## 13.8 Answers to the Problem Set

1.(a) $$T(z,t)=\overline{T}+A(z)\sin[\omega t+B(z)]$$

Now $$T=\overline{T}+A(0)\exp(-z/D)\sin[\omega t-z/D]$$, substituting expressions (13.12) and (13.13) into (13.11). Taking partial derivatives as indicated by Equation (13.8): \begin{align*} \frac{\partial T}{\partial t}&=\omega A(0)\exp(-z/D)\cos[\omega t-z/D] \\ \frac{\partial T}{\partial z}&=-\frac{A(0)}{D}\exp(-z/D)\sin[\omega t-z/D]-\frac{A(0)}{D}\exp(-z/D)\cos[\omega t-z/D] \\ \frac{\partial^2 T}{\partial z^2}&=\frac{A(0)}{D^2}\exp(-z/D)\sin[\omega t-z/D]+\frac{A(0)}{D^2}\exp(-z/D)\cos[\omega t-z/D] \\ &+\frac{A(0)}{D^2}\exp(-z/D)\cos[\omega t-z/D]-\frac{A(0)}{D^2} \exp(-z/D)\sin[\omega t-z/D] \\ \end{align*}

Notice the first and fourth terms in the expression for $$\frac{\partial^2T}{\partial z^2}$$ cancel. Substituting in Equation (13.8): $\omega A(0)\exp(-z/D)\cos[\omega t-z/D]=K(2A(0)/D^2)\exp(-z/D)\cos[\omega t-z/D]$ or $$\omega A(0)=K(2A(0)/D^2)$$ Canceling the $$A(0)$$’s, and rearranging the terms, we get $$D=(2K/\omega)^{1/2}$$, the identity of Equation (13.14).

1. The first boundary condition, $$T(\infty,t) = \overline{T}$$ (Equation (13.9)) is a constant, so it trivially satisfies Equation (13.8) (0 = 0).

The second boundary condition (Equation (13.10)) represents a special case of the solution (Equation (13.11)) when $$z = 0$$. Hence part (a) of this problem constitutes proof that the second boundary condition satisfies this problem.

2.(a) $$$T(z,t)=\overline{T}+A(0)\exp(-z/0)\sin[\omega t-z/D] \tag{13.15}$$$ We know from Equation (13.3) that
\begin{align*} G(z,t)&=-k\frac{\partial T}{\partial z} \\ G(z,t)&=-k\{-A(0)/D\:\exp(-z/D)\sin[\omega t-z/D]-A(0)/D\: \exp(-z/D)\cos[\omega t-z/D]\} \\ &=k A(0)/D \: \exp(-z/D)\{\sin[\omega t -z/D]+\cos[\omega t-z/D]\} \\ \end{align*} The surface heat flux is then
$G(0,t)=kA(0)/D\{\sin\omega t+cos\omega t\}$ Using the identity $$\sin(x+y)=\sin x\cos y+\cos x\sin y$$
setting $$y=\pi/4$$; $$\sin(x+\pi/4)=\frac{1}{\sqrt2}(\sin x+\cos x)$$, and substituting in the expression for $$G(0,t)$$, we have $G(0,t)=\sqrt{2}kA(0)/D\: \sin(\omega t+\pi/4)$ The maximum value of $$G(0,t)$$ occurs when $$\sin(\omega t+\pi/4)=1$$, so $G(0,t)_{max}=\frac{\sqrt{2}kA(0)}{D}$

