the gas turbine cycle is essentially the same cycle, but the block diagram differs. We have the compressor, but the compressor takes in ambient air at 1 instead of a heater you have a combustion chamber in which certain amount of fuel is also injected, it burns and up to here you have compressed air. Now, you have a gas that is products of combustion and hence the turbine which follows is known as the gas turbine and now this gas is exhausted at 4 this is 3 this is 2. Again you have a w dot t coming out w dot c going in and w dot net will be w dot t minus w dot. So, this is the first modification that we have studied of the Carnot cycle and it is a p s p s type of modification. So, first show them the T s diagram then emphasize that it has to be a multi equipment cycle. So, show the block diagram then do a standard analysis derive the efficiency and then show them the actual block diagram of a gas turbine cycle and say that here we have air, here we have air, but now here what you have is gas that is a gas turbine. Exhaust of combustion chamber and this is again the same gas which gets exhausted and because this gas has a different composition than the fresh air which is taken in, this is an open cycle. Now, if we see the Brayton cycle as a p s p s cycle another possibility is to consider a modification of the cycle which would be V s V s cycle. This V s V s cycle is traditionally known as the Otto cycle and the basic thermodynamics of the cycle will look like this. First thing you should remember is that heat is absorbed and rejected in the constant volume process. So, the Otto cycle will essentially be a cycle in a single equipment because a closed system is needed for constant volume processes and it is implemented in a cylinder piston arrangement and when you have a cylinder piston arrangement and expansion and constant volume processes it is possible to show this properly on a p v diagram. We will show it both on a T s diagram as well as the p v diagram. On the T s diagram we will now have two constant volume processes, a lower volume and a higher volume and we will have a isentropic compression, we will have constant volume heat absorption, we will have an isentropic expansion and a constant volume heat rejection 1, 2, 3, 4. So, the highest temperature point is at T 3, the lowest temperature point is T 1. No T 1, T 2, T 3, T 4, the lowest point is T 1. This is the lower volume, this is the higher volume. On the p v diagram it will look like this. There is a volume at which heat is added, there is a volume at which heat is rejected. So, this is V h, this is V l this is pressure and let us say we start at 1 and we make an isentropic compression 1 to 2, no heat is added. Then at during a constant volume process 2 to 3 heat is added and then 3 to 4 during another isentropic process expansion takes place and 4 to 1 is the constant volume heat rejection. So, you can show here this is Q 1, this is Q 2 and out here also this will be Q 1 and this will be Q 2. Thermodynamically, we do not have to do anything more for the basic explanation of the Otto cycle and you can show here this is the constant volume heat addition. So, this will be 2 to 3 expansion 3 to 4 heat rejection 4 to 1 and compression again 1 to 2. That way the cycle is completed and again you can do a standard analysis of the Otto cycle and show it to them that the standard efficiency of an Otto cycle as an expression very similar to that of a Brayton cycle exponent is different. And again bring it to their attention that one property of the fluid gamma is included in the right hand side. The second property is R v the volumetric ratio volumetric compression ratio which is defined as v higher divided by v lower. Now, at this stage students will ask and I have already been asked how this constant volume heat rejection takes place quickly and how this constant volume heat absorption that too quickly is created or implemented. So, at this stage it is necessary for us to explain to them the implementation of the Otto cycle. Now because we have a cylinder piston arrangement and a mechanism of the cylinder the piston the connecting rod sometimes a piston rod then the connecting rod then what they call the Gajan ping the crank the crankshaft and all that the problem becomes remains not only that of thermodynamics, but also that of mechanics. Add to that we need to have valves for that open cycle for taking in the charge and throwing out the products and the description becomes really complicated. You will have to bring in the idea of a 4 stroke machine at this time there is no need to confuse them with 2 stroke and 4 stroke. Tell them that there is a 2 stroke implementation, but explain to them at the basic level essentially of the 4 stroke. So, here you talk about all the details of mechanism you talk about the top dead center you talk about the bottom dead center you talk about stroke length you talk about the bore diameter and then you talk about clearance volume V C you talk about stroke volume V S and so on. And you talk about valves and how they are managed open closing that you can talk about the cam there is enough detail that one can worry about. In a thermodynamics course it is not worth spending time may be if you have a power point or an audio visual or a movie which you can show or a cut out model if you have in the lab that explains a lot. It is not worth spending in a thermodynamic course you have to spend about hour 2 hours in just explaining these technical details