<p>Solutions (Hints and Clues) for the practice 2</p><p>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</p><p>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.</p><p>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</p><p> du 1 2. Remind: acceleration a  , velocity u  u  at , distance s  ut  at 2 dt 0 2</p><p>(a) at depth z=-10m</p><p>PA  1 ghA  1 g(10m)</p><p>PB  1 ghB  1 g(10  0.05m) du 1 p p P  P   where  B A dt  x x 1000m</p><p> 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</p><p>(b) Longer. At depth z >-10m, the baroclinic pressure gradient is from left to right, the opposite direction of the barotrophic one.</p><p>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.)</p><p>PE=kinetic energy 1  (   )Vgh   Vu 2 2 1 2 2 Where V is the volume, h=20-4=16m</p><p> u=0.8m/s</p><p> s  ut  t=25s</p><p>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</p><p>(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.</p><p>5. The motion is a inertial flow. u  fv t v   fu t</p><p>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</p><p>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.</p><p>6. From the color figure, we can subtract the following figure</p><p>2N 8N</p><p>75m</p><p>125m</p><p>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.)</p><p>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.</p>