{"version":1,"lectureId":"01M19SKXPHE7GGRDJ7F1XEHZ6D","attempt":3,"publication":{"slug":"heat-transfer-lumped-body-fin-efficiency","title":"Lumped Bodies and Fins: Deciding Your Heat Transfer Model","subject":"engineering","summary":"A first course in heat transfer turns on two modelling decisions, and this lecture makes both of them. It starts with a small aluminium bracket cooling in a draught, writes the energy balance, separates the variables and integrates to the exponential decay and its thermal time constant. It then builds the Biot number, the ratio of conduction resistance inside the part to convection resistance at its surface, and uses it to say when a single temperature for the whole body is honest. The second half bolts on a fin, takes an energy balance on a slice to reach the fin equation, solves it for the temperature falling along the length, and reads off fin efficiency and fin effectiveness, including the case of a fin that removes less heat than the bare surface it covers.","metaDescription":"Exponential cooling, the Biot number that justifies one temperature, the fin equation, fin efficiency, and when a fin makes cooling worse.","transcript":"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.","watch":{"version":1,"scenes":[{"title":"A Body That Cools All Over At Once","start":0,"end":267.28106249999996,"objects":{"air_tag":"a Point [text] labelled \"T_infinity\" drawn in body (location=(4.9, 3.8), show_marker=False)","ambient":"a Line [gray] drawn in plot (start=(0.0, 20.0), end=(1600.0, 20.0), dashed=True)","ambient_tag":"a Point [gray] labelled \"T_infinity\" drawn in plot (location=(1420.0, 20.0), show_marker=False)","assume":"a Panel that says \"Treat the whole body as a single temperature $T(t)$, with no gradient inside it. 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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?","live":[],"does":[[0,"card is shown on the screen, written out."],[1.1802828054298644,"card: enter:write-left-to-right."],[14.0725,"card is hidden from the screen — left the board."]]},{"start":15.272499999999999,"say":"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?","live":null,"does":[[19.84,"question is shown on the screen, written out."]]},{"start":27.777499999999996,"say":"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.","live":["question"],"does":[[31.4,"question (the \"its temperature\" part) is emphasized."],[37.36,"question (the \"its temperature\" part) is no longer emphasized."],[41.80499999999999,"question is hidden from the screen — left the board."]]},{"start":43.004999999999995,"say":"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.","live":[],"does":[[43.004999999999995,"head_setup is shown on the screen, written out."],[43.004999999999995,"body is shown on the screen, written out."],[44.03999999999999,"block is shown on the screen, written out."],[44.40802848756074,"core is shown on the screen, written out."],[45.16,"out_up_a is shown on the screen, written out."],[45.362415668158405,"out_up_b is shown on the screen, written out."],[45.56483133631683,"out_left is shown on the screen, written out."],[45.76724700447524,"out_right is shown on the screen, written out."],[45.969662672633646,"out_down is shown on the screen, written out."],[46.28724700447523,"air_tag is shown on the screen, written out."],[46.599999999999994,"assume is shown on the screen, written out."]]},{"start":58.717499999999994,"say":"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.","live":["assume","body","head_setup","block","core","out_up_a","out_up_b","out_left","out_right","out_down","air_tag"],"does":[[61.12,"skin_tag is shown on the screen, written out."],[62.279999999999994,"loss is shown on the screen, written out."]]},{"start":74.65249999999999,"say":"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.","live":["assume","loss","body","head_setup","block","core","out_up_a","out_up_b","out_left","out_right","out_down","air_tag","skin_tag"],"does":[[77.32,"store is shown on the screen, written out."],[89.85749999999999,"body moves to a new place on the board."],[89.85749999999999,"assume is hidden from the screen — left the board."],[89.85749999999999,"head_setup is hidden from the screen — left the board."],[89.85749999999999,"loss is hidden from the screen — left the board."],[89.85749999999999,"store is hidden from the screen — left the board."]]},{"start":90.45749999999998,"say":"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.","live":["body","block","core","out_up_a","out_up_b","out_left","out_right","out_down","air_tag","skin_tag"],"does":[[90.45749999999998,"head_ode is shown on the screen, written out."],[99.32,"theta_def is shown on the screen, written out."]]},{"start":110.59749999999998,"say":"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.","live":["body","block","core","out_up_a","out_up_b","out_left","out_right","out_down","air_tag","skin_tag","theta_def","head_ode"],"does":[[112.11999999999999,"work_ode is shown on the screen, written out."]]},{"start":124.83999999999997,"say":"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.","live":null,"does":[[125.19999999999997,"work_ode is shown on the screen, written out."],[129.44,"work_ode (the \"- h A_s theta\" part) is emphasized."],[138.28749999999997,"work_ode (the \"- h A_s theta\" part) is no longer emphasized."]]},{"start":138.8875,"say":"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.","live":null,"does":[[139.64000000000001,"work_ode is shown on the screen, written out."],[147.95999999999998,"work_ode (the \"frac(h A_s, rho V c)\" part) is emphasized."],[153.89999999999998,"body is hidden from the screen — left the board."],[153.89999999999998,"block is hidden from the screen — body left the board."],[153.89999999999998,"core is hidden from the screen — body left the board."],[153.89999999999998,"out_up_a is hidden from the screen — body left the board."],[153.89999999999998,"out_up_b is hidden from the screen — body left the board."],[153.89999999999998,"out_left is hidden from the screen — body left the board."],[153.89999999999998,"out_right is hidden from the screen — body left the board."],[153.89999999999998,"out_down is hidden from the screen — body left the board."],[153.89999999999998,"air_tag is hidden from the screen — body left the board."],[153.89999999999998,"skin_tag is hidden from the screen — body left the board."],[153.89999999999998,"head_ode is hidden from the screen — left the board."],[153.89999999999998,"theta_def is hidden from the screen — left the board."],[153.89999999999998,"work_ode is hidden from the screen — left the board."],[153.89999999999998,"work_ode (the \"frac(h A_s, rho V c)\" part) is no longer emphasized."]]},{"start":154.5,"say":"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.","live":[],"does":[[154.5,"head_answer is shown on the screen, written out."],[154.5,"plot is shown on the screen, written out."],[155.19999999999996,"ambient is shown on the screen, written out."],[155.47605834222693,"ambient_tag is shown on the screen, written out."],[156.63999999999996,"work_sol is shown on the screen, written out."]]