 a diesel cycle is quite often known. Diesel cycle is an approximation of what happens in a diesel engine which is quite often known as an oil engine. And you say that look all these oils diesel petrol are petroleum fractions come out of refining of crude oil. And you say that petrol because most of our students have a reasonable background in chemistry including organic chemistry. Tell them that these are alkanes that is hydrocarbons. Petrol is typically C8 that is octane type whereas diesel is between C10 and C11, C10 or C12 the decaying type. So diesel has a higher molecular weight so it will have a higher boiling point, higher evaporation point and it will be a heavier fuel. Tell them that because petrol has a lower boiling point it will mix very easily with air. If you bring a can containing petrol and leave it open in the classroom within a minute or so the whole classroom will start smelling petrol but that is not true of diesel. You open a can of diesel it takes quite some time before people starts realizing that there is a diesel can open. And because of this it is not possible to do a premixing of petrol and diesel vapour it is not so easy to vaporize diesel. So what is done is the following. The idea is you take an auto type of engine that means a reciprocating engine what you do is compress only the air. So take in air compress it and when you compress it isentropically you will reach some state 2. If this is state 1 after compressing you will reach state 2. Now at this stage as it reaches states 2 inject a spray of diesel. This is a fine technology by itself earlier we used to have individual injectors on individual cylinders. Now we have what is known as a common rail injection system. A high pressure duct containing diesel oil is created using a high pressure pump and then you have nozzles with valves at the entry of every cylinder. Whenever is required you just open that valve opening the nozzle and spraying a very fine mist of diesel into the cylinder. So this spray of liquid spray of diesel is liquid first it will mix with air. The air is hot so it will evaporate and after evaporation the vapour will mix in appropriate proportion and then because of the high temperature the temperature here is pretty high it will burn. There is no spark needed and that is why the diesel engine quite often is known as a compression ignition engine and the petrol engine is known as a spark ignition engine. And then you say that all these processes dispersal of droplets in air their evaporation and mixing of vapour and air and the mixture finding itself at the correct temperature and the correct pressure, correct concentration takes some time. The piston does not have patience to wait it is connected to the crank so it starts moving. So what happens is combustion tries to increase the pressure, the piston movement tries to decrease the pressure, the net resultant is something in between and in a basic diesel cycle it is modelled as a constant volume process. So 2 to 3 is the constant volume heat addition by compression ignition of diesel and tell them that this constant volume is only an approximation approximate model sorry constant pressure combustion tries to increase the pressure expansion tries to decrease the pressure approximate constant pressure heat addition. This is a model for analysis in actual practice it does not happen like that after that the remaining stroke is used for expansion again exhaust valve opens and the remaining part take place including the exhaust as well as I am exaggerating this loop but you have a positive loop and negative loop the other part is almost like a petrol engine and at this stage you can say that you can do a standard analysis of a diesel engine assuming it to be a closed cycle and all that and you will get the standard slightly complicated formula for the efficiency of a diesel engine. The question then comes up is what is a dual cycle dual cycle or sometimes called dual combustion cycle there is no new cycle but remember that what we have done here is we have said that the model is approximate constant pressure heat addition but this is not a very good approximation it is a sort of acceptable but if you want to do a better approximation that is what is done in a dual cycle in dual cycle it is assumed that part of the combustion takes place at constant volume followed by the remaining part of the combustion which takes place at constant pressure you have the compression process from 1 to 2 then instead of 2 to 3 at constant pressure you assume that part of it is at constant volume part of it is at constant pressure let us say you can say this is 2 prime and then 3 to 4 as earlier and 4 to 1 exhaust process. So, here 2 to 2 prime is constant volume 2 prime to 3 is constant pressure this is just a new or a different model for combustion in a diesel engine at this stage we can come to the at least the power cycles based on gas and then you say that look we have not yet explored the option of vapour cycles and we tell the students that water is a ubiquitous material available in reasonable pure form we know all the properties of it it is rather easy to evaporate it at various pressures we know how to condense it we know all the properties of water it has good heat transfer characteristic so it should be a good working fluid for a power plant and yes it is a good working fluid for a power plant but then when we try to implement this we suddenly realize that if we take the Brayton cycle which is the PSPS cycle in which