 So, in this lecture, we are going to look at 3 applications involving psychrometric principles. The first one is adiabatic mixing of 2 streams and this probably may be considered as an application that comes in HVAC. The other 2 applications may be classified as other applications of psychrometric principles. In HVAC, we mentioned 4 unit processes namely heating slash cooling and we said humidification slash dehumidification as the 4 unit operations. Of course, mixing of 2 streams may also be considered as a psychrometric application because we can adjust the moisture content or temperature of an airstream by mixing it with another airstream at an appropriate with an appropriate moisture content and temperature. So, that is also possible without adding heat or removing heat from the airstream, without adding or removing water from the airstream, its temperature and humidity may be adjusted by simply mixing it with another stream and that is the application that we are going to consider first. So, here we have an insulated mixing chamber. So, this is the mixing chamber, we have seen this before. So, this is an insulated mixing chamber. So, we have an airstream coming in at state 1, another airstream coming in at state 2 and then exiting airstream. So, both 1 and 2 all the airstreams are actually moist airstreams. Let us say that our idea is to control the temperature and humidity of let us say stream 1 and we want to control it in such a way that we get a desired value for temperature and humidity ratio of the exiting stream that is generally the idea. So, if we do mass balance of dry air across the mixing chamber, we get m dot a1 plus m dot a2 equal to m dot a3. Water balance across the mixing chamber assuming that no condensation takes place. So, the amount of vapor that comes in through 1 plus amount of vapor that comes in through 2 is the amount of vapor that leaves at 3. And if we write this in terms of omega by using the definition of the humidity ratio, we may write this like this and if you rearrange, we may actually finally come up with an expression for omega 3 which is the humidity ratio of the exiting stream. So, if I know omega 1, omega 2 and the mass flow rates, omega 3 may be evaluated. On the other hand, if I desire a certain omega 3 and I want to adjust omega 2 so that I get that omega 3, then we may evaluate omega 2 from this expression. It is possible to do both. It is possible to predict the humidity ratio of the exiting stream or it is possible to come up with a value for the humidity of one of the incoming streams so that the exiting stream has a particular value. Application of SFE to the control volume gives rise to the following equation q dot and w x dot are both 0, q dot 0 because the mixing chamber is insulated. So, we can come up with an expression for h 3 star which is the specific enthalpy on a dry air basis for the exiting stream. Assuming that h 1 star, h 2 star and the mass flow rates are known, this allows us to evaluate h 3 star and hence along with omega 3, the exit temperature and the relative humidity may be evaluated using this. Now, we may rearrange the expression for omega 3 like this. So, if you look at the expression for omega 3, so if we rearrange this expression, we may write this as omega 3 minus omega 1 over omega 2 minus omega 1 to be m dot a 3 over m dot a 2 over m dot a 1 omega 3 minus omega 1 over omega 2 minus omega 3, I am sorry I made a mistake omega 2 minus omega 3. In the same manner, this may be rearranged as shown here, so this shows that omega 3 minus omega 1 over omega 2 minus omega 3 is equal to h 3 star minus h 1 star over h 2 star minus h 3 star and if we rearrange this like this, then this shows that in a psychrometric chart which has omega on the vertical axis and enthalpy axis like this, so this is h star. This shows that this is the equation for a straight line in this coordinate space, so points 1, 2 and 3 lie on the same straight line in the psychrometric chart for adiabatic mixing of two streams. If there is heat loss, then of course, this will not be true, this is true only for adiabatic mixing of two streams. Let us now go through a worked example. So, in the previous example, consider replacing the heating section with a mixing section to accomplish the same objective. So, if you look at the previous example, so here we had a heating section. So, this was the heating section, so we want to replace the heating section with the mixing section to accomplish the same objective which means that air at 100 percent relative humidity 7.5 degree Celsius enters the mixing section and we want air to leave at 22 degree Celsius and 40 percent relative humidity and also such that omega 2 is equal to omega 3. So, these are the conditions that we had in the heating