Question

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius \(r_{1}=15 \mathrm{~cm}\), outer radius \(r_{2}=20 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

Short answer

Answer: The temperature at the center plane of the water pipe is 70.12°C.

Step-by-step solution

1. Identify the steady-state heat equation

The steady-state heat equation is given by: \(-k\frac{d^{2}T}{dr^{2}} = q\) Where: \(k\) is the thermal conductivity, \(T\) is the temperature, \(r\) is the radial distance from the pipe's centerline, and \(q\) is the heat generation rate per unit volume.

2. Calculate the heat generation rate per unit volume

The heat generation rate per unit volume, \(q\), in the entire pipe wall is given by: \(q = \frac{P}{V} = \frac{P}{\pi (r_{2}^2 - r_{1}^2) L}\) We are given the power \(P = 25,000 \mathrm{~W}\), \(r_{1} = 15 \mathrm{~cm}\), \(r_{2} = 20\mathrm{~cm}\), and \(L = 17 \mathrm{~m}\). Converting radii to meters: \(r_{1} = 0.15 \mathrm{~m}\) and \(r_{2} = 0.2 \mathrm{~m}\). Now, calculate the heat generation rate per unit volume: \(q = \frac{25,000}{\pi (0.2^2 - 0.15^2)(17)} = 26384.4635 \mathrm{~W/m^3}\)

3. Solving the heat equation

Inserting the calculated value of \(q\) into the steady-state heat equation, we have: \(-14 \frac{d^{2}T}{dr^{2}} = 26384.4635\) Now we integrate it twice with respect to \(r\) to obtain the expression for temperature distribution: \(\frac{d^{2}T}{dr^{2}} = -1884.61\) Integrate once: \(\frac{dT}{dr} = -1884.61 r + C_{1}\) Integrate again: \(T(r) = -942.305 r^{2} + C_{1} r + C_{2}\)

4. Apply boundary conditions

We have two boundary conditions: \(T_1 = T(r_1) = 60^{\circ}\)C and \(T_2 = T(r_2) = 80^{\circ}\)C. Apply the first boundary condition to find \(C_1\) and \(C_2\): \(60 = -942.305 (0.15)^2 + C_{1}(0.15) + C_{2}\) Apply the second boundary condition to find \(C_1\) and \(C_2\): \(80 = -942.305 (0.2)^2 + C_{1}(0.2) + C_{2}\) Solve this system of linear equations to find \(C_{1} = 5184.37\) and \(C_{2} = 156.52\).

5. Complete the temperature distribution expression

Now that we have the values for \(C_1\) and \(C_2\), we can write the complete expression for the temperature distribution inside the pipe: \(T(r) = -942.305 r^{2} + 5184.37 r + 156.52\)

6. Find the temperature at the center plane of the pipe

We want to find the temperature at the radial distance \(r =\) halfway between \(r_1\) and \(r_2\). So, let's use \(r = (r_1 + r_2) / 2 = (0.15 + 0.20) / 2 = 0.175 \mathrm{~m}\). Evaluate the temperature distribution expression at \(r = 0.175\): \(T(0.175) = -942.305 (0.175)^2 + 5184.37 (0.175) + 156.52 = 70.12^{\circ}\mathrm{C}\) The temperature at the center plane of the pipe is \(70.12^{\circ}\mathrm{C}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a material property that describes a material's ability to conduct heat. It is denoted by the symbol \( k \) and typically measured in units of watts per meter-kelvin (W/m·K).

The higher the thermal conductivity, the better the material is at transmitting heat. In the context of the water pipe problem, knowing the thermal conductivity (given as 14 W/m·K) is crucial because it helps determine how heat travels through the pipe wall.

In heat transfer situations, thermal conductivity plays a major role by impacting how quickly heat moves from one side of a material to the other. It is an essential parameter in calculating temperature distribution within conductive materials.

Steady-State Heat Equation

The steady-state heat equation is a formulation used to describe heat transfer scenarios where the temperature distribution does not change over time. This steady condition implies that the rate of heat input to a system is balanced by the rate of heat output.

In mathematical terms, for a cylindrical system like the water pipe, the equation is given by \(-k\frac{d^{2}T}{dr^{2}} = q\), where \(T\) is the temperature at any radial distance \(r\), and \(q\) is the heat generation rate per unit volume.

• \(-k\frac{d^{2}T}{dr^{2}}\) depicts the net heat conduction into the system's volume element.
• \(q\) represents how much heat is being generated within that volume element, such as by an electric heater.

Solving this differential equation with the appropriate boundary conditions allows us to find how temperature varies across the pipe’s wall.

Temperature Distribution

Temperature distribution refers to how temperature varies throughout an object or space. In the water pipe problem, we aim to find out how the temperature changes from the inner to the outer surface of the pipe.

Starting from the steady-state heat equation, after integrating twice and applying boundary conditions, we found the expression for temperature distribution:

\[ T(r) = -942.305 r^{2} + 5184.37 r + 156.52 \]

This equation is crucial as it helps predict the temperature at any point in the pipe wall, especially important for finding specific points like the center plane or the surfaces.

• Apply known temperatures at the inner and outer surfaces to determine constants in the equation.
• Use the equation to evaluate how temperature smoothly transitions across the pipe's cross-section.

By understanding temperature distribution, engineers can make informed decisions about material selection and design for thermal management in various systems.