# 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 $^a$Assumed inverter efficiency 90%, mismatching loss 3%, dirt loss 4%

Question: 9.22

## Sizing an Array for a 150-ft Well in Santa Maria, California. Suppose that the goal is to pump at least 1200 gallons per day from the 150-ft well described in Example 9.21 using the Jacuzzi SJ1C11 pump. Size the PV array based on Siemens SR100 modules with rated current 5.9 A, mounted at an L + 15º ...

From the solar radiation tables given in Appendix ...
Question: 9.21

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The total length of pipe is 150 + 80 + 10 = 240 ft...
Question: 9.18

## Battery Sizing for an Off-Grid Cabin. A cabin near Salt Lake City, Utah, has an ac demand of 3000 Wh/day in the winter months. A decision has been made to size the batteries such that a 95% system availability will be provided, and a back-up generator will be kept in reserve to cover the other 5%. ...

With an 85% efficient inverter, the dc load is [la...
Question: 9.15

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a. With all 3109 Wh/day running on an 85% efficien...
Question: 9.14

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Using data from Table 9.10, we can put together th...
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The voltage across each PV cell will be about 12.5...
Question: 9.17

## Losses at High and Low Charging Rates. A 100-Ah, 12-V battery with a rest voltage of 12.5 V (at its current SOC) is charged at a C/5 rate, during which time the applied voltage is 13.2 V. Using a simple Thevenin equivalent: a. Estimate the internal resistance of the battery. b. What fraction of the ...

a. At C/5, the current is 100 Ah/5 h = 20 A. The i...
Question: 9.16