Question

A 600 MW pulverized coal-fired power plant of 40% efficiency uses coal with an ash content of 15% and a heating value of 26,700 kJ/kg. Assume 35% of the ash content goes up the stack as particulate matter in the flue gas. The PM emission is controlled by an ESP which has the following efficiencies and weight distribution in the given size ranges. Find the fly ash emitted in g/s.

Answer 1

Quantity of energy input by pulverized coal to operate power plant at full capacity = Power generation/ Efficiency                                                 

= 600 MW / 0.40

                                                 = 1500 MW

Using [1MW = 1,000 kJ per second]

    = 1500*103 (kJ/sec)

Quantity of Coal required = Energy supplied / heating value of coal

= 1500*103 (kJ/sec) / 26700 (kJ/kg)

=56.1797 (kg/sec)

Total ash generated = 15% of 56.1797 (kg/sec)

= 8.4269 (kg/sec)

Ash emitted as Flyash= 35% of Total ash generated

=8.4269*0.35 (kg/sec)

=2.9494 (kg/sec)

= 2949.4 (gm /sec) -- This will be collected by ESP based on efficiency.