 The first example that we are looking at is that of a mixing chamber. This was actually discussed in the previous course where we applied the steady flow energy equation to this device and then derived an expression for the ratio of the mass flow rates m dot 3 divided by m dot 2 and m dot 3 divided by m dot 1. So, that has already been done. So, now we want to calculate the rate of entropy generation in the mixing chamber. So, we want to calculate sigma dot i n t for this device. So, control volume is shown. This is our control volume. So, we apply our previous equation. Notice that since it is since the mixing chamber is insulated delta q naught is 0. So, the integral drops out. So, we are left with an expression like this. So, we can retrieve the following property values from the table. A compressed liquid state we simply approximate as S f of t comma p is simply approximated as the specific entropy of the saturated liquid at the same temperature. So, if you plug in the values, notice that we divide both sides by m dot and if you and m dot is nothing but. So, we divide through by m dot and these values have been derived in the in the previous course. So, if you substitute the values, you end up with sigma dot int divided by m dot to be equal to 0.2452. It is a positive number. It must be a positive number. Remember, sigma dot i n t is greater than or equal to 0. That is a good consistency check for our calculations. And the entropy generation in this case is entirely due to internal irreversibility from mixing. Remember, we said sigma dot int could be non-zero because of friction, dissipation, mixing and so on. So, in this case it is due to mixing. The next example that we are going to look at is this. Saturated liquid R134A at 30 degrees Celsius is throttled to 200 kilopascal inside a domestic refrigerator. So, the exit pressure is given 200 kilopascal is the exit pressure. Determine the rate of entropy generation neglecting any heat loss. KE and PE changes may also be neglected. So, this is our control volume. And we had shown earlier that for this process, if you neglect KE and PE changes and provided the change in specific volume is not very high, we can actually show that H2 is equal to H1. It is very important to remember that this is true only if the change in specific volume is not very large. So, we can retrieve these values from the property table. Now, the dryness fraction at the exit of the throttling valve may also be evaluated like this. And application of entropy balance to the control volume looks like this. The integral drops out because the device is insulated. So, if you substitute the values, we have retrieved S1 and S2 and we are doing the calculation on a per unit mass flow rate basis. So, if you do that, then we get sigma dot int over m dot to be equal to 0.0163 kilojoule per kg Kelvin. Once again, this is positive due to internal irreversibility associated with throttling the fluid. So, as the fluid flows through these small holes, it is throttled and it is a highly irreversible process. So, the entropy generation is due to throttling of the fluid in the small pores. The next example that we are going to look at is this. Calculate the rate of entropy generation inside the heat exchanger shown below. So, this is the heat exchanger and this was also analyzed in the previous course. So, we took this to be an intercooler in between two compressive stages. So, basically air came into a compressor at atmospheric pressure and was compressed to this pressure in a compressor. It was then taken to an intercooler which is this heat exchanger here, cooled to 300 Kelvin. As you can see here, pressure remains the same. So, it was cooled to 300 Kelvin before being taken to the next stage of compression. And the air stream was cooled using a water stream which came in at 30 degree Celsius and exited at 60 degree Celsius. So, we calculated the mass flow rate of water that would be required to accomplish this cooling and all these values are shown here. Now, we are asked to calculate the rate of entropy generation inside this heat exchanger. So, for steady state operation, the entropy balance equation becomes something like this m dot water times S x minus S y plus m dot air times S 2 minus S 3 plus sigma dot i n t for the control volume that we have shown here. Notice that the control volume includes both the air and the water stream. It is also given that there is no heat loss from the heat exchanger to the surroundings. So, the integral term drops out in the entropy balance equation. So, for water, we can retrieve the specific entropy values from the steam table. So, we simply approximate this S x to be S f at 30 degree Celsius and S y is approximated to be S f at 60 degree Celsius. Now, for the air stream, delta s may be evaluated from the TDS relations. Pressure remains constant. So, you may recall that from the TDS relations, you may write delta S for air equal to C p times natural log T 2 over T 1, where 2 and 1 denote the final and the initial state minus or natural log P 2 over P 1. Since there is no change in pressure here, that term drops out. So, we are left only with this term. And as you can see, because the air is cooled from 417 to 300 Kelvin, the specific entropy of the air decreases as this negative sign shows. And once again, that is a consistency check. So, now if I calculate sigma dot int for the entire heat exchanger, I get it to be a positive