Let's break down the problem step by step: ## **Given Data** - Thickness of strip, \( L = 6 \) mm \( = 0.006 \) m - Thermal conductivity, \( k = 21 \) W/m·K - Density, \( \rho = 8000 \) kg/m³ - Specific heat, \( C_p = 570 \) J/kg·K - Length of buffer zone, \( x = 5 \) m - Initial strip temperature, \( T_i = 500^\circ \)C - Final surface strip temperature, \( T_s = 45^\circ \)C - Air temperature, \( T_\infty = 15^\circ \)C - Convective heat transfer coefficient, \( h = 120 \) W/m²·K --- ## **Step 1: Lumped Capacitance or Not?** Check if Biot number (\( Bi \)) is less than 0.1: \[ Bi = \frac{h L_c}{k} \] Where \( L_c = \frac{V}{A} \) for a thin plate ≈ half thickness = \( \frac{0.006}{2} = 0.003 \) m \[ Bi = \frac{120 \times 0.003}{21} = 0.0171 < 0.1 \] **So, lumped capacitance method can be used.** --- ## **Step 2: Lumped Capacitance Equation** The temperature change is given by: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho V C_p}t\right) \] Let \( T(t) = T_s \) (final surface temperature), rearrange for time \( t \): --- ## **Step 3: Find Time Required for Cooling** \[ \frac{T_s - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho V C_p}t\right) \] Take natural log on both sides: \[ \ln\left(\frac{T_s - T_\infty}{T_i - T_\infty}\right) = -\frac{hA}{\rho V C_p}t \] \[ t = -\frac{\rho V C_p}{hA} \ln\left(\frac{T_s - T_\infty}{T_i - T_\infty}\right) \] For a strip (length does not matter since it's a continuous process), \( \frac{V}{A} = \) thickness = \( 0.006 \) m \[ t = -\frac{\rho \cdot 0.006 \cdot C_p}{h} \ln\left(\frac{T_s - T_\infty}{T_i - T_\infty}\right) \] --- ## **Step 4: Plug in the Values** - \( \rho = 8000 \) kg/m³ - \( C_p = 570 \) J/kg·K - \( h = 120 \) W/m²·K - \( T_i = 500^\circ \)C - \( T_s = 45^\circ \)C - \( T_\infty = 15^\circ \)C Calculate the log argument: \[ \frac{T_s - T_\infty}{T_i - T_\infty} = \frac{45 - 15}{500 - 15} = \frac{30}{485} = 0.06186 \] Now plug into the equation: \[ t = -\frac{8000 \times 0.006 \times 570}{120} \ln(0.06186) \] Calculate numerator: \[ 8000 \times 0.006 = 48 \] \[ 48 \times 570 = 27360 \] Divide by \( h \): \[ \frac{27360}{120} = 228 \] Now calculate \( \ln(0.06186) = -2.780 \): So, \[ t = -228 \times (-2.780) = 634.6 \text{ s} \] --- ## **Step 5: Maximum Speed of the Strip** The strip travels 5 meters in this time. So, \[ \text{Speed} = \frac{\text{distance}}{\text{time}} = \frac{5}{634.6} = 0.00788 \text{ m/s} \] --- ## **Final Answer** \[ \boxed{0.0079 \text{ m/s}} \] **(rounded to 2 significant digits)** --- ### **Summary of Steps** 1. Confirmed lumped capacitance is valid. 2. Used lumped capacitance cooling equation. 3. Solved for time to cool from \( 500^\circ \)C to \( 45^\circ \)C. 4. Calculated maximum speed as \( \frac{\text{distance}}{\text{time}} \). --- **If you have more questions, feel free to ask!**