大氣復(fù)點(diǎn)習(xí)資料含答案

1、大氣復(fù)習(xí)資料一、 概念解釋?zhuān)?） globe warming: global warming is the increase in the average measured temperature of the earths near-surface air and oceans since the mid-twentieth century, and its projected continuation.（2） temperature inversions：a temperature inversions is a thin layer of the atmosphere where the

2、 decrease in temperature with height is much less than normal (or in extreme cases, the temperature increases with height).（3） esp: electrostatic precipitator, which is like a gravity setter or centrifugal separator, but electrostatic force drives the particles to the wall.（4） hepa: high efficiency

3、particulate air（5） lean bum ：lean burn refers to the use of lean mixtures in an internal combustion engine .the air-fuel can be as high as 65:1 ,so the mixture has considerably less fuel in comparison to the stoichiometric combustion ratio (14.7 for petrol , for example ).（6） plume rise: the plume r

4、ising a distance h above the top of the stack before leveling out.（7） wet scrubber: a device that collects particles by contacting the dirty gas stream with liquid drops.（8） photochemical smog：（9） thermal no: thermal nox refers to nox formed through high temperature oxidation of the diatomic nitroge

5、n found in combustion air.（10） a/f ratio: air to fuel ratio for auto engines.（11） pm2.5:particle with the aerodynamic diameter less than 2.5 um, which is also called respirable particles.（12） alternative fuel: several other fuels except of conventional gasoline and diesel, which have been used for m

6、any years in slighutly modified automobile engines, for reasons of cost and availability.（13） vocs: volatile organic compounds are those organic liquids or solids whose room temperature vapor pressure are greater than about 0.01 psia(=0.0007atm) and whose atmospheric boiling points are up to about 5

7、00, which means most organic compounds with less than 12 carbon atoms.（1） scr and sncr:6no+4nh35n2+6h2o, 4no+4nh3+o24n2+6h2o 2no2+4nh33n2+6h2o.these reactions can be carried out over a variety of catalysts, the temperature is between 1600and 1800, once the temperature increases, the dominant reactio

8、n is nh3+o2no+1.5h2o, the catalytic processes are called scr, and the higher-temperature ones, without catalysts, call sncr.（2） aerodynamic diameter: airborne particles have irregular shapes, and their aerodynamic behavior is expressed in terms of the diameter of an idealized spherical particle know

9、n as aerodynamic diameter.（3） primary particles: particles found in the atmosphere in the form in which they were emitted, for examples, no,co,so2（4） point sources: small number of large sources that emit larger amounts per source, at higher elevations (power plants, smelters, cement plants, etc.) c

10、alled point sources二、 answer following questions (1) which are the main constituents for the ground level ozone formation?ozone is formed when the following constituents are present.nitrogen oxides, volatile organic, compounds, sunlight, high temperature(18 )no+voc+o2+sunlightno2+o3(2) please list f

11、ive major types of wet scrubbers.plate scrubber （板式）packed scrubber （填料式）preformed spray scrubber（噴霧式）gas-atomized spray scrubber （氣體霧化）centrifugal scrubber （離心式）impingement-entrainment scrubber（沖擊夾帶式）mechanically aided scrubber （機(jī)械輔助式）moving bed scrubber （移動(dòng)床式）(3) how to control vocs pollution by p

12、revention? two examples.substitution(代替), replacing gasoline as a motor fuel with compressed natural gas or propane is a form of substitution.process modification(過(guò)程修改), replacing gasoline-powered vehicles with electric-powered vehicles is a form of process modification.leakage（滲漏） control, storing

13、large amounts of gasoline in floating roof tanks.(4) please introduce the process of forced-oxidation limestone wet scrubbing briefly？(5) what are the most different points between scr and sncr?(6) can twc be applied in the treatment of diesel exhaust (=emission of diesel engine)? why？the characteri

14、stic of emission of diesel engine: ample pm and excessive o2, lower temperature;the difficulty in the reaction of solid-gas-solid.key: the mixture process of fuel and air for gas engines is distinguished from that for diesel engines, hence the character of one type engine is different from the other

15、 one: there are mainly five gases (nox, hc, co and o2, co2) in the exhaust of gas motors, while there are ample o2 and four other gases mentioned above in the tailpipe emission of diesel engine. and the presence of abundant o2 would inhibit the performance of twc.moreover, the contact and reaction o