section, we will impose the same conditions on the mixing section except that it is adiabatic. So, let us see what we have here, so this means that in the mixing section, if you go back to the illustration of the mixing section, this corresponds to inlet 1 corresponds to what we had here, let us see 30 degree Celsius 80 percent, I am sorry inlet 1 corresponds to 7.5 degree Celsius and 100 percent relative humidity and this corresponds to 22 degree Celsius and 40 percent relative humidity. So, what we want to find out is the following, I am going to send in ambient air ambient air at 30 degree Celsius, I do not know the relative humidity, I want to adjust the, I am sorry not ambient air, I want to send in air at 30 degree Celsius, I want to determine the volume flow rate and the relative humidity of this air, so that the exit conditions that I see may be accomplished. So, this means that, so air at 30 degree Celsius and unknown fee and unknown volume flow rate comes out, so that this can be accomplished and remember additional requirement here that imposed was that omega 1 equal to, I am sorry omega, omega 3 equal to omega 1, so no explicit addition or removal of water except whatever comes in through air stream, the moist air stream at 2. So, this is the norm and clature for this problem, so using this norm and clature, so we have m dot a at one point, I am sorry this also this information can also be included here 1.1236, so this is 1.1236 kg per minute. So, 1.1236 kg per minute of dry air at 7.5 degree Celsius comes in relative humidity 100 percent comes in and air at 30 degree Celsius unknown relative humidity, unknown volume flow rate is mixed with this, so that we have a mixture at 22 degree Celsius 40 percent relative humidity that leaves the mixing chamber. Notice that omega, we have explicitly stated, I am sorry we have explicitly constrained omega 1 to be equal to or omega 3 to be equal to omega 1, so that omega 2 is also equal to omega 1. So, if you look at this expression, if you look at this expression for omega 1 to be equal to omega 3, what we would require then is that omega 2 also be equal to omega 3, that is the only way this expression may be satisfied, so we know the humidity ratio at the inlet of the incoming air that is at inlet 2 and we also know the temperature there. So, from the psychrometric chart, we can get phi 2 to be equal to 25 percent. So, basically since we are requiring that omega 1 should be equal to omega 3, after adiabatic mixing the expression for omega 3 then suggests that omega 2 also has to be equal to omega 1, so that is what allows us to fix the omega 2, the unknown value for omega 2. So, temperature is available, omega 2 is available, so from the chart we can get phi 2 to be 25 percent and the specific volume of dry air to be 0.87 and H2 star to be 47. I am going to let you look it up from the psychrometric chart. Now, we apply SFEA to the mixing process as before and we now rearrange this so that we can get m dot A2 from this and all the other quantities in this expression are known. So, we get m dot A2 to be 1.9167 kg per minute and since we normally require values to be in units of meter cube per minute, this may be converted to 1.6676 meter cube per minute of dry air at inlet 2. So, this shows how we can actually replace the heating section with that mixing section to accomplish the same objective subject to the same constraints. So, in that sense mixing may also be considered as an additional unit process in HVAC. So, we can accomplish our end objective by more than 1 means, drying, I am sorry heating, cooling, humidification, dehumidification and mixing. The next example that we are going to look at involves drying and this would as I said be classified as other applications of psychrometric principles. So, wet paddy at 25 degree Celsius that contains 1 third moisture by weight. So, which means that if you have 1 kg of wet paddy, 1 third kg is water and the remaining two thirds is the actual dry paddy. So, this has to be dried completely before being stored to prevent rotting that is customarily done. The paddy is usually wet after harvesting and it has to be dried before it can be stored, otherwise it will start rotting. Now, hot air is to be used for this purpose. Ambient air at 25 degree Celsius, one atmosphere 50% relative humidity with the volume flow rate of 250 meter cube per minute is heated to 60 degree Celsius. The hot air is then sent to the drying section that the paddy is dried, determine the power required in the heating section and the maximum mass flow rate of wet paddy that can be dried, assume constant pressure no heat loss. So, we have two sections here. So, this is the heating section and this is the drying section. So, let us label this state 1, this