 The next application that we are going to look at is a diffuser. As I mentioned earlier the diffuser is an exact opposite of a nozzle. So in a nozzle basically the high enthalpy of the fluid at the inlet is converted to kinetic energy of the fluid at the exit or the enthalpy change across the nozzle is converted to an increase in kinetic energy. Now in the case of the diffuser the velocity of the fluid as it enters the diffuser usually is very high ok. So V is high. So what we would like to do is convert this momentum into static pressure or pressure of the fluid at the exit of the diffuser ok. So we would like the pressure to be high at the exit of the diffuser. What is that normally fluid will not flow from low pressure to high pressure. In this case because the velocity at the inlet is high we are forcing it to flow through the diffuser where the pressure at the exit is higher and that is possible because the momentum of the fluid is being converted to pressure. So the fluid decelerates as it flows through the as it flows through the diffuser ok. So here you can say that the velocity of the fluid at the inlet high velocity is converted to pressure at the exit of the diffuser or you can say the kinetic energy of the fluid is converted to an increase in enthalpy. Remember enthalpy of a fluid is u plus pv. So if the pressure increases for an ideal gas for instance if the pressure increases then other quantities also increase commensurately so the enthalpy increases. So the kinetic energy is converted to enthalpy in the case of a compressible fluid. But a more practical way of looking at it is to say that the high velocity is converted to the high momentum of the fluid is converted to high pressure. So what this basically allows you to do is that you know you can actually compress a fluid by a finite amount without using a compressor ok. Normally in order to compress a fluid you may use a compressor right we have to use a compressor. Now in this case we have managed to achieve compression of a fluid without a compressor without any moving parts simply by utilizing the momentum the high momentum of the incoming fluid. For this to work the important thing is that the velocity at the inlet should be very high or the momentum at the inlet should be very high ok. Now this is practically used in the case of supersonic missiles and supersonic aircraft such as the Concorde or the SR 71 or any other fighter aircraft they all have intakes where the air in a frame of reference where the aircraft is stationary the air seems to approach the intake at supersonic speed which is very very high. So this may be utilized to compress the air at the end of the diffuser or inlet or intake as it is called ok. Since the compression is achieved without any moving parts it can actually be a result in a lot of weight saving for the engine itself ok. So it is very extensively used in the intake of engines of aircrafts and missiles that normally fly at supersonic speeds ok. Let us work out an example involving a diffuser. So while cruising at an altitude of 24,000 meters and flight speed of 972 meter per second which is 3.2 times the local speed of sound. Notice that normal aircraft a commercial aircraft would usually fly at an altitude of 10,000 meters. So this is about 2 and a half times that much ok 10,000 meters or about 33,000 feet. So 24,000 meters works out roughly to about to an altitude of about 100,000 feet ok. So SR 71 cruises at this altitude and this speed which just out of interest is 3.2 times the local speed of sound ok. So the diffuser in the engine captures the outside ambient air which is at a pressure of 2.55 kilo Pascal which as you can imagine is very small and a temperature of minus 51 degree Celsius. So we used to accomplish a pressure ratio of 39 that means the pressure at the exit of the diffuser divided by the pressure at the inlet is 39 ok. So we assume steady state operation, neglect, pH changes and any heat loss and we are asked to calculate the final temperature assuming that the compression process obeys PV raise to constant and also the final speed of the air as it comes out of the diffuser. So for air we know gamma to be 775ths and Cp to be 1010.38 joule per kg Kelvin. Since the process obeys PV raise to gamma equal to constant we can easily evaluate the temperature to be at the exit to be 632 Kelvin. So the temperature goes from minus 51 degree Celsius to something like this. So we apply SFE to the control volume that was shown in the illustration and if you use the fact that for an ideal gas H is equal to Cp times, write H equal to Cp times T we can simplify this expression and then arrive at this and calculate the exit velocity of the air to be 341 meter per second. So the air has been decelerated from a speed of 972 meter per second to a speed of 341 meter per second approximately one-thirds in order to achieve a compression