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   INTRODUCTION

In section 1.2, a special case of sun shading by very long shading device is considered.  Here we will consider sun shading from general devices.  Vectorial relationships will be used.

            Consider Figure 1.3.1.


Figure 1.3.1    An inclined plane.







A coordinate (x1, x2, x3) is identified with the zenith, East and North direction. A unit vector representing the direction of the sun and a unit vector representing the normal to the inclined plane, based on the x-coordinate, are given as

                   Vsx  =   ,   solar vector                                 (1.3.1)

 

                   Vnx =   ,   plane normal                                  (1.3.2)

 

The angle between the solar vector and the plane normal can be obtained from the following relationships           

                               cosq          =          (Vsx, Vnx)

          =   sinas . cosb + cosas . sings . sinb . singp + cosas . cosgs . sinb . cosgp.

1.3.1  Sun Shading from a Shading Device of Regular Shape

For a shading device of regular shape, it suffices generally to consider the corner points of the shading device.

Shading by a Point Above the Inclined Plane

A shading device can be represented by its corner points.  Shading calculation is simplified by considering the shading points.

            Let Xp the vector of a point P above the inclined plane, in the x-coordinate.  Let the height of vector Xp above the inclined plane be h.

            Let S be the shadow of the point P casted by the sun on the inclined plane.  Vector Xs represents the point S.  The vector Xs, Xp and Vsx must be collinear, that is the vector     Xs-Xp is parallel to vector Vsx.  This relationship can be expressed mathematically as

                   Xs    =   Xp + l Vsx, 

or 

                   Xs - Xp    =   l Vsx, 

where   l is an arbitrary real number.

Multiplying Xs by Vn, the normal vector of the plane, gives

                   (Vnx, Xs)  =           (Vnx, Xp) + l (Vnx, Vsx)

But Vnx is perpendicular to Xs, so that (Vnx, Xs) = 0  and

                   (Vnx, Xp) = h,       the height of Xp from the plane.

Also,           Vnx, Vs) =   cosq.

Therefore,   l     =   -h / cosq   and

                          Xs =  Xp - h Vsx / cosq                                                              (1.3.3)

Example 1.3.1

The configuration is shown in Figure 1.3.2.


Figure 1.3.2    The configuration of a window with perpendicular overhang. The window faces North.






Consider a window facing north at 14:00 hour on 30 June in Bangkok.

Here            jd      =   181,

                   B     =   360 (jd - 81) / 364

                          =   98.9°, and

                   Eqt   =   -3.33 minutes.

For the solar time ts, it is given as

                   ts     =   14:00 - 4 (105° - 100.5°) + Eqt,

  =   13:38:40 or 13.6445 hours (decimal).

Therefore, we obtain

       w    =   24.675°,

       Lt    =   13.7° N, and

       d     =   23.184°.

From Equation (4.2.3), we obtain     

sinas=sin(13.7°).sin(23.184°)+cos(13.7°).cos(23.184°).cos(24.6675°)
  =   0.09324 + 0.8116 
  =   0.9048

This gives    as   =   64.8°

From Equation (1.2.4), we have      

       sings   =   cos(23.184°).sin(24.6675°) / cos(64.8º)

                            = 0.9011

This gives gs = 64.296° or 115.704° due west. The value 115.704° is correct for d = 23.2° which is greater than the latitude Lt =13.7° N of Bangkok. From the configuration of the window, we have

                   gp    = 180°,

                   b     =   90°.

The solar vector and the plane normal are obtained as

                   Vsx  =               and    Vnx  =  

The point P, a corner of the overhang, can be represented by Xp.  From the figure, Xp can be obtained as

Xp   =  

From Equation (1.3.3), the coordinate of the shadow is obtained as           

                   Xs   =   Xp  - h.Vsx/ cos q

Here,           cosq =   (Vnx,Vsx)  = = 0.1847

Therefore,  

Xs   =
=

Figure 1.3.3 shows the shade point S on the plane of the wall.  The shade of the shading device can be drawn from a consideration of symmetry.  The complete shade is also shown in the figure.

Figure 1.3.3    The configuration of the shade




In this example the shading device is rectangular.  For non symmetric shading device, more than one corner points of the shading device may have to be used.

