 Let me come back to where we were before we dispersed. We defined the work ratio as the second parameter. Now we go into that zone where things tend to become specific. But before that let me say that although many of these parameters or particularly the efficiency is a work, sorry it is a ratio and hence should be expressed only as a fraction or at best or another option is as a percentage. Sometimes it is expressed in terms of certain units. The dimension is still one but the units may be other than one. We will come to some illustrations later. The third parameter is specific output. Now this has something to do with the flow rate of the working fluid. Flow rate or circulation rate working fluid and let us say m dot WF is the this flow rate. Then for power plant the specific output is the net output divided by m dot WF. This is for power plants and the units would be this would typically be watt. This would be kg per second. So watt second per kg or watt per kg per second or kilowatt per kg per second or megawatt per kg per second is a proper unit. But since we know that watt second is joule. So this can also be expressed in terms of say joule per kilogram. For refrigerators and heat pumps for example refrigerators this can be expressed in terms of the specific output may be called specific refrigeration effect. For example this would be called specific power output if you want. Power is the optional word. For refrigeration plant this would be q dot ref divided by m dot working fluid. Units would be the same and this may be called the specific refrigeration effect. And what is the significance of this? You should notice that the larger plants will have larger flow rates of the working fluid. However when you compare two different technologies or two different modifications to a plant it is possible that you will have a plant with two plants with a lower specific output and another plant with a higher specific output. Lower specific output would mean that given a particular flow rate of the working fluid you have a lower power output. A higher specific output means for the same working fluid you have a higher power output. The one with the higher specific output would tend to be a more compact plant. As opposite to this or related to this you can write this as 3A or 4, choice is yours. This would be the essentially its reciprocal. This will be known as the specific flow rate and this is defined as nothing but the reciprocal of our previous specific output. This will be m dot working fluid divided by w dot net. So for power plant this will be m dot working fluid divided by w dot net. For refrigeration plant this will be m dot w h divided by u dot refrigeration. Again the unit, typical unit would be kg per second per watt or kilowatt or it could be kilogram per joule or kilogram per kilo joule because watt second is a joule. Since power is quite often converted into electricity and commercial unit for the sale and transfer of electricity so called simply unit. In fact that is one funny thing. Your electricity bill will say you have consumed 320 units in the last month or last two months that unit is actually a kilowatt hour. Kilowatt hour is the so called unit of electricity transfer or electrical energy consumption. So quite often this will be in terms of kilograms per kilowatt hour where kilowatt hour we know is the so called unit of electricity, electrical power consumed or transferred. Similarly here it could be kg per second per watt or kilo joule, kilogram per joule, kilo joule, mega joule or something like that. These things are more important and more common with power plants than with refrigeration effect. Coming to power plants by the way I should tell you that this thing specific flow rate quite often is known as steam rate or specific steam rate for steam power plant and one should remember that the if this is not rule of thumb but the typical this will be approximately of the order of 1 kg per second per megawatt. This is order that means I do not claim that it will be exactly 1 kg per second but if you see a typical steam power plant which works today with condensing in temperatures around 40 degree C and boiler inlet anywhere from say 50 bar, 60 bar saturated steam to 120 bar or 550 degree C or so. It will be around 1, if you get 0.7 or 1.3 or 1.4 it is okay but if you suddenly get a value of 10 kg per second per megawatt or say 0.1 kg per second per megawatt, do check your calculations again because you are likely to have made a mistake and this is for steam power plant with the typical parameter range of parameters if you use today. Remember that it is of the order of 1 kg per second per megawatt but it is an order that means around 1 is okay but if you go far beyond 1 say something like 10 or far below 1 something like 0.1 that quite often means that you have made a mistake. So do check your calculations again. Of course you can have for example a power plant based on low temperature waste heat where perhaps you will use if at all you use steam condensing temperature will be 40 degree C boiling temperature may be just 120 or 130 degree C but that is an atypical set of parameters and in that case perhaps you will have a steam rate much higher than 1 kg per second per megawatt. So if you are very far away from this particular what they say ballpark value or typical value do check out and see to it that you have some reasons to be so off from this. Now when it comes to power