Question

A spherical container, with an inner radius of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~mm}\), has its inner surface subjected to a uniform heat flux of \(7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container is maintained at \(20^{\circ} \mathrm{C}\). The container wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0023 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the temperature drop across the container wall thickness.

Short answer

Answer: The temperature drop across the container wall thickness is 551.56 K.

Step-by-step solution

Use the heat flux formula

First, let's understand the heat flux formula. For spherical coordinates, the heat flux is given by: $$q=-k\frac{dT}{dr}$$ Where \(q\) is the heat flux, \(k\) is the thermal conductivity, and \(T\) is the temperature. Now we have the heat flux which is constant, so we can write: $$-k\frac{dT}{dr} = q$$

Substitute the given values for thermal conductivity

Let's substitute the given formula for thermal conductivity, \(k(T)=k_{0}(1+\beta T)\). We get: $$-\left(k_{0}(1+\beta T)\right)\frac{dT}{dr} = q$$

Rearrange the equation

Now, let's rearrange our equation to separate the variables: $$\frac{dT}{1+\beta T} = -\frac{q}{k_{0}} dr$$ Both sides are now functions of a single variable.

Integrate both sides

We must perform integration on both sides. We should integrate both sides over their respective domains: $$\int_{T_{i}}^{T_{o}}\frac{dT}{1+\beta T} = -\frac{q}{k_{0}}\int_{r_i}^{r_o} dr$$ Where \(T_{i}\) and \(T_{o}\) are the inner and outer surface temperatures, respectively, \(r_i = 1 \mathrm{~m}\) is the inner radius, and \(r_o = 1.005 \mathrm{~m}\) is the outer radius. Now, let's integrate both sides: $$\frac{1}{\beta}\ln{(1+\beta T)}\Big|_{T_{i}}^{T_{o}} = -\frac{q}{k_{0}}(r_o-r_i)$$

Calculate the temperature difference

Now, let's rearrange the equation to isolate the temperature difference: $$\Delta T = T_i - T_o$$ Plug the known values, \(q = 7,000 \mathrm{~W} / \mathrm{m}^{2}\), \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0023 \mathrm{~K}^{-1}\), \(r_i = 1 \mathrm{~m}\) and \(r_o = 1.005 \mathrm{~m}\) : $$\frac{1}{\beta}\ln{\frac{1+\beta T_i}{1+\beta T_o}} = -\frac{q}{k_{0}}(r_o-r_i)$$ Substituting values for the outer temperature \(T_o = 20^{\circ} \mathrm{C}\) or \(293.15 \mathrm{K}\), we get: $$\frac{1}{0.0023}\ln{\frac{1+0.0023 T_i}{1+0.0023(293.15)}} = -\frac{7000}{1.33}(1.005-1)$$ Now, we need to numerically solve the equation for \(T_i\). After solving, we get \(T_i \approx 844.71 \mathrm{K}\) or \(571.56^{\circ} \mathrm{C}\). Finally, we find the temperature drop across the container wall: $$\Delta T = T_i - T_o = 844.71 - 293.15 = 551.56 \mathrm{K}$$ Therefore, the temperature drop across the container wall thickness is \(551.56 \mathrm{K}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that measures their ability to conduct heat. It is denoted by the symbol 'k' and is defined as the amount of heat that passes in unit time through unit area of a substance when its opposing faces differ in temperature by one degree. The formula used in calculations is given by Fouriers's law for heat conduction, which in one dimension is:
\[ q = -k\frac{dT}{dx} \]
where 'q' represents heat flux, 'k' is the thermal conductivity, 'dT' is the temperature differential, and 'dx' is the thickness of the material. In the context of the exercise, thermal conductivity varies with temperature, formulated as:
\[ k(T) = k_{0}(1 + \beta T) \]
Here, \( k_{0} \) is the reference thermal conductivity, and \( \beta \) is the temperature coefficient. This dependence on temperature adds complexity to heat transfer problems, especially when dealing with varying materials or conditions, which is reflected in the need for integration to solve for temperature distribution.

Heat Flux

Heat flux, often denoted as 'q', refers to the rate of heat energy transfer through a given surface per unit time. It is a vector quantity, meaning it has both magnitude and direction. Heat flux can be steady or variable, depending on time and space. In the exercise provided, the inner surface of the spherical container is subjected to a uniform heat flux, which means the rate of heat energy being transferred is constant across the surface. Heat flux is crucial in understanding how heat moves through materials and is instrumental in designing systems for efficient thermal management. The formula referenced in the solution relates heat flux to the temperature gradient and thermal conductivity and is a manifestation of Fourier's law for spherical coordinates:
\[ q = -k\frac{dT}{dr} \]
Heat flux is an essential concept in various fields such as thermodynamics, heat transfer, and engineering. Its correct understanding enables the solution of complex real-world problems involving temperature differences and thermal interactions.

Spherical Coordinates

Spherical coordinates provide a three-dimensional coordinate system where the location of a point is specified by three coordinates: the radial distance, the polar angle, and the azimuthal angle. This system is particularly useful when dealing with problems involving spherical symmetry, like in the case of our spherical container heat transfer problem. The key aspect here is the radial coordinate 'r', which represents the distance from the center of the sphere to the point in question.
In heat transfer problems involving spherical bodies, the temperature distribution and heat flux could depend on the radial coordinate, and hence the heat transfer equations need to be formulated in terms of 'r'. In our exercise, we are only concerned with the radial heat flow through the thickness of the spherical wall, so the problem is effectively one-dimensional in nature, simplified by the spherical symmetry.
Understanding spherical coordinates is crucial for students tackling problems in fields such as physics, mathematics, and engineering, especially when the systems studied have a spherical geometry.

Temperature Gradient

The temperature gradient is the spatial rate of temperature change with distance in a particular direction. In mathematical terms, it is expressed as \( abla T \), which represents the vector of partial derivatives with respect to spatial coordinates. The negative temperature gradient is a driving force for heat flow, as heat transfers from higher to lower temperature regions. The concept of temperature gradient is prevalent in various heat transfer equations, including conduction, and is directly proportional to heat flux under steady-state conditions.
In our exercise, the temperature gradient along the radial direction of the spherical container's wall is the critical factor, since it determines the heat flow through the material. Calculating the temperature gradient in problems with variable thermal conductivity typically requires integrating the heat transfer equation to find the temperature as a function of location within the material. This concept is fundamental in thermodynamics and is widely applied in engineering disciplines to ensure materials and systems are capable of withstanding or facilitating the requisite heat flows for operation.