Condition of Occurrence of a Shade

Shade will occur if the shading point is above a plane and the following conditions are met:

       a)  the sun must be up, sunrise time<solar time under consideration<sunset time,

b)      the sun must be facing the plane, i.e. the angle between the vector Vs and the plan  

      normal plane Vn must be positive, (Vs, Vn) = cosq  > 0.

The sunrise and sunset times can be found from earlier consideration. 

Example 1.3.2

For Bangkok at 30 June, we have

       jd      =   181,

       d     =   23.18°

       Lt    =   13.7° N, and

       tan Lt . tand =   0.10438

This gives ws,r = 96° º 1.675 radians  º 6.4 h from solar noon.

Sunrise, and sunset times    

=   12 ± 6.4 =   5.6 hour sunrise decimal, solar time,
  =   18.4 hour sunset
  º   5:36:00    sunrise time,    
  º   18:24:00  sunset time.    

Converting to clock time, from

       B     =   360° (181 - 81) / 364      

  =   98.9°,

we get         Eqt   =   9.87 sin(2 ´ 98.9°) - 7.53 cos(98.9°) - 1.5 sin(98.9°).

This gives local clock time        =   solar time + 4 (105° - 100.5°) - Eqt

                                               =   5:57:20         sunrise

                                               =   18:45:20       sunset.

Coordinate Transformation

Consider the coordinate (y1, y2, y3) which is on the inclined plane in Figure 1.3.1.  The direction y1 is coincident with the normal to the plane.  The coordinate y2 is on the line where the inclined plane intersects with the horizontal plane, and y3 is on the inclined plane itself.  There are occasions when it becomes necessary to transform the coordinate of a point in        x-coordinate into y-coordinate.  When two coordinates share the same origin as in Figure 1.3.1, it is simple to find the coordinate of a point in the alternative coordinate.

For a point represented by [x1, x2, x3], the corresponding coordinate values of  [y1, y2, y3] representing the same point can be found from

                   Y     =   [A] X ,      or

(4.3.4)

Example 1.3.3 

Following the previous example, for b = 90º, gp = 180º, the y-coordinate for a given point in  x-coordinate is obtained as

                 

So for Xs   =   ,  then

Sun Shading in Y-coordinate

In the situation when the inclined plane is not coincident with any of the planes in   x-coordinate, formulation of the problem in y-coordinate offers a simpler solution.  This point will be illustrated by an example.

Example 1.3.4   

Consider computing for the shade of an overhang in the following figure of a window facing southwest at 13:00 on 21 February.

Figure 1.3.4    The configuration of a shading problem involving a window facing southwest.





First, consider this using x-coordinate as in the followings.

The corner point P of the overhang in the x-coordinate is represented by Xp. From the configuration of the windows, the vector Xp can be obtained as (This step is challenging, the reader should try to verify the result given here.)

                   Xp    =  

In this example, the manner in which Xp is evaluated is not straight forward, because the plane of the window is not coincident with any plane formed by a pair of the x1, x2 and x3 coordinates.  This is in contrast to that in Example 1.3.1.

At 13:00 on 21 February in Bangkok, the following are obtained:

                   jd     =  52,

Lt     =  13.7º N,

Lgl    =  100.5° E,

Eqt    = -14.193 minutes,

d      = -11.226°,

ts      =   13:00 -  (18min) - (14.193min),

w     =   6.951°.

These give

sinas   =   0.8999,

as       =   64.138°,

sings    =   0.27213,

gs        =   15.791°.

The solar vector in x-coordinate is then obtained as

       Vsx    =  

                   cosq  =   (Vsx, Vnx)  =   0.38073

The position of the shade is obtained via (1.3.3) as

       Xs   =   Xp - h Vsx / cosq 

                   Xs   =    

This point is rather difficult to identify on the plane of the window. Let Ys be the y-coordinate for the point P. It is obtainable from

Ys    =   [A] X  =

This point is much simpler to identify on the y-coordinate.  It is in the plane formed by y2 and y3, where y2 is at the base of the window and y3 points to zenith. Figure 1.3.5 illustrates the location of shade point S, and the shape of the shade.

Figure 1.3.5    The position of the shade point, and the pattern of the shade.





