 So, this is 2 to the power 9 on thermodynamic cycles. So, we will see two power producing cycles, Rankine cycle and Brayton cycle and one power absorbing cycle that is the vapor compression refrigeration cycle. The first today problem is on basic Rankine cycle steam power plant cycle. The question is, a basic Rankine cycle operates between pressures of 10 mega pascals and 10 kilo pascals. So, you know that there is a pressure ratio for these cycles, highest pressure is 10 mega pascals, lowest pressure is 10 kilo pascals. So, in the boiler a fuel oil is burned with air that results in continuous production of odd gases. Steam with a degree of superheat of 288.9 degrees centigrade is produced in the boiler by using the heat from the hot gases. And it enters the turbine, the steam enters the turbine to undergo an adiabatic expansion process. Turbine power output is 40 megawatts and the quality of steam at the turbine exit is 0.9. So, these are the data given. So, it is given basic Rankine cycle. So, we have to understand what are the components in that and try to fit these what are given data into that and try to solve the problem. So, what is asked is, determine the isentropic efficiency of the turbine. Heat transfer to steam in the boiler, thermal efficiency of the plant neglecting the pump work. If the heating value or calorific value of the oil is 30 megadoules per kg and 85 percent of energy released during combustion of oil gets into the steam in the boiler. This is like a heat transfer efficiency. Water heat is generated during the burning of oil and air, producing hot gases. These hot gases exchange heat with water to produce steam. So, that efficiency we can say is 85 percent. So, if that is the case, Kalligewald is given and the efficiency of heat transfer is given, determine the amount of oil flow rate that is to heat the water to steam in the boiler. So, this is the problem. Now, if you see, let us first understand what is basic Rankine cycle, which we have already seen. This has basically four components, pump, boiler. So, in this boiler only what happens is basically the water flows through the seas of pipes and here this boiler has an inlet and an exhaust. Basically, here hot gases comes in, hot gases due to the combustion of oil and air comes in and it goes out, it exits, ok, hot gas in and hot gas out. Now, the boiler, the steam is produced and the steam expands in it, steam turbine produces the work. For the pump, we have to give some heat, sorry work. For the pump, we have to give some work input that is WP. Here this heat exchange happens. So, basically we can say here some amount of heat is given to the water and here this steam turbine and the exit of the steam is now connected to a condenser and here heat is lost. Now, this cycle continues. So, you can, you know this. Basically, here the additional thing what you would consider is the, the hot gases are coming in and going out, transfers the heat. So, what is the heat exchange happening here that you have to calculate? To in order to calculate the amount of oil required, ok. Now, let us say this is state 1, the entry to the pump is state 1 and entry to the boiler is state 2. So, this operates at higher pressure. So, pump actually increases the pressure. So, here pressure is P1 and here is pressure is P2, it is equal to the boiler pressure and this P1 is equal to condenser pressure. So, here the pressure ratio of this is P2 by P1, ok. Now, this is exit of the boiler or inlet to the steam turbine that is 3 and inlet to the condenser or exit of the steam turbine is state 4. So, this is the basic rank and cycle. When you say basic, see actually in real steam power plant there are other components like reheating, the regeneration, etcetera. Now, here only basic components are there and pressure ratio is P2 by P1 is the pressure ratio. So, this is given as 10 megapascals divided by 10 kilopascals, ok. So, condenser operates at a constant pressure, constant pressure of 10 kilopascals, ok. Then boiler operates at a constant pressure of 10 megapascals. So, this is the basic rank and cycle. So, now the data given, we have taken P1, P4 equal to, so I can say P1 equal to P4 equal to 10 kilopascals, then P2 equal to P3 equal to 10 megapascals. Now, what is given is degree of super heat is given. What is degree of super heat? T degree of super heat is T saturation, saturation temperature at the given pressure. So, in this case it is boiler pressure. So, I will say P boiler plus, so this degree of super heat we