 In today's lecture, we shall take consider further problems in which we shall apply Reynolds flow model for predicting mass transfer rates and the new problems are condensation, transpiration cooling, volatile fuel burning, drying and dissolution of solid in the liquid. These are the vast range of problems in which Reynolds flow model can be applied and I have selected a few so as to help you understand how easily one can estimate the mass transfer rates and arrive at very crucial engineering decisions in such problems. Let us take the first problem of condensation. The problem statement is as follows. Consider condensation of steam at one atmosphere on the outside of a copper tube. So, here is a copper tube. This is the symmetry line of the tube of radius r i and the wall thickness is r o minus r i through which cooling water is flowing and the steam is on the outside of the tube. The tube is made of copper, its diameter is 2.5 centimeter id and 2.9 centimeter od. The tube carries cooling water at 50 degree centigrade. Calculate the steam condensation rate on this tube when steam is pure and saturated and B steam is mixed with 20 percent air by mass. When steam is mixed with air as you know the condensation rate reduces and we have to estimate how much is the reduction. Data given are assume that condensate film thickness on the tube is 0.125 millimeters. The conductivity of copper is 300 watts per meter Kelvin that of water is 0.68 watts per meter Kelvin. The heat transfer coefficient on the tube side h of i is 4620 watts per meter square Kelvin. We are going to assume T ref equal to T w where lambda ref the latent heat is 2257 kilo joules per kilogram. Now, you are also given another information that if there was single phase flow of air in this case then the h coefficient without mass transfer would be 115 watts per meter square Kelvin. So, this is the value we shall use to estimate G star. Now, remember you can obviously solve this problem from heat transfer theory and you have always used heat transfer theory for solving problems of this kind, but we are going to treat this as a mass transfer problem. So, my first task would be to show equivalence between Reynolds flow model and the heat transfer theory of condensation and that is what I intend to do on the next slide. So, in our case B would be omega B infinity minus omega V w divided by omega V w minus 1 and that would equal h m infinity minus h m w h m w minus h T l plus q l by n w and that would equal n w by G. Now, as you said we are going to take T ref equal to T w then h T l will be of course, 0 and h m w will be lambda ref into omega V w there will be no sensible heat contribution to h m w because T ref is equal to T w, but h m infinity would equate to C P m that is the mixture specific heat multiplied by T infinity minus T w plus lambda ref into omega V infinity. So, if we substitute for h m infinity and h m w in this expression and also make use of the definition of B and n w equal to G B then substitution would show that n w can be written as q l that is the conduction heat transfer inside the transfer substance that is in the liquid film divided by C P m into T infinity minus T w divided by B minus lambda ref equal to G into B. So, this is our mass transfer or Reynolds flow model formula for calculating condensation rate, but from heat transfer theory we know that q l is actually written as h condensation into T w minus T s, where T s is the outside surface temperature of the tube and for pure steam omega V infinity will be equal to 1. So, B will be equal to 1 minus omega V w divided by omega V w minus 1 and therefore, B will be equal to minus 1 and hence our formula would simply read as n w equal to minus h condensation T w minus T s C P m into T w minus T infinity because B is minus 1 plus lambda ref. Now, this is precisely the formula you have used in heat transfer theory to calculate condensation rate when steam is pure. So, thus our mass transfer formula accords with the heat transfer formula with h condensation equal to condensation heat transfer coefficient and therefore, there is a complete equivalence between the two. So, let us turn to our part a of the problem when steam is pure. So, in our problem T s is the outside tube wall temperature is not known, but the cooling water temperature T c is known and therefore, we can invoke the notion of the total heat transfer coefficient and write the condensation heat transfer q l equal to divided by universal heat transfer coefficient u into T w I mean total heat transfer coefficient u into T w minus T c where simply what I have done is 1 over u is the total resistance is equal to the resistance due to inside heat transfer coefficient plus resistance due to a thickness of the copper wall and its conductivity. This is the resistance due to the thickness of the liquid film having conductivity k liquid that is the water film. So, there are three resistances adding up to 1 over u and the substitution if I substitute these values from the data given here, h kof i is 4620 k copper is 300 and k water 0.68 then it works out that