[VIP专享]大气复习资料(含答案)

大气复习资料一、概念解释（1）Globe warming:Global warming is the increase in the average measured temperature of the Earth's near-surfaceair and oceans since the mid-twentieth century, and itsprojected continuation.（2）Temperature inversions：A temperature inversions is a thin layer of the atmosphere where the decrease in temperaturewith 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 forcedrives the particles to the wall.（4）HEPA: High Efficiency 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 highas 65:1 ,so the mixture has considerably less fuel incomparison to the stoichiometric combustion ratio (14.7 forpetrol , for example ).（6）Plume rise: the plume rising 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 NO x refers to NO x formed through high temperature oxidation of the diatomic nitrogen found incombustion air.（10）A/F ratio: Air to fuel ratio for auto engines.（11）PM2.5:particle with the aerodynamic diameter less than 2.5um, which is also called Respirable Particles.（12）Alternative fuel: Several other fuels except of conventional gasoline and diesel, which have been used formany years in slighutly modified automobile engines, forreasons of cost and availability.（13）VOCs: Volatile organic compounds are those organic liquids or solids whose room temperature vapor pressure are greaterthan about 0.01 psia(=0.0007atm) and whose atmosphericboiling points are up to about 500℉,which means mostorganic compounds with less than 12 carbon atoms.（1）SCR and SNCR:6NO+4NH3→5N2+6H2O, 4NO+4NH3+O2→4N2+6H2O 2NO2+4NH3→3N2+6H2O.These reactions can be carried out over a variety ofcatalysts, the temperature is between 1600℉and1800℉,once the temperature increases, the dominant reaction isNH3+O2→NO+1.5H2O, the catalytic processes are called SCR,and the higher-temperature ones, without catalysts, callSNCR.（2）Aerodynamic diameter: Airborne particles have irregular shapes, and their aerodynamic behavior is expressed interms of the diameter of an idealized spherical particleknown 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 (powerplants, smelters, cement plants, etc.) called point sources 二、Answer following questions(1)Which are the main constituents for the ground level ozoneformation?Ozone is formed when the following constituents are present. Nitrogen oxides, Volatile Organic, Compounds, Sunlight, High temperature(>18 ℃)NO+VOC+O2+Sunlight→NO2+O3(2)Please list five major types of wet scrubbers.Plate Scrubber （板式）Packed Scrubber （填料式）Preformed Spray Scrubber（喷雾式）Gas-Atomized Spray Scrubber （气体雾化）Centrifugal Scrubber （离心式）Impingement-Entrainment Scrubber（冲击夹带式）Mechanically Aided Scrubber （机械辅助式）Moving Bed Scrubber （移动床式）(3)How to control VOCs pollution by prevention? Two examples. Substitution(代替),Replacing gasoline as a motor fuel with compressed natural gas or propane is a form of substitution. Process Modification(过程修改),Replacing gasoline-powered vehicles with electric-powered vehicles is a form of process modification.Leakage（渗漏）Control,Storing 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(=emissionof diesel engine)? Why？·The characteristic 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 one: there are mainly five gases (NO X, 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 of 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 concerned bypublic?a)Randomb)Combustion by-products1.CO, CO2, SO2,Formaldehyde,Hydrocarbons,NOx2.Particulates, polyaromatic hydrocarbonsc)Cigarettes d)Volatile organic compoundsf) Biological contaminants(8)List the technology strategy for the control of particles.(9)Give names of three typical kinds of combustion reactors.Whichhas lowest operator temperature among those reactors?1.Direct flame incineration2.Thermal incineration3.Catalytic incineration (has lowest operator temperature)(10)What are the major development problems of Forced-oxidationlimestone wet scrubbing?