Week 4: Lecture 17: First term approximation solution for radial systems

Let us continue our previous problem where we had taken just this is the plane wall. I will not again spend time on this and we just solved for the at x=0 what is the temperature? So we got the temperature as 42.9° C. So now what we will do is what is the heat flux? The next question is what is the heat flux? What is the heat flux in W per meter squared at the plane wall at t = to 8 minutes at the by convection by convection how much it is losing the what is the heat transfer rate it is losing here what is the heat transfer rate which is losing here that is given by Newton's law of cooling so I can write that qx X dash at X = L comma 480 that is 8 minutes 8 into 60 480 seconds is equal to QL dash is equal to H into T of L comma 480 seconds minus T infinity minus is t infinity. So I don't know the temperature at x = l. I need to find that. So theta at x t equal to t of x t minus t infinity divided by t i minus t infinity is equal to a1 e to the power of minus lambda 1^ 2 fer number cos of lambda 1 x by l for the same biode number and fer number we have already gotten the lambdas plus. So I can just substitute that and get so that is at x = theta 0 comma t that is I can get this so directly that is a1 what is my a1 a1 is or this can be thought of as we have just found that is theta of 0 comma t correct that is X equal to0 which we have found or otherwise you can just substitute also if you are getting confused there is no problem that is 1.047 e to the power of -.531 squared into fer number of 5.64 64 into cos of lambda 1.531 into x that is l by l that is 1 by x = l by l that becomes 1. Now cos this is in radians so I have to convert this into degrees to operate your calculator. So I'm multiplying that and dividing it by 180 by pi. So if you do this and this is t at l comma 8 minutes minus of 60 divided by -20 - 60 so I get t of l comma 8 as 8 minutes as 45.2° C. So the previous temperature at the center was 42.9 at the at the wall it is 45.2 which is expected because the hotness of the oil is felt at the wall first so it will be at the higher temperature. So now coming back if I substitute this here that is h is 500 into t of this is 45.2 2 minus of 60 that is equal to -7,400 W per m²ared. Okay. So this is what I get as the heat transfer rate. Now the last term what I need to get is the energy stored energy per meter pipe of the length pipe length per meter pipe length pipe length being transferred after t = 8 minutes. So for that I know Q by Q max equal to 1us theta wall into sin of lambda 1 by lambda 1. Okay. So that is equal to 1 minus of if you just go back and check this theta wall theta wall actually this this term alone this term alone reduces to 214 that's what I'm going to substitute here that is theta wall that is 214 theta wall is nothing but a1 e ^ of minus lambda 1^ 2 here number so that's what I substituted 214 into sin of lambda 1 that is 531 into 180 by you now know why I'm multiplying by 180 by pi divided by 531 so this comes down to8 now q max is equal to row v cp into t infinity minus ti So row is 7832 and volume volume is pi d into thickness volume of the cylinder. So how do I get that volume of the pipe? If I to just show quickly that is this is what it shows that is volume is given by by 4 d²us d² into l and I can rearrange like this per unit length that is per meter so l equal to 1 and volume I can uh this is pi into d by 2x2 and d + d by 2 here this is very small d and di are almost same so I can take it as D I so if I take that as D I so I can write this as P DI into 40 into 10 ^ minus 3. Okay. So that's what I have done. So if I substitute that here pi d into thickness that is 7832 into pi into d is 04 into di is 1 m and thickness is uh 04 into -20 - 60. So I get Q = -2.73 into 10 ^ of 7 Joule per m. So we solved the plane wall problem. We solved the plane wall problem and uh we are now convergent with this solution what we have adapted. So this is this was the exact solution for the plane wall. Now let us see how does it translate to when we do it for uh the cylinder and also for uh what the sphere. So how does it look like the temperature distribution for now? Let me take up the cylinder. The temperature distribution for a cylinder. So what is the figure for the cylinder? The cylinder is very large cylinder. This is I mean large means longer length and this is having R and this is radius R. Okay. And this temperature T equal to T I initially. Now the question is and again it is exposed to H comma T infinity. You can imagine this is the cylinder this is the pen. So over which the flow is taking place and we are interested in the temperature distribution T of R comma T. So again without derivation we will have to get the temperature distribution as a function of biote number for year number that's it. So how does it look like? So it looks like theta of r comma t is equal to t of r comma t minus of t infinity divided by t i - t infinity is equal to only one term series I'm writing a1 e to the power of minus lambda 1 2 fer number into bessel's function that is J KN this is a J not that is J KN is called zeroth order bessel function I will write that J not lambda 1 R by R not don't have to worry I will give the J not value for a given biot number and for number a biot number okay so this is the equation for theta