Conduction Heat Transfer

Derivation

Take a small element $dx\,dy\,dz$ in a solid of density $\rho$, specific heat $c$ and constant conductivity $k$. Let $\dot q_g$ be the heat generated per unit volume (W/m$^3$). Apply the energy balance:

(heat in) + (heat generated) = (heat out) + (increase of internal energy).

Heat conducted in the $x$ direction at face $x$ (Fourier's law):

$$Q_x = -k\,dy\,dz\,\frac{\partial T}{\partial x}$$

Heat leaving at $x+dx$ (Taylor series):

$$Q_{x+dx} = Q_x + \frac{\partial Q_x}{\partial x}dx = Q_x - k\,dy\,dz\,\frac{\partial^2 T}{\partial x^2}dx$$

Net heat gained in $x$ direction is $k\,\dfrac{\partial^2 T}{\partial x^2}\,dx\,dy\,dz$. Similarly for $y$ and $z$. Adding generation $\dot q_g\,dx\,dy\,dz$ and equating to the rate of increase of stored energy $\rho c\,\dfrac{\partial T}{\partial t}\,dx\,dy\,dz$:

$$k\left(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\right)+\dot q_g = \rho c\frac{\partial T}{\partial t}$$

Dividing by $k$ and defining thermal diffusivity $\alpha = k/\rho c$ (m$^2$/s):

$$\nabla^2 T + \frac{\dot q_g}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$

Reduced forms

Cylindrical coordinates $(r,\phi,z)$

$$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 T}{\partial \phi^2}+\frac{\partial^2 T}{\partial z^2}+\frac{\dot q_g}{k}=\frac{1}{\alpha}\frac{\partial T}{\partial t}$$

For one-dimensional radial steady flow without generation this becomes $\dfrac{d}{dr}\left(r\dfrac{dT}{dr}\right)=0$.

Fourier's law: the rate of heat conduction in a direction is proportional to the area normal to that direction and to the temperature gradient in it:

$$Q_x = -kA\frac{dT}{dx}\qquad q'' = -k\frac{dT}{dx}$$

The negative sign makes $Q$ positive in the direction of decreasing temperature (second law of thermodynamics).

Thermal conductivity

It is the heat flow per unit time through a unit area of a material of unit thickness when the temperature difference across it is 1 K: $k = \dfrac{Q\,L}{A\,\Delta T}$, in W/m K. It is a property of the material and measures how well it conducts heat.

Variation with temperature

• Pure metals: $k$ is high (free electrons) and decreases slightly as temperature rises, because lattice vibration scatters the electrons. Alloys and impurities reduce $k$ strongly; for alloys $k$ may rise with temperature.
• Gases: $k$ rises with temperature (about as $\sqrt{T}$), since molecular speed and collision energy exchange increase. Pressure has little effect.
• Liquids: $k$ generally falls with temperature (water is the exception, rising to about 130 °C).
• Insulating solids: $k$ increases with temperature and with moisture or density, because pore conduction and radiation across the voids increase.

Typical values (about room temperature)

Thermal circuit

 Gas     Firebrick   Ins.brick   Steel     Air
1200C --/\/\/--/\/\/--/\/\/--/\/\/--/\/\/-- 30C
        1/hi    L1/k1   L2/k2   L3/k3   1/ho

Area = 1 m$^2$. Resistances per m$^2$ (K m$^2$/W):

(a) Heat loss

$$q = \frac{T_{gas} - T_{air}}{\sum R} = \frac{1200-30}{1.1294} = 1036\ \text{W/m}^2$$

(b) Temperatures

Subtract $qR$ step by step:

$$\begin{aligned} T_{1} &= 1200 - 1036(0.02) = 1179.3\ ^\circ\text{C (hot face of firebrick)}\\ T_{2} &= 1179.3 - 1036(0.2212) = 950.2\ ^\circ\text{C (firebrick/insulating brick)}\\ T_{3} &= 950.2 - 1036(0.8214) = 99.2\ ^\circ\text{C (insulating brick/steel)}\\ T_{4} &= 99.2 - 1036(0.00013) = 99.1\ ^\circ\text{C (outer steel surface)} \end{aligned}$$

Check: $T_4 - 1036(0.0667) = 30.0\ ^\circ$C, which equals the air temperature.

