91Ó°ÊÓ

In steady-state heat transfer, the temperature across a material does not change with time. This implies that the rate at which heat enters a material is equal to the rate at which it leaves.

In the given exercise, the heat generated by the electric heater inside the hollow aluminum sphere is constant at 80 W. This steady energy production ensures that the thermal gradients, or temperature differences, remain constant throughout the system. The inner surface of the aluminum sphere is kept at a steady temperature of 250°C.

This specific condition allows us to use mathematical models that do not account for changes in time, simplifying our calculations. It lets us accurately determine values like the thermal conductivity of the insulation material.

Spherical geometry is essential in understanding how heat is distributed in this problem. Unlike flat surfaces, heat transfer in spheres (two concentric spherical surfaces in this case) involves radial symmetry.

The aluminum sphere has different inner and outer radii: 0.15 m and 0.18 m, respectively. The thickness of the insulation adds another layer of complexity, creating a third spherical surface at a radius of 0.30 m.

Given the geometry, the equations for heat transfer must account for the changes in the surface area with respect to the radius. The mathematical expressions involve logarithms, as you might have noticed with the fraction terms in the conduction equation: \[ Q = \frac{4\pi k(T_1 - T_2)}{\frac{1}{r_2} - \frac{1}{r_3}} \]
These expressions represent how the curvature of the surfaces affects the heat transfer processes, making spherical geometry a crucial concept for solving the problem correctly.

In our exercise, we have two main types of heat transfer: conduction and convection.

• **Conduction** involves heat transfer through a solid material, in this case, through the aluminum and the insulation. It relies on the material properties, specifically thermal conductivity. The equation for conduction, \[ Q = \frac{4\pi k(T_1 - T_2)}{\frac{1}{r_2} - \frac{1}{r_3}} \], shows how the thermal conductivity \( k \) influences the rate of heat transfer.


• **Convection** occurs at the outer surface of the insulation. Here, heat is transferred away to the surrounding air, facilitated by the movement of fluid. The convection heat transfer is described using \[ Q = h \cdot 4\pi r_3^2 (T_2 - T_∞) \], where \( h \) is the convective heat transfer coefficient, describing the efficiency of this process.

The crucial insight is that, at steady-state, the rates of conduction and convection are in balance. Knowing this allows us to equate the conduction model to the convection model to solve for unknown parameters, such as \( k \). This linked approach underlines the importance of understanding both conduction and convection principles to accurately find solutions to thermal problems.