plants quite often we define the heat rate. The heat rate is in principle defined as 1 over the thermal efficiency of a power plant. So that means this is q dot supplied divided by the power output. The unit should be well one dimensionless. However the typical units used are in principle the unit should be joules again joule but since this is used mainly for electrical power this will be say kilo joule or mega joule per kilo watt hour. Again that is because kilo watt hour is the typical unit of electricity transfer for electricity consumption. So this is the heat rate. Then just the way that the specific flow rate and the specific output are measures of compactness. For example specific flow rate compare two plants with the same W dot net. A plant with a lower m dot W f will have smaller component because smaller flow rate means smaller pipeline sizes, smaller pipeline diameter, smaller component sizes. So any plant with a specific flow rate or steam rate lower compared to some other plant is definitely a more compact plant. So that is why you will find large capacity plants tend to be more compact for their size because they will have a lower steam rate. Then the although these are applicable to general plant there are similarly just similar to heat rate for power plants again we have what is known as the specific fuel consumption SFC. Now we know that the energy which is supplied comes out of burning some fuel. If it is a power plant which causes fuel like petrol, diesel, coal, oil, natural gas, whatever. So the specific fuel consumption is based on the fuel flow rate and the specific fuel consumption is defined as the fuel flow rate per unit power output and the units of this would be kg per second per watt or in the commercial terms for electrical output of plant it could be kg per kilowatt hour of course this can also be replaced by kg per joule. So these are the various units which are used. Finally there are of course my list is not going to be exhaustive. So this particular parameter is defined for reciprocating engines and this is known as the mean effective pressure. For its definition it is good for us to look at the PV diagram of such a machine. We have already sketched a PV diagram earlier when we talked of the work ratio instead of going back to that. Let us say that the piston moves between two extremes say V1 and V2 reciprocates bar back and forth and let us say that the actual PV diagram is something like this. So we know that this area represents work done per cycle. We also know that the power output, this is the network done per cycle, notice that it is only the loop area. This will be equal to work done per cycle multiplied by let me say N dot cycle which is what is known as the cyclic rate. The number of cycles executed per unit time it can be cycles per second typically or hertz if you want to use the electrical electronic terminology. But you can relate it to the speed of the engine in terms of rpm and the number of strokes in a cycle. So this is the thing and now the idea of a mean effective pressure is this. Suppose I consider this work done per cycle to be made up of some uniform pressure acting on the piston as it moves from V1 to V2. So if I put this area equal to this area and force this to be a rectangle then this by definition is PME. This has the nomenclature PME or mean effective pressure MEP. So that means this has to be equal to PME into V2 minus V1 and this V2 minus V1 in IC engines terminology is known as Vs stroke volume. It is the volume set by the piston as it moves from one extreme position to the other extreme positions. So that means the work done per cycle is PME into Vs. So in terms of this the net power output turns out to be PME into Vs into n dot cycle. So this is the definition for PME. You can write PME to be the power output per cycle or the energy produced, the work done, net work delivered per cycle per unit stroke volume. That is what you can define it as. But remember that this tells you something more about the engine. First thing it tells you that if you want to increase the power of an engine you should either change the stroke volume of the engine. That means make it a bigger engine, larger volume enclosed. So this is the size relation proportional to the size of the engine or you can execute more number of cycles in an engine, more number of cycles in a given time. So that means run the engine at a higher speed. Difficult to increase Vs because that is a basic design parameter. Maybe it is difficult to increase the speed beyond a certain limit because there are mechanisms involved. There are flow rates, lubrication rates and heat transfer rates involved. So given n dot cycle, given Vs, if you want to improve the power output you have to improve the mean effective pressure. That means you have to improve the area of this enclosed diagram. So that is the significance of the mean effective pressure. At this time you should tell them that for a typical IC engine it is possible if you instrument it properly to measure and plot such diagrams. Earlier we used to have mechanical devices, now we have simply electronic sensors and the diagram is shown on a computer screen. Such a diagram when plotted is known as an indicator diagram and this power which is calculated using this diagram is quite often known as indicated power. So we can call it IP or a better thermodynamic nomenclature is W dot net I indicating that it has come out of