# Question 9.22: Sizing an Array for a 150-ft Well in Santa Maria, California......

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 tilt.

Step-by-Step
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From the solar radiation tables given in Appendix E, the worst month is December, with an insolation of 4.9 kwh/m²-day. From (9.60)

$Q\left(gpm\right) \ = \ \frac{\text{Daily demand} \ \left({gal}/{day}\right)}{\text{Insolation}\left({h}/{day \ @ \ 1-\text{sun}}\right) \ \times \ 60 \ {min}/{h}}$ (9.60)

$Q \ = \ \frac{1200 \ {gal}/{day}}{4.9 \ \left({h}/{day \ @ \ 1-\text{sun}}\right) \ \times \ 60 \ {min}/{h}} = \ 4.1 \ gpm$

From Fig. 9.60, at 4.1 gpm the total dynamic head is about 170 ft and the pump efficiency is about 34%. From (9.59), the estimated pump input power is

$P_{in} \ \left(W\right) \ \text{to pump} \ = \ \frac{\text{Power to fluid}}{\text{Pump efficiency}} \ = \ \frac{0.1885 \ \times \ H\left(ft\right) \ \times \ Q\left(gpm\right)}{\eta_{p}}$ (9.59)

$P_{in}\left(W\right) \ = \ \frac{0.1885 \ \times \ H\left(ft\right) \ \times \ Q\left(gpm\right)}{\text{Pump efficiency}} \ = \ \frac{0.1885 \ \times \ 170 \ \times \ 4.1}{0.34} \ = \ 386 \ W$

From Fig. 9.60, at 4.1 gpm and 170 ft of head, the pump voltage is a little under 45 V, which means that at 15 V per module, three modules in series should be sufficient.

Using (9.62), we can decide upon the number of parallel strings of modules

$\# \ \text{strings} \ = \ \frac{\text{Pump input power} \ P_{in}\left(W\right)}{\# \ \text{mods in series} \ \times \ 15 \ {V}/{mod} \ \times \ I_{R}\left(A\right) \ \times \ \text{de-rating}}$ (9.62)

$\# \ \text{strings} \ = \ \frac{386 \ W}{3 \ \text{modules string} \ \times \ 15 \ {V}/{\text{module}} \ \times \ 5.9 \ A \ \times \ 0.80} = \ 1.8$

so choose two parallel strings.

From (9.63), estimated delivery in January with two strings of three modules would be

$\begin{matrix} Q\left({gal}/{day}\right) & = \ 15 \ {V}/{mod} \ \times \ I_{R} \ \left(A\right) \ \times \ \left(\# \ \text{mods}\right) \ \times \ \left(\text{Peak} \ {h}/{day}\right) \ \times \ 60 \ {min}/{h} & \\ & \times \ \text{de-rating} \ \times \ {\eta_{p}}/{\left[0.1885 \ \times \ H\left(ft\right)\right]} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \left(9.63\right) \end{matrix}$
$\begin{matrix} Q & \approx \ \frac{15 \ V \ \times \ 5.9 \ A \ \times \ 6 \ \text{modules} \ \times \ 4.9 \ {h}/{day} \ \times \ 60 \ {min}/{h} \ \times \ 0.80 \ \times \ 0.34}{0.1885 \ \times \ 170 \ ft} \\ & = \ 1325 \ {gal}/{day} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}$

A system diagram is shown in Fig. 9.61.

Question: 9.21

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