16、f solid-solid-gas resulting in the difficulties for catalysts to oxidation the particulate in the exhaust of diesel engine, while the contact and reaction of solid-gas-gas occur in the exhaust of gasoline engine and the latter reaction is easier.(7) what kinds of indoor air pollutants are mostly con

17、cerned by public?a) random b) combustion by-products1. co, co2, so2,formaldehyde,hydrocarbons,nox2. particulates, polyaromatic hydrocarbonsc) cigarettes d)volatile organic compounds f) biological contaminants(8) list the technology strategy for the control of particles.(9) give names of three typica

18、l kinds of combustion reactors. which has lowest operator temperature among those reactors?1.direct flame incineration 2.thermal incineration 3.catalytic incineration (has lowest operator temperature)(10) what are the major development problems of forced-oxidation limestone wet scrubbing?(11) what a

19、re primary air pollutants and secondary air pollutants? any example?primary air pollutants are directly from the sources, for examples, no, co, so2. the secondary air pollutants are from the primary pollutants, such as no2, no3, fine particles.no+ch+o2+sunlight no2+o3(12) basic strategy of control f

20、or particulate pollutants (three aspects )?impaction 碰撞 interception 截留 diffusion 擴(kuò)散by forcing the individual particles to contact each other,by contacting them with drops of water,by preventing the emission of gaseous pollutants.(13) how to reduce the formation of nox in flue gas by modifying the c

21、ombustion processes?p459(14) please explain the formation of acid rain?sulfur oxides and nitrogen oxidesno no2 hno3 nitric acid so2 h2so4 sulfuric acid(15) what are basic principles of electrostatic precipitators (six activities)? ionization - charging of particlesmigration - transporting the charge

22、d particles to the collecting surfacescollection - precipitation of the charged particles onto the collecting surfaces charge dissipation - neutralizing the charged particles on the collecting surfacesparticle dislodging - removing the particles from the collecting surface to the hopperparticle remo

23、val - conveying the particles from the hopper to a disposal point三、calculation 1) 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 the wind. for all of the strips the wind velocity e

24、quals 3 m/s. the properties of each of the strips are described in the following table,name of striplength, kmemission rateq, g/skm2mixing heighth, mupwind suburbs5100400downtown2500500downwind suburbs5100400solution： c1=0+(100*5000)/(3*400)=416.7 g/m3 c2=416.7+(500*2000)/(3*500)=1083.4g/m3 c3=1083.

25、4+(100*5000)/(3*400)=1500.1g/m3 2) an esp is designed to treat 540,000 acfm（actual cubic feet per minute） with 99% efficiency. assuming an effective drift velocity of 0.12 m/s, calculate the required plate area and the number of plates. the plate size is 5.4m tall by 3m long(a =q(1)/p, acfm1ft3/min,

26、 1m=3.28ft, 1m21550 in210.7639 ft2). solution1ft3/min=(1/3.28)3m3/min=0.0283 m3/min 540000acfm=540000*0.028315282 m3/min=254.7 m3/sa= -(254.7/0.12)ln(1-0.99)=9774.47m2n=9774.47/514.3=6043) we wish to treat an airstream containing 0.005mol fraction (0.5%, 5000ppm) toluene, moving at a flow rate 2240m

27、3/h at 0 and 1 atm, so as to remove 99% of the toluene by water absorption. estimate the required water flow rate. here henrys law constant is 10000 atm. solution：4）a power plant flue gas contains 1000ppm of so2 and is emitted at a rate of 224m3/s at 546k and 1 atm. a forced-oxidation limestone wet

28、scrubbing system is to be used to achieve 90% removal of the so2. calculate the amount of caso42h2o contained in the final solid product in t/d.solution： =(101325*224*1000*10-6)/(8.315*546)=4.999mol/s so2 caso42h2o so n（caso4）=n（so2）=4.999mol/s*90%=4.4991mol/sm=4.4991mol/s *172g/mol *3600*24s/d *10-6 t/d=66.86 t/d5）a power plant emits 36 kg/h of so2 at height h=120m and the wind speed is 2 m/s. dispersion coefficients: y=40m and z=30m, estimate the ground-level concentration of so2 from this source at a distance 1km dire