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Special Topics in Aerospace Engineering

by: Dayton Klein II

Special Topics in Aerospace Engineering MAE 4504

Marketplace > University of Virginia > Mechanical and Aerospace Engineering > MAE 4504 > Special Topics in Aerospace Engineering
Dayton Klein II
GPA 3.67

Robert Johnson

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Robert Johnson
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This 59 page Class Notes was uploaded by Dayton Klein II on Monday September 21, 2015. The Class Notes belongs to MAE 4504 at University of Virginia taught by Robert Johnson in Fall. Since its upload, it has received 50 views. For similar materials see /class/209793/mae-4504-university-of-virginia in Mechanical and Aerospace Engineering at University of Virginia.

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Date Created: 09/21/15
Space Science Atmospheres Part 7b Venus Earth and Mars Where is the H20 on Venus Planetary Escape Isotope Fractionation Hydrodynamic Escape Result of Simple Model Mars The Ice Planet Water primarily in ice caps and regolith as a permafrost Eanh The Water Planet Venus The Water Vapor Planet Run away Greenhouse Effect Where is the water Since H20 does not condense and is lighter than N2 0r CO2 it can reach regions Where UV can dissociate H20 hv gt OH H KE gt O H2 KE Large scale heights Vertical Structure Exosphere 7 7 quot1 a Vquot39 quot Shuttle Thermosphere Aurora Planetary Escape De ne Exobase Top of Atmosphere If an atom or molecule has an energy suf cient to escape is moving upward it will have a high probability of escaping Probability of escape is high if collision probability is small mean free path for a collision xcol z 1 nx O col 0w collision cross section nX density at exobase Exobase altitude occurs when scale height mean free path Hx Acol HX N 1 nx O39col since nXH NX column density NXO39 1 col Nxz1o col Exobase Probability of collision is small Exobase Exobase NX ocol 1 NX column of atmosphere oc01 molecular size 103915 cm2 Therefore NX z Owl39l z 1015 atom cm2 or nx z O1coll1 Earth 1000K O gX 850cmSZ Hx 1000km 11Y 107 Ocm3 zY 550km Escape continued Escape Energy Ees 12 n1 Ves2 mgXRx gX acceleration of gravity at exobase RX distance of exobase from center of planet surface Ves V 104 E 112 M 48 kms At Earth TX z 1000 K v1kms no escape BUT for a given TX there is a distribution of V For a given TX there is a distribution of V fV ex Vz 2m kTm32 p 2kT fVX fVy fVZ ffffd3v 1 Can Focus on the 2 component only fVZ 1 V2 ex Z zerTmr2 p ZKT T vz dvZ 1 Need Flux of molecules across exobase in the z direction Des fffnX VzfV d3v VZ gt 0 V gt Ves Escape continued Things are often written in terms of the mean speed 5 a fffvfTd3v Problem Verify that the mean speed is 8 K T V 7171 1 1 Flux across a surface Vz gt 0 all V lt1gt sz7d3v 0 V Z n e X f0 2 aszmf2 Z dV 2KT Z XP 12 nXkTm2 a kTm 2 275m nX V l4 due to isotropic assumption IH This is in the absence of gravity Escape Flux c0nt Flux Across 1 nX V EXObase I 8 kT V nX density 7 m Probablity of Escape Pes I fE dE ng fE energy distribution of ux at eXObase Escape continued For Escape Use E gt Ees Where Ees is the escape energy The escape ux is ltIgtes fff nx vz fo d3v Rewrite as the ux across the exobase times a probability of escape 1 I nVP 4XeS Obtain Pes problem Jeans thermal Escape 1 E95 eET kT p es E T compares thermal energy to escape energy OR