},{"start":167.3375,"say":"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.","live":["plot","head_answer","ambient","ambient_tag"],"does":[[168.87999999999997,"work_sol is shown on the screen, written out."]]},{"start":180.445,"say":"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.","live":null,"does":[[180.79999999999993,"work_sol is shown on the screen, written out."],[184.15999999999994,"curve is shown on the screen, drawn."]]},{"start":193.51,"say":"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.","live":["plot","head_answer","ambient","ambient_tag","curve"],"does":[[194.7599999999999,"solution is shown on the screen, written out."],[205.9999999999999,"tau_def is shown on the screen, written out."]]},{"start":210.14999999999998,"say":"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.","live":["solution","tau_def","plot","head_answer","ambient","ambient_tag","curve"],"does":[[210.47999999999985,"rider is shown on the screen, written out."],[211.99999999999986,"rider is redrawn as the numbers it depends on change."],[211.99999999999986,"clock ticks to 1550.0."]]},{"start":222.95999999999998,"say":"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.","live":["solution","tau_def","plot","head_answer","ambient","ambient_tag","curve","rider"],"does":[[223.7999999999998,"tau_line is shown on the screen, written out."],[223.7999999999998,"tau_tag is shown on the screen, written out."],[227.2399999999998,"tau_dot is shown on the screen, written out."],[230.75999999999982,"three_tau_dot is shown on the screen, written out."]]},{"start":236.08499999999998,"say":"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.","live":["solution","tau_def","plot","head_answer","ambient","ambient_tag","curve","rider","tau_line","tau_tag","tau_dot","three_tau_dot"],"does":[[238.1599999999998,"tau_def (the \"frac(rho c L_c, h)\" part) is emphasized."],[245.3599999999998,"tau_def (the \"frac(rho c L_c, h)\" part) is no longer emphasized."],[246.15999999999977,"point is shown on the screen, grown."],[248.34257057138936,"point is hidden from the screen."]]},{"start":252.91249999999997,"say":"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.","live":null,"does":[[253.79999999999976,"A box is drawn around solution."],[266.23939583333333,"head_answer is hidden from the screen — left the board."],[266.23939583333333,"plot is hidden from the screen — left the board."],[266.23939583333333,"ambient is hidden from the screen — plot left the board."],[266.23939583333333,"ambient_tag is hidden from the screen — plot left the board."],[266.23939583333333,"curve is hidden from the screen — plot left the board."],[266.23939583333333,"rider is hidden from the screen — plot left the board."],[266.23939583333333,"tau_line is hidden from the screen — plot left the board."],[266.23939583333333,"tau_tag is hidden from the screen — plot left the board."],[266.23939583333333,"tau_dot is hidden from the screen — plot left the board."],[266.23939583333333,"three_tau_dot is hidden from the screen — plot left the board."],[266.23939583333333,"solution is hidden from the screen — left the board."],[266.23939583333333,"tau_def is hidden from the screen — left the board."],[266.23939583333333,"work_sol is hidden from the screen — left the board."]]}]},{"title":"When One Temperature Is a Lie","start":267.28106249999996,"end":464.2968124999999,"objects":{"bi":"a VariableNumber (initial_value=0.05, format_spec='.2f')","bi_def":"a Math [text] that says \"$upright(\"Bi\") = frac(h L_c, k)$\"","bi_note":"a Panel that says \"Conduction resistance inside the body, over convection resistance at its surface. Below about $0.1$, a single temperature is right to within a few percent.\"","drop_film":"a Line [blue] labelled \"Delta T_2\" drawn in wall (start=(1.9, 20.0), end=(1.9, (20.0 + (100.0 / (1.0 + bi)))))","drop_solid":"a Line [red] labelled \"Delta T_1\" drawn in wall (start=(1.06, (20.0 + (100.0 / (1.0 + bi)))), end=(1.06, 120.0))","face":"a Line [gray] drawn in wall (start=(1.0, 0.0), end=(1.0, 132.0), dashed=True)","film":"a Line [blue] drawn in wall (start=(1.0, (20.0 + (100.0 / (1.0 + bi)))), end=(1.8, 20.0))","fluid_tag":"a Point [text] labelled \"upright(\"fluid\")\" drawn in wall (location=(1.5, 7.0), show_marker=False)","free_stream":"a Line [gray] drawn in wall (start=(0.0, 20.0), end=(2.05, 20.0), dashed=True)","head_bi":"a Heading that says \"The Number That Compares Them\"","head_test":"a Heading that says \"Three Real Parts\"","head_two":"a Heading that says \"Two Resistances in Series\"","inside":"a Line [red] drawn in wall (start=(0.0, 120.0), end=(1.0, (20.0 + (100.0 / (1.0 + bi)))))","lc_def":"a Math [text] that says \"$L_c = frac(V, A_s)$\"","r_cond":"a Math [text] that says \"$R_upright(\"cond\") = frac(L_c, k A_s)$\"","r_conv":"a Math [text] that says \"$R_upright(\"conv\") = frac(1, h A_s)$\"","ratio":"a Math [text] that says \"$frac(R_upright(\"cond\"), R_upright(\"conv\")) = frac(h L_c, k)$\"","readout":"a Point [yellow] labelled \"upright(\"Bi\") = 0.05\" drawn in wall (location=(1.5, 143.0), show_marker=False)","solid_block":"a Polygon [gray] drawn in wall (vertices=((0.0, 0.0), (1.0, 0.0), (1.0, 132.0), (0.0, 132.0)), fill_opacity=0.16)","solid_tag":"a Point [text] labelled \"upright(\"solid\")\" drawn in wall (location=(0.5, 7.0), show_marker=False)","surface_dot":"a Point [yellow] drawn in wall (location=(1.0, (20.0 + (100.0 / (1.0 + bi)))))","survey":"a Table [text] that says \"Part $k$ $L_c$ $h$ $upright(\"Bi\")$ Aluminium $180$ $5$ $25$ $0.0007$ Steel $15$ $50$ $500$ $1.7$ Plastic $0.2$ $10$ $20$ $1.0$\" (rows=(('Part', '$k$', '$L_c$', '$h$', '$upright(\"Bi\")$'), ('Aluminiu…, header=True)","units":"a Text [text] that says \"Conductivity in watts per metre kelvin, $L_c$ in millimetres, and $h$ in watts per square metre kelvin.\"","wall":"an Axes (x_range=(0.0, 2.1), y_range=(0.0, 150.0), include_ticks=False)"},"beats":[{"start":267.28106249999996,"say":"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.","live":[],"does":[[267.28106249999996,"head_two is shown on the screen, written out."],[267.28106249999996,"wall is shown on the screen, written out."],[273.7610625,"solid_block is shown on the screen, written out."],[274.0646897485992,"face is shown on the screen, written out."],[274.2770872714538,"free_stream is shown on the screen, written out."],[274.62516158581536,"solid_tag is shown on the screen, written out."],[275.0972111287231,"fluid_tag is shown on the screen, written out."]]},{"start":281.93356249999994,"say":"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.","live":["wall","head_two","solid_block","face","free_stream","solid_tag","fluid_tag"],"does":[[285.84106249999996,"inside is shown on the screen, drawn."],[286.1663512151041,"surface_dot is shown on the screen, written out."],[291.6410625,"film is shown on the screen, drawn."]]},{"start":297.8285625,"say":"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.","live":["wall","head_two","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film"],"does":[[300.0010625,"wall moves to a new place on the board."],[300.0010625,"r_cond is shown on the screen, written out."],[306.48106249999995,"r_conv is shown on the screen, written out."]]