Let's break down the problem and solve each part step by step. --- ## **Given Data** - **Plate material:** Steel (AISI 1010) - **Plate thickness,** \( \delta = 6 \text{ mm} = 0.006 \text{ m} \) - **Plate side length,** \( L = 1 \text{ m} \) - **Initial plate temperature,** \( T_i = 300^\circ C \) - **Air temperature,** \( T_\infty = 20^\circ C \) - **Quiescent atmospheric air (still air)** - **Neglect radiation effects** ### **Properties of AISI 1010 Steel** (Approximate at high temperatures) - **Density,** \( \rho = 7832 \text{ kg/m}^3 \) - **Specific heat,** \( c_p = 477 \text{ J/kg}\cdot\text{K} \) - **Thermal conductivity,** \( k = 63.9 \text{ W/m}\cdot\text{K} \) ### **Properties of air (at ~160°C)** - **Kinematic viscosity,** \( \nu \approx 3.83 \times 10^{-5} \text{ m}^2/\text{s} \) - **Prandtl number,** \( Pr \approx 0.69 \) - **Thermal conductivity,** \( k_{air} \approx 0.036 \text{ W/m}\cdot\text{K} \) --- ## **Part (a): Initial Rate of Heat Transfer** ### **Step 1: Find the initial heat transfer coefficient, \( h \):** For a vertical plate in still air: \[ Nu = C (Ra_L)^n \] where, - \( Nu \) = Nusselt number - \( Ra_L \) = Rayleigh number \[ Ra_L = \frac{g \beta (T_s - T_\infty) L^3}{\nu \alpha} \] Where: - \( g \) = gravity \( = 9.81 \text{ m/s}^2 \) - \( \beta \) = \( 1/T_{film} \) (K\(^{-1}\)), \( T_{film} = (T_s + T_\infty)/2 \) in K - \( \alpha \) = \( k_{air}/(\rho_{air} c_{p,air}) \) - \( \nu \) = kinematic viscosity #### **Estimate \( Ra_L \):** - \( T_s = 300^\circ C = 573 \text{ K} \) - \( T_\infty = 20^\circ C = 293 \text{ K} \) - \( T_{film} = (573 + 293)/2 = 433 \text{ K} \) - \( \beta = 1/433 = 0.00231 \text{ K}^{-1} \) \[ Ra_L = \frac{9.81 \times 0.00231 \times (573-293) \times 1^3}{3.83 \times 10^{-5} \times \alpha} \] But let's estimate \( \alpha \) for air at \( T_{film} \): - \( c_{p,air} \approx 1007 \text{ J/kg}\cdot\text{K} \) - \( \rho_{air} \approx 0.7 \text{ kg/m}^3 \) at high temperature - \( k_{air} = 0.036 \text{ W/m}\cdot\text{K} \) \[ \alpha = \frac{k_{air}}{\rho_{air} c_{p,air}} = \frac{0.036}{0.7 \times 1007} \approx 5.13 \times 10^{-5} \text{ m}^2/\text{s} \] Now, plug in values: \[ Ra_L = \frac{9.81 \times 0.00231 \times 280 \times 1}{3.83 \times 10^{-5} \times 5.13 \times 10^{-5}} \] \[ = \frac{9.81 \times 0.00231 \times 280}{1.964 \times 10^{-9}} \] \[ = \frac{6.34}{1.964 \times 10^{-9}} \] \[ Ra_L \approx 3.23 \times 10^9 \] For \( 10^4 < Ra_L < 10^9 \), use: \[ Nu = 0.59 Ra_L^{1/4} \] Since \( Ra_L \) is slightly above \( 10^9 \), let's use: \[ Nu = 0.1 Ra_L^{1/3} \] \[ Nu = 0.1 \times (3.23 \times 10^9)^{1/3} \] \[ = 0.1 \times (3.23 \times 10^9)^{0.333} \] \[ = 0.1 \times (1.468 \times 10^3) \] \[ = 146.8 \] \[ h = \frac{Nu \cdot k_{air}}{L} = \frac{146.8 \times 0.036}{1} = 5.28 \text{ W/m}^2\cdot\text{K} \] --- ### **Step 2: Initial Rate of Heat Transfer** The plate has two sides: \[ A = 2 \times L \times L = 2 \times 1 \times 1 = 2 \text{ m}^2 \] \[ Q = h A (T_s - T_\infty) = 5.28 \times 2 \times (300 - 20) = 5.28 \times 2 \times 280 = 5.28 \times 560 = 2,956.8 \text{ W} \] **Initial rate of heat transfer: \(\boxed{2,960 \text{ W}}\) (rounded to 3 significant digits)** --- ## **Part (b): Time to Cool from 300°C to 150°C** We can use the lumped capacitance method if Biot number \( Bi < 0.1 \): \[ Bi = \frac{h L_c}{k} \] Where \( L_c = \frac{V}{A_{surf}} = \frac{\delta}{2} = 0.003 \text{ m} \) \[ Bi = \frac{5.28 \times 0.003}{63.9} = \frac{0.01584}{63.9} \approx 0.000248 \] This is much less than 0.1, so lumped capacitance is valid. The lumped capacitance equation: \[ \frac{T - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h A}{\rho V c_p} t \right) \] Solve for \( t \): - \( T = 150^\circ C \) - \( T_i = 300^\circ C \) - \( T_\infty = 20^\circ C \) - \( V = L \times L \times \delta = 1 \times 1 \times 0.006 = 0.006 \text{ m}^3 \) - \( A = 2 \text{ m}^2 \) - \( \rho = 7832 \text{ kg/m}^3 \) - \( c_p = 477 \text{ J/kg}\cdot\text{K} \) \[ \ln \left( \frac{T - T_\infty}{T_i - T_\infty} \right) = -\frac{hA}{\rho V c_p} t \] \[ \frac{150 - 20}{300 - 20} = \frac{130}{280} = 0.464 \] \[ \ln(0.464) = -0.769 \] \[ \frac{hA}{\rho V c_p} = \frac{5.28 \times 2}{7832 \times 0.006 \times 477} = \frac{10.56}{22.46 \times 477} = \frac{10.56}{10,719} \approx 0.000985 \text{ s}^{-1} \] \[ -0.769 = -0.000985 t \] \[ t = \frac{0.769}{0.000985} \approx 781 \text{ s} \] --- ## **Final Answers** ### **(a) Initial rate of heat transfer:** \[ \boxed{2,960 \text{ W}} \] ### **(b) Time to cool from 300°C to 150°C:** \[ \boxed{781 \text{ s}} \] If you need more precise property values or a step-by-step substitution, let me know!