and since there is no basic principle involved only implementation is involved this is best left to the students themselves. But what we should emphasize that is that the real implementation of auto cycle as in a petrol engine you tell them the following is that it is an open cycle it is an internal combustion cycle. In fact that you this we can do right in the beginning when we come to Brayton cycle the standard Brayton cycle here it is shown as an external combustion cycle whereas the real implementation of the Brayton cycle is an internal combustion cycle. Similarly, the standard scheme of the auto cycle is the standard analysis means external combustion and closed cycle whereas the real implementation of the auto cycle is internal combustion cycle. Then you say that the working fluid to begin with is a mixture of air come petrol vapor and in early days this air and petrol vapor mixture was created using a mixing device called carburetor. Nowadays the petrol vapor is sprayed or injected into the air stream before it enters the cylinder. Then you say that because of this it is internal combustion so it begins as a mixture of air and petrol vapor the process of combustion is triggered by a spark and then you say after that the working fluid is our exhaust gas which is nothing but product of combustion. And then at this stage it is necessary to tell them that we are actually not the heating process. It is not really implemented it is a constant volume remember that the heating process is by combustion triggered by the spark and because it is a premixed and already preheated because we have already compressed it. If it is a premixed air fuel mixture everything is already in the vapor form so the combustion is very fast. So fast that the movement of piston which is near TDC at that time can be neglected because it is very fast the delta T for combustion is so small that the movement of piston can be neglected. Hence the process can be approximated as a constant volume mind you it is not executed as a constant volume process it is approximated as a constant volume process. And if you want you can show them that the real process if say this is the ambient pressure and you start with the clearance volume this is the stroke volume V s plus V c then the actual process of combustion does not really take place like this it takes place something like this. There is a small movement of the piston but for a quick calculation and modeling it can be neglected. But if you use an indicator you can really trace this curvature of the PV diagram. Similarly the second question arises is how is the heat rejected at constant volume and there the explanation is heat is not rejected at constant volume. As the piston approaches this is the bottom dead center this is the top dead center. When the piston approaches the exhaust sorry the bottom dead center somewhere near here the exhaust valve opens and when the exhaust valve opens the high pressures in high pressure gas in the cylinder is suddenly exposed to the ambient of course through the exhaust valve the exhaust duct the silencer and may be any other piece of equipment you have. But the fact remains that the high pressure gas not very high pressure but reasonably high pressure gas suddenly find itself into a in a situation of free expansion and left to itself this will create a gunshot like sound what is known as a report a very sharp very uncomfortable sound and that is why all the petrol engines and all IC engines have a silencer or a muffler in that exhaust line. Good job is to reduce this effect of this sudden pressure reduction sudden expansion and reduce the noise so that the people surrounding that engine nearby they do not come to any harm their ears do not come to any harm. If you find that your car engine does not have a silencer or if the silencer has rusted and corroded off you will find this noise coming from the engine those of you who have a motorcycle can do a simple experiment remove the silencer and then just try to start the motorcycle the sound will be so horrible that you will never touch the motorcycle without a silencer again. The same thing is true of a car in fact any car which has a leaky silencer will immediately be known and heard by everybody else as a problem car. So when the exhaust valve opens the pressure suddenly drops and then in the final stroke the whatever is remain is thrown out at this stage you can show them an actual indicated diagram and show them that during the exhaust stroke the pressure will be slightly higher and during the induction stroke the pressure will be slightly lower exaggerate this and show them that the actual indicator diagram of an auto cycle will have a positive work loop and will also have a negative work loop and then at this stage it is good to tell them that the mean height of this diagram positive added and negative subtracted from the positive is the mean effective pressure of the cycle. Now that comes that brings us to the end of the basic auto cycle then comes the question immediately a student will ask what is the diesel cycle the thermodynamically the diesel cycle is a cycle in which heat is added in a constant pressure process that is this is the Q1 process expansion in a constant entropy expansion or almost constant entropy expansion heat rejection during a constant volume process and heat sorry compression again in a constant entropy process at this stage since we would have already talked about the mechanism of an auto cycle.