Heat Conduction in a Semi-Infinite Bar

Abstract

Impulsive heat conduction in a laterally-insulated semi-infinite bar is considered. This problem is solved by a similarity solution method that methodology being introduced in a pastoral interlude. That yields a temperature distribution involving the complementary error function the asymptotic behavior of which is deduced by means of Watson’s Lemma introduced in pastoral interludes. Then this problem is solved by using an approximation solution method and the results compared with the exact similarity solution. Finally a periodic boundary condition is imposed and the solution to that problem is used to examine heat conduction underground in the Earth’s crust from which it is deduced that when it summer on the surface it is winter 4.44 meters underground. The problems consider heat conduction in a laterally insulated infinite bar and in an axially insulated planar layer with a one-time introduction of heat at the respective center lines of each configurations, both of which are solved by similarity solution methods; and a real solution method for semi-infinite bar heat conduction in contact with a periodic bath which is solved by a complex solution method in this chapter.

Problems

6.1.

Consider an infinite laterally insulated bar of constant cross-sectional area $\mathcal {A}$ as depicted in Fig. 6.8. At time $t = 0$ a supply of heat of amount $Q_0$ is instantaneously introduced at position $s=0$ (this is not a constantly replenished source). Heat conduction in the bar is assumed to be dependent only on the s coordinate.

Fig. 6.8

Schematic diagram of heat conduction in the laterally insulated infinite bar.

• (a) Consider a finite region, R, of the bar between $s = \pm x$. Show that the amount of heat in this region is given by

  $$\begin{aligned} Q(x, t) = \rho _0 c_0 \mathcal {A}\int _{-x}^{x}{T(s,t)\, ds} + Q_e, \end{aligned}$$

  (P1.1)

  where $\rho _0\equiv$ density and $c_0\equiv$ specific heat of the bar, assumed to be constant, while $T(s, t)\equiv$ deviation of temperature in the bar from the ambient $T_e$ and $Q_e\equiv$ heat in the bar associated with that ambient temperature.

• (b) For $t > 0$ deduce that the heat conservation law governing this situation reduces to

  $$\begin{aligned} \frac{\partial Q}{\partial t}(x,t) = -\iint \limits _S{\varvec{J}^{(Q)}(s, t)\bullet \varvec{n}\, d\sigma }, \end{aligned}$$

  (P1.2)

  where $S\equiv$ bounding surface of R, $\varvec{J}^{(Q)}(s, t)\equiv$ heat flux vector, and $\varvec{n}\equiv$ unit outward-pointing normal to S, by explaining why there is no source term in (P1.2). Depict S and $\varvec{n}$ for this problem by means of a diagram analogous to Fig. 6.6 for the semi-infinite bar.

• (c) Either by evaluating the surface integral in (P1.2) directly or by applying the divergence theorem to it in conjunction with the fundamental theorem of the calculus and then making use of Fick’s First Law of Diffusion or, equivalently, Newton’s Law of Cooling for this situation

  $$\begin{aligned} \varvec{J}^{(Q)}(s, t) = J_1(s, t)\varvec{i} \text{ with } J_1(s, t) = -k_0 T_1(s, t) \text{ and } T_1(s,t)\equiv \frac{\partial T}{\partial s}(s, t) \end{aligned}$$

  where $k_0\equiv$ thermal conductivity and $\varvec{i}\equiv$ unit vector in the s-direction, conclude that

  $$\begin{aligned} \frac{\partial Q}{\partial t}(x, t) = k_0 \mathcal {A}[T_1(x, t) - T_1(-x, t)]. \end{aligned}$$

  (P1.3)

• (d) Using the fundamental theorem of the calculus and the chain rule prove Leibniz’s rule

  $$\begin{aligned} \frac{d}{dx}\int _{g(x)}^{h(x)}{f(s)\, ds} = f[h(x)]h^\prime (x) - f[g(x)]g^\prime (x). \end{aligned}$$

  (P1.4)

• (e) Using (P1.1) and (P1.4) show that

  $$\begin{aligned} \frac{\partial Q}{\partial x}(x, t) = \rho _0 c_0 \mathcal {A}[T(x,t) + T(-x, t)]. \end{aligned}$$

  (P1.5)

• (f) Differentiating (P1.5) with respect to x and comparing this result with (P1.3) show that Q(x, t) satisfies the following equation for $t > 0$ sometimes called Fick’s Second Law of Diffusion:

  $$\begin{aligned} \frac{\partial Q}{\partial t} = \kappa _0 \frac{\partial ^2 Q}{\partial x^2}, \, 0< x < \infty , \end{aligned}$$

  (P1.6)

  where $\kappa _0 = k_0/(\rho _0c_0)\equiv$ thermometric conductivity or thermal diffusivity.