(11)What are primary air pollutants and secondary airpollutants?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 for particulate pollutants (three aspects )?Impaction 碰撞 Interception 截留 Diffusion 扩散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 modifyingthe combustion 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(sixactivities)?·Ionization - Charging of particles·Migration - Transporting the charged particles to thecollecting surfaces·Collection - Precipitation of the charged particles onto the collecting surfaces·Charge dissipation - Neutralizing the charged particles on the collecting surfaces·Particle dislodging - Removing the particles from thecollecting surface to the hopper·Particle removal - Conveying the particles from the hopper to a disposal point三、Calculation1) 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 equals 3 m/s . The properties of u each of the strips are described in the following table,Name of strip L ength, km Emission rateq , g/s·km 2Mixing height H , m Upwind suburbs 5100400Downtown 2500500Downwind suburbs 5100400solution ： c 1=0+(100*5000)/(3*400)=416.7 μg/m 3uHqL b c += c 2=416.7+(500*2000)/(3*500)=1083.4μg/m 3uHqL c +=1C c 3=1083.4+(100*5000)/(3*400)=1500.1μg/m 3uHqL c +=2C 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－η)/ωp , acfm ＝1ft 3/min, 1m=3.28ft, 1m 2＝1550 in 2＝10.7639 ft 2). solution1ft 3/min=(1/3.28)3m 3/min=0.0283 m 3/min540000acfm=540000*0.028315282 m 3/min=254.7 m 3/sA= -(254.7/0.12)ln(1-0.99)=9774.47m 2N=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 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 Henry’s law constant is 10000 atm .Solution ：4）A power plant flue gas contains 1000ppm of SO 2 and is emitted at a rate of 224m 3/s at 546K and 1 atm . A Forced-oxidation limestone wet scrubbing system is to be used to achieve 90% removal of the SO 2. Calculate the amount of CaSO 4·2H 2O contained in the final solid product in t/d.Solution ： =(101325*224*1000*10-6)/(8.315*546)=4.999mol/s RTPV n =SO 2~ CaSO 4·2H 2Oso n （c a so 4）=n （so 2）=4.999mol/s*90%=4.4991mol/s m=4.4991mol/s *172g/mol *3600*24s/d *10-6 t/d=66.86 t/d5）A power plant emits 36 kg/h of SO 2 at height H=120m and the wind speed is 2 m/s. Dispersion Coefficients: σy =40m and σz =30m, Estimate the ground-level concentration of SO 2 from this source at a distance 1km directly downwind?()⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡⎥⎦⎤⎢⎣⎡+-+⎥⎥⎦⎤⎢⎢⎣⎡⎥⎦⎤⎢⎣⎡--⎥⎥⎦⎤⎢⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡-=22221exp 21exp 21exp 2,,z z y z y H z H z y u Q z y x C σσσπσσSolution ：u=2m/s , H=120m, Q=36kg/h=10g/s , y=0 , z=H=120 , x=1000m =0.40⎭⎬⎫⎩⎨⎧+⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡+-=130********exp 30*40*3.14*10*210C 6）A power plant flue gas contains 1000×10-6 （1000ppm ）of NOx ，and is emitted at a rate of 89.6m 3/s at 546K and 1 atm. The NOx is 90%mol NO, balance NO 2. A selective catalytic reduction system is to be used to remove the NOx. Calculate the minimum of ammonia needed in kg/h.Solution: 4NO+4NH 3+O 2→4N 2+6H 2O 2O+4NH 3+O 2→3N 2+6H 2OPV=nRT n=(101325*89.6*1000*10-6)/(8.315*546)=2.0mol/sn(no)=0.9*2mol/s=1.8mol/s so n(no2)=2.0-1.8=0.2mol/sn(NH3)= n(no)+2* n(no2)=1.8+0.4=2.2mol/sm=2.2mol/s * 17 g/mol =37.4g/s=134.64kg/h7）The efficiency of an ESP is 98%. The efficiency of the ESP drops to 93% as a result of flow rate changes. Calculate the ratio of flow rates for the above situation. Use appropriate assumptionSolution: A =－Q×㏑(1－η)/ωp SO Q=-A*ωp/㏑(1－η)Assume the plate area A and the effective drive velocity ωp are unchanged.ThenQ1/Q2=(-A1*ωp/㏑(1－η1)/ (-A2*ωp/㏑(1－η2)=㏑(1－η2)/ ㏑(1－η1)=㏑(1－0.93)/ ㏑(1－0.98)=0.68。