 In the previous lecture, we saw how Kuwait flow model can be applied to discover effect of property variations through the ratio of molecular weight of the mixture in the infinity and w states. In as much as the result we obtained was approximate, it did showed the correct tendency that g over g star of the variable property varies directly with m w divided by m infinity and raise to some power of course. Today, we are going to look at Reynolds flow model and its applications. In fact, in the next two lectures, I will be dealing with the Reynolds flow model. The first of which today, we will deal exclusively with air water vapor system, which requires use of psychrometry and steam tables. The problems that I shall consider are shown here on this slide, the wet bulb thermometer, the measurement of relative humidity in a ducted flow, where we shall estimate the true vapor concentration or the R H allowing for effect of radiation and without effect of radiation. Then we will consider evaporation from a porous surface and then evaporation from a lake and then whereas, all these problems are of external flow, we will turn to internal flow in which we will try to humidify air, which is entering into a tube whose walls are maintained wet. So, all these problems will require use of psychrometry and or steam tables as we shall go on to see. Let us look at the first problem, here is a problem. A wet bulb thermometer records 15 degree centigrade, when the dry bulb temperature is 27 degree centigrade. Calculate A, the relative humidity of air and B, compare with carrier correlation. Assume Lewis number equal to 1 and take C P vapor equal to 1.88 and C P air equal to 1.005. Now, of course, this problem can be solved using psychrometric tables, it can be solved by using what is the well known carrier correlation, but we shall apply our Reynolds flow model. So, here T w is 15 degree centigrade and T infinity is T dry bulb equal to 27 degree centigrade and we have said Lewis number equal to 1 and therefore, B m which is omega b infinity minus omega v w over omega v w minus omega t which is 1 will equal spaulding number calculated from mixture enthalpies as given here. Now, of course, wet bulb records its temperature when the heat conduction to it is 0 and therefore, Q L in our formula will be set to 0. If we take T ref equal to 0 degree centigrade, then lambda ref the latent heat is 2503 kilo joules per kg of and from steam tables, we would readily discover from partial pressure the value of omega v w. The way to do this is the following. So, for example, since you know T w is equal to 15 degree centigrade, you will read P sat 15 from steam temperature tables and then you will evaluate the specific humidity which is 0.622 times P sat divided by P total minus P sat and then you will evaluate omega as w over 1 plus w. So, if we carry out these calculations, you will see omega v w turns out to be 0.01068 as I have shown on the slide here. So, if we calculate now the mixture enthalpy in the infinity state, it will be 1.0025 into 27 plus 1.88 minus 1.005 into 27 plus 2503 into omega v infinity. H m w will be likewise 1.005 into 15 plus 0.875 into 15 plus 2503 into the omega v w which is 0.01068 and that will give you 0.41.95. Here, we did not know omega v infinity and therefore, it is left at this. H T l is the enthalpy of the transfer substance which is water in that l state that is just inside the w w surface and that would be C p of the liquid multiplied by T w. So, 4.187 multiplied by 15 gives you 62.805. Therefore, equating d m with b h, we will have omega v infinity equal to minus 0.0168 divided by 0.0168 minus 1 equal to all this 27.135 plus 2562.2 omega v infinity minus 41.95 divided by 41.95 minus 62.805. If we solve now for omega v infinity, we would get omega v infinity equal to 0.005936 which gives us specific humidity equal to omega over 1 minus omega equal to 0.00594 at 27 degree c. Now, from psychrometric chart, this value corresponds to relative humidity of 27 percent which is our answer A. Also, b m is b h and very, very small. If I substitute this value here, then you will see b m and b h would be equal to 0.00479 which is very, very small. Now, if we use carrier correlation which you have used in your undergraduate work, it gives p v infinity equal to p sat w minus p sat w t wet bulb minus t dry bulb 155 minus t wet bulb where t's are given in degree centigrade. Now, in our case, p dot is 1 bar, w w is of course, 0.0168 minus this value which is 0.0108 and p sat w is 0.01736 bar which can be read from the steam tables as well. Then, substitution gives p v infinity equal to 0.0097 and therefore, I can evaluate omega v infinity as p v infinity into 1.61 into p dot minus 0.61 p infinity equal to 0.00604 which is answer B. Now, you will see this value of 0.00604 is very, very close to the value 0.005936 that we applied obtained from Reynolds flow model. Therefore, we can say that Reynolds flow model is very well captures the psychometric relationships. Let us now turn to the second problem. It is given like this. Moist air flows through a duct whose