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# AIR POLLUTION ATM S 212

UW

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This 5 page Class Notes was uploaded by Ms. Imani Mante on Wednesday September 9, 2015. The Class Notes belongs to ATM S 212 at University of Washington taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/192397/atm-s-212-university-of-washington in Atmospheric Science at University of Washington.

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Date Created: 09/09/15

ATM S 212 10152008 Iquot 39 2 Kev Discuss the questions among yourselves and ask questions to the TA Atmospheric Stability 1 Draw the temperature curve that a balloon with a temperature of 20 C at the ground will follow if we lift it to 15 km altitude for two cases a Assuming that the air is dry without condensation what is its temperature A cu 3913 43 a as b Assuming that the air contains water vapor with condensation what is its temperature 10 15 hint the curve is a straight line temperature C note dry adiabatic lapse rate Fa 10 DCkm wet adiabatic lapse rate Fw 6 c Ckm 2 By drawing the temperature curves describe what will happen to dry pollution plumes that we release with a temperature of 21 C and 25 C at the ground The curve in the graph describes the environmental lapse rate Describe in words the behavior of these pollution plumes altitude km The 21 C pollution plume will rise and cool at the dry adiabatic lapse rate until its temperature is the same as surrounding air That altitude is 06 km 155 C This is an inversion layer and the parcel will stay there l 0 l 5 temperature C The 25 C plume will rise adiabatically until 09 km where it temperature will be equal to the surrounding air This part of the pro le is neutral and the air parcel will stay at that altitude with a temperature of 16 C Discussion 2 ATM S 212 Page 1 of3 Atmospheric Stability Fall 2008 3 Using the environmental lapse rate on the graph answer the following question Calculate the lapse rates for each part of the curve For each layer state whether the air is stableneutralunstable with respect to the dry adiabatic lapse rate a V b If we push the dry air parcel A sitting at 06 km altitude with a temperature of 16 C up or down is it going to sink rise or remain at this altitude Why c Now let s cool offthe dry air parcel sitting at 06 km by 2 C down to 14 C Ifwe push this air parcel B up or down what will happen Why altitude km 10 15 temperature C a Lapse rates for each palt 0f the curve are determined by looking at the rate of decrease of temperature with increasing height 0 06 km altitude The temperature decreases by 4 C over 06 km altitude so the lapse rate is lquot 4 C06km 66 Ckm lt10 Ckm The air is stable 06 11 km altitude The temperature increases by 225 C over 05 km altitude so the lapse rate is F 225 C05km 45 Ckm lt10 Ckm The air is stable 11 2 km altitude The temperature decreases by 5 C over 05 km altitude so the lapse rate is lquot 5 C05km 10 Ckm 1quota The air is neutral b Air parcel A will remain at 06km altitude If we push it up it will c001 at a rate of 10Ckm and will be colder than surrounding air it will thus sink back to its initial position If we push it down it will c001 less rapidly than the surrounding air it will be warmer so it will go back up c Air parcel B will sink to the surface If we push the air parcel down it will be colder and denser than surrounding air it will thus keep sinking following the dry adiabatic lapse rate red dashed line until it reaches the surface If we push the air parcel up it will still be colder than surrounding air and will thus sink Discussion 2 ATM S 212 Page 2 of 3 Atmospheric Stability Fall 2008 ATM S 212 10152008 Discussion 3 Quiz Kev Name Solution Please hand this sheet to the TA at the end of the discussion section Atmospheric Stability 1 The temperature at the top of the inversion is 13 C What is the minimum temperature a parcel must have at the surface to rise past the inversion if it rises at the wet adiabatic lapse rate PW of 6 Ckm altitude km Since the top of the inversion is at 15 km and 10 15 t tu C the saturated adiabatic lapse rate is 5 Ckm empera re the parcel would have cooled by 75 C by the time that it reached the inversion top Since the inversion is at 13 C at the surface the parcel would need to be at 205 C Discussion 2 ATM S 212 Page 3 of 3 Atmospheric Stability Fall 2008 ATM S 212 10292008 Box Model Practice Kev S 139 V Box Model equation I 1 The emissions rate of NO in downtown Metropolis is 58X1027 molecules of NO per hour If the lifetime of NO in downtown is 1A day calculate the concentration moleculescm3 of NO in the downtown area Use a volume of 47X1016 cm3 s 58x1027 molecules NOday V volume of the basin 47gtlt1016 cm3 1 14 day We can now calculate the concentration of NO by applying the above formula 58x1027 molecules N0y day q day 4 309 1010 molecules N0 v 47 x 1016 cms cm 2 Convert your answer from 3 to a mixing ratio ppbv Assume that 1 cm3 of air contains 25 X1019 molecules molecules N0 309x1010 Concentration cm3 123x 104 W AirDensity 9 mOIECules air molecule air 3 cm Mixing Ratio 123 ppbv N0 25x 101 3 The city of Metropolis has an approximate volume of 15X1018 cm3 The total SO emissions in that basin are 1X1030 molecules of SO per day and the residence time of S0 is 12 a day What is the concentration in moleculescm3 of SO in the city s 1gtlt1030 molecules SOday v volume ofthe basin 1875x1018 em3 1 12 day We can now calculate the concentration of SO by applying the above formula 1x103 molecules SO2 day S 1 day 2 molecules S0 q 18 3 333x10 32 v 15x10 cm cm Extra Practice Problems ATM S 212 Page 1 of 2 Box Model Fall 2008 4 The mixing ratio of carbon monoxide in Metropolis is 300 ppb What is its concentration in moleculescm3 Remember that 1 cm3 of air contains 25X1019 molecules and that 1 ppb means one part per billion billion109 300 ppbv of CO means 300 molecules of CO in 109 molecules of air MR300x10 9 Given that there are 25x1019 molecules of air per cm3 and we know that ConcentrationMixing Ratio Air Density Concentration 9 molecules C0 19 molecules air 12 molecules C0 300x10 25x 10 m3 75x 10 cms molecule air c 5 Let s assume that the residence time 5 of carbon monoxide in the Metropolis basin is 1 day and that 14 Million people live in the basin How many CO molecules does each person emit per day The volume ofthe basin is 1875X1018 cm3 To solve this question transform the steady state box model equation to calculate the source rate S from the concentration q the volume V and the residence time 5 5 r hu 139 tI q CO concentration 75gtlt1012 moleculescm3 V volume 0fthe basin 1875x1018 cm3 1 1 day We can now calculate the source rate of CO by applying the above formula BANZW 1s75x10 cm3 S q V cm 1 4x10 molecules C0 1 1 day day Each person the emits s Spopulation 14gtlt1031 molecules COday14gtlt107 persons 1gtlt1024 molecules COday per person Extra Practice Problems ATM S 212 Page 2 of 2 Box Model Fall 2008

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