Ocean 40 Physical Processes in the Ocean Project 1: Hydrostatic Balance, Advection and Diffusion Answers 1. Hydrostatic Balance a) Set all of the levels on one of the coluns to the lowest possible density. Click the Calculate Pressure button. What is the pressure at the botto? 501.7815 db b) Set all of the densities on one of the coluns to the highest possible density and find their pressures. What is the pressure at the botto? 504.4 db A coon approxiation in oceanography is that one decibar of pressure is about equal to one eter of water. In this case, how any eters off would you be and in which direction? Using this approxiation, we think that we are 504.4 eters deep when we are only 500 eters deep. This eans we think we are about 4 eters deeper than we actually are. c) Experient with the densities and find two different stable profiles that have the sae botto pressure. Record the densities in all levels for both of the. There were any correct answers one is given below. The point of the exercise was to realize that knowing the pressure at a given depth tells you nothing about the stratification. 10 kg/ 105 kg/ 104 kg/ 105 kg/ 105 kg/ 105 kg/ 106 kg/ 105 kg/ 107 kg/ 105 kg/ d) Given the following two profiles, calculate the pressure gradient between the (at each level) if they are 100 kiloeters apart. The pressures fro the given profile are: 100.56 db 100.405 db 00.7617 db 00.9579 db 01.16 db 01.7065 db 401.8176 db 40.504 db 50.566 db 50.511 db 1 p p p The horizontal pressure gradient is given by =. To calculate this based on our x x data, we subtract the right pressure fro the left pressure and divide by 100,000 eters. To put this in standard units, convert db into Pa. Layer 1: 100.405db 0.56db 7 db 4 = 4.9 Pa db = 0.0049 Pa N or
00.9079db 00.7617db Layer : = 1.96 = 0.0196 or 01.7065db 01.16db Layer : = 4.905 = 0.04905 or 40.504db 401.8176db Layer 4: = 6.867 = 0.06867 or 50.511db 50.566db Layer 5: = 7.848 = 0.07848 or If you left the pressure gradients in db/k, the answers are 10 ties saller. Note: if you calculated the pressure based on density, the hydrostatic balance ust be taken into account. That is, to calculate the pressure correctly, you ust add up the different densities ultiplied by the depth that they occur. If you have ore questions, coe by and ask e. e) Click on the Perturbation radio button. This shows the densities as a perturbation fro the standard value of 105 kg/. Input the two profiles in d) by subtracting 105 kg/ fro each of the values. Now again calculate the pressure gradient at each level assuing 100 kiloeter separation. How are these different fro the answers to d)? Layer 1: Layer : Layer : Layer 4: Layer 5: 100.405db 0.56db 7 = 4.9 = 0.0049 or 00.9079db 00.7617db = 1.96 = 0.0196 or 01.7065db 01.16db = 4.905 = 0.04905 or 40.504db 401.8176db = 6.867 = 0.06867 or 50.511db 50.566db = 7.848 = 0.07848 or These are no different than those in part d). If you found differences, it was likely due either to rounding or slight offsets when inputting the densities (i.e. 10.51 instead of 10.5). The perturbation does not change the density; it is just a different way of expressing it. When using the hydrostatic balance to copute the pressure, you use the total density which is a su of the reference density and the perturbation. The total density is = 0 +, with 0 being the reference density of 105 kg and the density perturbation.
f) For the pressure gradients that you calculated in d), calculate the size of the current at each level that would result after one day of the pressure gradient being applied. The ost coon istake in the proble was with the units. Using db/k does not give the current in /s without converting! The equation we use, in siplified for is u 1 p =. 0 x 1 p Solving for u, u =. We know the current is in /s, so let s check that the units 0 x work out right. 1 N kg s = s =. kg kg s s. Advection In this exercise we will exaine the effects of advection. Go to the Advection-Diffusion exercises, and choose Siple Advection. a) Initially set the velocity to be unifor across the channel with speed.0. Describe how the oveent of the particles evolves. Does the distance between particles change over tie? The particles ove uniforly across the page at a constant rate of.0. No, the distance between particles does not change. b) The initial set up has a velocity that is linearly sheared across the channel. Exaine the particle oveent over tie. How does the distance between particles change over tie? Does the change in distance depend on location? The particles stay in a straight line, but this line slopes ore and ore fro the vertical. The distance between particles increases over tie, but this does not depend on location, which can be seen in the equal spacing of the final line.
