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Chapter 11: Problem 13

Equilibrium temperature in a spherical shell. The equilibrium temperature inside a spherical shell, of inner radius \(a\) and outer radius \(b\), satisfies the differential equation $$ \frac{d}{d r}\left(r^{2} \frac{d U}{d r}\right)=0 $$ (a) Find the general solution of the differential equation. (b) Find the equilibrium temperature if the inner surface \(r=a\) is maintained at temperature \(u_{1}\) and the outer surface \(r=b\) is maintained at temperature \(u_{2}\).

Short Answer

Expert verified
The equilibrium temperature is \( U(r) = u_1 \frac{b-r}{b-a} + u_2 \frac{r-a}{b-a} \).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \( \frac{d}{d r} \left( r^{2} \frac{d U}{d r} \right) = 0 \). This indicates that the derivative of \( r^{2} \frac{d U}{d r} \) with respect to \( r \) is zero, which implies that \( r^{2} \frac{d U}{d r} \) is a constant. Let's denote this constant as \( C \). So we have \( r^{2} \frac{d U}{d r} = C \).
02

Solve the First Order Differential Equation

From \( r^{2} \frac{d U}{d r} = C \), we can express this as \( \frac{d U}{d r} = \frac{C}{r^{2}} \). Integrate this equation with respect to \( r \): \[ U(r) = \int \frac{C}{r^{2}} \, dr = -\frac{C}{r} + D \] where \( D \) is the constant of integration.
03

Evaluate Constants using Boundary Conditions

We have boundary conditions: at \( r = a \), \( U(a) = u_1 \) and at \( r = b \), \( U(b) = u_2 \). Substitute these into the general solution \( U(r) = -\frac{C}{r} + D \). For \( r = a \), \( u_1 = -\frac{C}{a} + D \). For \( r = b \), \( u_2 = -\frac{C}{b} + D \).
04

Solve for Constants C and D

Subtract the two equations derived from the boundary conditions to eliminate \( D \): \[ u_2 - u_1 = -\frac{C}{b} + \frac{C}{a} \] Re-arrange to get \( C \): \[ u_2 - u_1 = C \left( \frac{1}{a} - \frac{1}{b} \right) \] Hence, \( C = (u_2 - u_1) \frac{ab}{b-a} \). Substitute \( C \) back into one of the boundary condition equations, \( u_1 = -\frac{C}{a} + D \), to find \( D \): \[ D = u_1 + \frac{(u_2 - u_1)b}{b-a} \].
05

Form the Final Solution for U(r)

Substitute \( C \) and \( D \) into the general solution \( U(r) = -\frac{C}{r} + D \) to get the specific solution: \[ U(r) = - \frac{(u_2 - u_1) \frac{b}{b-a}}{r} + u_1 + \frac{(u_2 - u_1)b}{b-a} \]. Simplify it further for clarity if necessary.

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Key Concepts

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

Spherical Shell
A spherical shell is like a hollow sphere with two concentric surfaces: an inner surface at radius \( a \) and an outer surface at radius \( b \). This structure is common in many physical systems, such as in the design of pressure vessels or celestial bodies.
The main focus in problems involving spherical shells often involves calculating the characteristics within the hollow region between the two surfaces, such as temperature distribution or stress. Here, our task is to understand the distribution of temperature or equilibrium temperature inside a spherical shell.
By fixing temperatures at the inner and outer surfaces \( u_1 \) and \( u_2 \), respectively, we can determine how temperature behaves within the shell. Analyzing this requires foundational concepts like differential equations and boundary conditions.
Differential Equation
Differential equations are mathematical tools used to describe changes in functions with respect to one or more variables.
In our scenario, we deal with the differential equation \( \frac{d}{d r}\left( r^{2} \frac{d U}{d r}\right)=0 \). This equation helps us express the balance of heat flow within the spherical shell.
  • It suggests that the quantity \( r^{2} \frac{d U}{d r} \) is constant with respect to \( r \), offering a balance in the distribution of temperature.
  • Solving this equation involves integration, a process of finding a function whose derivative is the given function. Here, it gives us insights into how temperature \( U \) varies with radius \( r \).
Understanding differential equations allows us to explore complex systems systematically, making them crucial for mathematical modeling of physical phenomena.
Boundary Conditions
Boundary conditions are critical in solving differential equations because they provide necessary information to determine the specific solution that applies to a given physical situation.
In our exercise, we have two boundary conditions:
  • At \( r = a \), the temperature is \( U(a) = u_1 \).
  • At \( r = b \), the temperature is \( U(b) = u_2 \).
These conditions allow us to solve for the constants arising from the integration of our differential equation. They help determine the specific temperature distribution inside the shell that matches the given physical scenario.
Applying boundary conditions is like setting limits in real-world problems, ensuring the solutions we derive are both meaningful and accurate to the situation we intend to model. This process is vital in tailoring mathematical solutions to match physical realities.
Mathematical Modelling
Mathematical modeling involves using equations and concepts to represent real-world systems and predict their behavior.
In the context of our spherical shell scenario, mathematical modeling allows us to calculate the temperature distribution, a vital aspect for numerous engineering applications.
  • The model simplifies complex physical behavior into manageable equations.
  • By applying differential equations like \( \frac{d}{d r}\left( r^{2} \frac{d U}{d r}\right)=0 \), it enables precise predictions.
The power of mathematical modeling lies in its ability to represent and analyze systems, leading to solutions that can be practically implemented. It's a foundational tool in science and engineering, providing insights and guiding design and decision-making processes.

