# Question 9.18: Battery Sizing for an Off-Grid Cabin. A cabin near Salt Lake......

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%. The batteries will be kept in a ventilated shed whose temperature may reach as low as −10ºC. The system voltage is to be 24 V, and an inverter with overall efficiency of 85% will be used.

Step-by-Step
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With an 85% efficient inverter, the dc load is

$\text{DC load} \ = \ \frac{\text{AC load}}{\text{Inverter efficiency}} \ = \ \frac{3000 \ {Wh}/{day}}{0.85} \ = \ 3529 \ {Wh}/{day}$

With a 24-V system voltage the batteries need to supply

$\text{load} \ = \ \frac{3529 \ {Wh}/{day}}{24 \ V} \ = \ 147 \ {Ah}/{day} \ @ \ 24 \ V$

From Appendix E the following monthly insolation data are found for Salt Lake City: (Table 1)

Even though annual insolation is highest for a tilt angle equal to the local latitude, we’ll use an L + 15 tilt to help meet winter needs. December has the lowest insolation so that will be our design month. From Fig. 9.46 at 95% availability and 3.1 peak sun hours in December, it looks like we need about 4.6 days of storage.

$\text{Usable storage} \ = \ 147 \ {Ah}/{day} \ \times \ 4.6 \ \text{day} \ = \ 676 \ Ah$

We’ll pick deep discharge lead-acid batteries that can be routinely discharged by 80%. But, we need to check to see whether that depth of discharge will expose the batteries to a potential freeze problem. From Fig. 9.39, at −10ºC the batteries could be discharged to over 95% without freezing the electrolyte, so an 80% discharge is acceptable.

Batteries that are nominally rated at C/20 and 25ºC will be operated in much colder conditions which degrades storage capacity, but they will be discharged at a much slower rate, which increases capacity. Figure 9.42 suggests that at −10ºC and a C/72 rate (the 5-day rate for the cabin is even slower, so this is conservative), storage capacity would be about 0.97 times the nominal capacity. Applying the 0.80 factor for maximum discharge and the 0.97 factor for discharge rate and temperature in (9.32) gives

$\text{Nominal (C/20, 25ºC) battery capacity} \ = \ \frac{\text{Usable battery capacity}}{\left(\text{MDOD}\right) \left(T, \ DR\right)}$ (9.32)

$\text{Nominal (C/20, 25ºC) battery capacity} \ = \ \frac{676 \ Ah}{0.80 \ \times \ 0.97} \ = \ 871 \ Ah \ \left(\text{at} \ 24 \ V\right)$

Table 9.15 can help us pick a battery. None of those batteries comes close to providing that many Ah, so we’re going to have to have parallel strings. Partly to keep the weight of each battery low enough to be able to handle, and partly to keep the number of batteries under control, suppose we choose the Trojan 6 V, 225 Ah, T-105. Three parallel strings would give us 675 Ah, four would give 900 Ah, and our goal is 871 Ah. Suppose we slightly oversize with four strings.
To get 24 V, we need each string to have four batteries, so the total battery bank will have four parallel strings of four batteries each as shown in Fig. 9.47.

Table 1

 Tilt Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Year Lat − 15 2.9 4.0 5.0 5.9 6.6 7.2 7.3 7.0 6.3 5.0 3.3 2.5 5.2 Lat 3.2 4.3 5.2 5.8 6.2 6.6 6.7 6.7 6.4 5.4 3.7 2.9 5.3 Lat + 15 3.4 4.4 5.1 5.4 5.5 5.6 5.8 6.1 6.1 5.5 3.9 3.1 5.0
 TABLE 9.15 Example Deep-Cycle Lead-Acid Battery Characteristics BATTERY Voltage Weight (lbs) Ah @ C/20 Ah @ C/100 Concorde PVX 5040T 2 57 495 580 Trojan T-105 6 62 225 250 Trojan L16 6 121 360 400 Concorde PVX 1080 12 70 105 124 Surette 12CS11PS 12 272 357 503

Question: 9.22

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From the solar radiation tables given in Appendix ...
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Question: 9.14

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From Fig. 9.39, to avoid freezing, the maximum dep...
Question: 9.13