 Let us work out an example. An ideal air standard diesel cycle operates with the compression ratio of 18.75 and a cutoff ratio 2.4283. The temperature and pressure at the beginning state 1 are given where as to determine thermal efficiency, mean effective pressure and the second law of efficiency ok. So, in the same manner as what we did before we will walk through the cycle and determine property values at each state point and then carry out the analysis. So, state 1 is the ambient or dead state. So, this is easy to get. So, using this value of temperature we retrieve these values specific internal energy u, s 0 and v r from the table. Now s is equal to s 0 because this is the ambient state. Now at state 2, v r is equal to v r 1 divided by the compression ratio because it is an isentropic process. So, using this value of for v r we enter the tables and we retrieve these values from the table. Now the entropy remains the same because it is an isentropic process. I will explain in a minute how this value is computed. Now at the end of the heat addition process temperature is the temperature T 3 is equal to T 2 times r c because it is a constant pressure process. Let us see how this comes about. So, between state 2 and 3 what is that if I apply equation of state at state 2 and 3 p 2 v 2 is equal to r T 2 and p 3 v 3 equal to r T 3. So, if we divide these two expressions we get T 3 over T 2 equal to v 3 over v 2 what is that p 3 is equal to p 2. So, v 3 over v 2 is nothing but r c which is why we have done this here. So, T 3 is equal to T 2 times r c. So, with this value of temperature we go into the table and retrieve the air table and retrieve u s 0 and v r. What is that s here has to be evaluated using our expression s of t comma p equal to s 0 of t minus r times. Here we can use an expression that involves v. So, it can be calculated using the tedious relationship. Now for state 4 we evaluate v r like this. Notice that for state 4 the compression ratio the compression ratio is v 3 over v 4 for the expansion process the compression ratio is v 3 over v 4. So, v 3 over v 4. Now we may write this as v 3 over v 2 times v 2 over v 4 v 3 over v 2 is nothing but r c and v 2 over v 4 notice that v 4 is equal to v 1. So, v 2 over v 4 may be written as 1 over r. So, v 4 over v 3 is equal to r over r c and that is what we have done here. So, the effective expansion ratio is nothing but r over r c and that is what we have done here. So, with this value for v r we can then go to the table retrieve t u s 0 is not required because this is an isentropic process. So, the entropy remains constant. So, notice that we evaluated p 2 or we have not evaluated p 2. So, this value for p 2 has been evaluated using this expression. So, we have the equation of state at state 1 and 2 if we divide one by the other we can show that p 2 is equal to r times p 1 times t 2 over t 1. So, this we have already seen now p 4 may be evaluated in this manner again by writing equation of state at state 3 and state 4 and then dividing one by the other we can eventually get p 4 equal to p 3 times t 4 over t 3 times r c over r. So, now we can actually plug in these numbers and get the values that we are looking for the work input during the compression stroke u 2 minus u 1 is it may be evaluated as 466.71 kilo joule per kilogram. And the work done during the cycle which is nothing but the displacement work during the heat addition process plus the expansion work comes out to be this 1334.46. The network produced during each cycle is this 867.75. Now heat added during the cycle again we have applied first law to process 23. So, heat added during the cycle is 1564.5 and heat rejected during the cycle comes out to be equal to this the thermal efficiency of the cycle this comes out to be 55.47 percent I am sorry this should be 867.75. Let us see how this compares with the value that we obtained or value that we would have obtained using a cold air standard analysis. So, let us see. So, here the compression ratio is 18.75 cutoff ratio is 2.4283. So, 18.75 comes out to be somewhere over here cutoff ratio 2.4283 may be something like this. So, if I 18.75 I am sorry 18.75 comes like this. So, 2.4283 RC 2.4283 line may be drawn approximately like this. So, we can see that the specific work in this case is about 3. something and the efficiency RC 2.42 somewhere here 18.75 and this. So, you can see that the thermal efficiency from cold air standard analysis is about 60 percent or so approximately and we are getting 55.47. So, this also shows the power of the cold air standard analysis the numbers are not that different. Mean effective pressure may be calculated in the same manner as done for the Otto cycle and we get this to be 1071.59 kilopascal much higher than what we saw for the Otto cycle because the specific power is higher. Now, exergy supplied may be evaluated directly like this where we have taken the temperature Th to be 2200 Kelvin. This is the peak temperature that is seen in the cycle which would be at T3 at state 3. So, that is the peak temperature. So, we have taken that to be the maximum temperature in the cycle and used it as the temperature for the heat source. So, exergy recovered comes out to be like this and so the second law efficiency for the cycle comes out to be 72.16 percent and if we compare this with the value from the cold analysis air standard analysis we get. So, 18.75 and RC 2.4283. So, we get this to be approximately about 75 or so and we get the value actual value to be 72.16 remarkably close to the cold air standard value. Now, the air standard dual cycle is a combination of the Otto cycle and the diesel cycle where part of the heat release takes place at constant volume part of the heat release takes place at constant pressure. The real cycle for example, the real Otto cycle is probably better represented by the dual cycle than the Otto cycle itself. Let us just go back and take a look. So, you can see here that the real Otto cycle which is shown using a chain line probably is better represented as a dual cycle than the Otto cycle, but convention