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When was any one liter of Long Island Sound water last in the Atlantic Ocean? The Sound for this model is:

Area: 1,268 mi²

Mean depth: 63 feet

Length 110 miles

Eastern opening into the Atlantic is 9 miles wide

An average tide of 6 feet, twice a day.

As a sailor and swimmer in the Sound, I picture it as a relatively closed polluted body refreshed by the Atlantic. But I started wondering just how often the water near me, 50 miles from Plum Gut, turns over. Tidal currents are 1 knot, or about a mile an hour. In other words, each day the water at my location flows west for 6 hours as the tide rises, then back east for 6 miles as it drains, at the rate of one mile an hour.

The most simple model in my mind pictures a “plug” of Atlantic Ocean water traveling 6 miles west into the Sound before draining out - but of course, there’s turbulence and mixing – but how much mixing? And how much mixing at any one place lengthwise along the body? My hope is that someone here commands a better understanding of how to model this.

xray0
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Byterbit
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  • There are hopelessly too many variables to be able to determine when any one liter of water in the Long Island Sound (where I live but two miles from) to answer this question. – Bob D Jul 21 '19 at 22:20
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    [earthscience.se] would likely be a much better home for this question. – Emilio Pisanty Jul 21 '19 at 23:26

2 Answers2

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How frequently the contents of Puget Sound turns over is a little different from how long it takes a liter of water to get from the Atlantic Ocean into the Sound. Every last molecule of an original liter never exits the sound.

I calculated once that if a teaspoon of water were dumped into the ocean and the ocean were stirred thoroughly, any teaspoon of water then extracted from the ocean would have several molecules of that original teaspoon of water (which tells us by the way that we have water in our tissues that has passed through practically every fish, tree, and dinosaur that has ever existed).

So: turnover. Considering that fresh water flows continuously into the Sound via several rivers, and that the salinity of the Sound (2.9%) is not very low compared to that of the Pacific Ocean (3.4%) (see this link), the mixing rate must be substantial. A reasonable estimate of turnover rate could be based on the salinity ratio and the ratio of the annual inflow volume of fresh water into the Sound, to the volume of the Sound. Modeling turnover rate using historical records of ocean currents, tides, climate, etc., coupled with models of turbulence, would be possible but probably would not give a much more reliable value.

S. McGrew
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Continuing on McGrew's answer, we can estimate the residence time of water as $$\tau=\frac{V}{F}=\frac{\text{water volume } (m^3)}{\text{water inflow rate }(m^3/s)}$$ Were the water well mixed this means that the amount of water remaining from a starting volume declines as $\exp(-t/\tau)$.

So if the daily tide moves $F=2\text{[area][tide height]/[length of a day]}$ and the overall volume is $V=\text{[area]}\text{[average depth]}$, so the residence time becomes $\tau = 2\text{[average depth]}\text{[length of a day]}/\text{[tide height]} = 21 $ days.

(One can model the turnover in more sophisticated ways, including returning water and inflows.)

The problem is the well-mixed assumption. I do not think this is easily quantified without an elaborate hydrodynamic model: some estuary shapes no doubt keep some water "trapped" in bays more efficiently than others, not to mention temperature and salt layering in the depth.

Anders Sandberg
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