Let's break down the problem step by step: --- ## **Step 1: Understanding Reynolds Number** The Reynolds number (\( Re \)) is given by: \[ Re = \frac{\rho V L}{\mu} \] where: - \( \rho \) = density of air - \( V \) = velocity of air - \( L \) = characteristic length (e.g., car length) - \( \mu \) = dynamic viscosity of air For **dynamic similarity**, the Reynolds numbers of the model (\( Re_m \)) and prototype (\( Re_p \)) must be equal. \[ Re_m = Re_p \] --- ## **Step 2: Setting Up the Ratio** For a **1/10-scale model**: \[ L_m = \frac{1}{10}L_p \] Assume both are tested at **standard air pressure**, but possibly different temperatures. Let \( V_m \) and \( V_p \) be the velocities for the model and prototype, respectively. \[ \frac{\rho_m V_m L_m}{\mu_m} = \frac{\rho_p V_p L_p}{\mu_p} \] Assuming both use **air at standard pressure**, but possibly different temperatures, so \( \rho \) and \( \mu \) can change due to temperature. --- ## **Step 3: Rearranging for Model Testing** \[ \Rightarrow \frac{\rho_m V_m L_m / \mu_m}{\rho_p V_p L_p / \mu_p} = 1 \] \[ \Rightarrow \frac{\rho_m}{\rho_p} \cdot \frac{V_m}{V_p} \cdot \frac{L_m}{L_p} \cdot \frac{\mu_p}{\mu_m} = 1 \] We know: - \( L_m/L_p = 1/10 \) To make test practical, you want \( V_m \) not to be excessively high. --- ## **Step 4: Effect of Temperature on Air Properties** At **lower air temperatures**: - **Density \( \rho \)** increases (air is denser). - **Viscosity \( \mu \)** decreases (air is less viscous). At **higher air temperatures**: - **Density \( \rho \)** decreases. - **Viscosity \( \mu \)** increases. --- ## **Step 5: What Temperature Is Better?** We want to **match Reynolds numbers**. To do this with a smaller model (\( L_m < L_p \)), we need to compensate by increasing \( V_m \) or by changing fluid properties. \[ \frac{V_m}{V_p} = \frac{L_p}{L_m} \cdot \frac{\rho_p}{\rho_m} \cdot \frac{\mu_m}{\mu_p} \] If we **lower the temperature**: - \( \rho_m \) increases, \( \mu_m \) decreases. - Both effects **increase** the Reynolds number for the same velocity. Thus, for the **model** at **lower temperature**, we can **achieve the required Reynolds number at a lower velocity** (which is more practical and safer for wind tunnel operation). --- ## **Step 6: Final Answer** ### **Should the wind tunnel air be colder or hotter?** - **It should be colder than standard sea-level air temperature of 15°C.** ### **Why?** - **Colder air** has **higher density** and **lower viscosity**, which increases the Reynolds number for the same velocity and length. - This helps you **match the prototype Reynolds number with a lower wind tunnel speed**, making the test more practical. --- ## **Summary Table** | Temperature | Density (\( \rho \)) | Viscosity (\( \mu \)) | Reynolds Number (\( Re \)) | |-------------|----------------------|-----------------------|----------------------------| | Lower | Higher | Lower | Higher | | Higher | Lower | Higher | Lower | --- ## **Final Statement** > **It is better for the wind tunnel air to be colder than standard sea-level air temperature of 15°C, because lower temperature increases air density and decreases viscosity, allowing you to more easily achieve the required Reynolds number for dynamic similarity at a lower, more manageable wind tunnel speed.** --- If you have access to Table B4, you can verify the actual values of air properties at different temperatures to see this effect numerically!