University Physics — Year 2 · Bachelor Year 2

25Heat Conduction

Touch a metal rail and a wooden bench on the same cold morning: the rail feels colder, though a thermometer gives both the same temperature. Put a pan on the stove and its handle warms only after a while; dig two metres down in summer and the soil is still cold from the winter; a processor that dissipates a hundred watts on a square centimetre does not melt because a block of finned aluminium sits on it. All of this is heat conduction, the transport of internal energy through matter by the agitation of its molecules, without any flow of matter — and all of it obeys one law, Fourier’s, and one equation, the heat equation, which is the diffusion equation of the last chapter with temperature in place of density. This chapter sets up the law and the equation from an energy balance, solves them in the cases that matter (stationary walls and their thermal resistances, periodic heating and the thermal wave, cooling fins), introduces the exchange with a fluid (Newton’s law and the Biot number), and computes the entropy that conduction creates — for heat flows only downhill.

A finned heat sink on a processor: a hundred watts conducted out of a square centimetre of silicon, spread through a metal base, and handed to the air by fins that multiply the surface thirty times.

25.1 Fourier’s law

Definition 25.1 (Heat current density)

The heat current density $\vect{j}_Q$ (in $\mathrm{W}/\mathrm{m}^{2}$ ) is the vector such that the energy transferred by conduction through an oriented surface element $\dd\vect S$ during $\dd t$ is $\vect{j}_Q\cdot\dd\vect S\,\dd t$ ; the thermal flux (a power, in watts) through a surface $S$ is $\Phi = \iint_S\vect{j}_Q\cdot\dd\vect S$ .

Theorem 25.2 (Fourier’s law)

In a medium at rest whose temperature is not uniform,

\vect{j}_Q = -\lambda\,\vect{\operatorname{grad}}\,T ,

where the thermal conductivity $\lambda > 0$ (in $\mathrm{W}/\mathrm{m}/\mathrm{K}$ ) depends on the material and, weakly, on the temperature. Heat flows down the temperature gradient, from hot to cold — the second law built into a linear law.

Proof. Phenomenological, like Fick’s law, and with the same microscopic justification: the molecules (or the free electrons of a metal, or the lattice vibrations) carry their energy on a random walk, and more of it comes from the hot side. ∎

Example 25.3 (Conductivities)

Metals (electrons carry the heat as well as the current — conductivities track each other): copper $400\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , aluminium $240\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , steel $50\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ . Insulating solids: concrete $1.5\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , glass $1\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , brick $0.8\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , wood $0.15\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ . Liquids: water $0.6\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ . Gases: air $0.026\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ — and the best insulators, glass wool or foam at $0.04\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , are mostly still air, trapped so that it cannot convect. Four orders of magnitude from copper to air.

25.2 The energy balance and the heat equation

Theorem 25.4 (Local energy balance)

In a solid (or a fluid at rest) of density $\rho$ and specific heat $c$ , with a power $p$ released per unit volume (Joule heating, a chemical or nuclear reaction, absorbed radiation),

\rho c\,\frac{\partial T}{\partial t} = -\operatorname{div}\vect{j}_Q + p , \qquad\text{in one dimension}\quad \rho c\,\frac{\partial T}{\partial t} = -\frac{\partial j_Q}{\partial x} + p .

Proof. First law for the slab between $x$ and $x + \dd x$ (section $S$ , fixed volume, so no work): its internal energy $\rho c\,T\,S\,\dd x$ changes in $\dd t$ by the heat received, $[j_Q(x) - j_Q(x+\dd x)]S\,\dd t = -\partial_x j_Q\,\dd x\,S\,\dd t$ , plus $p\,S\,\dd x\,\dd t$ . In three dimensions the heat entering a fixed volume through its closed surface is $-\iint\vect{j}_Q\cdot\dd\vect S = -\iiint\operatorname{div}\vect{j}_Q\,\dd\tau$ . ∎

Theorem 25.5 (Heat equation)

For uniform $\lambda$ , $\rho$ , $c$ ,

\frac{\partial T}{\partial t} = a\,\Delta T + \frac{p}{\rho c}, \qquad a = \frac{\lambda}{\rho c} \quad\text{(the \emph{thermal diffusivity}, in $\mathrm{m}^{2}/\mathrm{s}$)} .

Everything said of the diffusion equation holds: linearity, irreversibility, smoothing, and the scales $L \sim \sqrt{at}$ , $t \sim L^2/a$ .

Proof. Insert Fourier’s law in the balance: $\rho c\,\partial_t T = \lambda\Delta T + p$ . ∎

Example 25.6 (Diffusivities and times)

Copper $a = 1.1 \times 10^{-4}\,\mathrm{m}^{2}/\mathrm{s}$ , air $2 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$ , concrete and brick $\approx 5 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ , water $1.4 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ , wood $1 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ . Heat crosses a centimetre of copper in a second, a centimetre of brick in three minutes, a $20\,\mathrm{cm}$ wall in a day (which is why thick walls smooth the day–night cycle), the $1.5\,\mathrm{cm}$ half-thickness of a steak in a quarter of an hour (cooking times scale as the square of the thickness), and a kilometre of rock in $10^{12}\,\mathrm{s}$ , thirty thousand years — the Earth is still cooling from its formation.

