Let's break down the problem step by step. --- ## **Given Data** - Strip thickness, \(\delta = 12 \, \text{mm} = 0.012 \, \text{m}\) - Density, \(\rho = 7900 \, \text{kg/m}^3\) - Specific heat, \(c_p = 640 \, \text{J/kg·K}\) - Thermal conductivity, \(k = 30 \, \text{W/m·K}\) - Emissivity, \(\epsilon = 0.7\) - Initial temperature, \(T_i = 300^\circ \text{C}\) - Final temperature, \(T_f = 600^\circ \text{C}\) - Furnace wall (and combustion gas) temperature, \(T_\infty = T_w = 850^\circ \text{C}\) - Convection coefficient, \(h = 100 \, \text{W/m}^2\cdot\text{K}\) - Furnace length, \(L = 105 \, \text{m}\) --- ## **Step 1: Lumped Capacitance Check** To use the lumped capacitance method, check the Biot number: \[ \text{Bi} = \frac{h \delta/2}{k} \] \[ \text{Bi} = \frac{100 \times 0.012/2}{30} = \frac{100 \times 0.006}{30} = \frac{0.6}{30} = 0.02 \] Since Bi < 0.1, the lumped capacitance method is valid. --- ## **Step 2: Lumped Capacitance Equation** For a body heated by convection: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho V c_p} t\right) \] For heating from \(T_i\) to \(T_f\): \[ \frac{T_f - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho V c_p} t_f\right) \] Take the natural log: \[ \ln \left( \frac{T_f - T_\infty}{T_i - T_\infty} \right) = -\frac{hA}{\rho V c_p} t_f \] \[ t_f = -\frac{\rho V c_p}{hA} \ln \left( \frac{T_f - T_\infty}{T_i - T_\infty} \right) \] Let’s assume the strip is very wide and long compared to its thickness, so we can use area per unit width and volume per unit width: - \(A\): Surface area per unit width \(= 2 \times \text{length}\) (top + bottom) - \(V\): Volume per unit width \(= \delta \times \text{length}\) So, \(\frac{V}{A} = \frac{\delta}{2}\) \[ t_f = -\frac{\rho c_p \delta}{2h} \ln \left( \frac{T_f - T_\infty}{T_i - T_\infty} \right) \] --- ## **Step 3: Plug in the Values** - \(\rho = 7900\, \text{kg/m}^3\) - \(c_p = 640\, \text{J/kg·K}\) - \(\delta = 0.012\, \text{m}\) - \(h = 100\, \text{W/m}^2\cdot\text{K}\) - \(T_i = 300^\circ \text{C}\) - \(T_f = 600^\circ \text{C}\) - \(T_\infty = 850^\circ \text{C}\) \[ t_f = -\frac{7900 \cdot 640 \cdot 0.012}{2 \cdot 100} \ln \left( \frac{600 - 850}{300 - 850} \right) \] Calculate the log argument: \[ \frac{600 - 850}{300 - 850} = \frac{-250}{-550} = 0.4545 \] Now calculate the coefficient: \[ \frac{7900 \cdot 640 \cdot 0.012}{2 \cdot 100} = \frac{7900 \cdot 640 \cdot 0.012}{200} \] \[ 7900 \cdot 640 = 5,056,000 \] \[ 5,056,000 \cdot 0.012 = 60,672 \] \[ \frac{60,672}{200} = 303.36 \] Now, the time: \[ t_f = -303.36 \cdot \ln(0.4545) \] \[ \ln(0.4545) = -0.7894 \] \[ t_f = -303.36 \cdot (-0.7894) = 239.4 \, \text{seconds} \] --- ## **Step 4: Calculate Strip Speed \(V_s\)** The strip spends \(t_f\) seconds in the furnace of length \(L\): \[ V_s = \frac{L}{t_f} \] \[ V_s = \frac{105}{239.4} = 0.438 \, \text{m/s} \] --- ## **Final Answers** \[ \boxed{t_f = 239 \;\text{s}} \] \[ \boxed{V_s = 0.44 \;\text{m/s}} \] --- ### **Summary Table** | Quantity | Value | |-----------------|--------------| | Time (\(t_f\)) | 239 s | | Speed (\(V_s\)) | 0.44 m/s | --- **If you need to consider radiation as well, let me know!**