value as I should equal to 0.743 watt per Kelvin. Notice that the specific entropy of the air stream decreases, but the specific entropy of the water stream increases because it is being heated. So, now the sigma dot int in this case is due to internal irreversibility associated with heat transfer across a finite temperature difference. Notice that the water stream has a varying temperature from inlet to outlet, which is different from that of the air stream, which also has a varying temperature from inlet to outlet. So, heat transfer across this finite temperature difference is what is causing entropy generation due to internal irreversibility. So, now you can realize that this equation that we derived is useful specifically for calculating sigma dot int for certain special cases, where the device is either insulated or it operates at a constant temperature. So, knowing the inlet and the exit states, this may be evaluated. So, we can actually evaluate sigma dot int using this entropy balance equation. There is no closed form for evaluating sigma dot int, it has to be calculated in an indirect manner like this. So, the next example that we are going to look at involves steam turbine. So, steam enters an insulated turbine, so Q dot equal to 0, steam enters an insulated turbine operating at steady state at 60 bar 400 degree Celsius and exits at 10 bar 190 degree Celsius. So, the steam as you can see based on the property values given, the steam is superheated at the inlet and superheated at the exit, determine the power developed and rate of entropy generation if the mass flow rate is 10 kg per second, K e and P e changes may be neglected. So, we apply steady flow energy equation to the steam turbine. So, basically, so this is what the situation looks like. So, we apply steady flow energy equation, drop all the other terms and we end up with something like this. So, if you retrieve the values for H 1 and H 2 from the tables, we can evaluate W x dot to be 3751.5 kilowatts for the given mass flow rate of 10 kg per second. Now, entropy balance equation for this case becomes after setting the integral equal to 0, we get this. So, rate of entropy production may be evaluated as 968 watt per Kelvin. One thing that we need to keep in mind here is that this entropy generation is due to internal irreversibility arising from friction, both mechanical friction as well as fluid friction or viscous effects. So, this is the cause for generation of entropy internally within this device. Now, you may recall that we wrote down two expressions at the beginning of this lecture. So, one was change in entropy of the system which involved sigma int, another one was change in entropy of the universe which involved sigma. So, change in entropy of the universe which involved sigma. So, what we are going to do now is develop an expression for calculating change in entropy of the universe as a result of a flow process. So, what we will do is start from the same control volume that we had before. So, we start with the same control volume and we use the same system as we did before. Only thing is now interaction with the surroundings also has to be broad inside. So, for this system if I write, so for this system I may write the change in entropy of the system between time instant t plus delta t and t, t like this. So, the system itself is comprised of two pieces at time t plus delta t, one within the control volume, one which is just about to leave. Similarly, at time t the system comprises of two pieces, one within the control volume, one within which is just about to enter. Now, during this time interval the change in entropy of the surroundings is given simply as delta q surrounding over t naught. Notice that there is no integral here because the surroundings are the temperature t naught and there is a heat interaction to the amount of delta q between the system and the surroundings. So, the change in entropy of the surroundings is simply delta q surroundings over t naught. So, the change in entropy of the universe during this time interval is ds system plus ds surroundings. So, delta sigma equal to ds system plus ds surroundings which is equal to this. Please bear in mind that delta q surroundings is equal to minus delta q for the system. So, if I rearrange this expression and take the divide by delta t and take the limit as delta t goes to 0, then I end up with the following expression for rate of entropy generation in the universe as a result of a flow process. So, this takes into account both entropy generation due to both sigma int I am sorry due to both internal irreversibility as well as external irreversibility. So, this is the rate of change of entropy within the universe itself. And as we mentioned earlier, this is a very very important performance metric for the device or the process that working substance executes inside the device. So, let us work out a few examples involving this concept. So, refrigerant 134A is compressed steadily in a compressor from 140 kPa minus 10 degree Celsius to 700 kPa 120 degree Celsius. Volumetric flow rate at the inlet is given, input power is also given. We are asked to calculate the rate of entropy generation. Ambient is assumed to be at 27 degree Celsius. So, if we apply steady flow energy equation to this, we end up with this expression. Notice that it is not explicitly mentioned in the problem statement that the device is insulated, which means that there is