ratio of 39. Notice that a compression ratio of 39 is quite high. Typically the sort of compression ratio that is achieved in using a compressor in an aircraft engine ok. It is a very high compression ratio. This illustrates that by decelerating the air we can actually accomplish compression like this. But in an actual SR 71 engine of course you know we cannot this is a ideal situation the compression is not accomplished exactly in the same manner because this sort of compression is difficult to accomplish you require a very long diffuser which adds to the weight of the engine. So it is done slightly differently but the basic idea is the same that the high momentum of the intake air or air coming into the engine is converted to pressurize across the diffuser. So the next application that we look at is that of a mixing chamber. So basically a mixing chamber as the name suggests mixes different streams and sends out perhaps one or two streams on the exit. That may take two streams or more than that at the inlet and send out one or maybe more than that at the exit. So here we have a we have a mixing chamber where steam at 10 bar 460 degree Celsius which is superheated comes in and we also have liquid water at 45 degree Celsius and 10 bar these are mixed together and what we want from the mixing chamber what we want to do is adjust the mass flow rates of these two so that we have saturated liquid at 10 bar exiting the mixing chamber that is the design requirement. So we neglect KE and PE changes assume steady operation and note that this is also insulated so Q dot is equal to 0. So if we try to show the states on a PV diagram let us go to the pressure table. So corresponding to 10 bar the you can see that corresponding to 10 bar the saturation temperature is about 80 degree Celsius. So let us say this is 10 bar. So the saturation temperature is 80 degree Celsius let us show the isotherm using a different color 80 I am sorry this is 180 degree Celsius I am sorry. So this is 180 degree Celsius. So 10 bar 460 degree Celsius the state at 1 will then lie here so this is state 1 that is a superheated state because this temperature is higher than the saturation temperature corresponding to 10 bar and this one is a compressed or subcool liquid because this temperature is less than the saturation temperature corresponding to 10 bar so that means state 2 will lie here and state 3 is saturated liquid at 10 bar. So these are the states of the incoming and exiting streams. So if you apply SFE to the control volume that is shown all the velocity terms and the elevation terms drop out because we have been told that they are negligible Q dot is also equal to 0 because it is the mixing chamber is insulated obviously there is no work interaction so Wx dot is also 0. One thing that you must remember when we say that work interaction is 0 is that we mean that Wx dot is 0 in the steady flow energy equation flow work has already been accounted for in the enthalpy term. So this is a doubt that students have all the time whether we have when I say Wx dot equal to 0 it means Wx dot equal to 0. Remember flow work also has to be accounted for and that has been accounted for in the enthalpy term. So if you rearrange this we get the ratio of the incoming mass flow rates m dot 1 over m dot 2 to be equal to this. H1 is a superheated state so we can directly retrieve the value from the property table no difficulties there but H2 is a compressed liquid state and if you go back and review the approximation procedure that we laid out for calculating enthalpy of compressed liquid you will realize that H is equal to U plus PV. So H2 is equal to U2 plus P2 V2 and we approximated U2 at this pressure and temperature as Uf at 45 degree Celsius plus P2 times V2 is nothing but Vf at 45 degree Celsius. So that comes out to be 189.42 and H3 is saturated liquid at 10 bar so that is H3 is equal to Hf at 10 bar so we get the required ratio to be 0.2181. The next application that we are going to look at is throttling application. Throttling literally indicates what it means you know that a fluid is throttled so that it goes from state 1 to state 2 without with very little change in velocity. So in general application this is what is intended V2 is approximately the same as V1 we will look at the conditions under which this is satisfied as we go along as a result of which we get H2 to be approximately equal to H1. So what we wish to accomplish is an isenthalpic process but a change in state so we usually go from high pressure and high temperature to a low pressure low temperature fluid so just the state alone changes with all other things remaining the same. So normally this is accomplished by making the fluid go through a very tortuous passage like for example in a porous plug which is indicated here or alternatively we may also use a valve which is partially open so that there is a steep