Let's solve the problem step by step: --- ## **Step 1: Gather Given Data** - Wall height, \( H = 2.5 \) m - Wall area, \( A = 6 \) m² - Surface temperature, \( T_s = 300 \) K - Ambient temperature, \( T_\infty = 280 \) K - Find temperature at \( y = .5 \) m from ceiling and \( x = .05 \) m from wall. - Air properties at average temperature (let's use \( T_{film} = (300+280)/2 = 290 \) K): From the table for \( T = 290 \) K: - \( \nu = 15 \times 10^{-6} \) m²/s - \( \alpha = 21.18 \times 10^{-6} \) m²/s - \( Pr = .7096 \) --- ## **Step 2: Calculate Grashof Number (\( Gr_x \))** The local Grashof number is: \[ Gr_x = \frac{g \beta (T_s - T_\infty) x^3}{\nu^2} \] Where: - \( g = 9.81 \) m/s² (acceleration due to gravity) - \( \beta = \frac{1}{T_{film}} \) (for ideal gas, in K\(^{-1}\)) So, \[ \beta = \frac{1}{290} = .00345 \, \text{K}^{-1} \] \[ x = \text{vertical distance from top} = .5 \text{ m} \] \[ \nu = 15 \times 10^{-6} \text{ m}^2/\text{s} \] \[ T_s - T_\infty = 20 \text{ K} \] Calculate: \[ Gr_x = \frac{9.81 \times .00345 \times 20 \times (.5)^3}{(15 \times 10^{-6})^2} \] Calculate numerator: \[ 9.81 \times .00345 \times 20 \times (.125) = .008467 \] Denominator: \[ (15 \times 10^{-6})^2 = 225 \times 10^{-12} = 2.25 \times 10^{-10} \] So, \[ Gr_x = \frac{.008467}{2.25 \times 10^{-10}} \approx 3.76 \times 10^7 \] --- ## **Step 3: Calculate η (eta)** The similarity variable, \[ \eta = \frac{y}{x} \left( \frac{Gr_x}{4} \right)^{1/4} \] Where: - \( y = .05 \) m (distance from wall) - \( x = .5 \) m (distance from top) Plug in: \[ \eta = \frac{.05}{.5} \left( \frac{3.76 \times 10^7}{4} \right)^{1/4} \] \[ = .1 \times (9.4 \times 10^6)^{1/4} \] Calculate \( (9.4 \times 10^6)^{1/4} \): - \( \log_{10}(9.4 \times 10^6) \approx 6.97 \) - \( \frac{6.97}{4} \approx 1.74 \) - \( 10^{1.74} \approx 55 \) So, \[ \eta \approx .1 \times 55 = 5.5 \] --- ## **Step 4: Use the Similarity Solution Chart** - \( Pr \approx .71 \), use the curve for \( Pr = .72 \). - For \( \eta \approx 5.5 \), from the chart, \( \theta \) (dimensionless temperature) is nearly **zero**. Recall, \[ \theta = \frac{T - T_\infty}{T_s - T_\infty} \] From the chart, at \( \eta = 5 \) or \( 5.5 \), \( \theta \approx \). --- ## **Step 5: Calculate the Local Temperature** \[ \theta = \frac{T - T_\infty}{T_s - T_\infty} \implies T = \theta (T_s - T_\infty) + T_\infty \] If \( \theta \approx \): \[ T \approx \times 20 + 280 = \boxed{280~\text{K}} \] --- ## **Step 6: Final Answer** ### **The temperature of air at .5 m down from the ceiling and .05 m from the wall is approximately:** \[ \boxed{280~\text{K}} \] --- ### **Summary of Steps** 1. Found average properties of air at 290 K. 2. Calculated local Grashof number \( Gr_x \). 3. Computed similarity variable \( \eta \). 4. Used the provided graph for \( \theta \) at calculated \( \eta \). 5. Determined local air temperature is approximately the ambient temperature (\( 280 \) K).