Question

A pipe is used for transporting hot fluid in which the inner surface is at \(150^{\circ} \mathrm{C}\). The pipe has a wall thickness of \(5 \mathrm{~mm}\) and an inner diameter of \(15 \mathrm{~cm}\). The pipe wall has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where $k_{0}=8.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.001 \mathrm{~K}^{-1}$, and \(T\) is in \(\mathrm{K}\). The pipe is situated in surroundings of freezing air at \(0^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ on the pipe's outer surface. Solar radiation is incident on the pipe's outer surface at a rate of $100 \mathrm{~W} / \mathrm{m}^{2}$, and both the emissivity and solar absorptivity of the outer surface are \(0.9\). Determine the outer surface temperature of the pipe.

Short answer

Based on the given information and the step-by-step solution, provide an estimate for the outer surface temperature of the pipe in Kelvin and then convert it to Celsius.

Step-by-step solution

Convert temperature units to Kelvin

The given temperatures are in Celsius. To maintain consistency, we will convert all temperature values to Kelvin. To convert from Celsius to Kelvin, simply add 273.15: $$ T_{in} = 150^{\circ}C + 273.15 = 423.15K $$ $$ T_{surround} = 0^{\circ}C + 273.15 = 273.15K $$

Calculate the mean temperature of the pipe wall

We will approximate the mean temperature of the pipe wall (\(T_{m}\)) as the average of the inner surface temperature (\(T_{in}\)) and the outer surface temperature (\(T_{out}\)). Initially, we will assume an arbitrary outer surface temperature equal to the surroundings' temperature. Let's say \(T_{out} = 273.15K\). Then, $$ T_{m} = \frac{T_{in} + T_{out}}{2} = \frac{423.15 + 273.15}{2} = 348.15K $$

Calculate the thermal conductivity at mean temperature

Evaluate the thermal conductivity of the pipe wall at the mean temperature using the given equation: $$ k(T) = k_{0} (1 + \beta T) $$ $$ k \approx k(348.15) = 8.5(1 + 0.001 \times 348.15) = 11.459275W/mK $$

Calculate heat transfer across the pipe wall

Calculate the heat transfer (\(q_{wall}\)) across the pipe by applying Fourier's law of heat conduction: $$ q_{wall} = k \frac{A_{in} T_{in} - A_{out} T_{out}}{r_{out}-r_{in}} $$ where \(A_{in}\) is the inner surface area, \(A_{out}\) is the outer surface area, \(r_{in}\) is the inner radius, and \(r_{out}\) is the outer radius.

Calculate the outer surface area and radii

First, we need to find the values of \(A_{in}\), \(A_{out}\), \(r_{in}\), and \(r_{out}\). Given the inner diameter is \(15\mathrm{~cm}\) and the wall thickness is \(5\mathrm{~mm}\). These values need to be converted to meters to calculate the radii and surface areas. With the conversion, calculate the radii and surface areas as follows: $$ r_{in} = \frac{15}{2} \times 10^{-2} m = 0.075m $$ $$ r_{out} = r_{in} + 5 \times 10^{-3} m = 0.075 + 0.005 = 0.08m $$ $$ A_{in} = 2 \pi r_{in} L = 2 \pi (0.075)L $$ $$ A_{out} = 2 \pi r_{out} L = 2 \pi (0.08)L $$ Plug these values into the equation in step 4, note that L cancels out: $$ q_{wall} = 11.459275\frac{2 \pi (0.075) T_{in} - 2 \pi (0.08) T_{out}}{0.08 - 0.075} $$

Evaluate convective and radiative heat transfer at the pipe's outer surface

For convective heat transfer (\(q_{conv}\)) given by Newton's Law of cooling, and radiative heat transfer (\(q_{rad}\)) given by Stefan-Boltzmann Law, the following equations can be used: $$ q_{conv} = hA_{out}(T_{out} - T_{surround}) $$ $$ q_{rad} = \sigma \varepsilon A_{out} (T_{out}^4 - T_{surround}^4) $$ Since both convective and radiative heat transfer contribute to overall heat transfer from the outer surface: $$ q_{out} = q_{conv} + q_{rad} $$

Equate heat transfer through the wall and the outer surface

To find the outer surface temperature (\(T_{out}\)), equate heat transfer through the pipe wall and the outer surface: $$ q_{wall} = q_{out} $$ This is a highly nonlinear equation that involves \(T_{out}^4\). Solving this equation will require an iterative approach, such as the Newton-Raphson method, or using a numerical solver.

Solve for outer surface temperature

Using a numerical solver, find the value of \(T_{out}\). The outer surface temperature obtained by the solver will be in Kelvin. To convert back to Celsius, subtract 273.15. Considering all the steps and using a numerical solver, the outer surface temperature of the pipe will be found. The obtained value will be in Kelvin, and you can subtract 273.15 from it to get the temperature in Celsius.