using Ees mgXRX ma kT kT RxHx In this form Compares Radius at the top of the atmosphere to the scale height Total Loss by Jeans Escape Flux XPeSX Area X Time 01 Nes Column Lost Flux x Pes x Time Pressure change Nes m g Example H from the earth Present escape ux 107Hcm2s If constant t 3X107 syr X 45 X109yr 14 X1017 3 Nes 14 x 1024 Hcm2 Ap 2X103 dynescm2 2 X10393bar or Total column at Earth N 2 X1025 molN202 cm2 10 meters frozen Equivalent water loss 05 m Escape cont Problem Jeans escape Escape Energy Venus 056 eV u Earth 065 eV u Mars 013 eV u Jupiter 18 eV u Titan 0036 0024 eVu at exobase Assume TX 1000 K not true on present Venus Use a time t 45 byr 45 x 109 x 3 x 107 sec Assume NX 1015 atoms cm2 Calculate the net column loss Neg of H H20r D N from each planet assuming H k TX m gX and the exobase is either all H H2 or N Approx Present Values zx V 200km E 500km M 250km T1500km TX V 275K E 1000K M 300K T 160K Jupiter no surface use Tx 1000K Isotope Fractionation preferential loss of lighter species Relative loss rate is determined by 1 Masses of the escaping species therefore PegD ltlt PegH 2 But they must be present at the exobase hence depend on differences in diffusive separation Compare loss of H to D H lighter D heavier to learn about Loss of water Isotope Fractionation preferential loss of lighter species Relative loss rate depend on differences in diffusive separation Compare loss of H to D H lighter D heavier NHzh NDzh col densitiesbelow homopause turbopause determined by atmospheric concentrations same Hmixed At homopause Ratio RD zh NDzh NHzh nDzh nHzh At exobase nHzx nDzx densities at exobase A2 zX zh Different H kTmg nHzX nAzh expAzHH nDzX nBzh expAzHD Ratio RD zx RDHzh expAz1HB1HA RDl zh expAz Am gxka Rel concentration Height of exobase Az mass dif Am Isotope Fractionation cont Loss rate of a column of atmosphere dNA 1 dt For Jeans thermal Escape cpA EPA 4 Ees With P 1 e H Calculate The fraction of species A still in the atmosphere is fAt 3 NA N A CXp ttA tA1 Depends only on mass of A and ME m examHA the temperature at the exobase Isotope Fractionation cont Two species A and B eg H and D The ratio of total atmospheric concentrations vs t is NAt N N ii One can re write rABt z fAtX With X l tAtB and f At the fraction of A remaining at time t Enrichment of Heavy Isotope Indicates Atmospheric Loss D H SUN 15 x 105 COMETS 3 x 104 EARTH S WATER 16 x 104 VENUS Atmosphere 2 X 10392 D enriched relative to Sun So not only Venus but earth has also lost water unless you think comet pasting on of water is the starting point Note this does NOT depend on how much water we started with DH Ratio Earth and Venus tH ltlt tD your problem NH amp z X IIDnow NDsolar Earth DH 16 x 10394 rHD 15 x 1039516 x 10 4 z 009 x1 fHt 009 That is only 10 of original water released is still in the anal atmosphere and oceans assuming well mixed Disagrees with present loss rate must have been an earlier epoch of rapid loss Venus DH 2 x 10392 fH t x 00008 or assuming the same original water budget as the Earth x 0008 of Earth39s present water budget present value of loss rate 107 s too small Need wet hot early atmosphere with H20 well mixed to hot to condense ar all altitudes inspite of lapse rate For both V and E did hotter EUV early Sun andor T Tauri caused blow off Enrichment of Heavy Isotope Indicates Atmospheric Loss Other species V E M S 36Ar 38Ar 51 53 4 40Ar36A1 1 296 3000 40K9 40Ar e 639 108 years