 So, the next example involves dehumidification. So, the problem statement reads like this moist air at 35 degree Celsius one atmosphere and 80 percent relative humidity enters an air conditioning duct where it is cooled and dehumidified as shown in the figure both the condensate and the air that is saturated that is very important. So, the air leaves fully saturated at 22 degree Celsius. The volume flow rate of air at the inlet is 120 meter cube per minute assuming steady state operation and neglecting heat loss determine the mass flow rate of the condensate the required cooling capacity in tons. Assume that the pressure remains constant. So, the scenario is depicted here. So, we have a cooling coil. So, air enters the duct at 35 degree Celsius one atmosphere 80 percent relative humidity and the volume flow rate of 120 meter cube per minute. So, it leaves saturated which means phi equal to 100 percent and the pressure remains the same. So, the mixture pressure remains the same at one atmosphere temperature is 22 degree Celsius. Now, as a result of heat removal from the duct some of the water vapor condenses and it is collected as saturated liquid at 22 degree Celsius. So, at the inlet relative humidity is given as a 0.8 80 percent or 0.8 and from the temperature table we can retrieve the saturation pressure corresponding to 35 degree Celsius and that is 5.627 kilo Pascal. So, the partial pressure of water vapor at the inlet works out to 4.5016 kilo Pascal. And once we have this value we may evaluate omega the humidity ratio at the inlet omega 1 using this expression and that comes out to be 0.0289 kg vapor per kg dry air. Now, volume flow rate of air is given at the inlet and as I mentioned in the previous example we are using the Dalton's model. So, volume flow rate of air dry air as well as vapor are all the same. So, we may take this to be the given value to be the volumetric flow rate of dry air. So, m dot a will be evaluated using the ideal gas equation of state like this and the partial pressure of dry air is what needs to be used in the equation of state for dry air. So, if you substitute the values we get the mass flow rate to be 131.447 kilogram per minute of dry air. At the exit the air is saturated. So, the partial pressure of water vapor is equal to the saturation pressure corresponding to the mixture temperature which is 22 degree Celsius. So, from the temperature table the partial pressure of water vapor at exit then becomes equal to 2.645 kilo Pascal. And omega 2 at the exit may be evaluated in the same manner as before after noting that the mixture pressure remains the same. So, we get this to be 0.0167 kg vapor per kg dry air. The mass flow rate of dry air is the same between inlet and outlet, but the amount of water vapor has decreased as you can see from here I am sorry. So, at the inlet the humidity ratio was 0.0289 kg vapor per kg dry air and at the exit it is 0.0167. So, it is almost halved between inlet to exit. If you look at the duct itself. So, we have certain amount of mass flow rate of water that is coming in through the inlet a certain mass flow rate of water that leaves through the exit whether in vapor form or liquid form. So, we do not care about that we are only looking at mass of water or mass flow rate between inlets and outlets. So, here saturated liquid leaves at 20 degree Celsius. So, the mass balance applied to this control volume gives the following mass flow rate of water liquid water that leaves the control volume is equal to mass flow rate of vapor that comes in minus mass flow rate of vapor that leaves. And if you use the definition of omega remember omega is equal to m dot vapor divided by m dot dry air. So, we may rewrite this and the mass flow rate of dry air remains the same or remains constant between inlet and exit. So, we can get the mass flow rate of liquid water to be 1.6037 kg per minute simple mass balance of water across the control volume. Now, S of E E applied to the control volume shown gives the following q dot minus W x dot which is 0 because there is no work input or output from the control volume m dot 1 h 1 minus m dot 2 h 2 minus m dot 3 h 3. So, we may rewrite this as follows in terms of dry air and water vapor and liquid water. So, this works out to these three terms may be combined to read like this m dot a times h a 2 minus h a 1 enthalpy change of dry air plus m dot v 2 h v 2 minus m dot v 1 h v 1 that is the enthalpy change of the water vapor plus m dot 3 times h 3. So, we have basically taken everything to the right hand side that is why this sign has become plus. And since dry air is being modeled as an ideal gas we may write h a 2 minus h a 1 as C p a times T 2 minus T 1. And just like before we use the definition of omega to replace m dot v in terms of m dot a the mass flow rate of dry air. So, expression reduces to something like this and h v 2 is approximated as h g of T 2 h v 1 is approximated as h g of T 1 and enthalpy of the liquid water is equal to the enthalpy of the saturated liquid at temperature T 3. So, we can retrieve these values from the temperature table h g of T 2 h g of T 1 and h f of T 3. And if you plug these values into the above expression we get q to be equal to this which when converted to tons comes out to be at 27.17 tons of refrigeration. So, you can see how the relative humidity in the air is