Consider February in Ithaca, New York, United States. Calculate the expected daily insolation to fall on a south-facing solar collector panel having a tilt angle of 45°. The ground reflectance is 0.7 (snow more than 1 in deep). Solar data for February in Ithaca are $\bar{H}$ =8.61 MJ/m $^2$ and $\bar{K}_{T}$ =0.435; these parameters are derived from the analysis of local insolation data over time. ${ }^{1}$ Assume isotropic sky conditions as a first approximation.

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From geographic data sources, the latitude of Ithaca is 42.3° north. The middle of February is day number 45. The solution sequence is to calculate solar declination, the sunset hour angle, and then the terms as defined through Eqs. (9-22), (9-26), (9-27), (9-28), and (9-29).

[latex]\bar{H}_{d} / \bar{H}=1.39-4.03 \overline{K_{T}}+5.53 \bar{K}_{T}^{2}-3.11 \bar{K}_{T}^{3}[/latex] (9-22)

[latex]\omega_{\mathrm{s}}=\cos ^{-1}(- an (L) an (\delta))[/latex] (9-26a)

[latex]\omega_{\mathrm{s}}^{\prime}=\min \left[\omega_{\mathrm{s}^{\prime}} \cos ^{-1}(- an (L-\beta) an (\delta)) ight][/latex] (9-26b)

[latex]\bar{R}_{b}=\frac{\cos (L-\beta) \cos (\delta) \sin \left(\omega_{s}^{\prime} ight)+\omega_{s}^{\prime} \sin (L-\beta) \sin (\delta)}{\cos (L) \cos (\delta) \sin \left(\omega_{s}+\omega_{s} \sin (L) \sin (\delta) ight)}[/latex] (9-27)

[latex]\bar{R}=\left(1-\frac{\bar{H}_{d}}{\bar{H}} ight) \bar{R}_{b}+\left(\frac{\bar{H}_{d}}{\bar{H}} ight)\left(\frac{1+\cos (\beta)}{2} ight)+\frac{ ho(1-\cos (\beta))}{2} [/latex] (9-28)

[latex]\bar{H}_{T}=\bar{H} \bar{R}[/latex] (9-29)

δ = −13.6° = 0.237 radians

ω[latex]_s[/latex] = 77.3°, equivalent to 77.3/15 = 5 h and 9 min after solar noon and ω'[latex]_s[/latex] = ω[latex]_s[/latex]

[latex]\bar{R}_{b}[/latex] =1.973, or 97.3% more beam insolation than on the horizontal surface

[latex]\bar{H}[/latex]= 1.598, or 59.8% more total insolation than on the horizontal surface

[latex]\bar{H}_{T}=1.598 imes 8.61=13.5 \mathrm{MJ} / \mathrm{m}^{-2}[/latex]

Components of H[latex]_{T}[/latex] are

direct: 1.130 × 8.61 = 9.73 MJ/m[latex]^2[/latex]

diffuse: 0.365 × 8.61 = 3.14 MJ/m[latex]^2[/latex]

reflected: 0.103 × 8.61 = 0.89 MJ/m[latex]^2[/latex]

[latex]^1[/latex] One approach is to approximate local insolation data by looking at national weather data websites such as http://rredc.nrel.gov/solar/old_data/nsrdb/bluebook/state.html and estimating local values using nearby cities. In this case, nearby cities such as Syracuse, NY, and Binghamton, NY, could be used

Question 9.4: Consider February in Ithaca, New York, United States. Calcul...

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