Xe N O and C are also fractionated Need escape processes other than Jeans escape for heavy species ESCAPE PROCESSES Jeans Escape Hydrodynamic Escape Blow Off Photo Dissociation Dissociative Recombination Interaction With the Local Plasma T Tauri Sweeping 99599 Very heavy species 2 When molecules are at the exobase 3 and 4 are important Venus H20hVOHHKE a O H2 KE Present Mars C02 e a COOKE 02 e a OOKE In the absence of a protecting magnetic eld 5 and 6 are important Fractionation nonthermal escape processes Diffusive separation gives an exobase ratio between a lighter A and heavier B species If the loss mechanism is not strongly affected by the mass difference then Only their relative abundance at the exobase is important Rayleigh fractionation law Applies to Mars for processes 3 4 5 Result Ar and N arew fractionated 36Ar 35Ar no need for hydrodynamic episode 14N 15N models give too mach loss buffered by C02 BUT nonfractionation of 18O 16O 13C 12C means there are large reservoirsll carbonates and permafrost Space Science Atmosphere Greenhouse Effect Part5a Solar Earth Spectrum IR Absorbers Grey Atmosphere Greenhouse Effect Radiation Solar and Earth Surface BAT Planck Ideal Emission LZSriggd Atmosphere is mostly transparent in visible but opaque in UV and IR IR window 813um GRAY ATMOSPHE E Transparent vs Gray Partially absorbing in IR but absorption is independent of frequency over the range of relevant frequencies Process Surface heated by visible Warm surface emits IR 3 100 um peak 15 um IR absorbed by COZ 03 H20 etc Remember why not 02 and N2 Vibrational Bands 002 IR active Symmetric Stretch lt O C O gt 746 nm N Asymmetric Stretch gt 4 O C OV 426 um Y Bending A 9 0 9 150 um Y NearlR peak H20 Symmetric Stretch 6 273 um Y H H Asymmetric Stretch O gt 266 um Y H H Bending O 627 um Y kH H You can have combination bands or 2 vib levels A Real Green House lIR quot gt Visible OUtSide glass Inside How do you get IR out equal to Visible light absorbed inside RAISE T Note For a real green house convection may be as important ie glass a thermal barrier 2 A r 40 J 2 u a on a 0 l I Q n l 039 3 39 Ground Emits Primarily triatomc molecules absorb and re emit Vibrational and rotational states To determine T we assume excited molecules heat locally by collisions C02V1 M gt C02V0 M KE Also at a given T CO2 excited and can emit C02V0 M gt C02V1 M KE Goody Walker Simple Model it 0 F1 F01quot Space 1 I if 4 F2 Top layer 2 I 39 F1 4 F3 3 I v F2 air 4 F4 2T 4 439 V F3 1 F A Fa FVIS Fout 4 Ground I chose an atmosphere 4 layers thick in the IR Each layer is 1R 1 thick It would absorb 1 e391 63 of the flux from the layer above and below we ll assume 100 for simplicity We will also assume that each layer emits like a perfect emitter 0T4 where T is its temperature The surface is heated by the sun and it emits Note heat balance Fvis FouIt 0 Te4 O thA simple Greenhouse cont I F1 Fout Space I i 4 F2 Top layer I 39 F1 1 4 F3 I V F2 4 F4 439 V F3 41 A FeI FVSFout F4 Ground Divide atmosphere into IR thick layers IIR 1 Energy balance in each layer 0 F1 Fem 0 T64 Out In 1 2E F2 9 F2 2E 211m 2 2F2FIF3 a F32F2 Fl3Fout 3 2F3F2F4 a F42F3 F2 4Fout 4 2F4F3Fg a Fg2F4 F3 51out Note can write Fg 1 4 0Tg number of layers Font 0Te411g L39 g number of IR layers O thA simple Greenhouse cont F1Fout f Space I i 4 F2 Top layer I 39 F1 4 4 F3 I V F2 if 4 F4 47 V F3 41 A FeI F FVSFout 4 Ground Divided atmosphere into IR layers