 So, the next example that we will look at is a compressor. So, in the case of a compressor as you can imagine air is taken in at a steady rate, it is compressed. So, the pressure increases across the compressor. So, the input work is utilized to increase the enthalpy of the fluid ok, whereas in a turbine the enthalpy of the fluid is converted to output work whereas here the work that is put in is converted into to increase the enthalpy of the fluid and most importantly to increase the pressure of the fluid ok. So, the compressor usually works with the prescribed pressure ratio between the outlet and the inlet ok. Once again what we have illustrated what I have illustrated here is an axial flow compressor with similar moving blades and fixed blades and in fact, let us just take a look at the picture of a compressor here. So, here you can see the moving blades and you can see that the moving blades are more numerous than the turbine blades. There is a reason for this which will be discussed later on in a in a turbomissionary course. So, you can see the moving blades and you can see the height of the blades decreasing as we go along. So, the cross sectional area of the passage decreases as the fluid is compressed and its density increases. So, m dot is equal to rho times A times V at any at any cross section. Since the density increases and we wish to keep velocity almost constant because the idea here is to convert the input work into increase in density not increase in kinetic energy or not a change in kinetic energy. So, we want to keep this approximately constant which means we allow the area to decrease so that m dot is a constant at any cross section that is the basic idea in compressor design. Let us look at couple of worked examples involving compressor. So, air at 100 kPa 27 degree Celsius enters an insulated compressor. So, q dot is equal to 0 steadily at a rate of 1 meter cube per minute where it is compressed to a pressure of 1 mPa. So, in compression process to obey PV raise to gamma equal to constant determine the required compressor power k e and P e changes may be neglected. So, our control volume for analyzing the compressor is shown here. So, this is our control volume and if you go ahead and apply SFE and then simplify it for this problem with the q dot equal to 0 k e and P e changes neglected this is what we end up with. And since air is an ideal gas we may write this as m dot Cp T1 minus T2. Now, m dot itself has to be evaluated, we may evaluate m dot at the inlet. So, m dot is equal to rho times the volume flow rate. So, what we are saying here is m dot is equal to the density at the inlet times the volume flow rate at the inlet. This is the value of mass flow rate at inlet it will also be equal to the mass flow rate at the exit because it is the compressor is operating at steady state. However, this expression if you apply it at the outlet the density and the volume flow rate may be different because the thermodynamic state is different. So, we calculate m1 dot using this expression and we say that that is equal to m dot through the compressor which is correct. So, we are given the volume flow rate to be 1 meter cube per minute and we convert that to meter cube per second divided by 60 and we can evaluate the mass flow rate through the compressor as 0.01924. Since the process obeys PV raise to gamma equal to constant we may evaluate T2 as 579 Kelvin. So, if you substitute these values in the expression that we had before m dot cp times T1 minus T2 we get this to be 5423 watts. Notice that this expression technically will yield a negative number, but we say that the compressor power is 5423 watts the understanding is compressor requires input power. So, the negative sign is dropped here with that understanding. The next example involves a refrigerant R134A. So, refrigerant 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 0.06 meter cube per minute. The heat loss to the ambient is 7 percent of the input power determined the power required ke and pe changes may be neglected. Again when volumetric flow rate is given you must bear in mind that mass flow rate at inlet and exit are the same because it is operating at steady state. But volumetric flow rate at inlet and exit will not be the same because the thermodynamic state is different. Densities are different. So, volumetric flow rate will be different. So, the volumetric flow rate at the inlet is given. So, we can calculate the mass flow rate at inlet by saying that it is equal to density times volumetric flow rate at inlet just like what we did in the previous problem. So, the control volume remains the same. So, we simplify the steady flow energy equation for this with ke and pe changes neglected q dot is actually given in this problem. So, we cannot neglect it. It is obvious from the given information 140 kPa minus 10 degree Celsius since pressure and temperature are given it is going to be superheated or compressed liquid. In this case we can easily show that it is a superheated state and get h2 from the superheated refrigerant table both at the inlet and the exit. The mass flow rate m dot may be evaluated as so, this is equal to rho 1 times v1 dot or if I write it in terms of a specific volume I may write this as v1 dot over v1. So, the specific volume at inlet has been retrieved from the refrigerant table. So, we can evaluate the mass flow rate to be 0.4107 kilogram per minute. Now, you have to be a little bit careful with the numbers and the signs for q dot and w x dot in this case remember w x dot is negative because it is a compressor we have to supply power to the compressor q dot is also negative in this case because there is heat lost to the ambient. So, we have explicitly taken this into account by writing in the q dot term as minus 0.07 times w x dot because that is 7 percent of the input power and this has also been taken into account by changing this to a plus sign and using an absolute sign around w x dot. So, once I do all this I get the absolute value of w x dot to be 49.68 kilo joule per minute. So, the compressor input power is 49.68 kilo joule per minute. So, you have to be careful with this equation when you actually plug in the numbers. Saturated liquid water at 45 