the indicator diagram. So this is the mean effective pressure. If you want at this stage you can introduce them to indicated power, introduce them to the brake power and if you do that you should notice that it is important for us that the indicated power is P mean effective multiplied by Vs multiplied by n dot cycle where remember that this comes out of geometry of the engine, this comes out of the operation of the engine whereas this is what is measured. So this is computed. We cannot directly measure it. A similar equation can be written down for the brake power. This will be PMe based on the brake power. Multiply it by Vs multiplied by n dot cycle and again Vs comes out of geometry, n cycle comes out of operation but the further things are different. It will be the brake power which will be measured using an appropriate dynamometer and the PMe based on the brake power will be computed. At this stage you can if you want introduce them to the mechanical efficiency of such reciprocating engines for that matter for any engine. This will be W dot net delivered as measured. This is the final delivery out of the engine divided by W dot net output as we can say developed as indicated by thermodynamic analysis, indicated diagram. And in terms of these two for reciprocating engines you can show that it is a ratio of PMe based on the brake power or the PMe as indicated or as determined or as measured using the indicated diagram. You may think that many of these are not properly definable because we have not described different engines and different cycles and I have said earlier that look that is a dilemma we have to face because either we define these things in general and then go to cycles or go on defining them as we explain the cycles. Now let us spend some time on looking at the thermodynamics. You know a cycle will be made up of a number of processes and we have a choice of these processes as constant volume, constant pressure and adiabatic. These are the three typical processes which we will have. Once in a while we will have an isenthalpic process or something slightly different but these are the three typical processes we have. We also have a choice of implementing these processes in an open system or in a closed system. Cloth system would be implementable in say a cylinder piston arrangement where the valves are closed. Open system would be implemented either in a cylinder piston arrangement with a valve open, at least one valve open or in other direct open systems like ducted open systems like a turbine, compressor, combustion chamber, pipe, seat exchangers, pumps, blowers etc. Again we have a choice of the working fluid which could be a gas or a vapor and in vapor apart from heating and cooling we also have evaporation and condensation as a choice. And let us see the result of this or the import of this. If you look at the processes let say constant volume process. This can be implemented in a closed system with valves closed and piston at one place. If I fix the piston in one place and keep the valves closed then it can be implemented in a closed system. But then all that we can do is either heating or cooling and unless it happens to be an internal combustion process, external heating cooling will always be external. It is likely to be a very slow process because heat will have to be transferred across the valve. So this is not a very common process unless we have an internal heat generation in many of the engines. In an open system it is very very difficult to execute a constant volume process because in a flow system keeping the volume constant or the density constant will be a very very difficult task. And pressure process is difficult in a closed system but can be implemented in an open system because we have the model of flow in a pipe. Arrange the velocity to be small so that the pressure drop is small. In which case we essentially have an isobaric process and if you heat the fluid during that process or cool the fluid during that process that is if it is a tube in a heat exchanger or a heater or a cooler or a condenser. We have a good practical model of an almost isobaric process and that is why this we will find is a very common process which we will come across. Another advantage of this constant pressure process is we should realize that a constant pressure process implemented with a vapor in which we have a process of boiling or evaporation or condensation with delta P small essentially implies that we have a constant temperature heating or cooling. And this is important because we have our background in the second law of thermodynamics saying that you know the Carnot cycle our ideal one with the best efficiency is something with the heating and cooling under isothermal condition. And this is a situation where without significantly worrying about the technology of simultaneous expansion etc expansion and heating we can have a flow system which is getting heated or cooled. So it is and if we arrange the flow situation in such a way that the pressure is small then in principle we have a isobaric system and in that if the flowing fluid is a vapor which is changing phase that means if during heating it is evaporating or cooling it is heating condense then this constant pressure process or almost constant pressure process means an almost isothermal process which is a very attractive proposition. And you will find that this also is reasonably common in practice. So indirectly