SIO 210 Due November 21, 2016 - University of California...
Transcript of SIO 210 Due November 21, 2016 - University of California...
SIO 210 Problem Set 3 Answerkey November 9, 2016 Due November 21, 2016 (20 points)
1. Use the wind curl map here and Sverdrup balance to predict the transport of the Kuroshio at 30°N. Assume that the wind curl is uniform and equals the maximum value of curl at 30°N. Look at the map of Sverdrup transport from lecture or the textbook to compare with your answer.
To assist with answer, draw a line along 30°N. You will be doing a crude integration along this line, from east to west. For illustration, I’ve put this map in powerpoint and drawn the line, put it back into the problem set. The expression for Sverdrup transport is:
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βV (x) = f curl(τ /ρf )dxx
xeast∫ For full theory, would carry through the derivatives of the f term within the curl, but here just set it to a constant and cancel out with f outside the integral.
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βV (x) ~ curl(τ /ρ)dxx
xeast∫
V (x) ~ 1βρ
curl(τ)dxx
xeast∫
V (x) ~ 1βρ
ave(curl(τ))Δx
where I’ve used “ave” to mean the average value, and have also used a constant density. β = 2Ω cos(latitude)/Re where
Ω = Earth angular rotation, and 2Ω = 1.454 x10-4 sec-1 Re is Earth’s radius = 6371 km = 6.371x106 m At 30°N, cos(latitude) = 0.87
Therefore β = 1.98 x 10-11 sec-1m-1
Estimating the other parts roughly: Curl (τ) from map is about
Curl (τ) ~ -0.8 x 10-7 N/m2 = -0.8 x 10-7 (kg m/ sec2)/m2 = -0.8 x 10-7 kg m-1 sec-2 ρ = 1025 kg/m3
Δx: approximately 120 ° longitude. At 30°N, 1° longitude = 111km *cos(30°) = 111 km * 0.86 = 96 km = 96 x103 m Δx ~ 120 (96 x103 m)
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V =1
1.98x10−11 sec−1m−11
1025kgm−3 (−0.8x10−7kgsec−1m−1)(120°)(96x103m /°)
V = 45.6x106m3/sec = 45.6 Sv This number should be rounded given all the approximations. If I had used a curl(tau) of -1.0x10-7 N/m3 instead, then the value would be more like 56 Sv. So the ballpark estimate is around 50 Sv, which is similar to the Sverdrup transport in the map from lecture based on this wind stress curl.
(16 points) 2. The California Current flows southward along the west coast of the U.S. For this question, consider it as a separate phenomenon from the subtropical gyre. The California Current is driven by a southward alongshore wind. The wind causes an offshore Ekman transport. (It is also driven by upwelling due to offshore strengthening of the southward wind, but that is not part of this problem. Assume here that the wind is uniform in latitude and longitude.) Assume that the total offshore Ekman transport is 1 Sv. (1 Sv = 1x106 m3/sec.)
(a) Explain how the California Current itself arises from this forcing. (short
answer) The alongshore wind stress drives offshore Ekman transport, as stated. This creates coastal upwelling. This (plus the broad upwelling driven by the wind stress curl arising from strongest winds offshore) causes the thermocline to rise towards the coast. It also piles water up offshore. The pile of water offshore creates an eastward pfg which drives a southward geostrophic flow which is the California Current. The tilted pycnocline causes this current to decay with depth. (b) Along the coast, there is upwelling in a strip that is about 10 km wide. Assume that it occurs over a 1000 km length of the coast. Make a reasonable assumption for the thickness of the upwelling layer based on what is causing the upwelling. ______100 to no more than 200_________m Assuming that the Ekman transport of 1 Sv occurs out of this box, calculate the offshore Ekman velocity (using the transport and dimensions of the box) The cross-‐sectional area for the transport is Areavert = 1000 km x 100 m = 1x106 m x 1x102 m = 1x108 m2 The average velocity is the Ekman transport divided by area. uave = 1 Sv / 1x108 m2 = (1 x 106 m3/sec)/( 1x108 m2) = = 0.01 m/sec = 1 cm/sec (c) Calculate the upwelling velocity into the box (using the transport and dimensions of the box). This requires the surface area of the box which is Areasurface = 1000 km x 10 km = 104km2 = 1010m2 W = 1 Sv / 1x1010 m2 = (1 x 106 m3/sec)/( 1x1010 m2) = 1x10-‐4 m/sec = 10-‐2 cm/sec (25 points) 3. Equatorial undercurrent speed. A plot of the surface height from west to east along the equator in the Pacific, shown in class, is shown here. (The axis is labeled ‘dynamic height’ but assume that it is actual physical height.) Ignore the inset diagram.
