In one city, concentrations of the three major air pollutants are shown as the
following:
SO2:1.20mg/Nm3,NO2:0.160mg/Nm3,PM10:0.25mg/NM3
What is the API of the city, and which type of air quality of this city is?
The concentration of the three air pollutants are shown as the following:
SO2:0.03ppm, O3:157μm/m3, CO: 9ppm
Please finish the unit’s conversion between ppm andμg/m3?
We have tested a plants emission of partials’ and found the following data:
Sample number Stack velocity,ftls Particle concentration,g/m3
1
50
0.50
2
60
0.30
3
40
0.70
Each sample represents the average over a 20-min period. What was the average
particle
Concentrations in this stack over the whole 60 min of this test?
In the textbook Example 4.1, we have now discovered that the velocity measuring
device was improperly calibrated. We have recalibrated it and find that the
velocity. In order, should be been 8, 12,16and 20m/s.The concentration of the gas
flowing in the stack?
Our cyclone separator is operating with Dcut=5u,it is new necessary to increase the
flow rate
to the cyclone (and hence the inlet velocity)by 25 percent (i.e,the new velocity will be
1.25times
the old velocity0.Nothing else will change. The cyclone is believed to obey Eq.
(9.18).Estimate the
new cut diameter.
A cyclone separator has an inlet width of 0.25ft and Vc=60ft/s.
(a) Estimate the cut diameter for this cyclone separator.
(b) Estimate the collection efficiency of this cyclone separator for particles with
diameter 1μ.
Please explain the particles collection mechanisms of the surface filter, depth filter
and scrubbers.
Answer:
Surface filter: these filters have many small holes, so when dirty
gas stream through them, the big particles are caught on the side of the
hole of the filters. Then they tend to bridge over the holes and make them
smeller. Thus as the amount of collected particles increase, the cake of
material becomes the filters and the filter medium that originally served
as a filter to collect the cake now serves only to support the cake ,and
no longer as a filter . This cake of collected particles will have average
pore sizes smaller than the diameter of the particles. In the oncoming
gas stream and thus will act as a sieve for them. This particle collects
on the front surface of the growing cake.
Depth filter: the particles, if it moves directly to the right, will
run into the target. The force moving it upward is given by the stocks
drag force. this is important and depth filter collect particles mostly
by impaction. Then small particles move in gases by diffusion. In depth
filter that diffusion leads to particles collection to that compacted
above by impaction. Then the behavior is interception. the first term on
the right is for diffusion collection. Whereas the second is for
collection by noinertial contact. In final term takes into account the
fact that the particles have finite diameter and hence will aontact the
target if their center passed Dok of it.
Scrubber : Most fine particles will adhere to a liquid drop if they
contact it.so it makes the drop and the particles touch each other ,the
particle will be caught on the drop.
What are the alternatives for control of vocs?
Answer: Control by prevention: 1) subsititution:2)process
mod-ification;3)leakage control
Control by concentration: 1) condensation 2) adsorption 3) absorption
Control by oxidation: 1) combustion 2) biological oxidation
example 7.7 shows that the stoichiometric air-fuel ratio(A/F) for methane is 17.3
lb/lb.
(a) Compute the stoichiometric A/F for hydrogen and for carbon.
(b) Write an equation for the stoichiometric A/F of fuel as a function of its medullar
(hydrogen/carbon) ratio: assuming that the fuel consists only of hydrogen and carbon.
The typical coal in example 7.10 is being burned under under the conditions in that
example, at a rate of 1000 kg/h of dry coal ,the combustion produets enter a treatment
device at 400F,what is the gas flow rate in scfm and acfm?
The typical coal in example 7.10 is being burned under under the conditions in that
example, at a rate of 1000 kg/h of dry coal ,the combustion produets enter a treatment
device at 400F,what is the gas flow rate in scfm and acfm?
air at 2200F and 1 atm is
flowing at a rate of 1000 lb/min. air of the afterburner on a hazardous waste
incinerator.
(a) How many scfm is this?
(b)
How many acfm is this?
Estimate the concentration of carbon monoxide at the downwind edge of a city. The
city may be considered to consist of three parallel strips, located perpendicular to
wind. For all of the strips the wind velocity u equals 3 m/s. The properties of each of
the strips are described in the following table:
Name of strip
Length, km
Emission rate
Mixing
height
Upwind suburbs
5
100
400
Downtown
2
500
500
Downwind suburbs
5
100
400
Assume that the fixed-box model applies to each of the strips. The background
concentration b in the air entering the upwind suburbs is 1mg/m3.
6.3
For Eq. (6.29), show the simplifications that result in each of these cases:
(a) We are only interested in concentrations on the centerline of the plume, i.e., at
y=0, z=H. Here assume that (z + H) is very large.
(b) We are only interested in concentrations at the same elevation as the plume
centerline, i.e., at z=H. Here assume that (z + H) is very large.
(c) We are only interested in concentrations directly below the plume centerline at
ground level, i.e., at z=0, y=0.
(d) We are only interested in ground-level sources, i.e., H=0.
6.6
A large, poorly controlled copper smelter has a stack 150m high and a plume
rise of 75m.
It’s currently emitting 1000g/s of SO2. Estimate the ground-level concentration of
SO2 from this source at a distance 5km directly downwind when the wind speed is
3m/s and stability class is C.
A factory emits 20g/s of SO2 at height H. The wind speed is 3m/s. At the distance of 1
km downwind. The value of σy and σz are 30m and 20m, respectively, what is the
SO2 ground level concentration at the downwind distance 1 km? (H=50m)
A gas stream contains two sizes of particles. 50wt% large and 50wt% small. We pass
this gas stream through two collection devices in series. The collection efficiencies
are show in the following table. What overall weight fraction of the particles is
collected?
First
collection device
Second collection device
Collection efficiency for small particles
0.50
0.25
Collection efficiency for large particles
0.75
0.40
Answer : if we consider that the mass of gas that contains 1.0kg of particles. we can
compute the following:
Particle size
incoming amount
*
penetration
=
outgoing
amount
Large
0.50kg
0.25*0.60
0.075kg
Small
0.50kg
0.50*0.75
0.1875kg
Total
1.00kg
0.2625kg
And the overall weight fraction of the particles is:
1-P=1-(0.2625/1.00)=0.7375=73.75%