25.3 Stationary regime: thermal resistances

Proposition 25.7 (Thermal resistance of a slab)

In the stationary regime without sources, a slab of thickness $e$ , area $S$ , conductivity $\lambda$ , between the temperatures $T_1$ and $T_2$ , has a linear profile and carries the flux

\Phi = \frac{T_1 - T_2}{R_{\text{th}}}, \qquad R_{\text{th}} = \frac{e}{\lambda S}\quad\text{(in $\mathrm{K}/\mathrm{W}$)} :

its thermal resistance. Temperature plays the role of potential, the flux that of current: resistances in series (layers of a wall) add, resistances in parallel (wall and window) add their inverses. For a cylindrical shell (a pipe’s insulation) between the radii $r_1 < r_2$ and of length $\ell$ , $R_{\text{th}} = \ln(r_2/r_1)/2\pi\lambda\ell$ ; for a spherical shell, $(1/r_1 - 1/r_2)/4\pi\lambda$ .

Proof. $\dd^2T/\dd x^2 = 0$ : affine profile, $j_Q = \lambda(T_1 - T_2)/e$ uniform, $\Phi = j_QS$ . In cylindrical geometry $\Phi = -\lambda\,2\pi r\ell\,\dd T/\dd r$ is the same at every $r$ (no accumulation), so $T = A - (\Phi/2\pi\lambda\ell)\ln r$ ; in spherical geometry, $\Phi = -4\pi r^2\lambda\,\dd T/\dd r$ gives $T = A + \Phi/4\pi\lambda r$ . ∎

Proposition 25.8 (Exchange with a fluid: Newton’s law)

A solid surface at $T_{\text{s}}$ in contact with a fluid at $T_{\text{f}}$ (away from the surface) loses the flux density

j_Q = h\,(T_{\text{s}} - T_{\text{f}}) ,

where the heat transfer coefficient $h$ (in $\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ) sums up the conduction through the thin fluid layer that sticks to the wall and the convection that renews it: $h \approx 5$ – $25\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ for air in natural or light convection, $50$ – $500\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ for forced air, $10^3$ – $10^4$ for water. The surface then adds a film resistance $1/hS$ in series with the wall.

Proof. Phenomenological: the fluid’s motion is not computed, it is summarised in $h$ (measured, or given by the correlations of fluid mechanics). ∎

A composite wall and its thermal circuit: film, brick, wool, film in series. Right: the temperature profile — almost all the drop falls across the layer of largest resistance, the wool, though it is the thinnest.

Example 25.9 (A wall and a window)

Per square metre: brick $20\,\mathrm{cm}$ at $0.8\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , $R = 0.25\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ ; the two films, $1/8 + 1/25 = 0.165\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ : total $0.415$ , i.e. $U = 1/R = 2.4\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ and, for a $20\,\mathrm{K}$ difference, $48\,\mathrm{W}/\mathrm{m}^{2}$ . Add $10\,\mathrm{cm}$ of glass wool, $R = 2.5\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ : the total is $2.9$ , $U = 0.34\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ — seven times less. A single pane of glass, $4\,\mathrm{mm}$ at $1\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , is all film: $R = 0.004 + 0.165$ , $U \approx 6\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ; a $12\,\mathrm{mm}$ layer of still air between two panes adds $0.012/0.026 = 0.46\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ and divides the loss by more than three.

Definition 25.10 (Biot number)

For a solid of size $L$ exchanging with a fluid, the Biot number

\mathrm{Bi} = \frac{hL}{\lambda} = \frac{\text{internal resistance } L/\lambda S}{\text{film resistance } 1/hS}

compares the two. If $\mathrm{Bi} \ll 1$ the solid is uniform in temperature at every instant and cools as a whole, exponentially with the time constant $\tau = \rho c V/hS$ ; if $\mathrm{Bi} \gg 1$ the surface is at the fluid temperature and the interior conducts at its own pace $L^2/a$ .

25.4 Periodic heating: the thermal wave

Proposition 25.11 (Thermal wave in a half-space)

If the surface $x = 0$ of a half-space is held at $T(0,t) = T_0 + \theta_0\cos\omega t$ , the temperature inside is, in the established regime,

T(x,t) = T_0 + \theta_0\,\eu^{-x/\delta}\cos\Big(\omega t - \frac{x}{\delta}\Big), \qquad \delta = \sqrt{\frac{2a}{\omega}} :

a wave that propagates inward at the speed $\omega\delta = \sqrt{2a\omega}$ and is damped over one penetration depth $\delta$ per radian of phase — by $\eu^{-2\pi} = 2 \times 10^{-3}$ per wavelength. Fast oscillations penetrate less than slow ones.