walls are maintained at 50 degree centigrade and wet bulb thermometer placed in the duct records the dry and wet bulb thermometers located in the duct show 70 degree centigrade and 25 degree centigrade respectively. So, we have a duct whose wall is at 50 degree centigrade and moist air is flowing through it. We have the dry bulb equal to 70 and we have the wet bulb equal to 25. Now, it is also given that the heat transfer coefficient is h cos v w equal to 0 for this situation of the bulb is measured as 17.5 watts per meter square Kelvin. So, under this situation you are asked to calculate relative humidity of air without radiation and B with radiation. So, in the first part we assume that whatever the wet bulb and dry bulb thermometers are reading are correct values and they are not influenced by radiation between the tube wall temperature and the bulb temperatures. So, straight away we can see that in this case corresponding to 25 degree centigrade, we will get w w the specific humidity as 0.02 and therefore, the mass fraction at the surface of the wet bulb thermometer would be 0.0196 which gives us mixture enthalpy as 74.61, h t l will be 4.186 into 25 giving 105.7 and h infinity would be 70.352564.25 into omega v infinity and therefore, equating B m with B h we would simply get that relationship from which we get solving we would get omega v infinity equal to 0.001446 and B itself would be 0.0185 and since omega v infinity is 0.001446 we can readily recover B v infinity from the relationship I have shown on the previous slide as 0.02328 bar and from the steam tables B v sat corresponding to T infinity equal to 70 degree centigrade which is the dry bulb temperature would be 3.119 bar and therefore, relative humidity would be the actual value partial pressure divided by the saturation partial pressure and which gives us 0.746 percent as the relative humidity when radiation was neglected that means, the moist air has moisture contained 0.746 percent, but now we wish to allow for radiation does that make any difference to the estimate of relative humidity. Let us look at that in the next slide in this part at the wet bulb as usual q l will be 0 and h t l will be h t plus q rad by n w you recall we account for q rad in the neighboring phase and therefore, h t will be q rad divided by n w and therefore, B m would be still omega v infinity by omega v w omega v w minus omega v t equal to h m infinity minus h m w over h m w minus h t which would transform to h m infinity minus h m w over h m w minus h t l plus q rad by n w. Now, this relationship you already know we have derived this when discussing the Reynolds flow model. Now, the first thing we must do is we must determine the true air temperature that is the driable temperature of the air because the driable temperature itself is influenced by the effect of radiation. So, we will say the heat transfer by convection to the driable thermometer which is h cough into T a true minus T driable will equal radiation from the thermometer bulb to the wall. So, sigma into T w T d b raise to 4 minus T w raise to 4 of course, we have assumed the emissivity of the bulb to be 1. Now, h cough has been given to you at 17.5. So, T a true minus 343 which was 70 degree centigrade equal to 5.67 into 10 raise to minus 8 which is the Stefan Boltzmann constant into 343 raise to 4 minus the wall temperature which is 50 degrees or 323 raise to 4. So, which gives us T a true equal to 352.6 k equal to 79.6 degree centigrade. So, remember in this case the driable temperature itself was not reading correctly and the true air temperature is in fact 79.6 degree centigrade. The correspondingly then what would be the q rad q rad will be sigma at the q rad at the wet bulb thermometer will be T w raise to 4 minus T wet bulb raise to 4 and that would be equal to 5.67 into 10 raise to minus 8 into 323 raise to 4 minus 298 raise to 4 which is wet bulb temperature of 25 and that will give you 170 watts per meter square. So, now we must evaluate b m and b h freshly because now we are going to introduce this q rad in here. Now, we already know that b is going to be very very small. So, I can replace n w equal to g b and that is what I have done on the next slide. So, if we take T ref equal to T wet bulb equal to 25. So, that lambda ref will be 2442.3 then H m infinity will be c p a into T a true minus T ref c p v minus c p a into T a true minus T ref plus lambda ref omega v infinity and that would be 54.52 plus 2490 omega v infinity plus 2490 omega v infinity. H m w of course, because these specific heat sensible heat terms go out and you will have omega v w into lambda ref equal to 0.0196 2442.347.87 and the liquid enthalpy would be c p l into T wet bulb minus T ref equal to 0 because we have taken reference temperature at the wet bulb itself. Now, taking omega mean equal to 0.0196 plus 0.01446 remember we had evaluated here omega v w equal to omega v infinity equal to 0.01446. So, we assume that that will still remain the same as far as omega mean calculation is concerned. So, that c p m is not too seriously affected by this choice