Question

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

Short answer

Answer: The approximate surface temperature of the spherical radioactive material in steady operation is 494°C.

Step-by-step solution

Identify the given variables

Diameter of the sphere, D = 3 cm Heat generation rate, Q_gen = 15 W/cm³ Surrounding medium temperature, T_infinity = 25 °C Heat transfer coefficient, h = 120 W/m²K

Calculate the heat generation volume and energy

Calculate the volume of the spherical material: \(V = \dfrac{4}{3}\pi \left(\dfrac{D}{2}\right)^{3} = \dfrac{4}{3}\pi \left(\dfrac{3}{2}\right)^{3} cm^{3}\) Calculate the total heat generation energy in Watts: \(Q_{total} = Q_{gen} \cdot V = 15 \cdot \dfrac{4}{3}\pi \left(\dfrac{3}{2}\right)^{3} W\)

Calculate the surface area of the sphere

Calculate the surface area of the sphere: \(A_{s} = 4\pi \left(\dfrac{D}{2}\right)^{2} = 4\pi \left(\dfrac{3}{2}\right)^{2} cm^{2}\) Convert the surface area from \(cm^{2}\) to \(m^{2}\): \(A_{s_m} = A_{s} \cdot \dfrac{1}{10000}=\dfrac{4\pi \left(\dfrac{3}{2}\right)^{2}}{10000} m^{2}\)

Apply energy balance and solve for the surface temperature

For steady operation, the energy generated equals the heat transfer rate: \(Q_{total} = h \cdot A_{s_m} \cdot (T_s - T_{\infty})\) Solve for the surface temperature, \(T_s\): \(T_s = T_{\infty} + \dfrac{Q_{total}}{h \cdot A_{s_m}}= 25 + \dfrac{15 \cdot \dfrac{4}{3}\pi \left(\dfrac{3}{2}\right)^{3}}{120 \cdot \dfrac{4\pi \left(\dfrac{3}{2}\right)^{2}}{10000}}\) \(T_s \approx 494^{\circ} \mathrm{C}\) The surface temperature of the material in steady operation is approximately \(494^{\circ} \mathrm{C}\), which corresponds to option (c).

Key concepts

Spherical Heat Generation

Understanding the concept of spherical heat generation is crucial when dealing with materials that release or absorb heat in all directions. This heat generation can be uniform or vary within the volume of the material. In our exercise, we encounter a spherical radioactive material with a uniform heat generation rate, which simplifies our calculations.

To illustrate, imagine a tiny heater placed at every point inside the sphere, each producing heat at the same rate. The total heat generated by the sphere can be calculated by multiplying the volume of the sphere with the given heat generation rate per unit volume. Here, the key takeaway is that the heat generation within a sphere is directly proportional to its volume, considering a uniform rate.

In practical applications, spherical heat generation is common in nuclear reactors, isotropic light bulbs, and biological cells undergoing metabolic processes.

Steady State Heat Transfer

The concept of steady state heat transfer is a fundamental principle in thermodynamics, which implies that the temperature within the system does not change with time. This means that the amount of heat generated within the sphere is equal to the amount of heat leaving the surface to the surrounding environment.

In steady state conditions, all temperatures within the system are constant and there is no accumulation of heat. In our exercise, we focus on the surface temperature of the sphere, which remains constant during steady operation. Understanding steady state conditions helps in simplifying the calculations of heat transfer problems and is essential for designing thermal management systems in engineering applications.

Heat Transfer Coefficient

The heat transfer coefficient is a measure of the heat transfer rate per unit area and per unit temperature difference between the solid surface and the surrounding fluid. It is denoted by 'h' and typically has units of W/m²K. This coefficient is influenced by various factors including the type of fluid, fluid velocity, surface roughness, and temperature difference.

The higher the heat transfer coefficient, the more efficient the heat transfer from the surface to the fluid. In the exercise, we are given a heat transfer coefficient of 120 W/m²K, which we use to calculate how efficiently the thermal energy is dissipated from the radioactive material to the surrounding medium, ultimately affecting the surface temperature of the material.

Thermal Energy Balance

Thermal energy balance involves equating the rate of heat generation within an object to the rate of heat loss to its surroundings, ensuring that the energy within the system remains constant over time. In other words, it is an application of the first law of thermodynamics to heat transfer.

When we apply the thermal energy balance to our exercise, it allows us to solve for the unknown surface temperature. We express the heat generation of the sphere in terms of its geometry and given heat generation rate, and set it equal to the heat loss calculated using the heat transfer coefficient, surface area, and temperature difference. The thermal energy balance is crucial in predicting the thermal behavior of objects and is widely used in engineering to ensure the safe and efficient operation of thermal systems.