Chapter 18, Problem 79P

To determine

The soil temperature at distances from 0, 10, 20 and 50 cm from the earth’s surface at the end of the $\text{10}\text{h}$ period.

Explanation of Solution

Given:

Initial uniform temperature of the furnace $\left(T_i\right)$ is $10°\text{C}$.

Temperature of the area $\left(T_{\infty }\right)$ is $-10°\text{C}$.

Convection heat transfer coefficient $\left(h\right)$ is $40\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Time period $\left(t\right)$ is $\text{10}\text{h}$.

Calculation:

Write the given properties of the soil.

  $\begin{array}{l}k=0.9\text{W}/\text{m}\cdot \text{K}\\ \alpha =1.6\times {10}^{-5}{\text{m}}^2/\text{s}\end{array}$

Calculate the one-dimensional transient temperature distribution in the ground.

  $\begin{array}{c}\frac{T\left(x,t\right)-T_i}{T_{\infty }-T_i}=\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)-\text{exp}\left(\frac{hx}{k}+\frac{h^2\alpha t}{k^2}\right)\left[\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}+\frac{h\sqrt{\alpha t}}{k}\right)\right]\\ =\left\{\begin{array}{c}\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)-\\ \text{exp}\left(\begin{array}{l}\frac{hx}{k}+\\ \frac{{\left(40\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}^2\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}{{\left(0.9\text{W}/\text{m}\cdot \text{K}\right)}^2}\end{array}\right)\\ \left[\text{erfc}\left(\begin{array}{l}\frac{x}{2\sqrt{\alpha t}}+\\ \frac{\left(40\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\sqrt{\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}}{0.9\text{W}/\text{m}\cdot \text{K}}\end{array}\right)\right]\end{array}\right\}\\ =\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)-\left[\text{exp}\left(\frac{hx}{k}+1138\right)\left[\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}+33.7\right)\right]\right]\end{array}$

It is concluded that the last term in the temperature distribution relation should be zero irrespective of distance x as the exponential term tends to infinity.

Write the reduced form of one-dimensional transient temperature distribution in the ground.

  $\begin{array}{l}\frac{T\left(x,t\right)-T_i}{T_{\infty }-T_i}=\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)\\ T\left(x,t\right)=T_i+\left(T_{\infty }-T_i\right)\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)\end{array}$

Calculate the soil temperature at distance from 0 cm from the earth’s surface.

  $\begin{array}{c}T\left(0,10h\right)=10°\text{C}+\left(-10°\text{C}-10°\text{C}\right)\text{erfc}\left(\frac{0}{2\sqrt{\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}}\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\text{erfc}\left(0\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\left(1\right)\\ =-10°\text{C}\end{array}$

Thus, the soil temperature at distance from 0 cm from the earth’s surface is $-10°\text{C}$.

Calculate the soil temperature at distance from 10 cm from the earth’s surface.

  $\begin{array}{c}T\left(0.1\text{m},10h\right)=10°\text{C}+\left(-10°\text{C}-10°\text{C}\right)\text{erfc}\left(\frac{0.1\text{m}}{2\sqrt{\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}}\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\text{erfc}\left(0.066\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\left(0.9257\right)\\ =-8.5°\text{C}\end{array}$

Thus, the soil temperature at distance from 10 cm from the earth’s surface is $-8.5°\text{C}$.

Calculate the soil temperature at distance from 20 cm from the earth’s surface.

  $\begin{array}{c}T\left(0.2\text{m},10h\right)=10°\text{C}+\left(-10°\text{C}-10°\text{C}\right)\text{erfc}\left(\frac{0.2\text{m}}{2\sqrt{\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}}\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\text{erfc}\left(0.132\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\left(0.8519\right)\\ =-7.0°\text{C}\end{array}$

Thus, the soil temperature at distance from 20 cm from the earth’s surface is $-7.0°\text{C}$.

Calculate the soil temperature at distance from 50 cm from the earth’s surface.

  $\begin{array}{c}T\left(0.5\text{m},10h\right)=10°\text{C}+\left(-10°\text{C}-10°\text{C}\right)\text{erfc}\left(\frac{0.5\text{m}}{2\sqrt{\left(1.6\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(10\times 3600\text{s}\right)}}\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\text{erfc}\left(0.329\right)\\ =10°\text{C}+\left(-20°\text{C}\right)\left(0.6418\right)\\ =-2.8°\text{C}\end{array}$

Thus, the soil temperature at distance from 50 cm from the earth’s surface is $-2.8°\text{C}$.