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Thread: pressure of fresh water......?

  1. #1 pressure of fresh water......? 
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    I am trying to do some math on something I have made.

    It is a coolant system using water configured in a circuit. I am trying to figure out how much pressure is created at the point of the thing I am trying to cool.

    This question has 2 parts - how would i best calculate the pressure excerted by the water against the object (there is 20cm of height)?

    Does the mass make a difference? I have been using pipes that grow larger in diameter as they go higher - should this not affect the pressure or the rate of flow in anyway?

    Much obliged.


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    I think we need a whole lot more information on your design. Is it open to the atmosphere or is it a closed, pressurized system? The size of the pipe would not affect the static head, but could affect the flow.
    The static pressure exerted by a column of fresh water is 0.43 psi per foot. In metric, it's 9804 pascals per meter of head.


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    Quote Originally Posted by Harold14370
    I think we need a whole lot more information on your design. Is it open to the atmosphere or is it a closed, pressurized system? The size of the pipe would not affect the static head, but could affect the flow.
    The static pressure exerted by a column of fresh water is 0.43 psi per foot. In metric, it's 9804 pascals per meter of head.
    thanks for the reply.

    It is a closed system that will pressurize.

    I have calculated the static head pressure & wasn't pleased - at 0.43psi per foot, 20 cm (sorry but I am in UK) means 0.1591 psi. So at room temperature it would seem the system excerts a total of 14.85 psi on the object I am cooling.

    With thermal expansion - if water expands at a rate of 0.00021 per deg.C at 20 deg.C then if the system reches 35 deg.C it would expand by 0.00315 - is this correct?

    If so my total static pressure at that point is roughly 14.9 psi.

    How would I best calculate the flow rate?
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    Is this a thermosiphon? You need to have a heat sink as well as a source. If you have both you can make the water circulate. Simply pressurizing the whole system won't produce any flow. A density difference between the warm side and the cool side will produce flow.
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    Quote Originally Posted by Bunbury
    Is this a thermosiphon? You need to have a heat sink as well as a source. If you have both you can make the water circulate. Simply pressurizing the whole system won't produce any flow. A density difference between the warm side and the cool side will produce flow.
    yep got both of those - yes this is a thermosiphon.

    I need to find a way of calculating flow.....................?

    Anyone able to help?
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    Quote Originally Posted by fatman57
    ...
    I have calculated the static head pressure & wasn't pleased - at 0.43psi per foot, 20 cm (sorry but I am in UK) means 0.1591 psi. So at room temperature it would seem the system excerts a total of 14.85 psi on the object I am cooling...
    I'm missing something here - a foot* is ~30cm (0.43 psi) -
    how did you arrive at 20cm (~8 inches) >>> 0.1591 psi**?

    *meaning 12 inches, not the thing attached to the leg ...

    ** which is what I'd estimate for ~12cm (~5 inches)...
    Nature abhors perfection; cats abhor a vacuum.

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    Quote Originally Posted by Cran
    Quote Originally Posted by fatman57
    ...
    I have calculated the static head pressure & wasn't pleased - at 0.43psi per foot, 20 cm (sorry but I am in UK) means 0.1591 psi. So at room temperature it would seem the system excerts a total of 14.85 psi on the object I am cooling...
    I'm missing something here - a foot* is ~30cm (0.43 psi) -
    how did you arrive at 20cm (~8 inches) >>> 0.1591 psi**?

    *meaning 12 inches, not the thing attached to the leg ...

    ** which is what I'd estimate for ~12cm (~5 inches)...
    yep - google says 1 foot = 30.48 cm.
    20/30(100)=66.667 (20 cm as a percentage of 30)
    0.43-66.667%=0.1433319 (which is lower then my result last night - i have it all written down so I will check what i did)

    I think 0.1591 is in the right ball park though.

