Answer:
fluid nozzle that is too large
Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6
<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 %