Question

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an 80% efficient cyclone. A recent law requires that the emissions from this stack be limited to 10.0 lb/hr, and the company is considering adding a wet scrubber after the cyclone. What is the required efficiency of the wet scrubber

Answer 1

Solution and Explanation:

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter =
`1000 \mathrm{cfm} * 400 \mathrm{gr} / \mathrm{tt} 3 * .000142857 \mathrm{lb} / \mathrm{ft} 3 * 60=3428.568 \mathrm{lb} / \mathrm{hr}`

Cyclone is to 80 % efficient

So particulate remaining =
`0.20 * 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136`

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %