Conduction

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Picture a house wall: twenty centimetres of brick, plus one thin layer of insulation, just five centimetres thick. Which layer does more to stop the heat from leaking out? The thick brick, surely, right? Hmm, hold that thought. Today: conduction, how heat flows through a solid, and by the end you will see the surprising answer. Problems on this come up often in JEE, so let's begin.

0:31Fourier's law: heat flow through a rod

So, take a metal rod, one end hot and the other end cold. Heat flows from the hot end to the cold end, through the metal itself: that is conduction. The rate of heat flow, d Q by d t, is the heat Q crossing per unit time; we call it the heat current, capital H, okay? Now think. Same metal, same end temperatures, but make the rod twice as long. Does more heat flow, or less? Less! Yes, exactly! The heat has a longer path. So Fourier's law puts the length L at the bottom: H equals K A delta T, over L, right? Capital A is the area of cross-section, and delta T, with the little triangle, the Greek capital delta, is the temperature difference between the ends. This holds in the steady state, once the temperatures along the rod stop changing. With me? Capital K is the thermal conductivity, a property of the material: copper has a large K, insulation a tiny one. Its unit is watt per metre per kelvin; the minus one powers mean per. Clear? One more fact. In the steady state, along a uniform rod with insulated sides, the temperature, capital T, falls linearly from the hot end to the cold end. Got it?

1:47Thermal resistance: Ohm's law for heat

Now, a beautiful idea. Call L over K A the thermal resistance, capital R, okay? Then H equals delta T over R, as simple as that! Look at this shape: a flow equals a difference over a resistance. Where have you seen a law like this before? Ohm's law! Correct! Delta T plays the role of the voltage V, and the heat current H plays the role of the electric current, capital I, right? That is what the double-headed arrow on screen means. So whatever you know about resistors works for heat. Slabs one after another are in series: R equals R one plus R two, where the small numbers below R say which slab. And the same H flows through each. Clear? Rods side by side, between the same two temperatures, are in parallel: one over R equals one over R one, plus one over R two. With me? Where rods meet at a junction, in the steady state, heat flowing in equals heat flowing out, like Kirchhoff's current law. Got it?

2:50Equivalent conductivity, and a trap

Now, sometimes a question wants one equivalent conductivity: K with small e q below it. For slabs of equal area in series, K e q equals L one plus L two, over the sum L one over K one, plus L two over K two. Okay? For rods of equal length in parallel, K e q equals K one A one plus K two A two, over A one plus A two. Right? Now, a trap. Two slabs of equal thickness, in series. Surely K e q is just the plain average of K one and K two? No, no, careful! Put L one equal to L two in the series formula, and you get two K one K two, over K one plus K two. That is the harmonic mean, not the average. Clear? The plain average, K one plus K two over two, belongs to rods of equal area in parallel. So: series, harmonic mean; parallel, arithmetic mean. Don't mix them up! Got it?

3:47Topper's trick: share the temperature drop

Now a topper's trick for series problems. The same H flows through every slab, so the temperature drop across slab one is H times R one, its own resistance. Okay? So tell me: which slab takes the bigger share of the total temperature drop? The one with the bigger resistance! Yes! The drop is shared in proportion to the resistances, like voltage across resistors in series. So, for slabs of the same size, the better conductor has the smaller temperature drop. This saves a lot of time. Right?

4:20Example 1: rate of heat flow through a copper rod

Now, Example one. Here is a copper rod, K equal to four hundred watt per metre per kelvin, one metre long, four square centimetres in cross-section. Its ends are at one hundred degrees Celsius and zero degrees Celsius; the small circle means degrees. Find the rate of heat flow. Pause the video, and try it first. Done? Okay, let's check. Area first. One centimetre is ten to the power minus two metre, so four square centimetres is four times ten to the power minus two square metre, right? Oh no, careful! Square the conversion too. So capital A is four times ten to the power minus four square metre. And delta T is one hundred degrees Celsius, the same as one hundred kelvin, because it is a difference. With me? Now Fourier's law: four hundred, times four times ten to the power minus four, times one hundred, over one metre. Okay? Four hundred times one hundred is four times ten to the power four. Times four times ten to the power minus four: the powers cancel, leaving four times four, sixteen. So H is sixteen watts: sixteen joules every second. Clear?

5:38Example 2: heat loss and interface temperature of a wall

Now Example two, our puzzle from the start. Here is a ten square metre wall: twenty centimetres of brick, K equal to zero point eight, lined with five centimetres of insulation, K equal to zero point zero five. Inside is twenty five degrees Celsius, outside five, and the brick is on the inside. Find the heat loss, and the temperature where the brick meets the insulation. Pause and try. Hint: two resistors in series. Done? Okay, let's check. The heat crosses the brick, and then the insulation. So, series or parallel? Series, yes! Brick first: R one equals L over K A, with twenty centimetres as zero point two metre, so zero point two, over zero point eight times ten. That is zero point zero two five kelvin per watt; in this unit, capital K means kelvin, not conductivity, and W means watt. Okay? Insulation: R two is zero point zero five, over zero point zero five times ten: zero point one kelvin per watt. Now look! The thin insulation has four times the resistance of the thick brick. Puzzle solved: the thin layer does far more to keep the heat in. Right? In series, the resistances add: zero point one two five kelvin per watt. The temperature difference is twenty five minus five, twenty kelvin, so H equals twenty over zero point one two five: one hundred and sixty watts. Clear? Now the interface. It sits at fifteen degrees, the average of twenty five and five, right? No, no, wait! The average works only for equal resistances. Use our trick: the drop across the brick is H times R one, one hundred and sixty times zero point zero two five: four kelvin. Got it? So the interface is at twenty five minus four: twenty one degrees Celsius, and the insulation takes the other sixteen kelvin. The brick has the smaller resistance, so it takes the smaller drop, just as the trick said. With me?

7:49Quick revision: Fourier's law and thermal resistance

So, quick revision. Fourier's law: H equals K A delta T over L, in the steady state. Thermal resistance R is L over K A, H is delta T over R, and you combine resistances in series and parallel just like resistors. And in series, the same H flows, and the bigger resistance takes the bigger temperature drop. Now go and try the practice problems on this page. All the best!