degree Celsius is pumped to 20 bar pressure steadily by a pump. So, we neglect heat losses so, q dot equal to 0 K e and P e changes determine the power required per unit mass flow rate. So, for this particular problem so, here is the diagram of a pump. So, our control volume in this case would look like this. So, this is our control volume for analyzing just the pump. Power is supplied water comes in at a certain thermodynamic state it leaves at a higher pressure. So, that is and we are asked to calculate the power required per unit mass flow rate. So, if you apply S of e to this control volume, q dot may be taken to be 0, K e and P e changes are neglected. So, the equation reduces to something like this. Now, if I expand the enthalpy term I may write this as u 1 plus p 1 v 1 and h 2 may be written as u 2 plus p 2 v 2. Since there is no substantial change in the temperature of the liquid as it goes through the pump u 1 is approximately taken to be equal to u 2. Furthermore, since it is a liquid it is practically incompressible. So, we may also take v 2 equal to v 1. So, with this simplifications the above equation reduces to W x dot equal to m dot v 1 times p 1 minus p 2. And v 1 is v f at 45 degree Celsius because it is given that the liquid water is saturated at 45 degree Celsius. So, v 1 is equal to v f at 45 degree Celsius and p 1 is equal to p sat at 45 degree Celsius because it is a saturated liquid that is also given. So, we now substitute these values into this expression and we get the pump power to be 2010.11 watt per unit mass flow rate. What is how we have used the fact that it is an incompressible liquid to simplify this expression and write it in terms of pressure rise while keeping the density constant. The next example is this. Here we are actually using a pump to lift water from sump to an overhead tank like this. So, basically the power that we are putting in is being converted to potential energy of the water as it is lifted from the sump at the ground level to an overhead tank. So, the control volume for this would look something like this. So, this is the control volume for this problem. All the other information that is required for the analysis is given and we are asked to determine the power required to run the pump. So, for the control volume shown, we take q dot equal to 0 and v 2 equal to v 1 since the pipe diameters are the same. We also assume that there is no significant change in the temperature of the liquid as it goes through the pump. Hence, u 2 is equal to u 1. So, with this simplifications, the steady flow energy equation assumes this form. And if you look at the illustration, since this water flows out like this, it is at atmospheric pressure. So, p 2 is equal to p atmosphere. But p 1 is equal to p atmosphere plus the height of the liquid column above the pump inlet. So, p 1 is equal to p atmosphere plus rho g times 2 meters which is the height of the water column above the pipe inlet, I am sorry, above the pipe inlet. So, with m dot equal to rho times v dot, where v dot is the volume flow rate, we may evaluate the pump power to be 2.06 kilowatts. So, one of the most interesting things that this example and the previous example illustrate is how the same equation takes on different forms depending upon the control volume that we have. So, for the control volume in blue, we ended up with something like this that there was no potential energy change. Whereas, for the control volume in red, there is a potential energy change and that is because of the power that we are putting in. In the case of the control volume with blue, the power that was being put in is converted to increase in enthalpy of the fluid. Whereas, with the red control volume, the input power is being used to increase the potential energy of the water as it is lifted from state point 1 to state point 2. The last example that we are going to look at under the category of steady flow devices is the heat exchanger. Now, functionally, heat exchanger is almost the same as the mixing chamber that we saw earlier in the sense that both of them accomplish transfer of enthalpy from one stream to another. However, in the case of a heat exchanger, the transfer takes place without physical mixing. Now, in the case of the mixing chamber, because the transfer takes place by virtue of mixing, the device operates at a constant pressure. So, the incoming streams and the outgoing stream or streams all have to be at the same pressure. Now, in the case of the heat exchanger, because there is no physical mixing of the streams, the streams can be at different pressure. So, basically the stream to be heated or cooled flows in a separate line and the stream that is heating or cooling flows in a separate line. So, they can be at different pressures. And the fluids are brought into contact indirectly and heat transfer takes place from one stream to the another or enthalpy transfer takes place from one stream to another. Now, heat exchangers are used extensively either for heating a stream or for cooling a stream. So, the example that we are going to look at next actually involves cooling a stream, but it can be used for heating a stream as well as cooling a stream. Now, many different types of heat exchangers or in use in practical applications probably three of the most commonly seen would be the parallel flow heat exchanger, where the two streams flow in the same direction. So, this is the heated stream or cooled stream and this is the heating stream or cooled stream and they both flow in the same direction. Now, in the case of counter flow heat exchangers, the streams move in opposite directions. So, for instance, in a parallel flow heat exchanger, one stream moves in this direction and the other stream also moves in the same direction. So, this is a parallel flow heat exchanger. Now, in the case of a counter flow heat exchanger, one stream moves in this direction and the other stream moves in this direction. So, they move in opposite direction. So, this is the counter flow heat exchanger. Now, the cross flow heat exchanger as the name suggests is a device where the streams flow in perpendicular direction feature. So, one stream flows like this and the other stream flows like this, but