• (g) From (P1.1) and the stated physics of the problem, respectively, conclude that for $t > 0$:

  $$\begin{aligned} Q(0,t) = Q_e, \, Q(x, t) \rightarrow Q_e + Q_0 \text{ as } x \rightarrow \infty . \end{aligned}$$

  (P1.7)

• (h) Using the idea of nondimensionalization and the concept of similarity solutions conclude that Q(x, t) satisfying (P1.6) and (P1.7) is of the form

  $$\begin{aligned} Q(x, t) = Q_e + Q_0 F(\eta ) \text{ where } \eta = \frac{x}{2\sqrt{\kappa _0 t}}. \end{aligned}$$

  (P1.8)

  [Note that here, as well as in parts (i) and (j), any result demonstrated previously in this chapter may be employed, rather than having to be reproduced.]

• (i) Substituting the similarity solution of (P1.8) into (P1.6) and (P1.7) show that F satisfies the following differential equation and boundary conditions:

  $$\begin{aligned} F^{\prime \prime }(\eta ) + 2\eta F^\prime (\eta ) = 0, \, 0< \eta < \infty ; \, F(0) = 0, \, F(\eta ) \rightarrow 1 \text{ as } \eta \rightarrow \infty . \end{aligned}$$

  (P1.9)

• (j) Solve (P1.9) for $F(\eta )$ and hence obtain

  $$\begin{aligned} F(\eta ) = \text{ erf }(\eta ) = \frac{2}{\sqrt{\pi }}\int _{0}^{\eta }{e^{-z^2}\, dz}. \end{aligned}$$

  (P1.10)

• (k) Using the fact that $T(x,t) = T(-x, t)$ [why?] in conjunction with (P1.4), (P1.5), (P1.8), and (P1.10), finally deduce that

  $$\begin{aligned} T(x, t) = \frac{Q_0e^{-x^2/(4\kappa _0t)}}{2\mathcal {A}\sqrt{\rho _0c_0k_0\pi t}}. \end{aligned}$$

  (P1.11)

• (l) By direct differentiation of (P1.11) show that this T(x, t) satisfies the same diffusion equation as Q(x, t). Namely

  $$\begin{aligned} \frac{\partial T}{\partial t} = \kappa _0 \frac{\partial ^2 T}{\partial x^2}. \end{aligned}$$

  (P1.12)

• (m) Schematically plot T versus x for different values of t and discuss the implications of these results.

• (n) Examine the behavior of T with t for a fixed value of $x\equiv x_0 > 0$ and represent this graphically.

6.2.

Consider an axially insulated solid layer of depth $\ell$ that is infinite in planar extent. At time $t = 0$ a supply of heat of amount $Q_0$ is instantaneously introduced along the $s = 0$ axis (this is not a constantly replenished source). Heat conduction in the layer is assumed to be dependent only upon s, the radial coordinate.

Fig. 6.9

Schematic diagram of the finite cylindrical region of the axially-insulated solid layer of infinite extent.

• (a) Consider a finite cylindrical region of the layer between $0\le s \le r$, $0\le \theta \le 2\pi$, and $0 \le z \le \ell$ as depicted in Fig. 6.9 where $\theta$ and z are the circumferential and axial coordinates, respectively. Show that the amount of heat in this region is given by

  $$\begin{aligned} Q(r, t) = 2\pi \rho _0c_0 \ell \int _{0}^{r}{T(s,t)s\, ds} + Q_e, \end{aligned}$$

  (P2.1)

  where $\rho _0 \equiv$ density and $c_0 \equiv$ specific heat of the layer, assumed to be constant, while $T(s, t)\equiv$ deviation of temperature in the layer from the ambient $T_e$ and $Q_e\equiv$ heat in the layer associated with that ambient temperature.

• (b) For $t > 0$ deduce the following conservation law:

  $$\begin{aligned} \frac{\partial Q}{\partial t} = -\iint \limits _S{\varvec{J}^{(Q)}(s, t)\bullet \varvec{n}\, d\sigma }, \end{aligned}$$

  (P2.2)

  where $S\equiv$ bounding surface of R, $\varvec{J}^{(Q)}(s, t)\equiv$ heat flux vector, and $\varvec{n}\equiv$ unit outward-pointing normal to S, by explaining why there is no source term in (P2.2). Depict S and $\varvec{n}$ for this problem by means of a figure analogous to Fig. 6.6 for the semi-infinite bar.