1. Comparing our expression for $$G(0,t)$$ with Equation (13.15) for $$z = 0$$, we find that temperature lags the heat flux by a phase angle of $$\pi/4$$ or 1/8 of a cycle. This corresponds to 3 hours and 1-1/2 months for the diurnal and annual case, respectively. Returning to our basic definition of heat flow in Equation (13.2), we write $G=-k\frac{\Delta T}{L}$ Comparing our result from 2(a), we note that $$L$$ corresponds to $$D/\sqrt{2}$$ if were to consider heat conduction in a slab, thickness $$L$$, with a temperature difference maintained between the faces of A. The amount of heat flowing $$(Q)$$ per unit area of soil during 1/2 cycle can be found by integrating $$G(0,t)$$. \begin{align*} Q&=\frac{\sqrt{2}A(0)k}{\omega D}\int_{-\pi/4}^{3\pi/4}\sin(\omega t+\pi/4)d(\omega t) \\ &=-\frac{\sqrt2A(0)k}{\omega D}\cos(\omega t+\pi/4)\int_{-\pi/4}^{3\pi/4} \\ &=-\frac{\sqrt2A(0)k}{D\omega}[-1-(+1)] \\ &=\frac{2\sqrt2A(0)k}{D\omega} \\ \end{align*} Recall that $$D=(2K/\omega)^{1/2}$$ and $$K=k/G$$; $Q=\frac{2K}{\omega}\frac{\sqrt2A(0)G}{D}=\sqrt2DA(0)G$

3.(a) Forming a ratio using Equation (13.12) at two depths, we find that $D=\frac{z_2-z_1}{\ln[A(z_1)/A(z_2)]}$ Using the 30 and 15 cm curves, we estimate amplitudes of 2 and 8°C, respectively. So $D=\frac{30-15}{\ln(8/\varepsilon)}\cong 11cm$ Recalling the definition of $$D=(2K/\omega)^{1/2}$$(Equation (13.14)), we can write the ratio $\frac{\mbox{D annual}}{\mbox{D diurnal}}=\left( \frac{\omega \mbox{ diurnal}}{\omega \mbox{ annual}} \right)^{1/2}$ $\mbox{D annual}=(365)^{1/2}\mbox{ D diurnal}=(19.1)11\cong210cm$ (b) Using D from part (a), we have \begin{align*} A(0)&=A(z)exp(+z/D) \\ A(0)&\cong 31cm \mbox{ using } z=15cm,\:D=11cm,\:A(z)=8°C \\ &\cong 31cm \mbox{ using } z=30cm, D=11cm, A(z)=2°C \\ \end{align*}

1. Solving Equation (13.14) for $$K$$, we get, using the results from 3(a): $K=\frac{\omega}{2}D^2=\frac{\omega}{2}\frac{z_2-z_1}{[lnA(z)_1)/A(z_2)]}^2$

Or, knowing that the maximum temperature occurs when $$\sin(\omega t_{max}-z/D)=1$$, or $$\omega t_{max}-z/D=\pi/2$$, we can subtract such expressions for different depths to get $\omega\Delta t_{max}=\frac{z_2-z_1}{D}$ or $D=\frac{z_2-z_1}{\omega \Delta t_{max}}$ so $K=\frac{\omega}{2}D^2=\frac{\omega}{2}\frac{(z_2-z_1)^2}{\omega^2\Delta t_{max}^2}=\frac{1}{2\omega}(\frac{z_2-z_1}{\Delta t_{max}})^2$ Using the first of these expressions for $$K$$, one gets $$8.8 \times 10^{-3} cm^2 sec^{-1}$$, while the latter gives $$5.9 \times 10^{-3}cm^2 sec^{-1}$$ if $$\Delta t_{max}$$ is taken to be 4.5 hrs.

Some divergence of answers is expected here due to variation in the thermal properties of the soil, which we have assumed negligible.

The soil was probably dry, since the soil in question is a sandy loam. This is shown most clearly by the figure in problem 4, where at high moisture contents $$K$$ doesn’t fall much below $$10 \times 10^{-3} cm^2 sec^{-1}$$.

1. The initial rapid increase in $$K$$ is due to the rapid increase of $$K$$ at lower water contents. This is because the water displaces air in the dry soil, and the water conductivity is approximately two orders of magnitude greater than that for air.

At higher water contents, heat capacity keeps increasing at a constant rate with addition of water, but the rate of increase of $$k$$ is continually decreasing. Hence a point is reached when the quotient $$k/G$$ peaks and decreases with further addition.