91Ó°ÊÓ

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Step by step solution

01

Express the differential equation for steady one-dimensional heat conduction through the container

The heat conduction equation for a steady one-dimensional heat conduction in a spherical coordinate system is given by: $$\frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0$$ where \(T\) is the temperature and \(r\) is the radial distance.

02

Express the boundary conditions

There are two boundary conditions for this problem. The first one is at the inner surface of the container, with an inner radius of \(r_1 = 2\) m, and a temperature of \(0^{\circ} \mathrm{C}\): $$T(r_1) = 0^{\circ} \mathrm{C}$$ The second boundary condition is at the outer surface of the container, with an outer radius of \(r_2 = 2.1\) m. Here, the heat transfer by convection is given by \(q = hA(T - T_{\infty})\), where \(A\) is the surface area of the sphere. Since the heat transfer due to conduction through the material is equal to the heat gained by convection, we can write: $$-kA\frac{dT}{dr}\Big|_{r=r_2} = hA(T(r_2)-T_{\infty})$$

03

Solve the differential equation for the temperature

To find the temperature variation in the container, we first need to solve the heat conduction equation: $$\frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0$$ Integrating with respect to \(r\), we get: $$r^2\frac{dT}{dr} = C_1$$ Dividing both sides by \(r^2\) and integrating again, we obtain: $$T(r) = -\frac{C_1}{r} + C_2$$

04

Apply the boundary conditions and find the constants

Now, we apply the boundary conditions to find the values of \(C_1\) and \(C_2\): For the first boundary condition, \(T(r_1) = 0^{\circ} \mathrm{C}\): $$0 = -\frac{C_1}{r_1} + C_2$$ For the second boundary condition, we have: $$-k \frac{dT}{dr}\Big|_{r=r_2} = h(T(r_2)-T_{\infty}) \Rightarrow -k\left(-\frac{C_1}{r_2^2}\right) = h\left(-\frac{C_1}{r_2} + C_2 - T_{\infty}\right)$$ Solve this system of two equations to find the values of \(C_1\) and \(C_2\) and substitute them back into the expression for the temperature: $$T(r) = -\frac{C_1}{r} + C_2$$

05

Evaluate the rate of heat gain to the iced water

Finally, the rate of heat gain to the iced water can be evaluated using the heat transfer coefficient and the temperature of the outer surface. We will use the relation: $$q = hA(T(r_2) - T_{\infty})$$ Use the obtained temperature function \(T(r)\) with the substituted values of \(C_1\) and \(C_2\), and plug in the values for \(h\), \(A\), and \(T_{\infty}\) to calculate the rate of heat gain to the iced water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steady-State Conduction

In steady-state heat conduction, the temperature distribution within the material does not change over time. This means that any heat added to one part of the system is balanced by an equal amount removed from another part. In this exercise, we have a steady-state scenario inside a spherical container, implying that the temperature field remains constant over time.
The governing principle here is that thermal energy enters and exits the container at the same rate, maintaining a stable temperature profile. This makes calculating the heat transfer rate more straightforward, as it can be done without considering transient effects or time-dependent changes in temperature.

Boundary Conditions

Boundary conditions are crucial for solving heat conduction problems because they define how the system interacts with its surroundings. For our spherical container, we have two main boundary conditions:

• At the inner surface ( _1 = 2 ext{ m} ), the temperature is held constant at 0°C, as it is in contact with iced water.
• At the outer surface ( _2 = 2.1 ext{ m} ), heat is transferred to the surrounding air by convection. This process is described by the equation -q = hA(T - T_{ ext{infinity}}) , where -h is the heat transfer coefficient.

These conditions let us link the mathematical model (the differential equation) to the physical system, ensuring the solution reflects the actual thermal behavior.

Differential Equation

The differential equation for one-dimensional steady-state heat conduction in spherical coordinates is critical for understanding how temperature varies across the container's shell. The equation is given by:
\[ \frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0 \]
This form arises because of the geometry involved; heat spreads radially, meaning it travels outward from the center. Solving this equation tells us how the temperature changes with radial distance, giving a clear thermal profile within the spherical object. It involves integrating twice to yield a general solution, \( T(r) = -\frac{C_1}{r} + C_2 \), where \( C_1 \) and \( C_2 \) are constants determined using boundary conditions.

Thermal Conductivity

Thermal conductivity ( -k ) is a measure of a material's ability to conduct heat. In this exercise, the spherical shell's material has a thermal conductivity of 30 W/m·K. This value indicates how effectively heat is transferred through the material when subjected to a temperature gradient.
High thermal conductivity means the material will quickly transfer heat, while a lower value would suggest slower conduction. Knowing the thermal conductivity is essential when calculating rates of heat transfer and solving the differential equation because it influences how much heat is transferred across the material's thickness under a given temperature difference.