 Now, steady flow devices that you usually encounter in mechanical engineering may be broadly classified in the following manner ok. So, devices for which the work interaction is identically 0 and the heat interaction may be approximately taken to be 0, in some cases it may not be 0 in which case we can account for it, but generally heat interaction is very very small and so may be neglected. So, work interaction is identically 0, but heat interaction is negligibly small. So, it may be neglected. Devices that fall into this category again may be classified like this nozzle and diffuser that these are devices which obey this. We also have mixing chambers and heat exchangers which may be categorized in this category and throttling devices may also be grouped into this category. The next set of devices or ones for which WX is WX dot is not equal to 0, but again QX dot may be taken to be equal to may be assumed to be equal to 0, very small negligibly small. So, we take it to be 0. Normally QX dot may be much much less than WX dot in these cases that is generally what happens ok. So, for these types of devices WX dot may be of the order of let say mega watts and QX dot may be of the order of say less than 1000 watts or something like that. So, it is very very small compared to WX dot. So, we may neglect it in the analysis safely. Now, there are two types of devices which fall into this category. First one is the turbine for which WX dot is positive, it is power producing and compressors and other devices which are power absorbing. So, turbine is power producing and then these devices are power absorbing. So, under the category of turbine we can we may have steam turbines, gas turbines or hydraulic turbines and in this category we may have compressor, fan, blowers and pumps. Notice that steam and gas turbine handle compressible substances whereas hydraulic turbines handle water which is an incompressible liquid. Similarly, pumps handle liquids which are incompressible whereas these three devices compressor fans and blowers handle compressible gases. So, this handles typically air, fans and blowers typically move large amounts of air without imparting huge pressure changes whereas a compressor compresses air. So, there is a large increase in pressure across a compressor whereas with fans and blowers these typically tend to move large quantities of air without imparting big pressure change to them. The third category of devices or ones for which the work interaction is identically zero but the heat interaction is not equal to zero. Devices that fall into this category are boilers and combustors where heat is supplied and condenses where heat is extracted from the device. So, this is the broad classification. Now, let us take a closer look at how these devices operate. Now, devices in the first category basically convert enthalpy to kinetic energy in a nozzle. So, if I look at the steady flow energy equation. So, for the nozzle we have said that this is approximately zero, this is identically zero. Assuming that there is no elevation change we can knock out these two terms. So, in a nozzle basically you can see that enthalpy incoming enthalpy incoming velocity is usually small incoming enthalpy is usually large. So, the enthalpy change of the fluid is actually converted to kinetic energy change these two terms. So, enthalpy change results in kinetic energy change in a nozzle. So, in a nozzle we convert enthalpy to kinetic energy and in the case of a diffuser which is the exact opposite of a nozzle the kinetic energy is converted to enthalpy. So, we basically use the momentum of the incoming stream and convert that to a high pressure or we convert the kinetic energy of the incoming stream to enthalpy of the fluid. Now, the case of mixing chambers and heat exchangers we have a hot stream and a cold stream the enthalpy from the hot stream is converted to is transferred to cold stream or vice versa depending upon what we want to do. So, there is enthalpy transfer from one stream to another in the case of mixing chambers and heat exchangers. So, kinetic energy and potential energy changes may be negligible in this case. So, there is an interplay between between the enthalpy of the different streams in these devices. Now, in the case of throttling there is no change in enthalpy also and typically the changes in kinetic energy are negligibly small. So, what happens is there is a change in the state of the fluid between inlet and exit without any change in enthalpy or kinetic energy that is a very special device we will take a look at that also. Now devices in the second category namely turbines or compressors convert enthalpy to power. Although I am saying work here this is typically in these cases this would be power. If you talk about this on a rate basis, but typically enthalpy is converted to useful work in a steam and gas turbine. Potential energy of the stored fluid in a dam is converted to work in the case of a hydraulic turbine. So, you can see that the steady flow equation has different terms q dot, w x dot, enthalpy, specific enthalpy, kinetic energy and potential energy. So, each one of these device represents an interplay between different terms of the steady flow energy equation. So, we knock out terms which are not relevant, keep terms which are relevant and then complete the analysis. So, that is what analysis of a control volume using the steady flow energy equation implies. We look at the interplay between the terms, keep the ones that are important, throw out the ones which are not important, complete the analysis. So, in the case of steam or gas turbine, enthalpy is converted to work. In the case of a hydraulic turbine, potential energy of the water that is stored in the dam is converted to work. In the case of a compressor, work is converted to enthalpy. In the case of a compressor or a pump and work is converted to flow work in the case of a fan or blower. Remember, fan or blower does not cause appreciable change in pressure, but they move large amounts of fluid. So, that means the power that you are putting into the device or work that you are putting in is converted to flow work. Remember, flow work is the amount of work that needs to be spent to push the device in and out of to push the fluid in and out of a device. So, in these devices, the work is converted to flow work. And in the case of a domestic pump, for example, one that you would use in your house to pump water from a sump in