},{"start":310.6835625,"say":"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.","live":["r_cond","r_conv","wall","head_two","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film"],"does":[[310.92106249999995,"ratio is shown on the screen, written out."],[317.8810625,"ratio (the \"frac(h L_c, k)\" part) is emphasized."],[320.96106249999997,"ratio (the \"frac(h L_c, k)\" part) is no longer emphasized."]]},{"start":321.56106249999993,"say":"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.","live":["r_cond","r_conv","ratio","wall","head_two","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film"],"does":[[324.0810625,"drop_solid is shown on the screen, written out."],[324.35708589181286,"drop_film is shown on the screen, written out."],[330.72106249999996,"readout is shown on the screen, written out."]]},{"start":336.14606249999997,"say":"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.","live":["r_cond","r_conv","ratio","wall","head_two","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film","drop_solid","drop_film","readout"],"does":[[341.4810625,"inside is redrawn as the numbers it depends on change."],[341.4810625,"surface_dot is redrawn as the numbers it depends on change."],[341.4810625,"film is redrawn as the numbers it depends on change."],[341.4810625,"drop_solid is redrawn as the numbers it depends on change."],[341.4810625,"drop_film is redrawn as the numbers it depends on change."],[341.4810625,"readout is redrawn as the numbers it depends on change."],[341.4810625,"bi ticks to 1.0."]]},{"start":349.60606249999995,"say":"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.","live":null,"does":[[353.0810625,"inside is redrawn as the numbers it depends on change."],[353.0810625,"surface_dot is redrawn as the numbers it depends on change."],[353.0810625,"film is redrawn as the numbers it depends on change."],[353.0810625,"drop_solid is redrawn as the numbers it depends on change."],[353.0810625,"drop_film is redrawn as the numbers it depends on change."],[353.0810625,"readout is redrawn as the numbers it depends on change."],[353.0810625,"bi ticks to 20.0."],[358.96106249999997,"drop_solid is indicated — a transient flash."],[363.18356249999994,"head_two is hidden from the screen — left the board."],[363.18356249999994,"r_cond is hidden from the screen — left the board."],[363.18356249999994,"r_conv is hidden from the screen — left the board."],[363.18356249999994,"ratio is hidden from the screen — left the board."]]},{"start":364.3835624999999,"say":"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.","live":["wall","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film","drop_solid","drop_film","readout"],"does":[[364.3835624999999,"head_bi is shown on the screen, written out."],[365.84106249999996,"bi_def is shown on the screen, written out."],[372.0810625,"lc_def is shown on the screen, written out."]]},{"start":377.27356249999997,"say":"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.","live":["wall","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film","drop_solid","drop_film","readout","bi_def","lc_def","head_bi"],"does":[[378.3610625,"bi_note is shown on the screen, written out."],[381.6410625,"inside is redrawn as the numbers it depends on change."],[381.6410625,"surface_dot is redrawn as the numbers it depends on change."],[381.6410625,"film is redrawn as the numbers it depends on change."],[381.6410625,"drop_solid is redrawn as the numbers it depends on change."],[381.6410625,"drop_film is redrawn as the numbers it depends on change."],[381.6410625,"readout is redrawn as the numbers it depends on change."],[381.6410625,"bi ticks to 0.08."],[391.97356249999996,"wall moves to a new place on the board."],[391.97356249999996,"bi_def is hidden from the screen — left the board."],[391.97356249999996,"bi_note is hidden from the screen — left the board."],[391.97356249999996,"head_bi is hidden from the screen — left the board."],[391.97356249999996,"lc_def is hidden from the screen — left the board."]]},{"start":392.5735625,"say":"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.","live":["wall","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film","drop_solid","drop_film","readout"],"does":[[392.5735625,"head_test is shown on the screen, written out."],[392.5735625,"readout is hidden from the screen."],[394.8810625,"survey is shown on the screen, written out."],[396.0810625,"survey is shown on the screen, written out."],[400.6410625,"inside is redrawn as the numbers it depends on change."],[400.6410625,"surface_dot is redrawn as the numbers it depends on change."],[400.6410625,"film is redrawn as the numbers it depends on change."],[400.6410625,"drop_solid is redrawn as the numbers it depends on change."],[400.6410625,"drop_film is redrawn as the numbers it depends on change."],[400.6410625,"bi ticks to 0.0007."]]},{"start":404.54856249999995,"say":"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.","live":["wall","solid_block","face","free_stream","solid_tag","fluid_tag","inside","surface_dot","film","drop_solid","drop_film","head_test"],"does":[[405.2410625,"survey (the \"$0.0007$\" part) is emphasized."],[410.84106249999996,"drop_solid is indicated — a transient flash."],[414.81356249999993,"survey (the \"$0.0007$\" part) is no longer emphasized."]]},{"start":415.41356249999995,"say":"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.","live":null,"does":[[416.3210624999999,"survey is shown on the screen, written out."],[424.3210625,"inside is redrawn as the numbers it depends on change."],[424.3210625,"surface_dot is redrawn as the numbers it depends on change."],[424.3210625,"film is redrawn as the numbers it depends on change."],[424.3210625,"drop_solid is redrawn as the numbers it depends on change."],[424.3210625,"drop_film is redrawn as the numbers it depends on change."],[424.3210625,"bi ticks to 1.7."],[427.1210625,"survey (the \"$1.7$\" part) is emphasized."],[432.3885624999999,"survey (the \"$1.7$\" part) is no longer emphasized."]]},{"start":432.98856249999994,"say":"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.","live":null,"does":[[433.80106249999994,"survey is shown on the screen, written out."],[441.7210624999999,"inside is redrawn as the numbers it depends on change."],[441.7210624999999,"surface_dot is redrawn as the numbers it depends on change."],[441.7210624999999,"film is redrawn as the numbers it depends on change."],[441.7210624999999,"drop_solid is redrawn as the numbers it depends on change."],[441.7210624999999,"drop_film is redrawn as the numbers it depends on change."],[441.7210624999999,"bi ticks to 1.0."],[442.7610625,"survey (the \"$1.0$\" part) is emphasized."],[446.9535625,"survey (the \"$1.0$\" part) is no longer emphasized."]]