although we do not know this value and this is simply to avoid iterative solution because c p m will not be influenced too much by the value of omega v in the infinity state. So, omega mean is taken as 0.0105 and therefore, c p mean is taken as 1.041 g star would be H cough divided by c p m which is 70.5 divided by 1041 equal to 0.01725 kg per meter square second. So, now since b is expected to be small q rad by n w will be q rad by g star b. So, that we get b equal to omega v infinity minus omega v w minus omega v T which is 1 equal to H m infinity minus H m w minus q rad by g star over H m w minus H T l. Substitution will show on the next slide q rad by g star is 0.17 as we calculated here 170 watts per meter square. So, in kilo watts per meter square it would be 0.17 into divided by 0.01725 equal to 9.885 kilo joules per kg and therefore, our balance equation would read like that. If you solve this equation you will get omega v infinity equal to 0.001693 instead of the previous value of 0.001446. So, as I said our value of c p m is not going to be seriously affected which gives b equal to 0.01826. So, our assumption of small b is verified from this value of omega v infinity we calculate specific humidity as that and therefore, p v infinity equal to 0.00264 bar, but p sat at 79.58 is 0.4739. And therefore, the relative humidity now is 0.557 percent instead of the earlier value of 0.746 percent as you can see here. So, the effect of radiation is actually suggest that this the moist air is in fact, drier than what ignoring radiation would predict. So, the true h r h is lower than that predicted by neglecting effect of radiation. Now, this kind of calculations are very important because often when the so called dry air is supplied it is never really completely dry. There are very small fractions of moisture and one wants to know what that small fraction is and you can see by allowing for effect of radiation we have shown that there is in fact, significant effect of radiation from 0.746 percent down to 0.557 percent which would be predicted radiation was accounted. So, with this I turn now to the next problem and here it is this is of evaporation and a case of high b as we shall see air at one bar and 27 degree centigrade and 90 percent r h flows over a porous flat surface which is kept wet by supplying water. The plate temperature is to be maintained at 82.5 degree centigrade. Now, of course, when you supply water the cooling would take place and if you want to maintain the temperature of the plate at 82.5 degree centigrade then we may have to supply some heat from the rear side of the plate. So, we want to calculate the evaporation rate and heat flux to be supplied. Now, problems of this type arise in food processing typically where some material processing where you want to maintain the surface at a given temperature for whatever the subsequent processing is. So, we are given here in this problem a plate whose length is 0.33 meters and whose width is 1 meter. The air is flowing over this plate and it is at 27 degree centigrade and 90 percent r h and the wall temperature T w is to be 82.5 degree centigrade and we want to find out what should be the q l to be supplied and what will be the evaporation rate N w from this plate. So, that is really the problem now the u infinity is given as 3 meters per second. The plate is porous and we shall calculate now first of all we must ensure whether the boundary layer is going to be laminar or turbulent or transitional and so on so forth. So, first of all we calculate the plate Reynolds number from the properties that are given here 3 into the plate length divided by the kinematic viscosity 0.103 divided by 3600 nu is given as 0.103 which will give 34600 and also which is of course, less than 3 lakhs and therefore, we expect this to be a laminar boundary layer with Prandtl number equal to 0.103 divided by 0.153 equal to 0.673 the Schmidt number will be 0.636 and therefore, the Stanton number relationship for a laminar boundary layer is 0.664 R e l to the minus 5.5 Prandtl to the minus 0.67 you will recall this was the solution we developed from similarity solutions and that would be equal to 0.00465 and since Prandtl number is very very close to Schmidt number we can readily say G star will be H cough divided by C p m which is H rho mu infinity into Stanton number and that rho m is 1.37 in this case u infinity is 3.00465 is the Stanton number and that gives us 0.0191 kilograms per meter square second. So, this is the value of the G star which is operating at the surface. So, now in order to calculate the mass transfer rate we must calculate the value of B. So, from the data given P V infinity will be 0.9 into P sat at 27 degree centigrade which is 0.9 into 0.03567 bar and that would be equal to 0.0321 bar and omega V infinity therefore, can be readily calculated from our formula and that would be equal to 0.02018. Similarly, a P sat 82.5 degree centigrade which is at the wall 0.5261 bar and therefore, omega V w will be 0.4081 and therefore, B will be simply omega V infinity which is 0.02018 minus 0.4081 