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Atmospheric Dynamics 1

by: Cordie Miller

Atmospheric Dynamics 1 ESCI 342

Cordie Miller

GPA 3.65


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This 7 page Class Notes was uploaded by Cordie Miller on Thursday October 15, 2015. The Class Notes belongs to ESCI 342 at Millersville University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 33 views. For similar materials see /class/223520/esci-342-millersville-university-of-pennsylvania in Earth Sciences at Millersville University of Pennsylvania.


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Date Created: 10/15/15
ESCI 342 Atmospheric Dynamics I Lesson 5 The Total Derivative Reference An Introduction to Dynamic Meteorology 3ml edition Holton Reading Holton Chapter 2 sections 21 through 23 THE TOTAL DERIVATIVE Meteorological variables such as p T 7 etc can vary both in time and space They are therefore functions of four independent variables x y z and t The differential of any of these variables eg T has the form dT a7Tdta7dea7Tdya7sz dt dx dy dz 0 Dividing through by the differential of time gives dT 6T 61 61a 61a alt at ax dt dy dt dz dt 0 By definition dx dy dz 7 E u 7 E v 7 E w dt dt dt so that d a 9 9 9 dt dt dx dy dz which can also be written as d7T a7T 7 VT dt 9 This shows that in general the partial derivative is not equal to the full derivative We refer to the full derivative with respect to time as the total derivative or material derivative and give it the special notation of DDt so that the total derivative operator is d 7 d d d d V0V7u7v7w7 dt dt dx dy dz In the example using temperature we therefore have a7T n a7T v7 W7 dt dx dy dz 0 The total derivative DDt represents the change relative to a reference frame attached to the air parcel and moving with it I This is referred to as a Lagrangian coordinate system The term ddt represents the change from a coordinate system fixed x y and z coordinates This is called the local derivative The term 7 39 V is called the advective term and represents that part of the local change that is due to advection transport of a property due to the mass movement of the uid s7 vT unlally xed In pace Theletole unng temperature a an example we wnte 7 i 7 l7 V T o n J o It 1 Important to undentand that the change we meaxule wlth out mxtrument may be due to elthel a change wlthln the am ltxelf reprexented by the DTDxtelm or due 7 m nt A h the V VJT term 0 Example The tempelatule at out tatlon ha been decreaxlng Th1 may be due the entlle all max 10mg heat due to ladlatlon OI conductlon DTDt 0 due to the VJT wlnd blowlng colder all lnto out area ADVECTION DE A SCALAR VERSUS A VECTOR VV7Vgt V V A V VA became you can t take the gradient of a vectol o Fol a vector MORE ON ADVECTION o gradlent vectol lt l therefore Ieadlly evaluated 0 ample The wlnd l from 330 at 25 m The lxothelm ale onented north outh a hown n the plctule below and are 100 km apart 30 C 20 C 10 C Ln component form the two Vector ale 397 125 mfr216 ms T 7 00001 mm The advectlon I theletole 47 tVT 125 ms 5 216 ms j 00001 Cm f 000125 Cs The advection would cause the temperature at a fixed point to increase by 45 C in one hour independent of any other temperature increase or decrease due to radiation or conduction Another way to solve this problem would be to find the angle between the two vectors in this case is it 120 and use the formula that 7 VT Vlvrl cos 0 25 m s00001 Cmcos120 000125 Cs O O Advection itself is defined asi 7 0 Vs where sis any scalar property eg u v w T o The minus sign ensures that if the velocity and the gradient are opposite then the advection is positive since the property would be increasing with time ADVECTION OF MOMENTUM AND THE CURVATURE TERMS O The equation of motion in vector form using total derivative notation is le 2 2X 7 VV2 7 1 Dr 0 0 Expanding the total derivative yields alwwilvp2 2xmgyv2v a 0 O The term 7 Vx7 represents the advection of momentum by the wind itself Expanded out it has the form av av av VIV Vuiviwi 2 ax By dz The derivatives on the right hand side of 2 expand as follows a 7 alfalA39all uaivai we 3y 3y 3y 3y 3y 3y 3y av duo aw de 05 a 312 ii1 7j7kuiviwi dz dz dz dz dz dz dz If the Earth were at then the terms involving derivatives of the unit vectors would all be zero Unfortunately the Earth isn t at so the directions of our unit vectors change with position We therefore need to evaluate all of the following derivatives ii if 9i ii if 9i ii if 9 9 ax ax ax ay ay ay az az az This is tedious but not difficult with the following results Biftan 4llg Bitan l 81A 3x a J a 3x a Bx a i0 l2 3y 3y a By a i0 i0 aizo 3z 3z Bz so that we have aljllqalj lg Mjl VMK 3x 3x 3x 3x a a a a 8 7 Bu 811A BWA VA WA ii1 7J7k 7ki 3y 3y 3y By a a 87V Bl 8v A 3w A az 31 31 31 and putting these into equation 2 gives the advective term in component form a a a A a 2 A a 142v2 A vVV VVu W i VVvutan ijjVVw k a a a a a O The full momentum equation in component form on a spherical earth is then f alt7Vu W l29vsin 29wcos vvzu at a a pax 2 j a7V 7VVWM l 2 2usin VV2v at a a pay A a u2v2 k alvvwg lal29ucos gVV2w at a paz 0 If we really want an ugly set of equations we can expand the advective terms further and get du du du du uvtan uw ldp uiviwi 7 7 2 av 81 aWEWM ii 2 2usin vvzv dz a a pdy aw 2 lap z 7 7 772 2 V at ax 8y waz a paz ucos g V w 7 2 2vsin 2owcos VVzu dt dx dy dz a a p dx Full Set of Momentum Equations on a Spherical Earth 0 The table below summarizes the various terms in the three momentum equations local advective terms curvature terms pressure Coriolis viscous derivative gradient terms d1 udlvdlwdl uvtan idl 29vsin 2 2wcos W214 dt dx dy dz a a 0 dx a ma a uztaw 4gp 29usm vvzv dt dx dy dz a a 0 dy 2 1w w 491 W w dt dx dy dz T 0 dz VECTOR FORM VS COMPONENT FORM OF MOMENTUM EQUATION 0 Notice that even for a spherical Earth the momentum equation has the simple form of DV lvp 2 2gtltt7gvv2x7 Dr 0 Of 31vwilvp2 xmgyv2v a 0 when written in vector form The curvature terms don t appear until we start writing the momentum equation in component form It is much easier to memorize the momentum equation in vector form If you know it s vector form you can always then expand it into it s components using your knowledge of vector calculus EXERCISES 1 Show that if the wind is blowing parallel to the isotherms that the temperature advection is zero 2 Holton problem 22 3 On the map below indicate an area where a The temperature would be increasing due to advection b The temperature would be decreasing due to advection c An area where temperature advection is very weak I 7 V 2850 2880 I e IE 391 K v 9 4 K W uf I A 2 a x r FRI re b 14 203 g 39 700 pres 7em FRI Feb 14 2003 3 win a 0 e 4 33f Jc 700 pres opoy ntial height 4 Showthat V39V 739Vaxi 739Vay3 7Vaz12 7Vujf 7Vvj 70vw 5 Show that Dr at at at s a For the followmg pro le of u explmh whether a downdm would came an andyuunz Bu Bu Bu 3 7 vW Hmt 9 By Hz 2 r 0 u I Do the mm a m 61 only for the followmgpm le z 7 vae the followmg ldentltlexi t A g klk 1 x a a a Lo 713 y a io r 0 z z


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