a. Compute the pressure gradient force (“pgf”) across the whole width of the Pacific. You’ve done a similar calculation on other homework. From graph, the height difference is Δz ~ 45 cm = 0.45 m. From graph, the distance is 50° longitude. At the equator, 1° = 111 km, so the distance is Δx = 50 x 111 km = 5550 km = 5.55 x 106 m. The pgf is (-1/ρ)(Δp/Δx). Δp comes from hydrostatic balance: 0 = -(Δp/Δz) –ρg, so
Δp = –ρgΔz and therefore
pgf = (-1/ρ)(Δp/Δx) = gΔz/Δx = (9.8 m sec-2) (0.45 m)/( 5.55 x 106 m) = 7.9x10-7 m sec-2
b. Assume the ocean response to the pgf is acceleration. What direction will the pressure gradient force accelerate the flow (east, west, north, south)? To the east Why? Because at the equator, the response to the pgf is direct, since there is no Coriolis force.
c. Calculate the eastward acceleration. It is equal to the pgf = 7.9x10-7 m sec-2 d. On the chart, SSH is nearly uniform between the western side and 160°W. If the
flow starts accelerating here, how fast will it be going when it reaches the boundary? Assume that there is just one value for the pgf. (Authors have attached two slopes to the data.) This is an approximate answer as well. We know the acceleration from (c). We do not know the time. With some manipulation of the expressions for velocity and acceleration, we can approximately eliminate the time and find the approximate velocity. Assume acceleration a is constant.
To get velocity, integrate acceleration from time 0 to t at the eastern side.
Assume initial velocity u(0) = 0.
u(t) – u(0) = u(t) = aΔt so a = u(t)/Δt Because accelearation is constant, u changes uniformly, and the velocity at the eastern side u(t) is twice the average velocity uave.
uave = Δx/Δt and uave = u(t)/2 so Δt = Δx/ uave = 2Δx/ u(t) Substitute this Δt into the expression for acceleration: a = u(t)/Δt = u2(t) /(2Δx) and u(t) = sqrt(2a Δx) Substitute in values: a = du/dt = Δu/Δt = 7.9 x10-7 m/sec2
Δx = 5.55 x 106m So u(t) = sqrt(2a Δx) = sqrt(2 * 7.9 x10-7 m/sec2 * 5.55 x 106m) = 2.96 m/sec
This is actually not too different from the actual maximum EUC speed. From the graph, you can see that the maximum acceleration actually ends around 140°W, which would give about half this velocity at that point, hence about 1.5 m/sec. The actual EUC has reduced acceleration in the eastern half (see graph)
e. Explain why you don’t see this speed current at the sea surface along the equator.
Because the trade winds blow the surface water westward (viscous) and this effect is stronger than the eastward flow. This does seem pretty improbable, because it means the steady winds should be creating a westward flow of at least 100 cm/sec. Should calculate the direct effect of the wind.
f. Equatorial upwelling processes under equatorial trade winds: briefly explain two
ways that water upwells along the equator, one for which Earth rotation is not important, and one for which rotation is important. Draw a diagram.
