Introduction to Physical Oceanography Homework 5 - Solutions

Transcription

1 Laure Zanna //5 Inroducion o Phsical Oceanograph Homework 5 - Soluions. Inerial oscillaions wih boom fricion non-selecive scale: The governing equaions for his problem are This ssem can be wrien as where u,v and u f v = Ju v + f Jv u J f A = f J = A u I will solve he ssem a lile bi more rigorousl han wha was done in class, bu we should all ge he same answer a he end! The general soluion o his equaion is given b. α a e λ + β a 2 e λ 2 2 We can find he eigenvalues λ,2 of he ssem b calculaing he deerminan of he marix A λi. This leads o λ 2 + 2λJ + = λ,2 = J ± i f 3 This is equivalen o he coefficien a found in class. We can look for he eigenvecors a and a 2 corresponding o he eigenvalues λ = J + i f and λ 2 = J i f respecivel. The eigenvecors saisfies A a i = λ i a i. I found a =,i and a 2 =, i. We can wrie he veloci componens as αe J+i f J i f + βe v = iαe J+i f iβe J i f The coefficiens α and β can be found from he iniial condiions, I will assume ha = u and v = = leading o such ha α = β = u /2. u = α + β = iα iβ αe J+i f + βe J i f u e J cos f v = iαe J+i f iβe J i f v = u e J sin f

2 You recognize he u veloci found in class. The veloci is oscillaing wih a frequenc equal o he Coriolis parameer as expeced from inerial oscillaions bu also decas exponeniall in ime due o he addiion of boom fricion. The deca of he veloci depends onl on he fricion coefficien non-selecive scale fricion. Figure a and c show ha he veloci oscillaes and decas as funcion of ime. As, u,v. Now ha we have found he velociies, we can look for he Lagrangian rajecories x;x and ; of a fluid parcel which was a he locaion x, a =. We need o inegrae he veloci: Z x x dx = Z Z = u e J cos fd x = x + u [ J + e J f 2 + J 2 Jcos f + f sin f ] d = Z = u e J sin f = + u f 2 + J 2 [ f + e J f cos f + Jsin f ] For he rajecories of a fluid parcel, we see in figure b and d ha he rajecories oscillae and deca oo. Bu as, we have x x + u J u f v a u v b Trajecories in he x plane u x..5 c Veloci as fc of ime u v 5 d Posiion as fc of ime x v x x 5 Figure : a u as funcion of v, b he rajecor of a fluid parcel in he x- plane, c u and v as funcion of ime, d x and as funcion of ime. 2

3 2. Coasal upwelling: Srong seasonal winds are displacing warm surface waer awa from he coas leaving a space ha is filled in b waer coming from he deep ocean. This rise of waer is called upwelling. The waer upwelling is cold, rich in nurien and comes from below he hermocline. Example : Californian coas see figure 2. In a srip near he coas, we see ha he sea surface emperaure is colder han awa from he coas. We conclude ha his region is an upwelling zone. If we look a saellie picures, we will see ha his region is characerized b greenish color due o he chlorophll. We expec he wind o have a srong souhward componen, causing a ne Ekman ranspor o he righ of he wind in he Norhern Hemisphere, such ha he waer ranspor is wesward awa from he coas and as a resul we have upwelling of cold waer rich in nurien. Figure 2: Sea surface emperaure along he cenral Californian coas Example 2: Coas of Senegal see figure 2. The upwelling mechanism is he same han he one described above. 3. Scale selecive fricion: The veloci a = is given b,x, = U coskx + l; v =,x, = U coslx + k. and he governing equaions are simpl u = K h u xx + u ; v = K h v xx + v a The balance in hese equaions is represened b he linearized acceleraion where Du/D u/, here we have negleced he nonlinear advecive erms such ha he acceleraion is now equal o he local rae of change and a scale selecive fricion. b The veloci field a = is shown in figure 4. I have ploed he conours of each of veloci componen a = and he vecor field using quiver. c We wish o find u and v as funcion of ime. Because of fricion, we expec he ampliude of he veloci o decrease exponeniall wih ime while he oscillaions in he 3

4 Figure 3: Sea surface emperaure along he coas of Senegal 2 x 4 a u 2 x 4 b v [m] [m] x [m].5 2 x 4.5 x [m].5 2 x 4 2 x 4 c Veloci field [m] x [m].5 2 x 4 Figure 4: Conours of u and v, and veloci field. 4