    I still need to figure out the flow. I'm thinking of using the relative densities between the heated and cool parts of the fluid, the difference should allow me to calculate the pressure excerted by the more dense fluid on the less dense one.
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    I design thermosiphons but they are on a much larger scale and the density difference is much greater (two-phase mixture in the hot side and liquid in the cold side. You won't get much circulation with just a few cm of height and a few degrees of temperature difference. However the principle is you calculate head difference due to density by the difference in height times density (cross sectional area doesn't matter for this part). Then you do an iterative calculation to find the flow rate where flow resistance balances pressure difference due to static head.

    You can find online calculators like this one:
    http://www.engineeringtoolbox.com/ha...ter-d_797.html

    You need to also find the equivalent length of all bends and fittings in your circuit and add these to the straight pipe lengths. The standard methods all assume turbulent flow but at low velocities you will probably have laminar flow and the usual formulas will be inaccurate. I recommend you start with the link and check out a few other similar ones till you find one you like.
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    Quote Originally Posted by fatman57
    yep - google says 1 foot = 30.48 cm.
    20/30(100)=66.667 (20 cm as a percentage of 30)
    0.43-66.667%=0.1433319 (which is lower then my result last night - i have it all written down so I will check what i did)

    I think 0.1591 is in the right ball park though.
    there's the bit I'm missing;
    why are you subtracting 2/3 of the total from the total?
    1 - 2/3 = 1/3 ...
    instead of multiplying - 1x0.66 ...

    because you want the 2/3 ... ie, ~ 0.2867 psi (for 20cm) ... don't you?
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    Quote Originally Posted by Bunbury
    I design thermosiphons but they are on a much larger scale and the density difference is much greater (two-phase mixture in the hot side and liquid in the cold side. You won't get much circulation with just a few cm of height and a few degrees of temperature difference. However the principle is you calculate head difference due to density by the difference in height times density (cross sectional area doesn't matter for this part). Then you do an iterative calculation to find the flow rate where flow resistance balances pressure difference due to static head.

    You can find online calculators like this one:
    http://www.engineeringtoolbox.com/ha...ter-d_797.html

    You need to also find the equivalent length of all bends and fittings in your circuit and add these to the straight pipe lengths. The standard methods all assume turbulent flow but at low velocities you will probably have laminar flow and the usual formulas will be inaccurate. I recommend you start with the link and check out a few other similar ones till you find one you like.
    wow thanks - sounds like you are the man!

    What I have found so far is since my system is so small I have very small margins - like you say the usual formulas will be inaccurate. I calculated my Reynolds number as 2 yesterday so I concluded I should have laminar flow! lol

    I am expecting my flow resistance to be low as well due to the small size of my system - I would like to be able to get some estimation as to what flow I can expect before turning the thing on. I will try and do some math tonight and post it for scrutiny.
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    Quote Originally Posted by Cran
    Quote Originally Posted by fatman57
    yep - google says 1 foot = 30.48 cm.
    20/30(100)=66.667 (20 cm as a percentage of 30)
    0.43-66.667%=0.1433319 (which is lower then my result last night - i have it all written down so I will check what i did)

    I think 0.1591 is in the right ball park though.
    there's the bit I'm missing;
    why are you subtracting 2/3 of the total from the total?
    1 - 2/3 = 1/3 ...
    instead of multiplying - 1x0.66 ...

    because you want the 2/3 ... ie, ~ 0.2867 psi (for 20cm) ... don't you?
    oops - yes you are right - call it a senior moment or something............

    awesome - 14.9867 psi static pressure then!

    EDIT...................

    right, done some math. It doesn't look good!

    First using the equation given by Bunbury I first calculated the difference in head due to density change - illustrated here as H x D (height x density):

    @24 deg.C - HxD = 20 x 0.997 = 19.94 cm
    @32.2 deg.C - HxD = 20 x 0.995 = 19.9 cm

    I then used these answers to recalculate the static pressure:

    @24 deg.C - 19.94 = 66.466% of 0.43psi = 0.2858 psi
    @32.2 deg.C - 19.9 = 66.333% of 0.43psi = 0.2852 psi

    Even though I havent yet done the flow rate it currently looks like I only have a pressure decrease due to heat of:

    @24 deg.C: this is my basic room temperature.
    @32.2 deg.C: 14.9867-14.9852=0.0015 psi

    So @32.2 deg.C (coolant temperature at object being cooled with a room temperature to be 24.4 deg.C) ) there will be a pressure difference at the object being cooled of 0.0015 psi - does this seem correct to anyone?

    Sorry about all the numbers - i tried to keep it simple!
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  13. #12  
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    You need more height, or more temperature difference, or preferably both. Don't have time to look at your numbers as I'm going to the Nuggets game.
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    thanks - don't worry i'm simply checking I haven't made any obvious mistakes and that the figures are what they are.

    Hope the game is good!
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    a further question................

    what happens if I increase the atmospheric pressure on the water?

    how would I best calculate what effect this has on the pressure excerted by the water on the object at the lowest point?
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    The pressure of just the water above, added to the pressure of just the air above.
    Wise men speak because they have something to say; Fools, because they have to say something.
    -Plato

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    Quote Originally Posted by Arcane_Mathematician
    The pressure of just the water above, added to the pressure of just the air above.
    thats what i was thinking but was getting confused...........

    @ 1 atm water pressure increases by 0.43 psi per 30.48 cm.

    With this in mind do I increase 0.43 by the same percentage as 1 atm is increased?
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    no. it's simply addition of pressures and thats it iirc
    Wise men speak because they have something to say; Fools, because they have to say something.
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    Quote Originally Posted by Arcane_Mathematician
    no. it's simply addition of pressures and thats it iirc
    if thats the case then the pressure difference created by a change in density due to heat remains the same no matter what atmospheric pressure there is?

    is this because water is considered an incompressible medium (unless put in the centre of a star for example)?
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  20. #19  
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    Are the questions here getting dummer or am I getting more cynical?
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    Quote Originally Posted by fizzlooney
    Are the questions here getting dummer or am I getting more cynical?
    there are many dumb questions asked in this forum, I simply ask for advice........dumb as it may be.............
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    Quote Originally Posted by fatman57
    Quote Originally Posted by Arcane_Mathematician
    no. it's simply addition of pressures and thats it iirc
    if thats the case then the pressure difference created by a change in density due to heat remains the same no matter what atmospheric pressure there is?

    is this because water is considered an incompressible medium (unless put in the centre of a star for example)?
    forgot about that little bit... No, water is compressible

    The overall pressure is really a measure of the amount of mass above the point you are considering for pressure. Highly pressurized gas means there will be more matter per unit volume, and consequently more mass and therefore more force pressing down upon the point you are considering.
    Wise men speak because they have something to say; Fools, because they have to say something.
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  23. #22  
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    Quote Originally Posted by Arcane_Mathematician
    Quote Originally Posted by fatman57
    Quote Originally Posted by Arcane_Mathematician
    no. it's simply addition of pressures and thats it iirc
    if thats the case then the pressure difference created by a change in density due to heat remains the same no matter what atmospheric pressure there is?

    is this because water is considered an incompressible medium (unless put in the centre of a star for example)?
    forgot about that little bit... No, water is compressible

    The overall pressure is really a measure of the amount of mass above the point you are considering for pressure. Highly pressurized gas means there will be more matter per unit volume, and consequently more mass and therefore more force pressing down upon the point you are considering.
    exactly................the problem I have is now to calculate it..........

    every 30.48cm of water = 0.43 psi increase of pressure.
    1 atm = 14.7 psi
    14.7 + 0.43 = 15.13 psi.

    if I increase the atm to 16 psi then the pressure excerted by the column of water will be greater then 0.43............................how do I calculate how this value will change relative to the atm?

    this is why I said I would calculate the percentage change of 14.7 to 16 psi then increase the value of the pressure excerted by the column of water (0.43 psi per 30.48cm) by the same ratio..............
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