without mixing. So, they move in perpendicular direction. So, this is the cross flow heat exchanger. The example that we are going to look at next is a counter flow heat exchanger. So, let us see what this does. So, basically what we are going to do here is we are going to accomplish the compression that we saw in one of the earlier examples. Let us go back and look at that. So, you may recall that we looked at a compressor where we had air that comes in. So, air at 100 kPa 27 degree Celsius enters the compressor and it leaves at the pressure of 1 MPa and the compressor is insulated. So, we looked at the steady flow operation of this compressor. So, what we are going to do in this example is to accomplish this compression in two stages. So, rather than going from 100 kPa to 1 MPa in one go, we will first compress the air from 100 kPa to 316.22 kPa in the first stage and then compress from 316.22 to 1000 kPa in the second stage. Now, in between the two stages, the air is taken to a heat exchanger where it is cooled down. So, it enters at a higher temperature after being compressed to 316.22 kPa. It is then cooled in the heat exchanger back to its original temperature, initial temperature of 27 degree Celsius before being sent to the next stage compressor. So, basically the arrangement would look something like this. So, this is the first stage compressor. So, this air is then taken into this. So, here the air comes in at 27 degree Celsius. So, it is at a higher temperature when it comes out, it is then taken to this one, the heat exchanger where it is cooled down to 27 degree Celsius. 27 degree Celsius, the air is then taken to the next stage compressor which looks like this. So, this we label as state 3 and the air leaves at 1 MPa as state 4. Now, water enters at 30 degree Celsius and after cooling the air, the water leaves at 60 degree Celsius. We are asked to calculate the total compression work and compare with the single stage compression work and also determine the required mass flow rate of water. So, the mass flow rate of water that is required to accomplish the set cooling which is to cool the air from whatever temperature it comes in to 27 degree Celsius which is the same as the inlet temperature here. So, we apply SFEE to the first stage compressor, we take u dot to be 0 because it is insulated, KEP changes are neglected. So, we end up with an expression like this and since air is an ideal gas, we can replace H as CP times T. So, we do this and the mass flow rate is the same as before. I remember that the volume flow rate at the inlet to the compressor was given and we calculated the mass flow rate from that. So, the mass flow rate remains the same as before at 0.01924 kg per second and we also know that the compression process obeys PV raise to gamma equal to constant. So, the temperature at the end of the first stage compression is 417 Kelvin. And we can compute the compressor power for the first stage by substituting the values. So, we know the exit, we know the exit temperature, we know the inlet temperature, this is 300 Kelvin, inlet temperature is 300 Kelvin. So, we can calculate the power for the first stage to be 227.45. Now, at the second stage, if you see the air enters at 27 degree Celsius, same as in the first stage and then it is compressed to a pressure of 1000 k part. Now, if you look at the pressure ratios p4 over p3 for the second stage and p2 over p1 for the first stage, you get the pressure ratio to be the same. You see that the pressure ratio is the same and the inlet temperature to the compressor is also the same, which means that the power required in the second stage compressor is also the same. So, the total power required is 4548.9 watts and this represents a 16 percent reduction in power when compared with single stage compression. Obviously, when you take the air out, cool it down before compressing it further, the power required will definitely go down and that is what we are seeing here. There is a 16 percent reduction in power when compared to single stage compression. This is why in many practical applications, this strategy is almost always employed. So, such a device is usually called, although we are calling it a heat exchanger in our example, such a device is usually called an intercooler to denote the fact that cooling is employed in between stages of compression. So, normally there is a two-stage compressor and we use intercooling in between the stages to reduce the compression work. So, we now apply SFE to the heat exchanger. So, this is the control volume that we are using. We assume that there is no heat loss from the heat exchanger as a whole. So, we have two streams, so two inlets and two outlets. So, we apply SFE to the heat exchanger, q dot is 0, wx dot is 0, no significant ke and pe changes. So, we get 0 equal to m dot water times this, hx and hy are the enthalpies of the water at the inlet and exit through the heat exchanger. Since pressure, since information relating to the pressure of the water is not given, we simply assume hx equal to hf at 30 degree Celsius and that are the saturated liquid and hy equal to hf at 60 degree Celsius. If the pressure is given, we can probably fix the thermodynamic state more precisely and evaluate hx and hy but probably not required for this problem. We will simply take hx and hy to be the enthalpy of the saturated liquid at the corresponding temperature and if you substitute the numerical values, we get the mass flow rate of water to be 0.0181 kilogram per second. So, one important aspect that you should notice is that we are saying q dot from the heat exchanger is 0. However, there is enthalpy transfer that is taking place between the streams. So, there is enthalpy transfer that is taking place between the streams. So, that is always there but our control volume encloses both the streams. So, the q dot that we say is the q dot that we are saying is equal to 0 is actually heat loss from the heat exchanger to the surroundings. So, whatever happens inside the control volume is accounted for through the enthalpy change in the two streams. So, this completes the examples that we have been looking at on steady flow analysis. What we will do next is look at unsteady flow analysis.