Question

Consider a chilled-water pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Water flows in the pipe at a temperature \(T_{f}\) and the heat transfer coefficient at the inner surface is \(h\). If the pipe is well-insulated on the outer surface, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe and \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation.

Short answer

Answer: As the radial distance (r) increases, the temperature of the pipe decreases.

Step-by-step solution

Formulate the differential equation for heat conduction

For steady one-dimensional heat conduction in the radial direction, the heat conduction equation can be written as: \(\frac{d}{dr} (k \frac{d T}{dr}) = 0\) where \(T\) is the temperature and \(r\) is the radial coordinate. Since the pipe's thermal conductivity \(k\) is constant, the equation simplifies to: \(\frac{d^2 T}{dr^2} = 0\)

Determine the boundary conditions

We need to consider two boundary conditions for this problem: one at the inner surface (r = \(r_{1}\)) and one at the outer surface (r = \(r_{2}\)). At the inner surface (r = \(r_{1}\)), the boundary condition is given by the heat transfer coefficient \(h\). We have: \(-k\frac{dT}{dr} \Big|_{r=r_{1}} = h(T_{f}-T_{1})\) where \(T_{1}\) is the temperature at the inner surface of the pipe (\(r=r_1\)). At the outer surface (r = \(r_{2}\)), the boundary condition is given by the insulation, which implies no heat flow through the outer surface: \(\frac{dT}{dr} \Big|_{r=r_{2}} = 0\)

Solve the differential equation

Now, let's integrate the differential equation \(\frac{d^2 T}{dr^2} = 0\) with respect to \(r\): \(\frac{dT}{dr} = C_{1}\) Integrate again with respect to \(r\): \(T(r) = C_{1}r + C_{2}\)

Apply boundary conditions

We can find the constants \(C_{1}\) and \(C_{2}\) by applying the boundary conditions. First, apply the boundary condition at the inner surface \(r=r_1\): \(-k\frac{dT}{dr} \Big|_{r=r_{1}} = -kC_{1} = h(T_{f}-T_{1}) \implies C_{1}=-\frac{h(T_{f}-T_{1})}{k}\) Next, apply the boundary condition at the outer surface: \(\frac{dT}{dr} \Big|_{r=r_{2}} = -\frac{h(T_{f}-T_{1})}{k} = 0\) This implies that the temperature at the outer surface is the same as that of the fluid temperature, i.e., \(T_{2}=T_{f}\). Finally, plug the value of \(C_{1}\) into the temperature equation \(T(r) = C_{1}r + C_{2}\) and apply the boundary condition at the inner surface: \(T_{1} = -\frac{h(T_{f}-T_{1})}{k}r_{1} + C_{2} \implies C_{2} = T_{1} + \frac{h(T_{f}-T_{1})}{k}r_{1}\)

Obtain the temperature distribution

With the constants \(C_1\) and \(C_2\) found, we can now write the temperature distribution along the radial direction of the pipe: \(T(r) = -\frac{h(T_{f}-T_{1})}{k}r + T_{1} + \frac{h(T_{f}-T_{1})}{k}r_{1}\) This equation describes the variation of temperature in the pipe for given values of \(k\), \(h\), \(T_{f}\), and \(T_{1}\).

Key concepts

Steady One-Dimensional Heat Conduction

Understanding steady one-dimensional heat conduction is crucial for analyzing the thermal behavior of systems like pipes. Imagine heat moving through a pipe wall at a consistent rate, unaffected by time. This kind of heat transfer is typically described by a temperature gradient along a single spatial dimension, with no variations over time. Thus, 'steady' implies no change in temperature at any point as time progresses, and 'one-dimensional' indicates that the temperature changes only in one direction—in this case, radially across the pipe wall.

In heat conduction problems like our textbook example, we look for a temperature distribution that does not change with time. This concept is fundamental in calculating the heat loss or gain in a pipe, determining how materials will perform under thermal stress and designing systems for thermal efficiency.

Thermal Conductivity

Thermal conductivity (\(k\)) is a material-specific property that measures how well a material conducts heat. High conductivity materials, like metals, transfer heat quickly, while low conductivity materials, such as insulators, do so much more slowly. In the context of our pipe, the thermal conductivity determines how the temperature drops from the hot inner surface to the cooler outer surface. The value of \(k\) will affect how steep this temperature gradient is. Knowing thermal conductivity is essential for engineers and designers when selecting materials for construction or insulation purposes. A correct understanding of \(k\) helps in predicting how the material will respond to temperature changes and aids in optimizing energy efficiency in thermal systems.

Boundary Conditions

Boundary conditions are like the 'rules of engagement' for solving heat conduction problems. They specify the conditions at the boundaries of the system. In the pipe example, we have two boundaries: the inner surface where the water flows and the outer surface. At the inner boundary, we know the heat transfer coefficient, also known as the film coefficient or \(h\), and the temperature of the fluid, making it possible to relate the temperature gradient in the pipe to the heat being removed by the flowing water. At the outer surface, the fact that the pipe is well-insulated means no heat is leaving, leading to a zero temperature gradient at this boundary. These boundary conditions are essential in determining the constants in the temperature distribution equation, eventually allowing us to compute the temperature at any point in the pipe wall.

Differential Equation

When dealing with heat conduction, we often use differential equations to describe how temperature changes across a medium. The differential equation arises from applying the physical laws of thermal conduction, represented mathematically. In the case of the pipe, we derived the equation \(\frac{d^2 T}{dr^2} = 0\) based on the principle of energy conservation. Solving this equation, as seen in the exercise, allows us to find a general expression for the temperature profile. We then apply the boundary conditions to pin down the exact form of this profile, yielding the specific temperature at any point along the pipe's radius. Understanding and solving these differential equations are essential skills for scientists and engineers, as they are the key to predicting system behavior in countless applications.