will add. So, say degree of super heat Tds. So, Tds is degree of super heat. So, saturation pressure, temperature, saturation temperature at the boiler pressure plus degree of super heat, that will be the T3, that is the temperature of the steam at the entry of the steam turbine, ok. So, that is the given data. One more data is given, what is that? That is the quality, quality of the steam at the turbine exit, that is X4 is given as 0.9. So, the given data is given here and power output, this WT, W dash T is 40 megawatts. So, all the data we have written here. First one is the pressure ratio, 10 megapascals to 10 kilopascals, that is boiler operates at 10 megapascals and condenser operates at 10 kilopascals. Then degree of super heat, which will be used to determine the inlet temperature for the steam turbine T3, which is equal to saturation pressure at the boiler saturation, sorry saturation temperature at the boiler pressure. What is the saturation temperature at 10 megapascals? That plus the degree of super heat, which is given as 289.9. And the exit state of the steam in the steam turbine, from the steam turbine is 0.9 quality. And the pressure, you know, obviously is 10 kilopascals. So, with this, you have to fix the states, then apply the first law. So, let us apply, first of fix the states. Now, before doing that, one more thing we should understand, that is the how this operates and how to plot the Ts diagram for this. So, saturation lines we have drawn. So, say this is 10 kilopascals pressure, then this is say 10 megapascal pressure, isobar in a Ts diagram. Now, there are two conditions, which is not given. So, for example, state 1, state 1 pressure I know is 10 kilopascals, where that is the pump inlet. Go back, you can see that pump inlet pressure is condenser pressure. So, it is 10 kilopascals, 10 kilopascals, that is known, but second property is not known. So, basically pump operates liquid only. So, that means, your state 1 should be at least a saturated liquid, if not sub-cooled liquid. Okay, see condenser operates, the condenser exits state, condenser heat is lost. Okay, if you go back, condenser heat is lost to the ambient. Okay, so there is a steam with a quality of 0.9 exiting the turbine that enters the condenser. So, 90 percent of the steam going into the condenser will be in vapor phase, mass of that 90 percent of mass. Now, in the condenser heat is lost and finally, at exit of the condenser, the state 1, I should get only liquid. So, state 1 should be liquid only. That means, X1 should be 0, I will say at least. Why at least? Because I can get sub-cooled liquid also. See, this is the saturated liquid state at 10 kilopascals. Correct? I can go to this state also, sub-cooled states also. Same pressure, I can go to lower temperatures also by losing more heat in the condenser. Okay, I can lose more heat here. So, when this increases, I can get to low and low, lower. So, it is not given, but the data is not given. But for the cycle to operate properly, at least the state 1 should be a saturated liquid state at 10 kilopascals. So, I can say state 1 can be fixed as P1 equal to 10 kilopascals and X1 equal to 0. That is it. So, that is the first thing you do. Now, state 2. So, I fixed state 1 now. State 2, how the pump increases the pressure of the liquid, this is saturated liquid to from 10 kilopascals to 10 megapascals. Okay, now there is no information about the pump operating conditions here. Okay. So, let us see if the pump operates reversibly. Okay. And the heat loss from the pump is negligible. So, there are two things we are trying to see here. There is no heat loss from the pump. Whatever the work supplied is used to increase the pressure of the liquid, saturated liquid is coming in into the pump. And final pressure is the pressure of the boiler, 10 megapascals. Okay. Now, if the pump operates in a reversible manner and adiabatic, heat loss is negligible, then the operation is isentropic. So, we can say, see, T dS equal to dH minus V dP. So, now that is isentropic means dS equal to 0. So, that means I can say dH equal to V dP. And when you integrate this, 1 to 2 dH equal to 1 to 2 V dP. So, H2 minus H1 will be equal to V dP integrated from 1 to 2. Now, for integrating this, I need to know the relationship between P and V. But for liquids, V does not change much. So, let us go to the tables and see this. So, here go to the table, saturation table based upon pressure. So, 10 kilopascals. 