*4.1. Solar Angle Calculation*

At a specific latitude, the height of the sun depends on the hour. To establish the height of the sun-latitude relationship, basic knowledge of celestial physics is used, where the planet Earth is located in the centre (Figure 8). The equatorial plane of the celestial sphere (NS) is the equatorial plane of the Earth, where the azimuth is positive when viewed from the north (A), A\* = 360◦ − A, i.e., clockwise.

**Figure 8.** Spherical triangle model for the estimation of the sun's shadow at a specific latitude.

The angle of the astronomical meridian with the equatorial plane is the solar height (h), Φ is the latitude, and δ is the declination angle of the Earth. Therefore, to determine the solar height (h) at a particular time (H) at some location on Earth (latitude), for a spherical triangle, the corresponding equations of spherical trigonometry are employed logically, as presented in Figure 7.

If a spherical triangle is used, the Z point will be the origin of the coordinates. Note that the sides are the angles in radians. Therefore, in our case, the sides of the triangle are:


Using the law of sines to link all these variables in one system of equations, the first two sides and their angles can be replaced according to Equation (8) [28]:

$$\frac{\left(\Theta0^{\circ}-\delta\right)}{\sin(A\*)} = \frac{\left(\Theta0^{\circ}-\mathbf{h}\right)}{\sin\mathbf{H}}\tag{8}$$

$$\frac{\sin\left(90^\circ - \delta\right)}{\sin\left(360^\circ - \text{A}\right)} = \frac{\sin\left(90^\circ - \text{h}\right)}{\sin\text{H}} \implies \frac{\cos\delta}{-\sin\text{A}} = \frac{\cos\text{h}}{\sin\text{H}}\tag{9}$$

In the last two fractions, the sine of the 90◦ and 360◦ angles are 1 and 0, respectively. In addition, sin (90◦ − v) = cos v, sin (360◦ − r) = −sin (r), giving the next equation:

$$
\cos \delta \cdot \sin \mathbf{H} = -\sin \mathbf{A} \cdot \cos \mathbf{h} \tag{10}
$$

Using spherical trigonometry, i.e., the first law of cosines next to it (90◦ − δ), the consequent equation is:

$$
\sin \delta = \sin \Phi \cdot \sin \mathbf{h} + \cos \Phi \cdot \cos \mathbf{h} \cdot \cos \mathbf{A} \tag{11}
$$

Then, in spherical trigonometry, if the first law of cosines is applied for (90◦ − h), the next equation can be found:

$$
\sin \mathbf{h} = \sin \Phi \cdot \sin \delta + \cos \Phi \cdot \cos \delta \cdot \cos \mathbf{H} \tag{12}
$$

Thus, the value of h is obtained in terms of the variables (δ, Φ, H).

On the other hand, it is also necessary to know the measures and inclinations (β, α) of the PTC to calculate the shadow. To establish the shading end points, we must consider at first that the solar hours are used for the design of the PTC facility. The solar hours refer to the central peak hour (H0) and the hours that are equally divided backwards and forwards from this central hour. The central peak is assumed to be H0 = 0◦ for the Greenwich meridian, with each hour corresponds to 15 degrees, with the adding on the right of 15 degrees per hour of solar gain and the subtracting on the left of 15◦ per hour (with 0 assumed to be 360◦ to prevent values of angles as negative). e.g., in Spain, for a setting of four hours of sun, at the time of 10:00 h, there would be an H of 330◦, and at 12:00 h, it is reached an H of 30◦ (see Figure 9 as guidance).

**Figure 9.** Shadows. (**a**) Shadow at 10:00, (**b**) shadow at 12:00; and (**c**) Shadow at 14:00.

### *4.2. The Extent of the Shade*

For the estimation of the shadow, the four corners of the concentrator are chosen. So, with known values of δ, Φ, H, and using Equation (12), it can be calculated the projected shadow (see Figure 10), where the shadow is calculated at 10, 12, and 14 h in continuous time. The outer contour of both shadows will be the envelope, these are shown in Figure 10, where the contours of full shadow at 10, 12, and 14 h have been represented.

**Figure 10.** Range of the total projection of the collector's shadow.

Afterward, the distance (d) to the shading for every point is calculated based on Equation (12), linking the tilt angle of the PTC corner, the length, and the sun height.

$$\mathbf{d} = \mathbf{W} \cdot \frac{\cos \alpha}{\text{tg } \mathbf{h}} \tag{13}$$

Until now, the PTC's corners have been projected at shadow distances (di). To plot these shading distances on the floor, polar coordinates are applied, whereby the angle Hi is the hourly angle of the point i. The critical event is the shortest day of one year, where h represents the shortest day of the solar field design. The shading envelope of a PTC is obtained, and from this envelope, the shadow projected for each whole row of the PTC assembly can be drawn so that no shadow can be cast between the rows.

The data to be computed is the spacing between the rows of PTC lines or the distance between the pylon (see Figure 11), the triangle must be solved, whose vertexes are: shadow of the first hour, shadow of the last hour (third point), and the projection of the corner of PTC. The vertex angle P is (360 − (H1 − H3)), meaning the difference in the hour angles of the other two vertexes, and the sides from point P to first point and to third point are d1 and d3, respectively. Before for h1 and h3 are calculated the values of d1 and d3. Then, it is possible to determine an effective distance of shade considering the projection angle of the sun, as shown in Equation (14) (Figure 11),

$$\mathbf{d}''' = \mathbf{d} \cdot \sin \mathbf{H} = \mathcal{W} \left(\frac{\cos \alpha}{\text{tg } \mathbf{h}}\right) \cdot \sin \mathbf{H} \tag{14}$$

where the distance between the pylon can be obtained depending on the dimension's collector, the angle from the vertical of the collector, the solar hour angle, and height, as shown in the following Equation (16). In short, the distance depends the dimensions of the collector and the location of the solar field. 

$$\text{Distance between the polygon} = \text{d}''' + \text{d}' = \mathcal{W} \left( \frac{\cos \alpha \cdot \sin \text{H}}{\text{tg } \text{h}} \right) + \mathcal{W} \sin \alpha \tag{15}$$

$$\text{Distance between the polygon} = \mathcal{W}\left(\sin\alpha + \frac{\cos\alpha\cdot\sin\mathcal{H}}{\text{tg }\mathcal{h}}\right) \tag{16}$$

**Figure 11.** Diagram of d"' calculation polygons.

The hour angle (H) corresponds to the position of the observer with respect to the sun and the azimuth angle (A) is based on the position of the observer with respect to the north. Then, by using Equation (12), shadows at sunrise are calculated, Ortho, h = 0◦, that is, sin h = 0; where for the equations, the angles are used in radians, whereas the solutions are expressed as degrees (◦) to make it more user friendly (Equation (17)).

$$0 = \text{tg } \Phi \cdot \text{tg } \delta + \cos \text{H} \tag{17}$$

The result of the calculations in HORTHO◦ and HSUNSET◦.

Considering the opening plane size as standard one, *W* = 5.760 m and focal distance *f* = 1.710 m for the distance calculation. Known *W* and *f*, it is possible the calculation of the distance of the vertex of the collector perpendicular to the aperture plane according to Equation (18):

$$z = \frac{\left(\frac{W}{2}\right)^2}{4f} \tag{18}$$
