 Let us start with the first example, 2 kg of ice at minus 15 degree Celsius is dropped into 10 kg of liquid water initially at 27 degree Celsius in an insulator container. Simultaneously, 3500 kilo joules of work is transferred to the container by means of a paddle build. So, the situation looks like this, so the vessel is rigid vessel is insulated, so at T equal to 0, it contains water plus some amount of ice at T equal to 0, so this is ice, 2 kg of ice at minus 15 degree Celsius and 10 kg of water. So, by means of a paddle build we are also transferring some amount of work which is 3500 kilo joules, we are asked to calculate the final temperature and the entropy change. So our system consists of ice plus water and paddle work crosses the system boundary, so if you apply first law delta E equal to delta U no ke or pe change equal to q minus w, q is 0 because the vessel is insulated, w is simply paddle work, displacement work is 0 because the system boundary does not deform, so w is displacement work because it is transferred to the system, it is actually equal to minus 3500, so this becomes plus then, so the change in internal energy of the system is the sum of the change in internal energy of ice plus internal energy change in internal energy of water. Now we will assume that all the ice melts which is a reasonable assumption, so assuming that all the ice melts, so the ice goes from minus 15 degree Celsius to ice at 0 degree Celsius then undergoes a phase change and then reaches some final temperature T2, so the water is initially at 27 degree Celsius, so it starts from water at 27 degree Celsius then reaches the same final temperature T2, so if you plug in the numerical values we get the final temperature to be 77 degree Celsius. Now entropy change for this process has to be calculated like this, so this is entropy change for going from minus 10 degree, minus 15 degree Celsius to 0 degree Celsius, ice at minus 15 degree Celsius to ice at 0 degree Celsius remember it is a solid, so we use this expression. So this is minus 15 degree Celsius and this is ice at 0 degree Celsius, so that is what we have done, remember all the temperatures have to be in Kelvin, so this is the entropy change for that phase and the phase change takes place at constant temperature, takes place at constant temperature, so that is simply equal to the latent heat divided by the temperature q over T that is simply delta s equal to q over T, so that is what we have done here and then the mass of ice goes from ice at 0 degree Celsius, I am sorry so here it goes from ice at 0 degree Celsius to water at 0 degree Celsius then we go from water at 0 degree Celsius to water at the final temperature of 77 and since it is incompressible we use again this expression, but now we go from water at 0 degree Celsius to water at 77 degree Celsius. So if you do this, you do the math then you get the delta s for ice to be 4.7634 and the delta s for water again it is incompressible, so water at 27 degree Celsius to water at 77 degree Celsius works out to something like this, so the total entropy change of the system is 11.2377, vessel is insulated, so if I write our original expression, so that is very useful for interpretation, so 1 to 2 delta q over T for the system plus sigma i n T, vessel is insulated, so this term is 0, so the entropy change of the system in this case is entirely due to internal irreversibility, there are 2 internal irreversibilities here, one is mixing of the ice with the water, the second one is stirring work, that is why delta s comes out to be positive, it is entirely due to internal irreversibilities. Let us look at one more example, an insulated box is divided into 2 compartments A and B by a negligibly thin partition, so this is the insulated box, this is A, this is B, so there is a thin partition in between, so compartment A initially contains n 2 and compartment B initially contains h 2, so this is the initial state. In the final state it contains n 2 plus h 2, that is the final state, so we take the box as the system, so we can see that for this system there is no transfer of heat because the box is insulated, there is no transfer of work because the system boundary does not deform at all and there is no other form of work that is being transferred. So, application of first slot to the system gives delta E equal to delta U, no KEPE change equal to Q minus W, Q is 0, W is 0 and delta U itself is made up of internal energy is a sum of internal energy change of n 2 plus internal energy change of h 2, which we can write like this as taking the final temperature to be T 2, so if you substitute the numbers we get the final temperature to be 353 Kelvin. So, initial volume occupied by each of the gases may be evaluated, so the total volume of the box is equal to 27.9116 that is the sum of the volumes of the hydrogen and nitrogen in the initial state, so this is the final volume that is actually occupied by the gases. We are using Dalton's model here which means the mixture or each component in the mixture is at the same temperature as the mixture occupies the entire volume as the mixture, but is at a separate partial pressure. So, the partial pressure of hydrogen in the final state may be calculated by using equation of state like