in one constant pressure process heat is added in another constant pressure process heat is rejected and then if we say that during this constant pressure process of heat addition we make the fluid evaporate and during this constant pressure process if we make the fluid condense then because at constant pressure you have two phases together this would essentially be a constant temperature process similarly during heat rejection although we are doing it at constant pressure because condensation takes place this would also be a constant temperature process so if you include this evaporation and condensation of a vapor substance or a fate changeable substance like water then you can essentially implement a Brayton cycle but what finally you will end up with is effectively a TSTS cycle which is nothing but the Carnot cycle that is the attraction of vapor cycles for us it is easy for us to implement constant pressure heating and cooling we do that so we say that we are doing only constant pressure heating and constant pressure cooling but during heating we arrange that the fluid evaporates during cooling we arrange that the fluid condenses and this evaporation and condensation at constant pressures turn out to be essentially at constant temperature leading us to an indirect implementation of the Carnot cycle let us try to do this and see what happens what we have is naturally it will be a multi equipment cycle we will show it on the TS diagram and we will say this is the critical point this is the saturated liquid line this is the trisaturated vapor line and let us say that we decide to work our cycle between a TH and a TL and then we can arrange the cycle as a rectangle within this maybe the maximum rectangle can be something like this this is our PL and this is our pH so although we are doing a constant pressure heat addition and constant pressure heat rejection combined with two isothermal processes isentropic processes these two isobaric processes are also isothermal processes consequently if you implement this this is Brayton cycle with vapor and fully in the two phase zone this is nothing but a Carnot cycle and the efficiency of this will be 1 minus TL by TH however when you try to implement this in practice the block diagram will look like this if I have these as four states the block diagram will show a boiler in which heat is added a turbine 3 to 4 almost isentropic the work is done a condenser in which heat is rejected and a compressor in which this wet steam here will be compressed and made available to the boiler when you do this the problem that arises is this wet steam compressor we have had a problem with a compressor earlier but remember there we were compressing almost dry saturated steam dryness fraction 0.95 if you look up here the dryness fraction of one will be pretty low and it will start off as a compressor but end up like a pump because at the end we have only saturated liquid so this compressor is a technologically difficult thing to implement so what is done is such an implementation never takes place what we do is the following let me change the color of my pen so I can show the modification in a different color this one it shifted from wherever it is to the saturated liquid state and then at the saturated liquid state we pump it because now it is a liquid to the new state 2 which is at the higher pressure but at a temperature much lower than th and the heat addition now starts from this lower temperature to sub cooled heating first then evaporation and then we come to 3 the current implementation of this cycle and during the chat I have been asked whether it is a Brayton cycle or a Rankine cycle what I showed you is the Brayton cycle now as modified with those small red markers in the previous thing we will call it the Rankine cycle so this part of the cycle remains as it is 3 to 4 here this is the pH line high pressure line this is the low pressure line PL but we do the condensation right up to saturated liquid then we pump it and then we have part of the heating increasing temperature part of the heating is at constant temperature so from the Carnot cycle we have given up a corner it is not a rectangle anymore and this cycle the modified Rankine cycle used for it is actually a Rankine cycle because it is actually a Brayton cycle because it is a PSPS out of this the heat rejection turns out to be at constant temperature but the heat absorption is not really at constant temperature part of it is at constant temperature but part is not at constant temperature this cycle is now known as the Rankine cycle the block diagram now looks before as earlier but traditionally the Rankine cycle block diagram is shown like this we have the boiler in which heat is added Rankine cycle will always be external combustion then we have a turbine then we have a condenser where some cooling stream usually a water usually cooling water from some surrounding lake or river or sea is used to cool it so this is the condenser this is the pump this is the boiler this is the turbine states 1, 2, 3, 4 the pump W dot net will be W dot T minus W dot net however because it is a vapor now we cannot do any standard analysis for a given lower pressure upper pressure and the way it is implemented we will have to work it out with steam tables every time at this state we have come to 13.4 where the basic cycles Brayton Rankine auto and diesel have been explained in some order at this stage you may decide to go over to the refrigeration cycles before going to modification