the ground floor or on the ground to overhead tank. So, here the work is converted to potential energy. So, this is the opposite of a hydraulic turbine. The work or power that we are putting in is converted to potential energy of the fluid. Now devices in the third category, boilers, boiler or combustor and condenser accomplish the following. It converts heat into enthalpy. So, Q dot is converted to Mi dot times He. So, the heat comes from the burning of fuels such as coal and oil. So, we burn these fuels and the calorific value of these fuels is realized in the form of enthalpy change of the fluid in the case of a boiler and combustor. Or in the case of a condenser, we actually extract the heat from the fluid and usually we discard it into the ambient. Condensers are used for cooling fluids in order to execute a cyclic process. We will see that later on. So, this heat is usually rejected into the ambient, which actually causes thermal pollution of the ambient and then overall increase in the temperature of the ambient. So, enthalpy is converted to heat and then discarded into the ambient in these devices. So, what we will do in the next lecture is to take an example from each one of this application from each one of this application and then apply the steady flow energy equation, remove terms that are not relevant, keep terms that are relevant and carry out an analysis of each one of this device. In this lecture, we will do steady flow analysis of few devices from the first category. So, we will look at steady flow through a nozzle, we will look at steady flow through a diffuser, we will also look at mixing chamber and an example on throttling. So, these are the four examples or four devices that we will analyze today using the steady flow energy equation. Let us now start with the first one. So, here we have a convergent nozzle. So, the area is decreasing from inlet to exit and the fluid flows through the nozzle because the pressure at the exit is less than the pressure at the inlet. So, the fluid naturally flows through the nozzle. Now, if I take any cross section of the nozzle, then the mass flow rate that passes through the through this cross section is given as density times cross sectional area times the velocity. Now, if the density is constant, let us say we are dealing with the liquid in case the density is a constant, then the reduction in the area causes. So, when the area decreases, the velocity has to increase to keep the mass flow rate the same. So, this is in the case of a liquid which is incompressible say like water. Now, in the case of a compressible substance such as gas or steam or R134 vapor, the acceleration due to the reduction in areas even higher because in these cases, since the pressure decreases from inlet to exit, the density continues to decrease in this case. So, let us denote this with a slightly different color for the compressible substance. So, in this case, since P2 is less than P1, the density decreases. So, rho 2 is less than rho 1. So, the fluid undergoes an expansion process. So, since the fluid undergoes an expansion process, the density decreases. So, now the density decreases in addition to the reduction in the area. So, the increase in velocity is even more in this case. So, density decreases, area also decreases. So, the velocity increases even more to keep the mass flow rate the same at each section. In fact, the acceleration in the case of a compressible fluid may be so high that the velocity at the throat or in this case, the exit of the nozzle may even become equal to the speed of sound. And most cases, it does become equal to the speed of sound because the acceleration is so high. All right. So, the working equation for analyzing problems involving the nozzle read like this. So, the mass flow rate at each section is a constant. So, the mass flow rate at each section is a constant. And application of the steady flow energy equation looks like this. We assume the heat loss from the nozzle to be negligibly small. There is no work interaction. So, the W dot x term is identically equal to 0. And so, we end up with an equation that looks like this. So, basically this says that the change in enthalpy is converted to a change in kinetic energy or increase in kinetic energy. Now, in case we are dealing with working substance which is an incompressible liquid, then we actually can assume the density to be constant. And notice that here we have used the factor h is equal to u plus pv. So, if I take a differential of this, I get dh equal to du plus using the product rule I get pdv plus vdp. Now, there is no significant change in temperature of the liquid as it flows through the nozzle. In other words, it does not really get heated up or it does not become cold. So, which means that the change in internal energy may be taken to be 0. Furthermore, since it is an incompressible liquid, change in density is also negligibly small. So, we may take this also to be 0. So, change in enthalpy dh may be written as may be approximated very well as vdp. So, that means, this equation may be further simplified into something that looks like this. So, the so, h1 minus h2 term may be written as so, what this leads to is that h1 minus h2. If I integrate both sides of this expression, I get h1 minus h2 equal to the specific volume which remains constant times p2 minus I am sorry. Let me write this as let me just erase this rewrite this. So, basically we have dh equal to vdp and if I integrate both sides, I get h2 minus h1 is equal to the specific volume times p2 minus p1. Since the specific volume remains constant. So, if I take it, I mean if I use the fact that the specific volume is the reciprocal of the density, then I may write this as p2 minus p1 divided by rho and that is what this equation tells us. Notice that this is identical to the very familiar Bernoulli's equation. So, basically, we will use these two equations in the case of compressible fluids and use simplified forms of these two equations in case the fluid is incompressible. So, the first example involves steam. So, we have dry saturated steam basically dry refers to the factor dry saturated refers to the factor x is equal to 1. So, if I show the state or process on a PV diagram, we have something like this. So, this is state 1. Notice that the inlet state is now denoted as state 1 and this is 5 bar. So, this is dry saturated vapor at 5 bar enters an adiabatic nozzle with negligible velocity. It leaves at 3 