Proof. Seek $\underline T = \theta_0\eu^{\iu(\omega t - \underline k x)}$ : $\iu\omega = -a\underline k^2$ , so $\underline k^2 = -\iu\omega/a$ , $\underline k = (1 - \iu)\sqrt{\omega/2a} = (1 - \iu)/\delta$ (the root that decays for $x > 0$ ). Take the real part. ∎

The thermal wave at three instants: the surface oscillation penetrates as a damped wave, its amplitude falling as -x/ (dashed) and its phase lagging by x/ — at x = π the temperature is opposite to the surface’s. — The thermal wave at three instants: the surface oscillation penetrates as a damped wave, its amplitude falling as $\eu^{-x/\delta}$ (dashed) and its phase lagging by $x/\delta$ — at $x = \pi\delta$ the temperature is opposite to the surface’s.

Example 25.12 (The soil, the cellar and the wine)

Soil, $a \approx 5 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ . Daily cycle, $\omega = 2\pi/86\,400\,\mathrm{s}$ : $\delta = 12\,\mathrm{cm}$ — half a metre down the day is gone. Annual cycle: $\delta = 2.2\,\mathrm{m}$ ; at that depth the $10\,\mathrm{K}$ summer–winter swing is down to $3.7\,\mathrm{K}$ and lags by one radian, two months: the cellar is coldest in March and warmest in September, and at $5\,\mathrm{m}$ it is $1\,\mathrm{K}$ from constant — the cellar that keeps wine. The same mathematics gives the skin depth of electromagnetism (Chapter 14): same equation, same $\sqrt{2/(\text{diffusivity} \times\omega)}$ .

25.5 Fins

Proposition 25.13 (The cooling fin)

A thin rod or plate (section $S$ , perimeter $p$ , conductivity $\lambda$ ) attached at $x = 0$ to a wall at $T_0$ , in a fluid at $T_{\text{f}}$ with the coefficient $h$ , has the excess temperature $\theta = T - T_{\text{f}}$ obeying

\frac{\dd^2\theta}{\dd x^2} = m^2\theta, \qquad m = \sqrt{\frac{hp}{\lambda S}} ;

for a long fin, $\theta = \theta_0\eu^{-mx}$ and the fin evacuates $\Phi = \sqrt{hp\lambda S}\,\theta_0$ — as much as a bare surface $\sqrt{\lambda S/hp} = 1/m$ long, with a temperature that has fallen to $\theta_0/\eu$ at $x = 1/m$ . For a fin of finite length $L$ (tip adiabatic), $\theta = \theta_0\cosh m(L-x)/\cosh mL$ and its efficiency — the flux over that of an isothermal fin at $\theta_0$ — is $\eta = \tanh(mL)/mL$ .

Proof. Stationary balance of the slice $[x, x + \dd x]$ : conducted in, $-\lambda S\theta'(x)$ ; out, $-\lambda S\theta'(x + \dd x)$ ; lost to the fluid, $hp\,\theta\,\dd x$ : $\lambda S\theta'' = hp\theta$ . The flux at the base is $-\lambda S\theta'(0)$ : $\lambda Sm\theta_0$ for the long fin, $\lambda Sm\theta_0\tanh mL$ for the finite one, to be compared with $hpL\theta_0$ . The Biot number across the fin’s thickness, $ht/\lambda$ , must be small for the one-dimensional treatment. ∎

Left: temperature along a fin, for a long fin and two finite ones (adiabatic tip) — beyond mx 2 a fin adds little. Right: the balance of a slice: conduction along, convection out through the perimeter. — Left: temperature along a fin, for a long fin and two finite ones (adiabatic tip) — beyond $mx \approx 2$ a fin adds little. Right: the balance of a slice: conduction along, convection out through the perimeter.

Example 25.14 (A heat sink)

Aluminium fins $1\,\mathrm{mm}$ thick ( $p/S \approx 2/t = 2000\,\mathrm{m}^{-1}$ ), $30\,\mathrm{mm}$ high, in forced air $h = 50\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ : $m = \sqrt{2h/\lambda t} = 20\,\mathrm{m}^{-1}$ , $mL = 0.6$ , $\eta = 0.89$ — the fins work at $89\%$ of their area. Thirty of them, $60\,\mathrm{mm}$ wide, offer $0.11\,\mathrm{m}^{2}$ where the bare $60\,\mathrm{mm}$ square offered $0.0036\,\mathrm{m}^{2}$ : a resistance $1/\eta hA = 0.2\,\mathrm{K}/\mathrm{W}$ instead of $5.6\,\mathrm{K}/\mathrm{W}$ ; $100\,\mathrm{W}$ raise the base by $20\,\mathrm{K}$ instead of $560\,\mathrm{K}$ . Longer fins gain little ( $\tanh$ saturates), thinner ones lose efficiency: $mL \approx 1$ is the engineer’s compromise.

25.6 Entropy created by conduction

Proposition 25.15 (Entropy production)

A flux $\Phi$ conducted from a body at $T_{\text{h}}$ to a body at $T_{\text{c}} < T_{\text{h}}$ through a stationary wall creates entropy at the rate

\dot S_{\text{c}} = \Phi\Big(\frac{1}{T_{\text{c}}} - \frac{1}{T_{\text{h}}}\Big) > 0 ;

locally, conduction creates $\sigma_s = \lambda\,(\vect{\operatorname{grad}}\,T)^2/T^2 \ge 0$ per unit volume and time — zero only where the temperature is uniform. Heat conduction is irreversible: the same energy, delivered at a lower temperature, can do less work.

