Week 4: Lecture 16: First term approximation solution for plane wall

So let's get started from where we stopped. So we started if you just recall we started from plain wall. So where we said that the temperature is varying along the space and also time. So that's why we had a temperature variation term with space and the temperature variation term with time in the heat diffusion equation and this was the initial condition and this is the conveictive boundary condition and this is the symmetric boundary condition symmetric boundary condition and we try to non-dimensionalize this we try to non-dimensionalize and what are the advantages of that non-dimensionalization how does the non-dimensional equation look like Let us write that. So our equation was d² t by d x² = 1x alpha do t by d t when non-dimensionalized reduced to do ^² theta by dx² equal to do theta by do fer number. What was theta? Theta of x t equal to t of x t minus t infinity divided by t i minus t infinity and capital x = small x by capital L and fer number is alpha t by l² and bi number is h l by k. Remember this L is the thickness of the half thickness of the wall. Okay? So not the characteristic length here for this case. So T of X comma 0 is equal to TI this is the initial condition. Now theta of X0 is equal to 1 which is what we got and do T by dox at X =0 is zero. And here do theta by dox at x =0 is zero. So minus k do t by dox at x = l is equal to h into t of l comma t minus t infinity reduces to do theta by do x capital x at x = 1 equal to minus biot number into theta at 1 comma real number. So what is the beauty? Why did we do this? Why did we do this non-dimensionalization? Because the temperature if you just take the temperature is a function of heat transfer X. It's a function of X. It's a function of X. It's a function of T time thermal conductivity. Heat transfer coefficient. Thickness of the wall. Alpha that is thermal diffusibility. So TI initial temperature T infinity so many parameters. But now if you reduce theta now is only a function of capital X biot number and fer number. So my temperature distribution is more generic more generic. when I get the temperature distribution in terms of theta. So now the next question is what is the physical significance before we move on to the plane wall and try to solve the problem. What is the physical significance of uh fer number? Physical significance of fer number. Physical significance of fer number. Fer number is equal to alpha t by l². So I can rewrite this as k l². Remember alpha is k by row cp. So, K L² deltat T by L divided by row CP L cq deltat T by T. I've just done nothing. Multiplied and divided by delta T and multiplied and divided it by L. Okay. So when I do that, when I do that, uh, when I do that, so this is nothing but the Q dot conducted within the body. And what is this row? This is nothing but row VCP dot T by D. This is Q dot stored. Q dot stored. So it is that is if you are to show this in the form of a uh beautiful figure you can see that the the Q dot whatever you are giving is how much of that is actually getting stored how much is getting stored within the body compared to how much is getting conducted. If the fer number is very high means that Q conducted is more than what it is stored and if the fer number is low means it is storing more compared to that of Q conducted. So that's what is the physical significance of this fer number. Now what really constitutes this is infinitely large plate. I'm taking first large plate. Actually after the large plate we shall take up the cylinder and also the sphere. But right now we are in a large plane wall. Large plane wall. When I say that it only means that the thickness of the wall is small compared to height and width of the plane wall. If you just take the wall of your room that would satisfy this condition. So that is what we will. Now the question is how to get this temperature distribution mathematically without proof we are going to give you the solution of this temperature distribution for a plane wall. So how what is that exact solution for a plane wall is our next question. So the exact solution exact solution of for a plane wall for a plane wall okay that is theta this is without derivation I'm giving in advanced courses you'll get this deration ation. If you are interested, you can email us. We can I can supply this. T of X comma T minus T infinity divided by T I minus T infinity is given by sigma of N = 1 to infinity A N E to the power of minus lambda n² fer number cos of lambda n x by L and how what is a n? Fer number you know fer number is nothing but alpha t by l² and now what is a n? A n is 4 sin lambda n upon 2 lambda n plus sin of 2 lambda n. Now who gives me lambda n? Lambda n is obtained if you know the biote number you will be knowing the biote number for your problem. So lambda n tan of lambda n this is a and what is the definition of the biote number? Budd number is h l by k. K is the thermal connectivity of the plane wall. Now this is what is called as transcendental equation and this will have transcendental transcendental equation and this will have multiple roots. So we don't have to get these roots they are all given as a table. So if you really see if you really see uh for a given biote number for a given biot number you can see that you there are multiple roots there are multiple roots but in this table I'm just giving you lambda 1 lambda 2 lambda 3 lambda 4. So if you know the biote number you can get these roots and once you get these roots once you get the lambda n you can go back and calculate the a n using this relation and once you get a n at any fer number