A certain power station produces an electrical output of 700 MW, operating at 42

Question

A certain power station produces an electrical output of 700 MW, operating at 42% efficiency. This plant uses a closed-circuit "dry" cooling system in which heat is expelled by cycling the cooling water through pipes which are in direct contact with air taken in from outside the plant. The fuel for the plant is lignite, which has a typical energy content of 15 MJ/kg.

(a) What is the mass of lignite consumed in 1.00 hour of operation of the plant?

(b) If the cooling water operates between a temperature of 90

Explanation / Answer

700 / .42 = 1670 MW = 1.67 * 10E9 Watts    energy input at 42% efficiency
1670 MW - 700 MW = 970 MW    energy wasted as heat
1.67 * 10E9 J/s * 3600 s/hr = 6 * 10E12 J    energy input in 1 hr
6 * 10E12 J / 1.5 * 10E7 J/kg = 400,000 kg or about 400 metric tons
Heat energy = 9.7 * 10E8 * 3600 = 3.5 * 10E12 J
3.5 * 10E12 J / 4.19 J/cal = 8.3 * 10E11 cal    heat output
1 cal / gm-deg * M * 60 deg = 8.3 * 10E11 cal
M = 1.4 * 10E10 gm = 1.4 * 10E7 kg    cooling water needed

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