of cycles and combination of cycles so let us first look at refrigeration again just the way we have the major cycles are Brayton Rankine gas and vapor there are corresponding cycles in the refrigeration scheme which you can say inverted Brayton cycle and inverted Rankine cycle the inverted Brayton cycle is quite often known as the Joule cycle it is a gas cycle which is used for refrigeration the inverted Rankine cycle with its modification does not have the name of a person but is known as the vapor compression cycle it is a good idea to start with the Joule cycle because that is almost an unmodified version of a Brayton cycle whereas vapor cycle is a modified version of a Rankine cycle so when it comes to a Joule cycle bring it to their attention that we need to have a T ref temperature of the refrigerator space and T ambient where we can reject the heat and a Brayton cycle is made up of two isentropics and two isobarics so we lay out the two isentropics like this and the two isobars are like this after doing this we tell them that look the cycle is laid out like this and now let us see how it is made to work we have at a low pressure an isobaric process in which heat is absorbed from the cold space for refrigerated space then we have an isentropic process 1 to 2 in which the temperature of the working fluid is raised by compression to a temperature higher than the ambient because notice 4 to 1 is a isobaric heating process where the temperature is below T ref except at T 1 it will just reach T ref we want to reject it to ambient so we must raise the temperature of this to a level above the ambient so we raise it by doing this compression 1 to 2 then we reject the heat to the ambient in an isobaric process at pH 2 to 3 and again reduce the temperature below that of the refrigerated space by doing an isentropic expansion so this is made up of 4 processes an isobaric heat addition 4 to 1 isentropic compression 1 to 2 isobaric heat rejection this is where q dot 1 is rejected to the surroundings and isentropic reduction in pressure or expansion 3 to 4 the block diagram will be seen like this you have the heat exchanger in the cold space going from 4 to 1 then you have the compressor which will take this low pressure gas typically air at state 1 and compress it to state 2 this is the compressor power will need to be supplied to the compressor then you have the heat exchanger which does heat rejection here q dot 2 is absorbed here q dot 1 is rejected and then you have a turbine you have a turbine which produces some power output this is state the net power required will be the power required to drive the compressor minus the power which comes out of the turbine you can do a standard analysis and determine COP and any other parameters you feel like now at this stage it is necessary to tell them may be going back or even otherwise that the work ratio of a Brayton cycle work ratio of a Brayton cycle Otto cycle or diesel cycle is typically of the order of 0.25 0.3 which is good it is nowhere near 1 but when you come to Rankine cycle tell them that and demonstrate it using a illustrative diagram that for a Rankine cycle the power consumed by the pump is a very small fraction and usually fraction of a percent of the power delivered by the turbine consequently here the work ratio is very small less than 0.01 and which is a very attractive thing for any Rankine cycle plant when it comes to implementation we find that for the gas turbine cycles the compressor and the turbine are invariably mounted on the same shaft because so much a significant fraction of the turbine power something like 20 30 percent needs to be made available to the compressor and we do not want to lose anything of that in you know bearings gear boxes and things like that so we have a direct drive on the same shaft compressor and turbine are mounted but in a steam plant the power required by the pump is so small that we do not directly couple it to the turbine either directly or through a gear box or a belt drive we let the turbine produce whatever the power it wants to produce and provide a separate drive may be a very small turbine or a small electric pump to take care of the power required by the pump so remember Rankine cycle as a very small pressure ratio Carnot cycle as a very large pressure ratio almost one Rankine cycle typical pressure ratio is 0.01 sometimes even lower Brayton cycle depending on the parameters of pressures and temperatures is typically 0.2 0.25 point now because the Joule cycle is an inversion of the Brayton cycle the turbine will produce may be 30 40 percent of the power which the compressor needs and hence to reduce the external power supply to the compressor here also the turbine and compressor are arranged on the same shaft now you come to the inversion of the Rankine cycle but now remember if you look at the Rankine cycle if you invert it you will find that the heat rejection will now take place just consider this Rankine cycle going in the opposite direction 1 to 4 4 to 3 3 to 2 and back to 1 and because it will be absorbed from 1 to 4 you will have the refrigerated space temperature nearly equal to 1 to 4 and what about heat rejection heat rejection will take place from 3 to 2 and the temperature of 2 is not significantly higher than 1 so that means the 0.1 the cold space temperature or the heat absorption temperature will not be significantly below the ambient that will be a disadvantage so what we do is we go back to our original Brayton cycle idea where we worked