controlled. So, you can see that when we when we cool the air so at the inlet the temperature is 35 degree Celsius. So, this is a typical atmospheric conditions in a place like Chennai 35 degree Celsius 80 percent relative humidity. So, you take this air and you first reduce the temperature by using a cooling coil when you do that the water vapor condenses. So, you can see that the temperature is reduced however the relative humidity has increased which is not nice. So, we want the temperature to be comfortable relative humidity also to be comfortable. So, this air will then be taken to another unit operation duct where we may heat the air slightly and adjust the relative humidity so that both the temperature and the relative humidity are comfortable. See 22 is somewhat on the lower side temperature of 28 degree Celsius is on the lower side it can probably be adjusted so that we get the relative humidity and the temperature to a comfortable value. So, this is how the ambient environment in large buildings is controlled. The next example is about a domestic evaporative cooler. So, it is quite popular in many dry or in regions where the climate is dry. As I said you know this cannot be used in Chennai because relative humidity is already high. But it can be used in cities like Hyderabad or other places where the relative humidity of the air is somewhat on the low side. So, basically you have a wick which is wet and you blow air through this wick the water evaporates and the evaporation of the water draws energy from the air and the air cools down. So, this is an evaporative cooler. However, notice that the relative humidity of the air in the room increases as a result of the water that evaporates into the room which may not be a bad thing where the climate is dry because relative humidity already is very low. So, additional moisture is good and reducing temperature is also good. So, it is used in it is used as sort of like a poor man's air conditioner in places where the air is dry and hot. So, it is a cheaper alternative to air conditioner where the air is hot but relatively dry. So, hot air at 40 degree Celsius 1 atmosphere 15 percent relative humidity enters the cooler at a rate of 1.344 meter cube per second. Saturated liquid water at 25 degree Celsius enters the wick where it evaporates a fan driven by an electric motor rated at 250 watt pulls the air through the cooler and sends it out at 25 degree Celsius. Assuming steady state operation calculate mass flow rate of liquid water and relative humidity of the air at the exit the pressure remains constant. So, a sketch of the domestic cooler looks like this. So, we have a wick which is soaked in water. So, air hot and dry air comes in. So, as a result of passing through this wick the water evaporates and temperature of the air that is going through it comes down and then the humidity of the air goes up this is then pulled through this by a fan and it is then sent into the room. So, here is where the room is. So, the relative humidity is available at the inlet. So, we look up P sat corresponding to 40 degree Celsius from the temperature table which is 7.381 kilo Pascal. So, the partial pressure of water vapor at the inlet is 1.107 kilo Pascal and the humidity ratio at the inlet comes out to be 6 grams 6.8 grams of vapor per kilogram of dry air. Notice that the amount of water vapor in the air is very low since the relative humidity is only 15 percent. So, mass flow rate of dry air may be evaluated in the same manner as before we are given that the volume flow rate of air at the inlet is 1.344 meter cube per second. So, if you convert this in the same manner as before this comes out to be 1.5 kg per second. Notice that mass flow rate of water in this case is the exact opposite of what we did before in the previous example water was condensing from the air and being collected and was leaving the control volume here water is being added to the control volume, but the balance remains the same. So, the mass flow rate of vapor at the exit is nothing but mass flow rate of vapor at the inlet plus how much ever water is being sent into the into the wick. Remember the wick is operating at steady state which means that the mass flow rate that is being sent into the wick is all evaporating while the wick itself remains soaked and contains the same amount of water. So, we can work out the mass flow rate of water from this and we also use the definition of humidity ratio to write this in terms of omega. Now, SFE notice that we do not have enough data to evaluate this omega 2 is not known omega 1 is known, but omega 2 is not known. So, we will not be able to evaluate m dot w from this expression. So, SFE applied to the control volume looks reads like this q dot is 0 there is no heat loss, but w x dot is not 0 because there is a fan and electrical work is being supplied to the fan. So, if we rearrange the terms in this expression we get minus w x dot equal to this and again we have collected the terms for dry air and collected the terms for water vapor and then liquid water. And again in the same manner as before we treat the dry air as an ideal gas and approximate Hv as Hg of T2 and Hv1 again as Hg of T1 and so we can now so everything is known in this except for omega 2. So, if you rearrange we get an expression for omega 2 that reads like this and if we substitute