Proof. The wall’s state does not change, so its entropy is constant: the hot body loses $\Phi/T_{\text{h}}$ per unit time, the cold one gains $\Phi/T_{\text{c}}$ ; the difference is created. Locally, the entropy balance $\partial_t(\rho s) + \operatorname{div}(\vect{j}_Q/T) = \sigma_s$ with $\rho T\partial_t s = -\operatorname{div}\vect{j}_Q$ gives $\sigma_s = \vect{j}_Q\cdot\vect{\operatorname{grad}}(1/T) = -\vect{j}_Q\cdot \vect{\operatorname{grad}}\,T/T^2 = \lambda(\vect{\operatorname{grad}}\,T)^2/T^2$ . ∎

Method 25.16 (Conduction estimates)

(1) Stationary: build the thermal circuit — slabs $e/\lambda S$ , films $1/hS$ , shells $\ln(r_2/r_1)/2\pi\lambda\ell$ ; series and parallel. (2) Transient: $\mathrm{Bi} = hL/\lambda$ ; if small, $\tau = \rho cV/hS$ ; if large, $t \sim L^2/a$ . (3) Periodic: $\delta = \sqrt{2a/\omega}$ , damping $\eu^{-x/\delta}$ , lag $x/\delta$ . (4) Sources: $p$ per unit volume, parabolic profiles. (5) Fins: $m = \sqrt{hp/\lambda S}$ , $\eta = \tanh(mL)/mL$ . (6) Entropy: $\Phi(1/T_{\text{c}} - 1/T_{\text{h}})$ .

25.7 Exercises

Exercise 25.1 ★

A brick wall $20\,\mathrm{cm}$ thick ( $\lambda = 0.8\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ), $100\,\mathrm{m}^{2}$ , between $20\,{}^{\circ}\mathrm{C}$ inside and $0\,{}^{\circ}\mathrm{C}$ outside (surfaces at these temperatures): flux density, total flux; with $10\,\mathrm{cm}$ of glass wool ( $0.04\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ) added.

Solution

Solution of Exercise 25.1.

$j = \lambda\Delta T/e = 80\,\mathrm{W}/\mathrm{m}^{2}$ ; $8\,\mathrm{kW}$ ; with wool $R = 0.25 + 2.5 = 2.75\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ : $7.3\,\mathrm{W}/\mathrm{m}^{2}$ , $730\,\mathrm{W}$ .

Exercise 25.2 ★

Diffusivities: copper ( $\lambda = 400$ , $\rho = 8900$ , $c = 385\,\mathrm{J}/\mathrm{kg}/\mathrm{K}$ ), concrete ( $1.5$ , $2300$ , $900$ ), wood ( $0.15$ , $600$ , $2500$ ). Times for heat to cross $1\,\mathrm{cm}$ and $20\,\mathrm{cm}$ of each; the $30\,\mathrm{km}$ crust of the Earth.

Solution

Solution of Exercise 25.2.

$a$ : $1.2 \times 10^{-4}\,\mathrm{m}^{2}/\mathrm{s}$ , $7 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ , $1 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ . $1\,\mathrm{cm}$ : $0.9\,\mathrm{s}$ , $140\,\mathrm{s}$ , $1000\,\mathrm{s}$ ; $20\,\mathrm{cm}$ : $6\,\mathrm{min}$ , $15\,\mathrm{h}$ , $5\,\mathrm{days}$ ; crust ( $a \sim 1 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$ ): $10^{15}\,\mathrm{s}$ , thirty million years.

Exercise 25.3 ★

Single glazing ( $4\,\mathrm{mm}$ , $\lambda = 1\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ) with films $h_{\text{i}} = 8$ , $h_{\text{o}} = 25\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ; double glazing with a $12\,\mathrm{mm}$ layer of still air ( $0.026\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ): $U$ values and fluxes for $20\,\mathrm{K}$ . Why is the real double glazing about twice worse than this estimate, and what do argon and low-emissivity coatings do?

Solution

Solution of Exercise 25.3.

Single: $R = 0.169$ , $U = 5.9\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ , $118\,\mathrm{W}/\mathrm{m}^{2}$ ; double: $R = 0.63$ , $U = 1.6\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ , $32\,\mathrm{W}/\mathrm{m}^{2}$ . The gap convects and the panes exchange by radiation (about half the real transfer); argon conducts less and convects less, a low-emissivity coating suppresses the radiation.

Exercise 25.4 ★

Cooking: $a = 1.4 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ for meat. Time for the centre of a $3\,\mathrm{cm}$ steak to feel the pan; of a $12\,\mathrm{cm}$ roast. A turkey twice as heavy: how much longer (the dimension scales as the cube root of the mass)?

Solution

Solution of Exercise 25.4.

$L^2/a$ with the half-thickness: $27\,\mathrm{min}$ ; roast: $16\times$ , $7\,\mathrm{h}$ ; $t \propto L^2 \propto m^{2/3}$ : $\times 1.6$ .