it as a Carnot cycle and then you say that look if we implement it like this then since the processes 4 to 3 now it will be a process of compression and 2 to 1 process of expansion are adiabatic I can have heat rejection to the ambient at 3 to 2 a high temperature and heat absorption from the ambient 1 to 4 at a low temperature and when you do that let me sketch it and show you what happens let me now take Rankine inverted or I should not really say you should say Brayton inverted with a vapor and fully in the 2 phase zone now this is T ambient this is T refrigerated space now I will have compression like this I will have cooling like this I will have expansion like this I will have heat absorption like this so heat will be absorbed here Q2 in the so called evaporator heat will be rejected to the ambient Q1 in the so called condenser this will be the compressor which will compress it from a lower pressure to a higher pressure and this will be the turbine or expander which will expand it from the higher pressure back to the lower pressure the higher pressure isobar will go like this pH the lower pressure isobar will go like this PL well if you try to implement it like this there are 2 issues which come up in its implementation one issue is that the compressor is starting with wet vapor and although it is good to have it is not very difficult to have wet vapor we do not mind compressing it in fact we would like to evaporate it fully and then compress it like this this is the modification which we do second thing is rather than expanding this this turbine or expander here will be a very complicated device because it is starting with a liquid as it gets depressurized it will start evaporating and that evaporation also will have to be taken care of it is an adiabatic process but the change in volume during this expansion would be very very high and because of that this process is replaced by a throttling process and now the resultant cycle known as the vapor compression cycle looks like this this is the higher temperature but more important is this is the pH isobar this is the lower pressure isobar this is the lower temperature this is let me say th but this is not lower pressure isobar will go like this we start with dry saturated steam at the lower pressure compress it ideally isentropic compression if I say this is 1 this is 2 2 to 3 is heat rejection usually at the end of compression you find that the vapor is superheated so part of the heat rejection is desuperheating part of the heat rejection is condensing so 2 to 2 prime then to 3 at 3 we put it through a throttling valve or a capillary and come to state 4 this way we lose out the work which we could have perhaps obtained in this turbine or expander but that is a very small amount of work compared to what we will have to do in the compressor so even if we lose out on that there is no significant degradation in performance moving the evaporation from this point wet vapor point to dry saturated vapor point not only healthy for the compressor but it will although it will also make the compressor very long lasting I mean it can be designed to have a very long life this can be now shown in a block diagram as follows 4 to 1 the heat absorption process is implemented in a heat exchanger known as the evaporator because in it the refrigerant evaporates gets compressed then condenses first goes through the throttling device 1 2 3 4 in the throttling device if we arrange areas appropriately h 4 will be equal to h 3 so notice that this vapor compression cycle has an isentropic compression isobaric heat rejection part of it turns out to be isothermal isobaric evaporation most of it is isothermal or the way it is shown here it is definitely isothermal and this is a device throttling device should not call it isenthalpic but isenthalpic at 3 to 4 is the same so if you want to write it as a diagram cycle for example the joule cycle is a PSPS cycle just like a Brayton cycle but in the other direction the Rankine cycle is also a PSPS cycle but much of the P part is also becomes T because of evaporation and cooling whereas for the vapor compression cycle you can say that it is a PSPS cycle it is absorbed in one P it is rejected in another P and part of this P cycles in part are at constant temperature constant P processes happen to be at constant temperature because of change of phase now at this stage tell them that because two lines are constant pressure and we have a line with constant enthalpy instead of a TS diagram a diagram with P as one of the variables one of the coordinates is attractive it could be PS but remember that entropy is useful only to thermodynamic people like us whereas enthalpy is useful because the work done or the power required for the compressor would directly be related to the enthalpy difference between two and one similarly the heat absorbed in the evaporator the so called refrigeration effect is directly proportional to one and four and hence instead of a TS diagram a pH diagram is attractive and you tell them that in the industry when you practice vapor compression cycles are usually plotted on a pH diagram and if you go to your moodle you go to the stuff of interest in moodle and go to charts and you will find that it is not just charts but it charts and tables and you will find that on various pages in this makes it you have a pressure enthalpy diagram for ammonia then you have pressure enthalpy diagram for refrigerant 22 this is over a wider range of pressure and temperature there is another