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: To find the inner surface temperature of the pipe, follow these steps: 1. Convert the given wall thickness and inner diameter into SI units: 0.01 m and 0.1 m, respectively. 2. Calculate the pipe's inner and outer radii: 0.05 m and 0.06 m, respectively. 3. Write down expressions for the temperature gradient and heat transfer rate. 4. Write the heat transfer rate equation at the outer surface, and convert the outer surface temperature to Kelvin, obtaining 773.15 K. 5. Write the thermal conductivity equation at the outer surface. 6. Solve for the temperature gradient at the outer surface. 7. Integrate the temperature gradient with respect to radius to find the temperature distribution. 8. Solve for the inner surface temperature, and convert the result back to Celsius. By following these steps, the inner surface temperature of the pipe can be calculated.

Step-by-step solution

Convert given information to SI units

First, we need to convert the given quantities into SI units. In this case, we need to convert wall thickness and inner diameter to meters: Wall thickness: \({10 \mathrm{~mm}} = 0.01 \mathrm{~m}\), Inner diameter: \({10 \mathrm{~cm}} = 0.1 \mathrm{~m}\).

Calculate the pipe radius

Next, we'll calculate the pipe's inner and outer radii: Inner radius, \(r_i = \dfrac{\text{inner diameter}}{2} = \dfrac{0.1 \mathrm{~m}}{2} = 0.05 \mathrm{~m}\) Outer radius, \(r_o = r_i + \text{wall thickness} = 0.05 \mathrm{~m} + 0.01 \mathrm{~m} = 0.06 \mathrm{~m}\)

Write down the temperature gradient and heat transfer rate

Now, we will write the expressions for the temperature gradient and the heat transfer rate (using a cylindrical coordinate system): - Temperature gradient: \(\dfrac{dT}{dr} = \dfrac{\Delta T}{\Delta r}\), where \(\Delta T = T_o - T_i\) (outer surface temperature - inner surface temperature), and \(\Delta r = r_o - r_i\) - Heat transfer rate: \(q = 2 \pi r k(T) \dfrac{dT}{dr}\), where \(r_i \le r \le r_o\)

Write down the heat transfer rate equation at the outer surface

Substituting \(r=r_o\) and \(T=T_o\) into the heat transfer rate equation, we have: \(q_o = 2 \pi r_o k(T_o) \dfrac{dT}{dr}\), where \(q_o\) is the given heat flux at the outer surface, with a value of \(5\times10^3\mathrm{~W/m^2}\). Now, we know the outer surface temperature is \(500^{\circ} \mathrm{C}\). We must convert this to kelvin by adding 273.15 to the Celsius temperature. This gives us \(T_o = 773.15 \mathrm{K}\).

Write the thermal conductivity equation at the outer surface

Substitute the values of \(k_0\) and \(\beta\) and the outer surface temperature in the thermal conductivity expression: \(k(T_o) = k_0 (1+\beta T_o) = 7.5\mathrm{~W/m \cdot K} (1 + 0.0012 \mathrm{~K}^{-1} \times 773.15\mathrm{K})\)

Solve for the temperature gradient at the outer surface

Now, rearrange the heat transfer rate equation at the outer surface to express the temperature gradient in terms of the other known quantities: \(\dfrac{dT}{dr}=\dfrac{q_o}{2\pi r_o k(T_o)}\) Then substitute the known values of \(q_o\), \(r_o\), and \(k(T_o)\) to calculate the temperature gradient: \(\dfrac{dT}{dr}=\dfrac{5\times10^3\mathrm{~W/m^2}}{2\pi(0.06\mathrm{~m})\times k(T_o)}\)

Integrate the temperature gradient to find the temperature distribution

Now, integrate the temperature gradient expression with respect to the radius to obtain the temperature distribution. Note that the temperature distribution will involve a constant of integration, which we can solve for once we have the temperature at the inner surface. \(T_i=T_o+\int_{r_o}^{r_i} \dfrac{dT}{dr} dr\)

Solve for the inner surface temperature

Now, rearrange the temperature distribution equation to solve for the inner surface temperature, and substitute the known values from the previous steps to calculate the temperature: \(T_i = T_o+\left(\int_{r_o}^{r_i} \dfrac{dT}{dr} dr\right) = T_o+\left(\int_{0.06\mathrm{~m}}^{0.05\mathrm{~m}} \dfrac{5\times10^3\mathrm{~W/m^2}}{2\pi r k(T_o)} dr\right)\) After solving the integral, convert the temperature back to Celsius by subtracting 273.15 from the result in Kelvin. The inner surface temperature of the pipe is the final answer.