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Thread: Pressure Drop

  1. #1 Pressure Drop 
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    This isn't homework. I'm trying to find a formula (or collection of them) that will help answer the following problem.

    A tube is in a bowl of water much like a straw in a glass. The pressure above the water in the tube is the same as the pressure above the water in the bowl (atmospheric). My tube has a piston in it though and as I draw the piston upwards in the tube it begins to lower the pressure in the tube above the water. As it does so, the atmospheric pressure above the water in the bowl forces water up the tube in an attempt to equalize the pressure difference the piston attempted to create. The column of water in the tube rises and comes to rest a distance above the surface of the water in the bowl. At this point, the weight mg of the column of water is held in place by the pressure on the water surface around the tube in the bowl. But that same weight prevents the water in the tube from rising high enough to return the pressure difference between inside and outside of the tube to zero. That is, there would still be a slightly lower pressure in the tube above the water column than outside of the tube.

    I'm thinking that the pressure in the tube is P - mg where P is atmospheric pressure, but this can't be the correct form as the units aren't the same.

    Any help would be appreciated.


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  3. #2 Re: Pressure Drop 
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    Quote Originally Posted by Ledger
    This isn't homework. I'm trying to find a formula (or collection of them) that will help answer the following problem.

    A tube is in a bowl of water much like a straw in a glass. The pressure above the water in the tube is the same as the pressure above the water in the bowl (atmospheric). My tube has a piston in it though and as I draw the piston upwards in the tube it begins to lower the pressure in the tube above the water. As it does so, the atmospheric pressure above the water in the bowl forces water up the tube in an attempt to equalize the pressure difference the piston attempted to create. The column of water in the tube rises and comes to rest a distance above the surface of the water in the bowl. At this point, the weight mg of the column of water is held in place by the pressure on the water surface around the tube in the bowl. But that same weight prevents the water in the tube from rising high enough to return the pressure difference between inside and outside of the tube to zero. That is, there would still be a slightly lower pressure in the tube above the water column than outside of the tube.

    I'm thinking that the pressure in the tube is P - mg where P is atmospheric pressure, but this can't be the correct form as the units aren't the same.

    Any help would be appreciated.
    The height of the water in the tube, at full draw of the piston, will be such that the pressure at the level in the bowl will equal atmospheric pressure.

    mg is weight, not pressure. Pressure would be where is mass density and is column height.


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  4. #3  
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    Thanks for the response. Yes, mg is the weight of the column of water in the tube, not pressure. And I probably wasn't too clear but there was air above the water in the tube before the piston moved up. Like say the tube is in the water and there is an inch or so between the water level and the piston.

    And the water level in the tube and the bowl are the same and the air pressure is atmosphere above the water in both tube and bowl. Then the piston moves up and inch or two in the tube. Water would rise in the tube and come to rest. The water rose because the piston created a lower pressure than atmospheric in the tube and the water was pushed up the tube in an attempt to equalize the pressure. But due to the weight of the water, the pressure of the air in the volume between the piston and the water column is less than atmospheric. That's the pressure I would like to calculate.
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  5. #4  
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    The Doc has given you the correct formula. The air in the tube above the water column won't make much difference. Its density is much less than the water.
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  6. #5  
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    Quote Originally Posted by Harold14370
    The Doc has given you the correct formula. The air in the tube above the water column won't make much difference. Its density is much less than the water.
    But the air is a gas, and so exerts a force on the fluid column. The pressure at the base of the column will be where is the pressure of the air column, which will be determined by the gas law. The ideal gas law should be sufficient. will depend on imitial piston position, how far the piston is moved and the temperature, but in the limit of infinite piston retraction will be 0. Temperature will be time dependent but at equilibrium will be ambient temperature.

    Edit: corrected typo in pressure formula
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  7. #6  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by Harold14370
    The Doc has given you the correct formula. The air in the tube above the water column won't make much difference. Its density is much less than the water.
    But the air is a gas, and so exerts a force on the fluid column. The pressure at the base of the column will be where is the pressure of the air column, which will be determined by the gas law. The ideal gas law should be sufficient. will depend on imitial piston position, how far the piston is moved and the temperature, but in the limit of infinite piston retraction will be 0. Temperature will be time dependent but at equilibrium will be ambient temperature.
    This is all true, and you would need to know that if you don't know the height of the water column after the tube has been withdrawn and need to calculate it. If you can measure the height of the water column then you can determine the pressure at the top of the column. It is equal to the pressure at the bottom, i.e., atmospheric, minus rho*g*h. The pressure at the top of the water column will be a little more than at the piston, due to the weight of the air which is probably negligible.
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  8. #7  
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by DrRocket
    Quote Originally Posted by Harold14370
    The Doc has given you the correct formula. The air in the tube above the water column won't make much difference. Its density is much less than the water.
    But the air is a gas, and so exerts a force on the fluid column. The pressure at the base of the column will be where is the pressure of the air column, which will be determined by the gas law. The ideal gas law should be sufficient. will depend on imitial piston position, how far the piston is moved and the temperature, but in the limit of infinite piston retraction will be 0. Temperature will be time dependent but at equilibrium will be ambient temperature.
    This is all true, and you would need to know that if you don't know the height of the water column after the tube has been withdrawn and need to calculate it. If you can measure the height of the water column then you can determine the pressure at the top of the column. It is equal to the pressure at the bottom, i.e., atmospheric, minus rho*g*h. The pressure at the top of the water column will be a little more than at the piston, due to the weight of the air which is probably negligible.
    Right. The air pressure at the top of the water column is (which follows from what I said earlier now that I fixed the typo). The effect of gravity on the presssure gradient in the air column will be negligible unless the air column is gigantic.
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  9. #8  
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    Thanks for the help. I've worked this a bunch of different ways but I don't think I'm getting what I'm after.

