<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:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = = 25 s
since the reaction is in first order
X = 1 -
= 1 - X
kt = In
k = In / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 =
therefore
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
