Phase 1

OUT OF DATE

Phase 1 / Part 1:

Electrical

Vernon's Engineering Geek Moment.

Topic: Solar Panels

For an 800-watt solar array, you would mathematically need a battery bank with a capacity of 300-360 AH, or potentially more depending on your specific needs and the depth of discharge you want to allow. My equations show an 800-watt array can produce 240-256 AH's per day, with an average of 5 hours of peak sun exposure.

Below is my mathematical equation for the statement above:

Power (P) is related to voltage (V) and current (I) by the formula: \(P=V\times). Energy (E) is related to power (P) and time (t) by the formula: \(E=P\times). Battery capacity in Ah is related to energy in Wh and voltage by the formula: \(Ah=\frac{Wh}{V}\). Consider losses in the system (assume (80%) efficiency). Consider depth of discharge (DOD) for the battery (assume (50%) DOD for lead-acid or (80%) for lithium).

Note: This is assuming (5) hours of peak sunlight per day. If I calculate the daily energy production: \(E=P\times t\) \(E=800\{\ W}\times 5{\ h}=4000{\ Wh}\) E required = E efficiency cap E sub required end-sub equals the fraction with numerator cap E and denominator efficiency end-fraction. E required = E efficiency cap E sub required end-sub equals the fraction with numerator cap E and denominator efficiency end-fraction =\(E_{required}=\frac{E}{\{efficiency}}\) \(E_{required}=\frac{4000\text{\ Wh}}{0.8}=5000\{\ Wh}\)

Assume a (12) V system and \(50%) DOD (lead-acid). Calculate the required battery capacity: \(Ah=\frac{E_{required}}{V\times DOD}\) \(Ah=\frac{5000\{\ Wh}}{12\{\ V}\times 0.5}=\frac{5000}{6}\{\ Ah}\approx 833.33\{\ Ah}\) If using lithium with \(80%) DOD: \(Ah=\frac{5000\text{\ Wh}}{12\{\ V}\times 0.8}=\frac{5000}{9.6}\text{\ Ah}\approx 520.83\{\ Ah}\)     

Solution: The required battery capacity is approximately (833.33) Ah for lead-acid batteries or (520.83) Ah for lithium batteries, assuming (12) V system, (5) hours of peak sun, (80%) efficiency and (1) day of autonomy. For Anaheim California, the optimal tilt angle for solar panels equates to be around 30 degrees (Longitude + Latitude + Distance from the equator = 30 degrees). This angle maximizes sunlight exposure throughout the year and is a good balance for energy production. While the ideal angle can vary slightly depending on specific circumstances, 30 degrees is a solid starting point for fixed solar panels in Anaheim. 1 degree of latitude is approximately 111 kilometers (69 miles) apart. 1 degree of longitude is not constant; it's the longest at the equator and zero at the poles. The distance from the equator can be calculated using the formula: \[ \{Distance} = \text{Latitude} \times 111 \{ km/degree} \]. The Earth is not a perfect sphere, but an oblate spheroid. 

Phase 1 / Part 2:

Miscellaneous Wood Framing

(to house electrical grid)

Phase 1 / Part 3:

Plumbing