• (c) Adopting the constitutive relation

  $$\begin{aligned} \varvec{J}^{(Q)}(s, t) = -k_0 T_1(s, t) \varvec{e}_r \text{ with } T_1(s,t) \equiv \frac{\partial T}{\partial s}(s, t), \end{aligned}$$

  where $k_0\equiv$ thermal conductivity and $\varvec{e}_r \equiv$ unit vector in the radial direction, conclude by directly evaluating the surface integral in (P2.2) that

  $$\begin{aligned} \frac{\partial Q}{\partial t}(r, t) = 2\pi \ell k_0 r T_1(r, t). \end{aligned}$$

  (P2.3)

• (d) Using (P2.1) and the fundamental theorem of the calculus show that

  $$\begin{aligned} \frac{\partial Q}{\partial r}(r, t) = 2\pi \ell \rho _0 c_0 r T(r, t). \end{aligned}$$

  (P2.4)

• (e) Differentiating (P2.4) with respect to r and comparing this result with (P2.3) show that Q(r, t) satisfies the following diffusion equation for $t > 0$:

  $$\begin{aligned} \frac{\partial Q}{\partial t} = \kappa _0 \left[ \frac{\partial ^2 Q}{\partial r^2} - \left( \frac{1}{r}\right) \frac{\partial Q}{\partial r}\right] , \, 0< r < \infty , \end{aligned}$$

  (P2.5)

  where $\kappa _0 = k_0/(\rho _0 c_0)\equiv$ thermometric conductivity.

• (f) From (P2.1) and the stated physics of the problem, respectively, conclude that for $t > 0$:

  $$\begin{aligned} Q(0,t) = Q_e, \, Q(r, t) \rightarrow Q_e + Q_0 \text{ as } r \rightarrow \infty . \end{aligned}$$

  (P2.6)

• (g) Using the idea of nondimensionalization and the concept of similarity solutions conclude that Q(x, t) satisfying (P2.5) and (P2.6) is of the form

  $$\begin{aligned} Q(r, t) = Q_e + Q_0 F(\eta ) \text{ where } \eta = \frac{r}{2\sqrt{\kappa _0 t}}. \end{aligned}$$

  (P2.7)

  [Note that here as well as in part (h) any result demonstrated previously in this chapter may be employed rather than having to be reproduced].

• (h) Substituting the similarity solution of (P2.7) into (P2.5) and (P2.6) show that F satisfies the following differential equation and boundary conditions:

  $$\begin{aligned} F^{\prime \prime }(\eta ) + \left( 2\eta - \frac{1}{\eta }\right) F^\prime (\eta ) = 0, \, 0< \eta < \infty ; \end{aligned}$$

  (P2.8a)
  $$\begin{aligned} F(0) = 0, \, F(\eta ) \rightarrow 1 \text{ as } \eta \rightarrow \infty . \end{aligned}$$

  (P2.8b)

• (i) Solve the boundary-value problem (P2.8) by integrating (P2.8a) twice and employing (P2.8b) to obtain

  $$\begin{aligned} F(\eta ) = 1 - e^{-\eta ^2}. \end{aligned}$$

  (P2.9)

• (j) Using (P2.4), (P2.7), and (P2.9), finally deduce that

  $$\begin{aligned} T(r, t) = \left( \frac{Q_0}{\ell }\right) \frac{e^{-r^2/(4\kappa _0t)}}{4\pi k_0 t}. \end{aligned}$$

  (P2.10)

• (k) Now by direct differentiation of (P2.10) demonstrate that this T(r, t) satisfies the usual axisymmetric radially dependent heat equation in cylindrical coordinates (see Chapter 12): Namely,

  $$\begin{aligned} \frac{\partial T}{\partial t} = \kappa _0 \left[ \frac{\partial ^2 T}{\partial r^2} + \left( \frac{1}{r}\right) \frac{\partial T}{\partial r}\right] = \frac{\kappa _0}{r}\frac{\partial }{\partial r}\left( r\frac{\partial T}{\partial r}\right) . \end{aligned}$$

  (P2.11)

  Observe in this context that unlike for the previous exercise involving Cartesian coordinates the one-dimensional temperature and heat functions no longer satisfy the same diffusion equation.

($\ell$):

Extrapolating from this result of (P2.11) that

$$\begin{aligned} \nabla \bullet \varvec{J}^{(Q)}(s, t) = -\frac{k_0}{s}\frac{\partial }{\partial s}\left( s\frac{\partial T}{\partial s}\right) , \end{aligned}$$

deduce (P2.3) by applying the divergence theorem to the surface integral in (P2.2).

6.3.