going to be some heat interaction between the device and the surrounding, the compressor and the surroundings. So, if we retrieve the property values from the table, we get H1 specific enthalpy, specific entropy at the inlet, specific volume at the inlet is required because we are given volumetric flow rate at the inlet. So, the mass flow rate at the inlet is V1 dot divided by the specific volume at the inlet and that works out to 0.006845 kg per second. So, if you plug in the values, we get the heat interaction to be minus 0.23 kilowatts, which means heat is being lost from the compressor to the surroundings. Again, when you look at this problem statement, you can see why we mentioned that an integral would be required for calculating the heat transfer in the case of just the device. You may recall that we wrote this term integral CS delta Q over TB, that is illustrated very nicely in this particular example. Even in the other examples, it is well illustrated. For example, in this case, if I sketch the compressor, so this is the inlet to the compressor and notice that the temperature at the inlet to the compressor is minus 10 degree Celsius. This is our control volume and the temperature at the exit is 120 degree Celsius. Ambient itself is at 27 degree Celsius. So, you can see from here that there will be heat transfer from the ambient to the compressor because the fluid is at a much lower temperature than the ambient. So, there will be heat transfer from the ambient to the control volume or the device in say this part of the device until its temperature increases to the ambient temperature beyond which when the temperature has become very high, heat transfer is from the device to the ambient. So, heat transfer is from the device to the ambient in some part of the device and heat transfer is from the ambient to the device in certain other parts, which is why the integral is required when we want to evaluate the term for the device itself. But as far as the surroundings are concerned, the net heat interaction is a certain amount and the surroundings are at the same temperature. So, if you want to calculate this for the device, then it becomes very difficult to do unless the device is insulated or is at a constant temperature. So, you can see why that particular integral is required from this. So, if I want to calculate integral, I mean if I want to calculate delta Q over T for this device, I have to do this integral along the control surface. So, I have to do delta Q over the bound T B, which is the boundary temperature across the entire control surface. So, basically I have to so, I have to integrate this across the entire control surface because the heat transfer magnitude as well as sign is different in different parts of the control surface. So, that is why this integral is required in that case. But in the case of the surroundings, it is simple because the surrounding is at the same temperature. So, if you plug in the values, then we get the heat transfer to be minus 0.23 kilowatts and overall there is a loss of heat from the device to the surroundings. So, rate of entropy generation in the universe may be evaluated using this expression. We substitute the numbers and remember temperature has to be in Kelvin. So, we substitute the numbers, we get this to be 2.27 watt per Kelvin, positive as it should be and this is the rate that the entropy is generated in the universe due to both internal as well as external irreversibility. External irreversibility in this case is heat transfer across a finite temperature difference between the device and the surroundings. Internal irreversibility is due to friction, both mechanical as well as fluid friction inside the device. So, what we get here is the sum of both and once again we see the usefulness of the expression that we derived. Although the integral itself cannot be evaluated, but in this case, the integral drops out of the equation because we are looking at entropy change. So, the integral that we had earlier in the entropy balance equation drops out in the expression for rate of entropy generation of the universe because we are looking at the universe system plus surrounding. So, that drops out. Next example, steam enters a turbine steadily at 30 bar 440 degree Celsius at the rate of 8 kg per second exits at 3 bar 160 degree Celsius. So, it is superheated at the inlet, superheated at the exit. Heat is lost to surroundings at 27 degree Celsius at the rate of 300 kilowatts. This has been given to us. Ke and Pe changes are negligible, determine the actual power output and rate at which entropy is generated. So, steady flow energy equation for this device comes out to be like this. We can retrieve these property values from the table and if you plug in the numerical values, we get Wx dot power generated to be 4015.2 kilowatts. Notice that Q dot is not equal to 0 and that is said to be minus 300 because heat is lost to the surroundings. Rate of entropy generation in the universe may simply be evaluated like this for a steady flow process. Time derivative drops out. So, for a steady flow process, we end up with this expression. After substituting the values, you can see that we get this to be again a positive number as it should be. That is the rate at which entropy is generated in the universe. I think this is the last example. No, we have one more example. So, what we will do is discuss this and the next example in the next lecture.