Exercise 25.5 ★★

Internal source. A cylinder of radius $R$ releases $p$ per unit volume; surface at $T_{\text{s}}$ . (a) Show that $T(r) = T_{\text{s}} + p(R^2 - r^2)/4\lambda$ . (b) Copper wire, $R = 1\,\mathrm{mm}$ , $j = 1 \times 10^{7}\,\mathrm{A}/\mathrm{m}^{2}$ , $\gamma = 6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ : $p$ and the centre–surface difference. (c) Nuclear fuel rod, $R = 5\,\mathrm{mm}$ , $p = 3 \times 10^{8}\,\mathrm{W}/\mathrm{m}^{3}$ , $\lambda = 3\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ : same question. (d) Comment.

Solution

Solution of Exercise 25.5.

(a) $(rT')'/r = -p/\lambda$ with $T'(0) = 0$ . (b) $p = j^2/\gamma = 1.7 \times 10^{6}\,\mathrm{W}/\mathrm{m}^{3}$ ; $pR^2/4\lambda = 1 \times 10^{-3}\,\mathrm{K}$ . (c) $625\,\mathrm{K}$ . (d) A wire is isothermal; a fuel pellet’s centre is hundreds of kelvins above its surface — the power density of a reactor is limited by conduction in the fuel.

Exercise 25.6 ★★

The ground. (a) Derive $\delta = \sqrt{2a/\omega}$ . (b) Soil, $a = 5 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ : daily and annual penetration depths. (c) Depth at which the annual amplitude is a tenth of the surface’s; phase lag there. (d) A water pipe must not freeze where the winter surface temperature dips to $-10\,{}^{\circ}\mathrm{C}$ around a mean of $10\,{}^{\circ}\mathrm{C}$ : minimum depth (annual wave).

Solution

Solution of Exercise 25.6.

(a) Proposition 25.11. (b) $12\,\mathrm{cm}$ ; $2.2\,\mathrm{m}$ . (c) $\delta\ln 10 = 5.2\,\mathrm{m}$ ; lag $2.3\,\mathrm{rad}$ , $130$ days. (d) The $20\,\mathrm{K}$ amplitude must halve: $x = \delta\ln 2 = 1.5\,\mathrm{m}$ .

Exercise 25.7 ★★

Biot. (a) Copper sphere, radius $1\,\mathrm{cm}$ , in air ( $h = 10\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ): Biot number, time constant of its cooling. (b) A potato ( $\lambda = 0.5\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , $a = 1.4 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ , radius $3\,\mathrm{cm}$ ) in an oven with $h = 20\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ : Biot number; which time governs its cooking? (c) Why is a thermocouple bead made small? (d) Why does a cup of coffee cool faster when stirred or blown on — which resistance changes?

Solution

Solution of Exercise 25.7.

(a) $\mathrm{Bi} = hR/3\lambda \approx 10^{-4}$ ; $\tau = \rho cR/3h = 1100\,\mathrm{s}$ . (b) $\mathrm{Bi} = 1.2$ : neither limit; the internal time $R^2/a \approx 6000\,\mathrm{s}$ governs. (c) Small $R$ : tiny $\tau$ and small Biot number. (d) The film resistance $1/h$ , the dominant one, drops.

Exercise 25.8 ★★

A fin. (a) Derive $\theta'' = m^2\theta$ and the long-fin solution. (b) Aluminium fin, $1\,\mathrm{mm}$ thick, $30\,\mathrm{mm}$ high, natural convection $h = 20\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ : $m$ , $mL$ , efficiency. (c) By what factor does the fin multiply the exchange area of its footprint, and the flux? (d) Why is a very long or very thin fin a waste of metal?

Solution

Solution of Exercise 25.8.

(a) Proposition 25.13. (b) $m = \sqrt{2h/\lambda t} = 13\,\mathrm{m}^{-1}$ , $mL = 0.39$ , $\eta = 0.95$ . (c) Area $\times 2L/t = 60$ , flux $\times 57$ . (d) Beyond $mL \approx 2$ the extra length is at the fluid temperature; a thinner fin has a larger $m$ and a lower efficiency.

Exercise 25.9 ★★

Pipes. (a) Derive the resistance of a cylindrical shell. (b) A steam pipe, outer radius $5\,\mathrm{cm}$ at $150\,{}^{\circ}\mathrm{C}$ , wrapped in $5\,\mathrm{cm}$ of glass wool, air at $20\,{}^{\circ}\mathrm{C}$ with $h = 10\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ outside: loss per metre, compared with the bare pipe. (c) Show that the total resistance of insulation plus film has a minimum at the critical radius $r_{\text{c}} = \lambda/h$ . (d) For an electric wire of radius $1\,\mathrm{mm}$ with a plastic sheath ( $\lambda = 0.2\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ), does the sheath cool or warm the copper?

Solution

Solution of Exercise 25.9.

(a) Proposition 25.7. (b) Wool $\ln 2/2\pi\lambda = 2.76\,\mathrm{K}\,\mathrm{m}/\mathrm{W}$ , outer film $1/2\pi r_2h = 0.16$ : $45\,\mathrm{W}/\mathrm{m}$ ; bare, $2\pi r_1h\Delta T = 410\,\mathrm{W}/\mathrm{m}$ . (c) $\dd/\dd r[\ln(r/r_1)/2\pi\lambda + 1/2\pi hr] = 0$ at $r = \lambda/h$ . (d) $r_{\text{c}} = 2\,\mathrm{cm} > 1\,\mathrm{mm}$ : the sheath increases the loss — it cools the copper.