the known values into this expression we get omega 2 to be 0.0132 kg vapor per kg dry air remember it was 6.8 grams of vapor per kg of dry air now it has increased almost doubled. So, now we may calculate the required mass flow rate of water and that is 9.492 gram per second. So, based on this analysis you can you should be able to realize that you know we can size the size them the cooler itself based on this the power of fan that is required mass flow rate of air that can be drawn in amount of water remember water is a precious resource. So, we need to ensure that you know we get we use as little water as possible while getting the best performance. So, all these design variables can be identified from the analysis that we are doing. So, that is the power of the concepts that you have learned in the previous course which are being applied to practical applications here. So, the relative humidity of the air may be evaluated using this expression. So, here we know omega and we are actually calculating phi from omega. So, the familiar expression for omega is like this 0.622 pV divided by p minus pV. So, p is known at the exit that is one atmosphere pV is also known at the exit we evaluated pV at the exit. So, that is 25 degree Celsius. So, that is p sat of 25 degree Celsius. So, this is p sat of 25 degree Celsius. So, if you rearrange this we may be we should be able to evaluate the relative humidity from here which comes out to be 69.33 percent. So, the relative humidity was 15 percent at the inlet and at the exit it is 69 percent. So, the air the temperature of the air was 40 degree Celsius at the inlet. So, it was 40 degree Celsius at the inlet and it has become 25 degree Celsius at the exit and the relative humidity is 69 percent which is not bad. Now, so far we have defined terms such as humidity ratio relative humidity and we have actually worked out examples where these quantities were given or with the other quantities we were asked to calculate these quantities, but in a practical HVAC application we need to know the temperature of the air which is relatively easy to measure. We also need to know either the relative humidity or the humidity ratio in order to be able to control this. In other words, so in all these examples what we are asking now is how do we get this only when I know this or if you go to the previous example. So, if I am trying to design let us say an air conditioning system I need to know this. So, how do I evaluate phi or omega for the air in this room? So, once I have a method or technique by which I can measure it then I can measure and then control it and then check to see whether it is working properly or not. So, what we are going to do next is try to understand how the two terms one of the two terms if I have one then I have the other one of the two terms omega or phi how it can be measured. So, we use a device called a sling psychrometer for this purpose. So, the sling psychrometer has a handle and it has two thermometers one is the usual thermometer it is called the dry bulb thermometer because the bulb of the thermometer is dry normal. So, it is called the dry bulb thermometer and the temperature that this reads is the temperature of the air in the room as we understand it normally although in the refrigeration air conditioning community it is called the dry bulb temperature it is the normal temperature that we have been using so far. So, when we say air at 25 degree Celsius enters the duct that is the temperature that we are talking about. Now, the other thermometer is called wet bulb thermometer because it is a bulb is covered by a cloth which is wet. Now, the cloth should be moistened in such way that it should be soaked just soaked but not dripping. So, it should just be at the point of dripping. So, we keep adding water to the to the cloth so that it becomes more and more moist and when it is just about to start dripping meaning no more water can be held by the sock that is when we stop this. So, that is the state of the cloth sock or the wet bulb. So, now we hold the handle and then swing the two thermometers like this in a horizontal plane. Now, as the thermometers are swung in a horizontal plane the water begins to evaporate from the cloth due to convective heat transfer in the air it begins to evaporate. Now, and as it evaporates it cools the bulb and the temperature of the and the temperature reading in the thermometer goes down. So, this reads the dry bulb thermometer reads a constant value which is the temperature of the air in the room which remains constant. Now, the temperature reading in the wet bulb thermometer keeps going down as the water from the wick or the cloth I am sorry the cloth evaporates. Now, at some point the temperature after decreasing no longer decreases it remains the same. This happens when the air in the vicinity of the cloth has become saturated not the air in the entire room, but the air in the vicinity of the sock has become saturated. So, that no more water can evaporate from the from the wet cloth. So, this temperature is called the wet bulb temperature. So, now we have two quantities that we have measured the dry bulb temperature which is a usual temperature and a wet bulb temperature. So, what we need now is a thermodynamic model which we can use to evaluate either omega or phi from these two measurements. So, we need