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Question 9.20: PVs for the Cabin in Salt Lake City. The cabin from Example ......

PVs for the Cabin in Salt Lake City. The cabin from Example 9.18 needs 3000 Wh/day of ac delivered from an 85%-efficient inverter. For a 24-V system voltage, a 90% Coulomb efficiency, and 10% de-rating (de-rating factor = 0.90), size a PV array using Kyocera KC120 modules.

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From Table 9.3, the Kyocera KC120 is a 120-W module with its maximum power point at a current of 7.1 A and a voltage of 16.9 V. From Example 9.18, the worst solar month is December, with 3.1 peak hours of sunlight at a tilt of L + 15. Using (9.35), one string of modules will therefore deliver in December about

\text{Ah to load} \ = \ I_{R} \ \times \ \text{Peak sun hours} \ \times \ \text{Coulomb efficiency} \ \times \ \text{De-rating factor} (9.35)

\text{Ah to inverter} \ = \ 7.1 \ A \ \times \ 3.1 \ {h}/{day} \ \times \ 0.90 \ \times \ 0.90 \ = \ 17.83 \ {Ah}/{day} \ \text{per string}

For an 85% efficient inverter to deliver 3000 Wh/day of 120 V ac, it needs a 24-V dc input of

\text{Inverter dc input} \ = \ \frac{3000 \ {Wh}/{day}}{0.85 \ \times \ 24 \ V} \ = \ 147 \ {Ah}/{day} \ @ \ 24 \ V

Since these modules have a rated voltage of 16.9 V, they are nominally “12-V modules.” Therefore two modules are needed in series to provide a single 24-V string.

The number of parallel strings of modules needed is

\text{Number of parallel strings} \ = \ \frac{147 \ {Ah}/{day}}{17.83 \ {Ah}/{\text{day-string}}} \ = \ 8.25 \ \text{strings}

Suppose that we undersize it slightly and use eight parallel strings with two modules per string, for a total of 16 modules.

Including the 0.90 de-rating factor, the PVs will deliver

\begin{matrix} \text{PV output} & = \ 8 \ \text{strings} \ \times \ 7.1 \ {A}/{\text{string}} \ \times \ 3.1 \ {h}/{day} \ \times \ 0.90 \\ & = \ 158 \ {Ah}/{day} \ @ \ 24 \ Vdc \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \end{matrix}

The batteries with 0.90 Coulomb efficiency will deliver

\text{Battery output} \ = \ 158 \ {Ah}/{day} \ \times \ 0.90 \ = \ 142 \ {Ah}/{day} \ @ \ 24 \ Vdc

The 85% efficient inverter will deliver

\text{Inverter output} \ = \ 142 \ {Ah}/{day} \ \times \ 24 \ V \ \times \ 0.85 \ = \ 2900 \ {Wh}/{day} \ @ \ 120 \ Vac

So, the design is just a bit shy of its goal of 3000 Wh/day ac. The system diagram for this cabin, including some of the energy flows for December, is shown in Fig. 9.51.

TABLE 9.3 Annual Energy Production in Various Cities per kW (dc, STC) of Installed PV Capacity^a
Location South Facing, L-15 Fixed 1-Axis, Polar Mount Ratio 1-axis/ Fixed
Average High Temp. (◦C) Insolation (kWh/m²-d) Annual kWh/kW Insolation (kWh/m²-d) Annual kWh/kW
Seattle, WA 15.3 3.8 1006 4.7 1247 1.24
New York, NY 16.8 4.5 1195 5.6 1479 1.24
Madison, WI 13.2 4.5 1202 5.7 1519 1.26
Boston, MA 15.0 4.5 1209 5.7 1529 1.26
Atlanta, GA 21.8 5.0 1294 6.4 1639 1.27
Honolulu, HI 29.1 5.5 1373 7.4 1834 1.34
Boulder, CO 17.9 5.3 1404 7.2 1885 1.34
Los Angeles, CA 21.3 5.5 1420 7.0 1808 1.27
El Paso, TX 25.3 6.3 1583 8.6 2159 1.36
Albuquerque, NM 21.2 6.3 1618 8.5 2199 1.36
^aAssumed inverter efficiency 90%, mismatching loss 3%, dirt loss 4%
Figure 9.51

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