Now, consider this problem in y-coordinate   To derive a general relationship, examine    Figure 1.3.4 again but consider all the relevant quantities in y-coordinate.  Denoting Vsy and Vny the solar and plane normal vectors in y-coordinate, the vector Ys representing the shade point due to shading by point P and the vector Yp are related by

                   Ys    =   Yp - h . Vsy / cosq                                                               (1.3.5)

This relationship is completely identical to (1.3.3) in x-coordinate.  It can be derived using the same argument. To use (1.3.5), we must obtain Vsy from Vsx by coordinate transformation.  To solve the problem in Example 4.3.1, first Yp is obtained from Figure 1.3.4 as

       Yp    =

Next, Vsy is obtained from Vsx through coordinate transformation from the following       

Vsy     =
 [A] . Vsx     
Vsy     =
 . Vsx
=

Then using (1.3.5), Ys is obtained as

 
 
   

This result is identical to the previous one.

1.3.2  Solar Radiation on Shaded Surface

Shading devices are used to shade beam radiation from window.  The most common shading devices in use are illustrated in Figure 1.3.6.

Figure 1.3.6    Common shading devices.




If the unshaded area of a window is Au, and the total area of the window is A, solar radiation on the window (with inclination angle b) is given as 

(1.3.6)

where   Eeg   =   EeS sinas + Eed, total solar radiation on a horizontal plane (global radiation).

In the relationship (1.3.6) it is assumed that the diffuse radiation is uniformly distributed, and the ground reflects solar radiation equally in all direction (perfectly diffusive surface).

Example 1.3.5  

On 21 February, at 13:00, given that the solar radiation is uniformly distributed, and the solar radiation components are

EeS   =   400 W/m2,

Eed   =   300 W/m2.

For a vertical wall facing southwest in Figure 1.3.4, total solar radiation on the window is given as

                   , for the vertical plane.

From the previous results, the areas of the shaded and unshaded portions of the window are computable from the following figure.

Figure 1.3.7    The geometric of the unshaped area.





Using geometrical relationship of like triangles, the following is obtained

 

This gives, for     c   =   0.6 - 0.06     = 0.54 m,

                          a   =   0.106 m.

Similarly, b is obtained as

                   b     =   0.361 m.

From these results, the unshaded area is then computed as

       Au    =   1.2 (a + b)/2 =   0.280 m2,

also,            A     =   1.44 m2.

From these results, then total solar radiation on the window is obtained as

Eew   =   (0.280) (0.38073) (400) + (1.44) (300) / (2) + (329.97) (1.44) rg

        =   258.6 + 475.5 rg .

In built-up area the view factor of the window to ground can be obscured by adjacent buildings.  For simplicity then rg is taken to be zero. In case where there is a wish to include the second term, rg can be taken to be 0.1 < rg < 0.2, with the higher value corresponding to a brighter surface such as a concrete surface.  The view factor from window to sky or sun can also be fully or partially obstructed, in which case the obstruction can be considered a shading device. 

A Comparison between Shading by Device of Infinite and Finite Length

In Section 1.2, simplifying assumption is made so as to be able to calculate the area of the shade casted by shading device.  It is now time to compare such result with result from more accurate method.

For the problem in Example 1.2.3, with the window configuration of Figure 1.2.5, the result give hs = 0.782, and the shaded area As of the window is 0.532.

Suppose both the window width and the length of the overhang are identical at 1m.  The method in Section 1.2 is an approximate one and the corresponding result can be used as an approximation.

But using the vectorial method, the exact areas of the shaded and unshaded portions are obtained.  Locating the origin of the coordinate at the center of the base of window in Figure 1.2.5 the followings are obtained

Yp   =
Vsy
Ys

The pattern of the shade is illustrated in Figure 1.3.8.

Figure 1.3.8    The pattern of shade.





Using like triangles, the size of a can be calculated as

                   a     =   0.05916 .

The unshaded (Au), and  shaded (As) area are

                   Au   = (1) (0.4681) + [(0.05916 + 0.1421) / 2] (0.5319),

                          = 0.522, and

                   As     = 0.478 .

These results are very similar to that obtained in Example 1.2.3 for this case.  This is because the finite shade almost cover the upper part of the window.

The two methods will produce results which differ significantly if the sun is either on the far right of far left of the window.