},{"start":447.55356249999994,"say":"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.","live":null,"does":[[461.7610624999999,"survey (the \"column=5\" part) is emphasized."],[463.00514583333324,"survey (the \"column=5\" part) is no longer emphasized."],[463.25514583333324,"head_test is hidden from the screen — left the board."],[463.25514583333324,"survey is hidden from the screen — left the board."],[463.25514583333324,"wall is hidden from the screen — left the board."],[463.25514583333324,"solid_block is hidden from the screen — wall left the board."],[463.25514583333324,"face is hidden from the screen — wall left the board."],[463.25514583333324,"free_stream is hidden from the screen — wall left the board."],[463.25514583333324,"solid_tag is hidden from the screen — wall left the board."],[463.25514583333324,"fluid_tag is hidden from the screen — wall left the board."],[463.25514583333324,"inside is hidden from the screen — wall left the board."],[463.25514583333324,"surface_dot is hidden from the screen — wall left the board."],[463.25514583333324,"film is hidden from the screen — wall left the board."],[463.25514583333324,"drop_solid is hidden from the screen — wall left the board."],[463.25514583333324,"drop_film is hidden from the screen — wall left the board."]]}]},{"title":"Bolting On a Fin","start":464.2968124999999,"end":710.9158333333332,"objects":{"area_tag":"a Point [text] labelled \"A_c\" drawn in fin (location=(8.5, 2.6), show_marker=False)","base_tag":"a Point [red] labelled \"T_b\" drawn in fin (location=(1.5, 3.1), show_marker=False)","base_wall":"a Polygon [red] drawn in fin (vertices=((0.4, 0.7), (1.4, 0.7), (1.4, 4.5), (0.4, 4.5)), fill_opacity=0.35)","bc_base":"a Math [text] that says \"$theta(0) = theta_b$\"","bc_tip":"a Math [text] that says \"$theta'(L) = 0$\"","blade":"a Polygon [blue] drawn in fin (vertices=((1.4, 2.2), (8.4, 2.2), (8.4, 3.0), (1.4, 3.0)))","down_a":"a Vector [yellow] drawn in fin (start=(3.0, 2.2), end=(3.0, 1.4))","down_b":"a Vector [yellow] drawn in fin (start=(6.4, 2.2), end=(6.4, 1.4))","down_c":"a Vector [yellow] drawn in fin (start=(7.8, 2.2), end=(7.8, 1.4))","dx_line":"a Line [text] labelled \"dif x\" drawn in fin (start=(4.6, 1.85), end=(5.2, 1.85))","fin":"a Figure (x_range=(0.0, 10.0), y_range=(0.0, 5.2), aspect=(10.0, 5.2))","fin_law":"a Math [text] that says \"$frac(dif^2 theta, dif x^2) - m^2 theta = 0$\"","head_ode":"a Heading that says \"The Fin Equation\"","head_profile":"a Heading that says \"How the Temperature Falls\"","head_slice":"a Heading that says \"An Energy Balance on One Slice\"","head_why":"a Heading that says \"Three Levers, and Only One You Own\"","length_line":"a Line [text] labelled \"L\" drawn in fin (start=(1.4, 0.9), end=(8.4, 0.9), dashed=True)","long_fin":"a FunctionPlot [red] labelled \"m L = 4\" drawn in prof (function=<function>)","mid_fin":"a FunctionPlot [yellow] labelled \"m L = 1.5\" drawn in prof (function=<function>)","newton":"a Math [text] that says \"$q = h A_s (T_s - T_infinity)$\"","options":"a Block [text] that says \"Raise $h$: a bigger fan, a pump, a different fluid. Lower $T_infinity$: rarely yours to choose. Raise $A_s$: bolt on more surface.\"","point":"a Point [yellow] drawn in prof (location=(0.75, 0.05650605969567148))","prof":"an Axes (y_range=(0.0, 1.05), x_ticks_every=0.25, y_ticks_every=0.25)","profile_law":"a Math [text] that says \"$frac(theta(x), theta_b) = frac(cosh(m (L - x)), cosh(m L))$\"","q_in":"a Vector [green] labelled \"q_x\" drawn in fin (start=(3.5, 2.6), end=(4.55, 2.6))","q_off":"a Vector [red] labelled \"dif q_upright(\"conv\")\" drawn in fin (start=(4.9, 3.05), end=(4.9, 4.2))","q_out":"a Vector [green] labelled \"q_(x + dif x)\" drawn in fin (start=(5.25, 2.6), end=(6.3, 2.6))","short_fin":"a FunctionPlot [blue] labelled \"m L = 0.5\" drawn in prof (function=<function>)","slice_box":"a Polygon [yellow] drawn in fin (vertices=((4.6, 2.2), (5.2, 2.2), (5.2, 3.0), (4.6, 3.0)), fill_opacity=0.55)","up_a":"a Vector [yellow] drawn in fin (start=(3.0, 3.0), end=(3.0, 3.8))","up_b":"a Vector [yellow] drawn in fin (start=(6.4, 3.0), end=(6.4, 3.8))","up_c":"a Vector [yellow] drawn in fin (start=(7.8, 3.0), end=(7.8, 3.8))","work_ode":"a Derivation [text] that says \"$- frac(dif q_x, dif x) dif x &= h P dif x thin theta \\ k A_c frac(dif^2 theta, dif x^2) &= h P thin theta \\ m^2 &= frac(h P, k A_c)$\"","work_slice":"a Derivation [text] that says \"$q_x &= q_(x + dif x) + dif q_upright(\"conv\") \\ q_x &= - k A_c frac(dif theta, dif x) \\ q_(x + dif x) &= q_x + frac(dif q_x, dif x) dif x \\ dif q_upright(\"conv\") &= h P dif x thin theta$\"","work_sol":"a Derivation [text] that says \"$theta(x) &= C_1 e^(m x) + C_2 e^(- m x) \\ theta(x) &= B_1 cosh(m x) + B_2 sinh(m x) \\ theta(x) &= theta_b frac(cosh(m (L - x)), cosh(m L))$\""},"beats":[{"start":464.2968124999999,"say":"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?","live":[],"does":[[464.2968124999999,"head_why is shown on the screen, written out."],[464.2968124999999,"fin is shown on the screen, written out."],[466.5368124999999,"base_wall is shown on the screen, written out."],[468.13681249999985,"blade is shown on the screen, written out."],[468.46625202802346,"base_tag is shown on the screen, written out."],[470.8568124999999,"up_a is shown on the screen, written out."],[471.07375308154224,"up_b is shown on the screen, written out."],[471.2906936630846,"up_c is shown on the screen, written out."],[471.50763424462696,"down_a is shown on the screen, written out."],[471.72457482616926,"down_b is shown on the screen, written out."],[471.82148506637157,"down_c is shown on the screen, written out."]]},{"start":475.38181249999985,"say":"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.","live":["fin","head_why","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c"],"does":[[476.3368124999999,"fin moves to a new place on the board."],[476.3368124999999,"newton is shown on the screen, written out."],[478.8568124999999,"options is shown on the screen, written out."],[481.93681249999986,"options (the \"Raise $h$: a bigger fan, a pump, a different fluid.\" part) is emphasized."],[484.73681249999987,"options (the \"Lower $T_infinity$: rarely yours to choose.\" part) is emphasized."],[484.73681249999987,"options (the \"Raise $h$: a bigger fan, a pump, a different fluid.\" part) is no longer emphasized."],[489.69681249999985,"options (the \"Lower $T_infinity$: rarely yours to choose.\" part) is no longer emphasized."],[489.69681249999985,"options (the \"Raise $A_s$: bolt on more surface.\" part) is emphasized."]]},{"start":490.9143124999999,"say":"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.","live":["newton","options","fin","head_why","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c"],"does":[[495.17681249999987,"options (the \"Raise $A_s$: bolt on more surface.\" part) is no longer emphasized."],[502.5368124999999,"length_line is shown on the screen, written out."],[502.87320279046764,"area_tag is shown on the screen, written out."],[506.8343124999999,"fin moves to a new place on the board."],[506.8343124999999,"head_why is hidden from the screen — left the board."],[506.8343124999999,"newton is hidden from the screen — left the board."],[506.8343124999999,"options is hidden from the screen — left the board."]]