 So, we will look at application of first law to psychrometric processes where we will look at both flow as well as non-flow processes. Although flow processes are probably more common in HVAC because if you want to heat or cool a building normally you have the refrigeration power plant perhaps on the roof of the building and then the plant is connected to the rooms through ducts and so on. Flow examples are more common although non-flow examples are also non-flow situations or applications are also there. So, we will look at both of them. Now, application of first law to psychrometric process is quite straight forward. The simplicity is that you know we are simply treating moist air as a mixture of two components, two ideal gases dry air and water vapor. So, that is where the simplicity comes from. So, you know really application of first law to psychrometric process is just like application of first law using ideal gases, mixture of ideal gases that we have done before. So, it is no different from that. What the two points that we need to note is that the change in internal energy of dry air is calculated assuming that dry air is a calorically perfect gas. So, we assume CV to be constant and the specific internal energy of water vapor, this is for non-flow process specific internal energy of water vapor is usually approximated like this that it is just equal to the specific internal energy of saturated vapor at the same temperature. Although we can probably look up the value from the table for engineering purposes, this is simply not what the trouble and we simply approximate as u g of t. So, u v of t comma p is simply taken as u g of t. Even though we may know the partial pressure of the water vapor and the temperature, we simply just take it as specific internal energy as u g of t. So, we looked at an example before which was something like this where moist air at 25 degree Celsius and one atmosphere was compressed in a polytropic process up to state 2 to 5 times its initial pressure and then it was cooled at constant volume. So, what we are asked to determine now is the amount of heat removed during the cooling process. So, if we simply apply first law to the cooling process, we get delta e equal to delta u equal to q minus w. There is no change in there are no kinetic or potential energy changes. So, delta e equal to delta u and w is 0 because it is a constant volume process, no displacement work and there are no other forms of work also. So, w is 0. So, q is equal to delta u and delta u itself consists of delta u of dry air plus delta u of the water vapor. So, we may write this as delta u of dry air plus delta u of water vapor. And the delta u of dry air may be written as m a times c v a times t 3 minus t 2. And the delta u for water vapor may be written as mass for the water vapor times u 3 minus u 2. Notice that what we are doing here? So, basically delta u of water vapor would be mass of water times change in its internal energy, but mass of water remains the same and that is equal to m v 1. So, we write it as m v 1 times u 3 minus u 2. And by using the definition of omega 1, remember omega 1 is equal to m v 1 divided by m a. So, we may write this as m a times omega 1 and then we can pull out m a and write m a itself as m over 1 plus omega 1. So, then the expression becomes something like this. And u 2 is simply equal to u g at 138 degree Celsius. The temperature was 138 degree Celsius after the compression process. So, we can get this from the temperature table. And the dryness fraction was evaluated to be, so the dryness fraction was evaluated to be 0.3955. So, the specific internal energy of the mixture at state 3 may be evaluated as u f plus x times u g minus u f. So, that is 1016.04. What is that? The specific internal energy of the, if it is just vapor, then the specific internal energy of the vapor alone is approximated as u g at that temperature. If it is a saturated mixture, then we need to evaluate it in the same manner as we have always done. So, we take CV of dry air to be 717.86 joule per kg Kelvin throughout. And if you substitute the known values, we get q to be minus 101 kilojoules. So, that is the, that is application of first law to psychrometric problem, which as we can see is no different from what we have done before. Only important things that you need to keep in mind are these sorts of things, mass of vapor. So, for example, if you look at this expression, delta u for water vapor itself would have been equal to m of water, let us say H 2 O times delta specific of the water, which itself may be written as, say in the final state, we have mixture of liquid plus water, I am sorry liquid plus vapor. So, we say m H 2 O times u 3 minus m, now instead of writing m H 2 O, we can simply write this as mass of vapor, because initially we have only vapor. So, we write m v 2 times u 2. And this is only vapor. So, we approximate this as u g of T 2. So, you need to keep track of these sorts of things vapor, if the water condenses, then you have to take into account the mass of the condensate. The mass of the condensate may be small, but specific internal energy is of the order of few 1000 kilojoules per kilogram. So, that actually can be a significant quantity, mass times the specific internal energy. If it is just mass, then in certain applications, you may simply neglect the mass of the water vapor that has condense without any impact on the accuracy, but for specific internal energy, that will not be there. Or for internal energy, that will not be the case. So, you just need to keep track of these masses of vapor and liquid in the water quite accurately in psychrometric applications. Let us look at the next example, 2 kg of air at 25 degrees Celsius, one atmosphere, and 25 percent relative humidity is compressed isothermally until the volume becomes one-fifth of the initial volume. Determine the amount of water that condenses, if any, and the heat that is removed. So, basically you may think of a piston cylinder mechanism like