[MUSIC] Hello, I hope you've been learning from the lectures so far. In the previous lectures, you saw how light emitted by the sun is treated as radiance incident on a solar panel. Let us now consider this topic in more detail by working through an example question. The question is stated as follows. Calculate the total irradiance in watts per meter squared incident on a panel at solar noon for the following conditions. Direct normal irradiance of 800 watts per meter squared. Diffuse irradiance of 100 watts per meter squared. Panel tilt angle of 25 degrees, panel azimuth angle of 170 degrees, and solar zenith angle of 45 degrees. If you wish, at this point, you may stop the video and try to solve the problem on your own. If not, let's continue. The irradiance on the panel is defined as power per area, and is therefore measured in watts per meter squared. The total irradiance is made up of two components, the direct irradiance and the diffuse irradiance. So before we begin this problem, let us briefly define these two components of irradiance. Let's first consider this special case when the panel is directly facing the sun. In other words, when the panel normal is coincident with the vector pointing at the sun. The irradiance incident on the panel in this situation is the direct normal irradiance, or DNI. In obtaining this value, one must ignore all the light coming from the rest of the sky. One could imagine directly viewing the sun through an aperture. In contrast, the diffuse irradiance is composed of the light coming from the rest of the sky, excluding that coming directly from the sun. This is the sun with all the rays bouncing off of clouds and dust and water molecules. On a cloudy day, if the sun is hidden by cloud cover, this contribution can be quite important. Notice also that above the atmosphere this value would be zero, hence the blackness of space once one escapes Earth's grasp. It's not always true, however, that the panel will be pointing directly at the Sun. In such a case, we must therefore refer to the direct irradiance rather than the direct normal irradiance. The direct irradiance is always measured in watts per meter squared, but will depend on the angle between the panel normal and the vector pointing at the Sun. More accurately, the direct irradiance will be the direct normal radiance, I nought, times the cosine of the angle, chi, between the panel normal and the vector pointing at the sun. We calculate this angle by first describing the panel normal vector and the sun position vector, each with two values of angle and polar coordinates. The reference directions we will use to do so will be a vector pointing straight up and one pointing due north. First, we describe the orientation of the panel using two angles. We define the panel tilt, beta, which is measured between the panel normal and the vector pointing straight up. For a beta of zero, a panel will be laying flat on the ground, and for a beta of 90 degrees, the panel is resting on its edge, like a wall. The second angle is the panel azimuth angle, alpha. This angle is measured relative to due North and is defined by the component of the panel normal in the plane of the horizon. Put more simply, if the panel is tilted towards due North, the panel azimuth is zero, and if it's titled toward due South the panel azimuth is 180 degrees. We can do a similar exercise for the sun's position in the sky. If the sun's position vector is also described by two angles, the solar azimuth angle psi s, for which a value of zero means the sun positioned directly North relative to the viewer. And the solar zenith angle, theta s, which is the angle between the position vector of the sun and the reference vector pointing straight up. This can also be expressed as the elevation of the sun above the horizon, gamma s. The sum of theta s and gamma s must be 90 degrees. We can now use these four angles to calculate absolute angle between the panel normal and the position vector of the sun. This can be simply done using the expression shown here. In case you're interested, this expression has been derived by transforming the two vectors from polar into Cartesian coordinates, and taking the scalar products of the two vectors. However, we can just use this equation as it is to do our calculation for the direct irradiation. One last consideration should be made about the diffuse irradiance and the tilt of the panel. We assume that diffuse irradiance is collected equally from all points in the sky. However, if the panel is tilted, then part of the sky will be behind the panel. This unseen portion of the sky can be calculated as shown. Using the panel tilt angle and the total angle expanding the entire sky, 180 degrees. We can now calculate the total irradiance by adding up these two different components. First, we calculate the direct irradiance for our tilted panel. To do so, we apply the formula that we saw previously. It's an important one, so take care to note it. Now, let's plug in the numbers. The direct normal irradiance was given as 800 watts per meter squared, and we were given the panel tilt, beta, and the panel azimuth angle, alpha, in the question. The fact that it is solar noon means that the sun is due South, so the solar azimuth angle, psi s, is 180 degrees, and the solar zenith angle, theta s, is given as 45 degrees. The equation given allows us to calculate the absolute angle between the panel normal and the sun position vector. Plugging in all the values, we obtain a value of direct irradiance of 748 watts per meter squared. We also need to know the value of the diffuse irradiance. This value will be reduced by the panel tilt, beta, alone, and it is described by this equation. We can now plug in the numbers to get a numerical value. The diffuse irradiance from the entire sky was given as 100 watts per meter squared. For a panel tilt of 25 degrees, this gives us our diffuse irradiance, which is 86 watts per meter squared. Finally, the total irradiance is just the sum of these two, the direct irradiance and the diffuse irradiance. Adding these two quantities together brings us to our answer, 834 watts per meter squared. Through this work problem, I hope you've gained some insight into how to calculate the total irradiance on a panel from available solar irradiance data. Thank you for your attention, and see you again soon! [MUSIC]