 So, what are we starting today is transient heat conduction, in the transient heat conduction essentially we will be worried about temperature variation with time that is unsteady. Usually people call this as unsteady heat conduction also, but I think transient is more appropriate than unsteady. So, we have I am going to cover under the headings introduction that is what is transient heat conduction and what is lumped system analysis followed by one dimensional transient heat conduction. Again semi infinite solids we will be understanding how in earth or in the soil the heat transfer takes place and moving on to multi dimensional systems. So, first let us see what is transient conduction. So, we have been telling if we just remind ourselves if we just remind ourselves the heat diffusion equation what did we say in Cartesian coordinates the heat diffusion equation del square t by del x square plus del square t by del y square plus del square t by del z square plus q dot by k equal to 1 upon alpha del t by del t. This is what we saw we have derived in the Cartesian coordinates. Yesterday we took 1 D conduction initially and we always kept this term 0 if you remember. This term was always kept 0 because we said that our cases are always going to be steady. So far we have dealt with only steady heat transfer situations, but today what we are going to do is we are going to study unsteady situations that means this term is going to be important for today's lecture. Of course, we will not take heat generation we will initially take the transient conduction without any variation of the temperature in space that is what is called as lump. So, let us see what it is. So, what we are going to say is that time dependent conduction that is temperature is 3 in a conducting body can be dependent on x, y, z and t that is actually the reality. In real life the temperature is going to be function of space and time in space means it is usually x, y, z and t. We make these simplifications because I get the analytical solution. What is an analytical solution? We always use these words very easily, but we need to know what are these terms. Analytical solution when I mean what when I say analytical solution I mean that by writing the mathematical equation I can solve that equation using the appropriate boundary conditions. So, that means this is mathematically tractable. So, that is when I use the term analytical solution. Analytical solution means I can write the governing equation solve it with the basic mathematics whatever we have studied in engineering and get the solution of course in this case for temperature and the heat transfer rate then we say that that is analytical solution that is always also called as closed form solutions. That means known with the unknowns are manageable that is whatever unknowns are there with the within the equation we can solve the equation and get the unknowns. So, that is what we use these terms analytical and closed form. So, now coming to lumped system analysis what is lumped? Lumped is in lumped system analysis temperature is going to be function of in lumped system analysis temperature is a function of time only not a function of space not a function of space. So, it is only function of time. So, so x y z it does not come into picture, but when do we come across this sort of situations? We do come across this sort of situations for example, if I take a copper ball why copper ball I have taken because it is thermal conductivity is very high that is thermal diffusivity is very high. So, we will understand this little later, but for now I can say that if the thermal conductivity of my body is very high that is copper ball you can see that the temperature throughout is more or less same that is uniform. There are two things two words we use replacingly many a times that is I use the term uniform. When I say uniform it does not change with space that means it is same everywhere in space same everywhere in space constant the word constant I use I we usually mean that it would not is same with time it does not vary with time same with time. So, these two terms we need to differentiate quite clearly. So, what we are saying here in this case is that temperature is uniform that means it is same everywhere in space, but if I take a potato which is taken out from a cooker the temperature is not same everywhere. So, you can see that 60, 65, 75, 70, 60 why because the thermal conductivity of potato is not very high because thermal conductivity of potato is how much it is potato is made of water only. So, thermal conductivity of potato will be of that of water what is the thermal conductivity of water 0.6, but copper is 400. So, that is why the temperature in a copper ball you can expect it to be more uniform as opposed to the temperature distribution in a potato which is taken out from a cooker after boiling. So, that is the understanding what we get by lumped system analysis. So, what is one of the important derivations I am going to do I want all of you to derive along with me that is what am I doing in this case is that I am just taking a copper ball in which the temperature is same all over the place. If I take a lumped body and lumped a hot body and push that into cold water let us say how does the temperature of this hot copper hot body decrease with the marching of time that is the question what we are asking ourselves that is let us derive that that is I am having a hot body, but uniform temperature distribution that means there is no variation of temperature with space uniform temperature distribution. Now, I have a container a very large container very large container filled with water whose temperature is T infinity T infinity is constant T infinity is constant it does not vary to maintain it constant even when I put a hot body in it it