As a rough rule of thumb it takes 1 to 2 acres of cooling la

As a rough rule of thumb, it takes 1 to 2 acres of cooling lake per megawatt of installed electrical
capacity. (a) If one conservatively uses the latter figure, what is the area for a 1000-MWe plant?
(b) Assuming 35% efficiency, how much energy in joules is dissipated per square meter per
hour from the water?

Solution

Area required = capacity*area per MW = 1000*2=200 acres

Power wasted would be passed to cooling lake

so Power wasted= (1-efficiency)*capacity = 650MW

Power per area = total power to lake / area = 650MW/200 acres= 650*10^6 W/(200*4046.86 sqare meter) = 803.091 W/square meter

Energy in one hour per square meter = Power per area *1 hour = 803.091*3600 sec = 2891130.41 J