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Most popular questions from this chapter

Avascular tumour with phase-two growth. Consider a tumour that has already reached the phase-two growth stage. Assume that the proliferating layer absorbs oxygen at a constant rate \(A=A_{0} \mathrm{ml}\) per second per unit volume of cells, and let \(c_{p}(r)\) be the axygen concentration here. The governing equation is $$ \frac{D}{r^{2}} \frac{d}{d r}\left(r^{2} \frac{d C_{P}}{d r}\right)-A_{0}=0 . \quad R_{q}(t)

Proliferating tumour. For a spherical tumour of radius \(R\), in the first phase of growth, the equilibrium oxygen concentration \(C_{p}(r)\) satisfies $$ \frac{D}{r^{2}} \frac{d}{d r}\left(r^{2} \frac{d C_{p}}{d r}\right)-A_{0}=0, \quad 0

Formulating a boundary value problem. Write, in mathematical form, boundary conditions for the following. Express them in terms of temperature \(U\) or heat flux \(J\). (a) A \(1 \mathrm{~m}^{2}\) section of a furnace wall gains heat from the end \(x=0\) at a fixed rate of \(300 \mathrm{~W}\), while the other end is maintained at temperature \(30^{\circ} \mathrm{C}\). (b) The outside of a wall of a house loses heat according to Newton's law of cooling, to the surrounding air at temperature \(10^{\circ} \mathrm{C}\), while the inside gains heat according to Newton's law of cooling from the inside of the house, which is at a temperature of \(25^{\circ} \mathrm{C}\). (c) A slab of material has its right end held at temperature \(80^{\circ} \mathrm{C}\) and the left end gaining heat according to Newton's law of cooling from the surroundings at temperature \(100^{\circ} \mathrm{C}\).

Solving a boundary value problem. Consider the differential equation $$ \frac{d^{2} U}{d x^{2}}=1 $$ which satisfies the boundary conditions $$ U(0)=1, \quad U(2)=0 $$ (a) Find the general solution. (b) Apply the boundary conditions to find the solution. (c) Suppose the two boundary conditions are replaced with $$ U(0)=1, \quad \frac{d U}{d x}(2)=0 $$ Apply these boundary conditions and find the solution. (d) Suppose the two boundary conditions are replaced with $$ \frac{d U}{d x}(0)=1, \quad \frac{d U}{d x}(2)=U(2) . $$ Apply these boundary conditions and find the solution.

Heat flux and thermal resistance. Suppose we have two different materials with widths \(d_{1}\) and \(d_{2}\), conductivities \(k_{1}\) and \(k_{2}\), and which are joined together at \(x=0\). The equilibrium temperatures \(U_{1}(x)\) and \(U_{2}(x)\) both satisfy the basic equilibrium heat equations $$ \frac{d^{2} U_{1}}{d x^{2}}=0, \quad \frac{d^{2} U_{2}}{d x^{2}}=0 $$ The temperature on the inside \(\left(x=-d_{1}\right)\) is held (by a thermostat) at a temperature \(u_{i}\) and the temperature on the outside \(\left(x=d_{2}\right)\) is held (by a thermostat) at a temperature \(u_{p} .\) (a) Assuming the temperature and heat flux are continuous where the materials join at \(x=0\), what are the boundary conditions? (b) Show that the heat flux is given by $$ J=\frac{u_{i}-u_{0}}{\frac{d_{1}}{k_{1}}+\frac{d_{2}}{k_{2}}} $$ (c) Interpret this in terms of thermal resistances.
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