is required that my water pool size is very large it has to be a big tank. So, I put a hot body which whose temperature this is my hot body whose temperature will it starts getting cooled the question asked is how does this temperature of this hot body vary with time that is the question what we are asking that is this is a lumped body. So, for lumped body how does the temperature vary that is the question what we are asking. So, if I apply the conservation of energy that is E dot in minus E dot out plus E dot G is equal to E dot S T if I take the control volume my hot body as my control volume. So, what is happening is there anything which is getting in no there is no energy transfer into the body because there is body is already hot. So, there is no heat transfer into the body. So, E dot in is 0 E dot in is 0. Now, E dot G there is no generation inside the body I have somehow heated it earlier and now dumped into the cold water. So, E dot G is also 0. Now, E dot S T what is E dot S T equal to rho V C P D T by D T note I have used total derivative D by ordinary derivative not total derivative sorry ordinary derivative D T by D T because temperature is a function of time only that is the reason I have written D T by D T. Now, what is E dot out what is E dot out the mode of the heat transfer in which it loses it is losing heat is by convection. So, convection means I should be writing that as minus h A S note the subscript I am using A S that means surface area they think A area into T of T minus T infinity. So, this is the formulation of the problem. So, now, if I rewrite this equation if I rewrite this equation in little different form that is as yesterday we will write theta equal to T of T minus T infinity. So, if I write this should imply that D theta by D T is equal to D T by D T. So, this is essentially why we are doing this because like in fins we want to make this equation look little elegant and solvable. So, that we can get the closed form solution. So, now, I guess you can pick up my word closed form solution. So, now, let us write this in terms of theta. So, if I do that in terms of theta. So, what do I get yes minus h A S into theta is equal to rho V C P D theta by D theta by D T. Now, let me rearrange this that is let me rearrange that rho V that is rho V C P by h A S D theta by theta equal to D T of course, negative sign is there D T that is whatever done here whatever done here separation of variables. I have taken theta on one side and time on one side this is separation of variables this we have studied in mathematics separation of variables. So, now, let me have a look at this rho V C P by h A S or let us first integrate this and then let us take look at this rho V C P by h A S that is if I integrate this between the limits. So, let me write this little differently that is D theta by theta is equal to is equal to h A S upon rho V C P with the negative sign into D T. Let me integrate this between the limits theta i to theta and time 0 to some time t time 0 to some time t what is theta i what is theta i T i minus T infinity T i here please note T i here is the initial temperature of my body initial temperature of the body which was put into cold water. So, if I now integrate this what do I get log of theta by theta i is equal to minus h A S by rho V C P into T into T. So, now let me call h A S sorry rho V C P by h A S as tau let me call rho V C P by h A S as tau what will happen to this equation if I write in terms of theta by theta i theta by theta i would become now equal to e to the power of minus T by tau is that right. Let us rewrite this in the next transparency that is remember we have got theta by theta i as e to the power of minus T by tau. So, let me do that in the next transparency that is theta by theta i is equal to e to the power of minus T by tau. So, what is theta T minus T infinity T being temperature at any instant of time T upon T i minus T infinity T i minus T infinity T i minus T infinity. Now, let us have a look at tau tau what is tau tau equal to rho V C P upon h A S what is the unit of tau what is the unit of tau time how did you know that let us work it out that is unit of density is k g per meter cube unit of volume is meter cube C P unit is joule per k g Kelvin h is vat that is joule per second vat per meter squared Kelvin I made a mistake I think vat per meter squared. So, meter squared should come in the top vat per meter squared Kelvin into 1 upon A S that is 1 upon meter squared. So, you see here meter cube meter cube k g k g then Kelvin Kelvin then you have meter squared meter squared joule joule getting cancelled out and you are left out with seconds that means tau is having the unit of seconds. So, what is this tau tau is the time constant tau is the time constant. So, because it is having the unit of time. So, now let us just go back and see we can plot this for different tau. So, if you just take the recap of what derivation I did minus h A S T minus T infinity is equal to rho V C P D T upon D T I integrated that and I ended up with T minus T infinity upon T I minus T infinity equal to exponential of minus T by tau. See tau can be looked at in little different fashion like this as well tau equal to rho V C P upon h A S that is 1 upon h A S is nothing, but my resistance rho V C P is the thermal inertia capacitance it is thermal inertia. So, it is capacitance. So, you can write down that is tau is tau is equal to 1 upon h A S into rho V C P equal to resistance into capacitance thermal resistance and thermal capacitance. So, this is basically the inertia term this is what we will tell whether I am going to able to respond fast or slow. You would have heard the term time constant quite often the time constants in building metro train metro rail in Delhi was very less that means it is very fast it is going it