divided by 0.4081 minus 1 equal to 0.6554. So, this is a considerably large B in an evaporation problem. You normally evaporation problems have a very small B and therefore, we now follow the Reynolds flow model in which we are advised that G over G star ln 1 plus B should be multiplied by Prandtl by Schmidt raised to 0.33 and also the property correction M mix w divided by M mix infinity raised to 0.67. Now, here M mix w at the wall state knowing omega V w we can evaluate M mix w as 24.51. M mix infinity would be evaluated as from this value of 0.02018 as 28.76 and therefore, substitution would give G equal to G star into 0.7039 equal to 0.01344 kg per meter square second and hence the evaporation rate from the plate will be G times B into area of the plate which is 0.01344 into 0.6554 into 0.33 into 1 which is the width equal to 0.0029 kg per second or 10.646 kg per hour would be the rate of evaporation from this porous surface. The next thing is how do we keep the surface at 82.5 degree centigrade and to do that we will proceed to the next slide. Now, in this case Lewis number is equal to 1 and therefore, B which is equal to H m in B calculated from enthalpy will also be equal to 0.6554 and therefore, we have H m infinity minus H m w over H m w minus h t l plus q l divided by n w equal to 0.6554 and if I take t ref equal to 0 then H m infinity would be simply that 78.12 because I know omega v w as 0.02018 I can calculate sorry H m infinity and I can calculate H m w also as 1133.85 because I know omega v w and H t l will be 0.4 into 1.187 into 82.5 equal to 3.4. So, hence q l divided by n w will evaluate to minus 2400 kilo joules per kg and the negative sign here clearly indicates that heat must be supplied because we take q l away from the plate into the transfer substance phase is positive and therefore, negative sign implies that q l is to be supplied to the plate and therefore, if I now multiply by therefore, q supplied will be simply m w which was 0.0029 into 2400 will give me 6.98 kilo watts. So, for this plate of 0.33 meters by 1 meter we will need to supply about 7 kilo watts of heat to maintain it at 82.5 degree centigrade which is the requirement the process requirement. Let us now turn to the next problem a 10 kilometers per hour breeze at 40 degree centigrade and 20 percent R H a situation something like middle of India say Nakpur or somewhere in blows over a lake whose water temperature is at 30 degree centigrade. The lake also receives solar radiation of average let us say 500 watts per meter square average for the day. So, calculate the time required for the water level in the lake to drop by 1 centimeter you are told that as the wind blows over the lake assume it to be observable and boundary layer throughout. So, you can see that this is a typical problem say you have a power plant close to the lake and you are drawing water from the lake for cooling in the condenser and the hot water again comes back here which mixes with the lake, but the lake water is continuously dropping and therefore, you need to do topping up of this water from some other source and nearby and typically the ground water source from nearby. So, that the lake level does not drop too much and your power plant cooling operations are not hampered. So, you are told that the assume turbulent boundary layer throughout and take kinetic viscosity to 15 into 10 to minus 6, sheet number equal to 0.67 and transfer equal to 0.7. So, in this case corresponding to 40 degree centigrade and 20 percent R H we would evaluate p v infinity equal to 0.2 into saturation pressure which is 0.07384 bar and that will give us 0.01477 bar and therefore, omega v infinity would evaluate to 0.00919 and therefore, taking Lewis number as equal to 1 which is fair assumption to start with and if b is small then it does not really matter too much. Then omega v infinity divided by omega v w over omega v w minus 1 H m infinity minus H m w H m w minus H t l plus q l by n w. Now, here we do not know omega v w that is the mass fraction at the surface of the lake and therefore, we also do not know the mixture enthalpy at the surface of the lake nor do we know q l. So, we have three unknowns. So, n w H t plus q rad plus q l is equal to n w H t l this would be the balance in the in the neighboring phase as usual and therefore, b h would be H m infinity minus H m w over H m w minus H t divided by q rad by n w and n w will be simply g into b h and therefore, b h would be H m infinity minus H m w q rad by g divided by H m infinity by H t equal to this value. What I have done here is I have replaced n value equal to g b and then multiplied through so that b h b h gets cancelled and q rad appears now on in the numerator. So, if I now take t ref equal to 0 then H m infinity would be 1.005 into 40 plus 0.875 into 40 plus 2503 into that omega v infinity value which is 63.3 I do not know the surface of the lake temperature. Therefore, I just keep it that H 1 0.005 T w plus 0.875 T w plus 2503 omega v w and H t will be simply 4.187 into 30 which is the temperature deep inside the lake and that is 125.4. Now, here