},{"start":508.0343124999999,"say":"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.","live":["fin","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c","length_line","area_tag"],"does":[[508.0343124999999,"head_slice is shown on the screen, written out."],[509.17681249999987,"slice_box is shown on the screen, written out."],[511.3368124999999,"dx_line is shown on the screen, written out."],[516.4568124999998,"q_in is shown on the screen, written out."],[519.4168124999999,"q_out is shown on the screen, written out."],[520.1368124999999,"work_slice is shown on the screen, written out."],[520.5768124999998,"q_off is shown on the screen, written out."]]},{"start":522.5793124999999,"say":"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.","live":["fin","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c","length_line","area_tag","head_slice","slice_box","dx_line","q_in","q_out","q_off"],"does":[[526.2568124999999,"work_slice is shown on the screen, written out."]]},{"start":539.9418124999999,"say":"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.","live":null,"does":[[547.4968124999998,"work_slice is shown on the screen, written out."]]},{"start":553.2593124999999,"say":"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.","live":null,"does":[[553.7768124999999,"work_slice is shown on the screen, written out."],[557.0568124999999,"work_slice (the \"h P dif x thin theta\" part) is emphasized."],[564.2343124999999,"head_slice is hidden from the screen — left the board."],[564.2343124999999,"work_slice is hidden from the screen — left the board."],[564.2343124999999,"work_slice (the \"h P dif x thin theta\" part) is no longer emphasized."]]},{"start":565.4343124999998,"say":"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.","live":["fin","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c","length_line","area_tag","slice_box","dx_line","q_in","q_out","q_off"],"does":[[565.4343124999998,"head_ode is shown on the screen, written out."],[569.0568124999999,"work_ode is shown on the screen, written out."]]},{"start":578.2893124999998,"say":"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.","live":["fin","base_wall","blade","base_tag","up_a","up_b","up_c","down_a","down_b","down_c","length_line","area_tag","slice_box","dx_line","q_in","q_out","q_off","head_ode"],"does":[[578.9368124999999,"work_ode is shown on the screen, written out."],[586.6968124999999,"work_ode (the \"frac(dif^2 theta, dif x^2)\" part) is emphasized."],[596.5568124999999,"work_ode (the \"frac(dif^2 theta, dif x^2)\" part) is no longer emphasized."]]},{"start":597.1568124999999,"say":"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.","live":null,"does":[[597.5368124999999,"work_ode is shown on the screen, written out."],[604.7368124999998,"work_ode (the \"frac(h P, k A_c)\" part) is emphasized."],[615.2268124999998,"work_ode (the \"frac(h P, k A_c)\" part) is no longer emphasized."]]},{"start":615.8268124999998,"say":"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.","live":null,"does":[[617.0168124999998,"fin_law is shown on the screen, written out."],[621.6568124999999,"A box is drawn around fin_law."],[631.8643124999999,"fin is hidden from the screen — left the board."],[631.8643124999999,"base_wall is hidden from the screen — fin left the board."],[631.8643124999999,"blade is hidden from the screen — fin left the board."],[631.8643124999999,"base_tag is hidden from the screen — fin left the board."],[631.8643124999999,"up_a is hidden from the screen — fin left the board."],[631.8643124999999,"up_b is hidden from the screen — fin left the board."],[631.8643124999999,"up_c is hidden from the screen — fin left the board."],[631.8643124999999,"down_a is hidden from the screen — fin left the board."],[631.8643124999999,"down_b is hidden from the screen — fin left the board."],[631.8643124999999,"down_c is hidden from the screen — fin left the board."],[631.8643124999999,"length_line is hidden from the screen — fin left the board."],[631.8643124999999,"area_tag is hidden from the screen — fin left the board."],[631.8643124999999,"slice_box is hidden from the screen — fin left the board."],[631.8643124999999,"dx_line is hidden from the screen — fin left the board."],[631.8643124999999,"q_in is hidden from the screen — fin left the board."],[631.8643124999999,"q_out is hidden from the screen — fin left the board."],[631.8643124999999,"q_off is hidden from the screen — fin left the board."],[631.8643124999999,"fin_law is hidden from the screen — left the board."],[631.8643124999999,"head_ode is hidden from the screen — left the board."],[631.8643124999999,"work_ode is hidden from the screen — left the board."]]},{"start":633.0643124999999,"say":"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.","live":[],"does":[[633.0643124999999,"head_profile is shown on the screen, written out."],[633.0643124999999,"prof is shown on the screen, written out."],[636.2968124999998,"prof moves to a new place on the board."],[636.2968124999998,"bc_base is shown on the screen, written out."],[641.8968124999998,"bc_tip is shown on the screen, written out."]]},{"start":649.6368124999999,"say":"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.","live":["bc_base","bc_tip","prof","head_profile"],"does":[[650.2568124999998,"work_sol is shown on the screen, written out."],[655.1368124999998,"work_sol is shown on the screen, written out."]]},{"start":664.0893124999998,"say":"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.","live":null,"does":[[671.9768124999997,"work_sol is shown on the screen, written out."],[672.5648889104839,"profile_law is shown on the screen, written out."]]},{"start":679.8493124999998,"say":"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.","live":["bc_base","bc_tip","profile_law","prof","head_profile"],"does":[[682.4168124999997,"short_fin is shown on the screen, drawn."]]},{"start":695.0043124999999,"say":"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.","live":["bc_base","bc_tip","profile_law","prof","head_profile","short_fin"],"does":[[696.8968124999997,"mid_fin is shown on the screen, drawn."],[701.3368124999997,"long_fin is shown on the screen, drawn."],[705.5768124999997,"point is shown on the screen, grown."],[708.0319214588375,"point is hidden from the screen."],[709.8741666666665,"bc_base is hidden from the screen — left the board."],[709.8741666666665,"bc_tip is hidden from the screen — left the board."],[709.8741666666665,"head_profile is hidden from the screen — left the board."],[709.8741666666665,"prof is hidden from the screen — left the board."],[709.8741666666665,"short_fin is hidden from the screen — prof left the board."],[709.8741666666665,"mid_fin is hidden from the screen — prof left the board."],[709.8741666666665,"long_fin is hidden from the screen — prof left the board."],[709.8741666666665,"profile_law is hidden from the screen — left the board."],[709.8741666666665,"work_sol is hidden from the screen — left the board."]]}]},{"title":"Efficiency, and the Fin That Made It Worse","start":710.9158333333332,"end":957.1152499999998,"objects":{"copper_eps":"a Math [text] that says \"$epsilon_f = 23$\"","copper_eta":"a Math [text] that says \"$eta_f = 0.97$\"","copper_k":"a Math [text] that says \"$k = 400 thin upright(\"W/m K\")$\"","copper_ml":"a Math [text] that says \"$m L = 0.3$\"","copper_verdict":"a Tex [text] that says \"Worth every gram.