now next is theta of theta t is equal to so but before that how do I get this uh a1 and lambda 1 for a given uh biot number that is given in the table that's given in the table if you just go back and see here in my notes so if you just see here for a given biot number what is there here is that for a given biot number we had used this table for a plane wall but for cylinder it is biode number equal to for a given biote number you can get lambda 1 and a1 but please note for sphere or cylinder it is actually h by k so you can get lambda 1 and a1 for both cylinder and sphere okay and this is for different bard numbers it will go on okay so that is what you are going to get for this is what is going to be for cylinder and this is the equation for sphere. I will write both the equations and reduce it for the plane wall. Okay. So now the theta of 0 comma t theta at 0 comma t is t of 0 comma t minus t infinity divided by t i minus t infinity is equal to a1 e to the power of minus lambda 1^ 2 for number because bessels zeroth order bessel at zero is one. Okay, that you can check in the table as well. If you just go back and see in the table also you are going to get that. Now let us write for sphere. Similarly, we can write for sphere. For a sphere the uh this is my sphere and this is the radius r and this is my r and everywhere it is experiencing h comma t infinity. So theta of r comma t is equal to t of r comma t minus t infinity divided by t i - t infinity is equal to a1 e to the power of minus lambda 1^ 2 number into sin of lambda 1 r by r divided by lambda 1 r by r. You see something very interesting here. The temperature variation with time always is exponential whether it is plain wall or cylinder or sphere only the space variation function is varying. Please note that similar to here theta of 0 comma t is t of 0 comma t minus t infinity divided by t i minus t infinity is equal to a1 e to the power of minus lambda 1^ 2 for number. You know that in plus two you have studied limit as x r t tends to zero this becomes equal to 1. So that's why you get this temperature distribution. Now the next question is what is the heat transfer for cylinder and sphere. So for cylinder for cylinder it is Q by Q max equal to 1 - 2 theta cylinder into J1 lambda 1 divided by lambda 1. So J1 is the bessel functions first order bessel function. J1 is the first order bessel function. You don't have to worry about that function because the values are given in the form of a table. Now for a sphere Q by Q max is equal to 1 minus of 3 theta sphere into sin of lambda 1 minus lambda 1 cos lambda 1 divided by lambda 1 cq. So we have gotten the temperature distribution and also the energy transferred. So this one term solutions are valid under circum certain circumstances that is at t =0 temperature is ti and it is uniform throughout the body. Okay. And the heat transfer coefficient H is constant. And also T infinity is also constant. It doesn't vary with time and there is no energy generation term then only these solutions what we have given are valid. So what we will do is now we will take up a simple problem where we have to do with sphere that is I will take egg. I will take a problem of an egg that is an ordinary egg is can be approximated as a 5 cm diameter sphere. Okay. The egg is initially at a uniform temperature of 5°C and dropped into boiling water at 95°C. Why five? Because you take your egg from the fridge refrigerator. So the temperature typically is around 5°C and dropped into your boiling water which is around 95°C. Taking the convective heat transfer coefficient between the yug surface and the water boiling water as 1,200 watt per meter squar°C determine how long it will take for the center of the yug to reach 70° C. So although egg is not truly spherical, we are going to consider this yug as sphere. So let's see how do we solve this problem that is I have an egg and its initial temperature is 5°C and h is equal to,200 w per m²ared kelvin and t infinity is 95°C. Celsius. And now the question asked is to reach at the center t not as 70°C. What is the time taken? And we are approximating this egg as a sphere of diameter 5 cm. So with this let us see how do we get the uh properties the properties of the yug are like water properties only I should take any living substance if you don't have any properties including me human body if I don't have the properties I have to take the properties of the water because most of the volume of my or the mass of my body is made of water similarly with the egg also so egg is 74% is water content so The properties of the water are taken at 5 + 70 by 2 that is t i + t by 2 that is 75 by 2 that is 37.5°C. So at 37.5°C I get a K of 627 W per meter kelvin and an alpha of.151 into 10 ^ of -6 m²ared per second. Now I cannot calculate fer number because time is what I need. I need to calculate fer number because what is the time taken is what I need to find. So next thing I can do is buy number H R by K. H is given to be 1,200 R is diameter is 5 cm. So 5 into 10 ^ of -2 divided by 2 because the radius and the thermal conductivity is 627. So you get a biote number of 15.95. This also means that this is not lumped because biode number is pretty much greater than 0.1. Now in fact the you should if you have to calculate the lumped LC equal to volume by surface area volume by this is d by 6 you get for a sphere. So you get 5 into 10 ^ of -2 divided by 6 actually lumped criterion will still be lower than will be still greater than 15 95 even then that is biot number hlc by k if you put that is,200 into 5 into 10 ^ of -2 divided by 6 divided by 627 will be around 15.95. 