to construct a thermodynamic model which represents the process and that the water vapor near this cloth has undergone. So, we turn to something called this model is called the. In fact, this temperature as I said is called the wet bulb temperature, but the model that we build will give us what is called an adiabatic saturation temperature. And this adiabatic saturation temperature is a very close approximation very good engineering approximation of the wet bulb temperature. We do not really want to get caught up with the heat transfer, the details of heat transfer near the cloth and so on. That would be superfluous for a thermodynamic analysis. So, we built a thermodynamic model which approximates this process reasonably well and the temperature that we get out of that is called an adiabatic saturation temperature. So, the model that we have idealizes or visualizes what is called an adiabatic saturator. So, basically moist air at temperature T1 enters a duct which is insulated and there is a reservoir of water at a temperature T3. Now, T1 is greater than T3. So, the air is hotter than the water in the reservoir. So, as the air flows over the reservoir, the water begins to evaporate. And if we make the duct long enough, although I have shown it short in this illustration, if I make this duct very, very long like this, then the air when it leaves the duct say let us call this 2, if I make it very, very long. So, the air when it leaves is at the same temperature as the water and the relative humidity will be equal to 100 percent. The air is saturated because it is picking up more and more of the water vapor that is evaporating. Its moisture content increases and its temperature drops. Remember, so when it finally leaves, in fact it will leave at the same temperature and become saturated, T2 will become equal to T3. And as we have already said T1 is greater than T3. So, the air, the temperature of the air decreases because the enthalpy that is required for evaporating the water is drawn from the air stream. So, the temperature of the air stream drops and at the same time its moisture content increases. So, when it leaves the adiabatic saturator, two things are important. T2 is equal to T3 and the relative humidity is 100 percent. In fact, the temperature T2 would be an excellent engineering approximation. This is the adiabatic saturation temperature and there is an excellent approximation to the wet bulb temperature. So, this is a thermodynamic model which simulates or mimics the wet bulb in the wet bulb thermometer. So, the highlights are the ducts should be made sufficiently long and at the exit the temperature T2 is equal to T3 and T2 is equal to 100 percent as we have already said. And we can say that the mixture pressure remains the same, there is nothing to change the mixture pressure. So, that remains the same between inlet and exit. So, the temperature T2 is called the adiabatic saturation temperature and if you take P1 equal to 1 atmosphere, then the adiabatic saturation temperature is an excellent approximation of the wet bulb temperature. Although, may be not so for P2 equal to let us say 2 atmospheres. There are psychrometric applications where the pressure is not 1 atmosphere but 2 atmospheres. So, the departure between TAs and T wet bulb will be higher at the higher pressure, but for 1 atmosphere it is an excellent approximation and that is what we are going to use. So, first we start off with the mass balance of water m dot v2, mass of water vapor that leaves is equal to mass of water vapor that comes in plus mass flow rate of liquid water that is being supplied. So, we are at a steady state. So, make up water has to be supplied to the reservoir. Remember, make up water has to be supplied to the reservoir so that the level remains the same in the reservoir. So, the make up mass flow rate has to be adjusted so that the level remains the same and that is an indication that steady state has been attained. So, we may write like this the mass flow rate of dry air remains the same. So, we may write it like this m dot a times omega 2 minus omega 1. Now, SFE applied to the control volume that shown looks like this q dot is 0, w x dot is 0 and again we sort out or we separate the enthalpy change for the dry air water vapor and liquid water. And in the same manner as before we approximate this like this the calorically perfect gas assumption and again Hv2 is approximate as Hg of T2, Hv1 is approximate as Hg of T1 and the mass flow rate of dry air comes out everything is written in terms of omega. So, the mass flow rate of dry air comes out. So, since it is equal to 0 this drops out and we end up with an expression that looks like this. So, if you look at this expression T2 is known remember we are saying T2 is T wet bulb for P1 equal to 1 atmosphere. T1 is nothing but the dry bulb temperature that we measure. So, this is the air whose relative humidity or omega that we are trying to measure. So, T1 is the dry bulb temperature T2 is the wet bulb temperature. Now, this is known this is known Cp is known. Now the only quantity that is not known in this or at least it appears prime of AC that there are two quantities that are known both omega 1 and omega 2. But you have to bear in mind that omega that the air is saturated at exit. So, which means that omega 2 is equal to omega sat of T2. In other