5 x plane are no affeced. We can guess soluions of he form Ae γ coskx + l and v = Be γ 2 coslx + k. u = K h u xx + u γ = K h k 2 l 2 γ = K h k 2 + l 2 v = K h v xx + v γ 2 = K h l 2 k 2 γ 2 = K h l 2 + k 2 We have γ = γ 2 = K h k 2 + l 2 = γ, such ha Ae K hk 2 +l 2 coskx + l v = Be K hk 2 +l 2 coslx + k Using he iniial condiions, we can find A and B such ha = Acoskx + l = Ucoskx + l A = U v = = Bcoslx + k = Ucoslx + k B = U Therefore he velociies as funcion of ime are given b Ue K hk 2 +l 2 coskx + l v = Ue K hk 2 +l 2 coslx + k d The parameers k and l are he wavenumber. The wavenumber is proporional o he number of peaks per uni disance, and hen inversel proporional o he wavelengh he wavelengh is basicall i is he disance beween 2 cress or 2 roughs of he wave in a given direcion. The corresponding wavelengh for k and l will be λ k = 2π k λ l = 2π l For his specific problem, we have λ k = km and λ l = 4km. For he x-componen of he veloci u, λ k is he wavelengh in he x-direcion while λ l is he wavelengh in he -direcion. For he -componen of he veloci v, he siuaion is opposie, λ k is he wavelengh in he -direcion and λ l is he wavelengh in he x-direcion. e Since we found ha he soluion decas exponeniall, where γ = K h k 2 + l 2, he deca ime scale is given b τ = = γ K h k 2 + l 2 In his case, he deca of he veloci depends on he fricion coefficien K h and on he scale of he problem given b k and l. Increasing he fricion coefficien and he wavenumber equivalen o decreasing he wavelengh resuls in increasing he deca ime scale he deca is faser. 4 5

6 4. Challenge problem: Inerial oscillaions in presence of boom fricion and pressure gradien. Consider he following equaions u f v = p x Ju ρ v + f Jv assuming ha he pressure gradien p x is consan in space and ime. For simplici, I will define C = p x ρ since i is jus a consan. Using he x-momenum equaion, we can wrie v as funcion of u, such ha v = C f + J f u + f u 5 and plugging v ino he -momenum equaion, we obain a 2nd order ODE: u + 2Ju + f 2 + J 2 6 The soluion o his equaion is he homogeneous soluion sine and cosine plus a paricular soluion u p = / and is given b +C e J sin f +C 2 e J cos f 7 The coefficiens C and C 2 depend on he iniial condiions. B using = u, we ge u e J cos f 8 +C e J sin f + Now ha we have u, we can find v using Eq. 5, I found v = C f + u e J sin f +C e J cos f 9 Using he iniial condiion for v where v = =, I found ha C = C f /, such ha our soluions are + C f e J sin f + u e J cos f v = C f + u e J sin f + C f e J cos f Recall ha C = p x /ρ, if we se C =, we recover he limi of inerial oscillaions in presence of boom fricion from quesion. The veloci is oscillaing and decaing as in quesion bu due o he presence of a consan pressure gradien in he x-direcion, he sead sae achieved b he veloci is no zero! see figure 5. As, he veloci ends o Jp x ρ v = f p x ρ 6

7 v a u v 3.5 x b Trajecories in he x plane u x x 4 v c Veloci as fc of ime u v 3.5 x d Posiion as fc of ime x x x 5 Figure 5: a u as funcion of v, b he rajecor of a fluid parcel in he x- plane, c u and v as funcion of ime, d x and as funcion of ime. To beer undersand his problem, I have also calculaed he Lagrangian rajecories and found ha [ p x J 2 f 2 x = ρ J + + u Jρ 2J f e J cos f + u ] f ρ e J sin f = p x ρ [ f + p x u Jρ p x e J sin f u f p x ρ p x + 2J f ] e J cos f Due o he consan pressure gradien, in addiion o a decaing oscillaion, he rajecories also have a linear rend. Phsicall, we can compare he consan pressure gradien o he slope of an inclined plane. If a ball sars moving on his inclined planed, due o Coriolis i will be defleced o he righ. Due o fricion, he ball will never reach back o he op of he inclined plane. I will coninue o oscillae bu he ampliude of he oscillaion will decrease wih ime due o fricion see figure 5. 7