Answers (Hints and Clues) for the Practice 2

Solutions (Hints and Clues) for the practice 2

u v w 1. Continuity Equation is    0 x y z In this case, along-shore is y-direction, i.e. north-south component; cross-shore is x- direction, i.e. east-west component u v w  (  ) (eq. 1) x y z

v 10cm / s w u (a)  and  0 plug into eq.1, we get  5*105 s 1 y 2km z x Integrate it along x, we get u  5*105 * x  C . We know at x=0, u=0, thus C=0. i.e. u  5*105 * x . Then we can get the velocity at x=1km.

v 10cm / s w 0  0.1cm / s (b)  and   2*104 s 1 plug into eq.1, we get y 2km z 0  (5m) u  1.5*104 s 1 , repeat steps in (a), we can get the velocity at x=1km. x

du 1 2. Remind: acceleration a  , velocity u  u  at , distance s  ut  at 2 dt 0 2

(a) at depth z=-10m

PA  1 ghA  1 g(10m)

PB  1 ghB  1 g(10  0.05m) du 1 p p P  P   where  B A dt  x x 1000m

du Plug in the numbers, we can get a   5*104 m / s 2 , a is constant from B to A at dt z= -10m. 1 2s 2*(1000m) Then from s  at 2 , we know t   2 a  5*104 m / s 2

(b) Longer. At depth z >-10m, the baroclinic pressure gradient is from left to right, the opposite direction of the barotrophic one.

3. Key: Potential energy doesn’t change if you move a parcel with density 1 within the upper layer, the same for moving a parcel with density 2 within lower layer. When this parcel was raised from z=-30m to z=-4m, it’s density is heavier than that of surroundings, i.e. the parcel had a potential energy. So it would fall. When it reached to the depth z= -20m, there was no more density difference. The potential energy is converted to kinetic energy, and the parcel had a velocity. Thus it would continue to fall, but without acceleration. (Suppose there is no friction.)

PE=kinetic energy 1  (   )Vgh   Vu 2 2 1 2 2 Where V is the volume, h=20-4=16m

 u=0.8m/s

s  ut  t=25s

4. Momentum equation, assume x is east-west direction, y is north-south direction (a) du 1 P    fv dt  x dv 1 P    fu dt  y

(b) du  0 dt i) i.e. u, v are constant. At t=0, u is toward east, v=0. Thus the flow will keep dv  0 dt going east with a constant speed. du  0 dt ii) where i is the surface slope. Thus there is no acceleration at x- dv 1 P    gi dt  y direction, there is acceleration at y direction. At t=0, the flow is toward east. It will have a y-direction velocity toward north plus the east component. Thus the flow will be toward north-east with a increase total velocity. iii) The flow is balanced by the pressure gradient force and coriolis force. Thus the flow is geostrophic current.

5. The motion is a inertial flow. u  fv t v   fu t

Period of a initial motion at 30N is 2 2 2 2 2 T       1 f 2sin 2sin(30)  2 sidereal day 1sidereal 2r  UT , where U is velocity magnitude at each time at each location, U=u2+v2, UT U U U U U 0.25m / s r        *1sidereal  3.43km Thus, 2 2 f 2sin(30)  2 2 T 1sidereal X=2r=6.86km

If at time t, the flow is northward, then half sidereal day later the flow will be southward, but 6.86km east away the original position.

6. From the color figure, we can subtract the following figure

Approximately one latitude apart equals to 111 km. Z 125  75m The slope of 28C isotherm is   7.5*105 x 6*111km Thus the surface sea level slope is 7.5*10-7 The geostrophic current in the upper 50m is balanced by pressure gradient force and coriolis force. (suppose x is positive to the east, y is positive to the north.)

1 P 0    fu Thus, u= gi/f= 0.59m/s  y i.e. the magnitude is 0.59m/s, and direction is pointing out of the paper to the east.