10 kilopascals is how much bar? So, 10,000 pascals that is 0.1 bar. Okay. So, 0.1 bar, if you see, what is the VF? Saturated liquid at, okay, saturated liquid means x1 equal to 0, correct. So, now what is VF? 0.001010 meter cube per kg, correct. Now, 10 megapascals, let us take 10 megapascals equal to 10 into 10 power 6 pascals, which is equal to 100 bar. So, go to 100 bar. So, from 0.1 bar to 100 bar, 0.1 to 1 is 1 order of magnitude increase. 1 to 10 is second. So, 2 orders of magnitude increase if you go, go to 100 bar. Now, you see that the volume as VF equal to, again saturated liquid if you see, VF equal to what? Basically, 0.001453 meter cube per kg. So, but you can see in the diagram, okay, if we go back to the PPT, you can see that if I go here, isentropically, isentropic process is interdependent constant. So, you can go isentropically and reach somewhere here. So, this is the state 2 for me. A subcooled liquid for the 10 megapascals, if the temperature is known, you can do this. So, please understand, I have to get to this state. So, I will say 2s. Because isentropically, I am going from 1 to 2. Now, the volume here, the specific volume in this case itself is, so, let us say this is the saturated liquid at 10 megapascals, correct? The specific volume there itself is, go back to this, is only 0.001453 when compared to at 0.1 bar saturation saturated liquid state, the volume is 0.001010. So, if you go back here, the specific volume will be much lesser than this. So, I can say that the volume has not changed much, correct? So, we can take V as V1 which is equal to 0.001010 meter cube per kg. And then I can say H2 minus H1 will be equal to V1 because it is constant. So, I take it out into integral 1 to 2 dP which is equal to V1 into P2 minus P1. So, this is the way I fix this state too. So, what is the main thing? We can see that in this point itself, the volume has not changed much, but the subcooled state where the pump is going to leave the high-pressure liquid, this state will have much smaller specific volume compared to this state. So, this is comparable to this volume only. So, we can say that the volume will not change, it is a liquid. So, volume will not change much, pressure rises and so I can fix the volume as V1 itself. And not much big error will be induced due to this assumption. So, now this is the H2 minus H1 or I will say H2S because the isentropic state 2 is denoted by 2S. Okay, now this will be equal to 0.001010 into 10 into 10 power 3 minus 10 which is equal to 10.11 kilojoule per kg. This difference is only this 10.11 kilojoule per kg. So, what I put? This is 10 mega pascals. So, I converted into 10 mega pascals is 10 into 10 power 6. So, I converted into kilo pascals. So, 10 into 10 power 3 minus P1 is 10 kilo pascals. So, I kept in kilo pascals, both are in kilo pascals. So, kilojoules per kg will come then. So, you can see the difference between H2 and H1 is only 10.11 kilojoule per kg. So, now I can fix the state 2, state 2, H2S I can find. So, H2S will be equal to H1 plus 10.11. Okay, what is H1? H1 goes back to the tables. Here for 0.1, 0.1 what is HF? Correct? HF is 191.8 kilojoule per kg. So, this is 191.8 plus 10.11 which is equal to 201.9 kilojoule per kg. So, this is the value of H2S. So, state 2 is now fixed as pressure equal to 10 mega pascals and H2S equal to 201.91. So, state 1 is fixed by assuming that the condenser exit state is at least a saturated liquid. If not, see if I want to fix it as a subcooled liquid then I should know the temperature. Since the infusion is not given, at least it should be liquid. So, we can take it as a saturated liquid. So, that is fixed in that way. State 2, I have assumed the pump to operate in a reversible manner and heat loss is also negligible. Thus, it is isentropic. So, for isentropic pump D is equal to 0. So, from the previous relationship D h equal to integral, D h equal to V dp or delta h equal to integral V dp. Noting that for liquids, V will not change much. We can say V as fixed V as V1 itself and do the integration as V1 into integral 1 to 2 dp. So, that will give you the H2S H1 as 10.11 kilo joule per kg and H2S is fixed. So, state 2 is now fixed. Okay, so now state 3 is actually given. Go back here. State 3 is the pressure is 10 mega pascals and degree of super heat is given. So, I know that how to fix these T3. T3 is what the saturation temperature at 10 mega pascals plus the degree of super heat. So, here you go. See what is the saturation temperature for, see for 10 mega pascals that is 100 bar, 100 bar second column is temperature. So, T