In this exercise we will consider two coal-fired power plants and a solar thermal plant, in order to compare their power densities, i.e., their power output vs. the amount of land they occupy. This method of comparison of diverse forms of renewable and non-renewable energy systems was pioneered by Vaclav Smil of the University of Manitoba in Canada. The first coal plant is highly efficient and is optimally located, next to a mine that supplies high-quality coal from a thick seam, and a river that provides water for cooling. The second coal plant is less efficient, is distant from a mine and a river (and thus requires large cooling towers), and is supplied with lower quality, high-sulfur coal from a thinner seam. Note that in 2012, coal-fired plants provided 37% of US electricity, down from 45% in 2009.
Please answer the ten numbered questions (in bold type) in the space provided. Helpful definitions are provided at the end of this exercise.
Part 1. Coal-fired power plant #1 is located next to a coal mine, on a river in southern Utah. It has a rated maximum power output of 1.2 GW. The Utah plant operates at a capacity factor of 80% and has a conversion efficiency of 38%. Its energy output is therefore
1.2 GW x 0.8 capacity factor = 960 MW average power output.
960 MW x 24 h/day x 365.25 days/yr = 8.4 x106 MWh = 8.4 TWh per year.
8.4 TWh x 3600 TJ/TWh = 30,200 TJ = 30.2 PJ per year (petajoules = 1015 joules).
This plant would therefore require an annual primary energy input, in the form of coal, of
30.2 PJ/0.38 = 79.5 PJ per year.
The coal is provided by a strip mine that is digging up a 15 m thick seam that provides about 20 metric tons of coal per square meter of the mine, containing 20 GJ of energy per ton.
1) What area of the mine (in m2) would have to be stripped of its coal for each year the plant operates?
The power plant consists of boiler and turbogenerator halls, electrostatic precipitators, maintenance buildings, offices, roads, and parking, covering an area of about 75 hectares (750,000 m2). (A hectare is equal to a square 100 m on a side.) Adding the annual area stripped at the mine, the total area used each year by the entire facility is (199,000 + 750,000 =) about 950,000 m2 and the resulting overall power density is for the plant is
960 MW/950,000 m2 = 960 x 106W/950,000 m2 = 1010 W/m2
Energy required , in the form of coal , per year.
E = 79.5 * 1015 J
Given Data About Coal :
Amount of coal obtained per m2 , from digging 15 m thick seam = 20 metric tones / m2
Amount of energy released by coal (obtained from diging 1m2 area and 15m thick seam) = Tons of coal dug * Energy per ton of Coal = 20 ton* 20 GJ/ton = 400 GJ
This is the energy released from mining 1m2 area . So energy released per m2 can be given as ,
Energy obtained per m2 of field = 400 GJ / 1 m2 = 400 * 109 J/m2
Total energy required per year = 79.5 * 1015 J
Area of field to be dug out = Total energy required per year / Energy obtained per m2 of field = 79.5 * 1015 J/ 400 * 109 J/m2= 0.19875 * 106 m2
Area to be stripped off to obtain the required amount of coal = 0.19875 * 106 m2 = 198750 m2
Power Density = Total power produced / Total area used = 960 MW / ( 198750 + 750000 m2) = 1011.85 J/m2
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