    If I stick a pipe in water the most I can 'suck up' is about 34 feet. That's all the atmospheric pressure will support. If my piston continues to rise, the air chamber between the top of the water column and the piston just gets bigger. It's the pressure in this air chamber I'm interested in calculating.

    With the formula provided if I plug in 34 feet for h I would think I would get a number close to atmospheric pressure such that when I subtract my number the difference would be close to zero. But what I calculate gives me a number that's really small compared to atm (100kN/m^2).

    Any thoughts?
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  10. #9  
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    If you have 34 feet of water (or whatever level of water column corresponds to the atmospheric pressure) then you would have to have a perfect vacuum. Any air pressure above the water will result in a column of water less than 34 feet. How much air pressure you have above the water will depend on temperature, etc. and you could get some water vapor in the air bubble as well.
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  11. #10  
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    Ok, so let's say there's 20 feet of water in the pipe and a five foot air chamber above the water column and below the piston. In that five-foot high cylindrical air chamber above the water column there's some air pressure that's less than atmospheric.

    Would I use the formula given above to calculate the pressure in this air chamber?
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  12. #11  
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    Quote Originally Posted by Ledger
    Ok, so let's say there's 20 feet of water in the pipe and a five foot air chamber above the water column and below the piston. In that five-foot high cylindrical air chamber above the water column there's some air pressure that's less than atmospheric.

    Would I use the formula given above to calculate the pressure in this air chamber?
    You can use that formula, since you know the height of the water column and atmospheeric pressure. Or you can use the gas law if you know the temperature and volume of the air chamber.
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  13. #12  
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    So here's what I see as a problem. I compute rho*g*h to be

    1.2kg/M^3 * 6.1m * 9.8m/s^2 = 71.74, say 72 N/m^2. Now I need to subtract this from atmospheric pressure which is 101,325 N/m^2. Well, taking 72 from that number is like almost taking nothing away from it.

    So it suggests that after 20 feet of water has been pushed up, the atmospheric pressure above the water is almost the same as before any water was pushed up. If I use 33 feet of water instead of 20, it doesn't change the outcome much, but at 33 feet we've got to be almost pulling a vacuum above the water.

    Where am I going wrong?
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  14. #13  
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    Quote Originally Posted by Ledger
    So here's what I see as a problem. I compute rho*g*h to be

    1.2kg/M^3 * 6.1m * 9.8m/s^2 = 71.74, say 72 N/m^2. Now I need to subtract this from atmospheric pressure which is 101,325 N/m^2. Well, taking 72 from that number is like almost taking nothing away from it.

    So it suggests that after 20 feet of water has been pushed up, the atmospheric pressure above the water is almost the same as before any water was pushed up. If I use 33 feet of water instead of 20, it doesn't change the outcome much, but at 33 feet we've got to be almost pulling a vacuum above the water.

    Where am I going wrong?
    The density of water is 1 gram/cc or 1 kg/liter or 1000 kg/m^3
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  15. #14  
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    OK, but I want to know the air pressure in the volume above the water.
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  16. #15  
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    Quote Originally Posted by Ledger
    OK, but I want to know the air pressure in the volume above the water.
    So do the calculation with the correct density for water. Your calculation used a value that was low by three orders of magnitude.
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  17. #16  
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    Wow, it worked.

    I can see the math but I can't see the physics.

    It appears that I computed information about the water in the column and somehow that told me something about the air above it?
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  18. #17  
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    Quote Originally Posted by Ledger
    Wow, it worked.

    I can see the math but I can't see the physics.

    It appears that I computed information about the water in the column and somehow that told me something about the air above it?
    Your calculation involved the whole system -- the atmosphere, the water column, and the gas volume above the column.

    The physics is that the pressure in the air bubble, plus the pressure of the water due to column height equals atmospheric pressure. If you know 2 you can find the 3rd.
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  19. #18  
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    Amazing.

    Thank you very much.
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