 Let's solve a couple of questions on equivalent thermal conductivity. So for the first one we have a small cylinder of radius r which is placed inside a big shell having inner radius r and external radius 2r. So we can see the small cylinder which is in purple and the big cylinder outside of it which is in pink. The ends of the cylinders are maintained at different temperatures as it represented below t1 and t2. The conductivity of the inner cylinder is 4k and that of the outer cylinder shell is 2k. We need to figure out the equivalent thermal conductivity of the system in terms of k. Okay as always pause the video and first attempt this one on your own. Alright hopefully you gave this a shot. Now in the question we see that these two cylinders the small purple cylinder and the bigger pink cylinder they are connected across the same temperature. So that's t1 and t2. Here at both of their ends is t1 and t2 which means they are connected in parallel. And we can find the equivalent thermal resistance for rods connected in parallel which is given by this relation right here. It is 1 upon r effective which is equal to 1 upon r1 plus 1 upon r2. So r effective here is the equivalent thermal resistance and r1 and r2 is the thermal resistance for rod 1 and rod 2. And we also know how thermal resistance is related to thermal conductivity. That is really r this is equal to l divided by ka. But what does equivalent thermal conductivity for rods in parallel mean? So we can say that if the two rods if these two rods if they are replaced by one just one single rod having the same total effective area of cross section then for the heat current to be the same. The conductivity of the new rod that is required is what we call equivalent conductivity. So just like if we had when we had two resistances just like these which are connected in parallel and they were connected across a cell. So the current that flows in the circuit let's say that's i and we can replace an effective resistance in place of these two individual resistances and even then the current that will flow in the circuit it will still be i. It will still be i even if this even if these two individual resistances let's call them r1 r2 if they are replaced by r effective. Similarly if we over here if we replace these two rods with one rod having the equivalent thermal conductivity then the heat current that is flowing initially will be the same. So we need to figure out what is that equivalent conductivity and to do that we can place we can place this relation over here and then figure out the k effective. So let's do that. This becomes one upon this is one upon l divided by k effective into area effective which is equal to one upon l divided by k let's call it k 4k into and let's say r1 is for inner cylinder this is 4k 4k multiplied by the area of the inner cylinder. So that would be that will be pi r2 and we are adding then one plus l the length is the same divided by 2k that is the outer cylinder shells thermal conductivity 2k into now the area of the outer shell. So let's let's figure that out. Now the outer cylindrical shell we can we need to find out the area all of all of this area. So what we can do is we can we can find the total area of radius 3r and then subtract the area of the purple cylinder. So that would be pi 3r2 minus pi r2. This is 9 pi r2 minus pi r2 which becomes 8 pi r2. So that is the area of the shaded part we have excluded the purple the purple cylinder inside. So this is this is 8 pi r2 and if you think about a effective a effective is a total area that we can see from the left. So that is that is pi 3r2 that is pi 3r2 the area of this purple cylinder pi r2 plus the area of the outer cylinder. So in total this distance is 3r. So the total area that we can see from the left is pi 3r2 that is that will come in place of a effective. Now one thing gets cancelled off that is l l just gets and k into a for all of these all of these quantities that will go to the top because it is in the denominator off the denominator. So it goes to the top for all of them and then it becomes k effective into pi 3r2 this is equal to 4k into pi r2 plus 16 pi r2 into k. Now let's cancel pi so pi gets cancelled r2 gets cancelled r2 gets cancelled. What remains is k effective I am writing that over here k effective this is equal to 4k plus 16k divided by 9. So this is 25k by 9 this is 25k by 9 and that is the equivalent thermal conductivity of the system in terms of k. Okay let's look at one more question. Alright so for this one we have a combination of three metallic blocks we can see that over here we need to figure out the equivalent thermal conductivity of the combination in terms of k. Now over here we can see that these two these two blocks they are connected in series and we can say these are the top blocks they are connected in series and then their combination is connected in parallel with the block the green block of thermal conductivity 3k. So what we can do is first we can try and figure out the equivalent thermal conductivity of the top blocks and then we can find the equivalent thermal conductivity of the entire combination. So the first step to begin with we can find the equivalent thermal conductivity of the first two the top blocks which are connected in series and for series the thermal resistance the equivalent thermal resistance this is r1 plus r2. We know that r is related the thermal resistance is related with thermal conductivity as l divided by ka. So if we place this relation over here r effective this really becomes this becomes l. So while it won't be l right this will be 2l this will be 2l 2l divided by ka effective into area that is the effective area is a by 4 so this is a by 4 which is equal to r1. So for r1 we take l and l divided by ka by 2 that is the thermal conductivity into the area which is a by 4 plus for r2 we again take l divided by the thermal conductivity which is 2k into the area which is again a by 4. So many things get cancelled off l goes away and a by 4 goes away and what remains is what remains is 2 divided by ka effective which is equal to 2 by ka plus 1 by 2k and I encourage you to pause the video and work out ka effective from this equation. When you do that you should get ka effective as 4 by 5k. This is the equivalent thermal conductivity of the top two blocks. Now as a result the combination will look somewhat like this. Now you have the top block and you have the bottom block. So the conductivity for the bottom block that is 3k and for this one we just calculated it is 4 by 5k. The length is 2l and the areas remain the same that is a by 4 for this and 3a by 4 for the bottom block. Now these two are connected in parallel and for parallel we know the equivalent thermal resistance that is 1 upon r effective. This is equal to 1 upon r1 plus 1 upon r2. We can again use this relation and place it over here. So when we do that this becomes 1 upon 2l divided by ka effective into a effective. This is equal to 1 upon 2l divided by 4 by 5k into a by 4 plus 1 upon 2l divided by 3k into 3a by 4 ok 3k for the green block 3a by 4 and that is the area. Now a effective is the total area. So when we see this combination from the left the total area a effective that is really a by 4 a by 4 plus 3a by 4 this is plus 3a by 4 and this comes out to be just a because it is 4a by 4. So in place of a effective we can just write a ok again many things get cancelled off l goes away a goes away in fact 2l the entire 2l factor goes away so 2l goes away and then a goes away ok now we only have the equation in terms of k and all the numbers. So what remains is k effective this is equal to everything goes to the top and this becomes 4 by 5k into 1 by 4 plus 3k into into 3 by 4. So when you work this out this is 4k divided by 20 plus 9k divided by 4 this will be k by 5 so let's let's write that let's write that this will be k by 5 and now I encourage you to pause the video and work out this take LCM and figure out what is k effective ok when you work this out you should get k effective as 49k divided by 20 and that is the equivalent thermal conductivity of the combination of three metallic blocks ok you can try more questions from this exercise in the lesson and if you're watching on YouTube do check out the exercise link which is added in the description.