From Eqs. (10-22) and (10-23),
\frac{U_{L}\left(T_{c}-T_{a}\right)}{\alpha \tau}=\frac{\Sigma I_{i}^{2}}{\Sigma I_{i}} (10-22)
\frac{\Sigma I_{i}^{2}}{\Sigma I_{i}}=0.219+0.832 K_{T} (10-23)
\frac{U_{L}\left(T_{c}-T_{a}\right)}{\alpha \tau}=0.219+0.832(0.502)=0.637 \mathrm{~kW} \cdot \mathrm{m}^{-2}
From Eq. (10-24), for May, S_M = 42 − 22 = 20° and
C_{f}=1.0-0.000117\left(S_{M}-S\right)^{2} (10-24)
C_f= 1.0 − 0.000117(20 − 45)^2 = 0.927
Leading to
\left(T_{c}-T_{a}\right)=\frac{(0.927)(0.637)(0.88)(0.95)}{(0.02)}=24.7 \mathrm{~K}
Assume
\left(T_{a}-T_{M}\right)=3 \mathrm{~K}
From the data provided
\left(T_{M}-T_{r}\right)=15.3-20=-4.7 \mathrm{~K}
The average cell efficiency for the month is calculated from Eq. (10-18) as follows:
\eta=\eta_{r}\left[1-\beta\left(T_{c}-T_{a}\right)-\beta\left(T_{a}-T_{M}\right)-\beta\left(T_{M}-T_{r}\right)+\gamma \log _{10}\left(I_{i}\right)\right] (10-18)
η = 0.15[1 − 0.0045(24.7 + 3 − 4.7)] = 0.134(13.4%)
The daily average insolation integral is 4.97 kWh · m^{−2}, and May has 31 days, thus
E_{May} = 0.134(31)(4.97) = 20.6kWh ⋅m^{−2}
Extending this solution to an entire year requires weather data for a year and perhaps a spreadsheet or computer language such as Matlab to expedite calculations.