Exercise 25.10 ★★★

Entropy. The wall of Exercise 25.1 (bare) between $293\,\mathrm{K}$ and $273\,\mathrm{K}$ . (a) Entropy created per second. (b) Compute $\int\sigma_s\,\dd V$ with the linear profile and check it agrees. (c) Entropy created once insulated. (d) The work that a reversible engine could have extracted from the $8\,\mathrm{kW}$ between these temperatures, and what remains of it after the wall.

Solution

Solution of Exercise 25.10.

(a) $8000(1/273 - 1/293) = 2.0\,\mathrm{W}/\mathrm{K}$ . (b) $\sigma_s = \lambda T'^2/T^2$ with $T' = 100\,\mathrm{K}/\mathrm{m}$ ; $\int\sigma_s\,\dd x = \lambda T'(1/T_{\text{c}} - 1/T_{\text{h}}) = j(1/T_{\text{c}} - 1/T_{\text{h}})$ ; times $100\,\mathrm{m}^{2}$ : the same. (c) $0.18\,\mathrm{W}/\mathrm{K}$ . (d) $\Phi(1 - T_{\text{c}}/T_{\text{h}}) = 550\,\mathrm{W}$ ; nothing — the heat now sits at $T_{\text{c}}$ ( $T_{\text{c}}\dot S_{\text{c}} = 550\,\mathrm{W}$ ).

Exercise 25.11 ★★★

Why metal feels cold. (a) For the thermal wave, compute the surface flux $j_Q(0,t)$ and show it leads the surface temperature by $\pi/4$ , with the amplitude $\sqrt{\lambda\rho c\,\omega}\,\theta_0$ . The quantity $b = \sqrt{\lambda\rho c}$ is the effusivity. (b) Two half-spaces at $T_1$ and $T_2$ are brought into contact: admitting that each then follows an erfc profile with the common surface temperature $T_{\text{c}}$ , and writing the continuity of the flux at the contact, show that $T_{\text{c}} = (b_1T_1 + b_2T_2)/(b_1 + b_2)$ . (c) Skin ( $b \approx 1000\,\mathrm{J}/\mathrm{m}^{2}/\mathrm{K}/\mathrm{s}^{1/2}$ , $33\,{}^{\circ}\mathrm{C}$ ) touching copper ( $b = 37000$ ) or wood ( $b = 400$ ) at $20\,{}^{\circ}\mathrm{C}$ : contact temperatures. (d) Why does a tile floor feel colder than a carpet at the same temperature, and why is the sensation only transient?

Solution

Solution of Exercise 25.11.

(a) $j_Q(0,t) = -\lambda\partial_xT = \lambda\theta_0\sqrt2/\delta\,\cos(\omega t + \pi/4) = \sqrt{\lambda\rho c\omega}\,\theta_0\cos(\omega t + \pi/4)$ . (b) Each side’s flux at the contact is $b_i|T_i - T_{\text{c}}|/\sqrt{\pi t}$ ; equate: $T_{\text{c}} = (b_1T_1 + b_2T_2)/(b_1 + b_2)$ . (c) Copper $20.3\,{}^{\circ}\mathrm{C}$ , wood $29\,{}^{\circ}\mathrm{C}$ . (d) Tile $b \approx 1500$ , carpet $\approx 100$ ; the formula holds while both bodies look semi-infinite; once the heat fronts reach the blood supply and the object’s far side, the steady state depends on the heat actually supplied.

Exercise 25.12 ★★★

Cooling a slab. A slab $-L < x < L$ at $T_0$ is plunged at $t = 0$ into a bath at $T_{\text{f}}$ with $h \to \infty$ . (a) Show that $\theta = T - T_{\text{f}}$ admits the solutions $\cos(k_nx)\,\eu^{-ak_n^2t}$ with $k_n = (2n+1)\pi/2L$ . (b) Expand the initial condition in a Fourier series on these modes and write the full solution. (c) After the first instants, which mode survives, and what is the time constant? (d) Time for the centre of a $3\,\mathrm{cm}$ steak ( $a = 1.4 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ ) to get within $1\%$ of the bath.

Solution

Solution of Exercise 25.12.

(a) Substitute; $\theta(\pm L) = 0$ gives $\cos k_nL = 0$ . (b) $a_n = 4\theta_0(-1)^n/(2n+1)\pi$ ; $\theta = \sum a_n\cos(k_nx)\eu^{-ak_n^2t}$ . (c) $n = 0$ , $\tau = 4L^2/\pi^2a$ . (d) $\tau = 650\,\mathrm{s}$ ; centre $(4/\pi)\eu^{-t/\tau} = 0.01$ : $t = 4.8\tau \approx 52\,\mathrm{min}$ .

