Let's break down and solve the problem step by step: --- ## **Given Data** - **Duct air temperature (\(T_{in}\))**: \(180^\circ C\) - **Wall layers:** - **Steel:** \(k_1 = 45\) W/m·K, \(L_1 = .004\) m - **Insulation:** \(k_2 = .05\) W/m·K, \(L_2 = .05\) m - **Outside conditions:** - Air temperature (\(T_\infty\)): \(29^\circ C\) - Radiation temperature (\(T_{rad}\)): \(28^\circ C\) - Wind speed (\(v\)): \(.3\) m/s - Solar irradiation absorbed (\(G_{sol}\)): \(25\) W/m² - Emissivity (\(\epsilon\)): \(.85\) - Absorptivity (\(\alpha\)): \(.85\) - Measured outer surface temperature (\(T_{so}\)): \(33^\circ C\) --- ## **a) \(h_{ext}\) using vertical plate correlation** The external convective heat transfer coefficient for a vertical plate, **forced convection** (since wind is present): ### 1. **Find properties of air at film temperature:** \[ T_{film} = \frac{T_{so} + T_\infty}{2} = \frac{33 + 29}{2} = 31^\circ C \approx 304 \text{ K} \] - Dynamic viscosity (\(\nu\)) ≈ \(18.4 \times 10^{-6}\) m²/s - Prandtl number (\(Pr\)) ≈ .7 - Thermal conductivity (\(k_{air}\)) ≈ .026 W/m·K ### 2. **Calculate Reynolds number (\(Re_L\)):** Assume characteristic length (\(L\)) = 1 m (if not specified). \[ Re_L = \frac{v \cdot L}{\nu} = \frac{.3 \cdot 1}{18.4 \times 10^{-6}} = 16,304 \] ### 3. **Nusselt number (\(Nu\)):** For \(10^4 < Re_L < 5 \times 10^5\): \[ Nu = .664 \cdot Re_L^{.5} \cdot Pr^{1/3} \] \[ Nu = .664 \cdot (16,304)^{.5} \cdot (.7)^{1/3} \] \[ Nu = .664 \cdot 127.7 \cdot .887 = 75.12 \] ### 4. **Convective coefficient (\(h_{ext}\)):** \[ h_{ext} = \frac{Nu \cdot k_{air}}{L} = \frac{75.12 \cdot .026}{1} = 1.95 \text{ W/m}^2\cdot\text{K} \] --- ## **b) \(q''\) total to the exterior** The energy balance at the external surface: \[ q'' = h_{ext}(T_{so} - T_\infty) + \epsilon \sigma (T_{so}^4 - T_{rad}^4) - G_{sol} \] Where: - \(\sigma = 5.67 \times 10^{-8}\) W/m²·K⁴ - All temperatures in Kelvin \[ T_{so} = 33 + 273.15 = 306.15\,K \] \[ T_\infty = 29 + 273.15 = 302.15\,K \] \[ T_{rad} = 28 + 273.15 = 301.15\,K \] **Plug in values:** 1. **Convection:** \[ q''_{conv} = 1.95 \cdot (306.15 - 302.15) = 1.95 \cdot 4 = 7.8\,W/m^2 \] 2. **Radiation:** \[ q''_{rad} = .85 \cdot 5.67 \times 10^{-8} \cdot (306.15^4 - 301.15^4) \] \[ 306.15^4 = 8.771 \times 10^9 \] \[ 301.15^4 = 8.241 \times 10^9 \] \[ \Delta = .53 \times 10^9 \] \[ q''_{rad} = .85 \cdot 5.67 \times 10^{-8} \cdot .53 \times 10^9 = .85 \cdot 30.051 = 25.54\,W/m^2 \] 3. **Solar absorbed:** \(25\,W/m^2\) **Total:** \[ q'' = q''_{conv} + q''_{rad} - G_{sol} \] \[ q'' = 7.8 + 25.54 - 25 = 8.34\,W/m^2 \] --- ## **c) Temperature at the internal surface of the duct (\(T_{si}\))** Heat flux through the wall: \[ q'' = \frac{T_{si} - T_{so}}{R_{wall}} \] Where: \[ R_{wall} = R_{steel} + R_{insulation} = \frac{L_1}{k_1} + \frac{L_2}{k_2} \] \[ R_{steel} = \frac{.004}{45} = 8.89 \times 10^{-5}\,m^2K/W \] \[ R_{insulation} = \frac{.05}{.05} = 1\,m^2K/W \] \[ R_{wall} = 1.00009\,m^2K/W \] \[ T_{si} = q'' \cdot R_{wall} + T_{so} = 8.34 \cdot 1.00009 + 33 = 8.34 + 33 = 41.34^\circ C \] --- ## **Summary of Answers** ### a) External heat transfer coefficient \[ \boxed{h_{ext} \approx 1.95\, W/m^2K} \] ### b) Total heat flux to the exterior \[ \boxed{q''_{total} \approx 8.34\, W/m^2} \] ### c) Internal surface temperature of the duct \[ \boxed{T_{si} \approx 41.3^\circ C} \] --- **Let me know if you need clarification or further detail on any step!**