c) Now try a velocity profile with a large velocity in the iddle of the channel, and saller velocities on the edges. What happens to the line of particles over tie? The particles go fro being in a straight line to a v-shape on its side. This v gets steeper and steeper as tie goes on. The distance between particles increases over tie, but whether it depends on location depends on the exact profile.. Diffusion Now choose the Siple Diffusion exercise. In this case, we have as an initial condition a unifor teperature fluid. At t=0, the teperature at the surface iediately rises. Over tie, the heat diffuses into the interior. a) Do five siulations of the teperature field using values of diffusivity fro 1 through 5. Record the tie when the teperature at depth 6 reaches 1 unit. Graph this as a function of the diffusion coefficient. What is the relationship between tie and the diffusivity? Diffusivity ( /s) 1 4 5 Tie (s) 45 1 15 11 7 The relationship is roughly exponential decay.
b) Copare the traces for diffusivities 1 and 5. When does the 5-diffusivity run look like the tie 50 trace of the 1-diffusivity run? 10 seconds. Use diensional analysis to explain why this is so. (Reeber that the units of the diffusivity are length squared per second). L L 1 =, = Diensions of. T T 1 L = 1T1 = T Solving for L for both and setting the equal since the length is the sae for both (both started and ended in the sae trace). 1 T = T1 Solve for the tie it will take for. 1 10s = s 50s Plug in nubers to see that it works. 5 s 4. Advection and Diffusion Now choose the -D Diffusion-Advection exercise. In this case, you have three things you can vary: 1) Where the initial patch of tracer is located (in the fast Gulf Strea or in the slow interior flow), ) whether the diffusivity is sall or large and ) whether you want a siulation with advection only, diffusion only, or advection with diffusion. a) Exaine the siulations with advection only. For which siulation (fast or slow flow) does the tracer spread out ore? The tracer spreads out ore with faster flow. b) Which velocity profile fro part does the tracer ost reseble for the fast flow? The profile fro c) ost resebles the fast flow otion. c) Now try with diffusion only in the Gulf Strea and interior regions. Try both large and sall values of diffusivity. Does the tracer spread out ore or less as copared to the siulations in part (a)? The tracer spreads out ore, that is, it covers ore area as copared to part a). d) Iagine that an oil spill has occurred in the Gulf Strea and you are consulted to deterine whether the spill will reach Newfoundland. You perfor siulations with both large and sall diffusivities. Under what conditions will the oil spill reach Newfoundland? With advection and high diffusivity the oil spill will reach Newfoundland. If you thought the question was asking only about diffusion, the spill never reaches the coast of Newfoundland. e) In the slow current region, do five siulations, one with advection only, two with diffusion only for both values of diffusion, and two with advection and diffusion with both values of diffusivity. Record the tie that the tracer first crosses 0 N. For which siulation does this happen fastest and why? Run A D sall D large A & D sall A & D large Tie (days) 60 -- 60 180 40
This happens fastest for the run with advection and the large diffusion coefficient. Since the current is oving towards 0N, advection pushes the water towards its goal. The large diffusion coefficient spreads the tracer out ore, again helping speed the water to 0N. These two together cobine to ake the tracer reach 0N first. 5. Getting quantitative with advection and conservation a) We have easured the oxygen concentration in the water on a float that oves with the fluid at 100 depth. We observe oxygen concentration starts at a values of 4 l/liter and decreases by over the course of 10 days to l/liter and that the float oves a distance of 50 k. If we assue that there has been no discernable ixing over the 10 days, what would be the oxygen respiration rate? Since we are oving with the fluid, the only change in concentration is due to respiration. To calculate the respiration rate (l/l/day) we use C, with C the concentration and t the tie. Thus, we have C C C1 l 4l = = = 0.l / day t t1 10days This represents a sink of oxygen. Respiration rate is defined as the positive value, so it is 0.l/l/day. Both were accepted as correct. b) Next we easure oxygen at two fixed oorings that have sensors at 100 depth. The oxygen levels at the beginning of the experient are 7 l/liter at 40N and 140W and l/liter at 9N and 140W. We also easure a southward current of 5 c/s. After 10 days, we easure an oxygen concentration of l/l at the southern ooring. Estiate the respiration rate (l/l/day) for the southern ooring. We want to coplete this proble in two steps. First, figure out what the concentration will be after 10 days due only to advection. Then, any difference between that and the given value of l/l ust be due to respiration. First, the advection part. We go about this like the exaple in class. C C tie space = v, with C the concentration and v the current. y We want to know the final concentration, so we solve forc final, Cspace C final = Cinitial v. y We plug in our values for these, aking sure the units agree: v = 0.05 / s, y = 111, 000, and t = 864, 000s. 7l l C final = l + 0.05 / s 864,000s = 4.56l 111,000 C C C1 l 4.56l As above, = = = 0.56l / day t t 10days 1 Again, respiration rate is the positive of this. If you used received full credit. Ctie instead of C final, you still