 Now, for devices which operate adiabatically or close to adiabatic either Q dot is identically equal to 0 or Q dot is approximately equal to 0. For such devices, we can actually define a performance metric called the isentropic efficiency. Because for such devices, if you recall delta S is 1 to 2 delta Q over t plus sigma i n t. Now, if the process is adiabatic, then of course, the first term is entirely equal to 0 and in the absence of any irreversibility delta S would remain 0. So, if the process that is executed in the device is an isentropic process, then entropy change is 0 and that would indicate that the if it is a work producing process or if it is an enthalpy conversion process, it would indicate that it is operating in the best possible manner because there are no internal irreversibilities. So, an isentropic process in such a device would represent the best possible process with the actual process being worse due to internal irreversibilities. So, for example, if you look at say a nozzle or a turbine both of which expand a fluid from pressure p 1 high pressure to p 2 which is at a which is a low pressure. If the expansion process in the nozzle or turbine follows an isentropic path like this, then the conversion of the enthalpy to kinetic energy in the case of the nozzle will be the maximum possible and the conversion of enthalpy to work in the case of the turbine will be the maximum possible. In the actual case because there is an internal irreversibility, if we write this expression in differential form, you get the following because the process is still adiabatic, we get this to be 0. So, in the real case because delta sigma int is present, it is not equal to 0. Remember, in the case of internal irreversibility, this would be greater than 0. So, that means, the entropy increases during the process continuously. So, the entropy increases continuously during the process, it is a point no problem. Which means the state point starting from 1, since the entropy increases, the state point keeps shifting to the right and that is why you get the process to be like this. So, that is why you get the process to be to the right of the isotropic process because the entropy increases continuously during the process. We have indicated that it is an irreversible process by using a dash line. So, now you can see that the actual performance of the nozzle or turbine between states 1 and 2 may now be evaluated against the performance of the turbines between 1 and 2 s and this can be used to define what is called an isotropic efficiency. So, the enthalpy change between state 1 and 2 which is the actual performance h 1 minus h 2 divided by enthalpy change between the initial state and the isotropic state at the end of the process h 2 s may be used for defining an efficiency. So, the isotropic efficiency of a nozzle or turbine is defined as h 1 minus h 2 divided by h 1 minus h 2 s. Notice that it is very important to note this point that p 2 s equal to p 2. So, starting from the same initial state, the working substance is expanded to the same final pressure. In one case, it executes the actual internally irreversible process and in the other case it executes an isotropic process to the same final pressure. This is very important. So, the basis of the comparison is expansion to the same final pressure. And we may extend this concept to a diffuser or compressor where in the case of a compressor, we are putting in work. In the case of a diffuser, we are converting the kinetic energy to enthalpy. So, starting from state 1, if the diffusion process had taken place isentropically, we would have gone from 1 to 2 s to the to the final pressure same final pressure. If in the I am sorry in the actual process, we go from 1 to 2 which is at the same final pressure, but because of irreversibilities, the state point moves to the right because entropy increases and so we follow path 1, 2. So, in this case, the compressor work in the case of an isentropic process is less than the compressor work in the case of the actual process. So, the efficiency is defined as h 2 s minus h 1 divided by h 2 minus h 1. From the diffuser also, because the isentropic process is more efficient, I am sorry, because the isentropic process can convert the kinetic energy to pressure rise much better than the actual process, we write the expression in this fashion. The h 2 s minus h 1 goes to the numerator here and the h 2 s is the denominator in this case. So, let us take a closer look at these expressions and then see what they mean. So, in the case of a nozzle, we have h 1 minus h 2 divided by h 1 minus h 2 s and if we use the steady flow energy equation to rewrite this, I may write it like this. So, this basically shows the conversion of the enthalpy to kinetic energy change in kinetic energy in the case of a real process in the numerator and the isentropic process in the denominator. Since efficiency is supposed to be less than or equal to 1, equal to 1 is also alright, we know that v 2 s is greater than v 2. So, what this says is the conversion of the enthalpy to kinetic energy is better in the case of the isentropic process, so that at exit the velocity of the fluid is higher. That is what this statement says v 1 being the same and we expand to the same final pressure. Now, in the case of the turbine, again we may use steady flow energy equation to write this as the ratio of the actual work divided by the isentropic work. Since eta is less than 1, we know that the actual work is less than the isentropic work as a result of internal irreversibilities. Now, in the case of the compressor, we may write the isentropic efficiency as a ratio of the isentropic work divided by the actual work, because isentropic work in this case is less than the actual work. The isentropic compression process is more efficient in their due to the absence of internal irreversibilities. So, that is what we are saying here. Now, in the case of the diffuser, you can see that the numerator has v 1 square minus v 2 s square and the denominator has v 1 square minus v 2 square. So, this needs to be interpreted carefully. What is that? Since, eta is less than 1, v 2 s is greater than v 2. Just like here, here also v 2 s is greater than v 2, but here the interpretation is somewhat different. So, starting from the same initial velocity v 1 and with the same final pressure p 2 equal to p 2 s, an isentropic process because it is more efficient, it can actually attain the same pressure