is 2 and the exit state is 3. So, ambient air at 25 degree Celsius 50% relative humidity enters the heating section. So, we can retrieve from the chart omega 1 to be 0.01 kg vapor per kg dry air. Specific volume may also be retrieved from the table as well as H1 star and the mass flow rate at the inlet may be evaluated. So, let us see here. So, 25 degree Celsius and 50% relative humidity that is state 1. So, state 1 is here, choose a different color on this purpose. So, state 1 is here, 25 degree Celsius and 50% relative humidity. Notice that this is the line that corresponds to 50% relative humidity. So, that is state 1. So, we may retrieve omega specific volume and H star. So, this is H1 star, this is omega 1 from the chart. So, the air is heated to 60 degree Celsius and if you look at the schematic here. So, the temperature at the end of this is 60 degree Celsius, but there is no addition or removal of water in the heating section, which means omega 2 equal to omega 1. So, that means state 2 omega 2 is equal to omega 1 and the temperature is 60 degree Celsius. So, temperature 60 degree Celsius omega 2 equal to omega 1. So, this is state 2 and we may look up H2 star from the psychrometric chart. So, mass flow rate of air that is coming in may be evaluated like this. So, for the hot air omega 2 equal to omega 1 because there is no water that is added or removed, temperature is known. So, we can retrieve H2 star. So, the heat that is added in the heating section may be evaluated by applying SFE to the heating section that gives Q dot H to be 167.64 kilowatts. Now, we are asked to evaluate for the given volumetric flow rate of air we are asked to evaluate the maximum mass flow rate of wet paddy that can be dried. So, the maximum if we want the paddy to leave completely dry then the air the maximum and that too maximum mass flow rate of wet paddy this both these then dictate that the air that leaves must be saturated. That is the maximum amount of water vapor that the air can absorb. So, the maximum mass flow rate of wet paddy corresponds to that condition. So, at the exit we say that. So, at the exit we say phi 3 equal to 100 percent. So, the state at the exit of the exit of the drying section will be fully saturated. So, if we do a mass balance of water in the drying section we have m dot w that is picked up from the paddy to be equal to m dot a times omega 3 minus omega 2. So, this is the amount of water that is picked up from the wet paddy omega 3 is not known, but m dot a is known. So, we have one equation now and steady flow energy equation applied to the drying section gives the following. This is the this comes from the water that is picked up from the drying section. So, here we are assuming implicitly that there is no appreciable change in the enthalpy of the paddy. If properties of the paddy are known that can also be accounted for, but for the sake of simplicity here we are assuming that there is no appreciable change in the enthalpy of the dry paddy itself. So, if we substitute the known values h 3 star is not known. So, let me indicate this in in red here. So, h 3 star is not known and omega 3 is also not known, but we have one equation that describes omega 3 which we have already used. Remember m dot w is not known here. So, m dot w is not known omega 3 is also not known. So, we replace m dot w here with the previous equation that we have. So, now we end up with one equation which contains seemingly contains two unknowns. Remember one more information is available and that is phi 3 equal to 100 percent. So, that means that we know that the the outlet state or state 3 lies along the phi equal to 100 percent curve. So, strictly speaking we really do not have two unknowns. So, if I take any point let us say any state along the phi equal to 100 curve, then once I fix state it is h 3 and omega 3 are automatically known. So, once that state is fixed both these quantities are known. So, strictly speaking there is only one unknown which is the temperature of the exiting airstream. So, once I take the temperature of the exiting airstream both h 3 star and omega 3 are known. And so this equation actually has to be solved iteratively. So, what we can do is the best way to go forward is to guess a value for t 3 and locate the state on the psychrometric chart. So, guess a value for t 3 and then locate a state. So, let us say that you know we guess. So, we guess the temperature to be let us say 26, 27 let us say 25 degree Celsius which is physically not correct, but let us say that we choose it to be at 25 degree Celsius then we retrieve h 3 star from here and omega 3. And we plug these values into this equation and see whether the equation is satisfied. If it is not satisfied then we adjust the value of t 3 and keep doing