Question

A circular metal pipe has a wall thickness of \(10 \mathrm{~mm}\) and an inner diameter of \(10 \mathrm{~cm}\). The pipe's outer surface is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\) and has a temperature of \(500^{\circ} \mathrm{C}\). The metal pipe has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0012 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the inner surface temperature of the pipe.

Short answer

Answer: The inner surface temperature of the pipe is 210.82°C.

Step-by-step solution

Convert given data to proper units

Let's convert the given data to the appropriate SI units. The inner diameter and wall thickness must be in meters and the temperature in Kelvin. Inner diameter: \(10 \mathrm{~cm} = 0.1 \mathrm{~m}\) Wall thickness: \(10 \mathrm{~mm} = 0.01 \mathrm{~m}\) Outer surface temperature: \(500^{\circ} \mathrm{C} = 500 + 273.15 = 773.15 \mathrm{~K}\)

Calculate outer and inner pipe radii

We will now calculate the outer radius \(r_o\) and inner radius \(r_i\) of the pipe using the given inner diameter and wall thickness. \(r_i = \frac{\text{inner diameter}}{2} = \frac{0.1 \mathrm{~m}}{2} = 0.05 \mathrm{~m}\) \(r_o = r_i + \text{wall thickness} = 0.05 \mathrm{~m} + 0.01 \mathrm{~m} = 0.06 \mathrm{~m}\)

Calculate heat flux and temperature gradient

Next, we will calculate the heat flux \(q\) through the wall and the temperature gradient \(\frac{dT}{dr}\). Heat flux through the wall: \(q = -k(T) \frac{dT}{dr}\). Temperature gradient: \(\frac{dT}{dr} = -\frac{1}{k(T)}\cdot q\).

Integrate the temperature gradient to find temperature distribution

Now, we will integrate the temperature gradient and solve for the temperature distribution across the pipe wall. First, substitute the variable thermal conductivity \(k(T)=k_{0}(1+\beta T)\) into the temperature gradient equation: \(\frac{dT}{dr} = -\frac{1}{k_0(1+\beta T)}\cdot q = -\frac{q}{k_{0}(1+\beta T)}\). Now, integrate the temperature gradient equation with respect to \(r\), using the boundary conditions \(T(r=r_i)=T_i\) and \(T(r=r_o)=773.15 \mathrm{~K}\): \(\int_{T_i}^{773.15} \frac{-d T}{1+\beta T} = \int_{0.05}^{0.06} \frac{q \cdot d r}{k_0}\), Let \(\tau = 1 + \beta T\), then \(d\tau = \beta dT\): \(\int_{(1+\beta T_i)}^{(1+\beta773.15)} \frac{-d\tau}{\tau} = \frac{q\cdot0.01}{k_0}\). Now, apply the integral of \(\frac{d\tau}{\tau}\): \(-\ln{(1+\beta773.15)} + \ln{(1+\beta T_i)} = \frac{q\cdot0.01}{k_0}\).

Solve for inner surface temperature \(T_i\)

Finally, we will solve for the inner surface temperature, \(T_i\), using the equation obtained in step 4. Rearrange the equation to isolate \(T_i\): \(T_i = \frac{1}{\beta}\left( \left( 1+\beta773.15 \right)e^{\frac{q\cdot0.01}{k_0}} - 1 \right)\). Plug in the values of \(q = 5 \times 10^3 \mathrm{~W} \mathrm{m^{-2}}\), \(k_0 = 7.5 \mathrm{~W} \mathrm{m^{-1}} \mathrm{K^{-1}}\) and \(\beta = 0.0012 \mathrm{~K^{-1}}\): \(T_i = \frac{1}{0.0012}\left( \left( 1 + 0.0012 \cdot 773.15 \right)e^{\frac{5\times10^3\cdot0.01}{7.5}} - 1 \right) \approx 483.97 \mathrm{~K}\). Convert the temperature back to Celsius: \(T_i = 483.97 - 273.15 = 210.82^{\circ} \mathrm{C}\). So, the inner surface temperature of the pipe is \(210.82^{\circ} \mathrm{C}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It defines how quickly heat can pass through a substance. The greater the thermal conductivity, the faster heat will transfer through the material. In our exercise, the thermal conductivity is not a constant value but varies with the temperature of the metal pipe. This variation is expressed using the formula:

• \( k(T) = k_0(1 + \beta T) \)

Here, \( k_0 \) is the base thermal conductivity at absolute zero, and \( \beta \) is a small coefficient indicating how much the conductivity changes per degree of temperature.Understanding this concept helps us appreciate why different materials are used in different applications based on how well they conduct heat. For example, metals generally have high thermal conductivity, making them suitable for heat exchangers and cooking utensils.

Heat Flux

Heat flux is a measure of the rate at which heat energy passes through a surface. In our exercise, the heat flux is given as \(5 \, \text{kW/m}^{2}\). This value indicates how much heat is being transferred through each square meter of the pipe's surface every second. Heat flux is crucial for determining the efficiency of thermal systems. By measuring the heat flux, engineers can figure out if a system is distributing or removing heat as intended. It's like checking how efficiently a radiator is heating a room or how well a cooler is lowering temperatures. The heat flux value helps us calculate the heat transfer through the pipe wall and contributes to determining the temperature difference across the material.

Temperature Gradient

The temperature gradient describes how temperature changes across a material. Specifically, it measures the temperature difference between two points per unit of distance. For the metal pipe, it shows how the outer temperature of \(773.15 \, \text{K}\) changes as we move towards the cooler internal surface.The formula we use is:

• \( \frac{dT}{dr} = -\frac{1}{k(T)} \cdot q \)

Here, \( \frac{dT}{dr} \) represents the temperature gradient, \( k(T) \) is the variable thermal conductivity, and \( q \) is the heat flux. The negative sign indicates that temperature decreases as you move into the material.Understanding temperature gradients helps engineers design systems with the appropriate material thickness and thermal properties to achieve desired temperature distributions.

Temperature Distribution

Temperature distribution refers to how temperature is spread out within an object. In our problem, we calculated the temperature distribution across the metal pipe's thickness to find the inner surface temperature.To determine this, we started by integrating the temperature gradient:

• \( \int_{T_i}^{773.15} \frac{-d T}{1+\beta T} = \int_{0.05}^{0.06} \frac{q \cdot d r}{k_0} \)

This technique helps us predict how thermal energy behaves inside the material. Finding the correct temperature distribution is vital as it determines if the pipe can handle the stress caused by temperature differences without failing. By solving the temperature distribution equation, we established that the inner surface temperature is approximately \(210.82^{\circ} \text{C}\), ensuring understanding of heat flow dynamics.