Lumped Bodies and Fins: Deciding Your Heat Transfer Model

Two questions run through this whole lecture, and both of them are decisions you make before you write anything down. When may I treat a whole lump of metal as though it had a single temperature? And why does bolting a fin onto something help? Here is the first one, as concretely as I can make it. A small aluminium bracket comes off a machine at a hundred and twenty degrees, and we leave it in a draught of air at twenty. How fast does it cool? And there is a second question hiding inside the first, in the words its temperature. A real bracket has a hot middle and a cooler skin. We are about to pretend that it does not, so we had better find out when that pretence is safe. So, the bracket, sitting in the draught. And here is the assumption, said out loud: the whole thing is at one temperature, T of t, everywhere at once. That is the lumped capacitance model, and for now it is an assumption and nothing more. Heat leaves through the entire surface by convection. Newton's law of cooling prices that: h, times the surface area, times how far the surface sits above the air. That h is the convection coefficient, and it belongs to the draught rather than to the metal. And what pays for that heat is the bracket's own stored energy. Its mass is rho V, its specific heat is c, so the rate at which it stores energy is rho V c times d T d t. While it cools, that number is negative. Before we solve anything, one substitution that pays for itself all lecture. Stop tracking temperature and track excess temperature: theta is T minus T infinity, how far above the air we currently are. The air is at a fixed temperature, so d theta d t and d T d t are the same number. Now put the two rates together. Rho V c, d T d t, equals minus h A s times T minus T infinity. That single line is the model, and everything in this half of the lecture comes out of it. Rewrite it in theta. The left side is the same derivative, and the right side collapses to minus h A s theta. Everything on the right is now proportional to the very thing we are solving for, and that is what makes this easy. So separate. Divide by theta and by rho V c. On the left, one over theta times its own derivative. On the right, a group of constants and nothing else: every symbol over there is a fixed property of the bracket or of the draught. Set the stage for the answer, and then integrate both sides over the cooling. Theta runs from theta i, the excess we started with, up to whatever it is now, while the clock runs from zero to t. The left side is a logarithm and the right side is linear in time. So the log of the ratio of excess temperatures falls at a perfectly constant rate, however hot or cool the bracket happens to be. Exponentiate, and there it is. The excess temperature decays exponentially. Notice that the shape of that curve does not depend at all on how hot the bracket started. Only its scale does. So give that group a symbol of its own. The exponent cannot carry units, and rho V c over h A s does not: kilograms and joules cancel straight through, and seconds are what survive. It is a time. We call it tau, the thermal time constant. Watch the bracket run down that curve. It loses heat fastest at the start, when it is furthest above the air, and it slows as it approaches, because the driving temperature difference is the very thing being consumed. One time constant in, this much of the initial excess is left: thirty seven percent of it. Three time constants in, at this point here, five percent, and for most engineering purposes that is over. For our bracket, tau is rho c L c over h. Twenty seven hundred, times nine hundred, times five millimetres, over twenty five. Four hundred and eighty six seconds. Call it eight minutes to lose two thirds of the excess, and about twenty four to be done. That is a complete answer, and it rests entirely on one line we simply assumed. So the honest next question is not how fast the bracket cools. It is whether a bracket has a single temperature at all.

We assumed the bracket has one temperature. Let us find out when that is a lie. Heat leaving the middle of a body has to do two things, one after the other: conduct out to the surface, and then convect off the surface into the fluid. Draw that as a temperature profile through the wall. Inside the solid, heat is conducted, and conduction needs a gradient, so the temperature slopes. Across the thin film of fluid clinging to the surface, heat is convected, and the temperature falls the rest of the way. Each leg has a resistance. Conducting a distance L c through a material of conductivity k costs L c over k A. Convecting off that same area costs one over h A. Divide the first by the second. The areas cancel, and what is left is h L c over k. That group is the Biot number, and it is the whole answer to our question. Watch what it does to the picture. These two bars are the two drops: the red one inside the metal, the blue one across the film. At a Biot number of one twentieth, almost the whole drop is out in the fluid and the solid is nearly flat. That is a body with a single temperature, and our exponential is honest for it. Now push the Biot number up to one. The two resistances are equal now, so half the drop is inside the metal and half is in the fluid. Calling that one temperature is already a fiction. Push on to twenty and the picture inverts completely. The surface has practically reached fluid temperature while the core is still hot, and the interior gradient is the entire story. So here is the number, and here is what goes into it. L c is a characteristic length, and for a lumped body we take volume over surface area, which is roughly the average distance heat has to travel to get out. And the working rule is Biot below nought point one. At that value the gradient inside the part is a few percent of the total drop, so a single temperature is wrong by a few percent as well. It is a convention rather than a law, but it is a good one. Let us try it on three real parts. First our aluminium bracket: conductivity a hundred and eighty, characteristic length five millimetres, and h of twenty five for a draught of air. Seven ten thousandths. That is miles below the limit, the profile inside the metal is flat to within a tenth of a degree, and the exponential we derived is safe. Now a stainless steel shaft, fifty millimetres of characteristic length, quenched in water at five hundred. Conductivity only fifteen, and the Biot number comes out at one point seven. Seventeen times over the limit, so that shaft keeps a hot core for a long time. And a plastic housing, only ten millimetres thick, in the same gentle draught the bracket was in. But conductivity nought point two, and the Biot number is one. Same air, same rough size, and this one fails. Which is the real lesson of this number. It is never the material alone, and never the size alone. It is conduction inside set against convection outside, and you cannot know which model a part deserves until you have put all three of those together.

Second question. Here is a hot wall, and here is a slab of metal bolted onto it, sticking out into the fluid. Everybody calls it a fin. Why does that help? Go back to Newton's law of cooling. There are exactly three things in it you could change to move more heat. Raise h, with a bigger fan or a pump. Lower the fluid temperature, which is almost never yours to choose. Or raise the surface area. The third one is cheap, and a fin is how you buy it. But there is a catch, and the whole rest of this lecture is that catch. The extra area is not at the base temperature. Heat has to conduct out along the fin to reach it, and the fin cools as it goes. So take a slice of the fin, this thin one here, of thickness d x. In steady state it stores nothing, so whatever conducts in at the left has to leave. Some of it conducts out at the right, and the rest convects off the sides. Now price each term. The conduction into the slice is Fourier's law: minus k, times the cross-sectional area A c, times the temperature gradient. I am writing it in theta again, the excess over the fluid, because the fluid temperature is constant and drops out of every derivative. What leaves at the right face is what entered, plus however much the heat flow itself has changed across the slice. That is just a Taylor expansion, kept to one term, which is exact in the limit as d x goes to nothing. And the convection off the sides uses the area of those sides: the perimeter P, times the thickness d x. Times h, times how far the fin sits above the fluid at that station. Put those three into the balance. The q x on each side cancels, and what survives is this: minus the rate of change of the heat flow, times d x, equals the convection off that same d x. Now substitute Fourier's law for q x. Differentiating minus k A c theta prime gives minus k A c theta double prime, and the two minus signs cancel. The d x cancels as well, off both sides, which is what makes this a differential equation instead of a bookkeeping exercise. The group h P over k A c appears once and never leaves, so give it a name. Call it m squared. Convection off the sides sits on top, conduction along the fin sits underneath, and it is the same competition the Biot number was measuring, now pointed along the length instead of across it. And there is the fin equation. Second order, linear, constant coefficients, and about as friendly as differential equations get. Note what m carries in its units: one over a length. So one over m is the distance over which this fin cools. To solve it we need two boundary conditions. At the root, the fin is welded to the wall, so it sits at the base excess theta b. At the tip, we will say no heat comes off the end: the tip area is tiny compared with the sides, so the gradient there is zero. The general solution is a pair of exponentials, one growing and one decaying. It is tidier to write the same thing as a hyperbolic cosine and a hyperbolic sine, because those are the combinations our two conditions will actually pick out. Impose them. The zero gradient at the tip forces one particular combination, and the base condition fixes what is left. The answer is a cosine hyperbolic of m times the distance remaining to the tip, over a cosine hyperbolic of m L. Look at what that says. Here are three fins with the same base temperature, differing only in m L. This one, with m L a half, is barely cooler at the tip than at the root: nearly the whole fin is working at full strength. At m L of one and a half, the tip has lost well over half of the excess. And at m L of four, the outer half of the fin is at fluid temperature and doing essentially nothing. It is metal you paid for and it is not removing heat.

