Question

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated 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

The outer surface temperature of the pipe is approximately 317.76 K, which corresponds to 44.61°C.

Step-by-step solution

Convert the temperature values

Convert the given temperatures to Kelvin:\ \(T_{inner}=100^{\circ} \mathrm{C}+273.15=\mathrm{373.15 \ K}\)\ \(T_{ambient}=20^{\circ} \mathrm{C}+273.15=293.15 \ \mathrm{K}\)

Express thermal conductivity as a function of temperature

We are given \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in Kelvin.

Set up conservation of energy equation

Since we are considering steady-state heat transfer, the heat conduction through the pipe wall must be equal to the heat convection from the outer surface to the surrounding air, so we can set up a conservation of energy equation:\ \(Q_{conduction}=Q_{convection}\)

Apply Fourier's Law for Heat Conduction

Using Fourier's Law, \(Q_{conduction}=-k\frac{A_{c}}{t}\Delta T\), we have:\ \(Q_{conduction}=-k\frac{2 \pi r_{inner} L}{t}(T_{inner}-T_{outer})\),\ where \(r_{inner}\) is the inner radius of the pipe, \(t\) is the thickness of the pipe wall, and \(L\) is the length of the pipe.

Apply Newton's Law of Cooling for Heat Convection

Using Newton's Law of Cooling, \(Q_{convection}=h A_{s} \Delta T\), we have:\ \(Q_{convection}=h A_{s}(T_{outer} - T_{ambient})\),\ where \(A_{s}\) is the outer surface area of the pipe.

Define the coordinate system and express areas

Define a coordinate system with the radial distance to be \(r\). The areas are expressed as:\ \(A_c = 2 \pi r_{inner} L\)\ \(A_s = 2 \pi r_{outer} L\),\ where \(r_{outer} = r_{inner} + t\).

Solve for outer surface temperature

Using the conservation of energy equation, we now have:\ \(k\left(\frac{2 \pi r_{inner} L}{t}\right)(T_{inner}-T_{outer})=h \left(2 \pi r_{outer}L\right)(T_{outer} - T_{ambient})\) First, we can simplify the equation by removing the common term \(2\pi L\) on both sides:\ \(k\left(\frac{r_{inner}}{t}\right)(T_{inner}-T_{outer})=h r_{outer}(T_{outer} - T_{ambient})\) Now, we can rearrange the equation to solve for \(T_{outer}\). In order to do this, we will substitute \(T_{outer}\) in terms of \(r_{inner}\), \(r_{outer}\), \(T_{ambient}\), and \(T_{inner}\):\ \(T_{outer}=\frac{h r_{outer} T_{ambient} + k\left(\frac{r_{inner}}{t}\right)T_{inner}}{h r_{outer}+k\left(\frac{r_{inner}}{t}\right)}\) Finally, we will plug in the given values for \(k\), \(\beta\), \(r_{inner}\), \(t\), \(T_{inner}\), and \(T_{ambient}\), and solve for \(T_{outer}\):\ \(T_{outer}=\frac{50 \cdot 0.014 \cdot 293.15 + 1.5(1+0.003\cdot373.15)\cdot\frac{0.025}{2}\cdot373.15K}{50\cdot0.014+1.5(1+0.003\cdot373.15)\cdot\frac{0.025}{2\cdot0.003}}\) The calculated outer surface temperature of the pipe is approximately \(T_{outer} = 317.76 \ \mathrm{K}\), which corresponds to \(44.61^{\circ} \mathrm{C}\).