\"","criterion":"a Math [text] that says \"$frac(k P, h A_c) gt.eq 1$\"","criterion_note":"a Panel that says \"Below one the fin removes less heat than the bare base it covers. Designers usually want $epsilon_f gt.eq 2$ before fitting anything.\"","eff":"an Axes (x_range=(0.0, 4.0), y_range=(0.0, 1.05), x_ticks_every=1.0)","eta_curve":"a FunctionPlot [blue] drawn in eff (function=<function>, x_range=(0.0, 4.0))","eta_note":"a Panel that says \"Fin efficiency compares the heat the fin actually removes with the heat it would remove if all of it sat at the base temperature.\"","eta_rule":"a Math [text] that says \"$m L approx 1 quad => quad eta_f approx 0.76$\"","head_eps":"a Heading that says \"Was It Worth Fitting?\"","head_eta":"a Heading that says \"Fin Efficiency\"","head_rate":"a Heading that says \"All the Heat Goes In at the Base\"","head_two":"a Heading that says \"Two Pins of Exactly the Same Size\"","line":"a Line [yellow] drawn in eff (start=(1.0, 0.7615941559557649), end=(0.0, 0.7615941559557649), dashed=True)","line_2":"a Line [yellow] drawn in eff (start=(1.0, 0.7615941559557649), dashed=True)","ml":"a VariableNumber (initial_value=0.5, format_spec='.1f')","pin":"a Figure (x_range=(0.0, 10.0), y_range=(0.0, 5.2), aspect=(10.0, 5.2))","pin_blade":"a Polygon [blue] drawn in pin (vertices=((1.4, 2.2), (8.4, 2.2), (8.4, 3.0), (1.4, 3.0)))","pin_down_a":"a Vector [yellow] drawn in pin (start=(3.0, 2.2), end=(3.0, 1.4))","pin_down_b":"a Vector [yellow] drawn in pin (start=(5.2, 2.2), end=(5.2, 1.4))","pin_down_c":"a Vector [yellow] drawn in pin (start=(7.4, 2.2), end=(7.4, 1.4))","pin_up_a":"a Vector [yellow] drawn in pin (start=(3.0, 3.0), end=(3.0, 3.8))","pin_up_b":"a Vector [yellow] drawn in pin (start=(5.2, 3.0), end=(5.2, 3.8))","pin_up_c":"a Vector [yellow] drawn in pin (start=(7.4, 3.0), end=(7.4, 3.8))","pin_wall":"a Polygon [red] drawn in pin (vertices=((0.4, 0.7), (1.4, 0.7), (1.4, 4.5), (0.4, 4.5)), fill_opacity=0.35)","plastic_eps":"a Math [text] that says \"$epsilon_f = 0.57$\"","plastic_eta":"a Math [text] that says \"$eta_f = 0.02$\"","plastic_k":"a Math [text] that says \"$k = 0.2 thin upright(\"W/m K\")$\"","plastic_ml":"a Math [text] that says \"$m L = 42$\"","plastic_verdict":"a Tex [text] that says \"Worse than no fin at all.\"","point":"a Point [yellow] drawn in eff (location=(1.0, 0.7615941559557649))","q_law":"a Math [text] that says \"$q_f = sqrt(h P k A_c) thin theta_b tanh(m L)$\"","root_arrow":"a Vector [green] labelled \"q_f\" drawn in pin (start=(0.5, 2.6), end=(1.55, 2.6))","tex":"a Tex [text] that says \"Copper pin in air\"","tex_2":"a Tex [text] that says \"Plastic pin in water\"","walker":"a PlotPoint [yellow] labelled \"m L = 0.5\" drawn in eff (target='eta_curve', x=<VariableNumber ml = 0.6>)","work_eps":"a Derivation [text] that says \"$epsilon_f &= frac(q_f, h A_c theta_b) \\ epsilon_f &= sqrt(frac(k P, h A_c)) thin tanh(m L)$\"","work_eta":"a Derivation [text] that says \"$q_upright(\"max\") &= h P L thin theta_b \\ eta_f &= frac(q_f, q_upright(\"max\")) \\ eta_f &= frac(tanh(m L), m L)$\"","work_rate":"a Derivation [text] that says \"$q_f &= - k A_c thin theta'(0) \\ theta'(x) &= - m theta_b frac(sinh(m (L - x)), cosh(m L)) \\ q_f &= k A_c m theta_b tanh(m L)$\""},"beats":[{"start":710.9158333333332,"say":"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.","live":[],"does":[[710.9158333333332,"head_rate is shown on the screen, written out."],[710.9158333333332,"pin is shown on the screen, written out."],[713.0758333333332,"pin_wall is shown on the screen, written out."],[713.0758333333332,"pin_blade is shown on the screen, written out."],[720.7158333333332,"pin_up_a is shown on the screen, written out."],[720.7963418079095,"pin_up_b is shown on the screen, written out."],[720.8768502824857,"pin_up_c is shown on the screen, written out."],[720.957358757062,"pin_down_a is shown on the screen, written out."],[721.0438720742534,"pin_down_b is shown on the screen, written out."],[721.1358817594834,"pin_down_c is shown on the screen, written out."],[723.7558333333333,"root_arrow is shown on the screen, written out."]]},{"start":724.8483333333332,"say":"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.","live":["pin","head_rate","pin_wall","pin_blade","pin_up_a","pin_up_b","pin_up_c","pin_down_a","pin_down_b","pin_down_c","root_arrow"],"does":[[728.6358333333333,"pin moves to a new place on the board."],[728.6358333333333,"work_rate is shown on the screen, written out."]]},{"start":737.0658333333332,"say":"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.","live":null,"does":[[737.3158333333332,"work_rate is shown on the screen, written out."],[750.4358333333332,"work_rate is shown on the screen, written out."]]},{"start":751.6883333333333,"say":"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.","live":null,"does":[[751.9158333333332,"q_law is shown on the screen, written out."],[759.3558333333333,"q_law (the \"sqrt(h P k A_c)\" part) is emphasized."],[762.0758333333332,"q_law (the \"sqrt(h P k A_c)\" part) is no longer emphasized."],[762.0758333333332,"q_law (the \"tanh(m L)\" part) is emphasized."],[765.1158333333333,"q_law (the \"tanh(m L)\" part) is no longer emphasized."],[766.6483333333332,"head_rate is hidden from the screen — left the board."],[766.6483333333332,"pin is hidden from the screen — left the board."],[766.6483333333332,"pin_wall is hidden from the screen — pin left the board."],[766.6483333333332,"pin_blade is hidden from the screen — pin left the board."],[766.6483333333332,"pin_up_a is hidden from the screen — pin left the board."],[766.6483333333332,"pin_up_b is hidden from the screen — pin left the board."],[766.6483333333332,"pin_up_c is hidden from the screen — pin left the board."],[766.6483333333332,"pin_down_a is hidden from the screen — pin left the board."],[766.6483333333332,"pin_down_b is hidden from the screen — pin left the board."],[766.6483333333332,"pin_down_c is hidden from the screen — pin left the board."],[766.6483333333332,"root_arrow is hidden from the screen — pin left the board."],[766.6483333333332,"q_law is hidden from the screen — left the board."],[766.6483333333332,"work_rate is hidden from the screen — left the board."]]},{"start":767.8483333333332,"say":"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.","live":[],"does":[[767.8483333333332,"head_eta is shown on the screen, written out."],[767.8483333333332,"eff is shown on the screen, written out."],[772.1558333333332,"eff moves to a new place on the board."],[772.1558333333332,"eta_note is shown on the screen, written out."],[775.9158333333332,"work_eta is shown on the screen, written out."]]},{"start":783.5408333333332,"say":"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.","live":["eta_note","eff","head_eta"],"does":[[784.0758333333332,"work_eta is shown on the screen, written out."],[794.0358333333332,"work_eta is shown on the screen, written out."],[795.4758333333332,"eta_curve is shown on the screen, drawn."]]