Let's break down the problem step by step: ### **Given Data** - Dimensions of PCB: \( 16\, \text{cm} \times 21\, \text{cm} \) (\(0.16\, \text{m} \times 0.21\, \text{m}\)) - Power dissipated (\(Q\)): 8 W - Room temperature (\(T_\infty\)): 20°C = 293.15 K - Emissivity (\(\varepsilon\)): 0.8 - Surrounding temperature: Same as air, 20°C - Properties of air at film temperature (32.5°C, 305.65 K): - \( k = 0.02067 \) W/m·°C - \( \nu = 1.631 \times 10^{-5} \) m²/s - \( Pr = 0.7275 \) - \( \beta = 0.003273 \) K\(^{-1}\) --- ## **Step 1: Area Calculation** \[ A = 0.16\, \text{m} \times 0.21\, \text{m} = 0.0336\, \text{m}^2 \] --- ## **Step 2: Combined Heat Transfer** Heat loss is by convection and radiation, so: \[ Q = Q_{\text{conv}} + Q_{\text{rad}} \] \[ Q = hA(T_s - T_\infty) + \varepsilon \sigma A (T_s^4 - T_\infty^4) \] Where: - \( h \) = convection heat transfer coefficient - \( \sigma = 5.67 \times 10^{-8} \) W/m²·K⁴ (Stefan-Boltzmann constant) - \( T_s \) = surface temperature (in K) - \( T_\infty = 293.15 \) K --- ## **Step 3: Estimate Surface Temperature** Assume initially \( T_s \approx 40^\circ \text{C} = 313.15\, \text{K} \) (for film temperature). We'll iterate if needed. ### **3.1. Convection (Natural Convection for Horizontal Plate Facing Up)** The Rayleigh number: \[ Ra_L = \frac{g \beta (T_s - T_\infty) L^3}{\nu^2} Pr \] Where: - \( g = 9.81\, \text{m/s}^2 \) - \( \beta = 0.003273\, \text{K}^{-1} \) - \( L \) = characteristic length = area/perimeter = \( \frac{0.16 \times 0.21}{0.5 \times (0.16+0.21)} \approx 0.186 \) m (but for a flat plate, can use length or width; let's use the longer side, \( L = 0.21\, \text{m} \) ) \[ Ra_L = \frac{9.81 \times 0.003273 \times (T_s - T_\infty) \times (0.21)^3}{(1.631 \times 10^{-5})^2} \times 0.7275 \] Assuming \( T_s - T_\infty \approx 20\,\text{K} \): \[ Ra_L = \frac{9.81 \times 0.003273 \times 20 \times (0.21)^3}{(1.631 \times 10^{-5})^2} \times 0.7275 \] \[ = \frac{9.81 \times 0.003273 \times 20 \times 0.009261}{2.661 \times 10^{-10}} \times 0.7275 \] \[ = \frac{0.0059}{2.661 \times 10^{-10}} \times 0.7275 \approx 2.22 \times 10^7 \] #### **Nusselt Number (for horizontal plate facing up, \(10^4 < Ra < 10^7\))** Use Churchill & Chu correlation: \[ Nu = 0.15 Ra_L^{1/3} \] \[ Nu = 0.15 \times (2.22 \times 10^7)^{1/3} \] \[ (2.22 \times 10^7)^{1/3} \approx 281 \] \[ Nu = 0.15 \times 281 \approx 42.15 \] #### **Heat Transfer Coefficient** \[ h = \frac{Nu \cdot k}{L} \] \[ h = \frac{42.15 \times 0.02067}{0.21} \approx \frac{0.871}{0.21} \approx 4.15\, \text{W/m}^2\cdot\text{K} \] --- ### **3.2. Radiation** \[ Q_{rad} = \varepsilon \sigma A (T_s^4 - T_\infty^4) \] --- ### **3.3. Plug in Values and Solve for \( T_s \) Numerically** \[ 8 = hA(T_s - 293.15) + \varepsilon \sigma A (T_s^4 - (293.15)^4) \] Substitute values: - \( h = 4.15 \) W/m²·K - \( \varepsilon = 0.8 \) - \( \sigma = 5.67 \times 10^{-8} \) - \( A = 0.0336 \) - \( T_\infty = 293.15 \) \[ 8 = 4.15 \times 0.0336 \times (T_s - 293.15) + 0.8 \times 5.67 \times 10^{-8} \times 0.0336 \times (T_s^4 - (293.15)^4) \] \[ 8 = 0.13944 (T_s - 293.15) + 1.524 \times 10^{-9} (T_s^4 - 7.396 \times 10^9) \] Let’s make an initial guess, \(T_s = 313.15\,\text{K} (40^\circ \text{C})\): - \(T_s - 293.15 = 20\) - \(T_s^4 = (313.15)^4 = 9.629 \times 10^9\) - \((T_s^4 - (293.15)^4) = 2.233 \times 10^9\) Plug in: \[ 8 = 0.13944 \times 20 + 1.524 \times 10^{-9} \times 2.233 \times 10^9 \] \[ 8 = 2.7888 + 3.402 \] \[ = 6.19 \] This is less than 8 W, so \(T_s\) must be higher. Try \(T_s = 323.15\, K\) (\(50^\circ\text{C}\)): - \(T_s - 293.15 = 30\) - \(T_s^4 = (323.15)^4 = 10.888 \times 10^9\) - \((T_s^4 - (293.15)^4) = 3.492 \times 10^9\) \[ 8 = 0.13944 \times 30 + 1.524 \times 10^{-9} \times 3.492 \times 10^9 \] \[ = 4.183 + 5.322 \] \[ = 9.505 \] Now it's slightly more than 8 W. Let's interpolate: - At \(T_s = 40^\circ\text{C}\), \(Q = 6.19\) - At \(T_s = 50^\circ\text{C}\), \(Q = 9.5\) - Target: \(Q = 8\) \[ T_s = 40 + \frac{8 - 6.19}{9.5 - 6.19} \times (50-40) \] \[ = 40 + \frac{1.81}{3.31} \times 10 = 40 + 5.47 = 45.5^\circ\text{C} \] --- ## **Final Answer** \[ \boxed{45.5^\circ \text{C}} \] --- ### **Summary Table** | Temp (°C) | \( Q \) (W) | |-----------|-------------| | 40 | 6.19 | | 50 | 9.50 | So, the average temperature of the hot surface of the board is **45.5°C**.