. consider the replacement of the 400 mw san juan thermoelectric power plant in puerto rico, which uses heavy fuel oil to genera

Question

1 year ago

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. consider the replacement of the 400 mw san juan thermoelectric power plant in puerto rico, which uses heavy fuel oil to genera

te electricity, with a solar photovoltaic farm. the oil plant has a heat rate of 10,811 btu/kwh and a capacity factor of 53%. a. for installation of south-facing fixed tilt pv arrays, with tilt angle equal to the local latitude in san juan, how many acres of collector surface area would be required to replace the generation from the fuel oil plant? assume that the photovoltaic cells have an efficiency of 17%. b. by how many acres can the surface area in part (a) be reduced if single-axis tracking is added (with the collector tilt angle still equal to the local latitude)? c. what will be the annual fuel cost savings ($) if the oil power plant is retired, if heavy fuel oil costs $290 per metric ton and has a heating value of 39,500 btu/kg?

Physics

1 answer:

bearhunter [10] · Answer 1 · 2024-01-14T21:33:30+03:00

Annual saving energy produced *cost of oil/heating value is

$ 10020.46177.

<h3>Latitude and longitude: what are they?</h3>

Latitudes are lines that are horizontal and show how far a location is from the equator. In relation to the meridian in Greenwich, England, longitudes are vertical lines that show the east and west directions. Latitude and longitude are used in conjunction to locate points or places on the globe by cartographers, geographers, and others.

Puerto Rico's capital city's latitude and longitude are 18° 27' 58.7988" North and 66° 6' 20.5956" West.

Local San Juan latitude divided by tilt angle equals 18.46 WBAN No. 11641 = for year = 5.5 provides information on solar radiation for flat plate collectors facing south with a fixed tilt.

KWh/m2/day = 5.5 KWh/m2/day (1 day =3600h)

Formula used-

Solar radiation at that place* Collector Area *Collector efficiency = Heat rate * Output

power*Capacity factor

1 KW= 3,412.14 BTU/h

Heat rate* output power* capacity factor = 10,811*400*10^3*0.53 BTU/h = 671699.2855

KW

Collector Area = 671699.2855*3600/5.5*0.17

= 25862218.48 m^2= 6390 acres

C.

Cost of oil = 290$ /10^3 kg

Energy produced= 400*10^3 KW = 1364856 *10^3 BTU/h