we know how to under certain conditions execute an almost constant temperature process. Then comes the next one is the adiabatic process. Well adiabatic process only means the heat transfer should be small or heat transfer should be 0 or at least negligible and this can be reasonably well executed in either cloth systems by executing a process reasonably fast in a reasonably insulated cylinder. So you know a crude adiabatic process we execute with our hand pump by which we use to pressurize the tubes in our cycle, cycle tires similarly the compression of the charge in a petrol engine in a diesel engine is a reasonably adiabatic or approximately adiabatic situation. So is the expansion in a petrol engine and part of the expansion in a diesel engine. In open systems compressors and turbines are good candidates of equipment which executes adiabatic processes. After having done this let us look at the various equipment which we have. Let us say that what are the standard equipment which we have so that we can combine equipment and processes and create cycles. First we have the cylinder piston arrangement. In this constant v will be difficult, constant v possible but with internal combustion definitely possible. Others it is possible but may become very slow. Adiabatic yes possible, approximately at least to we have something called a turbine in which we have adiabatic expansion possible. Then we have a compressor open systems. This is the only closed system everything else beyond this is open system. Compressor well we have near adiabatic then we have a turbine a heater or a boiler we have constant pressure heating process and if it is a boiler we have constant pressure and temperature. And five if we have a cooler or a condenser we will again have a constant pressure heat exchange process and if it is a condenser we will have a constant pressure and temperature. And of course then we have some other pieces of equipment which we will need as needed. For example a throttling device may be used in a vapor compression refrigeration plant. Details we will explain when we come to that. Now notice that in any cycle let us say first look at the cycle for engines. We have seen Carnot which is TSTS and we have seen the purported advantages and the obvious disadvantages. We have also looked at Ericsson TPTP and Sterling TPTP. With a heat exchange Sterling which is TV TV again with a heat exchange and we have seen the advantages and disadvantages. But the real engines will generally be not of anything of this type. And the major reason for that is the constant temperature heat addition and constant temperature heat rejection apart from some other characteristics. So what we do is the real engines where begin with gas engines and in which if you see the compare either Ericsson or Sterling with the Carnot cycle you will notice that the isothermal heat addition and heat rejection are common to all three. What we have played around with is replacing the isentropic compression and expansion processes by either isobaric processes or constant volume processes. Why not replace give up the idea of isothermal absorption and rejection and replace it by either constant volume heat absorption and rejection or constant pressure heat absorption or rejection. So first one is we try to replace TSTS which is Carnot by TSTS that is heat absorption at constant pressure, expansion as in Carnot isentropically, heat rejection at constant pressure and compression like Carnot isentropically. I am told I am back on stream. So we were discussing replacement of the TSTS which is Carnot by PSPS. And this thing leads to the basic cycle which is the Brayton cycle assuming that the working fluid is a gas. And here we have heating at constant pressure which is done in a heater the working fluid flows through a pipe and externally supplied with heat we are just getting the idea. Similarly the cooling can be done by immersing the coil or exposing it to the environment so that the heat rejection takes place. And in between you have the isentropic expansion process and the isentropic compression process which is executed in two components the turbine and the compressor. Turbine, compressor, heater the turbine produces w dot t the compressor consumes compressor power w dot c and the net power is the power output of the turbine minus the power consumed by the compressor. If I show these four states like this fluid goes from state 1 to state 2 higher pressure, pressurized by the compressor 2 to 3 isobaric heating 3 to 4 adiabatic expansion. In the ideal case it could be reversible so isentropic expansion and then cooling completing the cycle. The if you neglect the pressure draw in the heater and the cooler on a T s diagram and on a P s diagram the cycle will look like this. Notice that first we should show the P s diagram because 2 3 is an isobaric process 4 1 is also an isobaric process. So it works between an upper pressure and a lower pressure P h and P l. You have 1 to 2 assuming reversible isentropic reversible adiabatic that is isentropic expansion this is the sorry isentropic compression. So this is the compression process followed by 2 to 3 isobaric heating process followed by 3 to 4 the expansion in the turbine adiabatic assumed reversible in the ideal case so isentropic and finally 4 to 1 isobaric cooling. The corresponding T s diagram will be between 2 isobars 1 P l 1 P h and the 4 corresponding states will look like this. You can