The wind on the equator blows the surface water directly westward, creating upwelling in the east and downwelling in the west. This does not involve rotation. The winds just off the equator create off-equatorial Ekman transport, which creates upwelling all along the equator. This does involve rotation (Coriolis).
g. From (f), explain why there is a cold tongue and warm pool along the equator, and why the cold tongue extends very far westward along the equator, not just at the eastern boundary.
The non-rotating part of the answer causes the thermocline to be shallower in the east than in the west. The rotating part creates upwelling all along the equator. This along-equatorial upwelling draws up colder water in the east (‘cold tongue’) than in the west (in the thick warm pool).
(24 points):
4. Very long swells generated by a distant storm are observed in the open ocean, which is assumed to be very deep (about 5000 m). Their period is T seconds.
a. What is the formula for (deep water) surface wave speed c in terms of wavelength
L? (This is the phase speed.)
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c =gk
=gL2π
!answer (can’t make equations
red)
b. Rewrite this: what is the wavelength L in terms of wave period T? Answer here, can’t highlight inserted equations in red:
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c =ωk
=2πT
L2π
=LT
c 2 =gk
=gL2π
=L2
T 2
L =gT 2
2π
c. Evaluate the wavelength L if the period T = 20 sec and g = 9.8 m/s2.
L = 624m
d. Calculate the speed of these waves. c = L/T = gT/2π = 31 m/sec As these waves pass by, water parcels at the surface carry out circular orbits (forward at the crest, then downward, backward at the trough, then upward). e. If the wave height (from crest to trough) of these waves is H, what is the diameter D of the orbit? It is the same as the height: D = H f. What is the expression for the average speed s with which water particles traverse the orbit? (remember the formula for circumference of a circle) Distance traveled by particle is the circle of the orbit (plus a tiny bit for forward motion of wave, don’t include). Circumference of the circle is πD = πH. Speed = s = distance/time = πH/T where T is the wave period g. Numerically evaluate the average speed s for waves of period T=20 s and height H=10 m T = L/c = 624m/(31m sec-1) = 20.1 sec s = πH/T = π(10m)/20.1sec = 1.56 m/sec h. Compare your answers for (d) and (g).
The speed of actual particles in the waves, going around their orbit, is much smaller than the speed of the wave through the water. s = 1.56 m/sec while c = 31 m/sec (15 points) 5. Short answer questions about tides. For each
(i) mark the most nearly correct answer (1 point) (ii) write a brief explanation of your answer. (2 points) (iii) draw a small diagram to illustrate your answer. (2 points)
a. Spring tides (times of large semidiurnal tidal range) occur twice a month
A. when the moon is in the earth's equatorial plane, B. when the moon is out of the earth's equatorial plane, C. at full or new moon, D. at the quarter moons, E. at lunar perigee.
During full and new moon, the moon and sun are aligned, which increase the gravitational attraction and therefore increases the size of the tide. I have inserted the diagram from the SIO210 CSP lecture that illustrates the spring tide, when moon and sun are aligned; the other spring tide is when the moon is opposite the sun because the tidal bulges are equal and opposite for both the moon and sun attraction.
b. The daily inequality (elevation difference between a high tide and its immediate successor) vanishes for lunar tides
A. when the moon is in the earth's equatorial plane, B. when the moon is out of the earth's equatorial plane, C. at full or new moon, D. at the quarter moons, E. at lunar perigee.
The daily inequality arises from the moon being out of the equatorial plan, so that when a given location rotates under the moon, say, one high tide bulge is closer to the plane of the moons’ orbit than the other. I have inserted the diagram from the SIO210 CSP lecture that illustrates the daily inequality that occurs when the moon is NOT in the equatorial plane.
c. Neap tides
A. occur near the times of an eclipse of the sun or of the moon, B. never occur near the times of an eclipse, C. ONLY occur at the times of an eclipse.
Neap tides are the lowest high tides. They occur when the sun and moon are perpendicularly aligned. During an eclipse, they are completely aligned. Therefore neap tides can never occur during an eclipse. Drawing is from lecture and DPO 6th chapter 8, showing alignment in neap tide.