sat at 100 bar equal to 311.1 degrees centigrade. So, that I use. Okay. So, here T3, T3 equal to 311.1 plus 288.9. That is what is given here. In the power, degree of super heat is 288.9. Okay, now using that I get 600 degrees centigrade. T3 is fixed. So, now I can put this. So, this is the state 3. This is 600 degrees centigrade. So, adding the T sat, saturation temperature at 10 mega pascals which is 311.1 degrees centigrade plus the degree of super heat which is given in the problem as 288.9. I can get the temperature at the inlet to the turbine. That is state 3. So, that is the state 3 here. So, state 3 values I can fix. So, state 4, the quality is given as 0.9. So, somewhere here it will be. Okay. So, here the quality is 0.9. So, I can fix the state 4 here. So, now that is the state 4. Okay. So, now I complete this. So, state 4 is also fixed. Okay, easily because state 4 is what? State 4 is here x4 equal to 0.9. So, state 4 is fixed by pressure being 10 kilo pascals and quality being 0.9. State 3 is fixed as pressure being 10 mega pascals and temperature being saturation temperature plus degree of super heat which is equal to 600 degrees centigrade. So, two properties are required to fix each state. So, how the states are fixed? I will repeat. First one is the first state 10 kilo pascals is the pressure. We have assumed that the condenser exit is at least a saturated liquid. So, x1 equal to 0 is taken. So, from that I can fix state 1. State 1 enthalpy h1 equal to hf at 10 kilo pascals. So, that is this value we have taken from the table 191.8. So, equal to 190, 1.8 kilo joule per kg. Similarly, state 2, we assume that the pump operation is isentropic. See, the pump delta h is very small. So, pump operation is isentropic. And from that, we can use the TTS relationship, find the DH value as VDP and integrate that. Now, noting that for the liquids, it is basically the volume physically remains constant. So, using V as V1, we can just integrate as V1 into delta P, which gives the delta h as 10.11. From that, state 2 is fixed. So, state 2 is fixed as pressure being 10 mega pascals and the enthalpy is 201.91. State 3, which degree of stability we have fixed and state 4 is given at 10 kilo pascals, the quality is 0.9. So, this is the representation of the cycle. You please understand here that in this, if you take a particular component, it is a control volume. There is a mass flow inside into the particular component phase of pump. There is a liquid water flowing into the pump and higher pressure liquid comes out of the pump. And for that, I have to give some work input W dot P. If you take this, basically the boiler, you can see that boiler, some heat input comes to the boiler. Liquid at higher pressure goes into constant pressure operation, then the same pressure superheated vapor comes out. And steam turbine again, as a control volume, it actually delivers work. It is actually adiabatic. So, it is given here that undergoes an adiabatic expansion in the turbine. So, it is adiabatic. So, W dot P is the work delivered by the turbine and again the superheated vapor at 10 mega pascals, 600 degrees enters and a steam at the quality of 0.9 at the pressure of 10 kilo pascals exits. And finally, in this control volume, the contents are the steam with the quality of 0.9 is cooled so that at least a saturated liquid at the constant pressure of 10 kilo pascal is coming out. So, these are four control volumes. Taking the group of all the control volumes, you get the system. So, power plant is a system basically. So, now in this slide, we have fixed all the states very carefully. Of course, with two assumptions, one is the inlet state of the pump taken as saturated liquid because no other information is available, but there is nothing wrong in taking this. In most of the cases, it will be saturated liquid or slightly subcooled that will not affect the results much. Similarly, the isentropic operation is assumed for the pump that is also fine. Now, we will go to the first law for each. So, now let us fix the enthalpies. So, H1 is known 191.8 kilo joule per kg. H2S also we have calculated 201.91, 201.91 kilo joule per kg. Then H3 is nothing but H at 10 mega pascals 600 degree centigrade. Now, we have to go to the table for this here. For 100 mega pascals that is 100 bar, 10 mega pascals 100 bar, the saturation temperature is 311. Obviously, a degree of superheated is added. So, that means the state is superheated vapor. So, go to the tables, superheated tables, superheated steam