A coal-burning power plant that uses an evaporating water tower to cool the gene

Question

A coal-burning power plant that uses an evaporating water tower to cool the generated steam has the following characteristics: uses about 2,000 metric tons/day of bituminous coal, has an overall efficiency of 35% in converting the energy from the combustion of coal to electricity, and 10% of the heat energy produced is lost to the surrounding environment as hot air through the chimney of the plant (stack). Determine how much water is required to run the plant, i.e., to cool the generated steam. For water: specific heat (15'C) = 4.100 kJ/kg'C; heat of vaporization (100'C) = 2,250 kJ/kg; heat of vaporization (15'C) = 2,500 kJ/kg. For coal, heat released by combustion, or H = 20 MJ/kg.

Explanation / Answer

Mass of coal burnt per day = 2,000 metric tons = 2,000 metric tons x (1000 Kg / 1 metric ton) = 2x106 Kg/day

Hence total heat released per day = 2x106Kg x (20 MJ/kg) = 4.0x107 MJ

= 4.0x107 MJ x (1000 KJ / 1 MJ) =  4.0x1010 KJ

Amount of heat lost to the surrounding = 4.0x1010 KJ x (10 / 100) = 4.0x109 KJ

Hence amount of heat used to evaporate water = 4.0x1010 KJ - 4.0x109 KJ = 3.6x1010 KJ

If 15 C is the initial temperature of water, then the amount of heat required vaporize 1 Kg of water

= mSdT + mHv

= 1 Kgx 4.100 kJ/kg'C x 85 C + 1 Kg x 2,250 kJ/kg

= 2599 KJ/ kg

Hence total mass of water required = 3.6x1010 KJ / 2599 KJ/ kg = 1.385 x 107 Kg = 13850 metric ton water

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