Question

Assuming that only CaSO3 is produced, calculate the daily production rate (in tons/day) of a 55%-solids sludge from a 90%-efficient limestone FGD system on a 600-MW power plant burning 3.5%-sulfur coal. The plant has a thermal efficiency of 35%, and the coal has a heating value of 12,000 Btu/lbm. Assume that the limestone is 95% CaCO3 and 5% inert substances, and that the ratio of the actual amount of limestone fed to the theoretical amount required is 1.15. Cooper, C. David. Air Pollution Control: A Design Approach (p. 519). Waveland Pr Inc. Kindle Edition.

Answer 1

The daily production rate is 191.6 ton/day

Step-by-step explanation:

The first step that is required to be carried out is by determining the thermal input:

The thermal input can be calculated via the expression:

`W_(in) = (W_(out))/(n_(th))`

`W_(in) = (600)/(0.35)`

`W_(in) = 1714 \ MW`

The feed rate is calculated as:

coal feed rate =
`1714*10^6 \ watt * (3.412 \ Btu)/(watt.hr )* (lb)/(12000 \ Btu)`

= 487347 lb/hr

The sulfur feed rate is :

= 487347 × 0.035

= 17057 lb/hr

Sulfur removal rate = 17057 × 0.9

= 15351 lb/hr

However, to determine the actual alkalinity; we have:

actual alkalinity = 1.15 × 239.57

= 275.5 lb . mol / hr

The total alkalinity is =
`275.5 \ lbmol/hr *(100.09 \ CaCo_3)/(lb mol \ CaCo_3)`

=
`27575 \ lb \ CaCo_3 / hr`

The limestone feed rate =
`(27575)/(0.95)`

= 29026 lb/hr

=
`29026 \ lb/hr * (24 \ hr)/( 1\ day)*(1 \ ton )/(2000 \ lb)`

= 348.3 ton/day

Finally, the daily production rate = 0.55*348.3 ton/day

= 191.6 ton/day