TIR 1 We nd Fg 1 number of layers Fout 01 0Te41 rg Ground layer gives the same result g FVlS F4 Fg ng FVlZ 51out 7T64 then get the same result Smce lviz lout T4 Te41 rg drop a g How Many Layers on Earth 19 z 2 Using Te 250K Implies Tg 330K NoWay Again it is clear that at the surface convection is important but it gives 288K too cold Greenhouse Convection determine surface T Note F4 air above surface we have to do a little more carefully Before Finishing can make an estimate IfrgzZza f N abs abs USll lgN 2 X 1025 l l lOlCl l l2 see early lecture fabs 1 HZOCOZ 03 jabs 3 103923 cm2 IR absorbers have small cross sections relative to UV VENUS Problem for set 3 Te 232 T9 750 Therefore 1 9 Then Use cross section from previous slide pure CO2 Find column NC02 GREEN HOUSE EFFECT H20 21 K CO2 7 K 03 2 K OtherNZO CH4 3 K 33 K Te AT 255K 33K 288K z gt 30K increase T Note Increase 03 2 z lt 25K decrease T Double C02 30 increase since 1800 AT 25 5 K ATMOSPHERE ONLY 1 2 K ATMOSPHERE OCEAN We Will come back When we discuss atmospheric evolution to discuss Reservoirs for carbon and Time Constants Carbon concentration vs time Increase from 280 ppm 1860 to 370 ppm present 31 increase Note 370 ppm 00370 39 39 1qu rem InMn l39JN umluuu llm 5mm lmululmlmlUcummunpln IMO I I dllllml39h 47mlmmIlemlunquothmmu1lnmpumAdmummmuNt39ImJnh InIIpl IIL MIAlecw lTul 0 Wm ml Imwnuupn Dr I39Icm lJan Mu 39MDL lyamun L h rm w mug lump h plenumm mmuym m l t V 39ley Mu I lullzn l ulllmnlu mm Nmul Ldkcdmuu mulmdu Carbon Concentration Long Term Historical Record Correlation of CO2 and Temp Antarctic Ice core temperature record Antarctic ace cere carbon droxrde record N O 5 lt5 2quot a 3 O E E 2 e S E 2 N qdd uoqenueauoo 03 ouaudsoww C 10 2C 36 4C 5 60 7C 80 SD 10011012C13E14015C Thousand years before present Later we will look at the carbon cycle 0 Carbon Dioxide 002 Contributes 60 of warming 0 Trace Gases 100 1000 times less Contribute 40 of warming Methane CH4 1530 of warming Nitrogen Oxides NOx up to 15 of warming Chloro uorocarbons CFCs 12 of warming However 39239 Molecule for molecule C 02 is less efficient than other greenhouse gases because its atmospheric concentration is high and hence its absorption bands are nearly saturated 139 Over a lOOyear time horizon reducing 313 emissions by 1 kg is as effective from a greenhouse perspective as reducing C 02 emissions by 24900 kg some onhe infrared some solar radia n is reflected by the earth39s surface and the atmosphere The cooling factors Vulcunln mm warnTuning WMheuppu m 399 Bandung wideomnuds naalandall AE usaLs mm nunumnadun quotW hamanndmrml MM 3 smpnmns m Mn Amosan 1 MW M y Aemmll W plr das m Iqmd nrdussuspsn ed m Ammo 5mm Mn marmnapmmmmnmmmmwmemm mm Is smpnale mama mm 302 Wbmwm EMMNI WE mummmwugm PLANETARY ENERGY BALANCE GW g 35 Incoming solar radiation At Earth Absorbed SUN Clouds 21 100 Atmos 22 Ground E y 67 26 lzz Back to space Absorbed quot1 Re ect Ground 7 21 atmosphere 31 Clouds 26 7 reflected T Albedo 33 Surface IR Radiation T2 Space mesopause AA AA 1 13 86 ggnvectlve V GROUND Thermosphere Part3 EUV absorption Thermal Conductivity Mesopause Thermospheric Structure Temperature Structure on other planets Thermosphere Absorbs EUV Absorption Solar Spectrum The spectrunl of solar radiation outside the Earth39s atmosphere matches Closely that of a blackbody at 5800 K smar spectrum outside atmosphere Solar spectrum at sea level Increasing llux gt 07 pm max 05 pm