Question

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

Short answer

Answer: The outer surface temperature of the pipe is approximately \(66.23 ^{\circ} \mathrm{C}\).

Step-by-step solution

Convert temperatures to Kelvin

First, we must convert the inner surface and ambient temperatures to Kelvin. \(T_{i} = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\) \(T_{\infty} = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\)

Calculate the heat transfer through the pipe using convection

Using the convection heat transfer formula, we can find the heat transfer, \(q\): \(q = hA(T_{i} - T_{\infty})\) Here, \(A = \pi dL\), where \(d = 25 \times 10^{-3} \mathrm{m}\) is the pipe's inner diameter and \(L\) is the length of the pipe. \(q = 50 \cdot \pi(25 \times 10^{-3})L (373.15 - 293.15)\)

Calculate the thermal conductivity at the average temperature

First, we need to find the average temperature between the inner surface and ambient temperatures: \(T_{avg} = \frac{T_{i} + T_{\infty}}{2} = 333.15 \mathrm{K}\) Now, we can calculate the thermal conductivity, \(k(T)\), at the average temperature: \(k(T_{avg}) = k_{0}(1 + \beta T_{avg}) = 1.5(1 + 0.003 \cdot 333.15)\)

Calculate the heat transfer through the pipe using conduction

Using the conduction heat transfer formula, we can find the heat transfer, \(q\): \(q = kA\frac{T_{i} - T_{o}}{t}\)

Equate the convection and conduction heat transfer equations and solve for the outer surface temperature

Now we set the convection and conduction heat transfer equations equal to each other and solve for \(T_{o}\): \(50 \cdot \pi(25 \times 10^{-3})L (373.15 - 293.15) = 1.5(1 + 0.003 \cdot 333.15) \cdot \pi(25 \times 10^{-3})L \frac{T_{i} - T_{o}}{3 \times 10^{-3}}\) Solve for \(T_{o}\): \(T_{o} = T_{i} - \frac{50(373.15 - 293.15)}{1.5(1 + 0.003 \cdot 333.15) \cdot \frac{1}{3 \times 10^{-3}}} = 339.38 \mathrm{K}\)

Convert the outer surface temperature back to Celsius

Our final step is to convert \(T_{o}\) back to Celsius: \(T_{o} = 339.38 \mathrm{K} - 273.15 = 66.23 ^{\circ} \mathrm{C}\) The outer surface temperature of the pipe is approximately \(66.23 ^{\circ} \mathrm{C}\).

Key concepts

Convection Heat Transfer

Convection heat transfer is a process where heat is transferred between a solid surface and a liquid or gas. This happens through the motion of the fluid or gas against the surface. The movement enhances the heat exchange rate, making it a crucial mode of heat transfer.
This transfer is quantified by the convection heat transfer coefficient, denoted as \( h \). This coefficient varies depending on the fluid properties, the surface area involved, and the temperature difference between the surface and the surrounding fluid.
In the context of the pipe example, the ambient air at 20°C cools the pipe's outer surface through convection. The formula used to calculate this transfer is:
\[ q = hA(T_i - T_{\infty}) \]
where:

• \( q \) is the heat transfer rate
• \( h \) is the convection heat transfer coefficient
• \( A \) is the surface area through which heat is being transferred
• \( T_i \) and \( T_{\infty} \) are the temperatures of the inner surface of the pipe and the ambient air, respectively

This equation underscores the importance of temperature differences in driving the rate of heat transfer through convection.

Conduction Heat Transfer

Conduction heat transfer involves the movement of heat within a solid or between solids in direct contact. It is a molecular process, where heat is transferred through collisions and diffusion of molecules inside the material.
This type of heat transfer is characterized by the material's thickness and thermal conductivity. In the case of the pipe, conduction takes place through the pipe's wall.
The basic formula to describe conduction heat transfer is:
\[ q = kA\frac{(T_i - T_o)}{t} \]
where:

• \( q \) is the rate of heat transfer
• \( k \) is thermal conductivity of the material
• \( A \) is the cross-sectional area
• \( T_i \) and \( T_o \) are the temperatures of the inner and outer surfaces of the pipe, respectively
• \( t \) is the thickness of the pipe wall

Solving the exercise, the conduction equation helps establish a connection between the thermal energy lost from the inner surface to the outer surface, allowing us to calculate parameters like the outer surface temperature.

Thermal Conductivity

Thermal conductivity is a property of a material that indicates its ability to conduct heat. This property varies based on the material's structure, temperature, and other factors.
In the given exercise, the thermal conductivity of the pipe's material is not constant. It changes with temperature according to the formula:
\[ k(T) = k_0(1 + \beta T) \]
Where:

• \( k_0 \) is the base thermal conductivity at a reference temperature
• \( \beta \) is the coefficient that represents how thermal conductivity changes with temperature
• \( T \) is the temperature in Kelvin

Calculating the average thermal conductivity is crucial as it influences how effectively the pipe material can conduct heat from the inner to the outer surface. In the solution, the knowledge of variable thermal conductivity, which depends on temperature, allows for a more accurate calculation of heat transmission through the pipe material.