Hello. Sorry for a probably stupid question, but how changes liquid density under pressure? My specific question is: What may be the specific gravity (weight) of 1 liter of fresh water under the pressure of 1000 B?
Thank you.

The volume of your 1 liter reduces and it's mass stays the same.

OR

Do you mean a volume of 1 liter of water at 1000 Bar has a mass of ?

http://en.wikipedia.org/wiki/Bulk_modulus

http://en.wikipedia.org/wiki/Compressibility

http://www.ajdesigner.com/phpp...us_volume_change.php

Try 1.049 liter's as your initial volume...when this is compressed by 1000 bar it becomes 1 liter...IF I have my calc correct.

One litre of water has a mass of almost exactly one kilogram...to find the density, divide the mass by the volume.


Regards Woody

This message has been edited. Last edited by: Woodygb,

Just to add what Woody wrote.

quote:

Originally posted by Nahum Goldenberg:
Just to add what Woody wrote.

The source of the Table
www.engeeniringtoolbox.com
Sorry.

I was wondering what should be the mass of apparatus they use to go down to the ocean depthes of some 10km and what might be water mass down there (to be displaced by the apparatus). The latter turned out not so frightfully big as I supposed - only some 5% more than that at the atmospheric pressure.
Thank you for interesting links you have provided.

Today I really learned smth new, thank you indeed very much.
BTW Interestingly, why scientists do not explore ocean bottom of such depthes - on the merit alone it to be the closest reacheable point to the core of the Earth.

challenge....I got curious...
How much higher would the sea level be if seawater was uncompressable...
calculate average dept 3800m covering 70.8% of earth's surface area, 510,000,000 km²
water density is 1kg/liter


I dont know the answer but I am working on it now...

quote:

Originally posted by AKKAMAAN:
How much higher would the sea level be if seawater was uncompressable...

http://data.piercecollege.edu/weather/water.html

You would need to know the topography of the ocean and use calculus. Obviously the densest water is at the bottom of the ocean so water in a shallow pool will not compress as much as water to a column of water.

I haven't bothered to check woodygb's number.
I would start with a 1 liter of sea water compressed to 1000 bar and then let it expand to 1 atm. The value can be approximated by using the equation ΔP=β*ΔV/V. The calculus solution involves natural logs.

BTW, ΔP=β*ΔV/V is one of the key equations I use when modeling systems.

So back to finding the level of the ocean if the water is incompressible. First I would do a more controlled calculation. How much does a column of water compress when it is 1000ft or 2000ft or 3000ft long? You don't need the volume only the height. Divide the ΔV and V by area you get ΔP=β*Δh/h. Now you must integrated that over x and y, the surface of the ocean but you must also know h(x,y) at each point. Oooh fun but I don't have h(x,y) at each point and I don't know who does. On top of that the water columns are smaller at the bottom as they get closer to the center of the earth.

So Akkamaan, you should be content just go compute use this to compute how much water compresses as a function of depth. ΔP=β*Δh/h You need to use calculus as h gets bigger because the rates of change change as a function of depth.

quote:

Originally posted by Peter Nachtwey:

quote:

Originally posted by AKKAMAAN:
How much higher would the sea level be if seawater was uncompressable...

http://data.piercecollege.edu/weather/water.html

You would need to know the topography of the ocean and use calculus.

Well thats why I tried to simplfy a little and set a flat bottom with a dept of 3800m....

quote:

Originally posted by AKKAMAAN:
Well thats why I tried to simplfy a little and set a flat bottom with a dept of 3800m....

In that case you only need h=3800m and you don't care about the surface area of the ocean. The problem can be solved but you still need to use calculus because the rate of change in pressure is changing as you go deeper.

Akkamaan, how to you solve these kinds of problems. I use Mathcad and wxMaxima. Woodygb uses a spread sheet. He can break the ocean depths in to 3800 rows assuming each row is 1 meter and do the calculation iteratively one meter at a time. If more accuracy is required he can use 38000 rows assuming each row is 0.1 meters. Eventually woodgb runs out of rows but his answers will be good enough given the information we have is not perfect either.

quote:

Originally posted by Peter Nachtwey:

quote:

Originally posted by AKKAMAAN:
Well thats why I tried to simplfy a little and set a flat bottom with a dept of 3800m....

In that case you only need h=3800m and you don't care about the surface area of the ocean. The problem can be solved but you still need to use calculus because the rate of change in pressure is changing as you go deeper.

Akkamaan, how to you solve these kinds of problems. I use Mathcad and wxMaxima. Woodygb uses a spread sheet. He can break the ocean depths in to 3800 rows assuming each row is 1 meter and do the calculation iteratively one meter at a time. If more accuracy is required he can use 38000 rows assuming each row is 0.1 meters. Eventually woodgb runs out of rows but his answers will be good enough given the information we have is not perfect either.

Finally I got some time at my Excel sheet.....how does this sound, a 3.086 meter raise of sealevel, if water did not compress.....I used the 38000 row method.....

Does the answer vary much from assuming what the average pressure is at 1900m and then applying
ΔP=β*ΔV/V?

quote:

Originally posted by Peter Nachtwey:
Does the answer vary much from assuming what the average pressure is at 1900m and then applying
ΔP=β*ΔV/V?

nope it's basically the same number :-(
I dont know, but that dont feel right?

quote:

Originally posted by AKKAMAAN:
I dont know, but that dont feel right?

That means you probably did the calculation correctly. The iterative method should provide a slightly bigger number. You must remember that water doesn't compress much.

From the web site I posted a link to.

quote:

This compressibility of water is so slight we could never actually see it with our own unaided eyes. We might think water is not compressible. However, if that were the case then the oceans would be about 30 meters higher than they are now, and therefore cover an extra 5 million square kilometers of Earth!

This 30 meters seems extreme. Your 3.8m solution sounds better. I haven't done the calculations but the pressure at 1/2 the depth provides a quick and dirty answer. Temperature will have a bigger effect on the difference.

quote:

Originally posted by Peter Nachtwey:

quote:

Originally posted by AKKAMAAN:
I dont know, but that dont feel right?

That means you probably did the calculation correctly. The iterative method should provide a slightly bigger number. You must remember that water doesn't compress much.

From the web site I posted a link to.

quote:

This compressibility of water is so slight we could never actually see it with our own unaided eyes. We might think water is not compressible. However, if that were the case then the oceans would be about 30 meters higher than they are now, and therefore cover an extra 5 million square kilometers of Earth!

This 30 meters seems extreme. Your 3.8m solution sounds better. I haven't done the calculations but the pressure at 1/2 the depth provides a quick and dirty answer. Temperature will have a bigger effect on the difference.

I didnt feel 100% well with my calculation so I have recalc'ed a few times and I found a decimal error....my answer is 30.8 meters.....

no concideration taken about that both density and bulkmodulus increase with pressure, but they might change proportionally....

What answer did you Woody get?