},{"start":798.5058333333333,"say":"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.","live":["eta_note","eff","head_eta","eta_curve"],"does":[[802.4758333333333,"walker is shown on the screen, written out."]]},{"start":809.5358333333332,"say":"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.","live":["eta_note","eff","head_eta","eta_curve","walker"],"does":[[810.6758333333332,"walker is redrawn as the numbers it depends on change."],[810.6758333333332,"ml ticks to 1.0."],[811.6358333333333,"point is shown on the screen, grown."],[811.6358333333333,"line is shown on the screen, drawn."],[812.1158333333333,"line_2 is shown on the screen, drawn."],[814.3503859694315,"point is hidden from the screen."],[814.3503859694315,"line is hidden from the screen."],[814.3503859694315,"line_2 is hidden from the screen."]]},{"start":818.1308333333333,"say":"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.","live":null,"does":[[819.1158333333333,"walker is redrawn as the numbers it depends on change."],[819.1158333333333,"ml ticks to 2.0."],[821.1958333333332,"walker is redrawn as the numbers it depends on change."],[821.1958333333332,"ml ticks to 3.0."],[827.5688333333333,"eta_rule is shown on the screen, written out."],[831.3008333333332,"eta_note is hidden from the screen — left the board."],[831.3008333333332,"eta_rule is hidden from the screen — left the board."],[831.3008333333332,"head_eta is hidden from the screen — left the board."],[831.3008333333332,"work_eta is hidden from the screen — left the board."]]},{"start":832.5008333333333,"say":"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.","live":["eff","eta_curve","walker"],"does":[[832.5008333333333,"head_eps is shown on the screen, written out."],[836.9158333333332,"walker is redrawn as the numbers it depends on change."],[836.9158333333332,"ml ticks to 0.6."]]},{"start":844.5458333333332,"say":"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.","live":["eff","eta_curve","walker","head_eps"],"does":[[848.1958333333332,"work_eps is shown on the screen, written out."]]},{"start":855.5808333333332,"say":"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.","live":null,"does":[[855.9158333333332,"work_eps is shown on the screen, written out."],[865.2758333333333,"work_eps (the \"sqrt(frac(k P, h A_c))\" part) is emphasized."],[870.9233333333332,"work_eps (the \"sqrt(frac(k P, h A_c))\" part) is no longer emphasized."]]},{"start":871.5233333333332,"say":"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.","live":null,"does":[[877.9558333333332,"criterion is shown on the screen, written out."],[878.369909178909,"criterion_note is shown on the screen, written out."],[884.6533333333332,"criterion is hidden from the screen — left the board."],[884.6533333333332,"criterion_note is hidden from the screen — left the board."],[884.6533333333332,"eff is hidden from the screen — left the board."],[884.6533333333332,"eta_curve is hidden from the screen — eff left the board."],[884.6533333333332,"walker is hidden from the screen — eff left the board."],[884.6533333333332,"head_eps is hidden from the screen — left the board."],[884.6533333333332,"work_eps is hidden from the screen — left the board."]]},{"start":885.8533333333332,"say":"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.","live":[],"does":[[885.8533333333332,"head_two is shown on the screen, written out."],[892.7958333333332,"copper_k is shown on the screen, written out."],[901.4758333333333,"copper_ml is shown on the screen, written out."]]},{"start":902.4983333333332,"say":"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.","live":["copper_k","copper_ml","head_two"],"does":[[902.7958333333333,"copper_eta is shown on the screen, written out."],[908.9558333333333,"copper_eps is shown on the screen, written out."],[913.1958333333333,"copper_verdict is shown on the screen, written out."]]},{"start":915.9508333333332,"say":"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.","live":["copper_k","copper_ml","copper_eta","copper_eps","copper_verdict","head_two"],"does":[[917.5558333333333,"plastic_k is shown on the screen, written out."],[929.3558333333333,"plastic_ml is shown on the screen, written out."]]},{"start":930.7133333333333,"say":"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.","live":["copper_k","copper_ml","copper_eta","copper_eps","copper_verdict","plastic_k","plastic_ml","head_two"],"does":[[930.9558333333333,"plastic_eta is shown on the screen, written out."],[932.7958333333332,"plastic_eps is shown on the screen, written out."]]},{"start":941.3158333333332,"say":"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.","live":["copper_k","copper_ml","copper_eta","copper_eps","copper_verdict","plastic_k","plastic_ml","plastic_eta","plastic_eps","head_two"],"does":[[941.7158333333332,"plastic_eps (the \"0.57\" part) is emphasized."],[943.1558333333332,"plastic_verdict is shown on the screen, written out."],[953.2758333333333,"plastic_eps (the \"0.57\" part) is no longer emphasized."],[956.0735833333332,"copper_eps is hidden from the screen — left the board."],[956.0735833333332,"copper_eta is hidden from the screen — left the board."],[956.0735833333332,"copper_k is hidden from the screen — left the board."],[956.0735833333332,"copper_ml is hidden from the screen — left the board."],[956.0735833333332,"copper_verdict is hidden from the screen — left the board."],[956.0735833333332,"head_two is hidden from the screen — left the board."],[956.0735833333332,"plastic_eps is hidden from the screen — left the board."],[956.0735833333332,"plastic_eta is hidden from the screen — left the board."],[956.0735833333332,"plastic_k is hidden from the screen — left the board."],[956.0735833333332,"plastic_ml is hidden from the screen — left the board."],[956.0735833333332,"plastic_verdict is hidden from the screen — left the board."]]}]},{"title":"Which Model Does This Part Deserve?","start":957.1152499999998,"end":1083.291,"objects":{"closing":"a Panel that says \"Both numbers weigh conduction inside the metal against convection at its surface. Across the body it decides whether a gradient matters. Along the fin it decides whether the far end is doing any work.