tables here for 100 bar. So, we have to go down. Here you can see for 100 bar superheated table, the temperature given is 600 degree centigrade. From that, I take the value of enthalpy under entropy. Both I can take 63625 and 6.903. Why I want entropy? Because it is asked in the question that isentropic efficiency of the turbine to be determined. So, that is why I need the value of entropy. Okay. So, now H3 equal to 3625 kilo joule per kg. From the same table, I have taken S3 also as 6.903 kilo joule per kg Kelvin. So, these are the values I have taken. Now, H4. So, now H4 is what? H4 is HF plus X4 into HG minus HF at 10 kilo Pascal. So, from 10 kilo Pascal table, saturation table, you have to find the values of HF and HG and X4 is known 0.9. So, substituting that we can get this. So, let us go to the tables again. Here, what is the value of HG HF? And I want to get the value of the SF and SG also. So, at this line, you can say this is HF, this is HG. 1363, 2004 through and this is 42 and this is the SF and SG values. Let us take this, HF. So, at, okay, I have to take it at 10 kilo Pascal. This is 100 kilo Pascal. For 10 kilo Pascal, you can see that this is the thing. HF is 191.8, HG is 2585, SF is 0.649 and SG is 8.1 for the 10 kilo Pascal. So, we will take these values. Here, HF is equal to 191.8 kilo Joule per kg, HG equal to 2585 kilo Joule per kg. Similarly, SF, SF equal to 0.649 kilo Joule per kg Kelvin and SG equal to 8.15 kilo Joule per kg Kelvin. So, you can just go back and see that 0.649, 8.15, 191.8, 2585. So, these are the values. Now, I calculate H4 as, so 191.8 plus 0.9 times 2585 minus 191.8, which is equal to 232345.68 kilo Joule per kg. So, now I have fixed all the states, basically. Now, what is the turbine power? So, q dot minus w dot equal to m dot into H3H4 minus H3. Neglecting the kinetic energy and potential changes. Now, this is 0 adiabatic operation. So, turbine power also is given to us 40 megawatts. You can go back and see in the problem 40 megawatt turbine power is given. So, that you substitute here, 40 into 10 power 3 kilowatts equal to m dot I do not know into H4 I have calculated 2345.68 minus H3. H3 is 3625. So, this implies m dot equal to what? Because this is negative, this is also negative, so negative will cancel. m dot equal to 40 into 10 power 3 divided by this difference. That will give you the mass flow rate of steam as 31.267 kilogram per second. This is the mass flow rate of steam. Steady state, steady flow. So, this much of steam is flowing through the turbine. Similarly, it will flow through condenser, but the phase will change. As it flows as a superheated vapor through the turbine, it becomes a saturated mixture of liquid and vapor in the exit of the turbine or in the inlet of the condenser. It becomes saturated liquid and in the pump entry and it becomes higher pressure liquid, subcooled liquid in the exit of the power inlet to the boiler, exit of the pump or inlet to the boiler. So, this is the steady flow rate, mass flow rate. So, mass flow rate will remain constant to the control volumes. So, this is done. Mass flow rate is calculated. Then we have to calculate isentropic efficiency. So, now go back to this figure. Isentropic efficiency. So, basically this means that you can see this basically H3, H4 basically are different in this case. So, basically I cannot draw this line because in the actual process it is irreversible. So, I have to draw only dashed line because I do not know exactly what happens. But on the other hand, if I draw a vertical line like this, this will be 4s. Isentropic state exists state of the turbine. That is 4s. Actual exit state of the turbine is 4. Now, I have to determine the isentropic exit state. Then calculate the work involved in this. From then I can calculate the isentropic efficiency. Do you understand? So, for that, if the turbine operation is isentropic, then exit state of turbine, exit state entropy of the turbine will be equal to this. And I know the value 6.903 kilojoule per kg Kelvin. So, now state 4s is what? p equal to p4 equal to 10 kilopascals and s 4s equal to what? 6.903 kilojoule per kg Kelvin. Now, I have taken the value of values of sf and sg and I find that s4s is less than sg because s4s 6.9 is 8.15 at 10 kilopascals, correct? So, that means the state is obviously from the figure itself you can understand. In this figure itself you can see that the quality has reduced much. State 4, quality is 0.9. State 4s, the quality is lesser than 0.9 as the figure indicates. So, I