at a given location x you can use this series it's a series solution next question is how many terms of this series I am supposed to take that is how many terms terms I have to take to get the correct solution. Fortuously or luckily uh this is for different B numbers I have given. So luckily what happens is that for a fer number of greater than.2 for a fer number greater than.2 2. Luckily if the fer number is greater than 02 one term is first term of the series is suff would suffice. First term would suffice. In that case what will happen for a plane wall? Theta of x t is equal to t of x t minus t infinity divided by t i - t infinity is equal to a1 e to the power of minus lambda 1^ 2 fer number cos of lambda 1 x by l. So one term is sufficient to solve the problem. Okay. So now incidentally there is a table you don't want to calculate A1 also using this complicated equation. So we are giving you a table. We are giving a table where in which for a plane wall for example you see here for a plane wall this is the biote number and for a plane wall for a given biote number for getting only the one term series you can get lambda 1 and a1. So biode number you will calculate using hl by k plane wall of thickness 2 l and you will substitute that a1 and lambda 1 at any given location x at any time t for real number equal to alpha t by l² you can compute the temperature at any given x t for a given t i and t infinity. So that's what we are going to do. We will solve a problem so that you will become comfortable with this methodology and then we will take up cylinder and sphere and we will solve a problem as well. Okay. So at the center at the center of the plane wall what will happen? At the center of the plane wall that is if you take up at x=0 at x=0 what will happen at x = 0. So cos of 0 will become 1. So you will have only theta at 0 comma t will be equal to a1 e to the power of minus lambda 1^ 2 fer number. If you really closely see this equation, if you recall for lumped body the temperature varied exponentially that is continuing here. This cos lambda this is the time varying parameter. Time variation is captured here with this term and the space variation is captured with this term. So the time variation continues to be exponential and space variation is varying with cos of lambda 1 x by l. Okay. So with this exact solution for temperature now the next question would be what is the total energy transferred? Total total energy transferred from the wall. That is the next question. from the wall. That is delta E ST that is the energy stored that is nothing but this is nothing but Q equal to row VCP that is row CP temperature variation that is T at X comma T minus of TI into DV. Okay, at any given instant of time, remember this is DV. So this should be this should be triple integral. Okay. So the integration is performed this is integration is performed over the volume of the wall. And here the negative sign in case if you end up with negative sign negative sign implies that heat is leaving the body. leaving the body. Okay. So just to uh get you acclimatized with what I'm saying is that so my body was at TI and which is having row VCP and it is exposed to H comma T infinity and what is the maximum Q dot one can have? What is the maximum Q dot is for a time tending to infinity my temperature will become uh t infinity my the temperature of the body will become will reach the fluid temperature eventually after time t infinity. So that is the maximum heat transfer I can get. So but the actual heat transfer is obtained by integrating. So the maximum heat transfer is Q maximum is MCP T infinity minus T I that is my body has reached T infinity. So row VCP Q maximum equal to T infinity minus T I. So Q maximum equal to row V CP T infinity minus T I now without derivation like temperature distribution I am going to give you the relation Q by Q max is equal to 1 minus theta KN wall that is for plane wall into sin of lambda 1 divided by lambda 1. I have taken only the first term series. And if you recall theta wall was equal to theta wall was equal to a1 e to the power of minus lambda 1 2 into 4 year number. Okay. So this is fetches you what is the total energy transferred from the wall. Okay. So this is what it will uh get you. So what is that we got? Let me just recall what is that we got. We we just started with the plane wall and we non-dimensionalized this equation and got this as a function of biote number fer number and x without doing maths we stated without solving the pd that is the partial differential equation we gave you the solution where the temperature is a function of freeer number and also the location that is the time and location. So here we have gotten the for a given biot number uh you can get the lambda 1 lambda 2 lambda 3 lambda 4 and we said that if the fer number is greater than 2 one term is sufficient. So for a given biote number from the table you will get lambda 1 and a1 and biote number being hl by k. So and after that we went to energy transferred and we got the relation for energy transferred for the plane wall as Q by Q max for the wall equal to 1 - theta wall sin lambda 1 by lambda 1. So for number of point 2 the infinite series solution can be approximated only by first term. So what we will do now is that uh we will solve a problem and then we will see how to get acclamatized with this uh problem that is in the problem what we are taking here is that in the problem so the there is a plain wall it's a simple plane wall so let me explain you this problem it's a simple plane wall actually it is a pipe okay let Let me read that problem and then we will come