is being built fast. So, this time constant term is being regularly used. So, I do not think there is any difficulty in getting attached or getting understanding in of the term time constant. So, if I plot this T minus T infinity upon T i minus T infinity verses time constant sorry verses time. So, you can see this is one graph which is having a time constant of 1 comma 4 and this is 1 comma 3 this is 1 comma T comma 2 and this is T comma 1. So, the body which is having lowest time constant is responding fast it is responding fast that means why is it responding fast if it is having lowest time constant means its h A S is large. So, the resistance is less and capacitance is also less, but if the time constant is large means the either resistance is large or rho V C P is large. So, it is having more inertia it takes more time to respond for the outside changes in the environment or the temperature. So, that is about the temperature distribution, but we are not just happy with the temperature distribution we always look at the heat transfer rate as well. So, let us see how do we calculate the heat transfer rate. So, now let us derive the heat transfer rate C Q equal to that is heat transfer rate net heat transfer rate equal to Q D T which is to be integrated between the limits 0 to T. What is this Q how do I get this Q that is equal to h A S into T of T minus T infinity D T between 0 to between 0 to T. So, now if I integrate this before integrating let me rewrite this in terms of theta because I have temperature distribution just now derived in terms of theta that means this is equal to Q equal to integral of 0 to T h A S h A S theta D T is that right. So, theta what is theta we are just now derived theta is equal to theta i into e to the power of minus T by tau. Let us substitute that here that is 0 to T h A S. Theta i e to the power of minus T by tau D take a while and integrate this for me I can take h A S theta i outside I am assuming that h is independent of time which is not an unreasonable assumption which is most of the time h is independent of time and theta i is constant and surface area is constant. So, definitely I can take them out of integral. So, that is h A S theta i if I integrate e to the power of minus T by tau that is e to the power of minus T by tau upon minus 1 by tau between the limits 0 to T. So, what do I get what do I get. So, if I substitute for tau what is tau tau is let us remind ourselves tau is rho V C P upon h A S. So, if I substitute rho V C P upon h A S. So, what will I get h A S h A S gets cancelled out. So, I am left out with minus rho V C P theta i e to the power of minus T by tau is that all minus T by tau minus 1 minus is there minus is there. So, how did I get this minus 1 because when I substitute for T equal to 0 I get e to the power of minus 0 e to the power of minus 0 is nothing but 1. So, let us rewrite this equation by absorbing that minus sign inside that is q equal to q equal to rho V C P theta i into 1 minus of e to the power of minus T by tau. So, this is the total energy transfer let me remind you this is total energy transferred from whom to whom. From hot body to cold water or cold liquid or cold fluid. So, we have now what did we do so far let us just take recap we have just found for a lumped body we have found the temperature distribution and also the total energy transferred up to some time T. So, now, let us ask ourselves a question what is the question to be asked the question asked is when can I say my body is lumped when is my body lumped because our derivation whatever we did so far is all for lumped body that means temperature is only a function of time not space. So, when can this occur this occurs when my biot number is very much less now the question is what is biot number biot number is h l c by k let us write that what is h l c by k biot number equal to h l c by k that is if I rewrite this as h into delta t upon k delta t by l c this is convective heat transfer convection at the surface of the body upon conduction within the body convection within the body what are we saying biot number should be as small as possible that means what that means what compared to compared to compared to convection conduction should be very small that means there should not be conduction within the body there should not be any conduction within the body. So, there should be conduction dominant so that is sorry conduction see what is what we are saying is I can understand this because I got into trouble because I tried to interpret in little different manner that is biot number let me rewrite this again as l c by k upon 1 by h that is conduction resistance within the body upon convective resistance at the surface of the body if the biot number is 0 means what if the biot number is 0 means conductive resistance is 0 when conductive resistance is 0 means everywhere it can conduct itself in no time is not it that means temperature if there is no conductive resistance means the temperature within the body everywhere should be same. So, that is what so that means the biot number can never be 0 it has to be as small as possible. So, all that we are saying is conductive resistance should be lower than the conductive resistance should be lower than the convective resistance that is all we are saying. So, typically the thumb rule thumb rule for saying a body lumped is that biot number should be around less than 0.1 whenever biot number is less than 0.1 we say that it is lumped the only one issue I need to tell here is what is this l c what is this l c is actually volume by surface area volume by surface area please note this I have not mentioned so far l c is volume by surface area. So, we have very nicely introduced biot number actually in transient conduction we are going to get introduced with two numbers that is biot number and one more number Fourier number for now let us keep the Fourier number as it is and let us focus on biot number. So, we have understood now what is biot number. So, if I take a spherical ball whose thermal conductivity is 401 which is the case which I took in the very first transparency of my class initially that is I told that if I take a copper ball the temperature is same everywhere is it true let us check it out how can I check I can check through biot number. So, biot number so typically if I put the copper ball exposed to atmosphere h is 15 h is 15 and volume let us take a diameter of the copper ball as 12 centimeters quite huge one 12 centimeters is almost as big as small scale little small scale. So, that is volume is pi d cube by 6 of a sphere and area is pi d square. So, that turns out to be d by 6 so that is 12 by 6 turns out to be 0.02 meters. So, h l c that is l c h is 15 l c is 0.02 and k is 401 you see the biot number point triple 075 naturally my copper ball has to have the temperature same everywhere. So, that is the point here we are trying to say. So, the point is biot number can be used as a criterion for checking whether my body is lumped or not if my body is lumped I just can go ahead and use this relation to get the temperature distribution of the body variation with time something wrong in that calculation. So, q equal to rho v c p theta i into 1 minus e to the power of minus 2 by 2. So, that makes our life easy and of course it goes without saying if I apply Fourier law of conduction here del t by del x the thermal conductivity is high mean for the same heat transfer rate my gradients within my body have to be very small very small. So, that is another way of looking at a lumped body that is precisely what here also it is being said if the convective heat transfer coefficient is high and k is low then the gradients are going to be high because biot number is high h l by k biot number is high. So, my gradients are going to be high that is what this figure is shown. Now, let us take up this interesting problem I would like you all of you to solve this problem along with me if you solve this problem we definitely understand what are we whatever we have derived so far. It is a simple thermocouple which we routinely use in our heat transfer labs. So, it is a thermocouple junction which is typically made of copper nickel or copper constantan. So, here we do not have to worry about the material it is given I guess this is copper constantan only. So, a thermocouple junction which may be approximated as a sphere is to be used for temperature measurement in a gas stream. The convection coefficient between the junction surface and the gas is known to be h equal to 400 watts per meter square Kelvin. Let us go on writing I have a sphere and h is given to be 400 watts per meter square Kelvin and the junction thermo physical properties are k equal to 20 watt per meter Kelvin c p is 400 and rho is 8500. You see here copper constantan copper is roughly around 400, but nickel thermal conductivity is around 80 or 70. So, that is why this is lower than yesterday we discussed for an alloy the thermal conductivity of the alloy is lower than the proper thermal conductivity of the individual material that is copper and nickel in this case. So, that is why the thermal conductivity is lesser. So, 20. So, k is let me write this k is equal to k of my thermocouple wire is bead is watt per meter Kelvin and c p c p is 400 joule per kg Kelvin rho is 8500 kg per meter. So, this is the thermocouple bead whose properties are given and the heat transfer coefficient is given to be 400. 400 is quite a high heat transfer coefficient that means really the gas is flowing around it then only my heat transfer coefficient can be so high. Determine the junction diameter needed for the thermocouple to have a time constant of 1 second given here is we want a time constant of 1 second time constant of 1 second. So, time constant is given to be 1 second if the junction is at 25 degree Celsius. So, what is this 25 degree Celsius T i T i is 25 degree Celsius and this junction is exposed is exposed to a gas stream that is a 200 degree Celsius. That means what is the temperature here T infinity T infinity is T infinity is T infinity is 200 degree Celsius 200 degree Celsius. So, what do I get what is the question asked so if this is 200 degree Celsius if my thermocouple time constant is 1 second what should be the diameter of my thermocouple that is the first question. So, this I guess this can be first let us write down we have found this is the this is the diameter and we are main assumption we are making is bead is assumed to be in spherical shape not always it is possible for us to make the thermocouple bead sphere, but that is the closest assumption one can make for a thermocouple. So, thermocouple bead so what is tau the time constant is given to be rho V C p upon h a s. So, tau equal to rho V C p upon h a s. So, what is rho here so before we say that will let us put V and a s rho C p upon h into volume is pi d cube by 6 area is pi d square. So, I get rho pi pi gets cancelled out rho C p d by 6 here h is there. So, that is tau is what tau is 1 second is equal to rho is found to be 8500 into 8500 into C p is 400 diameter is what I need to find out upon h is 400 into 6. So, what do I get diameter is I get that as 7.06 into 10 to the power of minus 4 meter 10 to the power 7.06 into 10 to the power of minus 4 meter. That means in terms of mm how much it would be 0.71 or 0.706 mm. You see the diameter of the thermocouple has to be so small if it is time constant is 1 second, but when I say time constant is 1 second means generally let me go back away from this problem because I forgot to mention one point here that is generally what is time constant. Here for example, if the temperature is dropping temperature is dropping one time constant is assigned here I forgot to mention this point in this graph the temperature the time required to drop the temperature to 0.368 times its initial temperature difference that is the that is what is called as time constant. If it is heating means what will be this value 1 minus 0.368 that means it will be 1 minus that is 0.6, 0.63, 0.63 or 0.64 if you want to round it off 0.63 or 0.64 is the time constant is the time required that is the time required to reach 0.63 times the initial temperature difference. So, or if you are not getting me let me write this let me write this so that you are convergent with this number of course, I am digressing from the main problem that is here tau equal to the time required to drop the temperature drop the value theta t minus t infinity upon t i minus t infinity equal to 0.368 in this case in this case what is happening my gas temperature is higher compared to that of the body temperature. So, that means I should be writing this as t infinity minus t upon t infinity minus t i equal to 0.63 1 minus 0.368. Would be what 3 yeah 0.63 time required for reaching this ratio 63 percent that is what is the time constant definition that is what is 1 time constant means 0.62. So, typically if it has to it takes 3 to 4 time constants before it can reach the maximum temperature fine. So, let us get back to our problem. So, in our problem we found the time constant, but what is the question in the there is second part of the question in the second part of the question what is it being told how long it will take for the junction to reach 199 degree Celsius that is the question. So, let us see how do we do that for that I need to go to the temperature distribution. So, let me write that that is what is the temperature distribution given by theta by theta i equal to e to the power of minus I think instead of writing on the board I will just try to because this is simple substitution. So, I would not like to spend time on that. So, if I just write that time equal to rho v C p upon h a s I have taken into the power of minus t by tau as logarithmic. So, logarithmic of t i minus t infinity upon t minus t infinity we should not be worried about t infinity minus t i or t infinity minus t because negative signs will get cancelled out. So, if I substitute all the values what I have got that is diameter density C p heat transfer coefficient and the initial temperature but I want the final temperature as 199 you see my t required is 5.2 second this is what I was saying it generally require 3 to 4 time constant before it reaches the maximum temperature. So, our time constant was 1 second now it took 5.2 seconds before it could reach 199 seconds. So, the point here is we need to understand several things what are the outcomes from this problem see what we realized here is that the time constant rho v C p upon h a s is not a function of thermocouple material time constant is not a function of is not dependent on thermocouple material. Yes it is dependent on thermocouple material but not only thermocouple material only but it is also dependent on h it is dependent on h it is dependent on h also. So, here in case of air if it is air if I am using thermocouple in case of air my h is going to be lower then my time constant would be higher but if I am using the same thermocouple in water my h would be higher my h would be higher and then my time constant would be lower. So, the point here is the thermocouple time constant is not is not just dependent on the material that is what usually the question asked by generally whenever you are going to the class the student ask oh my thermocouple is having a time constant of 1 second. But when we have to specify under what conditions where I am inserting my thermocouple in natural convection my h is going to be very low even if it is liquid then my time constant of my thermocouple will be less but if I am using the same thermocouple in air but flow is there then in that case my time constant would be low. So, thermocouple time constant is not dependent only on the thermocouple material but on the situation where we are using which decides my h my h that is very very important point to be noted through this problem because why I am emphasizing so much because in our heat transfer lab the major work horse for measuring the temperature is thermocouple. So, that is the reason why we need to emphasize on our students the importance of this time constant. So, with this we will move on to the next issue that is the spatial effect. So, I think before we move on to spatial effects we will take questions for next 10 minutes SVNIT Surat. Why we select the biot number less than 0.1 not 0.5 or not 0.05 that is a good question. The question asked is why is biot number less than 0.1 has been given as a criterion for lumped system analysis. So, that is a good question. So, actually you can always question this biot number less than 0.1 generally for lumped bodies for convective boundary condition generally it has been found that for biot number less than 0.1 there are no temperature gradients within the body. This has been experimentally proved and also numerically also one has tested this and found that biot number less than 0.1 indeed ensures that my body is lumped. Any questions? Best situation would be when biot number is equal to is equal to 0, but since we are not able to, but since that is not going to be practical you are going to have a small biot number that is of that is allowed and that is why 0.1 has been found anything smaller than that you are going to lose out on cases where actually lumped capacitance could have been used. So, if you are imposing a very stringent condition you will have to lose out on a simplified solution for a lot of problems. So, can