this. So, hydrogen exists at the same temperature as the mixture temperature and occupies the same volume as the mixture, so we can evaluate partial pressure of hydrogen and partial pressure of nitrogen. So, the change in entropy of hydrogen may be evaluated as we said before in this fashion, remember we wrote it down, we can actually calculate it as a sum or calculate individual values, so if you remember this is what we wrote, so what we are doing now is calculating each one of the term in this summation, we are calculating for hydrogen, nitrogen separately with the understanding that these are the partial pressures of hydrogen and nitrogen as appropriate. So, if you do that you get delta S of H2 to be this, delta S of N2 to be this and delta S for the system comes out to be 8.3858 and once again we may write this expression delta S for the system is 1 to 2 delta Q over T plus sigma INT, delta Q is 0 because the box is insulated, so the change in entropy in this case is entirely again due to internal irreversibility, but in this case the internal irreversibility itself is only due to mixing, there is no work transfer unlike in the previous case, so this is the result of mixing or increase in entropy which results from mixing which is highly irreversible. So you can see how we can use this expression and then this expression which comes from TDS relationship, this comes from TDS relationship, so we can use both the expressions, so this expression may be used to calculate the actual value and then the other expression may be used to evaluate or interpret the values and then identify where the entropy change is coming from. Had there been heat transfer then we would have said that the entropy change is partly due to heat transfer and perhaps partly due to internal irreversibility. So this expression is very useful for interpreting the value, the TDS relations are used for calculating TDS relation or the tabulated data or calculated for or used for calculating the actual value. So we will work out a few more examples in the next lecture. The next example we are going to look at reads like this, rigid insulated vessel of volume 1.5 meter cube, so let us see what this vessel looks like, we have a rigid insulated vessel like this, volume is given it initially contains water at 20 bar and 50 percent drainage fraction, a valve on the top of the vessel is now opened and the content of the vessel is allowed to escape slowly. So basically we have a valve here and so this contains liquid and saturated vapor, so the valve is opened and the saturated vapor is allowed to escape slowly until the pressure reduces to 200 kilo Pascal at which point the valve is closed. We are asked to determine the final state of the contents of the vessel and the mass that escapes. So we have seen this sort of problem earlier and where we have used a control volume analysis for, so this was the control volume that we used. So let us see how we proceed with this particular version of this problem. So the initial state we have 20 bar drainage fraction is given, so we know that it is a two phase mixture, so we can retrieve these values from the pressure table V f, V g, S f and S g, note that we need S f and S g also and based on these values we can calculate the specific volume at state 1 and the specific entropy at state 1. Total volume of the vessel is given and since the specific volume is known we can evaluate the mass that was initially in the vessel as 29.762 kilograms. Now when you look at the problem description we need two property values to fix the final state of the system. We know only one property value which is 200 kilo Pascal, the final pressure is 200 kilo Pascal, nothing more is given. So we now look at the process and see whether we can infer something from there. So the content of the vessel meaning the saturated vapor is allowed to escape slowly which means that we can say that the process is a fully resisted process and so the contents of the vessel have gone through a fully resisted process. So it is insulated and the vapor escapes slowly, we may assume that the contents of the vessel, the contents of the vessel to have undergone an isentropic process which changes to contents of the vessel to have undergone an isentropic process which means that S 2 is equal to S 1, notice that this is the specific entropy. One common difficulty that students have with this particular problem is that they take capital S 2 to be equal to capital S 1 which is not correct because the mass changes during the process, the mass of the vapor plus liquid in the vessel changes during the process. So it is not capital S 2 equal to capital S 1 but specific entropy at the final state being equal to specific entropy at the initial state and remaining constant throughout. So now at the final state we have two property values namely pressure and specific entropy. So we can retrieve from the pressure table S f and S g then evaluate the final dryness fraction as 0.515. So the final state continues to be a saturated mixture state but with a slightly higher dryness fraction. So the most important variation to this problem from what we did before is that now given the problem description or based on the problem description we infer that the contents have undergone