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

Answer:

The daily production rate  is 191.6 ton/day

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:

$W_{in} = \dfrac{W_{out}}{n_{th}}$

$W_{in} = \dfrac{600}{0.35}$

$W_{in} = 1714 \ MW$

The feed rate is calculated as:

coal feed rate = $1714*10^6 \ watt * \dfrac{3.412 \ Btu}{watt.hr }* \dfrac{lb}{12000 \ Btu}$

= 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 = $275.5 \ lbmol/hr *\dfrac{100.09 \ CaCo_3}{lb mol \ CaCo_3}$

= $27575 \ lb \ CaCo_3 / hr$

The  limestone feed rate = $\dfrac{27575}{0.95}$

= 29026 lb/hr

= $29026 \ lb/hr * \dfrac{24 \ hr}{ 1\ day}*\dfrac{1 \ ton }{2000 \ lb}$

= 348.3 ton/day

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

= 191.6 ton/day