(c) Share of insulating brick

$$\frac{0.8214}{1.1294}\times100 = 72.7\%$$

Answer: $q \approx 1036\ \text{W/m}^2$; interface temperatures 1179 °C, 950 °C, 99.2 °C, outer surface 99.1 °C; insulating brick gives about 72.7 % of the total resistance.

Data: $r_1 = 0.04$ m, $r_2 = 0.045$ m, $r_3 = 0.045+0.05 = 0.095$ m. For a cylinder, $R_{cond} = \dfrac{\ln(r_{o}/r_{i})}{2\pi kL}$ and $R_{conv} = \dfrac{1}{hA}$. Take $L = 1$ m.

Resistances (K/W per metre)

(a) Heat loss

$$Q/L = \frac{250-25}{1.90663} = 118.0\ \text{W/m}$$

(b) Temperatures

$$\begin{aligned} T_{\text{inner steel}} &= 250 - 118.0(0.03979) = 245.3\ ^\circ\text{C}\\ T_{\text{pipe/insulation}} &= 245.3 - 118.0(0.00042) = 245.26\ ^\circ\text{C}\\ T_{\text{outer insulation}} &= 245.26 - 118.0(1.6989) = 44.8\ ^\circ\text{C} \end{aligned}$$

Check: $44.8 - 118.0(0.16753) = 25.0$ °C.

(c) Critical radius

$$r_c = \frac{k_{ins}}{h_o} = \frac{0.07}{10} = 0.007\ \text{m} = 7\ \text{mm}$$

Since the pipe outer radius (45 mm) is already much greater than $r_c$, adding insulation always reduces heat loss here.

Answer: $Q/L \approx 118\ \text{W/m}$; interface 245.3 °C; insulation outer surface 44.8 °C; $r_c = 7$ mm.

(a) Hollow cylinder

For steady radial flow without generation the conduction equation reduces to

$$\frac{d}{dr}\left(r\frac{dT}{dr}\right)=0$$

Integrating twice: $T = C_1\ln r + C_2$. Boundary conditions: $T = T_1$ at $r = r_1$ and $T = T_2$ at $r = r_2$. Hence

$$C_1 = -\frac{T_1 - T_2}{\ln(r_2/r_1)},\qquad \frac{T - T_1}{T_2 - T_1} = \frac{\ln(r/r_1)}{\ln(r_2/r_1)}$$

The temperature varies logarithmically. Heat flow by Fourier's law with $A = 2\pi r L$:

$$Q = -k(2\pi rL)\frac{dT}{dr} = \frac{2\pi kL\,(T_1-T_2)}{\ln(r_2/r_1)}$$

The thermal resistance is $R = \dfrac{\ln(r_2/r_1)}{2\pi kL}$.

(b) Critical radius

When a cylinder of outer radius $r$ is covered with insulation, the outer radius increases, so the conduction resistance rises but the outer surface area, and so the convection resistance, falls. For an outer radius $r_o$ of insulation, with fluid at $T_\infty$ and coefficient $h_o$:

$$R_{total} = \frac{\ln(r_o/r_1)}{2\pi k L} + \frac{1}{h_o\,2\pi r_o L}$$

Heat loss is maximum where $R_{total}$ is minimum:

$$\frac{dR}{dr_o} = \frac{1}{2\pi kL\,r_o} - \frac{1}{2\pi h_o L\,r_o^2} = 0 \;\Rightarrow\; r_c = \frac{k}{h_o}$$

The second derivative is positive at $r_c$, so $R$ is minimum and $Q$ is maximum there.