have to calculate s, the quality is 4s, that is x4s, which is what? Which is 6.903 minus 0.649 divided by 8.15 minus 0.649, ok. This is nothing but what? s4s minus sf divided by sg minus sf, ok. So, that will give the quality as 0.8337. So, you can see that the quality has still reduced in the isentropic process. But the work output will increase. So, from this I can find h4s as hf plus x4s into hg minus hf. So, which is equal to h4s will be equal to 2187 kilojoule per kg, ok. Now, what will be the isentropic efficiency of the turbine will be equal to specific work, ok, actual divided by specific work isentropic. So, which is equal to what? h3 minus h4, actual specific work divided by h3 minus h4s. So, which is equal to 3625 minus 2345.68 divided by 3625 minus 2187. So, which is equal to 0.8896 or 8089 percentage. So, you can see this isentropic efficiency that is the actual operation you get only 89 percent of the ideal or isentropic operation, the work output. So, this is done. Then the third one is the boiler heat. So, in boiler heat added, how it is equal to m dot into go back here, boiler h3 minus h2, constant pressure operation, correct. First, some heat is added to principally heat the subcooled liquid at 10 megapascals to saturated liquid. Then latent heat is supplied for the total mass flow rate, latent heat supplied and it becomes saturated vapor at 10 megapascals. Then the super heat is added that is called degree of super heat. So, it is added to make it super heat vapor. So, now, you can say this is nothing but heat added q, q dot h equal to m dot into h4 sorry h3 minus h2 s. So, which is equal to 3 1.267 into 3625 h3 minus h2 s 21201.91. So, that will be equal to 100 and 7.03 megawatts. So, now what is the thermal efficiency work of the turbine divided by q dot h? Why? Because it is asked in the problem that calculate the thermal efficiency neglecting the pump work. What is the point in calculating the pump work not required? Because pump work is very small. So, you can see this change. See for example, h3 minus h4, h3 minus h4 is basically 3625 minus 2345. It is about 1300 plus correct 1200 something. But the pump work comes to be pump delta h is only 10.11. So, that means there can be neglected. So, this w dot t that is 40 megawatts that is the power of the actual actual power of the turbine delivered by the turbine divided by this 100 and 7.03. So, the thermal efficiency comes out to be 37.37 percentage thermal efficiency. So, we have solved here. First one is isentropic efficiency. Then heat transferred to the steam in the boiler that is 100 and 7.3 megawatts, then the thermal efficiency. So, finally, I need to calculate the oil amount of oil required. So, now please understand that I have to supply this much of heat, the 107.03 megawatt heat I have to transfer. So, for this let us say m dot o is the mass flow rate of the oil into is its heat of combustion or calorific value into efficiency of heat transfer. So, that will be equal to this correct. So, I know Cv. Cv is 30 mega joules per kg and this is 0.85 correct. I know this to go to the problem. The calorific value of the Cv, calorific value Cv is 30 mega joules per kg and 80 percent of the energy released. So, what is the energy released? Per kg of oil 30 mega joules will be released. So, if m dot kg per second of oil is flow is used for heating, then m dot into 30 mega watts will be released. In that, 85 percent only is used for heating. So, that is what I have written here. m dot oil which I have to determine, Cv is 30 mega joules per kg, efficiency of heat transfer is 0.85. That should be equal to 101.03. From this, I can evaluate the value of m dot O. So, m dot O will be 4.2 kg per second. So, you have to use 4.2 kilograms of oil per second in order to heat, provide heat to produce a steam from subcooled liquid state at 10 mega pascals to superheated steam at 600 degree centigrade. So, that is the answer. So, you can see this. All the things are now determined by oil flow rate based upon the calorific value given as 30 mega pascals and 80 percent heat transfer efficiency. So, this is about the problem. We can see that the basic rank and cycle, only four control volumes are used. Lot of other additions, you can see the efficiency here is only 37 percentage. So, in order to improve efficiency, several other additions are made to the rank and cycle. But in this problem, we are only concerned about the basic rank and cycle which has only four control volumes. So, this is about the first problem.