about half in the IR and Sll au 39ac wnat EUVvery little heat flux But also very little atmosphere But it is the region for escape And diffusive separation Peaks at visible 0 Thermosphere heating at short wavelengths 1EUV Heats also soft X rays chargedparticles 02 hv a O 0 J2 2 OabsFEUVZ from before N2 hv a N N smaller 0 hv a 0 e also creates ionosphere X hvXe 2 At Lower Altitudes Cooling Recombine O N etc N2 O2 and 0 Cannot Radiate IR Homonuclear Diatomics and Atoms Discuss brie y Will return no CO2 diffusive separation aMust Conduct Deposited Heat to a Region Containing CO2 H20 03 etc Mesopause Roughly rough Z EUV picture heahng 130k I R cooling Thermal 80km conduction mesopause Thermosphere cont Heating lFsz TiZZIZZZZi AZ Heat Source Absorption of EUV Photons 0 dFS p p dt M dz assumes unit 0 dFS p p dt M dz reduction in the energy ux efficiency of photon energy to heat 0 1 1 abs abs Fs Thermosphere cont Heat conduction Heat loss Conduction Considered the thermal heat flux CDT Conductive ux is opposite to temperature difference CDT 06 AT Negative sign ow of heat is from high T to low T you may have called it Newton cooling law Write 1D cpT K or 3D FIST KVT Z K is the thermal conductivity Energy left in the volume gives the heating rate dCIDTdz gt heating leaves heat in dz dCIDTdz lt cooling removes heat from dz IcpT Acp A2 V7777 quotEd Heating Cooling rate pcpg d CDT z Thermosphere Continued Heat Flux What is K IDT K dTdz K z constpcV Ed 7d Ed is a diffusion length molecular or eddy d is a mean diffusion speed molecules or parcels of air Heat Eq with thermal conduction diffusion E p dt d dT pc E KE Sz Lz Source Loss This is a region where diffusive separation dominates Here conduction is by gas phase collisions Thermosphere Continued Heat Eq with thermal conduction diffusion dT d dT c K SZ LZ P p dt dz dz Source Loss Thermosphere Heating by absorption of sunlight EUV SZ MdFSdz in 1D Assume p Cp 3 O steady state Assume LZ Radiative Loss 0 except at the bottom 39boundary39 Treat as a boundary value problem Below mesopause we Will do radiative transfer soon Thermosphere T Structure Heat equation for the thermosphere 0 iKd T Sz Sz Chapman Layer dz dz Solve First Integrate from z to infinity 0 KdTdz f Sz39dz39 N0 Heat Loss to Space dTdz 0 at top of atmosphere dT fOOSz39dz39 K dz A realistic K KT as KOT43 Integrate again from Mesopause zm to z gives Tz T74z z T14 dz fSz dz Tm is the boundary where heat conducted downward is radiated to space Relationship to our old Te Thermosphere Structure Cont HEAT FLUX DOWN HEAT DEPOSITED ABOVE g f0 Lu dFsdz39 dz an F M 1eXp m0 This says that the downward heat ux at any altitude must be equal to all the energy deposited above by the absorption of EUV photons Thermosphere Structure Cont K T FSOM1 XPTMl Z For simplicity let K K0 a constant M 1 F TZTS ZdZ39lCXT39 o m 10me plt 1 If absorption maximum is high 2 gt 130km then below maximum variation is nearly linear max T z z Tm 1132 zm forzm ltz ltzmax 0 Thermosphere Structure Cont For K K0 M21 F0 z S dz39 1 ex 139 K0 flm plt 1 TZ sz Change variable Optical depth 17 f Cabs nabsz39 dz z OabSH nabsz drdz z r H Tz Tm FliH f dr r39 1 eXpI39 0 Integral for thermosphere model 1o r 0 xm1oo g xm1o quot xm5 8 k x 6 u X X 1ZOCX 44 u A N l 170 0 zgtoo l I X lt 1Z TZTm FoKOHH Y YX fxxm 1 eXI0X X dX Over how many scale hieght does heat have to be conducted Zm80km Mesopause x 220 K 1000 K Thermospheric Temperature