reduction in pressure and also reduction in temperature or in practical applications a capillary tube which is a very long tube or of required length but very very small diameter that may also be used for this purpose throttling is quite extensively employed in refrigerators and domestic air conditioners because we want to achieve this change of state in as little distance as possible with as little added weight as possible. So capillary tubes are actually quite good porous plug may get clogged with time so the capillary tube is easier to replace and easier to service so capillary tubes are quite extensively used in domestic refrigerators and air conditioners. Now let us take a look and see under what conditions we are able to use this. So if we apply SFE to this control volume what is that the control volume is shown like this quite far away from the porous plug so that there is no so the fluid has sufficient time to attain equilibrium so this is an equilibrium state this is also an equilibrium state. So if you apply SFE to this assuming no heat loss steady state operation no KE or PE changes and obviously there is no external work interaction so we get something like this. Now if states 1 and 2 are such that there is no significant change in the specific volume then because the area is constant remember m dot is equal to a1 v1 over v1 or a2 v2 over v2. Since area is constant if v1 v2 is approximately the same as v1 then the velocities will remain the same. So now we know that if the states are such that there is no significant change in specific volume then v2 is approximately equal to v1 and it follows then from this equation that h2 is equal to h1. So throttling process is an isenthalpic process if and only if the change in specific volume is small otherwise there will be a change in the exit velocity because of the change in specific volume because the fluid expands as the pressure drops the fluid expands. If the expansion of the fluid is significant then v2 will be substantially different from v1 and the process will no longer be isenthalpic. The analysis can still be carried out using this equation but the process will no longer be isenthalpic. So here we have saturated liquid R134A at 30 degree Celsius it is a throttle to 200 kilo Pascal in a domestic refrigerator heat loss to the ambient is 2 kilowatt per unit mass flow rate of refrigerant we are asked to determine the final temperature and dryness fraction if saturated of the refrigerant. Ke and Pe changes may be neglected. So let us see what the initial pressure is just to get an idea of how much pressure drop is being accomplished in this particular throttling valve. So we go to for that purpose we go to the temperature table and look at the saturation pressure corresponding to 30 degree Celsius. So let us go to the temperature table for R134A so corresponding to 30 degree Celsius which is so the saturation pressure is 770.60 kilo Pascal. So the pressure drop across this throttling valve is from 770, 770 to 200 kilo Pascal or 500 about a 570 kilo Pascal drop which is actually quite high. So the specific enthalpy may be retrieved from the temperature table is 93.58 because it is saturated liquid. So this is equal to Hf at 30 degree Celsius. So we apply SFE after taking into account Q dot I am sorry this term should actually be Q dot over M dot. So we calculate the exit enthalpy to be notice that Q dot goes in with a negative sign because it is heat loss from the device. So the exit enthalpy is 91.58. Now corresponding to 200 kilo Pascal we have Hf as 38.41 Hg as 244.5 since H2 lies between Hf and Hg exit state after throttling is a saturated mixture. So the final temperature is the saturation temperature corresponding to 200 kilo Pascal which is minus 10 degree Celsius. So the temperature drops from 30 degree Celsius to minus 10 degree Celsius. So the pressure drops from 770 kilo Pascal to 200 kilo Pascal temperature drops from 30 degree Celsius to minus 10 degree Celsius. So that is the function of a throttling valve. So as we mentioned here high pressure high temperature low high pressure high temperature to low pressure low temperature. The temperature may not be high it is high compared to this but 30 degree Celsius is ambient temperature. So it is high pressure and certain temperature to low pressure and low temperature and that is what we are accomplishing here. So with the value of known value of H2 we may calculate the dryness fraction and exit to be 0.258. So if I illustrate the process on a PV diagram there is very little space here. So let us try to manage. So this is 770 kilo Pascal and this is 200 kilo Pascal. So we go from saturated liquid at 770 kilo Pascal to let me just erase this a little bit just change the color. So we go from saturated liquid at 770 kilo Pascal to a saturated mixture at 200 kilo Pascal with dryness fraction equal to 0.258. So this concludes