diagram which restricts itself to the right of the saturated vapor line if you see that second diagram is in this zone I do not know whether you see this on the yes I think you have seen that on the screen so that you can read it better you should not be under the impression that you always work with diagrams so for refrigerant 22 there are some properties saturation properties also in the form of tables and if you go to the library you will find much more detailed tables and diagram and finally this is the diagram for refrigerant 134A which is the current replacement for the old refrigerant 12 which was used till about 10 years ago 10 15 years ago in all household refrigerators these three fluids I have included because I think there are exercises in the cycle analysis part C A 1 onwards which use one or more of these fluids but if you go to the ashray handbook or you go to the internet you will find charts and tables for various refrigerants and various other fluids and if a fluid is used as a refrigerant be sure that the chart will be available in the pressure enthalpy diagram form in fact a PV diagram or a TS diagram may be difficult to find but if a fluid is a candidate as a refrigerant its pressure enthalpy diagram would definitely be available so we can use this the only disadvantage is these diagrams tend to be cluttered because well pressure and enthalpy are the coordinates but you have constant entropy lines you have constant temperature lines and you have constant volume lines so that is what makes the reading of this is a bit difficult if you print it out be sure that you get a good printer but try to print it out or get it enlarged on a A3 size sheet not on an A4 size sheet if you really want to use this for solving problems and A3 print out is necessary so that brings me to the end of the basic cycles part that is up to 13.5 we have talked about refrigeration cycles and joule and vapor compression cycles so now let me go back to where we left off I showed you the pH diagram on the printed charts but a few things about the pH diagram the vapor compression cycle will always be in two parallel isobars the pH and the PL the dome for most of the fluid goes like this with the critical point at the top and the cycle will start with whatever points let me see one is evaporator exit and it goes anticlockwise from that evaporator exit then we have an iso isentropic which brings it to 2 this is the compression process then you have the condensation fully isobaric but only part of it from 2 to 2 prime in the sorry 2 to 2 prime in the D super heating zone and 2 prime to 3 in the condensing zone so that part is isothermal then you have the throttling process at least if you are solving this in the course of thermodynamics you should show it by dotted line not by a continuous line because we do not know what is happening in between that brings it to 4 and 4 to 1 is the evaporation process and the thing to note here is this is H2 minus H1 and this enthalpy difference is proportional to the compressor power in this analysis we will generally be neglecting any change in potential energy and kinetic energy so W dot C will be m dot into this delta H the difference between H2 and H1 similarly the refrigeration effect or the heat absorb will be proportional to H1 minus H4 so this will be proportional to W dot C and since we have a throttling process instead of the turbine there is no other work transfer there is nothing like a turbine or an expander which will give you some work which you can compensate with the compressor work this is proportional to Q dot 2 which technically is known as the refrigeration effect and hence the ratio of this length if this is plotted to scale to this length this length is a this length is b the ratio of this length of a to length of b becomes H1 minus H4 divided by H2 minus H1 and which becomes proportional to or which is equal to Q dot 2 by W dot C so this straight away gives you the coefficient of performance if you plot it on a reasonable scale then you can even use a scale for a measuring strip to measure this out in length and may be almost graphically determine the coefficient of performance we did we have not yet talked about the Hs diagram you may as well introduce them to the Hs diagram on which Rankine cycles or turbine state spaces are often plotted unfortunately for steam the part of the Hs diagram which is the saturated liquid line becomes very very cluttered so you cannot really show what it is the reason for that is the if this is the triple point this is the critical point and the saturated liquid line is all here and the dry saturated vapor line goes like this so if you have a low pressure isobar like this and a high pressure isobar like this this particular part is very very small that just not seen anywhere so our diagram will be 1 to 2 isentropic 2 to 3 isobaric part of it isothermal 3 to 4 again isentropic and 4 to 1 isobaric again all of it isothermal this is how the cycle will look and again here notice that this is the power output of the turbine we may call it delta H across the turbine it will be proportional to the power output of the turbine this will be delta H across the condenser which is proportional to the thermal energy loss or cooling effect in the condenser this will be this is exaggerated this is about only 1 percent of delta H T this is delta H P proportional to the power absorbed in the condenser and this is delta H B proportional to the power for the energy heat transfer rate in the boiler.