the total heat transfer coefficient u will be 2663 watts per meter Kelvin. Now, for pure steam at p equal to 1 atmosphere T w T infinity will equal T saturation and that would equal 100 degree c and therefore, our formula is minus n w equal to g equal to h kof by C p of steam and h kof for natural convection was given as 115 watts per meter square Kelvin and therefore, we readily calculate g is equal to 115 divided by 1.88 into 10 raise to 3 equal to 0.06117 kg per meter square. So, this is a straight forward application of our formula to estimate n w which is negative n w equal to that and from our model writing q l equal to u into T infinity minus T c that is 2663 into 100 minus 50 into 1880 which is the specific heat of the mix of the vapor into T w minus T infinity both are 100 degrees and therefore, that term makes no contribution divided by the lambda ref which is 2257 into 10 raise to 3 and the result is 0.059 kg per meter square second. So, the two results are very very close 0.0617 and 0.059. Now, of course, the difference arises mainly because we assume h kof divided by C p v and assuming h kof given is of the right order of magnitude that the two results are extremely close and that verifies our Reynolds flow model for estimating condensation rate. Negative sign of course, in both these results minus n w equal to that and minus n w equal to minus 0.059 indicates condensation. We now turn to the part b of the problem where there is 20 percent air in the steam and therefore, omega v infinity will be 0.8 and q l as before will be u into T w minus 50, but T w is not known and therefore, our formula for n w will be 2663 T w minus 50 divided by C p m into 100 d minus T w divided by b minus lambda ref and that would equal h kof o by C p m l n 1 plus b that is the formula for where b lambda ref and C p m all functions of T w, but we do not know T w and therefore, we must do iterations. So, the thing is you assume T w and therefore, evaluate omega v w from saturation condition there and that gives us the value of b knowing omega v w we calculate mean specific heat as 1614 and also evaluate lambda ref equal to 2283.2 into 10 raise to 3. So, the left hand side of this relationship gives us minus 0.046 whereas, the right hand side on here gives us minus 0.05 when T w was 90. So, obviously, there is a difference in the between left and right hand side. So, we take next guess of 91 degree centigrade and we find that the left hand side is now minus 0.0472 and the right hand side is minus 0.044. So, obviously the result must be in between the two and we now take 90.5 and we find that the result is minus 0.0475 as the left hand side and we accept that result as nearly correct and take n w equal to minus 0.0473 kg per meter square second. Now, you will see that the our earlier result was minus 0.059 whereas, now the result is minus 0.0473 and therefore, the condensation rate has reduced because of the 20 percent air in steam and this is a well known problem in condensers because the condenser operates at a very low pressure. There is always a chance of air being ingress and therefore, the rate of condensation reduces and that is why air ejectors are used in condensers. This is just a problem of that variety which checks out what we normally do in order to prevent falling of condensation rate we always remove air from the steam which enters the condenser. I will now take up the problem of transpiration cooling which is always used to protect surfaces which are exposed to very high temperature gases. So, consider a problem as given here. A porous metal surface is swept by air at 540 degree centigrade that is very high temperature. Now, since metal oxidizes at 425 degree centigrade it is decided to keep the surface temperature down to 370. So, our T w we want is at 370 by blowing gases through the pores for this purpose 3 candidate gases available at 35 degree centigrade are considered. One is air itself that is you inject air in air you inject helium in air and thirdly hydrogen in air. Calculate the supply rate of each gas assuming operating g of 370 kg per meter square hour and we assume that in all three cases of injection g more or less remains constant. In case of air assuming constant specific heat we assume that between 370 and 540 there is not much difference in say specific heat of air. Then the problem becomes not that it cannot be handled with variable specific heat, but for simplicity C p is equal to T infinity minus T w C p divided by T w minus T t. So, that simply becomes 540 minus 370 370 minus 35 which is T t and that is equal to 0.5074 that is the b and therefore, n w of air would be g times b equal to 187.75 kg per meter square per hour. So, that is the answer for the part one of the problem. Now, we consider helium and hydrogen part b. So, in this case now specific heat of helium is 5.25 kilo joules per kg Kelvin and C p a infinity is 1.1 kilo joules per kg per Kelvin. So, therefore, if I take T ref equal to T w then you will see I do not have to calculate mixture specific heat because it will simply mean T w minus T ref gets