u infinity is 10 kilometers per hour that has been given which is essentially mean 2.78 meters per second and therefore, the Reynolds number would work out to be 18.53 into 10 raise to 6 and therefore, we calculate for 100 meter lake. We assume b tending to 0 and g tending to g star and then for a turbulent boundary layer the Stanton number would be 0.0365 R e l to the power of minus 0.2, Prandtl to the power of minus 67 and that would be equal to 0.001617. Now, rho m the mixture density t will be equal to rho of the vapor plus rho of the air which would be rho a into 1 plus specific humidity mean which is the kg of dry air due to the kg of the mixture and g star would be therefore, 0.0049 into rho m. So, I have multiplied here by u infinity which is 2.78. So, g star becomes 0.0049 into rho m and q rad over g star is 0.5 divided by 0.00449 this is q rad is 500 watts which is 0.5 kilowatts divided by rho m and therefore, this will be 111.36 divided by rho m kilo joules per kg and therefore, we have on substitution we will get b which is from the mass fraction would be this and from the mixture enthalpies it would work out to this value. So, we have to solve this equation iteratively. So, w v w corresponds to T w at saturation and rho m is evaluated from w mean equal to w w by T plus w infinity. Now, of course, what you do is you assume T w and therefore, calculate w w from which you calculate w mean and therefore, you can calculate rho mean knowing w w we can also calculate w omega v w and then check whether the left hand side equals the right hand side or not. So, iterative calculations are required and I have already given you a correlation for evaluating omega v w from T w. Alternatively, you can use steam tables or psychrometric chart anything will do to evaluate because we are only discovering trying to discover the saturation condition at the wall temperature. So, these two left hand side and right hand side must be balanced by iteration. So, this is the trial and error procedure and I have already explained to you how the trial and error is to be performed. So, in the present case it works out to be T w equal to 39.5 degree centigrade which is very very close to the T infinity. In fact, omega v w turns out to be 0.043 rho m turns out to be 1.16 giving b equal to 0.0354. So, our assumption of using n w equal to g b here is not too bad because b itself is very very small. So, n w will be simply g star into b is equal to and g star we have already calculated. So, therefore, it is 2.035 into 10 to the minus 4 kg per meter square second or 0.7324 kg per meter square hour. So, therefore, the lake water will drop in order to calculate the lake water drop what we do is rho water into delta h into surface area a surface area. This will be the divided by delta T will be equal to m dot w which is the surface area of the lake and that will give us rho w into delta h by delta T equal to m dot w divided by a lake and that will be equal to n w. So, we can now estimate since we know now what the n w is n w is we have evaluated as 0.7324. We take rho w equal to 1000 water delta h which is 1 centimeter divided by delta T. So, if I valid that delta T will be equal to 0.7324 divided by 10 or 0.07324. And that is what you get or in terms hours sorry bigger one here it will be 0.7324 is on the right hand side delta h is 0.01 into 1000. And therefore, delta T will be delta T will be 1000 into 0.01 divided by 0.7324 which gives you 13.65 hours. So, of course, delta what we have done here is that for the whole day over the entire period of this time we have said the solar radiation is constant which of course, is not the case the solar radiation varies during the day. But we have taken an average value and we find that time required for 1 centimeter drop of the lake would be about 13.65 which is assuming sunshine hours of about 10 hours we can say it is about one and a half day would be required to reduce the lake water by 1 centimeter. Such estimates are very useful for estimating the top of water required in the lake. So, as to maintain it in its lane level constant and not affect in any way the cooling water provision for the power plant. We now look at the final problem and this time it is an internal duct flow problem. So, moist air at inlet specific humidity of 0.003 at 24 degree centigrade enters a tube which is 2.5 centimeter diameters and 0.75 meters long I beg your pardon it is 0.75 centimeters but it is actually 0.75 meters long at the rate of 9 kg per hour. The tube wall is maintained at 24 degree centigrade. So, it is the same temperature as this calculate the specific humidity at exit from the tube and the heat supplied to the tube wall to maintain its temperature at 24 degree centigrade. Take D equal to 0.09 meter square per hour diffusivity and Schmidt number equal to 0.6 assume temperature remains constant throughout. So, even at exit the temperature is going to be at 24. So, in this case omega v w is constant with x because T w is constant with x and therefore, if we define B x as I said in internal flow omega v x will be the bulk value