 Good afternoon, welcome to the afternoon session. What we will do is maybe do a problem or two from psychrometric processes. So you will see that these handouts are already given to you in last week and another two problems in compressible flow and we will give you a general outline of what we are doing and we can go ahead with this. So I have put on this sheet on psychrometric processes and problem one, I mean I can read out a system containing moist air at a dBt of 30 degrees. So I will just write dBt of 30 degrees and the dew point is 20 degrees and two questions have been asked what happens if P total that is the system pressure is one atmosphere and if it is 0.9 bar. So based on these two pressures what we need to find out is omega, R H, H and V. So these are the quantities that need to be found out. So the total pressure has been given one atmosphere at 0.9 bar just to ensure that you just do not go ahead and look at the psychrometric chart rather than that you would rather look at the steam tables. So what you can do now is once you know the dew point temperature as I said you can find out the saturation pressure at the dew point temperature and that is what is needed to get PV. So let me just go to the next page here and redraw my TS diagram. I am on some pressure line here and the temperature is given as 30 degrees C. So this is the 30 degrees C line and if I cool at the same PV I reach this saturation point. So this overall is the 20 degrees C line. So what I need is the saturation pressure here and that is very straightforward. I just have to go to the steam tables and I will not project the steam tables. I will ask all of you to go to the steam tables that you may have and what we can do is go to the temperature part of the steam tables and at various temperatures the saturation pressure would be given. So if I look at the saturation pressure at 20 degrees C it is 0.02339 bar. So this is the saturation pressure that is given at 20 degrees centigrade. So that means the PV here is 0.2339 bar and I can express in Pascal or I can express this in mm of mercury but I should express it in a consistent unit. So what we will do is right now stick to Pascal's. So if I just multiply this number by 10 raise to 5 I will get my answer in Pascal's. So this would be 1, 2, 3, 4, 5. So 2339 Pascal's. So what I will now do is if I need omega it is very straightforward. We need 0.622 PV upon P total minus PV. So now depending on the total pressure this will either be 1 atmosphere or it will be 0.9 bar or 90,000 Pascal's and this would be 1.01325 into 10 raise to 5 Pascal's. So the moment I have the dew point temperature the PV gets fixed irrespective of the total pressure the PV gets fixed and what I really need to find out omega is what the total pressure is. So if the total pressure is different I get a different omega and that is what is needed. So I can get omega like this. Now if I want the relative humidity, relative humidity is just PV by PV sat and PV sat is to be calculated at 30 degrees C. So again we go to the steam table go to the temperature part of the steam table and we see that PV sat at 30 degrees C is 0.04246 bar or 4246 Pascal's. So this is just the ratio of 2339 upon 4246. So this is how you would get your relative humidity. Again it depends on the dry bulb temperature and PV and you will realize it has nothing to do with the total pressure at this time. And the next thing you would require is specific enthalpy H is again CP of dry air into T. T is 30 degrees and this we will express in kilojoules 1.005 kilojoule per kilogram Kelvin and then we will add omega. Now you realize that you have two values of omega one at one atmosphere and one at 0.9 bar and I will get 2501 plus 1.88 which is the CP for water vapor into T and this is 30 degrees C. So the only difference here in the calculations is in the omega which is different for these two different pressures and that will change. So this is calculated on per kg dry air basis. So how much omega is being carried per kg dry air is entirely dependent on the total pressure because the total pressure will determine what is the partial pressure of dry air and hence the amount of water vapor per kilogram of dry air changes. And finally if I want specific enthalpy again this is per kg of dry air and we said we will just say P dry air V dry air is equal to M dry air R dry air into T. T is 30 this is universal gas constant by 29. This we take as 1 kg and this is now going to be calculated as P minus P V and you will have two values. If P is one atmosphere you get a different value for pre dry air and if P is 0.9 bar you get a different value for P dry air. In both cases P V is the same but pre dry air changes and the specific volume would be just R dry air into T upon P dry air that we need to calculate. So this is how we would use only the steam tables and go ahead. So I will just go ahead with the second problem too and in this case various things are given either the temperature is given with omega or the temperature is given with WBT or the temperature is given with RH and the temperature is given with the dew point temperature. Now the temperature given with dew point temperature is exactly the same as what we did in problem 1. I use the dew point temperature get P sat at dew point temperature from steam table which means I know P V which means I know omega. Once I know omega I can get H and dry bulb temperature I will get the P sat at dry bulb temperature and using this I will calculate RH because this is just P V upon P sat at temperature. So I get all properties of essence that I need to calculate now if I have T and relative humidity what it means is I can using T I can get P sat at T. So I will use relative humidity because it just gives me gives me P V by P sat. So P V will be just relative humidity as a fraction multiplied by P sat. Once I know P V I can get omega and once I get no omega I can get H so I know everything. Similarly T and WBT I will use the carrier equation I will use the wet bulb temperature to get P sat at wet bulb temperature again I will use the steam table substitute and get my P V actual. Once I know P V actual I know omega I know H and using T I know P V sat and hence I get get relative humidity if I have given T and omega I will use the relation between omega and P V and get P V I can directly get using T and omega what is the enthalpy and using T I will get P V sat and with these two ratios I will now get my relative humidity. So given any combination here I would be able to get my necessary quantities and go ahead and this is what is required when we do the other problems forward. So I hope once this is straight forward and you have the simple mixing formula we can go ahead. So I will leave it to Professor Puranik to solve a problem or two regarding compressible flows and then we can take further questions on any of these. Thank you.