The purpose of this problem is to solve (6.8.3) by using a real variables approach. Toward that end we reconsider this problem for $u = u(x, t)$:

$$\begin{aligned} \frac{\partial u}{\partial t} = \kappa _0 \frac{\partial ^2 u}{\partial x^2}, \, 0< x < \infty ; \end{aligned}$$

(P3.1a)

$$\begin{aligned} u(0,t) = \cos {(\omega t)},\, u(x, t) \rightarrow 0 \text{ as } x \rightarrow \infty ; \end{aligned}$$

(P3.1b)

and seek a solution of it of the form

$$\begin{aligned} u(x, t) = f(x) \cos {(\omega t)} + g(x) \sin {(\omega t)}. \end{aligned}$$

(P3.2)

• (a) Substituting (P3.2) into (P3.1) and employing the linear independence of $\cos {(\omega t)}$ and $\sin {(\omega t)}$, show that f and g satisfy the coupled system

  $$\begin{aligned} f^{\prime \prime } = 2\alpha ^2 g; \, g^{\prime \prime } = -2\alpha ^2f \text{ where } 2\alpha ^2 = \frac{\omega }{\kappa _0} \text{ and } \alpha > 0; \end{aligned}$$

  (P3.3a)
  $$\begin{aligned} f(0) = 1, \, f(x) \rightarrow 0 \text{ as } x \rightarrow \infty ; \, g(0) = 0, \, g(x) \rightarrow 0 \text{ as } x \rightarrow \infty . \end{aligned}$$

  (P3.3b)

• (b) Eliminating g between the differential equations of (P3.3a), show that f satisfies

  $$\begin{aligned} f^{(4)} + 4\alpha ^4f = 0 \text{ where } f^{(4)} \equiv \frac{d^4f}{dx^4}. \end{aligned}$$

  (P3.4)

• (c) Letting

  $$\begin{aligned} f(x) = e^{rx}, \end{aligned}$$

  (P3.5a)

  substituting (P3.5a) into (P3.4), obtaining

  $$\begin{aligned} r^4 + 4\alpha ^4 = (r^2)^2 + (2\alpha ^2)^2 = 0, \end{aligned}$$

  (P3.5b)

  and noting that upon completing the square on its linear term (P3.5b) can be factored to yield

  $$\begin{aligned} (r^2)^2 + 2(r^2)(2\alpha ^2) + (2\alpha ^2)^2 - (2\alpha r)^2 = (r^2 + 2\alpha r + 2\alpha ^2)(r^2 - 2\alpha r + 2\alpha ^2) = 0, \end{aligned}$$

  (P3.5c)

  find that

  $$\begin{aligned} r_{1,2,3,4} = \alpha (\pm 1 \pm i) \end{aligned}$$

  (P3.5d)

  and hence conclude

  $$\begin{aligned} f(x) = e^{\alpha x}[c_1\cos {(\alpha x)} + c_2 \sin {(\alpha x)}] + e^{-\alpha x}[c_3\cos {(\alpha x)} + c_4\sin {(\alpha x)}]. \end{aligned}$$

  (P3.5e)

  Observe that the roots of (P3.5d) could have been determined by computing the four complex roots $(-4\alpha ^4)^{1/4} = \sqrt{2}\alpha (-1)^{1/4}$ using the formula for $z^{1/n}$ with $z = -1$, $n=4$, and $k=0,1,2$, and 3 but here except for the complex roots resulting from the application of the quadratic formula to (P3.5c) the analysis has been restricted to the reals.

• (d) Applying the boundary conditions for f in (P3.5b) deduce from (P3.5e) that

  $$\begin{aligned} f(x) = e^{-\alpha x}h(x) \text{ where } h(x) = \cos {(\alpha x)} + c_4\sin {(\alpha x)}. \end{aligned}$$

  (P3.6a)

  Now use (P3.3a) to show from (P3.6a) that

  $$\begin{aligned} g(x) = -e^{-\alpha x}\frac{h^\prime (x)}{\alpha }. \end{aligned}$$

  (P3.6b)

• (e) Applying the boundary conditions for g in (P3.3b) deduce from (P3.6b) that $h(x) = \cos { (\alpha x)}$ and hence finally obtain the result, equivalent to (6.8.16a), from (P3.6) and (P3.2)

  $$\begin{aligned} u(x, t) = e^{-\alpha x}[\cos {(\alpha x)}\cos {(\omega t)} + \sin {(\alpha x)}\sin {(\omega t)}] = e^{-\alpha x}\cos {(\alpha x - \omega t)}. \end{aligned}$$

  (P3.7)