Let's break down the problem into steps: **Given:** - Width of heater, \( W = 40 \, \text{cm} = 0.4 \, \text{m} \) - Air velocity, \( u_\infty = 20 \, \text{m/s} \) - Surface heat flux, \( q'' = 2000 \, \text{W/m}^2 \) - Maximum surface temperature, \( T_s = 150^\circ\text{C} = 423.15\,\text{K} \) - Ambient air temperature, \( T_\infty \) (Assume room temp, say \( 25^\circ\text{C} = 298.15\,\text{K} \)) - Need to find **length** of the heater, \( L \), so that surface temperature does not exceed \( 150^\circ\text{C} \) - Both **laminar and turbulent boundary layers** exist (so use appropriate correlations for average \( h \) on a flat plate). --- ## 1. **Find properties of air at film temperature** **Film temperature:** \[ T_f = \frac{T_s + T_\infty}{2} = \frac{423.15 + 298.15}{2} = 360.65\,\text{K} \approx 87.5^\circ\text{C} \] At \( T_f \approx 87.5^\circ\text{C} \) (use tables or approximate): - \( \nu \) (kinematic viscosity) ≈ \( 25 \times 10^{-6} \, \text{m}^2/\text{s} \) - \( k \) (thermal conductivity) ≈ \( 0.029 \, \text{W/m}\cdot\text{K} \) - \( Pr \) ≈ 0.7 --- ## 2. **Convective heat transfer** The surface heat flux is: \[ q'' = h (T_s - T_\infty) \] \[ h = \frac{q''}{T_s - T_\infty} \] \[ h = \frac{2000}{150 - 25} = \frac{2000}{125} = 16\,\text{W/m}^2\cdot\text{K} \] --- ## 3. **Relate \( h \) to the Nusselt number** For external flow over a flat plate (length \( L \)), average Nusselt number: \[ Nu_L = \frac{h L}{k} \] --- ## 4. **Combined Laminar and Turbulent Correlation** For a flat plate where both regions exist (from Incropera & DeWitt): \[ Nu_L = 0.664 Re_L^{1/2} Pr^{1/3} \quad (\text{Laminar, } Re_x < 5 \times 10^5) \] \[ Nu_L = 0.037 Re_L^{0.8} Pr^{1/3} \quad (\text{Turbulent, } Re_x > 5 \times 10^5) \] For a plate that has both laminar and turbulent flow: \[ Nu_L = 0.664 Re_c^{1/2} Pr^{1/3} + 0.037 (Re_L^{0.8} - Re_c^{0.8}) Pr^{1/3} \] where \( Re_c = 5 \times 10^5 \) (critical Reynolds number for transition). --- ## 5. **Express Reynolds Number** \[ Re_L = \frac{u_\infty L}{\nu} \] --- ## 6. **Set up the equation** Now, plug in all values: - \( h = 16 \, \text{W/m}^2\cdot\text{K} \) - \( k = 0.029 \, \text{W/m}\cdot\text{K} \) - \( Pr = 0.7 \) - \( u_\infty = 20 \, \text{m/s} \) - \( \nu = 25 \times 10^{-6} \, \text{m}^2/\text{s} \) - \( Re_c = 5 \times 10^5 \) \[ Nu_L = \frac{h L}{k} \] So, \[ \frac{h L}{k} = 0.664 Re_c^{1/2} Pr^{1/3} + 0.037 (Re_L^{0.8} - Re_c^{0.8}) Pr^{1/3} \] Let \( Re_L = \frac{20L}{25 \times 10^{-6}} = 8 \times 10^5 L \) (with \( L \) in meters) --- ## 7. **Plug in Numbers** \[ \frac{16L}{0.029} = 0.664 (5 \times 10^5)^{1/2} (0.7)^{1/3} + 0.037 \left[ (Re_L)^{0.8} - (5 \times 10^5)^{0.8} \right] (0.7)^{1/3} \] First, calculate constants: - \( (5 \times 10^5)^{1/2} = 707.1 \) - \( (0.7)^{1/3} = 0.887 \) - \( (5 \times 10^5)^{0.8} = e^{0.8 \ln(5 \times 10^5)} \approx e^{0.8 \times 13.1224} \approx e^{10.498} \approx 36346 \) So, \[ \frac{16L}{0.029} = 0.664 \times 707.1 \times 0.887 + 0.037 \left[ (Re_L)^{0.8} - 