Question

A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete $\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.94 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right.\(, \)\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)c_{p}=0.84 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ ) cooled to \(14^{\circ} \mathrm{C}\) during a cold night is heated again during the day by being exposed to ambient air at an average temperature of \(28^{\circ} \mathrm{C}\) with an average heat transfer coefficient of $14 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. Using the analytical one-term approximation method, determine \((a)\) how long it will take for the column surface temperature to rise to \(27^{\circ} \mathrm{C}\), (b) the amount of heat transfer until the center temperature reaches to \(28^{\circ} \mathrm{C}\), and \((c)\) the amount of heat transfer until the surface temperature reaches \(27^{\circ} \mathrm{C}\).

Short answer

Answer: It will take 4.77 hours for the column's surface temperature to rise to 27°C. The amount of heat transfer until the center temperature reaches 28°C is 85691 kJ, and the amount of heat transfer until the surface temperature reaches 27°C is 5652 kJ.

Step-by-step solution

Identify the given values and constants

We are given the following information about the concrete column: - Diameter: \(30 \,\text{cm}\) - Height: \(4 \,\text{m}\) - Thermal conductivity: \(k = 0.79 \, \frac{\text{W}}{\text{m} \cdot \text{K}}\) - Thermal diffusivity: \(\alpha = 5.94 \times 10^{-7} \, \frac{\text{m}^2}{\text{s}}\) - Density: \(\rho = 1600 \, \frac{\text{kg}}{\text{m}^3}\) - Specific heat capacity: \(c_p = 0.84 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}}\) - Initial temperature: \(14^{\circ} \mathrm{C}\) - Ambient temperature: \(28^{\circ} \mathrm{C}\) - Average heat transfer coefficient: \(h = 14 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}\)

Calculate the radius of the column

Since we are given the diameter of the column, we can calculate the radius as follows: \(r = \frac{D}{2} = \frac{30 \,\text{cm}}{2} = 15 \,\text{cm} = 0.15 \,\text{m}\)

Calculating the Biot number

The Biot number (Bi) can be calculated using the formula: \(\text{Bi} = \frac{h * L_c}{k}\) where \(L_c\) is the characteristic length. For a cylindrical column, the characteristic length is equal to half of the radius: \(L_c = \frac{r}{2} = 0.075 \,\text{m}\) Now we can calculate the Biot number: \(\text{Bi} = \frac{14 \frac{\text{W}}{\text{m}^2 \cdot \text{K}} \cdot 0.075 \,\text{m}}{0.79 \frac{\text{W}}{\text{m} \cdot \text{K}}} = 1.32\)

Calculating the Fourier number

We can now calculate the Fourier number (Fo) using the formula: \(\text{Fo} = \frac{\alpha * t}{L_c^2}\) We need to find the time \(t\). We know that for the one-term approximation method, the temperature difference between the center and surface of the cylinder can be given as: \(\Theta (\tau) = 1 - 0.5 \, \exp(-\text{FoBi})\) Where \(\Theta\) is the dimensionless temperature difference, and \(\tau\) is the dimensionless time. Since we know the desired surface temperature is \(27^\circ \text{C}\), we can find \(\Theta\) as: \(\Theta = \frac{28^\circ \mathrm{C} - 27^\circ \mathrm{C}}{28^\circ \mathrm{C} - 14^\circ \mathrm{C}} = \frac{1}{14}\) Now we can solve for \(\tau\): \(0.0714 = 1 - 0.5 \exp(-\text{FoBi})\) \(\tau = 0.0714\)

Solve for t (time)

We know that: \(\tau = \frac{\alpha * t}{L_c^2}\) Solving for \(t\), we get: \(t = \frac{\tau * L_c^2}{\alpha} = \frac{0.0714 * (0.075\,\text{m})^2}{5.94 \times 10^{-7} \frac{\text{m}^2}{\text{s}}} = 17 167 \,\text{s}\) or approximately \(4.77 \,\text{hours}\).

Calculate the amount of heat transfer (part b and c)

The dimensionless heat transfer can be given as: \(Q^* = 1 - \exp(-\text{FoBi})\) For part (b), since the center temperature is desired to be \(28^{\circ} \mathrm{C}\), \(\Theta = 0\), \(Q^* = 1 - \exp(0) = 1\) (as that would mean the temperature difference is zero) For part (c), using the \(\Theta\) calculated in Step 4: \(Q^* = 1 - \exp(-0.0714)\) Now, we can calculate the total heat transfer: \(Q = Q^* * (\rho * V * c_p * \Delta T)\) where \(V = \pi r^2 h\) is the volume of the cylinder and \(\Delta T = 28^\circ \mathrm{C} - 14^\circ \mathrm{C} = 14\,\text{K}\). For part (b), we get: \(Q = 1 * (1600\frac{\text{kg}}{\text{m}^3} * (\pi (0.15\,\text{m})^2 \cdot 4\,\text{m}) * 0.84 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} * 14 \,\text{K}) = 85691 \,\text{kJ}\) For part (c), we get: \(Q = (1 - \exp(-0.0714)) * (1600\frac{\text{kg}}{\text{m}^3} * (\pi (0.15\,\text{m})^2 \cdot 4\,\text{m}) * 0.84 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} * 14 \,\text{K}) = 5652 \,\text{kJ}\) So, the answers are: - (a) It will take \(4.77 \,\text{hours}\) for the column's surface temperature to rise to \(27^{\circ} \mathrm{C}\). - (b) The amount of heat transfer until the center temperature reaches \(28^{\circ} \mathrm{C}\) is \(85691 \,\text{kJ}\). - (c) The amount of heat transfer until the surface temperature reaches \(27^{\circ} \mathrm{C}\) is \(5652 \,\text{kJ}\).