Question

Assume that the temperature effects of glaciations can be represented by a periodic surface temperature with a period of \(10^{4} \mathrm{yr}\). If it is desired to drill a hole to a depth that the temperature effect of the glaciations is \(5 \%\) of the surface value, how deep must the hole be drilled? Assume \(\kappa=1 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).

Short answer

The hole must be drilled approximately 3300 meters deep.

Step-by-step solution

Define the Problem

We want to find the depth at which the temperature effect is reduced to 5% of the original periodic surface temperature, with a period given as 10,000 years. Here, thermal diffusivity \( \kappa \) is \( 1 \ mm^2/s \).

Convert Time Period

Convert the given period from years to seconds to match the units of \( \kappa \). Since 1 year has approximately 31.536 million seconds,\[ T = 10^4 \times 31.536\times10^6 = 3.1536\times10^{11} \text{ seconds} \]

Calculate the Angular Frequency

The angular frequency \( \omega \) can be calculated using the formula:\[ \omega = \frac{2\pi}{T} \]Substituting the value of \( T \) calculated in Step 2:\[ \omega = \frac{2\pi}{3.1536\times10^{11}} \text{ s}^{-1} \]

Set Temperature Reduction Condition

We want the temperature at the desired depth to be 5% of the surface value. For a periodic solution, the temperature decay with depth \( z \) is given by:\[ e^{-z\sqrt{\frac{\omega}{2\kappa}}} = 0.05 \]

Solve for Depth \( z \)

Take the natural logarithm on both sides of the decay equation:\[ -z\sqrt{\frac{\omega}{2\kappa}} = \ln(0.05) \]Solve for \( z \):\[ z = \frac{-\ln(0.05)}{\sqrt{\frac{\omega}{2\kappa}}} \]

Compute \( z \) using Known Values

Substitute the values of \( \omega \) and \( \kappa \) into the equation from Step 5:\[ z = \frac{-\ln(0.05)}{\sqrt{\frac{2\pi}{2 \, \cdot 3.1536\times10^{11} \, \cdot 1\times10^{-6}}} \approx 3300 \text{ meters} \]

Key concepts

Thermal Diffusivity

Thermal diffusivity is a material-specific property that measures the rate at which heat spreads through a material. It is defined as the thermal conductivity divided by the product of density and specific heat capacity. This means it essentially determines how quickly temperature changes can propagate through a medium. The higher the thermal diffusivity, the faster the heat is transferred through the material. A simple way to understand it is by thinking about how fast heat can spread through different materials. For example, metals typically have high thermal diffusivity, so they heat up and cool down quickly. In the context of our exercise, thermal diffusivity allows us to determine how deep the glaciation effects penetrate the ground. We use a value of \( \kappa = 1 \mathrm{\ mm}^2/\mathrm{s} \), which is equivalent to saying the material's rate of heat spread is 1 square millimeter per second.

Periodic Surface Temperature

A periodic surface temperature refers to a situation where temperature varies over time in a regular repeating pattern. This is often modeled as a sinusoidal function for simplicity, which means the temperature rises and falls in predictable cycles. In our exercise, glaciation events are modeled as periodic temperature changes with a period of 10,000 years. Such a long period corresponds to slow changes in temperature on the surface due to the repeated advance and retreat of glaciers over geological time scales. The periodic surface temperature is responsible for driving heat distribution into the ground. This cyclical pattern of warming and cooling translates into oscillations of temperature down below the Earth's surface over time. It means that the deeper you go, the less pronounced these temperature changes become.

Angular Frequency

Angular frequency is a measure of how rapidly a periodic function oscillates with time. It is denoted by \( \omega \), and measures in radians per second. Think of it as how many cycles or oscillations occur per unit of time. It links directly to the period of the oscillation through the equation \( \omega = \frac{2\pi}{T} \). In this formula, \( T \) represents the time it takes for one complete cycle of a wave (in our case, one full glaciation cycle).Having converted the period of 10,000 years into seconds, we find the \( \omega \) for our problem. This value helps calculate how deeply the surface temperature changes, or how heat propagates below the surface, affecting the geothermal gradient. In simpler terms, the angular frequency tells us how effectively temperature cycles percolate through the Earth.