Question

A 1000- MW power plant burns coalwith a heating value of 25000Btu/kg, a sulfur content of 2.1%, and an ash content of 8%.

The plant has an overall conversion efficiency of 30%. Calculate:

a. The coal consumption rate in tons/hour.

b. The ash generation rate in tons/hour.

c. If a wet scrubber is added to remove 98% of the SO2, find the CaO consumption rate, also in tons/hour.

1W = 3.411Btu/hr
1hp = 746W = 2545 Btu/hr
Molecular weights: S = 32 gr, H = 1 gr, and O = 16gr, and Ca = 40gr

Answer 1

Ans) Given,

Capacity of power plant = 1000 MW

Heating value of coal = 25000 BTU/kg

Let the coal consumed be 'x' kg, then ,

Heat generated in an hour = 25000x BTU/hr

= 7329.22x W

Since, capacity is 1000 x 10⁶ W, therefore,

7329.22 x = 10⁹

x = 136440.2 kg/hr or 136.440 ton/hr

Since , plant has overall efficiency = 30% , therefore,

Actual coal consumed = 136.440 / 0.30

= 454.80 ton/hr

Ans b) Given, Ash content = 8%

Therefore ash generated = 0.08 x 454.80

= 36.38 ton/hr

Ans c) SO2 released = 0.021 x 454.8 = 9.55 ton

  SO2 removed = 98%

= 0.98 x 9.55 = 9.36 ton

  Following reaction will occur :

CaO + H2O --> Ca(OH)₂

Ca(OH)₂ ---> Ca⁺ + 2OH^-

Ca⁺² + 2H⁺ + 2OH^-+ SO3^-  ---> CaSO3 + 2H2O

Therefore, we can see that ,

CaO --> CaSO₃

56 gm CaO removes 64 gm SO2,

Therefore, CaO required = 9.36 x 56 / 64 ton/hour

= 8.19 ton/hr