control of Primary Particulates

Introduction The control of primary particles is a major part of air pollution control engineering. Although primary particles are generally larger than secondary particles , but many of them are small enough to be respirable and are thus of health concern. If possible the collected particles are recycled to somewhere in the in the process that generates them.

Wall collection devices Three types of control devices will be consider are, gravity settlers Cyclone separators , and Electrostatic precipitators All function by driving the particles to a solid wall , where they adhere to each other to form agglomerates that can be removed from the collection device and disposed of.

gravity settlers Easy to construct Requires littlie maintenance Treating very dirty gases in some industries Contaminated gas passes slowly, allowing time for the particles to settle by gravity to the bottom.

Either the fluid going through is totally unmixed (block or plug flow model) or The total mixing in the entire device or in the entire cross section, perpendicular to the flow (backmixed or mixed model) For both cases the average horizontal gas velocity in the chamber is :

For the block flow model, assume : 1- the horizontal velocity of the gas in the chamber is equal to Vavg. Everywhere in the chamber 2- the horizontal component of the velocity of the particles in the gas is always equal to Vavg 3- the vertical component of the velocity of the particles is equal to their terminal settling velocity due to gravity Vt 4- if a particle settles to the floor , it stays there and is not re-entrained.

If the distance is greater than or equal to h, then it will reach the floor of the chamber and be captured. For the same size particles , what will happen ?

For mixed flow model, assume: 1- the gas flow is totally mixed in the z direction but not in the x direction. Most real gas flows are turbulent, leading to internal mixing in process equipment. This because mixing in x direction moves particles up and downstream with the little effect on collection efficiency. Whereas mixing in the z direction leads to a decrease in the collection efficiency.

Consider a section of the settler with the length dx. The fraction of the particles in that section that reach the floor will equal the vertical distance an average particle fall due to gravity divided by the height of the section The change in concentration passing this section is :

The time the average particle takes to pass through this section is : Integrate from the inlet (x=0) to outlet (x=L) finding

Or Substitute Vt from stake's law Comparing this result with the plug or block assumption

Example 1 Compute the efficiency-diameter relation for a gravity settler that has H=2m, L= 10 m and Vavg= 1 m/s Assuming stake's law for both models Particles density 2000 kg/m3 , Viscosity 1.8x10-5 Kg/m.s Particle diameter in micron 1,10,30,50,57.45,80,100,120

Centrifugal separators The disadvantage of gravity settlers is not effective in small particles. Therefore, it has to find a substitute that is more powerful then the gravity force to drive the particles to the collection surface.

Centrifugal separators If a body moves in a circular path with radius r and velocity Vc along the path then it has angular velocity Ɯ= Vc/r and

Centrifugal separators Gravitational settling velocity less than a hundredth of centrifugal one .

Cyclone separator

Centrifugal separators Ex: A particle is travelling in a gas stream with velocity of 18 m/s and radius of 30 cm , what is the ratio of centrifugal force to the gravity force acting on it?

Centrifugal separators Ex: compute the efficiency –diameter relation for a cyclone separator that has Wi = 15 cm , velocity 18 m/s and n=5. for both block and mixed flow assumptions, assuming stokes law

Electrostatic Precipitator ESP Gravity settlers and centrifugal settlers are devices driving the particles against a solid walls. For particles below 5 micron , the above devices are not effective. ESP is like gravity of centrifugal forces driving the particles to the wall and effective on much smaller particles than the previous devices

Electrostatic Precipitator ESP Which is better relation to diameter ?

Electrostatic Precipitator ESP The basic idea of ESP is to give the particles an electrostatic charges and then put them in electrostatic field that drives them to the collecting wall. Two steps charging and collecting process. Common known electronic air filter.

The gasses possesses between the plates and which are electrically grounded (zero voltage). Between the plates are rows of wires held at a voltage of 40 000 volts. This combination of the charged wires and grounded plates produces both the free electrons to charge the particles and the field to drive them against the plates.

For particles larger than 0 For particles larger than 0.15 micron, the dominant charging mechanism is field charging . As the particles become more highly charged , they bend the paths of the electrons away from them. Thus the charge grows with time reaching a steady state value of

q is the charge on the particle is dielectric constant of particle is dimensionless, 1 for vacuume, 1.0006 for air and 4 to 8 for typical solid particles. is the permittivity of free space 8.85 x 10-12 C/V.m. D is the particle diameter and Eo is the local field strength

Ex : a 1 micron diameter particle of a material with dielectric constant of 6 has reached its equilibrium charge in an ESP at a place where the field strength is 300 kV/m. how many electronic charges has it ?

The electrostatic force on a particle is Ep is the local electric field strength causing the force Why we do use Ep in this equation and Eo in the previous equation.