an isentropic process which means that S 2 equal to S 1 equal to S and based on that we can then proceed with solving the problem. So we have obtained X 2 now the final specific volume may also be evaluated we can retrieve V f 2 and V g 2 from the tables and once we have that dryness fraction is known so we can evaluate the final specific volume to be this and the mass that remains in the vessel may then be evaluated using the total volume. So total volume divided by V 2 gives me the mass that remains in the vessel. So the mass that escapes is that the initial mass minus the final mass so 26.458 kilograms escape during this process. So this is an important point you must bear in mind that the specific entropy remains constant during this process. So S 2 equal to S 1 equal to S throughout not only is the specific entropy the same between the initial and the final state it also remains same throughout. The next example is also familiar to us we have seen this example with air as the working substance before and now we have steam as the working substance. So steam is contained in a rigid insulated rigid tank of volume 2 meter cube initially at a pressure and temperature of 15 bar and 420 degree Celsius. So we can very quickly understand that or realize that the steam is superheated at the initial state the steam is slowly discharged through an insulated turbine into the atmosphere at 100 k power. It may be assumed that the steam is always expanded to the atmospheric pressure the mass of steam in the turbine at any instant and its energy may be neglected determine the work developed by the turbine. So for the steam that is initially in the tank remember it is superheated so we can retrieve the following property values from the superheated table specific volume specific internal energy and specific entropy. So the mass of the steam that is initially in the tank since we know the volume this may be evaluated as 9.5465 kilograms and just like what we did before we take the tank and the turbine together as the control volume and we simplify the unsteady energy and mass balance equation keeping in mind that q dot is 0 mi dot is 0 and we neglect the K e and P e terms steam after expansion to 100 kilo Pascal leaves the control volume. So me dot is not equal to 0 so we can write the unsteady energy equation like this unsteady mass balance equation like this and if we combine these two then this is the equation that we end up with. Now just like what we did in the earlier problem or just like what we did in the previous problem and just like what we did in the earlier problem with air as the working substance we assume that the steam has undergone an isentropic expansion process. So hence a specific entropy of the steam in the tank at any instant denoted as S and the specific entropy of the steam at the turbine exit denoted as S e or both equal to S 1. So in other words the specific entropy of the steam in the tank at any instant S is equal to S 1 the steam that is in the tank is assumed to have undergone an isentropic expansion so S equal to S 1. Now the steam that leaves the tank and goes through the turbine and is expanded up to the atmospheric pressure is also assumed to have undergone an isentropic process which means that S e which is the specific entropy of the steam at the exit of the turbine is also equal to S 1. So S e equal to S e equal to S 1 now when the process is completed the steam that remains in the tank based on this obviously based on this relationship would also have the same specific entropy so S 2 is also equal to S 1. So it is very important that you understand what we are doing here. So the steam that remains in the tank is assumed to have undergone an isentropic expansion so S 2 equal to S 1 at any instant during the process the steam in the tank that remains in the tank is assumed to have undergone an isentropic expansion which means S equal to S 1 and the steam that leaves the tank expands through the turbine and exits at 100 k power is also assumed to have undergone an isentropic expansion so S e equal to S 1. So at the exit of the turbine we have two property values pressure is 100 k power and specific entropy is also known S e is equal to S 1. So we can calculate the dryness fraction and exit to be 0.9955 and the specific enthalpy at the exit to be equal to this. Notice that the specific enthalpy of the steam that leaves the turbine or leaves the control volume remains a constant. So the specific enthalpy at the exit to the control volume is a constant we can actually integrate the energy equation and get the work output from the turbine like this. Also bear in mind that at the final state the steam in the tank is at the pressure of 100 kilopascal with the specific entropy S 2 equal to S 1. So with specific entropy S 2 equal to S 1 and the final pressure is also 100 kilopascal which means we can calculate the specific volume and the specific internal energy and the mass that finally remains in the tank may also be evaluated as 1.1847 kilograms. So the key to solving this problem is recognizing that the steam is undergoing an isentropic expansion process which allows us to calculate property values but it also allows us to integrate this equation because H e becomes a constant during the process.