 So the next thing that we'll be doing with this example problem, one where we cannot use a psychometric chart, is we will apply the first laws. We've applied mass flow rate conservation to both the dry air as well as water vapor. And with that we are able to determine the amount of water coming in. We know a specific humidity of 1, 2 and 3. The things that we're still trying to find, however, is temperature in the midsection at point 2. We don't know the relative humidity there either nor do we know the amount of heat flowing in between 1 and 2. So we'll now take a look at the first law and see what that can do for us. So we're going to begin by applying the first law from 1 to 2. So this is our heating section. So that's the full blown form of the first law there. Now for a heating section we're not doing any work. We're either just adding heat or not removing it. So the work terms disappear as well as does the heat out because we're putting heat in and what we remain or left with is Q in. And again I'm doing this thing here by expressing it in terms of the mass flow rate of dry air because remember we're dealing with enthalpy for h back and there we always express enthalpy per kilogram of dry air. So we have h2 and h1 in this equation. We already know m.a, the mass flow rate of dry air. So let's write out h2. So looking at our expression for enthalpy at state 2 and similarly enthalpy at state 1. Okay so when I look at this equation we know the inlet temperature t1. We know specific humidity 1. We know specific humidity at 2. However we do not know t2 and so consequently we do not know h2. So that's an unknown that we still have in our problem. So let's move along and let's take a look at applying the first law between points 2 and 3. Actually I'm going to go back to one slide here and given that we do know the conditions for h1 I am going to evaluate that. So I can plug in the values for t1 as well as specific volume of 1 and we get the enthalpy at 1 to be 24.3608. But what we can say is we do not know h2 at this point because we do not know t2. So let's go back to the first law looking between states 2 and 3 and see if there's anything more we can pull from this. And what we have here this is our system whereby we're injecting steam. So we basically just have mass flow in mass flow out multiplied by the enthalpy stream so let's expand that. Okay so we have this expression here. Now we know the mass flow rate of air at 2 is equal to the mass flow rate of air at 3. And the other thing that we can determine is the enthalpy of the inlet stream. We're told it's 100 degrees C saturated vapor and so from that you can go to your steam tables. So that's how we can get that. So what I'm going to do I'm going to isolate for h2 in this equation because remember we don't know t2. If we know h2 that will help us get t2. So in this equation we know a number of these different terms. But one term that we don't know however we still don't know what h3 is so if we can determine that we know everything else in this equation. That will enable us to get h2 and then t2 indirectly. So let's work towards getting h3. So we have that expression. Now we know t3. We know specific volume of 3 so we can directly calculate h3. h3 then turns out to be the following. And I put da there for kilograms of dry air. Now we'll go back to our last equation here. And we now have all the things enabling us to get h2. So let's go ahead and plug those values in. Oops sorry. So we get that for h2 by plugging in the values. It's 34. And remember what we're after here. We want to know the mid-temperature. So let's equate that to our definition for enthalpy at 0.2. And here I have expanded the second term. And so with this what we want to do we want to go ahead and calculate or solve for t2. So when we do that we get t2 is 14. Sorry 19. 19.49 degrees C. And that is answered apart A. Now they also want us to get the relative humidity at this temperature. So let's go ahead and solve for the relative humidity. So again there what we do we go to our steam tables for this temperature of 19.5 degrees C. And the specific volume at 2 is the same as the sorry specific humidity at 2 is the same as specific humidity at 1. And with that we can then directly evaluate the relative humidity at 0.2. So the relative humidity at 0.2 turns out to be 37.8 percent. And so that is answered to part A. So notice we relative humidity with heating. We went from 70 percent down to 37.8 percent so it was dropping as a result of the heating. The final thing they want us to determine is the heat transfer in our heating section. So let's take a look at that. For that we go back to the first law between 0.1 and 0.2. And that one we expressed in terms of the flow rate of dry air. And we have all the terms required to solve for this now. That's in kilojoules per second and given they gave us flow rate in per minute I'll express it in kilojoules per minute. And so that gives us the answer to part B. So that's a solution to this example problem. You can see it's a little bit laborious when you don't or are not able to use a sacrametric chart but nonetheless it does demonstrate all the different equations that you can use for solving these sorts of air conditioning processes. The next example we'll look at will be quite a bit easier because it will be at one atmosphere and consequently we'll be able to use a sacrametric chart.