# Step-by-Step Solution We are given: - **Cylindrical column diameter,** \( D = 30\ \text{cm} = 0.30\ \text{m} \) - **Height,** \( L = 4.250\ \text{m} \) - **Thermal conductivity,** \( k = 0.79\ \text{W}/(\text{m} \cdot \text{K}) \) - **Thermal diffusivity,** \( \alpha = 5.94 \times 10^{-7}\ \text{m}^2/\text{s} \) - **Density,** \( \rho = 1600\ \text{kg}/\text{m}^3 \) - **Specific heat,** \( c_p = 0.84\ \text{kJ}/(\text{kg} \cdot \text{K}) = 840\ \text{J}/(\text{kg} \cdot \text{K}) \) - **Initial (cold) temp,** \( T_i = 14^\circ \text{C} \) - **Ambient temp,** \( T_\infty = 28^\circ \text{C} \) - **Heat transfer coefficient,** \( h = 14\ \text{W}/(\text{m}^2\cdot \text{K}) \) - **Final center temp,** \( T_{center, f} = 28^\circ \text{C} \) ## 1. **Assumptions** - The column is long, so we can treat it as an **infinite cylinder** (1-D radial conduction). - The initial temperature is uniform. - The ambient temperature and \( h \) remain constant. ## 2. **Relevant Equations** The one-term analytical solution for transient heat conduction in a cylinder (from Incropera, Fundamentals of Heat & Mass Transfer): \[ \frac{T(r, t) - T_\infty}{T_i - T_\infty} = A_1 \exp \left( -\lambda_1^2 \frac{\alpha t}{r_0^2} \right) J_0 \left( \lambda_1 \frac{r}{r_0} \right) \] Where: - \( r_0 \) = cylinder radius = \( D/2 = 0.15\ \text{m} \) - \( J_0 \) = Bessel function of order zero - \( \lambda_1 \) and \( A_1 \) are constants determined from the Biot number At the center (\( r = 0 \)), \( J_0(0) = 1 \): \[ \frac{T_{center}(t) - T_\infty}{T_i - T_\infty} = A_1 \exp \left( -\lambda_1^2 \frac{\alpha t}{r_0^2} \right) \] ### **Biot Number** \[ \text{Bi} = \frac{h r_0}{k} \] ### **Characteristic Equation** \[ \lambda_1 \frac{J_1(\lambda_1)}{J_0(\lambda_1)} = \text{Bi} \] ### **Coefficient** \[ A_1 = \frac{2 \lambda_1 J_1(\lambda_1)}{\lambda_1 J_0(\lambda_1) + \text{Bi} J_1(\lambda_1)} \] ### **Total Heat Transfer** \[ Q = \rho V c_p (T_i - T_{avg, f}) \] Where \( T_{avg, f} \) is the final average temperature (must be found). --- ## 3. **Calculate Biot Number** \[ r_0 = 0.15\ \text{m} \] \[ \text{Bi} = \frac{14 \times 0.15}{0.79} \approx \frac{2.1}{0.79} \approx 2.66 \] --- ## 4. **Find \(\lambda_1\) (First Root Numerically)** From Biot number tables for cylinders, for \( \text{Bi} = 2.66 \): \[ \lambda_1 \approx 2.96 \] (You can interpolate from tables or use an online calculator; for this problem, 2.96 is a good approximation.) --- ## 5. **Find \(A_1\) Coefficient** First, we need the Bessel functions for \( \lambda_1 = 2.96 \): - \( J_0(2.96) \approx -0.181 \) - \( J_1(2.96) \approx 0.270 \) Now substitute: \[ A_1 = \frac{2 \times 2.96 \times 0.270}{2.96 \times (-0.181) + 2.66 \times 0.270} \] \[ = \frac{1.5984}{-0.536 + 0.718} = \frac{1.5984}{0.182} \approx 8.785 \] --- ## 6. **Solve For Time When Center Reaches \(28^\circ\)C** \[ \frac{T_{center} - T_\infty}{T_i - T_\infty} = A_1 \exp \left( -\lambda_1^2 \frac{\alpha t}{r_0^2} \right) \] When \( T_{center} = T_\infty \), left side is zero, but due to exponential decay, let's solve for when the center is just about to reach \( T_\infty \). Practically, let's say \( T_{center} = 27.99^\circ \)C (since exponential never reaches zero). \[ \frac{T_{center} - T_\infty}{T_i - T_\infty} = \frac{27.99 - 28}{14 - 28} = \frac{-0.01}{-14} \approx 7.14 \times 10^{-4} \] Set equal to \( A_1 \exp(-\lambda_1^2 \frac{\alpha t}{r_0^2}) \): \[ 7.14 \times 10^{-4} = 8.785 \exp(-2.96^2 \frac{5.94 \times 10^{-7} t}{0.15^2}) \] Take log of both sides: \[ \ln \left( \frac{7.14 \times 10^{-4}}{8.785} \right) = -2.96^2 \frac{5.94 \times 10^{-7} t}{0.0225} \] \[ \ln (8.13 \times 10^{-5}) = -8.76 \frac{5.94 \times 10^{-7} t}{0.0225} \] \[ -9.418 = -8.76 \frac{5.94 \times 10^{-7} t}{0.0225} \] \[ 9.418 = 8.76 \frac{5.94 \times 10^{-7} t}{0.0225} \] \[ \frac{9.418 \times 0.0225}{8.76 \times 5.94 \times 10^{-7}} = t \] \[ = \frac{0.2119}{5.202 \times 10^{-6}} \] \[ t \approx 40757\ \text{s} \] \[ t \approx 11.3\ \text{hours} \] --- ## 7. **Find Average Temperature \(T_{avg, f}\)** The average temperature is calculated using the one-term approximation: \[ \frac{T_{avg}(t) - T_\infty}{T_i - T_\infty} = \frac{2 A_1}{\lambda_1^2} \left[ J_1(\lambda_1) \right] \exp \left( -\lambda_1^2 \frac{\alpha t}{r_0^2} \right) \] But as \( t \) increases, \( T_{avg, f} \to T_\infty \). For practical purposes, at \( T_{center} \approx T_\infty \), \( T_{avg, f} \approx T_\infty \). --- ## 8. **Calculate Heat Transfer** **Volume of column:** \[ V = \pi r_0^2 L = \pi (0.15)^2 \times 4.25 \approx 0.301\ \text{m}^3 \] **Mass:** \[ m = \rho V = 1600 \times 0.301 = 481.6\ \text{kg} \] **Heat transfer:** \[ Q = m c_p (T_{avg, f} - T_i) \] \[ Q = 481.6 \times 840 \times (28 - 14) \] \[ = 481.6 \times 840 \times 14 = 481.6 \times 11,760 = 5,669,616\ \text{J} \] \[ Q \approx 5.67 \times 10^6\ \text{J} \] or \[ Q \approx 5.67\ \text{MJ} \] --- # **Final Answer** ## **The amount of heat transfer until the center temperature reaches \(28^\circ \text{C}\) is approximately:** \[ \boxed{5.67\ \text{MJ}} \] --- ### **Summary Table** | Parameter | Value | |--------------------------|------------------------------| | Volume | \(0.301~\text{m}^3\) | | Mass | \(482~\text{kg}\) | | Time to heat | \(11.3~\text{hours}\) | | Heat transferred, \(Q\) | \(5.67~\text{MJ}\) | --- **Note:** If you need the heat transfer over a specific time or at a different final temperature, or want to see the full temperature-time curve, let me know!