this. This contains moist air and it is compressed until the volume becomes one-fifth of the initial volume. We are asked to determine any water or the amount of water that may condense and the heat that is removed. So, the relative humidity is given as 25 percent at state 1 and the temperature is also known. So, at state 1, as you can see from here, 25 degrees Celsius, one atmosphere and 25 percent relative humidity. So, we may look up P sat of 25 from the temperature table and we get the partial pressure of water vapor to be 0.79 225 kPa. So, that state is indicated here. Now, it is being compressed isothermally. So, that means we are following this line here. Now, if you cross the saturated vapor line during the process, then obviously the water will condense. So, we need to find out the pressure, temperature is already known. So, we need to find out the pressure at which the process line crosses the saturated vapor line. Since the compression process is isothermal, the final pressure is 5 times 101.325 kPa. So, basically what we are saying is the final pressure of the contents of the cylinder is 5 times the initial pressure which is 101.325. So, 5 times 101.325 is the final pressure of the contents of the cylinder because it is an isothermal process. Now, the partial pressure of the water vapor is basically, notice that the partial pressure of water vapor can only go up to. So, here is the isobar. So, the partial pressure of water vapor can go only up to p sat of 25 which is equal to, as we just looked it up, p sat of 25 is 3.169. So, once the partial pressure of water vapor exceeds that, water begins to condense. Since we have said that, you know, the pressure is 5 times that much and the volume is also going to be one-fifth of that. So, we have to be careful. But the important point is the partial pressure of the water vapor can increase only up to 3.169 which is just about 4 times its initial pressure. So, at the at the instant when the process line crosses the saturated vapor line, the partial pressure of water vapor becomes 3.169 which is 4 times the initial value, initial partial pressure of the water vapor. So, once it reaches there, the partial pressure of the water vapor remains constant because phase change begins to take place. Water vapor begins to condense. So, the partial pressure of the water vapor remains at 3.169 kilopascal, while the mixture pressure continues to increase as the contents of the cylinder are compressed. So, that is a very, very subtle point. So, the water pressure, partial pressure of water vapor can go only up to 3.17 and then it remains constant. So, the pressure increases from here to here and then remains constant. Whereas, the mixture pressure continues to increase until it is 5 times the initial pressure. So, the specific volume of water vapor at the initial state, it is slightly superheated. So, we will assume that it is an ideal gas and use the ideal gas equation of state and calculate it to be 173.74. The final specific volume is one-fifth of the initial specific volume V1 divided by 5 and the initial specific volume itself is equal to VV1 because it is in the vapor form. So, the final specific volume comes out to be 34.748 meter cube per kilogram. And we may evaluate the dryness fraction, once this is known, we may evaluate the dryness fraction as 0.8014. Using the definition of dryness fraction, we may evaluate the mass of water that has condensed to be 1.94 grams, just like what we did before. Note that using the definition of the dryness fraction, we may write the mass of liquid water as mw equal to 1 minus x times mv1. Remember, dryness fraction is defined as mass of vapor divided by mass of the mixture and mass of vapor itself may be written as mass of the mixture minus mass of liquid divided by mass of mixture. So, if you rewrite this, you get mass of liquid to be m times 1 minus x where m is the mass of the saturated mixture. Now, remember, we had certain amount of water vapor initially and out of that water vapor some amount has condensed and some remains as water vapor. So, the mass of the mixture is equal to the mass of vapor that we started out with. So, m in this case is equal to mv1. So, that is how we end up with this expression. Application of first law to the isothermal compression process gives delta E equal to delta U, no ke or pe changes, q minus w and w is a displacement work. So, that is p1 v1 natural log v2 over v1 where p is the mixture pressure and v is the volume occupied by the mixture. And delta U itself is comprised of two things delta U for dry air plus delta U for water. Delta U for dry air is 0 because it is an isothermal process and we have assumed dry air to be calorically perfect gas. In delta U for water is nothing but so delta U for water is mass of water times U2 minus U1. And mass of water is mv1 because there is a total amount of vapor that was present and part of which is condensed now. So, this may be written as mv1 times U2 minus U1. So, this was initially superheated. So, we approximate this as Ug of T1 and U2 itself may be evaluated like this 1951.3 using the temperature table. And the initial volume, remember we are using the Dalton's model. So, the initial volume is written like this. The initial volume is written like this because we are using Dalton's model and the components occupy the entire volume that is available. They are different pressures, same temperature and same volume. So, we may evaluate the initial volume as m volume occupied by vapor divided by the specific volume of the vapor. So, if you take the mass of the vapor outside, we may write it in terms of omega and mass of the mixture and after substituting the values we get Q to be minus 4.4736 kilojoules. So, this again illustrates how application of first law to a psychrometric process is carried out. And once again you must keep track of the mass of different phases in these problems especially when condensation takes place. If condensation does not take place then it is relatively