Discussion

 Q ^        .---.
   |      /       \
   |    /           ` ----
   |  /
   +--------|--------------> r_o
           r_c

• If the bare pipe radius $r_1 < r_c$, adding insulation first increases heat loss until $r_o = r_c$, and further insulation reduces it. Used for small wires and thin pipes (for example electric cables, where this is desirable).
• If $r_1 \ge r_c$, any insulation reduces the heat loss. This applies to steam pipes of normal size.
• For a sphere the critical radius is $r_c = 2k/h_o$.

Expression for a hollow sphere

Steady radial conduction with no generation: $\dfrac{d}{dr}\left(r^2\dfrac{dT}{dr}\right)=0$, so $T = -C_1/r + C_2$. Using $Q = -k(4\pi r^2)\dfrac{dT}{dr}$ and the boundary temperatures $T_1$ at $r_1$ and $T_2$ at $r_2$:

$$Q = \frac{4\pi k\,(T_1 - T_2)}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}$$

and the temperature at any radius:

$$T = T_1 - \frac{Q}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r}\right)$$

(a) Heat flow

Given $r_1 = 0.1$ m, $r_2 = 0.2$ m, $k = 15$ W/m K.

$$\begin{aligned} \frac{1}{r_1}-\frac{1}{r_2} &= 10 - 5 = 5\ \text{m}^{-1}\\ Q &= \frac{4\pi(15)(300-100)}{5} = 7540\ \text{W} \end{aligned}$$

(b) Temperature at 150 mm

$$\begin{aligned} T &= 300 - \frac{7540}{4\pi(15)}\left(10 - \frac{1}{0.15}\right)\\ &= 300 - 40\,(10 - 6.667) = 300 - 133.3 = 166.7\ ^\circ\text{C} \end{aligned}$$

The temperature at r = 0.15 m (166.7 °C) is lower than the arithmetic mean of the surfaces (200 °C), because the area is smaller near the inside and the gradient is steeper there.

Answer: $Q \approx 7.54\ \text{kW}$; $T(0.15\ \text{m}) \approx 166.7\ ^\circ\text{C}$.

Derivation

Let the plate occupy $-L \le x \le L$ with $2L = 40$ mm, so $L = 0.02$ m. For steady one-dimensional conduction with generation $\dot q_g$:

$$\frac{d^2T}{dx^2} + \frac{\dot q_g}{k} = 0$$

Integrate: $T = -\dfrac{\dot q_g}{2k}x^2 + C_1x + C_2$. By symmetry $dT/dx = 0$ at $x = 0$, so $C_1 = 0$. At the surface $x = L$ the conduction equals convection:

$$-k\frac{dT}{dx}\Big|_{L} = h\,(T_s - T_\infty)\;\Rightarrow\;\dot q_g L = h\,(T_s-T_\infty)$$

so $T_s = T_\infty + \dfrac{\dot q_g L}{h}$ and

$$T(x) = T_s + \frac{\dot q_g}{2k}\left(L^2 - x^2\right)$$

The maximum is at the centre: $T_{max} = T_s + \dfrac{\dot q_g L^2}{2k}$.

(a) Surface temperature

$$T_s = 80 + \frac{(5\times10^6)(0.02)}{500} = 80 + 200 = 280\ ^\circ\text{C}$$

(b) Maximum temperature

$$T_{max} = 280 + \frac{(5\times10^6)(0.02)^2}{2(20)} = 280 + 50 = 330\ ^\circ\text{C}$$

It occurs at the mid-plane $x = 0$.

(c) Heat removed per face

By energy balance, heat generated in the half-plate equals heat leaving one face:

$$q'' = \dot q_g L = (5\times10^6)(0.02) = 1\times10^5\ \text{W/m}^2 = 100\ \text{kW/m}^2$$

Check: $h(T_s - T_\infty) = 500\times200 = 1\times10^5$ W/m$^2$.

Answer: $T_s = 280\ ^\circ$C; $T_{max} = 330\ ^\circ$C at the centre; $q'' = 100\ \text{kW/m}^2$ per face.