through diffusion with a smaller change in the kinetic energy, which means that there is more kinetic energy that is left in the stream, which can be further converted to a pressure rise if you want, which is why v 2 s is greater than v 2. So, with a smaller change in kinetic energy, an isentropic process can convert the kinetic energy to pressure, same final pressure starting with the same initial velocity and same initial pressure, it can reach the same final pressure with a smaller change in kinetic energy when compared to the real process. That is one way of looking at it. Other way of looking at it, let us say that we decelerate the fluid to the same velocity instead of trying to reach the same pressure, if you decelerate to the same final velocity, then what this would imply is that the pressure rise at the end of the isentropic process for deceleration to the same final velocity will be more than the pressure rise at the end of the actual process. That is another way of looking at this. So, we can either keep pressure, final pressure the same or final velocity the same. We have chosen to keep the final pressure the same as the basis of our comparison. That is what we have indicated here also. So, this is the basis of our comparison, same final pressure because initial state is the same. So, there is no dispute there. Final state is same final pressure. So, the explanation is little bit more involved in the case of the diffuser. You need to think a little bit more. Let us work out a few examples involving the concept of isentropic efficiency. Steam enters an insulated turbine operating at steady state at 60 bar 400 degree Celsius and exits at 10 bar 190 degree Celsius determine the isentropic efficiency. From the given information, we can easily infer that the steam is superheated at inlet and exit of the turbine. So, we can get H 1 and specific enthalpy, I am sorry specific entropy S 1 from the tables H 2 and from the superheated table. Now, if the expansion process is isentropic, then the exit pressure would have been the same 10 bar. We are expanding to the same exit pressure and S 2 S would be equal to S 1 because we are expanding along an isentropic process. So, S 2 S equal to S 1 equal to 6.5404. So, in the case of the isentropic process from the pressure table, we notice that state 2 S is in the mixture region. State 2 is superheated, but state 2 S is in the mixture region. So, we can evaluate X 2 S as 0.9897. So, it is just about to be saturated vapor. So, just on the two-phase region of the saturated vapor line. So, if we show the process on a T S diagram. So, let us say this is state 1. So, let us say this is state 1. The isentropic process, this is 2 S. So, remember the isobars look like this on a T S diagram. So, the actual state is superheated. So, that would look like this as an increase in entropy between 2 S and 2. So, the actual state looks like that. So, the dryness fraction at the end of the isentropic process is 0.9897. So, we may evaluate H 2 S using the data from the pressure table like this. So, the isentropic efficiency may be evaluated as 0.89. So, again you see that 2 S and 2 are at the same pressure. So, P 2 S equal to P 2, although 2 S as a consequence ends up in the being in the saturated mixture region while 2 is in the superheated region. So, the next example that we are going to look at involves a compressor or 134 A enters an adiabatic compressor a saturated vapor at minus 5 degree Celsius and leaves at 700 k power 40 degree Celsius. So, it is superheated at the exit and it is a saturated vapor at inlet. So, if you draw a T S diagram for this. So, it starts as a saturated vapor at minus 5 degree Celsius and it is superheated at the exit. So, this is state 2 and the isentropic process most likely will also end up as a superheated state at the end of the process. So, now we get H 1 equal to H g at minus 5 degree Celsius S 1 equal to S g at minus 5 degree Celsius. So, the specific entropy and specific enthalpy at state 1 are known. Now, specific enthalpy at state 2 and specific entropy at state 2 are also known because it is a superheated state. At the end of an isentropic compression process to the same pressure remember we want to go to the same pressure we have P 2 S equal to 700 kilo Pascal and specific entropy remains the same. So, S 2 S equal to S 1 equal to 0.9 3 4 5. So, this is also a superheated state as we had guessed earlier. So, from the superheated table we can retrieve H 2 S to be equal to this after interpolation. Most of the times such data may not be available directly in the table. So, we have to use interpolation or if you have software to calculate the properties you can use that also. So, we get H 2 S to be 269. So, we can evaluate the isentropic efficiency to be 0.71 in this case. So, the values that we are getting are typical of actual devices that turbine isentropic efficiencies are usually in the range of as we saw in the as we saw in the previous one 0.89. So, turbine efficiencies typically would be 0.9 to 0.95 and compressor efficiencies typically depending on the working substance could be anywhere between 0.75 to 0.9 or so. So, the last example that we are going to see on isentropic efficiency involves a diffuser that we discussed earlier in the SR 71 example. So, the isentropic efficiency of the diffuser is given and we are asked to calculate the final speed to which the air is decelerated for the same pressure rise. So, we are still keeping the exit pressure the same. So, for an isentropic process we have already calculated the values because remember in the earlier version of this problem, we had said that the air obeys P v raise to gamma equal to constant which is an isentropic process for air. So, for an ideal gas we may write H equal to C p times T and the isentropic efficiency may be written in terms of the temperatures. Since the value for isentropic efficiency is given we may calculate the exit temperature as 658 Kelvin and the final speed may be calculated in the same manner from the steady flow energy equation V 2 is equal to square root of V 1 square plus 2 C p times T 1 minus T 2 which gives us 252.45 meter per second. You may check this with what we calculated previously and make sure that V 2 is equal to comes out to be less than V 2 S. This is what we had said earlier also if you recall. So, we said V 2 should be less than V 2 S. So, you may actually go back and check the values and make sure that V 2 is indeed less than V 2 S.