# Problems: Solar and Photovolatics

In question 1 containing no more than about 150 words, for me to better understand and further my knowledge of solar and photovoltaic energy and then for me to compile a report I need to understand;

Question 1

a. If a large solar collector with an area of 9.0m² is mounted, facing south and tilted at 45° to the horizontal, on the roof of a house in the London area, where the household's annual energy requirement for hot water is 2750 kWh. The annual total radiation is 1053kWh m−² Using the relevant data show that, with an overall efficiency of 30%, this system could in principle meet that requirement.

b. Considering the contributions during the months of June 1043kWh m−² and December 29kWh m−², explain why the above reasoning could be misleading.

Question 2

A large PV power plant will consist of 350 000 solar panels, with a total area of one square kilometer. Its rated output will be 70 MWp, and its predicted capital cost is £150 million.

a. Calculate the capital cost per peak watt for the above system.

b. The annual insolation on the panels is 1500 kWh per m² per year. Assuming that two-thirds of the panel area is occupied by active PV surfaces and that their average efficiency is 10%, show that the annual output of the plant will be 100 GWh per year.

c. Given that the capital is to be repaid over 20 years with capital costs of £102 per year at an interest rate of 8%, calculate the expected cost per kWh of output from this plant.

https://brainmass.com/earth-sciences/energy-and-the-environment/problems-solar-photovolatics-625653

#### Solution Preview

In accordance with BrainMass standards this is not a hand in ready assignment but is only background help.

Step 1

If a large solar collector with an area of 9.0 m squared is mounted facing south and tilted at 45 degrees and the annual total radiation is 1053kWh m squared, then the total radiation of the entire area of 9.0 m squared will be 9477 kWh m squared (9.0 X 1053). Further, if the overall efficiency of the system is 30% then the total annual radiation will be 2843. 1 kWh m squared (9477 X 0.3). In this case 2843.1 kWh m squared, the energy produced is higher than the household's annual energy requirement for hot water which is 2750 kWh. In principle this system can meet the requirement of the household.

b)Considering the contributions during the months of June 1043 kWh m squared and December 29 kWh m squared, the above ...

#### Solution Summary

This posting gives you a step-by-step explanation of problems related to Solar and Photovolatics. The response also contains the sources used.