words, Pv2 is equal to P sat of T2. So, omega 2 is equal to 0.622 P sat of T2 which is known divided by P minus P sat of T2. Note that P is known P sat of T2 is known. So, omega 2 is also known only omega 1 is unknown. So, omega 2 is known only omega 1 is unknown. So, omega 1 alone is not known all the other quantities are known. So, if you rearrange the previous expression after taking omega 1 to the left hand side we end up with an expression like this and as I just mentioned omega 2 may be written like this. So, once we have measured the wet bulb and the dry bulb temperature which correspond to T2 and T1 in this expression omega 1 may be evaluated. So, once a wet bulb temperature is known omega 2 may be evaluated and then based on that omega 1 may be evaluated. So, that is how we measure omega once omega is known evaluating relative humidity is very easy to do. Normally in engineering applications omega is the quantity of interest because that is ratio of mass of water vapor to dry air. So, that actually tells us how much water needs to be added to the air or how much water should be removed from the air. So, omega is the quantity of engineering interest, phi is the quantity of practical interest. Relative humidity is what people say what is the relative humidity in this room it is high or low. But omega is the quantity of engineering interest because that is what we are going to control by adding mass water to the air or by removing water from the air. So, by measuring wet bulb and dry bulb temperature we can actually evaluate the humidity ratio in the given sample of air. And the important thing is the dry bulb temperature obviously is greater than the wet bulb temperature which itself is greater than the dew point temperature. The similarity probably some of you may have realized the similarity between the dew point temperature and the wet bulb temperature. Wet bulb temperature is not the dew point temperature that you should bear in mind because the bulb is still wet and it reaches a steady state it does not go below that. In fact, this is shown in our TV diagram like this. So, the given sample of water this is the so this is at a temperature given by the dry bulb thermometer and this is the temperature given by the wet bulb thermometer and this is the dew point temperature. So, you can see that T dry bulb is greater than T wet bulb and which itself is greater than the dew point temperature. So, the wet bulb temperature comes out it is on the saturated vapor line because the air in the vicinity of the wet bulb is saturated not the air in the room. So, it falls on the saturated vapor line for that reason. So, as you can see from this figure T dry bulb greater than T wet bulb greater than dew point temperature. Let us work out a couple of examples involving the dry bulb and wet bulb temperatures. So, the dry bulb and wet bulb temperatures in a room are measured to be 25 degree Celsius and 18 degree Celsius respectively assuming the pressure to be one atmosphere determine the humidity ratio relative humidity and dew point temperature. So, from the temperature table we can get P sat of 25 degree Celsius H v 1 which is H g of 25 degree Celsius P sat corresponding to the wet bulb temperature H f corresponding to wet bulb temperature and H g corresponding to wet bulb temperature. So, all these values can now be plugged into this expression. So, remember this is T wet bulb. So, we first calculate omega wet bulb which comes out to be like this. So, omega wet bulb expression for omega wet bulb is given here. So, omega wet bulb comes out to be like this and from which we can evaluate omega to be 0.01 kg vapor per kg dry air. So, once I have omega from the expression for omega we may evaluate the partial pressure of water vapor to be 1.603 kilo Pascal dry bulb temperature is 25 degree Celsius. So, the relative humidity may be evaluated as 50.6 percent. Now, from the temperature table T sat of 1.603 kilo Pascal is 14 degree Celsius and that is the dew point temperature. So, you can see that dry bulb temperature is 25, wet bulb temperature is 18, dew point temperature is 14. So, T dry bulb greater than T wet bulb greater than T dew point. So, in the next lecture what we plan to do is to introduce the psychrometric chart. Now, the analysis that we have done so far mass balance of water or mass balance of dry air SFEA applied to the control volume. These are actually quite adequate for solving any such problem in HVAC. But the engineering community actually uses a psychrometric chart for doing calculations in HVAC. That is why the chart is very handy and quite useful also. So, what we will do in the next lecture is to see what the chart is all about, why we need a chart and how the chart can actually make life easy for easy while doing such calculation. So, there is there are no shortcomings with the procedures that we have outlined. So, the chart simply makes it easy to do this calculation. In fact, we can do the calculations faster if you use the chart. So, what we will do is we will try to solve each one of this example using the chart. So, that you understand that there is nothing wrong with what we have done or there are no shortcomings in what we have done. It is simply easier and quicker to use the psychrometric chart which is for that reason it is quite extensively used by the refrigeration community.