Air Pollution Control
Viscosity is the resistance a material has to change in form. There are two types of absolute
viscosity. Dynamic, Kinematic viscosity.
- Motion under the influence of gravity.
The force F action on a particle.
F=mg=Πρpd3g/6
Motion under the influence of gravity in standard air terminal velocity vt=29609ρpd2
EXAMPLE: A settling chamber is to be used to collect particles of 50 µm diameter and 2000
k/m3 density from a stream of 10 m3/s of standard air. If the chamber is to be 1.5 m wide, 1.5
m high, and if it will have nine trays including the bottom surface, how long must it be to give
theoretically perfect collection efficiency?
a) Do the computations on the assumption that the laminar equations hold.
b) Compute the Reynolds number and decide if the laminar assumption is reasonable. If not,
how many trays need be included to make the flow laminar? Recalculate the length in this
case
c) What will be the collection efficiency of particles of 25 um diameter?
SOLUTION: The terminal velocity of the 50 µm particles
Vt =29609ρpd2=29609*2000*(0,5*10-4)2 =0,148m/s
When the efficiency is to be equal to unity L=Q/nWvt
L=10/9*1,5*0,148=5m
Reynolds number Re=2Q/ν(nW+H)=2*10/1,55*10-5(9*1,5+1,5)
This number is well beyond the laminar range. To estimate the number of trays required to
bring the Reynold number down to 2300, other quantities remaining the same, Eq. (6-8) may
be solved for n:
n=2q/vwRe=2(10)/1.55x10e-5(1.5)(2300)=374
Ans.
The required length then becomes
L=10/374(1.5)(0.148)=0.12 m
Ans.
Thus, it is clear that to have laminar flow in a settling chamber usually requires an awkward
design . However, a very large chamber combined with a small flow may well give laminar
flow in the chamber.
(c) For a particle of 25 nm diameter, the terminal settling veloticy is
Vt= 29,609(2000)(0.25x10e-4)=0.037 m/s
Then Eq.(6-10) gives the efficiency as
n=nWLVı/Q=374(1.5)(0.12)(0.037)/10=0.249 Ans.
EXAMPLE 6-2 Use the data of Example 6-1. Determine the length of settling chamber
required the achieve 99 percent efficiency for 50-mm particles. What length would be
required for 99,9 percent efficiency? If the first length is used, what would be the efficiency
with 25-mm particles?
SOLUTION Solving Eq.(6-18) for L and using the given data gives
L=-Q/(nWVı)xIn(1-n)=-10/9(1.5)(0.148)xIn(1-0.99)=23.0 m Ans.
For an efficiency
Suppose that r1=0,25 m and r2=0,5 m. For a flow rate with a Width of 1.0 m3/s.m, plot the
velocity profile and determine the pressure difference between the inside and outside radii.
Assume standard air
A water droplet with a diameter of 0.8 mm falls through standard air under the influence of
gravity. What will be its terminal velocity?
1- How does an electrostatic precipitator run? Explain it on a figure
2- Design a settling chamber of laminar flow for a stream of 100 m3/s of standard air
containing particles having a density of 1500 kg/m3. The chamber must not exceed 5
m in width or 6 m in height. It must collect particles of 70 µm diameter with perfect
efficiency. Determine the length of the chamber required and the number of trays.
3- Standard air is flowing a U bend at the rate of 2.25 m3/s.m width. The inner and outer
radii are 10 and 30 cm, respectively. Determine the x and y components of velocity at
a point of radius 20 cm located 30° downstream of the midsection of the U.