25.8 Problem: Insulating a house, cooling a processor

Problem 25.1

Weekend problem — conduction at the scale of a building and of a chip

Data: brick $\lambda = 0.8\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , $\rho = 1800\,\mathrm{kg}/\mathrm{m}^{3}$ , $c = 840\,\mathrm{J}/\mathrm{kg}/\mathrm{K}$ ; glass wool $0.04\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; glass $1\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; still air $0.026\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; aluminium $240\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ , $\rho = 2700\,\mathrm{kg}/\mathrm{m}^{3}$ , $c = 900\,\mathrm{J}/\mathrm{kg}/\mathrm{K}$ ; silicon $150\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; thermal paste $5\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; soil $a = 5 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ , $\lambda = 1\,\mathrm{W}/\mathrm{m}/\mathrm{K}$ ; films $h_{\text{i}} = 8$ , $h_{\text{o}} = 25\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ .

Part I — The house. A house has $100\,\mathrm{m}^{2}$ of walls (brick $20\,\mathrm{cm}$ ) and $20\,\mathrm{m}^{2}$ of single-glazed windows ( $4\,\mathrm{mm}$ ); inside $20\,{}^{\circ}\mathrm{C}$ , outside $0\,{}^{\circ}\mathrm{C}$ .

Resistance per square metre of the brick alone; flux density if its faces were at the two temperatures.
With the two films: $R$ and $U = 1/R$ per square metre; same with $10\,\mathrm{cm}$ of glass wool added.
$U$ of the single glazing; of a double glazing with $12\,\mathrm{mm}$ of still air.
Total loss of the house before and after insulating the walls and doubling the windows.
Over a heating season of $2500$ degree-days (the integral of the temperature difference over the season, in $\mathrm{K}\,\mathrm{day}$ ), energy lost before and after, in kilowatt-hours.
Inner surface temperature of the wall, bare and insulated; why does a cold wall feel uncomfortable and grow mould?
Diffusion time across the brick; what does it do to the day–night cycle?

Part II — The ground under the house.

Derive the penetration depth $\delta = \sqrt{2a/\omega}$ of a thermal wave.
Daily and annual $\delta$ in the soil.
Depth at which the daily swing is down to $1\%$ ; annual amplitude at $2.2\,\mathrm{m}$ and at $5\,\mathrm{m}$ for a $10\,\mathrm{K}$ surface swing.
Phase lag at $2.2\,\mathrm{m}$ , in days: when is the cellar coldest?
Surface heat flux of the annual wave (amplitude and phase); compare its amplitude with the geothermal flux $0.06\,\mathrm{W}/\mathrm{m}^{2}$ .
The geothermal gradient is $30\,\mathrm{K}/\mathrm{km}$ : check it against the flux and $\lambda$ , and explain why it is invisible in the cellar’s seasons.

Part III — The processor. A chip dissipates $100\,\mathrm{W}$ in a die of $1\,\mathrm{cm}^{2}$ , $0.5\,\mathrm{mm}$ thick; a $50\,\text{µ}\mathrm{m}$ layer of paste over $10\,\mathrm{cm}^{2}$ bonds it to an aluminium heat sink: base $60\,\mathrm{mm}$ $\times$ $60\,\mathrm{mm}$ $\times$ $5\,\mathrm{mm}$ , thirty fins $60\,\mathrm{mm}$ wide, $30\,\mathrm{mm}$ high, $1\,\mathrm{mm}$ thick; a fan gives $h = 50\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ; ambient $25\,{}^{\circ}\mathrm{C}$ .

Flux density through the die; resistance of the die.
Resistance of the paste; and if a $50\,\text{µ}\mathrm{m}$ air gap replaced it.
Fin parameter $m$ , $mL$ , efficiency; check the Biot number across the fin thickness.
Total fin area and the resistance of the fins; resistance of the base (conduction through $5\,\mathrm{mm}$ ); junction temperature.
Same without fins (bare $60\,\mathrm{mm}$ square): conclusion.
The fan stops ( $h = 10\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ): junction temperature; what does the processor do at $100\,{}^{\circ}\mathrm{C}$ ?
Mass and heat capacity of the sink; its time constant $RC$ ; and that of the die alone ( $0.1\,\mathrm{g}$ , $700\,\mathrm{J}/\mathrm{kg}/\mathrm{K}$ ).
Why do good sinks have a copper base or heat pipes?

Part IV — Entropy.

Entropy created per second by the bare house envelope (question 4) between $293\,\mathrm{K}$ and $273\,\mathrm{K}$ ; after insulation.
Entropy created by the chip’s $100\,\mathrm{W}$ passing from the junction to the ambient; and by the conversion of electrical work into heat in the die.
Show that for a slab $\int\sigma_s\,\dd V = \Phi(1/T_{\text{c}} - 1/T_{\text{h}})$ .
Summarise in five lines what governs a wall, a fin, a cellar and a chip.

Solution

Solution of Problem 25.1.

1. $0.25\,\mathrm{m}^{2}\,\mathrm{K}/\mathrm{W}$ ; $80\,\mathrm{W}/\mathrm{m}^{2}$ .

2. $R = 0.415$ , $U = 2.4\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ; insulated $R = 2.9$ , $U = 0.34\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ .

3. $5.9\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ ; $1.6\,\mathrm{W}/\mathrm{m}^{2}/\mathrm{K}$ .

