Question

Consider a steady-state heat conduction equation for the Earth’s crust where \( A \) is the heat source and \( k \) is the thermal conductivity. Given the boundary conditions: \( T = 0 \) at the surface and \( Q \) is the heat flux at the surface, Which one of the following options would be the temperature \( T \) at depth \( z \)?

• A. \( \frac{-(Az + 2Q)z}{2k} \)
• B. \( \frac{-(Az + 2Q)z}{k} \)
• C. \( \frac{-(Az + Q)z}{2k} \)
• D. \( \frac{-(A + 2Q)z^2}{2k} \)

Hint:

In heat conduction problems with given boundary conditions, ensure you account for both the heat source and flux when determining the temperature profile.

Answer 1

The Correct Option is A

Step 1: Heat conduction equation.
The steady-state heat conduction equation in one dimension is given by:
\[ \frac{d^2 T}{dz^2} = -\frac{A}{k} \] where A is the volumetric heat source (W/m³) and k is the thermal conductivity (W/m·K).

Step 2: General solution.
Integrating once with respect to z, we obtain:
\[ \frac{dT}{dz} = -\frac{Az}{k} + C_1 \] where C₁ is the first constant of integration.

Integrating again:
\[ T(z) = -\frac{Az^2}{2k} + C_1z + C_2 \] where C₂ is the second constant of integration.

Step 3: Applying boundary conditions.
Let the surface be at z = 0 with temperature T = 0, and surface heat flux be Q = -k \(\frac{dT}{dz}\) at z = 0.

Applying T(0) = 0 gives: \[ T(0) = 0 = -\frac{A \cdot 0^2}{2k} + C_1 \cdot 0 + C_2 \Rightarrow C_2 = 0 \] Using the heat flux at the surface: \[ Q = -k \left.\frac{dT}{dz}\right|_{z=0} = -k \left( -\frac{A \cdot 0}{k} + C_1 \right) = -k \cdot C_1 \Rightarrow C_1 = -\frac{Q}{k} \] Substituting back: \[ T(z) = -\frac{Az^2}{2k} - \frac{Q}{k}z \] or written as: \[ T(z) = \frac{-(Az + 2Q)z}{2k} \]

Step 4: Conclusion.
Thus, the temperature profile with internal heating and surface heat flux is:
\[ T(z) = \frac{-(Az + 2Q)z}{2k} \]