Question 7.12: One-Axis and Two-Axis Tracker Insolation. Compare the 40º la......

One-Axis and Two-Axis Tracker Insolation. Compare the 40º latitude, clear-sky insolation on a collector at solar noon on the summer solstice for a two-axis tracking mount versus a single-axis polar mount. Ignore ground reflectance.

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1. Two-Axis Tracker: To find the beam insolation from (7.21) I_{B} \ = \ Ae^{-km}, we need the air mass ratio m, the apparent extraterrestrial flux A, and the optical depth k. To find m, we need the altitude angle of the sun. Using (7.7) with a solstice declination of 23.45º,

\begin{matrix} \beta _{N} \ = \ 90º \ − \ L \ + \ \delta \ = \ 90 \ − \ 40 \ − \ 23.45 \ = \ 73.45º \\ \text{Air mass ratio} \ m \ = \ \frac{1}{\sin \ \beta} \ = \ \frac{1}{\sin \ 73.45º} \ = \ 1.043 \quad \quad \end{matrix}

From Table 7.6, or Eqs. (7.22), (7.23) and (7.28), we find A = 1088 W/m², k = 0.205, and C = 0.134. The direct beam insolation on the collector is therefore

A \ = \ 1160 \ + \ 75 \ \sin \ \left[\frac{360}{365} \left(n \ – \ 275\right)\right] \quad \left({W}/{m^{2}}\right) (7.22)

k \ = \ 0.174 \ + \ 0.035 \ \sin \ \left[\frac{360}{365} \left(n \ – \ 100\right)\right] (7.23)

C \ = \ 0.095 \ + \ 0.04 \ \sin \ \left[\frac{360}{365} \left(n \ – \ 100\right)\right] (7.28)

I_{BC} \ = \ I_{B} \ = \ Ae^{-km} \ = 1088 \ \left({W}/{m^{2}}\right) \ \cdot \ e^{−0.205x  1.043} \ = \ 879 \ {W}/{m^{2}}

Using (7.35) the diffuse radiation on the collector is

\begin{matrix} I_{DC} & = \ C I_{B} \ \left[\frac{1 \ + \ \cos\left(90º \ – \ \beta\right)}{2}\right] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ & = \ 0.134 \ \cdot \ 879 \ \left[\frac{1 \ + \ \cos\left(90º \ – \ 73.45º\right)}{2}\right] \ = \ 115 \ {W}/{m^{2}} \end{matrix}

The total is I_{C} \ = \ I_{BC} \ + \ I_{DC} \ = \ 879 \ + \ 115 \ = \ 994 \ {W}/{m^{2}}

2. One-Axis Polar Tracker: The beam portion of insolation is given by (7.38)

I_{BC} \ = \ I_{B} \ \cos \ δ \ = \ 879 \ {W}/{m^{2}} \ \cos \left(23.45º\right) \ = \ 806 \ {W}/{m^{2}}

The diffuse portion, using (7.39), is

\begin{matrix} I_{DC} & = \ C I_{B} \ \left[\frac{1 \ + \ \cos\left(90º \ – \ \beta \ + \ \delta\right)}{2}\right] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ & = \ 0.134 \ \cdot \ 879 \ {W}/{m^{2}} \ \left[\frac{1 \ + \ \cos\left(90 \ – \ 73.45 \ + \ 23.45\right)}{2}\right] \ = \ 104 \ {W}/{m^{2}} \end{matrix}

The total is I_{C} \ = \ I_{BC} \ + \ I_{DC} \ = \ 806 \ + \ 104 \ = \ 910 \ {W}/{m^{2}}

The two-axis tracker provides 994 W/m², which is only 9% higher than the single-axis mount.

TABLE 7.6 Optical Depth k, Apparent Extraterrestrial Flux A, and the Sky Diffuse Factor C for the 21^{st} Day of Each Month
Month: Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
A (W/m^{2}): 1230 1215 1186 1136 1104 1088 1085 1107 1151 1192 1221 1233
k: 0.142 0.144 0.156 0.180 0.196 0.205 0.207 0.201 0.177 0.160 0.149 0.142
C: 0.058 0.060 0.071 0.097 0.121 0.134 0.136 0.122 0.092 0.073 0.063 0.057
Source: ASHRAE (1993).

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