cancelled here C w minus T ref gets cancelled here this is h infinity and this is h t and that would simply be C p a into T infinity minus T ref there are minus C p helium into T t minus T ref. So, I will get 1.1 into 540 minus 370 divided by minus 5.25 into 35 minus 370 b is 0.1063. So, b for helium is smaller than that for air because of high specific heat and you will see n w helium is g times b will be about 40 kg per meter square hour. So, much reduced quantity of helium is required compared to air. Now let us take the case of hydrogen. In this case hydrogen specific heat is 14.5 and hydrogen is going to burn in such a hot environment. So, we assume a simple chemical reaction hydrogen plus half O 2 equal to H 2 O giving r s t is equal to 16 by 2 equal to 8 that is the stoichiometric ratio for this case is 16 by 2 equal to 8 and the latent heat of hydrogen is 118 mega joules per kilogram and therefore, taking H equal to C p m into T minus T ref plus del h c by r s t omega O 2 as you know we have associated the latent heat with oxygen and therefore, we divide this by r s t and if we take T ref equal to T w we have b equal to H infinity H w of course, will be 0 because omega O 2 cannot survive at the surface as well as T is equal to T ref and we will have minus C p H 2 into T t minus T w which will be the H t this is 1.1 C p a infinity is 1.1 into 540 minus 370 plus 118 mega joules 10.4 into 10 raise to 3 divided by 8 into 0.232 divided by minus 14.5 into 35 minus 70 and now the b increases to 0.745 and therefore, N w of hydrogen will be g into b 275.8 remember we are using hydrogen which is going to burn to keep the surface cool at 370 degree centigrade and you will see now that in the three applications we find that the amount of hydrogen required will be greater than that for air and that for heat. This sort of simple calculations enable us to select the right kind of gas to inject through the porous surface depending on our requirement. I will now take up the next problem and that is of missile cooling. Missiles as you know travel at very high altitudes at very very high speeds of the order of 5 to 7000 meters per second very similar to the kind of velocities that are encountered when the reentry vehicle enters the upper atmosphere at around 10000 meters per second. So, these are very very high velocity projectiles which have to be kept cool because of the viscous heating that takes place near the surface. So, the problem reads as follows consider axis symmetric stagnation point of a missile traveling at 5500 meters per second through air where static temperature is almost say 0 k. It is desired to maintain the surface temperature at about 1200 degree centigrade by transpiration cooling of hydrogen at 38 degree centigrade. Evaluate B and N w given G star in this case equal to 0.467 kg per meter square second. So, here the value of G star is again given. Now, because of the high velocity here we must account for the kinetic energy contribution and define H m equal to C p m t minus t ref plus delta H c divided by R s t omega O 2 which is as usual plus V square by 2000 as the kinetic energy contribution to enthalpy in kilo joules per kg. Therefore, V square by 2 into 1000 that is what in kilo joules per kg will be. So, taking t ref equal to t w so that H m w again is 0 then the first that numerator will be H infinity divided by minus H t will be C p H 2 into t t minus t w and you get C p a infinity equal to 1.1 minus 0 minus 1473 because t infinity is almost taken as 0. The wall temperature is 1200 degree centigrade or 1473 Kelvin plus 118.4 into 10 raise to 3 which is the delta H c of hydrogen divided by 8 as before into 0.232 which will the omega O 2 which is that which is the mass fraction of oxygen in air plus 5500 square divided by 2000 divided by minus 14.5 into 38 minus 1200. So, this gives us large V of value 1 and n w will be g star into ln 1 plus b will give you g star is 0.467 and therefore, 0.325 kg per meter square second. Missiles of this type have to carry a certain amount of gas with them so that in order to keep the surface cool the particularly the stagnation point of the missile has to be kept very cool because it is travelling at a very high velocity. Under such conditions if you know the time of flight knowing the mass transfer rate as we have calculated here we can calculate knowing the time of flight that is seconds and knowing the surface area over which you are going to inject the gas we can calculate the amount of hydrogen that the missile must carry with it. So, problems of this kind give you a first estimate of how much hydrogen to carry in a missile without solving any differential equation. Let us now look at the another problem and that is of burning of a volatile fuel. So, in a diesel engine liquid fuel C 12 H 26 with heat of power combustion 44 mega joules per kg and specific gravity 0.854 and latent heat 358 and boiling 0.425 degree centigrade is injected in the form of small droplets. After ignition delay part of the fuel vaporizes and burns abruptly and the remainder of the fuel burns as fast as possible.