This problem asks us to calculate the amount of solar energy reflected by a window, given the direct solar irradiance, the window's orientation, the sun's position, and the glass's reflectivity.

Irradiance Calculation Basics

The amount of solar energy reflected by a surface depends on two main factors:

• The intensity of the solar radiation hitting the surface (Incident Solar Irradiance).
• The reflectivity of the surface material.

The formula relating these is:

$$ G_{ref} = G_{inc} \times \rho $$

Where:

• $G_{ref}$ is the reflected solar irradiance (in W/m$^2$).
• $G_{inc}$ is the solar irradiance incident on the surface (in W/m$^2$).
• $\rho$ is the reflectivity of the surface (dimensionless).

The challenge here is that the given solar strength (1000 W/m$^2$) is the direct irradiance in space, not necessarily the irradiance on the tilted/oriented window surface. We need to account for the angle between the sun's rays and the window surface.

Coordinate System and Vectors

To calculate the angle of incidence, we need to define a coordinate system and the relevant vectors:

• Let's use a standard coordinate system: X-axis pointing East, Y-axis pointing North, and Z-axis pointing Up.
• Window Normal Vector ($n_w$): The window is vertical and faces South. The normal vector points perpendicular to the surface, outwards from the room. In our coordinate system, the South direction corresponds to the negative Y-axis. Therefore, the window normal vector is $n_w = (0, -1, 0)$.
• Sun Vector ($s$): The sun's position is described by its azimuth and altitude angles.
  • The sun is in the Southwest. Assuming azimuth is measured clockwise from North (where North=0°, East=90°, South=180°, West=270°), Southwest corresponds to an azimuth angle $\phi = 225^\circ$.
  • The sun is 30° above the horizon. This is the altitude angle, $\beta = 30^\circ$.
  The components of the sun's direction vector (pointing towards the sun) are calculated as:
  • $s_x = \cos(\beta) \sin(\phi)$
  • $s_y = \cos(\beta) \cos(\phi)$
  • $s_z = \sin(\beta)$
  Substituting the values:
  • $\sin(225^\circ) = -1/\sqrt{2}$
  • $\cos(225^\circ) = -1/\sqrt{2}$
  • $\sin(30^\circ) = 1/2$
  • $\cos(30^\circ) = \sqrt{3}/2$
  So, the sun vector components are:
  • $s_x = (\sqrt{3}/2) \times (-1/\sqrt{2}) = -\sqrt{6}/4 \approx -0.6124$
  • $s_y = (\sqrt{3}/2) \times (-1/\sqrt{2}) = -\sqrt{6}/4 \approx -0.6124$
  • $s_z = 1/2 = 0.5$
  The sun vector is $s = (-\sqrt{6}/4, -\sqrt{6}/4, 1/2)$. This is a unit vector ($|s|=1$).

Angle of Incidence Calculation

The angle of incidence ($\theta_i$) is the angle between the sun's direction vector ($s$) and the surface's normal vector ($n_w$). We can find the cosine of this angle using the dot product:

$$ \cos(\theta_i) = \frac{s \cdot n_w}{|s| |n_w|} $$

Calculating the dot product $s \cdot n_w$:

$$ s \cdot n_w = (-\sqrt{6}/4)(0) + (-\sqrt{6}/4)(-1) + (1/2)(0) = \sqrt{6}/4 $$

Since both $s$ and $n_w$ are unit vectors ($|s|=1$, $|n_w|=1$), we have:

$$ \cos(\theta_i) = \sqrt{6}/4 $$

$$ \cos(\theta_i) \approx 0.6124 $$

This value represents the orientation factor – how effectively the sun's rays hit the window surface.

Incident Solar Irradiance on Window

The direct solar irradiance is given as $G_{direct} = 1000$ W/m$^2$. The irradiance actually incident on the window surface ($G_{inc}$) is this value multiplied by the cosine of the angle of incidence:

$$ G_{inc} = G_{direct} \times \cos(\theta_i) $$

$$ G_{inc} = 1000 \text{ W/m}^2 \times (\sqrt{6}/4) $$

$$ G_{inc} \approx 1000 \times 0.6124 \approx 612.4 \text{ W/m}^2 $$

Reflected Solar Energy Calculation

Now we can calculate the reflected solar energy ($G_{ref}$) using the incident irradiance ($G_{inc}$) and the glass reflectivity ($\rho = 0.08$):

$$ G_{ref} = G_{inc} \times \rho $$

$$ G_{ref} \approx 612.4 \text{ W/m}^2 \times 0.08 $$

$$ G_{ref} \approx 48.99 \text{ W/m}^2 $$

Rounding this value to the nearest whole number gives 49 W/m$^2$.

Conclusion

Therefore, the estimated amount of solar energy reflected by the window is approximately 49 W/m$^2$.