4. $4800 + 2360 = 7.2\,\mathrm{kW}$ ; $680 + 640 = 1.3\,\mathrm{kW}$ .

5. $UA = 358\,\mathrm{W}/\mathrm{K}$ : $358 \times 2500 \times 86400 = 7.7 \times 10^{10}\,\mathrm{J} = 21\,500\,\mathrm{kWh}$ ; after, $66\,\mathrm{W}/\mathrm{K}$ : $4000\,\mathrm{kWh}$ .

6. $20 - 48/8 = 14\,{}^{\circ}\mathrm{C}$ ; $20 - 6.8/8 = 19.2\,{}^{\circ}\mathrm{C}$ ; a cold wall radiates less to the body and can reach the dew point.

7. $a = 5.3 \times 10^{-7}\,\mathrm{m}^{2}/\mathrm{s}$ ; $e^2/a = 7.5 \times 10^{4}\,\mathrm{s} \approx 21\,\mathrm{h}$ : the wall averages out the day (the daily $\delta = 12\,\mathrm{cm}$ is less than its thickness).

8. Seek $\theta_0\eu^{\iu(\omega t - \underline kx)}$ : $\underline k = (1 - \iu)/\delta$ .

9. $12\,\mathrm{cm}$ ; $2.2\,\mathrm{m}$ .

10. $4.6\delta = 54\,\mathrm{cm}$ ; $3.7\,\mathrm{K}$ ; $1.1\,\mathrm{K}$ .

11. One radian, $58$ days: mid-March (the surface minimum being mid-January).

12. Amplitude $\lambda\theta_0\sqrt2/\delta = 6.3\,\mathrm{W}/\mathrm{m}^{2}$ , leading by $\pi/4$ ( $45$ days); a hundred times the geothermal flux.

13. $\lambda \times 0.03\,\mathrm{K}/\mathrm{m} \approx 0.03\,\mathrm{W}/\mathrm{m}^{2}$ , the right order; it is a steady $0.1\,\mathrm{K}$ across the cellar’s height, buried in the seasonal swing.

14. $1 \times 10^{6}\,\mathrm{W}/\mathrm{m}^{2}$ ; $R = 0.033\,\mathrm{K}/\mathrm{W}$ .

15. $0.01\,\mathrm{K}/\mathrm{W}$ ; air: $1.9\,\mathrm{K}/\mathrm{W}$ , $190\,\mathrm{K}$ .

16. $m = 20\,\mathrm{m}^{-1}$ , $mL = 0.61$ , $\eta = 0.89$ ; $\mathrm{Bi} = ht/2\lambda = 10^{-4}$ .

17. $A = 0.108\,\mathrm{m}^{2}$ ; $R_{\text{fins}} = 1/\eta hA = 0.21\,\mathrm{K}/\mathrm{W}$ ; base $0.006\,\mathrm{K}/\mathrm{W}$ ; total $0.26\,\mathrm{K}/\mathrm{W}$ : $26\,\mathrm{K}$ , junction at $51\,{}^{\circ}\mathrm{C}$ .

18. $1/hS = 5.6\,\mathrm{K}/\mathrm{W}$ : $560\,\mathrm{K}$ — impossible.

19. $R_{\text{fins}} \approx 0.94\,\mathrm{K}/\mathrm{W}$ , total $\approx 1\,\mathrm{K}/\mathrm{W}$ : $125\,{}^{\circ}\mathrm{C}$ — it throttles, then shuts down.

20. $V = 7.2 \times 10^{-5}\,\mathrm{m}^{3}$ , $0.19\,\mathrm{kg}$ , $C = 175\,\mathrm{J}/\mathrm{K}$ ; $\tau = RC \approx 46\,\mathrm{s}$ ; die: $C = 0.07\,\mathrm{J}/\mathrm{K}$ , $\tau \approx 3\,\mathrm{ms}$ .

21. The heat must spread from $1\,\mathrm{cm}^{2}$ to $36\,\mathrm{cm}^{2}$ : copper halves that spreading resistance; a heat pipe moves heat by evaporation and condensation with an effective conductivity a hundred times copper’s.

22. $7200(1/273 - 1/293) = 1.8\,\mathrm{W}/\mathrm{K}$ ; $0.33\,\mathrm{W}/\mathrm{K}$ .

23. $100(1/298 - 1/324) = 0.027\,\mathrm{W}/\mathrm{K}$ ; $100/324 = 0.31\,\mathrm{W}/\mathrm{K}$ — ten times more: turning work into heat is the big irreversibility.

24. $\int\lambda T'^2/T^2\,\dd x = \lambda T'\,[-1/T] = j(1/T_{\text{c}} - 1/T_{\text{h}})$ .

25. Wall: resistances in series, the worst conductor rules. Fin: $m = \sqrt{hp/\lambda S}$ , $mL \approx 1$ . Cellar: $\delta = \sqrt{2a/\omega}$ , damping and lag together. Chip: a chain of resistances from die to air, the fins’ area doing the work. Everywhere: heat flows down, and creates entropy doing so.

Terms defined in this chapter

See all 431 terms in the glossary