bar with the dryness fraction of 0.97. That means that at the exit, it is a saturated mixture. So, we may indicate that state like this 3 bar and dryness fraction of 0.97. So, this is the exit state. So, this is the process. We can approximately show the process that it undergoes. We do not know anything more about the process. So, it would be erroneous to say anything more or approximately it looks something like this. Later on when we redo these examples using concepts from second law, we will say that the process is isentropic meaning entropy remains constant. But in this case, we can just indicate it like this. So, we are told that this operates at steady state and potential energy changes are negligible where as to calculate exit velocity and exit area given the mass flow rate. So, the working equation in this case reduces to this V1 is given to be approximately 0. So, we said V1 equal to 0 and we end up with an equation like this from which we can get V2 is equal to this. State 1, since it is saturated vapor, we can directly retrieve enthalpy values from the pressure table. Let us just quickly take a look at this pressure table for steam. So, 5 bar for steam is shown here like this. So, let us just scroll it up a little bit. So, Hg corresponding to 5 bar is 2748.6 which is what we used over over here 2748.6 and state 2 we can retrieve Hf and Hg from the pressure table and using the given drainage fraction we may calculate H2 equal to Hf plus x2 times Hg minus Hf. Specific volume at the exit is also required because we have given the mass flow rate and we are asked to calculate the area at the exit. So, specific volume at the exit may also be evaluated using the given drainage fraction. So, if I plug these values into the previous expression I get the exit velocity to be 420 meter per second. So, basically the fluid accelerates from almost 0 velocity or very small velocity to velocity of 420 meter per second at the exit of the nozzle. So, you can see that in the case of a compressible fluid the acceleration of the fluid is quite high. So, mass flow rate is at the exit may be written as a2v2 divided by the specific volume at exit and if you substitute the known values we get the exit area to be 28 centimeter square. The next example also requires us to do flow through a nozzle but now instead of steam we have a mixture of ideal gases. So, the mixture consists of three components component A, component B and component C whose molecular weights and ratio specific heats are given and this expands adiabatically in a nozzle from 700 kilopascals 727 degree Celsius to 350 kilopascals. And we are given that the nozzle operates at steady state and potential energy changes are negligible and we are asked to calculate exit temperature assuming that PV raised to gamma equal to constant or gamma is the ratio of specific heats of the mixture. We are also asked to calculate the exit velocity and the exit area given the mass flow rate. So, it is clear that we need to know the gamma for the mixture and we also need to know Cp for the mixture in order to calculate the enthalpy change. So, the mole fractions are given notice that this gives us 30 percent, 40 percent and 30 percent of component C by volume which means that the mole fractions are given to us. So, we may calculate the mixture molecular weight using the given mole fractions. You may recall that we you may recall from our previous lecture on mixture of ideal gases that Vi the partial volume of a mixture over the total volume is nothing but the mole fraction. So, since the partial volume is given here partial volume over total volume is given here as a percentage we infer from that that it is directly the mole fraction and we use that to calculate the mixture molecular weight. The mixture gas constant may also be evaluated as universal gas constant divided by the mixture molecular weight which is equal to this. The mixture molar specific heat may be calculated using this expression. This expression was also written down earlier. We use the molar form of the expression because the mole fractions are given and it is straight forward to use this. The Cp for each mixture may be written in terms of gamma like this and if you substitute the values we get the molar specific heat of the mixture to be equal to this and by using my S relation we can then get gamma over gamma minus 1 equal to 3.35 from which we get gamma equal to 1.4255 for the mixture. And the mixture specific heat on a mass basis may be evaluated as Cp bar divided by mixture molecular weight and this comes out to be 765.162. We need this for the process equation and we need the Cp for the valuation of the specific enthalpy of the mixture. What is that? Again we calculated gamma by evaluating Cp bar which itself uses the gamma of the individual mixture. So, notice that from this you can easily see that the mixture gamma is not sigma yi gamma i that is the most important point that you must understand here. Since we are given that pv raised to gamma equal to constant and since it is a mixture of ideal gases which itself is behaves as an ideal gas we may write for the mixture pv equal to RT for the mixture which means that I can use equation of state in this and finally evaluate my exit temperature, mixture exit temperature to be 813 Kelvin. Now if I if I apply steady flow energy equation for this case nozzle is insulated so q dot is 0 there is anyway no work interaction so w dot x equal to 0 p changes may be neglected. So, we get v2 to be 2 times h1 minus h2 just like the previous example but now because it is an ideal gas we may replace h as Cp times T. So, this may be written as 2 times square root of 2 times Cp times T1 minus T2 which gives me the exit velocity to be 535 meter per second. At the exit the mass flow rate is nothing but rho times A times V and if I use the equation of state to replace rho we get p equal to rho times R times T. So, I may replace rho 2 with p2 over RT2 and if you substitute the numbers we get the exit area cross sectional area to be 49.6 centimeter square. So, this example illustrates how to calculate mixture of molecular weight for a mixture of ideal gases specific gas constant from that and Cp for the mixture as well as gamma for the mixture. So, once you have Cp for the mixture and gamma for the mixture it is as if we have a single component gas we replace this mixture with a single component gas with this Cp and this gamma and then proceed with the analysis.