our discussion of devices in the first category that we mentioned here. We looked at nozzle, we looked at a diffuser, we looked at a mixing chamber and we also looked at a throttling process. What we will do next is to look at examples from devices in the second category. So in this lecture we will look at examples of devices in the second category for which Wx dot is not equal to 0. So Wx dot is not equal to 0 but for all intents and purposes we still assume Q dot to be equal to 0. First device that we will look at in this case is the turbine. Now just like in a nozzle in a turbine also the pressure decreases across the turbine and the fluid undergoes an expansion process but the similarity ends there. So thermodynamically we may say that you know the pressure decreases across the turbine also and the working fluid undergoes an expansion process in the turbine also just as in the nozzle. But you may recall that in the case of the nozzle, so in the case of the nozzle you may recall that M dot is equal to rho times A times the velocity V and since the objective here is to convert the enthalpy to kinetic energy, we force it to go through a narrowing passage or a passage of decreasing cross-sectional area. When we do that since the area decreases the velocity increases. Now in the case of a compressible fluid since the fluid also undergoes an expansion process because the pressure decreases from inlet to exit the density also decreases which causes the velocity to increase even more. So this is how we accomplish the conversion of enthalpy to kinetic energy. So by forcing it to go through a narrowing passage we accomplish conversion of enthalpy to kinetic energy. In the case of a compressible fluid for instance the fluid naturally wants to expand because it is undergoing an expansion process but by restraining it to go through a converging passage we are actually converting the enthalpy into kinetic energy of the fluid. Now if we go back to the turbine the thermodynamic process is the same but the objectives are different. So here also M dot at any cross-section of the turbine so if I take any cross-section like this M dot is equal to rho times A times V where V is the axial velocity that is the velocity which the fluid is flowing from inlet to exit. Now in this case the fluid is actually undergoing an expansion process which means the density is decreasing but the objective here is to convert the enthalpy of the fluid into work. So we allow the fluid to expand in this case which shows that the cross-sectional area actually is allowed to increase to accommodate the expansion of the fluid. But the conversion of the enthalpy into work is accomplished in the blade passages in the turbine. So the blade passages are shaped in such a way that the enthalpy of the fluid may be converted to work while allowing the fluid to expand. So turbines may be designed in many different ways but the simplest thing is as you can see the density decreases we allow the cross-sectional area to increase to accommodate the expansion of the fluid. In contrast to the nozzle where the cross-sectional area decreased and that was actually acting against the natural tendency of the fluid to expand. In this case we allow the area to increase to accommodate the expansion so consequently this velocity in the case of a turbine is usually a constant. So that is how these types of turbines are designed this is called an axial turbine as the fluid flows axially along from the inlet to the exit. So you can see how the blade passage plays a key role in accomplishing the conversion of enthalpy to work in this case. So when you later on take a course on turbo machines you will be taught how the blade passage is designed to accomplish this effect. So if you look at the but such details are not necessary in this course but this gives an idea of how different devices accomplish different things. So you can see here that different blades are shown. So we have shown a cross-sectional area. So these are blades let us just change color here. So these are blades which are fixed to the shaft as you can see. So this entire blade is in the circumferential direction. So this is fixed to the shaft this is fixed to the shaft this is fixed to the shaft and this is fixed to the shaft. So these are rotating blades. The blades which are in between them such as this one is fixed to the casing or the valve of the turbine and they are stationary. So the fluid moves from rotating blade through a stationary blade through another rotating blade and so on. So as it moves through these blades or blade passages the cross-sectional area of the blade passage accomplishes the conversion of enthalpy to work in the case of a turbine. That is why the turbine is illustrated schematically with an increasing cross-sectional area as you can see here. Whereas a nozzle you may recall was illustrated schematically using