back. Okay. Consider a steel pipeline that is 1 m in diameter and has having a wall thickness of 40 mm. The pipe is heavily insulated on the outside and before the initiation of the flow the wall of the pipe are at uniform temperature. You may be thinking that this is a cylinder but how can I use plain wall? I'm going to make an approximation that the thickness 40 mm is very small compared to 1 m diameter. So the curvature effect can be neglected and the pipe wall can be considered as a plain wall. Let me repeat the wall plane wall of the pipe wall is considered as a plane wall because the wall thickness is very small compared to the diameter. Hence the curvature effects are neglected and assumed this as a plain wall. And the rest is it is the walls of the pipe are at a uniform temperature before that is TI is 20°C. With the initiation of the flow the hot oil at 60°C that is T infinity is 60°C is flowing with a convective condition that is H equal to 50 W per me² Kelvin which is quite high. So now question is what are the appropriate biot and fer numbers at eth minute after the initiation of the flow and at 8th minute what is the temperature at the exterior pipe surface covered by the insulation that is at x= to l what is the pipe surface temperature and what is the heat flux to the pipe from the oil at t= 8 minute and how much is the energy per meter of the pipe length has been transferred from the oil to the pipe at 8th minute. So these are the this is the plane wall. Let me write this. So let's take a and let's solve this problem. So that is you have a plane wall. Okay. This side you have insulated and one side you have put the convective boundary condition. H = 500 W per m²ared kelvin and T infinity is 60° C and this is TI is at -20° C. TI is at -20°C and this length L equal to 40 mm. Now you might be wondering that we solved for a plane wall with at the center and this is this was x = l. We have the solution for this case where you have h comma t infinity and h comma t infinity. But is this case similar to this case? Yes. Because if it is insulated at this condition do t by dox okay at x =0 will be zero because there is no heat transfer rate. So it is insulated that is why. So this solution of what we had here also we had taken do t by dx by symmetry do t by dx at x=0 as zero. So both are same both are same. So that's why I can apply this solution to this case. Now I need properties at what temperature I should be taking. I should be taking the properties at average of -20 + 60 divided by 2 that is at 20°C or 293 kel. So if you take the properties here at this temperature, so you get row equal to 7832 kg per meter cube and CP is 434 jou per kg kelvin and alpha that is k by row cp is 18.8 into 10 ^ of -6 m²ared per second. Now the first thing is I need to check what is now time at t = 8 minutes what is the biote number if the biot number is very less I can consider this as lumped but let us just check by number is h l by k h is what is given to be 500 l is given to be 40 mm that is 40 into 10 ^ of -3 Three thermal conductivity is found to be the thermal conductivity is given to be 63.9 W per meter kelvin. So 63.9 so you get a biote number of 313 which is greater than.1 which is greater than.1 so not lumped at all. So I have to take the spatial variation. It is not lump. Now at 8th minute what is the fer number? Alpha t by l² alpha is 18.8 into 10 ^ of - 6 8 minute 8 into 60 divided by 40 into 10 ^ of -3²ared. So fer number I get as 5.64. 64. So this freer number is indeed greater than 0.2. So one term solution is good enough for me. Okay. So if that is the case my solution first question is center at the center of the plane wall what is the temperature distribution? So at the center of the temperature distribution that is theta 0 comma t that is t of 0 comma t minus t infinity divided by t i minus t infinity equal to a1 e to the power of minus lambda 1 2 fer number. Now for a biote number of 313 I need to get a1 and lambda 1 from table. So let us go back to the table. If you go to the table. Yeah, here is the table. Here is the table. So in this table, so we have the biot number. My biot number is 313. And here at 3 I have and 4 I have I have to interpolate between these two. I will show for one how to do the interpolation. So at 3 it is 0.5218 and at 4 it is 5932 lambda 1. So at a1 is 1.045 and 1.058. So let's see how do we do this interpolation and get the values that is you have at biot number three you have a lambda 1 as 5218 a1 as 1.045 at 4 you have 5932 and this is 1.058 058. So for interpolation so for lambda 1 is equal to 0 5218 plus 5932 -.5218 divided by 4 -.3 into 5 into sorry this is into 313 3 minus.3 this will fetch you of a lambda 1 of.531 I just did the interpolation similarly for a1 you can write this as 1.045 plus 1.058 - 1.045 045 divided by 4 -.3 into 313 -.3. So you get a1 equal to 1.047. So substituting this that is this implies that t at 0 comma t minus t infinity is 60 divided by -20 -60 equal to a1 is 1.047 e to the power of -.531 squared into fer number the fer number we know that it is we have calculated 5.6 64 for the 8th minute it is 5.64. If you just do this pressing the calculator you get t at 0 comma 8 minute is equal to 42.9° C. So it got heated up from -20°C to 42.9°C in 8 minutes at the exterior of the wall that is at x = uh zero at x=0. So you have that is at the insulated condition that is at here at the insulated condition you are getting the uh 42.9° C. So in the next class what we will do is we shall continue this problem and solve the rest of the that is the heat transfer rate and the energy stored we shall solve uh in the next class. Thank you.