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Question 9.15: Accounting for Inverter Losses. Suppose that a dc refrigerat......

Accounting for Inverter Losses. Suppose that a dc refrigerator that uses 800 Wh/day is being considered instead of the 1140 Wh/d ac one given in Example 9.14. Estimate the dc load that the batteries must provide if an 85% efficient inverter is used (a) with all loads running on ac and (b) with everything but the refrigerator running on ac.

Figure 9.36
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a. With all 3109 Wh/day running on an 85% efficient inverter, the dc load that the batteries must supply would be

\text{dc battery load} \ = \ \frac{3109 \ {Wh}/{day}}{0.85} \ = \ 3658 \ {Wh}/{day}

b. With the 1140-Wh/day refrigerator removed, the remaining ac load is

\text{ac load} \ = \ \left(3109 \ − \ 1140\right) \ {Wh}/{day} \ = \ 1969 \ {Wh}/{day}

accounting for the inverter efficiency, that ac load would be supplied by

\text{dc for ac loads} \ = \ \frac{1969 \ {Wh}/{day}}{0.85} \ = \ 2316 \ {Wh}/{day}

Adding in the 800-Wh/day dc refrigerator, the total dc load becomes

\text{dc battery load} \ = \ \left(2316 \ + \ 800\right) \ {Wh}/{day} \ = \ 3116 \ {Wh}/{day}

That’s a 15% decrease in energy needed. Figure 9.37 shows these data.

Figure 9.37

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