straightforward mass of the individual components remains the same and we can just go ahead. If phase change takes place, phase change for the water takes place then we need to keep track of the amount of water that condenses and the amount that is remaining in the vapor phase. So, the next set of examples that we are going to discuss or steady flow examples we have seen non-flow examples so far. Next we will look at steady flow example and as I mentioned earlier HVAC involves unit operations like heating, cooling or dehumidification. So, we will try to look at examples involving these operations. Since these are steady flow applications we will encounter the enthalpy or specific enthalpy of individual components and not specific internal energy and the enthalpy of the mixture is calculated like this. It is the sum of the enthalpy of dry air plus water vapor. And as just like what we did earlier for specific internal energy the specific enthalpy of the vapor is approximated as Hg of T. Earlier we said U of the vapor specific internal energy was Ug of T. This one is Hg of T and the specific enthalpy of the air itself is written as Cp times T where we take Cp to be 1.005. So, we are assuming as I said before the dry air to be calorically perfect. So, the specific enthalpy of dry air is written as Cp times T. So, the first example reads like this ambient air at 5 degree Celsius, 1 atmosphere pressure and 85 percent relative humidity is to be heated as it flows steadily in a duct. So, that the relative humidity becomes 25 percent. The volume flow rate is 100 meter cube per minute at the inlet determine the rate of heat addition and assume pressure remains constant that is mixture pressure remains constant at 1 atmosphere. So, the inlet state is completely fixed. We know the temperature, relative humidity, volume flow rate at the pressure at exit relative humidity is given. We are asked to determine how much heat is being added. So, relative humidity is given at the inlet temperature is also known. So, we may evaluate P sat of 5 degree Celsius from the temperature table and evaluate the partial pressure of water vapor at the inlet to be 0.74205 kilopascal. And the humidity ratio may also be evaluated as 4.59 grams of vapor per kilogram of dry air. At the outlet the mixture pressure remains the same and since no water is added or removed the partial pressure of water vapor also remains the same. So, Pv2 equal to Pv1 equal to this and omega2 is equal to omega1. So, because no water is added or removed and the relative humidity value is also given. So, using the definition of relative humidity we may evaluate P sat of T2 to be 2.9682 kilopascal and from the temperature table we can then work out that the exit temperature is 24 degree Celsius. Now, we apply steady flow energy equation to the duct and this gives us q dot minus Wx dot plus m dot times h1 minus h2 equal to 0 and there is no work in this case. So, Wx dot is 0 and we may write h1 is nothing but so m dot times h1 may be written as m dot A times h1 or we may write m dot times h1 minus h2 we can split that into two things and write it as m dot A times hA2 minus hA1 plus m dot B times hV2 minus hV1. So, we treat this as two streams and hA2 minus hA1 itself may be written as Cpa times T2 minus T1. What is that m dot B may be replaced using the definition of omega. So, omega equal to m dot V over m dot here. So, we may write m dot V equal to omega times m dot A. So, we write like this and hV2 because it remains in the vapor state and does not condense hV2 may be approximated as hG of T2 hV1 may be approximated as hG of T1. Now, m dot A remember it is given that the volume flow rate is 100 meter cube per minute at the inlet. So, volume mass flow rate of dry air may be evaluated using this information V1 dot divided by its own specific volume which we may evaluate using the ideal gas equation of state. So, this works out to 126.072. What is that the partial pressure of air is used here to calculate the mass flow rate of air. So, that is the correct way of doing this P a1 which is equal to P1 minus P V1. So, notice that again we are using Dalton's model. So, the volume flow rate that is given although it is not specifically given that is volume flow rate of dry air that is understood because we are using the Dalton's model. So, volumetric flow rate is given. Once I use the partial pressures correctly I should be able to use the same value of volume because the entire volume as I said is occupied by all components of mixture only the pressures are different temperature and volume are the same. So, we can get hG at the inlet state and hG at the exit state from the temperature table and if you substitute the value we get Q dot to be 40.458 kilowatts. The next example that we are going to look at involves moist air but in this case we are actually going to cool and dehumidify coolant. So, we are going to cool the air remove the water vapor. So, in the previous case we actually heated the air in this case we are going to cool the air. So, here we have written mass of liquid water to be equal to 1 minus x times mv1. So, where we have used the definition of the dryness fraction. So, dryness fraction is defined as mass of vapor divided by mass of mixture and mass of vapor itself may be written as m minus mass of water divided by mass of mixture. So, that mass of water is equal to 1 minus x times mass of the mixture. So, we started out with the certain amount of water as you can see from here. So, we started out with the certain amount of water vapor at state 2 and now part of it has condensed as liquid and part of it remains as vapor. So, the mass of the mixture at state 3 is equal to the mass of water vapor that we started out with. So, we may then write mass of the mixture as 1 minus. So, this may be written as 1 minus x times mass of vapor at state 2 or I am sorry mass of vapor at state 1 is also okay because the vapor quantity of vapor remains the same. So, we may write this as mass of vapor at state 1 also.