Sure! Let’s break down the problem step by step. --- ## **Given Data** - **Surface area of circuit board, \( A \):** \( .07 \, m^2 \) - **Air temperature, \( T_\infty \):** \( 300 \, K \) - **Circuit board surface temperature, \( T_s \):** \( 340 \, K \) - **Table 2.1 (forced convection, air):** - **Convection heat transfer coefficient, \( h \):** \( 25 - 250 \, W/(m^2 \cdot K) \) --- ## **Step 1: Write the Convection Heat Transfer Equation** The heat transfer by convection is given by: \[ Q = h \cdot A \cdot (T_s - T_\infty) \] --- ## **Step 2: Substitute Known Values** - \( A = .07 \, m^2 \) - \( T_s = 340 \, K \) - \( T_\infty = 300 \, K \) - \( h \) will be two cases: **smallest (25)** and **largest (250)** \[ Q = h \cdot .07 \cdot (340 - 300) \] \[ Q = h \cdot .07 \cdot 40 \] \[ Q = h \cdot 2.8 \] --- ## **Step 3: Calculate Smallest and Largest Heat Transfer Rates** ### **A. Smallest Value (\( h = 25 \, W/(m^2 \cdot K) \))** \[ Q_{min} = 25 \cdot 2.8 = 70 \, W \] --- ### **B. Largest Value (\( h = 250 \, W/(m^2 \cdot K) \))** \[ Q_{max} = 250 \cdot 2.8 = 700 \, W \] --- ## **Step 4: Final Answers** - **Smallest heat transfer rate:** \( \boxed{70 \, W} \) - **Largest heat transfer rate:** \( \boxed{700 \, W} \) --- ## **Summary Table** | Case | \( h \) \([W/m^2K]\) | \( Q \) (W) | |-------------|---------------------|-------------| | Smallest | 25 | 70 | | Largest | 250 | 700 | --- ### **Explanation** - The heat transfer rate depends directly on \( h \), the convection coefficient. - For forced convection over a surface, \( h \) can range widely depending on air speed and flow characteristics. - Using the lowest and highest values from the table covers the full possible range. --- If you have any questions about the steps, let me know!