Question

Water at \(100^{\circ}\) is flowing through a long pipe of radius 1 rapidly enough so that we may assume that the temperature is \(100^{\circ}\) at all points. At \(t=0,\) the water is turned off and the surface of the pipe is maintained at \(40^{\circ}\) from then on (neglect the wall thickness of the pipe). Find the temperature distribution in the water as a function of \(r\) and \(t .\) Note that you need only consider a cross section of the pipe. Answer: \(\quad u=40+\sum_{m=1}^{\infty} \frac{120}{k_{m} J_{1}\left(k_{m}\right)} J_{0}\left(k_{m} r\right) e^{-\left(\alpha k_{m}\right)^{2} t}, \quad\) where \(J_{0}\left(k_{m}\right)=0\).

Short answer

The temperature distribution is \[ u(r,t) = 40 + \sum_{m=1}^{\infty} \frac{120}{k_m J_1(k_m)} J_0(k_m r)e^{-(\alpha k_m)^2 t} \] where \( J_0(k_m) = 0 \).

Step-by-step solution

- Define the problem in terms of the heat equation

The problem involves solving the heat equation in cylindrical coordinates due to radial symmetry. The heat equation is \[ \frac{\partial u}{\partial t} = \alpha \left( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} \right) \] where \( u(r,t) \) is the temperature distribution and \( \alpha \) is the thermal diffusivity.

- Apply initial and boundary conditions

The initial condition is that the temperature is uniformly \(100^{\circ}\) throughout the pipe: \[ u(r,0) = 100 \] The boundary condition is that the surface temperature of the pipe is maintained at \(40^{\circ}\), i.e., \[ u(a,t) = 40 \] where \(a = 1\) is the radius of the pipe.

- Use separation of variables to solve the equation

Assume a solution of the form \( u(r,t) = T(t) R(r) \). Substituting into the heat equation, we obtain \[ \frac{1}{\alpha T} \frac{dT}{dt} = \frac{1}{R} \left( \frac{d^2 R}{dr^2} + \frac{1}{r} \frac{dR}{dr} \right) = -\lambda \] where \(\lambda\) is a separation constant. This gives two ordinary differential equations: \[ \frac{dT}{dt} + \alpha \lambda T = 0 \] and \[ r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + \lambda r^2 R = 0 \]

- Solve the radial part of the equation

The radial part is a form of Bessel's equation: \[ r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + ( k_m r)^2 R = 0 \] where \( k_m = \sqrt{\lambda} \). The solutions are Bessel functions of the first kind, so \[ R(r) = J_0(k_m r) \] Applying the boundary condition at \(r = a\), we obtain the eigenvalues \(k_m\) determined from \(J_0(k_m a) = 0 \).

- Solve the time part of the equation

For the time part, \[ \frac{dT}{dt} + \alpha k_m^2 T = 0 \] The solution is \[ T(t) = C_m e^{-\alpha k_m^2 t} \]

- Form the general solution

Combining the solutions from step 4 and step 5, the general solution is a series: \[ u(r,t) = \sum_{m=1}^{\infty} C_m J_0(k_m r) e^{-\alpha k_m^2 t} \]

- Apply initial condition to determine coefficients

Using the initial condition \( u(r,0) = 100 \), the coefficients are determined by \[ 100 = \sum_{m=1}^{\infty} C_m J_0(k_m r) \] Using orthogonality of Bessel functions, we find \[ C_m = \frac{2 \int_0^a r J_0^2(k_m r) dr}{a^2 J_1^2(k_m)} 100 \] Simplifying, \[ C_m = \frac{120}{k_m J_1(k_m)} \]

- Final temperature distribution expression

Substitute \( C_m \) back into the general solution to get the final temperature distribution: \[ u(r,t) = 40 + \sum_{m=1}^{\infty} \frac{120}{k_m J_1(k_m)} J_0(k_m r) e^{- (\alpha k_m)^2 t} \]

Key concepts

Bessel functions

Bessel functions are a type of special function that often appear in problems with cylindrical or spherical symmetry. They are solutions to Bessel's differential equation:
\[ r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + (k_m r)^2 R = 0 \]
For our problem, the solutions for the radial part are the Bessel functions of the first kind, denoted as \( J_0(k_m r) \). These functions are important because they help describe how temperature or other properties vary with radius in cylindrical coordinates.
Bessel functions are characterized by their oscillatory nature, and they have a series of zeros where they cross the x-axis. The specific modes or eigenvalues, \( k_m \), are determined by the condition \( J_0(k_m a) = 0 \), where \( a \) is the radius of the pipe in our case. These eigenvalues describe different vibration or thermal modes.

thermal diffusivity

Thermal diffusivity, represented by the symbol \( \alpha \), is a measure of how quickly heat spreads through a material. It is defined by the equation:
\[ \alpha = \frac{k}{\rho c} \]
where:

• \( k \) is the thermal conductivity
• \( \rho \) is the density
• \( c \) is the specific heat capacity

In our heat equation, thermal diffusivity plays a crucial role in determining how the temperature distribution evolves over time. The heat equation in cylindrical coordinates, given by:
\[ \frac{\partial u}{\partial t} = \alpha \left( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} \right) \]
shows that \( \alpha \) affects the rate at which temperatures change. Higher thermal diffusivity means heat will spread more quickly through the material. Understanding \( \alpha \) is vital for solving transient heat conduction problems as it impacts how fast the temperature in the pipe adjusts to the boundary conditions applied.

separation of variables

Separation of variables is a mathematical method used to solve partial differential equations (PDEs), like the heat equation. The technique involves assuming that the solution can be written as the product of functions, each dependent on a single coordinate or variable.
For the heat equation in cylindrical coordinates, we assume a solution of the form:
\[ u(r,t) = T(t)R(r) \]
Substituting this form into the heat equation allows us to separate the variables such that:
\[ \frac{1}{\alpha T} \frac{dT}{dt} = \frac{1}{R} \left( \frac{d^2 R}{dr^2} + \frac{1}{r} \frac{dR}{dr} \right) = -\lambda \]
where \( \lambda \) is a separation constant. This results in two ordinary differential equations (ODEs): one for \( R(r) \) and one for \( T(t) \).

• The time-dependent part:
  \( \frac{dT}{dt} + \alpha \lambda T = 0 \).
  The solution is an exponential decay: \( T(t) = C_m e^{-\alpha k_m^2 t} \).
• The radial part:
  \( r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + \lambda r^2 R = 0 \).
  This is Bessel's equation whose solutions are the Bessel functions: \( R(r) = J_0(k_m r) \).

This method simplifies complex PDE problems into more manageable ODEs, making it easier to find a solution by addressing each variable separately.