<u>Solution and Explanation:</u>
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 = 
Cyclone is to 80 % efficient
So particulate remaining = 
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 %
Answer:

Explanation:
We have to combine the following formula to find the mass yield:


The diffusion coefficient : 
The area : 
Time : 
ΔC: 
Δx: 
Now substitute the values

![M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]](https://tex.z-dn.net/?f=M%3D-%286.0%2A10%5E%7B-8%7D%20m%2Fs%5E%7B2%7D%29%280.25%20m%5E%7B2%7D%29%283600%20s%2Fh%29%5B%280.64-3.0kg%2Fm%5E%7B3%7D%29%283.1%2A10%5E%7B-3%7Dm%29%5D)

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