We have the temperature everywhere along the fin. What we actually want is the heat it removes, and there is a trick that saves us an integration. Every watt that leaves the sides of this fin got in through one place: the base. So instead of adding up the convection off the whole length, take Fourier's law once, at x equal to zero. Minus k A c, times the slope of the temperature profile right at the root. Differentiate the profile we derived. The cosine hyperbolic becomes a sine hyperbolic, and the chain rule brings down a factor of minus m. Put x equal to zero and sinh over cosh of m L is just the hyperbolic tangent. Tidy the front of that. k A c times m is the square root of h P k A c, so the whole rate is that group, times the base excess, times the hyperbolic tangent of m L. Every fin you will ever meet is that formula. Now, is that a lot? A number on its own tells you nothing, so compare it with the best the fin could conceivably do. That best case is the whole fin sitting at base temperature: h, times its entire side area P L, times theta b. The ratio of what you get to that best case is the fin efficiency. Divide one by the other and the theta b cancels, the h and the P cancel, and what is left is beautifully simple: hyperbolic tangent of m L, over m L. One number governs the whole design. At m L of a half the fin runs at ninety two percent efficiency, which is close to ideal: the tip is barely cooler than the root. Take m L to one and you are at seventy six percent. That is the usual sweet spot, and it is where most real fins are designed to sit. At m L of two, forty eight percent. At three, thirty three. Past about two you are bolting on metal that spends its life near fluid temperature, so the sensible move is not a longer fin, it is another fin. Efficiency is not the whole story though, and here is why. A fin that is ninety percent efficient can still be a mistake, because bolting it on covers up a patch of wall that was quietly convecting on its own. So compare the fin against that patch instead. Effectiveness is the heat the fin removes, divided by the heat the bare base area would have removed with no fin there at all. Substitute what we found. The h and one A c cancel down, and out comes the square root of k P over h A c, times the hyperbolic tangent. Conduction along the fin on top, convection off it underneath, for the third time this lecture. And now the embarrassing possibility is right there in the algebra. If k P over h A c is less than one, the effectiveness is less than one, and the fin removes less heat than the bare wall it replaced. Two pins, identical in every dimension: five millimetres across, thirty long. The first is copper in a draught of air. Conductivity four hundred, h of fifty, and m works out at ten per metre, so m L is nought point three. Efficiency ninety seven percent, because the pin is short compared with the distance over which it cools. And effectiveness twenty three: it moves twenty three times the heat that the patch of wall underneath it ever did. Now the same pin in plastic, in water. Conductivity nought point two, h of five hundred, and m jumps to about fourteen hundred per metre. The same thirty millimetres of length is now an m L of forty two. Efficiency two percent, and effectiveness nought point five seven. The pin is at water temperature a millimetre out from the wall, and everything past that is decoration. Less than one. Fitting that pin removes forty three percent less heat than leaving the wall bare, because a poor conductor in a vigorous fluid is a worse path out than the surface it is standing on. Check the effectiveness before you cut any metal.

So here is the part, with its fin on it, and here is the order I would work in when one of these lands on my desk. Four numbers, and none of them needs a computer. First, the characteristic length: volume over surface area, the average distance heat has to travel to get out of the body. It takes ten seconds and it is the input to everything else. Second, the Biot number, h L c over k. Below nought point one, the inside of the part is very nearly uniform and you may treat it as a single temperature. Above it, you cannot, and you need the conduction equation rather than an energy balance on a lump. Third, if the lumped model survived, you already have the answer: an exponential decay towards the fluid, with a time constant of rho V c over h A s. Thermal mass on top, ease of escape underneath. And fourth, before you fit any fin, check k P over h A c. If it is not comfortably above one, the fin is going to make things worse, and no amount of clever shaping will rescue it. If it is, size the fin so that m L is around one. Three questions, three numbers. Can I use one temperature for this body? Biot, and you want it small. Is a fin worth fitting here? k P over h A c, and you want it large. And is my fin the right length? m L, and you want it about one. Look at the middle column and notice what all of it is made of. Every one of those numbers weighs conduction through the metal against convection at its surface, and that is the only physics in this lecture. Across the body, that competition decides whether a temperature gradient inside the part matters at all. Along the fin, the very same competition decides whether the far end of it is doing any work. Same fight, two directions. So before you write anything down, work those numbers out. The modelling decision is cheap when you make it first, and expensive when you discover it halfway through an analysis that was never going to be true.