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EVERGREEN FTTS 2004 - Understanding the properties of water is central to understanding climate

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Paul’s Lecture Climate 11/17/04Understanding the properties of water is central to understanding climateWater is a dipolePhysical consequences of water’s dipole structureRelative humidityAtmospheric CompositionAtmosphere is a mixture  gases, aerosols, water dropletsresults from exchange between atmosphere and Earth surfacePhysical processesBiological,GeochemicalVertical Structure of AtmosphereThickness of Atmosphere No well-defined upper boundary - difficult to establish Density decreases with altitude: @ 16 km (10%), 50 km (1%) Thin layer enveloping Earth where weather occurs set 100 km as upper limit – 99.99% of atmosphere, 2% of Earth’s radius Relatively thin but with great mass (5.14 x 1015 kg)Paul’s Lecture Climate 11/17/04Atmospheric PressureAtmosphere – exerts large downward pressureGravity – compresses air near surface (increase density)Mass of atmosphere below 5500 mSea level pressure = 1kg/m2; 1013.2 mb; 101.32 kPaPermanent gases• Major gases: N2 and O2• Minor gases (Ar, Ne, Kr, Xe) plus Hydrogen (H) and Helium (He)Mean residence time (MRT)Atmospheric MRTsPermanent gases have long MRTsN2 13 million yearsO2 10,000 yearsMixing time of atmosphere is about 4.8 yearsGases with MRTs greater than mixing time are evenly distributedVariable gases have short MRTs Carbon Dioxide (CO2) MRT= 4-5 yearsWater vapor (H20)MRT= 10 daysOzone (O3) MRT= 3-7 daysAerosols (Particulate Matter)Small solid particles and liquid dropletsNormal conc. = 10,000 particles/cm3Associated with human and natural processesSmallest particles have radii = 0.1um from sulfate gas convert to liquid/solidPractice with some atmospheric calculationsDo these in groups of two or three. Check your answers with us before you’re done.1. Calculate the MRT for water in humans with the following examples. Water gains are about 40ml/kg per day body weight for both genders. Male: 70 kg, 60% water; Female: 56.6 kg, 50% water.2. Lifting of air masses cools them according to the adiabatic rate (dry adiabatic rate -10°C/ 1000m,wet rate is -6°C/ 1000m). Here in the PNW, orographic lifting (lifting of air due to topographyforcing it up) is responsible for much of the rainfall we receive on the west side. At what elevationwould clouds and rain begin to form for air that is at 16°C and 74% RH at sea level (dewpoint is11°C). What would the air temperature be at Paradise (1676 m)? Calculate the temperature on theeast side of the mountains for the same elevation where clouds formed on the west side. Assume allheating/cooling is due to elevation changes only. (1m = 3.28


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EVERGREEN FTTS 2004 - Understanding the properties of water is central to understanding climate

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