of the mass fraction minus omega v w divided by omega v w minus 1. So, if I take a differential d omega v x will be equal to omega v w minus 1 into d v x. If I now look at this small differential element of length d x then mass transfer N w multiplied by the perimeter which I take as d x the perimeter is pi d in multiplied by d x would be the area. So, N w d x will be equal to m dot into change in mass fraction but the perimeter value here is taken as pi d in the denominator here. So, N w d x into m dot by pi d d omega v x if I substitute for omega v x this expression I get an N w equal to g v d x then I get g v x times d x equal to m dot divided by pi d omega v w minus 1 d v. On the next slide you see d v x by b x would be equal to pi d g m dot w omega v w minus 1 into d x. If I integrate this equation from inlet to outlet then this will give me logarithm of b out over b in equal to that expression into L. So, that simply becomes pi d g L m dot omega v w minus 1 with an exponential sign. Of course, since omega v w is going to be less than 1 this quantity is negative and therefore, exponential of a negative is a fraction and therefore, b out will be less than b in and L is the length of the tube. So, in this problem corresponding to 24 degree centigrade omega v w is 0.01875, w w is 0.01632 and corresponding to w in equal to 0.003 w v omega v in will be 0.0029991 and therefore, b in is 0.01606. We now determine g. Now, at inlet the mean value of the specific humidity will work out to be 0.01088 because we know w w and we know w in. At 24 degree centigrade rho air is 1.189 and therefore, rho m will be 1 rho a into 1 plus w m equal to 1.202 and then nu m will be Schmidt number into d which is 0.054 meter square per hour or 1.5 into 10 to the minus 5 meter square per second and therefore, now the mass flow rate is m dot into u in into pi by 4 d square and therefore, u in will be 5.09 meters per second and therefore, I evaluate Reynolds number as 8488. Now, if I take Lewis number equal to almost 1 then I can evaluate Sherwood number from the Dittus-Bolter relationship as 0.023 Reynolds is to 0.8 Schmidt number raise to 0.4 as 26.06 which gives me g equal to 112.76 and therefore, the substitution in here. Now, I know pi I know the diameter of the tube I know g now which evaluated at 112.76 kg per meter square hour m dot is 9 l is 0.75 meters and omega v w is already known as 0.01875 which gives me b out equal to 0.00757 and we had evaluated b in equal to 0.01606. So, I can see that b out is actually smaller than b in. Now, omega v out will be omega v w plus b out omega v w minus 1 from the definition of b out and it will be 0.01137 and therefore, the specific humidity at the outlet would be 0.01145 because knowing w v omega v out I can evaluate w out. So, you can see whereas, w in was 0.003 as you can see the specific humidity at inlet was 0.003, but now the air has been humidified and the specific humidity at outlet is 0.01145. So, air has indeed picked up moisture from the surface. Now, if I do energy balance then you will see m dot into change in enthalpy would be equal to the mass transfer from the surface h t l minus q l which is the conduction heat transfer or the heat away from the wall into pi d dx which is the perimeter of the tube into dx. So, if I integrate now this expression for q l x then I will get minus q bar l equal to minus 1 over l 0 to l q l x dx which will be m dot by pi d 0 to l d h m by d x d h m by d x minus g h t l 0 to l b x d x which will give me m dot pi d into h m out minus h m in minus h t l omega v out minus omega v in. And therefore, you will see taking t ref equal to 0 h m out will work out to 52.82 because we know t out is also 24 degree centigrade we already know omega v out. So, we can work out h m out we can work out h m in which is the inlet condition and h t l is 4.187 into 24 therefore, 100.49 and that gives me q bar l equal to minus 0.646 kilo watts per meter square negative sign indicates that the heat must be supplied to the tube and so this is the second answer to our problem. So, we do find that in the tube the specific humidity is increased and the temperature has been given as constant there is no change in temperature of the air in which case you will have to supply heat to the wall so that it is maintained at 24 degree centigrade and the value has been found. So, you can see we have used psychrometry in variety of applications like wet bulb thermometer mass transfer from a porous surface which is which is to be kept at a specified temperature. We have looked at flow humidification of air in a tube which has to be kept at a specified wall temperature and we also looked at amount of water depletion that would take place due to evaporation convective evaporation due to wind as in the presence of solar energy. So, a variety of problems can be solved by using Reynolds flow model in our Reynolds flow model and you can see it is a very very convenient tool. We have not solved any differential equation we have simply made use of the available correlations to get the appropriate value of G star in each case.