Consider a residential PV solar system composed of 100 W mod

Consider a residential PV solar system composed of 100 W modules (i.e., panels). The cell efficiency is 17%. The installed cost of the system is $6000/kW including the PV panels, battery storage, inverters, and modifications to the house. The system has a solar access factor of 95%, a cleanliness factor of 98%, a temperature factor of 0.88, wiring and mismatch loss reduction factor 95%, an inverter efficiency of 90%, and a storage efficiency (i. e., round trip efficiency) of 95% based on input energy. All power delivered by the system is provided through the batteries. Solar insolation in the region average averages 250 W/m^2 year round. The system may be purchased with a 20 year payback period at an interest rate of 6%. O&M; costs are $0.01/kWh. The household has an annual electricity usage of 11,000 kWh. Each panel requires 0.72 m^2 of roof area. What is the expected power output of a 100 W module in this system at the rated solar insolation of 1000 W/m^2 (W)? What is the expected annual electricity production of a 100 W panel at this location (kWh/y)? What is the rated power of the system needed to provide 100% of the household power (kW)? What roof area is required for the PV solar system to meet the annual electricity needs of the household (m^2)? What is the expected cost of the household electricity generated by the system ($/kWh)?

Solution

a. Expected power output of 100W panel at solar insolation of 1000 W/m²

At solar insolation of 1000 W/m² and 0.72 m² area maximum power that be generated

= Insolation x Area x efficiency x losses at the panel

= 1000 x 0.72 x 0.17 x (0.95 x 0.98 x 0.88) = 100.28 W

Since Panel rating is 100 W it cannot exceed output of 100 W hence output will be 100 W

Answer: 100 W

b. Power generated at this location every second = Insolation x Area x efficiency x losses at the panel

= 250 x 0.72 x 0.17 x 0.95 x 0.98 x 0.88 = 25.07 W

= 25.07 x1 /1000 = 0.02507 kWh

For the year = 0.02507 x 365 x 24 = 219.6132 kWh/y (Assuming Year round Sun for 24 hrs)

Answer = 219.6132 kWh/y

c. Rated power to get 100% of 11000 kWh per annum

Power available in household for one 100 W panel = Insolation x Area x efficiency x losses

= 250 x 0.72 x 0.17 x 0.95 x 0.98 x 0.88 x 0.95 x 0.90 x 0.95 = 20.363 W

Per Annum = 0.020363 x 365 x 24 = 178.38 kWh/y

Number of 100 W systems required = Total consumption/production per annum by one panel = 11000/178.38 = 61.66

Hence 62 Panels are required which means rated power is 62x100 = 6200W or 6.2 kW

Answer: Rated power = 6.2 kW

d. Total roof area required to meet annual electricity need:

As calculated above total of 62 Panels are required for 11000 kWh/y of energy consumption

Area of 62 panels = 62 x 0.72 = 44.64 m²

Answer: 44.64 m² Is the roof area.

C. Expected cost of household electricity:

Installation cost = $6000/kW

Total system rating = 6.2 kW

Total initial cost = 6.2 x 6000 = $37200

Total interest for 20 years at 6% Simple Interest = 37200 x 0.06 x 20 = $44640 (principle x Interest rate x Years)

Total production cost = 37200 + 44640 = $81840

Total production cost/kWh = Total cost / Total consumption for 20 years = 81840/(11000x20) = $0.372/kWh

Cost per unit = Production costs + O&M Cost = 0.372 + 0.01 = $0.382/kWh

Answer: Cost per unit = $0.382/kWh