A Lesson in Physical Chemistry

bedouin said:
I hadn't thought of this before - but don't the lock gates at Chichester marina have a curved profile on the marina side so they open without displacing any water - hence won't have this problem?
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It's a widely used principle and presumably works well, though there will still be some added load if there is a gradient across the gate. I don't think that they had thought of this when Heybridge was built. Anyway, flat gates are prettier and more fun to operate.
 
johnalison said:
It's a widely used principle and presumably works well, though there will still be some added load if there is a gradient across the gate. I don't think that they had thought of this when Heybridge was built. Anyway, flat gates are prettier and more fun to operate.
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I know there are places that use curved gates even where there isn't a salinity gradient - allows you to open the gates earlier to supplement the sluces.
 
Interestingly the magnitude of this effect will depend on the vertical position of the sluice. The pressure acting on each side of the gate varies from 0 at the water surface, increasing linearly to a maximum at the base. The rate of increase depends on the water density, which as we know is greater for salt water than fresh. This pressure gradient can be thought of as a right angle triangle, with its apex at the water surface, and the slope of the hypotenuse depending on the density. The total force on each side of the gate is the area of the triangle for that side.

The reason that sluice position is relevant is that the pressure must be equal at that position, otherwise water would flow through the sluice.

If the sluice is very close to the water surface (unlikely in a real lock) then the level would be equal on both sides of the gate. Because salt water is denser its pressure gets greater faster as we go down the gate, so the triangle on that side is larger. This results in a force acting to push the gate towards the fresh water side (which would open the gates at Heybridge).

If on the other hand the sluice is at the bottom of the gate (a more likely scenario) then the effect will be reversed. The pressure at the base must be equal, so the water level on the salty side will be lower. The triangles will have equal base length but the salt one, being shorter, will have a smaller area. Thus there is a force acting to push the gate towards the salt water, holding it closed. When the gate is opened slightly the levels will attempt to equalise, resulting in rapid flow into the lock.

Of course at some sluice depth between the two extremes the forces will balance, though there will still be flow as the gates are opened. Fresh water will flow into the lock at the top, and salt water will flow out at the bottom.

The resultant force depends on difference in density, height of water (squared) and width of the gate. For a 4m deep lock, gate width 3.75m this could give a force of 10 kN ~ 1 Tonne which I guess most of us would notice.
 
johnalison said:
That's a great explanation. It's good to know that there is at least one intelligent person in East Anglia.
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The p = - g ρ z of alandalus11’s cryptic set of formulae at #18 did give the gist of it: the calculation of the pressure at depth z. The σt bit just refers to the way oceanographers subtract 1000 from the density ρ in kg/m3 to give easy-to-handle two-digit density numbers - but the Total Dissolved Solids (TDS) bit was confusingly irrelevant.

But I suspect you may perhaps have surmised that - and of course he may not be in East Anglia. :ambivalence:
 
Hydrozoan said:
The p = - g ρ z of alandalus11’s cryptic set of formulae at #18 did give the gist of it: the calculation of the pressure at depth z. The σt bit just refers to the way oceanographers subtract 1000 from the density ρ in kg/m3 to give easy-to-handle two-digit density numbers - but the Total Dissolved Solids (TDS) bit was confusingly irrelevant.

But I suspect you may perhaps have surmised that - and of course he may not be in East Anglia. :ambivalence:
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Oh Gawd - that makes three of you.
 
If it helps I don't live in East Anglia any more....

I would be interested to know whether there is actually a visible difference in the height of water on each side of the gate, and if so how do you know when the lock is full?
 
johnalison said:
Oh Gawd - that makes three of you.
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How potentially appropriate to the season – but I’m not a man from the east either.

TLouth7 said:
... I would be interested to know whether there is actually a visible difference in the height of water on each side of the gate, and if so how do you know when the lock is full?
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I broadly agree with your force difference (I get about 0.75 tonnes for a likely coastal salinity) so I assume that my calculations are reasonable. They also say that equal pressure at half gate height would be obtained with just 2mm less depth on the seaward side. :) Do yours concur?
 
I used a seawater density of 1.025 kg/m3 (for 32ppt salinity and 5 degrees C) and obtained a total pressure (including atmospheric) at 2m depth of 121436 Newton/m2 for SW and 120945 Newton/m2 for FW. That is a pressure difference of 491 Newton/m2 - equalized by a drop of ca. 2mm on the SW side.

I would not be surprised if I am wrong - but I did check and though 2mm doesn't sound much, I reckon it's only a 0.4% difference of pressure (and force) at 2m. Anyway, 2mm or 50mm - it's a test of the Major’s distance vision! ;)
 
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Hydrozoan said:
I used a seawater density of 1.025 kg/m3 (for 32ppt salinity and 5 degrees C) and obtained a total pressure (including atmospheric) at 2m depth of 121436 Newton/m2 for SW and 120945 Newton/m2 for FW. That is a pressure difference of 491 Newton/m2 or about 0.4% - equalized by a drop of ca. 2mm on the SW side.

Anyway, 2mm or 50mm - it's a test of the Major’s distance vision! ;)
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I am indebted to you all. It's clear that considerable effort has gone in to expressing why the gate was so bloody heavy and caused me to repair to the pub for several hours. I return to the boat on Saturday and will endeavour to supply the exact measurements for you all. My distance vision is fine, but I'm currently in the adjoining county making a mess in the kitchen.
 
Hydrozoan said:
I used a seawater density of 1.025 kg/m3 (for 32ppt salinity and 5 degrees C) and obtained a total pressure (including atmospheric) at 2m depth of 121436 Newton/m2 for SW and 120945 Newton/m2 for FW. That is a pressure difference of 491 Newton/m2 - equalized by a drop of ca. 2mm on the SW side.

I would not be surprised if I am wrong - but I did check and though 2mm doesn't sound much, I reckon it's only a 0.4% difference of pressure (and force) at 2m. Anyway, 2mm or 50mm - it's a test of the Major’s distance vision! ;)
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Did you remember to account for the difference in atmospheric pressure on either side caused by the height difference :)
 
I agree on the 490 N/m^2, but I think this would be balanced by a drop of 50mm. 50kg/m^2 / 1025kg/m^3 = 49 mm

The difference in density is 2.5%, so difference in surface height (from the equilibrium height) will be in the same ratio. 2000mm * 0.025 = 50mm
 
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