a decreasing cross-sectional area. So notice that for our purposes how exactly this conversion of enthalpy to work is accomplished is immaterial. So which is why we have just shown a passage of increasing cross-sectional area a shaft with external work interaction W x dot which is positive in this case because it is a turbine. So the inner details are unnecessary and this is our control volume. So we apply the steady flow energy equation assuming steady flow we can apply the steady flow energy equation to the turbine eliminate terms that are not relevant or equal to 0 and then proceed with the analysis. So the thermodynamic process is the same in a nozzle and a turbine fluid undergoes an expansion process but the objectives are different between the nozzle and the turbine. Let us look at a worked example involving a turbine. So steam at 60 bar in fact before we do that let us also take a look at a turbine picture of a turbine to see how so you can see that this portion is the turbine portion and you see how the cross-sectional area of the turbine increases like this. So you can see how the cross-sectional area of the turbine blades increases like this and in the case of the compressor which is what we are going to look at next we can see how the cross-sectional area decreases. So here we can see the increase in cross-sectional area of the turbine as the fluid expands and what are shown here or the rotating blades that I mentioned before. These are the blades which are fixed to the shaft and so they rotate. So, the stator blades which are the stationary blades are located in between these rows of rotating blades, they are not shown here for the sake of clarity. So, you can see how the cross sectional area increases axially as we move along the turbine. So, we now move on to a worked example involving steam. So, steam at 60 bar 680 degree Celsius enters an insulated turbine. So, this means that q dot equal to 0 that is operating at steady state. So, we are allowed to use the steady flow energy equation. Part of the steam is extracted from the turbine at 10 bar 460 degree Celsius. The rest is expanded completely and leaves at 45 degree Celsius and the dryness fraction of 94 percent. Determine the power developed if the extracted mass flow rate is 17.9 percent of the total. KE and PE changes may be neglected. So, we showed the control volume here. So, this is our control volume and we simplify the steady flow energy equation applied to this control volume with the q dot equal to 0 and KE PE changes neglected. So, we end up with an expression like this and so, inlet 1 which is which would be this one in this particular example in addition to inlet 1 and 2 we also have I am sorry in addition to inlet 1 and exit 2 we also have an additional exit 3 where the steam is extracted. So, we label this exit as 3. So, at 1 the steam is superheated at 60 bar and 680 degree Celsius. The enthalpy values may be retrieved from the superheated table quite easily. At exit 3 again the steam is superheated. So, the enthalpy at exit 3 may also be retrieved from the superheated table. Now, at exit 2 steam leaves at 45 degree Celsius and the dryness fraction of 0.94. So, we retrieve Hf and Hg from the temperature table and evaluate the enthalpy at exit 2 as 2438.67. So, the power developed by the turbine specific power that is per unit mass flow rate developed by the turbine comes out to be 1236.95 kilojoule per kilogram. Notice that the mass flow rate of the extracted steam was given to be 17.9 percent of the total mass flow rate. So, since we have taken m1 dot to be 1 kg per second we take this to be 0.179 and this is the mass flow that is expanded completely and this is the mass flow rate that is taken out at exit 3 itself. Now, in the absence of extraction exit 3 will not be present. So, we go from inlet 1 to exit 2 and the power developed per unit mass flow rate is simply H1 minus H2 and that comes out to be 1407.63 kilojoule per kilogram. Obviously, in the absence of extraction the power produced is more. However, in many power plants some of the steam is usually extracted for other purposes. In fact, we saw if you look at this particular case this is coming out at 10 bar and 460 degree Celsius. So, the steam is extracted at 10 bar, 460 degree Celsius and you may recall the mixing chamber example that we did where the steam came in at 10 bar, 460 degree Celsius. This is intentional. So, to demonstrate what is actually done with extracted steam. So, although you lose some power when you extract steam from a turbine that steam is usually used for heating other liquids and that is actually a very good way of doing that. So, it is taken to a mixing chamber mixed with compressed liquid like this which then leaves at a high temperature as a saturated liquid. So, the steam can be used for other purposes. So, you lose some power, but you also gain something else.