# Step-by-Step Solution We are given: - **Copper rod**: - Diameter, \( d = 10 \) mm \( = 0.01 \) m - Length, \( L = 100 \) mm \( = 0.1 \) m - **Temperature at both ends**: \( T_1 = T_2 = 90^\circ \)C - **Ambient air temperature**: \( T_\infty = 25^\circ \)C - **Air velocity**: \( U_\infty = 25 \) m/s - **Rod material**: Copper (\( k \approx 400 \) W/m·K) We are to find: 1. **Temperature at the midplane** (\( T_{mid} \)) 2. **Rate of heat transfer from the rod** (\( Q \)) --- ## 1. **Model the Problem** This is a classic **finned rod problem** (convection from a horizontal cylinder, 1D conduction along the rod). The rod is exposed to air along its length (except at the ends), which are maintained at \( 90^\circ \)C. Assumptions: - Steady-state - 1D conduction along the rod axis - Uniform convection around the rod circumference --- ## 2. **Find the Convection Heat Transfer Coefficient (\( h \))** ### **Step 1: Find Reynolds Number (\( \text{Re}_D \))** \[ \text{Re}_D = \frac{U_\infty D}{\nu} \] - \( U_\infty = 25 \) m/s - \( D = 0.01 \) m For air at \( \sim 25^\circ \)C: - Kinematic viscosity \( \nu \approx 1.5 \times 10^{-5} \) m²/s \[ \text{Re}_D = \frac{25 \times 0.01}{1.5 \times 10^{-5}} = \frac{0.25}{1.5 \times 10^{-5}} \approx 16,\!667 \] ### **Step 2: Find Nusselt Number (\( \text{Nu}_D \))** For flow across a cylinder (\( 10^4 < \text{Re}_D < 10^5 \)), use the **Hilpert correlation**: \[ \text{Nu}_D = C \cdot \text{Re}_D^m \cdot \text{Pr}^{1/3} \] From tables (Incropera & DeWitt): - \( C = 0.193, \ m = 0.618 \) - For air, \( \text{Pr} \approx 0.71 \) \[ \text{Nu}_D = 0.193 \times (16,\!667)^{0.618} \times 0.71^{1/3} \] \[ (16,\!667)^{0.618} \approx 384 \] \[ 0.71^{1/3} \approx 0.892 \] \[ \text{Nu}_D \approx 0.193 \times 384 \times 0.892 \approx 66.1 \] ### **Step 3: Calculate \( h \)** \[ \text{Nu}_D = \frac{h D}{k_{air}} \] - \( k_{air} \) at 25°C \( \approx 0.026 \) W/m·K \[ h = \frac{\text{Nu}_D \cdot k_{air}}{D} = \frac{66.1 \times 0.026}{0.01} \approx 172 \text{ W/m}^2\cdot\text{K} \] --- ## 3. **Fin Parameter** The rod acts as a **straight fin** of uniform cross-section, with both ends at the base temperature. ### **Fin parameter \( m \):** \[ m = \sqrt{\frac{h P}{k A_c}} \] Where: - \( P \): Perimeter = \( \pi D \) - \( A_c \): Cross-sectional area = \( \frac{\pi D^2}{4} \) \[ m = \sqrt{\frac{h \pi D}{k \cdot (\pi D^2/4)}} = \sqrt{\frac{4h}{k D}} \] Plug in values: - \( h = 172 \) W/m²·K - \( k = 400 \) W/m·K - \( D = 0.01 \) m \[ m = \sqrt{\frac{4 \times 172}{400 \times 0.01}} = \sqrt{\frac{688}{4}} = \sqrt{172} \approx 13.12 \ \text{m}^{-1} \] --- ## 4. **Temperature Distribution** For a fin of length \( L \) with both ends at \( T_b \) and convective cooling along the length, the temperature at distance \( x \) from one end: \[ \frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh\left(m \left(\frac{L}{2} - x\right)\right)}{\cosh\left(\frac{mL}{2}\right)} \] At the **midplane** (\( x = L/2 \)), the numerator becomes \( \cosh(0) = 1 \): \[ \frac{T_{mid} - T_\infty}{T_b - T_\infty} = \frac{1}{\cosh\left(\frac{mL}{2}\right)} \] \[ T_{mid} = T_\infty + (T_b - T_\infty) \frac{1}{\cosh\left(\frac{mL}{2}\right)} \] Plug in values: - \( T_\infty = 25^\circ \)C - \( T_b = 90^\circ \)C - \( m = 13.12 \) m\(^{-1}\) - \( L = 0.1 \) m \[ \frac{mL}{2} = \frac{13.12 \times 0.1}{2} = 0.656 \] \[ \cosh(0.656) \approx 1.217 \] \[ \frac{1}{1.217} \approx 0.821 \] \[ T_{mid} = 25 + (90 - 25) \times 0.821 = 25 + 65 \times 0.821 \approx 25 + 53.36 = 78.4^\circ \textrm{C} \] --- ## 5. **Total Heat Loss from the Rod (\( Q \))** For a fin with both ends at base temperature: \[ Q = 2 \cdot \left[ \sqrt{h P k A_c} \cdot (T_b - T_\infty) \cdot \tanh\left( \frac{mL}{2} \right) \right] \] - \( P = \pi D = 0.0314 \) m - \( A_c = \frac{\pi}{4} D^2 = 7.85 \times 10^{-5} \) m² \[ \sqrt{h P k A_c} = \sqrt{172 \times 0.0314 \times 400 \times 7.85 \times 10^{-5}} \] Calculate inside the square root: - \( 172 \times 0.0314 = 5.4008 \) - \( 400 \times 7.85 \times 10^{-5} = 0.0314 \) - \( 5.4008 \times 0.0314 = 0.1697 \) - \( \sqrt{0.1697} = 0.412 \) \[ Q = 2 \times 0.412 \times (90 - 25) \times \tanh(0.656) \] \[ \tanh(0.656) \approx 0.576 \] \[ (90-25) = 65 \] \[ 0.412 \times 65 = 26.78 \] \[ 26.78 \times 0.576 = 15.42 \] \[ Q = 2 \times 15.42 = 30.84 \ \textrm{W} \] --- # **Final Answers** ## **1. Temperature at the midplane of the rod** \[ \boxed{T_{mid} \approx 78.4^\circ \textrm{C}} \] ## **2. Rate of heat transfer from the rod** \[ \boxed{Q \approx 30.8 \ \textrm{W}} \] --- ### **Summary Table** | Quantity | Symbol | Value | |-------------------------|------------|---------------| | Midplane Temperature | \( T_{mid} \) | \( 78.4^\circ \)C | | Heat Transfer Rate | \( Q \) | \( 30.8 \) W | --- **Diagram (for reference):**  *Alt: Diagram of a horizontal copper rod with ends at 90°C, length 0.1 m, diameter 0.01 m, exposed to air crossflow at 25°C and 25 m/s.*