 Let us look at a few work examples which will illustrate the concepts that we have studied so far before moving on to the actual application. First one is a simple example ambient air at 25 degree Celsius in one atmosphere is observed to have a dew point temperature of 15 degree Celsius, determine the relative humidity and the humidity ratio. So we are given that dew point temperature T sat of PV is 15 degree Celsius. So we go to the steam table, temperature table and we can retrieve PV to be 1.706 kilo Pascal. So this is the saturation pressure corresponding to, so I am sorry, this is the saturation pressure corresponding to 15 degree Celsius, let us just quickly go back and take a look at this. So T sat of PV equal to 15 degree Celsius. So what we do is we go to the table, we go to the table and then we look at the pressure for which the saturation temperature is 15 degree Celsius. Now P sat of 25 degree Celsius can be retrieved from the table again from the temperature table as 3.169 kilo Pascal. So we can evaluate relative humidity as PV over PV P sat of 25 degree Celsius and this comes out to be 53.83 percent. Now we can also calculate the humidity ratio using this expression, since we know PV the relative I am sorry humidity ratio comes out to be 0.01065 kg vapor per kg air. So you can see that the amount of water vapor is not very high. It is usually of the order of few grams. So it is important that you sort of have this diagram in mind all the time so that it is helpful when you are trying to solve problem. So for this problem what we could have done is we could have drawn for example, a TV diagram like this. So water vapor is at 25 degree Celsius. So this is 25 degree Celsius and this is the isobar. So the actual state of the water vapor let us say is over here. So this is the isobar that passes through this given state and this temperature is given to be 15 degree Celsius. So we go to the temperature table and determine the pressure for which 15 degree Celsius is the saturation temperature and then we retrieve. So this is 1.706 kPa and P sat is 3.169. So this is the isobar corresponding to 3.169 kPa. So now relative humidity and other quantities may be evaluated. So please remember this TV diagram and try to use this when you are solving problem. It will be very, very helpful. Next example reads like this ambient air at a geographic location is 25 degree Celsius, one atmosphere and 90% relative humidity. Determine the maximum amount of liquid water that can be extracted per unit volume of ambient air. You may know nowadays it is sort of becoming popular to extract water from the air and use it for drinking purposes. So there are machines which now nowadays do this. It extracts water, it is a pure water that is taken from the air and so on. So let us try to find out how much water can realistically be extracted. In fact this example uses 90% relative humidity that is relatively high. So we must be able to extract a lot of water, one more thing let us try to calculate. So saturation pressure corresponding to 25 degree Celsius is 3.169 kPa from the table. So for phi equal to 90 from the definition of relative humidity we can evaluate the partial pressure of water vapor to be 2.8521 kPa. And since we are assuming the water vapor to be an ideal gas the amount of if you assume the total volume, mixture volume to be 1 meter cube we can actually evaluate the amount of water vapor to be 20.72 grams or since density of water is 1000 kg per meter cube we can say that we can extract from this air 20.72 cc of liquid water from 1 meter cube of air. Even with such a high relative humidity we are able to get only about 21 cc of liquid water under ideal condition, practical situation it is likely to be much less than this. So you know extracting water from air to generate drinking water is not really a very practical way of you know producing drinking water plus you have to keep in mind that as you extract the moisture from the air the air becomes dry and as I said before dry air is also not very comfortable that can also cause lots of health issues. So this my opinion is not a very practical or practicable way of producing drinking water. The last example that we are going to look at is this 1 kg of air at 25 degree Celsius 1 atmosphere and 70 percent relative humidity is compressed to 5 atmosphere in a polytropic process. It is then cooled at constant volume to 25 degree Celsius determine the humidity ratio and the relative humidity of air after the compression process also determine the amount of water if any that condenses as a result of the cooling and the final pressure. So in the initial state relative humidity is given. So we know P sat water at 25 degree Celsius so we can evaluate the partial pressure of water vapor initially to be 2.283 kilopascal and the humidity ratio in the beginning may be evaluated using this expression it comes out to be 0.0139 kg vapor per kg dry air. So this is the initial state the water vapor so it is at 25 degree Celsius and as I said an amount of relative humidity. Now it is heated in a polytropic process so it goes from state 1 to state 2. So this is a polytropic process illustrated just qualitatively. So the temperature at the end of the compression may be evaluated using the given expression for the polytropic process and that comes out to be 411 Kelvin. Now after the compression remember no water is added or removed so the humidity ratio remains the same as 0.0139 kg vapor per kg dry air and since omega 2 is known we may evaluate the partial pressure of water vapor using this expression remember we