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Limestone scrubbing is used to remove SO2 in a flue gas desulfurization (FGD) system. Relevant reactions are given below. A limestone mix is used that consists of 94% CaCO3 and 6% inert material. The actual feed rate of the limestone mix is 36,000 lb/hr. The SO2 in the flue gas is 20,314 lb/hr and the FGD efficiency is 97%. If the resulting sludge has 58% solids, determine the total sludge production rate (lb/hr). Round your answer to the nearest whole number.

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Answer:

hello some data related to your question is missing attached below is the missing data

answer : 63700 Ib/hr

Step-by-step explanation:

Given data :

Limestone mix : consists of 94% CaCO3 and 6% inert material

Actual feed rate = 36,000 Ib/hr

SO2 in flue gas = 20,314 Ib/hr

FGD efficiency = 97%

resulting sludge contains 58% solids

Calculate the Total sludge production rate

First : determine SO2 removed in sludge

= 0.97 * 20314

= 19704.58 Ib/hr

next : moles of SO2 removed

= 19704.58 / 64 Ib/ Ib mol

= 307.88 Ib mol / hr

also moles of CaSO3 produced = 307.88 Ib mol / hr

mass of CaSO3 = 307.88 * 120 = 36946.09 Ib/hr

Therefore Total sludge production rate

= 36946.09 / 0.58

= 63700.15 Ib/hr

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