For both E will apply the average and find the drift velocity Since proportion to particle diameter, one would compute larger values for the larger particles present in the stream. Proportion to the square of E which is approximately equal to the wire voltage divided

By the wire to plate distance By the wire to plate distance. How to achieve unlimited drift velocities?

Ex : a 1 micron diameter particle of a material with dielectric constant of 6 has reached its equilibrium charge in an ESP at a place where the field strength is 300 kV/m. how many electronic charges has it ? Calculate the drift velocityfor the particle.

If now consider the section between the row of wires and one plate, its collection area is : The volumetric flow through the section is :

Ex : compute the efficiency diameter relation for an ESP that has particles with dielectric constant of 6 and (A/Q) = 0.06 min/m for mixed flow only

Ex : ESP has measured efficiency of 90 percent Ex : ESP has measured efficiency of 90 percent. The target is to achieve 99 percent, by how much must increase the collecting area ?

Scrubber It is a device for collecting small fine particles on liquid drops. The main duty of the device is to collect particles by contacting the dirty gas stream with the liquid drops. Most of the fine particles will adherse to a liquid drop if they contact it and the particle will be caught on the drop.

Consider a space of gas with dimensions Δx, Δy and Δz Consider a space of gas with dimensions Δx, Δy and Δz. The concentration of particles in the gas of this space is C (kg/m3)

Scrubber The volume of space swept out by the drop is the cylindrical hole as shown in the figure. The mass of particles transferred from gas to the drop is :

Scrubber For the x,y,z region, the concentration of particles in the air changes during the rainstorm. Multiply top and bottom of the equation by the volume of a single spherical drop and simplify to obtain.

Scrubber The total liquid volumetric flow going to the scrubber is QL

Example : A rainstorm is depositing 2 Example : A rainstorm is depositing 2.54 x 10-3 m/h, all in the form of spherical drops 1 mm diameter. The air through which the drops are falling contains 3 micrometer diameter particles at an initial concentration of 100 microgram per cubic meter. What will the concentration be after one hour?

Cross flow Scrubber The gas is assumed to move through the scrubber in uniform at a total volumetric flowrate QG

Cross flow Scrubber The linear velocity of the gas is (QG / Δy Δz) Cross flow Scrubber The linear velocity of the gas is (QG / Δy Δz). Therefore, the time it takes a parcel of gas to pass through is the length of the scrubber divided by the linear velocity

Counter flow Scrubber Liquid enters the top of the scrubber through series of spray nozzles that distribute it uniformly and falls by gravity. The gas enters the bottom of the scrubber and flows upward in uniform flow.

Counter flow Scrubber The drop at its terminal settling velocity relative to the gas surrounds it. The gas is moving upward with velocity VG= QG/ΔX ΔY The velocity of the drop relative to the fixed coordinates of the scrubber is VD-fixed= Vt- VG

Counter flow Scrubber

Counter flow Scrubber To compute the quantity (volume swept by drops /time) it must compute instantaneous number of drops per unit volume. The liquid flow into the system is QL and this consists ND drops /time, each of volume (∏/6 DD3). The average time each drop spends in the scrubber is the vertical distance divided by vertical velocity

Counter flow Scrubber The volume of gas that these drops swept out per unit time is their number times their cross-sectional area times the velocity at which they move relative to the gas.

Counter flow Scrubber

Example : The throat velocity for a scrubber is 122 m/s Example : The throat velocity for a scrubber is 122 m/s. the particles to be collected have diameters of 1 micron, and the droplet diameter is 100 micron. 10/1000 cubic meter of liquid is feeding per cubic meter of gas . At a point where the settling velocity is 0.9 gas velocity. What is the rate of decrease in particle concentration in the gas phase?

Problem 1 : We whish to design a particle sampling device that will have a greased , horizontal microscope slide 75 mm long over which a contaminated gas stream will flow in a narrow slit 2 mm high. What velocity should we choose if we want 90% of the 1 micrometer diameter particles to fall (and stick ) to the surface of the microscope slide.

Problem 2 : Cyclone separator is operating with Dcut= 5 micron Problem 2 : Cyclone separator is operating with Dcut= 5 micron. It is now necessary to increase the flow rate to the cyclone by 25 percent. Estimate the new cut diameter .

Problem 3 : At the equator, what is the ratio of centrfugal force due to the earths rotation to the gravitational force? The radius of the earth is about 6400 km and its rotational velocity at the equator is about 1600 km/hr.

Problem 4 : An ESP is treating a particle air stream, collecting 95 percent of the particles. We now double the air flow rate keeping the particle loading constant. What is the new percent recovery. Assume mixed flow.

Problem 5 : Exhaust gas from a low efficiency dry ESP at a coal-fired electric power plant. They report average wet ESP penetrations of 5% with A/Q=0.035 min/m. estimate the drift velocity of this wet ESP