already had this expression so omega equal to 0.622 times PV over P minus PV. So if you rearrange this expression since omega is known and the mixture pressure final pressure is also known remember it is compressed to a pressure of 5 atmospheres so that pressure is also known so you may evaluate PV from this as 11.07 kilopascal. The temperature at the end of compression is 411 Kelvin or 138 degree Celsius so P sat corresponding to 138 degree Celsius is 341.92 so the relative humidity comes out to be 3.24 percent. So some of these numbers are entered here so T sat corresponding to I am sorry T sat is the final mixture temperature so here is the isotherm corresponding to that and this is the saturation pressure corresponding to 138 so this is the corresponding isobar. So you can see that the relative humidity has decreased considerably although no water has been removed but the humidity ratio remains the same because no water has been added or removed. So now the water undergoes constant volume cooling process until it reaches the final state 3 where it is temperature is 25 degree Celsius. Now we have illustrated this qualitatively but what we do not know is whether state 3 lies state 3 may also lie here above the saturated vapor line or it can lie below the saturated vapor line as indicated here. So we need to find out the temperature at which condensation will begin in other words we need to find out the temperature of this state point on the saturated vapor line. Let us try to do that so cooling takes place at constant volume. So condensation begins at that temperature when the specific volume of the vapor becomes equal to the specific volume of the vapor becomes equal to specific volume of the saturated vapor. So you can see here that at this state specific volume of the vapor V is also equal to specific volume of the saturated vapor at that temperature. So at this temperature the specific volume of the saturated vapor is equal to specific volume of the vapor that is present in the air. Now, the specific volume of the vapor after the compression process may be evaluated using ideal gas equation of state. We know the temperature, we know its partial pressure, so we may evaluate the specific volume as 17.15. So, we go to the temperature table for water and establish that for a temperature of 42.8 degree Celsius, Vg is equal to 17.15. So, condensation begins at 42.8 degree Celsius, which means that for the final temperature of 25, definitely the state is going to be on this side and some water vapor will definitely condense. So, the final specific volume again remains the same at 17.15, so we may evaluate the dryness fraction at state 3. So, the dryness fraction at this state may be evaluated as 0.3955 and based on the, remember the definition of dryness fraction is mass of vapor divided by mass of water or mass of mixture which contains both liquid and vapor. So, basically this is mv divided by m liquid plus m vapor. Actually, we may, for the present purpose, we may write this as m minus m liquid divided by m, which itself may be written as 1 minus m liquid over m. So, m liquid is what we are denoting here as mw. So, you can see that m liquid is nothing but 1 minus x times the mass of water, but the water was present initially entirely as vapor. So, that means that this mass of water is equal to the m here is equal to mv1 and that is what we have used here because the water is entirely present as or present entirely as vapor at the beginning of the process. So, from which we may evaluate mass of liquid as 1 minus x times mass of vapor and mass of vapor itself by definition may be written like this. Omega is nothing but mass of vapor divided by mass of dry air. So, omega 1 times mass of dry air and we may then simplify this further using the relation to the total mass of the mixture and finally get this as 8.287 grams. So, that is the amount of water that condenses. Now, the amount of water that condenses is only a few grams. So, its volume is going to be negligibly small. So, we neglect the volume occupied by the liquid water. So, the final pressure may then be evaluated like this because state 2 and state 3 are connected by a constant volume process. So, Pa3 V3 over R times T3 equal to Pa2 V2 over R times T2. So, R cancels out, V3 equal to V2 constant volume process. So, this also cancels out. So, we may evaluate the final pressure of the dry air to be 359.308 kilo Pascal. And the partial pressure of the water vapor at the final state is equal to the saturation pressure as that you can see from here. So, the partial pressure of the water vapor is equal to saturation pressure because it is a saturated mixture. So, the pressure of the dry air is known, pressure of the water vapor is also known. So, the mixture pressure is the sum of the two and that is equal to 362.477 kilo Pascal. So, these three examples illustrate the use of the concepts or terms humidity ratio, relative humidity, partial pressure of water vapor, saturation pressure, saturation temperature and so on. So, these all these are interlinked and these three examples indicate how they are interlinked and how they can be evaluated in practical situations. So, what we will do in the next lecture is apply first law to psychrometric process. So far, we have only looked at equation of state and definitions of omega, humidity ratio and relative humidity and saturation pressure and saturation temperature. Now, we will apply first law because we said psychrometric applications involve heating cooling which means first law and humidifying dehumidifying is addition or removal of water vapor.