Let's break down the problem and solve it step by step. --- ## **Given Data** - **Thermal conductivity of blade, \( k \) = 17 W/m·K** - **Length of blade, \( L \) = 5.3 cm = 0.053 m** - **Perimeter, \( P \) = 11 cm = 0.11 m** - **Cross-sectional area, \( A_c \) = 5.13 cm² = 5.13 × 10⁻⁴ m²** - **Hot gas temperature, \( T_\infty \) = 973°C** - **Convection heat transfer coefficient, \( h \) = 538 W/m²·K** - **Base temperature, \( T_b \) = 450°C** - **Tip is adiabatic (no heat loss at the tip)** --- ## **Step 1: Recognize the Problem Type** This is a **fin (extended surface)** problem, where heat is conducted through the blade and lost to the surroundings by convection. The tip is adiabatic. --- ## **Step 2: Calculate the Fin Parameter (\( m \))** \[ m = \sqrt{\frac{hP}{kA_c}} \] Plug in the values: \[ m = \sqrt{\frac{538 \times 0.11}{17 \times 5.13 \times 10^{-4}}} \] Calculate numerator and denominator: - \( hP = 538 \times 0.11 = 59.18 \) - \( kA_c = 17 \times 5.13 \times 10^{-4} = 0.008721 \) So, \[ m = \sqrt{\frac{59.18}{0.008721}} = \sqrt{6788.19} \approx 82.41\, \text{m}^{-1} \] --- ## **Step 3: Fin Heat Transfer Rate (Adiabatic Tip)** For a fin with an **adiabatic tip**: \[ Q_{fin} = \sqrt{hPkA_c} (T_b - T_\infty) \cdot \tanh(mL) \] First, calculate \( \sqrt{hPkA_c} \): - \( hPkA_c = 59.18 \times 0.008721 = 0.5159 \) - \( \sqrt{0.5159} = 0.7188 \) Now, calculate \( mL \): - \( mL = 82.41 \times 0.053 = 4.3677 \) Now, \( \tanh(4.3677) \approx 0.9996 \) (since tanh approaches 1 for large values). So, \[ Q_{fin} = 0.7188 \times (450 - 973) \times 0.9996 \] \[ Q_{fin} = 0.7188 \times (-523) \times 0.9996 \] \[ Q_{fin} = -376.2 \text{ W} \] The negative sign indicates heat flows **from the hot gas to the blade base** (which makes sense since \( T_\infty > T_b \)). The **magnitude** is the heat transfer rate. \[ \boxed{Q_{fin} = 376\, \text{W}} \] --- ## **Step 4: Temperature at the Tip (\( T_L \))** For adiabatic tip: \[ \frac{T_L - T_\infty}{T_b - T_\infty} = \frac{1}{\cosh(mL)} \] Calculate \( \cosh(mL) \): - \( \cosh(4.3677) \approx 39.16 \) \[ \frac{T_L - 973}{450 - 973} = \frac{1}{39.16} \] \[ T_L - 973 = \frac{450 - 973}{39.16} \] \[ T_L - 973 = \frac{-523}{39.16} = -13.36 \] \[ T_L = 973 - 13.36 = 959.6^\circ\text{C} \] --- ## **Final Answers** ### **1. Heat transfer rate to the blade:** \[ \boxed{376\, \text{W}} \] ### **2. Temperature at the tip:** \[ \boxed{960^\circ \text{C}} \] --- ### **Summary Table** | Quantity | Value | |-------------------------|------------| | Heat transfer rate, \(Q\) | 376 W | | Tip temperature, \(T_L\) | 960°C | --- If any step needs further explanation, let me know!