at this stage analyze the cycle and do a standard analysis of the gas turbine cycle sorry of the Brayton cycle or maybe I should use another page for this. Now the Brayton cycle which we have seen on the T s and P s diagrams uses a gas and because it uses a gas we can do what is known as a standard analysis. The standard analysis assumes the following it assumes that the gas is ideal with constant C p C v it assumes that all processes are ideal that means the gas is ideal. Wherever an isobaric process is present data p is 0 and it also assumes that whenever there is an adiabatic process it is also assumed to be reversible. So it can be replaced by an adiabatic reversible or isentropic process and because of this assumption this becomes a rectangle on the P s diagram and has such a nice shape on the T s diagram. But because of this under these assumptions if we do a standard analysis we can show that eta standard of a Brayton cycle standard means under the standard analysis assumptions will be 1 minus 1 over r p raise to gamma minus 1 by gamma where gamma is the ratio of specific heats of an ideal gas assumed constant and r p is what is known as the pressure ratio sometimes called the compression ratio for such a Brayton cycle and this is P h divided by P l and things to notice immediately are two parameters number one higher the r p better is the efficiency and number two higher is the gamma better is the efficiency and you can set up illustrative examples saying for the same r p if you use air which has a gamma which can be approximated well by the value 1.4 you will get one efficiency but instead of air if you use helium which has a gamma of 1.67 you will have a higher efficiency given the same pressure ratio r p. Now at this stage or at some other stage you can show them the actual Brayton cycle what we have seen so far is the ideal Brayton cycle remember that the ideal Brayton cycle was considered a closed cycle that is why we could consider helium as a working fluid but the disadvantage of this is that we have to supply heat from outside and we have to reject heat to the outside in a heat exchanger and the heat exchanger would become large bulky components not only costly prone to the technical problem but there will be significant heat transfer across there will be heat transfer large amount of heat transfer across temperature significant temperature differences which is not thermodynamic. The actual Brayton cycle compared to ideal Brayton cycle is an open cycle there is no cooler so one less equipment two it is an internal combustion cycle so air and fuel interact and burn react with each other inside the plant so the actual Brayton cycle plant will look like this you will have a compressor will take in air from the ambient so this arrow comes from nature the compressor compresses it to some pressure and then instead of a heater you have a combustion chamber in which fuel is supplied it is sprayed appropriate flow situation is created to have a stable flame so what goes in is air so at two also we have air but what comes out of the combustion chamber is gas that is air plus remaining air plus the products of combustion. This goes through the turbine so this turbine is given the name gas turbine and this cycle or this plant is known as the gas turbine plant or gas turbine power plant at four the gas is exhausted and the job of converting that gas back into the input air is left to nature so this is the actual implementation of a Brayton cycle plant you can sketch it on a TS diagram and if you want you can take care of some component efficiency for example you can determine what happens if the turbine does not have an isentropic efficiency of one but has a finite isentropic efficiency. Similarly you can determine what happens if the compressor has an isentropic efficiency less than one those details are there I think in some of the problems in the cycle exercise sheets but now let us look at what happens if we have Brayton cycle with vapor with a vapor as a working fluid let us go back to this PS diagram this PS diagram is good when the Brayton cycle is a gas but because the two processes are isobaric heating and cooling 2 to 3 for heating and 4 to 1 for cooling and the two processes adiabatic 3 to 4 and 1 to 2 are adiabatic or in ideal case adiabatic reversible processes this is also true this diagram when the thing is a vapor working fluid is a vapor the TS diagram is good only when the working fluid is a gas it is not correct when the working fluid is a vapor. Now let us see what would happen if the working fluid is a vapor on the TS diagram now we have to first sketch the vapor dome and the two isobars now this is the critical point the two isobars let them be shown like this this is the low pressure isobar PL and although I am not able to show it correctly here let me show the high pressure isobar now we have a choice of putting our two isentropic processes one of expansion and one of compression anywhere between the isobars we can start from here and go down we can start from here go down here can start from here go down here and start from here go down here for the expansion and we can have the compression either here or here or here or here if you notice that if I put the compression here and the expansion here creating this rectangle purely in the two phase zone as the rectangle representing my vapor Brayton cycle then you will realize that we have created an essentially Carnot cycle because although this process is an isobaric process and this process is also an isobaric process because during this process the pressure drop is neglected either boiling or condensation is occurring the isobaric process is also an isothermal process so inadvertently or by some trick we have created we have implemented a PSPS cycle but ended up being executing indirectly a TSTS cycle which is a very attractive thing now we realize that I can have the expansion process beginning from dry saturated steam and the compression process ending at saturated liquid we had I am just showing the earlier processes here I can go as far as this for my expansion process and as far as this for my compression process so my Carnot cycle the largest Carnot cycle which I can fit between a pH and PL is at shown here so I will show it as 1, 2, 3, 4 now when we implement this in actual practice what happens is we have no issue with the processes 2 to 3, 3 to 4 and 4 to 1 we can have a boiler which does this exactly for us 2 to 3 a turbine which does 3 to 4 and a condenser which does part of the condensation which is from 4 to 1 however when it comes to 1 to 2 what we are doing is we are compressing a two phase mixture of steam and air and after compression it is dry saturated sorry it is saturated liquid at a higher pressure. If you take some typical values for pH and PL we will find out that the compressor or the pump which we will be using we will need to tackle a extremely large reduction in volume of the working flow because one will have a reasonable amount of vapor in it which has a very large volume whereas at 2 everything is in a liquid state so the volume occupied is small. Such a pump or such a compressor is very difficult to design and even if it is designed we will have a very poor life in actual operation and hence what is done in practice is this process is removed instead of this 1 to 2 we modify the pump and when it comes to condensing the vapor we go right up to the saturated liquid state at the lower pressure use a pump to pump it to the higher pressure and then complete the cycle. Now because of this notice that the isobaric heating process is not fully isothermal it is isothermal from 2 to say 3 prime and then 3 prime to 3 it is isobaric 2 to 3 prime to 3 but it is isothermal only from 3 prime to 3 this. This earlier part 2 to 3 prime is heating at constant pressure of sub cooled water which will not be an isobaric process. If you look at the cycle this cycle is known as the Rankine cycle but it is nothing but the Brayton cycle with a vapor as a working fluid with the condition that at the end of the cooling process which is a condensation process we bring the liquid to the saturated liquid state we bring the condensate to the saturated liquid state then pump it up to the boiler pressure 2 to 3 prime is isobaric heating of sub cooled liquid 3 prime is saturated liquid 3 to 3 3 prime to 3 is evaporation. Now it turns out that since the critical point is only something like 375 degree C we still have the capacity to increase the temperature of the steam further along this line and it is very common for us to replace this process by a process at a higher entropy so you keep on heating it till you go to state 3 which is in the superheated zone making it still further away from the corresponding Carnot cycle expand it in a turbine. So quite often the Rankine cycle and this is a very common occurrence now looks like this so the isobaric heating process is partly 2 to 3 prime which is sub cooled heating up to saturated liquid state 3 prime to 3 double prime which is isobaric and hence almost isothermal heating to dry saturated steam and then 3 prime 3 double prime to 3 which is some amount of superheating so 3 is superheated steam. The advantage of superheating we need not emphasize this in a course in thermodynamics is to even show to them that apart from improving the thermal parameters like efficiency and steam rate it also increases the dryness fraction at the exit of the turbine which is good for the turbine from a technical point of view. The schematic for the block diagram for the Rankine cycle plant looks like this we have a boiler, we have a turbine, we have a condenser and we have a pump. State 1 is exit of condenser 1 to 2 is the pumping process the pump water goes into the boiler goes from 2 to 3 prime 3 to 3 double prime and then to 3 which is usually superheated where it is ducted to the turbine in the turbine it expands to 4 at which stage it goes into the condenser comes out as a condensate as 1. The energy interactions are heat supplied by combustion of fuel in the boiler or say Q dot B heat rejected in the condenser where cooling water is made available to extract the heat that will be Q dot C turbine power output W dot T pump power input W dot T. We can use our first law tricks to analyze all these and relate it to M dot which is the water or steam circulation rate and you will get W dot net equal to W dot T minus W dot C which should also be equal to W dot T minus Q dot C the efficiency would be cycle efficiency would be W dot net divided by Q dot B using the second form you can write it if you feel like as 1 minus Q dot C by Q dot B the work ratio if you want will be W dot C divided by W dot E and the specific output can be written down in terms of W M dot and W dot net. It is almost 1245 not only I am a bit exhausted by Q for 3 and half hours but I think up here now.