dictates that we use Otto cycle and that is what we have done. So, it is possible to carry out a similar sort of analysis for the for the dual cycle. We will not do that here. I leave that to you to look at it in the textbook and then work out some examples involving the dual cycle, but it is relatively straight forward to do that. Now, just to give a context, I said earlier that in this course we are actually interested in the ideal cycles and as they are utilized for the actual engines. So, the relevance of looking at IC engines, particularly at this time is probably important to discuss. It has become quite popular to say that IC engines will soon be replaced by electric vehicles in the near future and so on. So, the relevance of studying these cycles and IC engines is becoming very, very important. In fact, it is important to establish the relevance of studying these engines. If you actually look at the issues objectively and in a quantitative manner, here I am referring to this article that appeared in International Journal of Engine Research. So, all the text here are directly quoted from that article. I have not written these sentences myself. These are directly taken from the article. So, the article has a lot of interesting things to say. It is written by a handful of authors about a dozen authors or so. Most importantly, it says that it is possible to reduce fuel consumption and hence CO2 emission by as much as 50 percent in SI engines. Now, in the case of CI engines, notice that high filtration efficiency in diesel and gasoline engines, particulate filters, urea injection and selective catalytic reduction can reduce in extremely low notes that is 15 to 20 milligram per kilometer. In fact, there are vehicles which have tailpipe unburned hydrocarbon emissions, which is actually less than the hydrocarbon that are present in the ambient air. In other words, these engines take in air which has a higher amount of hydrocarbon than what comes out of the exhaust. So, they are actually negative emission vehicles. So, the technology of selective catalytic reduction, urea injection and filtration have become so good that for hydrocarbon emissions, some of these vehicles are actually emissions negative. Now, in terms of particulate matter, remember we said that there will be soot that comes out of diesel engine which is particulate matter and also unburned hydrocarbon. So, we have dealt with unburned hydrocarbon here in terms of particulate emission. It is actually important to recognize that today the particulate emission from these engines is of the order of 5 milligrams per kilometer, which is less than the particulate emission from what you get from tire wear as you drive on the road, both the tire and the brake linings wear out and they give out solid particles or particulate emissions. So, the tire wear and brake line emission gives about and brake line particulates is about 50 milligrams per kilometer compared to particulate emission of 5 milligrams per kilometer from the exhaust pipe. So, emissions and other things from these engines are all well under control and there is scope for probably improving them even more. The technology is mature, but still improvements are being made. So, in fact, I invite you to read this article. It has lots of other additional details. Quantitative numbers which actually discuss the true impact of internal combustion engines and what they have to offer today. In fact, we discussed in the beginning of this module I started by talking about this Mazda engine. This engine is an example of what is possible today, what kind of improvements in efficiency and emissions are possible today. So, this sky active G where the G stands for gasoline. This engine operates at a compression ratio of 14 to 1. What is that? We said that auto cycle we cannot go beyond 10, because you encounter the knocking problem auto ignition or knocking problem. Now, this engine uses in cylinder sensors and onboard computers and other things to continuously monitor the equivalence ratio amount of fuel and the engine detect knocks actively and then adjust the fuel injected or the air fuel ratio so that the knocking is never allowed to progress. So, the auto ignition is never allowed to progress into a knock. So, it constantly monitors in cylinder parameters and then controls the amount of fuel air and so on and adjust other things so that the engine although it is operating at a compression ratio of 14, it still operates in a stable manner. So, the computer continuously does this. It also uses additional techniques like cylinder deactivation where when the load on the engine is less come some of the cylinders are automatically deactivated only one or two cylinders may be functioning the other two may just be motoring idly going back and forth. So, this reduces emissions and also improves the thermal efficiency of the engine because the thermal efficiency of these engines typically tend to be high at high load and somewhat poor at part load. So, by deactivating the cylinder the remaining operational cylinders are actually operating at high load. So, their efficiency continues to be high. Now, Skyactive D is diesel engine version of this Skyactive engine and it also uses advanced technologies it uses a lower compression ratio in fact startlingly it uses a compression ratio of 14 again, but it has much higher fuel efficiency much less emissions and so on. So, there is still a lot of scope for improving efficiency and emissions of internal combustion engines. So, the point that I am making here is that it is too early to write off the internal combustion engine without actually doing a fair objective and quantitative comparison of the electric vehicle and the internal combustion engine that is very very important that makes what we have studied so far discussed so far all the more relevant you need to have quantitative numbers and concepts and facts so that a fair comparison between IC engines and electric vehicles can be made when it comes to emissions per unit power or emissions per seat kilometer, passenger kilometer traveled.