\"","fin_tag":"a Point [text] labelled \"m L\" drawn in part (location=(5.3, 2.95), show_marker=False)","head_close":"a Heading that says \"One Competition, Read Twice\"","head_steps":"a Heading that says \"Four Numbers, in Order\"","lump":"a Polygon [red] drawn in part (vertices=((0.6, 1.1), (3.0, 1.1), (3.0, 3.9), (0.6, 3.9)), fill_opacity=0.35)","lump_tag":"a Point [text] labelled \"upright(\"Bi\")\" drawn in part (location=(1.8, 2.3), show_marker=False)","off_a":"a Vector [yellow] drawn in part (start=(1.4, 3.9), end=(1.4, 4.6))","off_b":"a Vector [yellow] drawn in part (start=(2.4, 3.9), end=(2.4, 4.6))","off_c":"a Vector [yellow] drawn in part (start=(4.4, 2.9), end=(4.4, 3.6))","off_d":"a Vector [yellow] drawn in part (start=(6.6, 2.9), end=(6.6, 3.6))","off_e":"a Vector [yellow] drawn in part (start=(4.4, 2.2), end=(4.4, 1.5))","off_f":"a Vector [yellow] drawn in part (start=(6.6, 2.2), end=(6.6, 1.5))","part":"a Figure (x_range=(0.0, 9.0), y_range=(0.0, 5.0), aspect=(9.0, 5.0))","spine":"a Polygon [blue] drawn in part (vertices=((3.0, 2.2), (7.6, 2.2), (7.6, 2.9), (3.0, 2.9)))","steps":"a Block [text] that says \"Work out $L_c = V slash A_s$, the average distance heat must travel out. Form $upright(\"Bi\") = h L_c slash k$. Below $0.1$, one temperature will do. If it will, the answer is $theta = theta_i e^(- t slash tau)$ with $tau = rho V c slash h …\"","summary":"a Table [text] that says \"Question Number You want One temperature for the body? $upright(\"Bi\") = h L_c slash k$ $< 0.1$ Is a fin worth fitting? $k P slash h A_c$ $>> 1$ Is the fin the right length? $m L$ $approx 1$\" (rows=(('Question', 'Number', 'You want'), ('One temperature for the …, header=True)"},"beats":[{"start":957.1152499999998,"say":"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.","live":[],"does":[[957.1152499999998,"head_steps is shown on the screen, written out."],[957.1152499999998,"part is shown on the screen, written out."],[957.8352499999999,"lump is shown on the screen, written out."],[958.7952499999998,"spine is shown on the screen, written out."],[959.0352499999998,"off_a is shown on the screen, written out."],[959.1272814253646,"off_b is shown on the screen, written out."],[959.2674385521884,"off_c is shown on the screen, written out."],[959.5435328282827,"off_d is shown on the screen, written out."],[959.819627104377,"off_e is shown on the screen, written out."],[959.9329267676766,"off_f is shown on the screen, written out."]]},{"start":967.1977499999998,"say":"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.","live":["part","head_steps","lump","spine","off_a","off_b","off_c","off_d","off_e","off_f"],"does":[[967.1977499999998,"part moves to a new place on the board."],[967.1977499999998,"steps is shown on the screen, written out."],[972.1952499999999,"steps (the \"Work out $L_c = V slash A_s$, the average distance heat must travel out.\" part) is emphasized."],[978.6227499999999,"steps (the \"Work out $L_c = V slash A_s$, the average distance heat must travel out.\" part) is no longer emphasized."]]},{"start":979.2227499999998,"say":"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.","live":["steps","part","head_steps","lump","spine","off_a","off_b","off_c","off_d","off_e","off_f"],"does":[[980.3952499999998,"steps (the \"Form $upright(\"Bi\") = h L_c slash k$. Below $0.1$, one temperature will do.\" part) is emphasized."],[986.8752499999998,"lump_tag is shown on the screen, written out."],[995.2727499999999,"steps (the \"Form $upright(\"Bi\") = h L_c slash k$. Below $0.1$, one temperature will do.\" part) is no longer emphasized."]]},{"start":995.8727499999999,"say":"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.","live":["steps","part","head_steps","lump","spine","off_a","off_b","off_c","off_d","off_e","off_f","lump_tag"],"does":[[1000.4752499999998,"steps (the \"If it will, the answer is $theta = theta_i e^(- t slash tau)$ with $tau = rho V c slash h A_s$.\" part) is emphasized."],[1009.3227499999998,"steps (the \"If it will, the answer is $theta = theta_i e^(- t slash tau)$ with $tau = rho V c slash h A_s$.\" part) is no longer emphasized."]]},{"start":1009.9227499999998,"say":"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.","live":null,"does":[[1012.8752499999998,"steps (the \"Before fitting a fin, check $k P slash h A_c$, then size it with $m L approx 1$.\" part) is emphasized."],[1023.1152499999998,"fin_tag is shown on the screen, written out."],[1025.9702499999999,"head_steps is hidden from the screen — left the board."],[1025.9702499999999,"part is hidden from the screen — left the board."],[1025.9702499999999,"lump is hidden from the screen — part left the board."],[1025.9702499999999,"spine is hidden from the screen — part left the board."],[1025.9702499999999,"off_a is hidden from the screen — part left the board."],[1025.9702499999999,"off_b is hidden from the screen — part left the board."],[1025.9702499999999,"off_c is hidden from the screen — part left the board."],[1025.9702499999999,"off_d is hidden from the screen — part left the board."],[1025.9702499999999,"off_e is hidden from the screen — part left the board."],[1025.9702499999999,"off_f is hidden from the screen — part left the board."],[1025.9702499999999,"lump_tag is hidden from the screen — part left the board."],[1025.9702499999999,"fin_tag is hidden from the screen — part left the board."],[1025.9702499999999,"steps is hidden from the screen — left the board."],[1025.9702499999999,"steps (the \"Before fitting a fin, check $k P slash h A_c$, then size it with $m L approx 1$.\" part) is no longer emphasized."]]},{"start":1026.5702499999998,"say":"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.","live":[],"does":[[1026.5702499999998,"head_close is shown on the screen, written out."],[1026.8352499999999,"summary is shown on the screen, written out."],[1031.5952499999999,"summary is shown on the screen, written out."],[1034.5952499999999,"summary is shown on the screen, written out."],[1040.4352499999998,"summary is shown on the screen, written out."]]},{"start":1044.1452499999998,"say":"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.","live":["head_close"],"does":[[1044.8752499999998,"summary (the \"column=2\" part) is emphasized."],[1049.1952499999998,"closing is shown on the screen, written out."],[1055.5727499999998,"summary (the \"column=2\" part) is no longer emphasized."]]},{"start":1056.17275,"say":"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.","live":["closing","head_close"],"does":[[1056.4352499999998,"summary (the \"row=2\" part) is emphasized."],[1062.1952499999998,"summary (the \"row=2\" part) is no longer emphasized."],[1062.1952499999998,"summary (the \"row=3\" part) is emphasized."],[1069.77525,"summary (the \"row=3\" part) is no longer emphasized."]]},{"start":1070.97525,"say":"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.","live":null,"does":[[1073.71525,"summary is indicated — a transient flash."],[1073.922281888798,"summary is indicated — a transient flash."],[1074.776345666394,"summary is indicated — a transient flash."],[1082.2493333333332,"closing is hidden from the screen — left the board."],[1082.2493333333332,"head_close is hidden from the screen — left the board."],[1082.2493333333332,"summary is hidden from the screen — left the board."]]}]}]},"durationSeconds":1083,"chapters":[{"title":"A Body That Cools All Over At Once","startSeconds":0,"narration":"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."},{"title":"When One Temperature Is a Lie","startSeconds":267.28106249999996,"narration":"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."},{"title":"Bolting On a Fin","startSeconds":464.2968124999999,"narration":"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."},{"title":"Efficiency, and the Fin That Made It Worse","startSeconds":710.9158333333332,"narration":"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."},{"title":"Which Model Does This Part Deserve?","startSeconds":957.1152499999998,"narration":"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."}]}}