# Step-by-Step Solution: Heat Flux Through Furnace Wall ## 1. **Given Data** - **Wall thickness, \( L \)**: \( 5 \) cm \( = .05 \) m - **Wall thermal conductivity, \( k \)**: \( 7 \) W/m·°C - **Inside hot gas temperature, \( T_{\text{hot}} \)**: 402°C - **Inside convective heat transfer coefficient, \( h_{\text{in}} \)**: 37 W/m·°C - **Outside air temperature, \( T_{\text{out}} \)**: 27°C - **Outside convective heat transfer coefficient, \( h_{\text{out}} \)**: 20 W/m·°C - **Outside surface emissivity, \( \varepsilon \)**: .8 - **Surrounding temperature for radiation, \( T_{\text{sur}} \)**: 27°C \( = 300.15 \) K (convert to Kelvin for radiation) - **Stefan-Boltzmann constant, \( \sigma \)**: \( 5.67 \times 10^{-8} \) W/m²·K⁴ ## 2. **Thermal Circuit Schematic** ``` T_hot --[Convection, h_in]-- T1 --[Conduction, k, L]-- T2 --[Convection, h_out] AND [Radiation, ε]-- T_sur ``` - **T_hot**: Bulk gas temperature (402°C) - **T1**: Inner wall surface temperature - **T2**: Outer wall surface temperature - **T_sur**: Surroundings (27°C) ## 3. **Thermal Resistances** ### a) **Inside convection** \[ R_{\text{conv,in}} = \frac{1}{h_{\text{in}}} = \frac{1}{37} = .027 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ### b) **Wall conduction** \[ R_{\text{wall}} = \frac{L}{k} = \frac{.05}{.7} = .0714 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ### c) **Outside convection** \[ R_{\text{conv,out}} = \frac{1}{h_{\text{out}}} = \frac{1}{20} = .050 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ### d) **Outside radiation** \[ R_{\text{rad}} = \frac{1}{\varepsilon \sigma (T_2 + T_{\text{sur}})(T_2^2 + T_{\text{sur}}^2)} \] We need an estimate for \(T_2\). Start with an assumption and improve as needed. Assume \(T_2 \approx 100^\circ\)C \(= 373\) K for first estimate. \[ R_{\text{rad}} = \frac{1}{.8 \times 5.67\times10^{-8} \times (373 + 300.15) \left[ (373)^2 + (300.15)^2 \right] } \] Calculate: - \( (373 + 300.15) = 673.15 \) - \( 373^2 = 139,129 \) - \( 300.15^2 \approx 90,090 \) - \( 139,129 + 90,090 = 229,219 \) So, \[ \varepsilon \sigma (T_2 + T_{\text{sur}}) (T_2^2 + T_{\text{sur}}^2) = .8 \times 5.67\times10^{-8} \times 673.15 \times 229,219 \] Calculate: - \( .8 \times 5.67\times10^{-8} = 4.536\times10^{-8} \) - \( 673.15 \times 229,219 \approx 154,324,436 \) - \( 4.536\times10^{-8} \times 154,324,436 \approx 7. \) So, \[ R_{\text{rad}} \approx \frac{1}{7.} = .143 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ## 4. **Combined Outside Resistance** Radiation and convection are parallel: \[ \frac{1}{R_{\text{out,eq}}} = \frac{1}{R_{\text{conv,out}}} + \frac{1}{R_{\text{rad}}} \] \[ \frac{1}{R_{\text{out,eq}}} = \frac{1}{.05} + \frac{1}{.143} = 20 + 6.99 = 26.99 \] \[ R_{\text{out,eq}} = \frac{1}{26.99} = .037 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ## 5. **Total Resistance** \[ R_{\text{total}} = R_{\text{conv,in}} + R_{\text{wall}} + R_{\text{out,eq}} = .027 + .0714 + .037 = .1354 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] ## 6. **Heat Flux Calculation** \[ q = \frac{T_{\text{hot}} - T_{\text{sur}}}{R_{\text{total}}} \] \[ q = \frac{402 - 27}{.1354} = \frac{375}{.1354} = 2,770 \ \text{W/m}^2 \] ## 7. **Check Outer Surface Temperature (\(T_2\))** Let \(q\) = 2,770 W/m². Compute \(T_2\): \[ T_2 = T_{\text{sur}} + q \cdot R_{\text{out,eq}} = 27 + 2,770 \times .037 = 27 + 102.5 = 129.5^\circ\text{C} \] Convert \(T_2\) to Kelvin: \(129.5 + 273.15 = 402.65\) K. Recalculate \(R_{\text{rad}}\): - \(T_2 + T_{\text{sur}} = 402.65 + 300.15 = 702.8\) - \(T_2^2 + T_{\text{sur}}^2 = (402.65)^2 + (300.15)^2 = 162,128 + 90,090 = 252,218\) - \(.8 \times 5.67\times10^{-8} = 4.536\times10^{-8}\) - \(702.8 \times 252,218 = 177,318,170\) - \(4.536\times10^{-8} \times 177,318,170 = 8.04\) So, \[ R_{\text{rad}} = \frac{1}{8.04} = .124 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] \[ \frac{1}{R_{\text{out,eq}}} = \frac{1}{.05} + \frac{1}{.124} = 20 + 8.06 = 28.06 \] \[ R_{\text{out,eq}} = \frac{1}{28.06} = .0356 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] Update total resistance: \[ R_{\text{total}} = .027 + .0714 + .0356 = .134 \ \text{m}^2\cdot^\circ\text{C}/\text{W} \] \[ q = \frac{375}{.134} = 2,798 \ \text{W/m}^2 \] This is close to previous estimate, so **final answer is:** --- ## **Final Answers** ### **1. Thermal Circuit** - **Series:** Inside convection → Wall conduction → [Parallel: Outside convection + Outside radiation]  *Alt text: Thermal resistance circuit showing convection, conduction, and parallel convection+radiation elements* ### **2. Heat Flux Through Wall** \[ \boxed{q = 2,800 \ \text{W/m}^2} \] (Rounded to two significant digits) --- ### **Summary Table** | Resistance | Value (\(\text{m}^2\cdot^\circ\text{C}/\text{W}\)) | |--------------------|:--------------------------------------------------:| | Inside convection | .027 | | Wall conduction | .071 | | Outside convection | .050 | | Outside radiation | .124 | | Outside equivalent | .036 | | **Total** | **.134** | --- ### **3. Key Steps** 1. Calculate resistances for convection, conduction, and radiation. 2. Find parallel combination for outside convection and radiation. 3. Add all resistances to get total. 4. Use temperature difference and total resistance to find heat flux. --- ### **Assumptions** - 1D heat transfer - Steady-state - Large room, so surroundings at uniform temperature - All properties constant --- **Final heat flux through the furnace wall is approximately \(2,800~\text{W/m}^2\).**