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Chapter 10

Q. 10.4

Assume a PV array tilted at 45° to the south at a location where the latitude is 42° north. Determine the total May kWh output from the array based on the following parameter values:

\begin{aligned} U_{L} &=0.02 \mathrm{~kW} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1} \text { (standard assumption) } \\ a &=0.88 \\ t &=0.95 \\ \eta_{r} &=0.15 \\ T_{r} &=20^{\circ} \mathrm{C} \\ \beta &=0.0045^{\circ} \mathrm{C}^{-1} \\ K_{T} &=0.502 \\ \bar{H}_{d} &=4.97 \mathrm{kWh} m^{-2} \text { day }^{-1} \text { (irradiating the tilted PV array) } \\ T_{M} &=15.3^{\circ} \mathrm{C} \end{aligned}

 

 

Step-by-Step

Verified Solution

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.