Question

The exponential depth dependence of heat production is preferred because it is self-preserving upon erosion. However, many alternative models can be prescribed. Consider a two-layer model with \(H=\) \(H_{1}\) and \(k=k_{1}\) for \(0 \leq y \leq h_{1},\) and \(H=H_{2}\) and \(k=k_{2}\) for \(h_{1} \leq y \leq h_{2} .\) For \(y>h_{2}, H=0\) and the upward heat flux is \(q_{m} .\) Determine the surface heat flow and temperature at \(y=h_{2}\) for \(\rho_{1}=2600 \mathrm{~kg} \mathrm{~m}^{-3}\), \(\rho_{2}=3000 \mathrm{~kg} \mathrm{~m}^{-3}, k_{1}=k_{2}=2.4 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) \(h_{1}=8 \mathrm{~km}, h_{2}=40 \mathrm{~km}, \rho_{1} H_{1}=2 \mu W \mathrm{~m}^{-3}, \rho_{2} H_{2}=\) \(0.36 \mu W \mathrm{~m}^{-3}, T_{0}=0^{\circ} \mathrm{C},\) and \(q_{m}=28 \mathrm{~mW} \mathrm{~m}^{-2}\).

Short answer

Surface heat flow is 60 mW/m² and temperature at y=h2 is 625.81°C.

Step-by-step solution

Calculate Heat Production Rates

The heat production rate within each layer must be determined. Use the given densities and heat production values:\[A_1 = \frac{\rho_1 H_1}{\rho_1} = 2 \mu W/m^3\] \[A_2 = \frac{\rho_2 H_2}{\rho_2} = 0.36 \mu W/m^3\] This confirms the given values for heat production in each layer.

Determine Temperature Gradient for First Layer

Using Fourier's Law, calculate the temperature gradient in the first layer:\[ \frac{dT}{dy} = -\frac{q + A_1 \cdot h_1}{k_1} \] \[ \frac{dT}{dy} = -\frac{28 \times 10^{-3} + 2 \times 10^{-6} \times 8000}{2.4} \] \[ \frac{dT}{dy} = -\frac{44}{2.4} \approx -18.33 \] degrees per km.

Determine Temperature at Interface h1

Using the temperature gradient from Step 2 and the boundary condition at the surface (\(T_0 = 0^{\circ}C\)), calculate the temperature at the interface \(y = h_1\):\[T_1 = T_{surface} - \frac{44}{2.4} \times 8 \approx 146.67 ^{\circ}C.\]

Determine Temperature Gradient for Second Layer

Apply the same principle for the second layer:\[ \frac{dT}{dy} = -\frac{q_m + A_2 \cdot (h_2 - h_1)}{k_2} \] \[ \frac{dT}{dy} = -\frac{28 \times 10^{-3} + 0.36 \times 10^{-6} \times 32000}{2.4} \] \[ \frac{dT}{dy} \approx -\frac{39.52}{2.4} \approx -16.47 \] degrees per km.

Calculate Temperature at y=h2

Using the gradient from Step 4, find the temperature at \(y = h_2\):\[T_2 = T_1 - \frac{39.52}{2.4} \times 32 \approx 625.81 ^{\circ}C.\]

Determine Surface Heat Flow

The surface heat flow \( q_0 \) is the sum of conductive heat flow and internally generated heat:\[q_0 = q_m + A_1 \cdot h_1 + A_2 \cdot (h_2 - h_1)\] \[q_0 = 28 \times 10^{-3} + 2 \times 10^{-6} \times 8000 + 0.36 \times 10^{-6} \times 32000\] \[q_0 = 60 \times 10^{-3} \] W/m².

Key concepts

Heat Production

Heat production refers to the generation of heat within the Earth’s interior due to radioactive decay of isotopes like uranium, thorium, and potassium. In this exercise, heat production differs in two layers of rocks beneath the Earth's surface. Layer 1 produces heat at a rate of \(2 \mu W/m^3\) and Layer 2 at \(0.36 \mu W/m^3\). This variation in rates illustrates how geological formations with different compositions contribute differently to the overall heat budget. To see how these rates impact temperature distribution, it's important to note:- Heat production influences how rapidly temperatures increase with depth.- More heat production typically means a steeper geothermal gradient.- Understanding heat production is crucial for geophysical studies and energy calculations.
This exercise precisely calculates these heat production rates and uses them to determine thermal characteristics of the Earth’s subsurface.

Fourier's Law

Fourier's Law helps explain how heat conducts through materials. It's crucial for calculating temperature gradients with given heat flux values in subsurface layers. The law can be stated mathematically as \( q = -k \frac{dT}{dy} \), which tells us:- \(q\) is the heat flux.- \(k\) is the material's thermal conductivity.- \(\frac{dT}{dy}\) is the temperature gradient.In the first layer, using Fourier’s Law, the calculation involves substituting known values to find the temperature gradient of \(-18.33\) degrees per km. Fourier's Law essentially tells us how efficiently heat is transferred.In practical terms:- Higher conductivity indicates easier heat flow.- Changes in heat flux and geothermal gradients can signal geological transitions.
This forms a basis to determine subsequent temperature changes throughout the layers in the exercise.

Thermal Conductivity

Thermal conductivity quantifies a material's ability to conduct heat. In the exercise, the rocks of both layers share the same thermal conductivity value of \(2.4 \text{ W m}^{-1} \, \text{K}^{-1}\). This means each layer conducts heat with the same efficiency despite having different heat production rates.Considerations about thermal conductivity include:- It helps determine how much temperature rises as you move deeper.- Materials with higher conductivity usually mean lower temperature gradients.- Experiments often measure conductivity to analyze subsurface energy transfer.
In sum, by knowing the thermal conductivity, the exercise calculates heat flow and temperature changes accurately across the two geologic layers.

Surface Heat Flow

Surface heat flow accounts for both conductive heat from below and heat produced within subsurface layers. It's indicative of the Earth's heat coming through to the surface. In this scenario, it combines conductive heat flow and heat released from radioactive decay.Surface heat flow is calculated as follows:\[ q_0 = q_m + A_1 \, h_1 + A_2 \, (h_2 - h_1) \]This expression sums:- \(q_m\): the mantle's upward heat flux (\(28 \text{ mW m}^{-2}\)).- \(A_1 h_1\) and \(A_2 (h_2 - h_1)\): internal heat production in layers.The combined heat flow amounts to \(60 \text{ mW m}^{-2}\) at the surface. Understanding surface heat flow is significant for:- Estimating geothermal energy resources.- Investigating tectonic and volcanic activities.
By understanding surface heat flow, geologists better predict energy distributions and thermal states within the Earth.