5 divided by 3. So still this will be very high. So point is bode number will be higher than 0.1 and it is lumped. So if this is lumped then I have to take the one turn series solution for the center of the sphere. If you recall we have theta of 0 comma t equal to t of 0 comma t minus t infinity divided by t i minus t infinity is equal to a1 e to the power of minus lambda 1 2 fian number okay so I have to get for biot number 15.95 I have to get the uh values of lambda 1. So if you just come down so this is this is indeed 15.95. This is indeed 15.95. So this is 15 this is this is LC this is LC LC. So you get this is 5 into 10 ^ of -2 / 6 you get a bout number greater than.1 but now for this calculation I have to calculate biot number as h by k so h is200 into 5 5 into 10 the 5 into 10 ^ of -2 divided by 2 divided by k of 627. So the biote number happens to be 47.8. For this 47.8 I need to get the uh lambda 1 and a1. So if you really see here so for a biote number of 47.8 if you do the interpolation between 40 and 50. I'm not going to do the interpolation because I have shown that in the previous class. So you get lambda 1 as 3.0753 which is in between these two a1 as 1.9958. So if I take that value so you will get lambda 1 biot number at 47.8 from the table you get lambda 1 equal to 3.0753 0753 and a1 equal to 1.9958. So you get at the center theta of 0 comma t is equal to t of 0a t minus t infinity divided by t i minus t infinity is equal to a1 e ^ of minus lambda 1 2 for number. So if you just substitute this is 70 center of the egg should be 70 and this is 95 boiling water divided by initial is 5 - 95 is equal to 1.9958 e to the power of - 3.0753 into fer number so I get fer number of.209 209 luckily fer number is greater than 02 only one term is sufficient one term is sufficient so now freer number equal to I need time alpha t / l² so alpha what is alpha alpha we have found 151 into 10 ^ of -6 t divided by L² that is 2 that is here in this case it is going to be R² R is 5 into 10 ^ of -2 / 2 so it is 2.5 into 10 ^ -2² okay is equal to 209 that fetches me t = 865 seconds that is approximately 14.4 minutes. So that means for the boil for the egg to boil I have to keep under this condition for at least 14.4 minutes. So so that's how we could solve this problem. So uh so we have covered in today's lecture what did we cover is that if you just recall so what we gave is the temperature distribution for the we said that exact solution will be of this form okay and we gave for plain wall cylinder and a sphere and the temperature distribution as you can see what is so beautiful here is that the temperature variation with time irrespective of plane wall, cylinder and sphere is exponential which is the same as what we had gotten in lumped body. Now in a plane wall it varies as a cost function in a cylindrical wall in a cylindric in a cylinder it consider it takes a bessel function and in a sphere it is s by lambda. So this is how we get the temperature distribution. to get this a1 and lambda 1 I need to get the biote number. Once I know the biote number, I can get the uh lambda 1 a1 for plane wall, cylinder and sphere. The most important thing I want to re-emphasize is that for biote number for plane wall it is hl by k and for cylinder and sphere it is hr by k. Remember that this biote number is different from the biote number what we defined for lumped body. For a lumped body it is H LC by K where LC is volume by surface area. You are not supposed to take volume by surface area when you do this plane wall or cylinder or sphere for spatial variations in time. Okay. So and for one term approximation is valid for fer numbers greater than 02 and at the center of the plane wall the all these at the center of the plane wall it is only going to vary exponentially for all the cases. Okay. So now very interesting case we come across is that if what if if you go to this table if you go to this table for a biote number infinity also there is a value given if the biote number that is how can I realize biote number infinity as an engineering as an engineer biote number infinity implies that biote number infinity implies that heat transfer coefficient is very large. So when will this occur? When will the heat transfer coefficient is very large? If you recall in my introduction slides, we have said that for boiling the heat transfer coefficient will be of the order of 50,000 60,000. So in that case the one by number that is K by HL will become zero. K by HL will become zero that is I can consider that biot number as infinite. So in that case for bot for boiling or if the heat transfer coefficient is very large what does that mean? So let's understand this little slowly. So actually it if the heat transfer coefficient is very large 1 by number is zero it actually becomes constant wall temperature. Okay. So this I will explain in the next class in more detail. What is the physical implication of biot number infinite and subsequent to that we will take up uh in the next class what is called as the uh heat transient heat conduction in semi-infinite solids. Thank you.