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 ΔH = 20 MJ/kg.

Answer 1

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