we consider 5 by 5 meter plate as a lumped body? Lumped body when? Question asked is can we consider 5 by 5 meter as a lumped body? I would say the question is incomplete why because we having given the criterion biot number h l c by k I need to know h, I need to know l c, I need to know k put these numbers and check out your cell. Sir, infinitely long case of the fin or plate body, sir how can we find the heat transfer coefficient in the case of the winglet fin? If there is a case in the infinitely long case? If I understand the question the question asked is for infinitely long a section or a long body how can I get the heat transfer coefficient is that the question? Winglet fin sir winglet fin winglet is that winglet you mean? Again I will put the same question back to you see the question asked is in case of winglet is the heat transfer how do how can one get the heat transfer coefficient? I think this question is little unrelated. However, one thing I would suggest is that for a winglet also I know the winglets are generally used as a rib as enhancers of heat transfer coefficient. So, this question is little unrelated that is why I said the winglets are used for enhancing the heat transfer locally in internal flow situations. But however, the question if I rephrase that for our situation within the winglet whether the my body is lumped or not again I would go back and say I have to put h l c by k and check whether my biot number is less than 0.1. Now, I guess I will have to move away from S V N I T Surat having taken three questions. The next question I am going to take it from Nirma University Ahmadabad. Sir the biot number equation is h l by k and in convection case the Nusselt number is also h l by k. What is the basic difference between to identify the biot and Nusselt number? Very good question. It is little premature. It is the question asked is biot number definition is also h l c by k. Nusselt number is also h l c by k. So, what is the difference between biot number and Nusselt number? I think I would I would just say that the thermal conductivity here is thermal conductivity of my material. In Nusselt number it is thermal conductivity of the fluid flowing over my body. However, this is question is little premature. I think I have answered in a very concise manner we will deal with this when we deal with convection. Biot number refers to the resistances on the surface and the convective in conduction and convective resistance. Whereas, if you look at Nusselt number we will talk about how it is related to the thermal conductivity of the fluid later on. Okay, over you sir. Yeah, any questions from Amrita Koyambattur? Sir, the time constant the definition you told for the thermo cup lumped system you told when it is cooling the theta value comes down by 36 percent. Sorry, comes down by 63 percent and when it is heating it is the other way round. So, is it because how is the time constant defined sir? Is it defined for heating separately and cooling separately or is it the same? No, no, no, no it is not that way. See it is 63.7 percent of the initial temperature difference. That is all it is. The time constant definition is 63.2 percent of the initial temperature difference whether it is cooling or heating. It so happens that our ratio is if you see here in that case if I have to write theta by theta i equal to 1 minus of e to the power of minus t by tau I will get. So, here in this case I will get theta by theta i equal to e to the power of minus t by tau. So, time constant is 63.2 percent only but heating or cooling will decide the temperature distribution that is all. But the time constant is 63.2 percent only. Is that ok? Fine. Thank you sir, thank you. Is that ok? Yes sir, yes sir, over to. Over to Bharamati, VPCOE. Sir, my question is regarding an answer. In an answer there is increase in convective heat transfer coefficient it is at a cost of friction factor. I have seen a review paper in which I have observed that increase in convective heat transfer coefficient is always less than that of friction factor. Is it always true or is it possible to build a roughness geometry in which we can have a convective heat transfer coefficient increase will be more than increase in friction factor. The question asked is usually the convective heat transfer coefficient increases with by placing rib by placing enhancer. But enhancer by putting enhancer the friction factor also increases that is the pressure drop also increases. It is like life. In life also we have to give more than what we receive. So, that is true in heat exchangers also, that is true in enhancers also. So, always generally 99 percent of the times the what you gain in terms of heat transfer coefficient is lower than what you spend in terms of pressure drop or the pumping power or the friction factor. The quest is always to try to minimize this friction factor to the extent possible. So, to go ahead because you refer to review paper to go ahead and answer this the general enhancement criterion used in the literature is N u by N u naught that is when I say N u naught is with reference to Nusselt number in smooth surface upon F by F naught to the power of 1 by 3 should be of the order of 2. People have not achieved anything greater than 2 till today. So, always the quest is for going above 2, but I think your review paper what you are referring is professor Ligrani's review paper. So, in that review paper also if you go back and see none of the rib none of the enhancers have gone above 2. So, they are just hovering around 2.1, 2.2, 2.3. So, to answer your question in one sentence we have to give more than what we gain that is always the philosophy that is true for even heat transfer. We will stop this question answer session and move ahead with special effects with professor Arun.