36346 \right] \times 0.887 \] \[ \frac{16L}{0.029} = 416.7 + 0.0328 \left[ (Re_L)^{0.8} - 36346 \right] \] Now, \( Re_L = 8 \times 10^5 L \): \[ (Re_L)^{0.8} = (8 \times 10^5 L)^{0.8} = (8 \times 10^5)^{0.8} \cdot L^{0.8} \] Calculate \( (8 \times 10^5)^{0.8} \): - \( \ln(8 \times 10^5) = \ln(8) + \ln(10^5) = 2.0794 + 11.5129 = 13.5923 \) - \( 0.8 \times 13.5923 = 10.874 \) - \( e^{10.874} \approx 52,753 \) So, \[ (Re_L)^{0.8} = 52,753 \cdot L^{0.8} \] So equation becomes: \[ \frac{16L}{0.029} = 416.7 + 0.0328 \left[ 52,753 \cdot L^{0.8} - 36,346 \right] \] \[ \frac{16L}{0.029} = 416.7 + 0.0328 \times 52,753 \cdot L^{0.8} - 0.0328 \times 36,346 \] \[ \frac{16L}{0.029} = 416.7 + 1,728.3 L^{0.8} - 1,191.1 \] \[ \frac{16L}{0.029} = 1,728.3 L^{0.8} - 774.4 \] \[ \frac{16L}{0.029} + 774.4 = 1,728.3 L^{0.8} \] \[ \frac{16L}{0.029} + 774.4 = 1,728.3 L^{0.8} \] --- ## 8. **Solve for \( L \) (Iterative/Numerical solution)** Let \( L = x \). \[ \frac{16x}{0.029} + 774.4 = 1,728.3 x^{0.8} \] \[ 551.72 x + 774.4 = 1,728.3 x^{0.8} \] Let’s try \( x = 1 \) m: \[ 551.72 (1) + 774.4 = 1,728.3 (1)^{0.8} \] \[ 1,326.12 = 1,728.3 \] So, left side is less than right. Try \( x = 1.5 \): \[ 551.72 \times 1.5 + 774.4 = 827.58 + 774.4 = 1,601.98 \] \[ 1.5^{0.8} = e^{0.8 \ln 1.5} = e^{0.8 \times 0.4055} = e^{0.3244} = 1.383 \] \[ 1,728.3 \times 1.383 = 2,390.7 \] Still left < right. Try \( x = 2 \): \[ 551.72 \times 2 + 774.4 = 1,103.44 + 774.4 = 1,877.84 \] \[ 2^{0.8} = e^{0.8 \ln 2} = e^{0.8 \times 0.6931} = e^{0.5545} = 1.741 \] \[ 1,728.3 \times 1.741 = 3,009.1 \] Still left < right. Try \( x = 3 \): \[ 551.72 \times 3 + 774.4 = 1,655.16 + 774.4 = 2,429.56 \] \[ 3^{0.8} = e^{0.8 \ln 3} = e^{0.8 \times 1.0986} = e^{0.8789} = 2.408 \] \[ 1,728.3 \times 2.408 = 4,159.8 \] Still left < right. Try \( x = 5 \): \[ 551.72 \times 5 + 774.4 = 2,758.6 + 774.4 = 3,533.0 \] \[ 5^{0.8} = e^{0.8 \times 1.6094} = e^{1.2875} = 3.623 \] \[ 1,728.3 \times 3.623 = 6,261.5 \] Still left < right. Try \( x = 10 \): \[ 551.72 \times 10 + 774.4 = 5,517.2 + 774.4 = 6,291.6 \] \[ 10^{0.8} = e^{0.8 \times 2.3026} = e^{1.842} = 6.313 \] \[ 1,728.3 \times 6.313 = 10,903 \] Still left < right. Try \( x = 20 \): \[ 551.72 \times 20 + 774.4 = 11,034.4 + 774.4 = 11,808.8 \] \[ 20^{0.8} = e^{0.8 \times 2.9957} = e^{2.3966} = 10.98 \] \[ 1,728.3 \times 10.98 = 18,981 \] Still left < right. Try \( x = 30 \): \[ 551.72 \times 30 + 774.4 = 16,551.6 + 774.4 = 17,326 \] \[ 30^{0.8} = e^{0.8 \times 3.4012} = e^{2.721} = 15.209 \] \[ 1,728.3 \times 15.209 = 26,292 \] Still left < right. Try \( x = 50 \): \[ 551.72 \times 50 + 774.4 = 27,586 + 774.4 = 28,360.4 \] \[ 50^{0.8} = e^{0.8 \times 3.9120} = e^{3.1296} = 22.873 \] \[ 1,728.3 \times 22.873 = 39,543 \] Still left < right. Try \( x = 100 \): \[ 551.72 \times 100 + 774.4 = 55,172 + 774.4 = 55,946.4 \] \[ 100^{0.8} = e^{0.8 \times 4.6052} = e^{3.6842} = 39.78 \] \[ 1,728.3 \times 39.78 = 68,772 \] Now, left is closer to right. - At \( x = 100 \), left: 55,946, right: