8Currents without Friction:Geostrop

8
Currents without Friction:
Geostrophic Flow
IN THIS chapter we will discuss some of the characteristics of motion which we
can deduce from the equations of motion when it is assumed that the F terms
in equations (6.1) or (6.2) (i.e. friction, gravitation of the moon and sun, etc.) are
zero and that there is a steady state, that is the velocities at any point do not
change with time (i.e. du/dt = dv/dt = êw/dt = 0). Except for the example of
inertial oscillations we shall also assume that the advective acceleration terms
may be neglected. These approximations for the large-scale mean circulation in
the ocean's interior were justified in the previous chapter.
8.1 Hydrostatic equilibrium
As a preliminary to discussing moving fluids, let us first look at stationary
ones. Let us assume that (1) u = v = w = 0, i.e. that the fluid is stationary,
(2) dV/di = 0, i.e. the fluid remains stationary, and (3) all the F terms are zero.
Then, from equations (6.2) we are left with only
«^ = 0, 4 = 0, 4=-g. (8.1)
ox oy oz
The first two mean that the isobaric (constant pressure) surfaces are
horizontal, i.e. there is no pressure term, in fact no force at all in this case, to
cause horizontal motion. The third can be written as
dp = -pgdz, (8.2)
which is the hydrostatic (pressure) equation in differential form, i.e. it gives the
pressure dp due to a thin layer dz of fluid of density p. If p is constant
(independent of depth) it becomes p = — pgz. This is not a very exciting
result—really all that it does is confirm that the equations of motion do give a
previously known answer (as shown from first principles in Appendix 1) when
the fluid is stationary. As we showed in Section 7.3, this equation remains an
excellent approximation even for flows at typical ocean speeds.
63
64 INTRODUCTORY DYNAMICAL OCEANOGRAPHY
The reason for the minus sign on the right is because we take the origin of
coordinates at the sea surface with z positive upward. Measurements up into
the atmosphere are given as, for example, "the masthead is at + 10 m", while
measurements down into the sea are given as, for example, "the fish is at a depth
of 50 m, i.e. at z = - 50 m". The pressure at this depth is (taking
p = 1025kgm"3)
p50 = - (1025 x 9.8 x - 50) = + 5.02 x 105 Pa = + 502 kPa.
An increase in depth of 1 m yields an increase in pressure of about 10 kPa.
8.2 Inertia! motion
We first assume that (1) dp/dx = dp/dy = 0 (i.e. there is no slope of the sea
surface and all the pressure surfaces inside the fluid are also horizontal; we shall
look at the situation when these terms are not zero presently), (2) that we can
ignore the F terms as stated above, and (3) that w = 0 (i.e. that there is only
horizontal motion). Then the x- and y-equations of motion become
—du = 2Ù sin dt; ö v and — = - 2Ù sin ö u. (8.3)
at at I
The equations (8.3) have solutions
u = VH sin (2Ù sin ö ß),
v= FHcos(2Qsini), (8.4)
where V = u2 + v2 and t here stands for time. Now these are the
equations of motion for a body in the northern hemisphere travelling
clockwise in a horizontal circle at constant linear speed VH and angular speed
2Ù sin ö. If the radius of the circle is B9 then V%/B = 2Ù sin ö VH, i.e. the
centripetal acceleration VjB is provided by the Coriolis acceleration '2Óßýçö
VH (Fig. 8.1). Physically, such motion might be generated when a wind blows
steadily in one direction for a time, causing the water to acquire a speed VH, and
then the wind stops and the motion continues without friction (to a first
approximation) as a consequence of its "inertia" (properly its momentum),
hence the term "inertia! motion". Flow variations of inertial period are often
present in records from current meters. The amplitudes vary depending on the
strength of generating mechanisms and they decay due to friction when the
generation stops.
Note that equations (8.3) are non-linear but do have solutions, equations
(8.4), so non-linear equations can sometimes be solved explicitly. Note also,
however, that if we regard the equations as Lagrangian equations for a fluid
parcel, they are linear and the terms which would be non-linear in Lagrangian
terms (friction) have been assumed to be small, making solution easy.

For a speed VH = 0.1 m s - 1 at latitude ö = 45°, then B ~ 1 km. For VH
= 1 ms"1, then B ~ 10 km. The period of revolution is
2ð _ 2ð _ 11 sidereal day _ Tf
angular speed 2Ù sin ö 2 sin ö 2
because Ù = 2ð/1 sidereal day. The quantity Tf = (1 sidereal day/sin ö) is
called "one pendulum day" because it is the time required for the plane of
vibration of a Foucault pendulum to rotate through 2ð radians. The value of
0.5 77 (one-half pendulum day) is 11.97 h at the pole, 16.93 h at 45° latitude and
infinity at the equator.
The direction of rotation in the inertial circle is clockwise viewed from above
in the northern hemisphere and anticlockwise in the southern hemisphere. If
one thinks of observing the motion in the southern hemisphere by looking
down through the earth from the northern hemisphere then the motion also
appears clockwise. However, the observer in the southern hemisphere is
upside-down relative to the observer in the northern hemisphere and he calls
the motion anticlockwise. Likewise, he says that the Coriolis force acts to the
left of the velocity in the southern hemisphere. It is a matter of point of view. In
the terms used by meteorologists, the motion is anticyclonic in both
hemispheres. The term cyclonic comes from cyclone, a storm with low pressure
at its centre about which the winds are anticlockwise in the northern
hemisphere and clockwise in the southern hemisphere. An anticyclonic system
has high pressure at its centre and winds circulate in the opposite direction. The
reason for this behaviour will become clear when we discuss geostrophic flow.
Equivalent terms contra solem and cum sole were occasionally used by
oceanographers in the older literature meaning, respectively, against and with
the direction of motion of the sun as seen by an observer facing the equator.
These terms are related in Fig. 8.2.

8.3 Geopotential
In preparation for the discussion of the geostrophic method for calculating
currents we must introduce the concept of geopotential. The quantity àw
= M gdz is the amount of work done ( = potential energy gained) in raising a
mass M through a vertical distance dz against the force of gravity (ignoring
friction). We then define a quantity called geopotential (Ö) such that the change
of geopotential dO over the vertical distance dz is given by
ÌÜÖ = dw = Mgdz (joules),
or dO = gdz (joules kg"1 = m2s"2) (potential energy change/unit mass)
= — ccdp (from equation (8.2)).
Integrating from æã to z2 we have
Ç2 Ç2 Ã2
Now writing a = a35 0 p +

Hasil (Bahasa Indonesia) 1:

8Arus tanpa gesekan:Aliran geostrophicDALAM bab ini kita akan membahas beberapa karakteristik dari gerak yang kitadapat simpulkan dari persamaan gerak ketika diasumsikan bahwa kata Fdalam persamaan (6.1) atau (6.2) (yaitu gesekan, gravitasi bulan dan matahari, dll) yangnol dan bahwa ada kondisi mapan, yang adalah kecepatan pada setiap titik tidakberubah dengan waktu (yaitu du/dt = dv dt = êw/dt = 0). Kecuali untuk contohosilasi inertia kita akan juga beranggapan bahwa persyaratan advective percepatandapat diabaikan. Pendekatan ini untuk sirkulasi berarti skala besar dilaut interior dibenarkan dalam bab sebelumnya.8.1 hidrostatik kesetimbanganSebagai pendahuluan untuk membahas bergerak cairan, mari kita lihat di stasionerorang-orang. Mari kita asumsikan bahwa (1) u = v = w = 0, yaitu bahwa cairan stasioner,(2) dV/di = 0, yaitu sisa cairan stasioner, dan (3) semua istilah-istilah F adalah nol.Kemudian, dari persamaan (6.2) kami yang tersisa hanya«^ = 0, 4 = 0, 4 =-g. (8.1)sapi oy ozDua yang pertama berarti bahwa permukaan isobaric (tekanan konstan)horisontal, yaitu ada tidak ada istilah tekanan, pada kenyataannya tidak ada kekuatan sama sekali dalam hal ini, kemenyebabkan gerakan horisontal. Ketiga dapat ditulis sebagaiDP = - pgdz, (8.2)yang adalah persamaan (tekanan hidrostatik) dalam bentuk diferensial, yaitu memberikantekanan dp karena lapisan tipis dz cairan kepadatan p. Jika p konstan(independen dari kedalaman) menjadi p =-pgz. Ini sangat tidak menyenangkanresult—really all that it does is confirm that the equations of motion do give apreviously known answer (as shown from first principles in Appendix 1) whenthe fluid is stationary. As we showed in Section 7.3, this equation remains anexcellent approximation even for flows at typical ocean speeds.6364 INTRODUCTORY DYNAMICAL OCEANOGRAPHYThe reason for the minus sign on the right is because we take the origin ofcoordinates at the sea surface with z positive upward. Measurements up intothe atmosphere are given as, for example, "the masthead is at + 10 m", whilemeasurements down into the sea are given as, for example, "the fish is at a depthof 50 m, i.e. at z = - 50 m". The pressure at this depth is (takingp = 1025kgm"3)p50 = - (1025 x 9.8 x - 50) = + 5.02 x 105 Pa = + 502 kPa.An increase in depth of 1 m yields an increase in pressure of about 10 kPa.8.2 Inertia! motionWe first assume that (1) dp/dx = dp/dy = 0 (i.e. there is no slope of the seasurface and all the pressure surfaces inside the fluid are also horizontal; we shalllook at the situation when these terms are not zero presently), (2) that we canignore the F terms as stated above, and (3) that w = 0 (i.e. that there is onlyhorizontal motion). Then the x- and y-equations of motion become—du = 2Ù sin dt; ö v and — = - 2Ù sin ö u. (8.3)at at IThe equations (8.3) have solutionsu = VH sin (2Ù sin ö ß),v= FHcos(2Qsini), (8.4)mana V = u2 + v2 dan t di sini berdiri untuk waktu. Sekarang ini adalahpersamaan gerak untuk tubuh di belahan bumi utara perjalanansearah jarum jam dalam lingkaran horisontal kecepatan konstan linier VH dan kecepatan sudutÖ dosa 2Ù. Jika jari-jari lingkaran B9 maka V%/B = 2Ù dosa Umlaut VH, yaitupercepatan sentripetal VjB disediakan oleh percepatan Coriolis ' 2ÓßýçöVH (Fig. 8.1). Secara fisik, seperti mungkin dihasilkan ketika angin bertiupterus di satu arah untuk waktu, menyebabkan air untuk mendapatkan kecepatan VH, danlalu angin itu berhenti dan gerak terus tanpa gesekan (untuk pertamaperkiraan) sebagai akibat dari yang "inersia" (benar momentum),maka istilah "inersia! gerak". Variasi aliran inertia periode yang seringhadir dalam catatan dari sekarang meter. Amplitudo bervariasi tergantung padakekuatan menghasilkan mekanisme dan mereka membusuk karena gesekan ketikagenerasi berhenti.Perhatikan bahwa persamaan (8.3) non-linear tetapi memiliki solusi, persamaan(8.4), sehingga non-linear persamaan kadang-kadang dapat diselesaikan secara eksplisit. Perhatikan juga,Namun, bahwa jika kita menganggap persamaan sebagai persamaan Lagrangian untuk cairanparsel, mereka linear dan syarat-syarat yang akan menjadi non-linear di Lagrangianpersyaratan (friction) telah diasumsikan menjadi kecil, membuat solusi yang mudah. Untuk kecepatan VH = 0,1 m s - 1 di latitude Umlaut = 45°, maka B ~ 1 km. Untuk VH= 1 ms "1, maka B ~ 10 km. Periode revolusi2ð _ 2ð _ 11 hari sideris _ Tfkecepatan sudut 2Ù dosa Umlaut 2 dosa Umlaut 2karena Ù = 2ð/1 sideris hari. Kuantitas Tf = (1 hari sideris dosa Umlaut) adalahdisebut "satu pendulum hari" karena itu adalah waktu yang diperlukan untuk bidanggetaran pendulum Foucault untuk memutar melalui 2ð radian. Nilai0,5 77 (setengah pendulum hari) adalah 11.97 h di kutub, 16.93 h 45° lintang danInfinity di khatulistiwa.Arah rotasi dalam lingkaran inertia searah jarum jam dilihat dari atasdi belahan bumi utara dan separa di belahan bumi selatan. Jikaorang berpikir tentang mengamati gerak di belahan selatan dengan melihatturun melalui bumi dari belahan bumi utara kemudian gerak jugamuncul searah jarum jam. Namun, adalah pengamat di belahan selatanterbalik dibandingkan pengamat di belahan bumi utara dan dia memanggilgerakan separa. Demikian juga, ia mengatakan bahwa Angkatan Coriolis bertindak untukkiri kecepatan di belahan bumi selatan. Ini adalah masalah sudut pandang. Dalamistilah yang digunakan oleh Meteorologi, gerak anticyclonic di keduabelahan. Istilah cyclonic berasal dari topan, badai dengan tekanan rendahdi pusatnya yang angin separa di Utarabelahan bumi dan searah jarum jam di belahan bumi selatan. Sistem anticyclonicmemiliki tekanan tinggi di pusat dan angin beredar dalam arah yang berlawanan. Thealasan untuk perilaku ini akan menjadi jelas ketika kita membahas aliran geostrophic.Equivalent terms contra solem and cum sole were occasionally used byoceanographers in the older literature meaning, respectively, against and withthe direction of motion of the sun as seen by an observer facing the equator.These terms are related in Fig. 8.2. 8.3 GeopotentialIn preparation for the discussion of the geostrophic method for calculatingcurrents we must introduce the concept of geopotential. The quantity àw= M gdz is the amount of work done ( = potential energy gained) in raising amass M through a vertical distance dz against the force of gravity (ignoringfriction). We then define a quantity called geopotential (Ö) such that the changeof geopotential dO over the vertical distance dz is given byÌÜÖ = dw = Mgdz (joules),or dO = gdz (joules kg"1 = m2s"2) (potential energy change/unit mass)= — ccdp (from equation (8.2)).Integrating from æã to z2 we haveÇ2 Ç2 Ã2Now writing a = a35 0 p + <5 from Section 2.23 we getÖ2 - *i = 0 (z2 - *i) = - aas.o.p dp -= -AThe quantity (Ö2—Ö÷) is called the geopotential distance between the levelsz2 and Zj where the pressures will be p2 and pv The first quantity on the right of> getCURRENTS WITHOUT FRICTION 67equation (8.5) is called the ''standard geopotential distance" (AOstd, a functionof p only) while the second is called the "geopotential anomaly" (ÄÖ, a functionof S, Tand p). In size, the second term is of the order of one-thousandth of thefirst.The reader is reminded that although (Ö2—Ö^ is called the geopotential"distance" in océanographie jargon, it really has the units of energy per unitmass (Jkg"1 or m2s"2) and for g = 9.8ms"2 and <5z=lm, then dO= 9.8 Jkg"1. For numerical convenience, oceanographers in the past haveused a unit of geopotential called the "dynamic metre" such that 1 dyn m= 10.0 Jkg"1. To indicate that this unit is being used, it is usual to use thesymbol D for geopotential. The geopotential distance (D2 — Dl) is thennumerically almost equal to (z2 — zx) in metres, e.g. relative to the sea surface:SI units Mixed unitsat a geometrical depth in the sea = +100 m + 100 mthen z2 = -100 m -100 mthe pressure will be about p = + 1005 kPa + 100.5 dbarand the geopotential distancerelative to the surface Ö2 - Ö1 = - 980 J kg" D2 - ûã = - 98 dyn m.It is because of its use in the calculation of geopotential "distance" that tables ofa as a function of 5, T and p are more common than tables of p.8.31 Geopotential surfaces and isobaric surfacesA surface to which the force of gravity, i.e. the plumb line, is everywhereperpendicular is called a geopotential surface because the value of thegeopotential must be the same everywhere on the surface. The term "levelsurface" is taken to mean the same thing. An example of such a surface is thesmooth surface of a lake in which there are no currents and where there are nogelombang, atau dari bilyar meja set up dengan benar. Alasan untuk menentukan "tidakarus"akan dijelaskan dalam bagian berikutnya.Permukaan isobaric adalah salah satu di mana tekanan di mana-mana adalah sama. DalamDanau stasioner di atas permukaan air akan isobaric permukaan p = 0(tekanan atmosfer yang diasumsikan konstan dan diabaikan). Isobaric permukaanuntuk tekanan yang lebih tinggi akan lebih dalam di danau dan akan geopotential(tingkat) permukaan selama Danau masih.Isobaric permukaan harus tingkat dalam keadaan stasioner. (Jangan membingungkan ini"stasioner negara" ketika Anda = v = w = 0 dengan "kesetimbangan" Kapan Anda, v dan wmungkin bukan nol tetapi tidak berubah dengan waktu, yaitu ketika du dt = dv dt = dw dt= 0, lihat Apendeks 1.) Misalnya, untuk saat ini bahwa permukaan isobaric (berlaribaris dalam Fig. 8.3(a)) cenderung permukaan (garis penuh di 8.3(a) gambar). The tekanan kekuatan pada partikel air A unit massa akan adp dn seperti yang ditunjukkan.(d/dn berarti gradien sepanjang normal, yaitu tegak lurus, ke permukaan dandi pesawat kertas.) Selain itu, gravitasi bertindak pada partikel. Ini adalahsituasi tidak stabil karena kedua kekuatan tidak keseimbangan, karena mereka tidakpersis menentang, tetapi harus memiliki resultant ke kiri. Situasi ditampilkan dalamlebih detail untuk partikel B whe