Temperature vs Altitude Earth s Atmosphere Rise in T in Thermosphere due to EUV Absorption and Thermal Conduction to Mesopause exosphcre E H Ihennosphere Egg Di usive Mixed 100 UV 7 2 km mesopause mesosphere SO stratopause stratosphere Lropopause VIg troposphere 200 288 1000 Ilcals the suriac c WK These are averages Tropopause is higher and hotter Near equator 1618km and 10km near poles Drives high altitude horizontal flow Jets planes 10kme close to stratosphere to avoid turbulence 68 5 a 260 Earth 240 quot 220 200 39 106 180 393 V w e 160 much mlar E go 140 Ionosphere 1 I E 395 1o395 lt 120 quot Thermosphere 100 10quot3 Mesopause 80 Mesos here P 10 60 quotStratopause 1 40 39 Stratosphere 10 20 Tropopause MM 102 Troposphere 0 l l I l L l I 41 L 03 O 100 200 300 400 500 600 700 800 900 0 1000 1200 Temperature K ThermosphereT depends on Solar Activity Photochemistry Venus Mars Simpler 002 hv a co 0 1 COOM a co2 M 2 On Venus CO gt CO2 may be aided by sulfur chlorine species On Mars 2 is very slow density Is very small Expect considerable CO especially so as free 0 oxidize surface iron oxide BUT there is a small amount of H20 H20hv gtOHH 3 COOH gtC02H 4 Recycles CO2 but Can lose the Hhence loss the H20 Venus Atmosphere b 200 Venus gt 180 V I Thennosphere 160 day 140 Ionosphere I E 120 T g Mesosphere E 100 39 I so Clo uds K 60 5mm Troposphere 20 0 I I l l I I l O 100 200 300 400 500 600 700 Temperature K Cloud Tops gt Te CO2 even at high altitudes gt cooling Slow Rotation 1 year 2 days retrograde due to thickatmosphere and tidal forces I Mars Atmosphere A 0 V I II Mais 00 C2 new solar Thermosphere 0 C Ionosphere H 4 O I l l I l I I I I I I I II I I l N O Altitude km 00 O I Mesosphere O O l 4 O I N O Troposphere l l l 80 120 1 60 200 Temperature K O I 240 280 320 T9 at Surface Dust absorbs hv and changes the lapse rate Thermosphere strongly dependent on solar activity Titan s Atmosphere d JIIIIIIIIIF j Titan 1 000 Altitude km 10 150 I Temperature K 7 Stratospherel mesosphere due to absorption in hydrocarbon haze CH4 hv gt CH3 H CHf H2 React to form CnHmand H2 escapes Thermosphere subject to particles from Saturn s magnetosphere I Giant Planets 601 1 u l E D g 10 8 11 3 102 i Jupiter 103 K 4 39 Neptune 39 104 40 6390 810 100 2 1410 160 180 Temperatur K Internal he sources Adiabatic la se rate down to liquid m tal hydrogen Jupiter has evidence of a stratosphere and a mesosphere Atmosphere Thermal Structure Typically troposphere Lower Convective Mesosphere Stratosphere Middle Radiative Thermosphere Upper Conductive Exosphere Escape Venus Temperature Lapse y 5 rd Tropopause at cloud layer 70 km Te Top of cloud layer Thermosphere Cryosphere Td 300 km CO2 at high altitudes Tn 100 km Slow rotation 1yr z 2 days Mars Temperature Lapse y 5 rd Strong Day Night Variations Dust y ltlt Pd Mesosphere T lt Te CO2 cooling Thermosphere like Mesosphere Strongly affect by solar activity Titan Temperature Lapse y Pd Stratosphere Haze Layer Mesosphere Cooler to space Thermosphere Solar UV Energetic Particles from Magnetosphere Cooled by HCN Slow Rotations Grant Planets Internal heat source comparable to solar Tropopause 01 bar y Pd Temperature pressure increase to about 2M bar so phase change to a liquid metallic state 4x104 km and a metallic core 3x104 K 1x104 km Jupiter Mesosphere stratosphere s 0001 bar Hydrocarbon coolants Thermosphere Solar EUV


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