this until this equation is satisfied. So, if you do this after about one or two iterations you should be able to show that t 3 is equal to 27.5 degree Celsius and h 3 star is 86.5. And omega comes out to be 0.0233 that is what is shown in this chart. So, this actually comes after. So, this actually comes after a few iterations. The most important thing is that phi 3 is equal to 100 that is very important, phi 3 equal to 100 percent. So, the mass flow rate of wet paddy may then be evaluated as 3 times m dot w. m dot w here is the water that is absorbed from the wet paddy. And so the and we are given that for in every kg of wet paddy one third is water two thirds remaining two thirds is dry paddy. So, the mass flow rate of wet paddy maximum that can be dried is 3 times the mass flow rate of water that is picked up. So, that comes out to be 11.633 kg per minute. So, you can see I can see that psychrometric principles are useful not only in HVAC predominantly in HVAC, but also in other applications which are very practical and real life sort of applications like the one that we saw just now. The next application that we are going to look at is that of a cooling tower. Now, you may have seen this let us see recall seeing this when we discussed the Rankine cycle. So, in the Rankine cycle I stated that we had looked at all the components in this in this diagram except the except the cooling tower and that is what we are going to look at now. So, basically the cooling tower is used in the condenser to cool the water that comes from the turbine. So, it as you can see it follows its own circuit. So, the cooled water from the condenser enters I am sorry cooled water from the cooling tower enters the condenser it then picks up the heat from the saturated mixture that comes from the low pressure turbine and then leaves at a higher temperature. Notice that this is in a separate circuit and the water from the turbine runs in a separate circuit actually that is at a sub atmospheric pressure as I had mentioned before. So, basically we have water from the cooling tower which enters the condenser at about ambient temperature and when it leaves usually it leaves at temperatures around 50 or 60 degree Celsius. So, now when we look at the cooling tower itself the basic purpose is to take water at 60 degree Celsius and cool it to 30 degree Celsius and then send it back to the condenser. So, that is what we are going to look at now. So, here is water hot water from the condenser as I said typically at 60 degree Celsius or so and the cooled water is then returned back to the condenser. So, this goes back to the condenser as you can see here. Now, the cooling tower that I have shown here is called an induced draft cooling tower because it has a fan at the top and the fan forces the ambient air through the cooling tower like this. So, there are openings on the side of the cooling tower and the ambient air is drawn in to the cooling tower and forced to flow upwards like this by the fan. So, it is called an induced draft cooling tower. So, basically the hot water from the condenser comes down through these types of shower nozzles. So, that the hot water comes down the ambient air flows up. So, as the ambient air flows up because it is cooler than the water that is coming down the water evaporates. The water evaporates and the temperature of the air then increases and then it goes up and the remaining water is cooled down. As the water evaporates it draws enthalpy from the air and it cools down and collects at the bottom as liquid water. Now, some other water that has evaporated is carried out by the air stream. So, we require make up water to make up for the water that leaves as water vapor. And so, what we typically have to do in this is to determine the amount of make up water for a given mass flow rate of water from the condenser and temperature difference that has to be maintained. So, we cannot allow the temperature of the water to become too high because normally the water is drawn from a lake or a river or ocean and thermal pollution is to be avoided. So, normally we cannot allow the water temperature to go beyond 60, 70 degree Celsius also 60 would be I think what the upper limit. So, we want to maintain that. So, given these quantities how do we estimate the amount of air that is required or any other design quantity that we are interested in. So, let us summarize what we have said so far. So, this is an induced draft cooling tower because the fan draws the atmospheric air through the tower. There is not forced push through the cooling tower it is rather pulled through the cooling tower. So, the warm water from the condenser comes downwards in the shower and it meets the upward draft of the atmospheric air which is drawn in through the