Science, asked by ponnadajyothi5651, 9 months ago

For the home in the previous question, what size array (in kW to the nearest tenth 0.1) would be needed to meet their target of 60% of their annual energy if their home receives 2008 full sun hours annually you assume an 85% system efficiency.

Answers

Answered by Anonymous
11

Answer:

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Explanation:

Answered by muthuvijaykannan
0

Answer:

4.7

Explanation:

The previous requirement was 7980 kW.

Efficiency  - 0.85

Full sun hours - 2008

System size - 7980/2008/0.85   = approx. 4.7

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