Finite-difference method

The region is covered by a grid of nodes spaced $\Delta x$ and $\Delta y$. The Laplace equation $\dfrac{\partial^2T}{\partial x^2}+\dfrac{\partial^2T}{\partial y^2}=0$ is replaced using central differences:

$$\frac{\partial^2T}{\partial x^2}\approx\frac{T_{m+1,n}-2T_{m,n}+T_{m-1,n}}{\Delta x^2},\qquad \frac{\partial^2T}{\partial y^2}\approx\frac{T_{m,n+1}-2T_{m,n}+T_{m,n-1}}{\Delta y^2}$$

For $\Delta x = \Delta y$ this gives the interior-node equation

$$T_{m,n} = \frac{1}{4}\left(T_{m+1,n}+T_{m-1,n}+T_{m,n+1}+T_{m,n-1}\right)$$

so each node is the average of its four neighbours. Writing it for every unknown node gives a set of linear equations, solved by matrix inversion or iteration (Gauss-Seidel).

Nodal equations

   500   500   500   500
 100 [ 1 ]-[ 2 ] 100
      |      |
 100 [ 3 ]-[ 4 ] 100
   100   100   100   100

Node 1 neighbours: top 500, left 100, node 2, node 3.

$$\begin{aligned} 4T_1 &= 500 + 100 + T_2 + T_3\\ 4T_2 &= 500 + 100 + T_1 + T_4\\ 4T_3 &= 100 + 100 + T_1 + T_4\\ 4T_4 &= 100 + 100 + T_2 + T_3 \end{aligned}$$

Solution

The conditions are symmetric about the vertical centre line, so $T_1 = T_2$ and $T_3 = T_4$:

$$\begin{aligned} 3T_1 - T_3 &= 600\\ 3T_3 - T_1 &= 200 \end{aligned}$$

From the second, $T_1 = 3T_3 - 200$. Substitute in the first: $9T_3 - 600 - T_3 = 600$, so $T_3 = 150$ °C and $T_1 = 250$ °C.

Check in node 1 equation: $(500+100+250+150)/4 = 250$. Correct.

Answer: $T_1 = T_2 = 250\ ^\circ$C; $T_3 = T_4 = 150\ ^\circ$C.

Derivation

For steady one-dimensional flow, $Q$ is constant along $x$:

$$Q = -kA\frac{dT}{dx} = -k_0(1+bT)A\frac{dT}{dx}$$

Separate the variables and integrate from $x = 0$ ($T_1$) to $x = L$ ($T_2$):

$$Q\int_0^L dx = -k_0A\int_{T_1}^{T_2}(1+bT)\,dT$$ $$QL = k_0A\left[(T_1 - T_2)+\frac{b}{2}\left(T_1^2-T_2^2\right)\right]$$ $$Q = \frac{k_0A}{L}\left[1+\frac{b}{2}(T_1+T_2)\right](T_1-T_2) = \frac{k_mA\,(T_1-T_2)}{L}$$

So the heat flow is the same as for constant conductivity equal to the mean value $k_m = k_0\left[1+b\,\dfrac{T_1+T_2}{2}\right]$, the value of $k$ at the average temperature.

Shape of temperature profile

Integrating only up to a general $x$ gives a quadratic in $T$: $T + \dfrac{b}{2}T^2 = T_1+\dfrac{b}{2}T_1^2 - \dfrac{Q x}{k_0A}$. Since it is quadratic in $T$, $T$ is not linear in $x$. Where $k$ is large the gradient is small, and where $k$ is small the gradient is steep. For $b > 0$ the hot side has the larger $k$, so the profile is flatter near the hot face and steeper near the cold face (concave downward).

Numerical value

$$\begin{aligned} k_m &= 0.8\left[1 + 0.0006\times\frac{800+100}{2}\right] = 0.8(1.27) = 1.016\ \text{W/m K}\\ q &= \frac{k_m\,(T_1-T_2)}{L} = \frac{1.016\times700}{0.25} = 2845\ \text{W/m}^2 \end{aligned}$$

Answer: $q \approx 2.85\ \text{kW/m}^2$ with $k_m = 1.016$ W/m K.