openings provided along the sides of the cooling tower or circumference of the cooling tower because it is usually circular and cross section. Since the air is at a lower temperature some of the liquid water evaporates and the moisture laden air then goes up. So, the evaporate the water vapor is then carried up by the by the ambient air and it leaves through the tower. So, the rest of the water cools down as a result of evaporative cooling and collected the bottom of the cooling tower. Makeup water as I said is to be added to account for the additional water vapor that leaves through the top. So, basically the cooling tower design from a psychrometric perspective will involve two things. One is mass balance of water, another one is mass balance of dry air. So, let us look at an example. Cooling water enters the condenser of a stream plant at 30 degree Celsius and leaves at 45 degree Celsius with the mass flow rate of 3.75 times 10 raise to 7 kg per hour. You may recall that when we did the examples in the module on Rankin cycle we took the condenser temperature to be 45 degree Celsius. So, I have used the same value in this example also. It is then taken to a cooling tower where it is cooled before it is returned. Ambient air at 25 degree Celsius, 45 percent relative humidity enters the cooling tower and leaves at 31 degree Celsius fully saturated. Makeup water at 25 degree Celsius is also provided. Determine mass flow rates of dry air and make up water. Let us enter some of the information that is given here. So, let me erase some of this. So, it comes from the condenser at a temperature of 45 degree Celsius, leaves at temperature of 30 degree Celsius. Makeup water at 25 degree Celsius is also provided. So, the ambient air, let us go back and just quickly check this. Ambient air at 25 degree Celsius and 45 percent relative humidity. So, 25 degree C, 45 percent relative humidity is also provided. So, the air leaves fully saturated. So, that means, phi is equal to 100 percent here. And the mass flow rate of the water is also given 3.75 times 10 raise to 7 kg per hour. So, 3.75 times 10 raise to 5 kg per hour. And this is also 3.75 times 10 raise to 5 kg per hour. I am sorry, 3.75 correct and correct. So, this is the information that is available. The state points are all marked here. So, notice that for state one, for state one and state two and state five, we need to go to the steam tables and retrieve HF. So, HF has to be retrieved from the steam table for these states. For the remaining states, we will use the psychrometric chart. 25 degree Celsius, 45 percent. So, ambient air 25 degree Celsius, 45 percent relative humidity, we can get H3 star and omega 3. So, here we have retrieved these values from the steam table. Now, mass balance of water across the cooling tower. So, liquid water comes in at 1 and 2, but the mass flow rates are the same. So, these two terms cancel out. We have makeup water coming in at 5, water leaving in the form of, I am sorry, entering in the form of vapor at 3 through the ambient air and water leaving in the form of vapor through the top of the cooling tower. So, if you rearrange, we get m dot w5, which is the mass flow rate of makeup water to be m dot a times omega 4 minus omega 3. 31 degrees, I am sorry, that information, yeah, V equal to 100 percent and 31 degree Celsius. So, that means omega 4 is known, omega 3 is also known. So, omega 4 and omega 3 are both known. So, here we are okay. Now, SFE applied to the cooling tower gives us the following and if you rearrange, then we may write it like this. So, omega 4, omega 3, H5, everything is known except m dot a. So, we may write and we may evaluate m dot a to be 4.21 times 10 raise to 7 kg per hour. So, let us just, let me just show it in a different green color. So, H5 is known, omega 4, omega 3, H1, H2, H3 star, H4 star are all known. We have retrieved these from the tables and listed here. We have retrieved from the tables and listed here. So, m dot a may be evaluated and once this is known, using this equation, we may evaluate the mass flow rate of makeup water and that comes out to be 8.42 times 10 raise to 5 kg per hour. So, this completes our discussion of psychrometry and psychrometric principles as applied to HVAC and a couple of other applications. Again, bear in mind that no new concepts have been introduced, only new terms which are specific to psychrometric applications have been introduced. Terms like humidity ratio, relative humidity, dew point temperature and so on, wet bulb temperature and so on. The fundamental principles or application of first law either for non-flow process or steady flow process and then mass balance, water either liquid or vapor phase and dry air, mass balance of dry air. So, you should keep that in mind.