1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work done by a siphon.

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the work done by air, when it pushes water out of a barrel (cylindrical), that is 1m in diameter. The starting water level was 80 cm from the bottom of the barrel and the water level at the end was 20 cm from the bottom. The water was pushed out through an inverted-U-shaped tube.

    2. Relevant equations
    [tex]A=PV[/tex]
    3. The attempt at a solution
    Since [itex]A=PV[/itex] and I know that [itex]V=π*0,5^{2}*0,6=0.15π[/itex], all I need to do is calculate the average pressure. What I find problematic is that neither the height of the barrel, nor the length of the tube is given. Am I correct in assuming that those variables affect the pressure that is needed to pump the water out of the barrel and therefore also affect the answer? Or is there some sort of a siphon effect that makes these variables unnecessary?

    I also assumed that the pressure on the outer end of the inverted-U shaped tube is air pressure. First the gas needs to push the water up the tube, work equaling the volume of the tube to the top times the pressure created by the water column half-way to the top (half-way to get average pressure) plus air pressure. Then as the water goes down the tube, I would have to subtract the height of the water column in the tube from air pressure to which I add the height of the water column half-way from the current water level to the 20cm level, all of which I multiply by the volume of water needed to be pumped out to reach 20 cm water level.

    Is my reasoning correct for the steps needed to solve the problem, and that without being given the height of the barrel or the tube, an answer could not be calculated?

    A solution to the problem was also posted, which I don't think is correct:

    Work done by the air pressure is equal to the increase of potential energy of the center of mass of the water (0,3 m below starting level).

    [tex]A=E=mgh=ρVgh[/tex]
    [tex]A=1000*0,15π*10*0,3=1410J[/tex]
     
  2. jcsd
  3. Dec 6, 2011 #2
    Other than some some slight multiplication error, I agree with the posted answer of 1410J. The work done is the change in potential energy of the moved weight of water. If you assume the lower end of the siphon is at 0.2 meters, the cg of the moved water went from 0.5 m to 0.2 m.

    Your statement: "I also assumed that the pressure on the outer end of the inverted-U shaped tube is air pressure."

    What you state is correct. If it were not true, a water hose would not squirt out a cylinder of water whose outside diameter equals the inside diameter of the exit orifice. If the pressure were above atmospheric, the water would disperse radially. It doesn't.
     
  4. Dec 6, 2011 #3
    If the exit end of the siphon tube is lower, then the differences in potential energy should also be lower. Does that mean that more work needs to be done to pump the water out? But if the exhaust end of the siphon is lowered, doesn't that mean that the suction force is stronger and less pressure is needed to get the water moving?
     
  5. Dec 6, 2011 #4

    Delphi51

    User Avatar
    Homework Helper

    I find this question really strange! Maybe you guys can straighten me out. I know you have to suck on the hose to get it started, so initially a differential air pressure is required. But once you get water into the down tube, I don't think air pressure pushes the water out. Reduced pressure in the water at the top of the inverted U pulls it up, but that pressure drop is caused by the weight of the water column outside the barrel. Energetically, isn't it the same as making a hole in the bottom of the barrel? The water falls out due to the pressure of the water caused by its weight. The same thing would happen if the barrel was filled with sand. In a vacuum.

    No work is done when something falls, is it? We would say potential energy is converted into kinetic energy or something like that.

    A Google search for "work done by a siphon" returns only this thread!
     
  6. Dec 6, 2011 #5
    "If the exit end of the siphon tube is lower, then the differences in potential energy should also be lower. Does that mean that more work needs to be done to pump the water out? But if the exhaust end of the siphon is lowered, doesn't that mean that the suction force is stronger and less pressure is needed to get the water moving?"

    If the exit end is lower then there is a larger difference in potential energy and the water will flow faster. What drives this process is the difference in potential energy between the water surface and where the water exits the siphon. That potential energy is directly proportional to the difference in altitude. The pressure at the surface of the barrel is 1 atmosphere. The pressure at the exit end of the siphon is 1 atmosphere. This ONLY is true when the water is flowing at constant speed.

    Excluding the inertia of the moving water, the flow of the siphon will cease when the surface of the water in the barrel reaches the same level as the exit end of the siphon. If you had a longer siphon and put the entrance end deeper below the surface of the water in the barrel, the flow would still cease when the level of the water surface reaches the level of the exit end of the siphon.
     
  7. Dec 6, 2011 #6
    That potential energy is directly proportional to the difference in altitude.

    I should have said: That potential energy difference is directly proportional to the difference in altitudes.
     
  8. Dec 6, 2011 #7
    The pressure drop cannot pull the water out (neglecting attraction between water molecules), the pressure drop allows the air pressure on the other side to push it out.

    I'm not sure if the pressure on the surface has to be 1 atm. If so, couldn't I just multiply the volume of water pushed out by 1 atm and get the answer? That would give me an answer of 471000 joules if I calculated correctly. (0,15pi * 10^5 Pa)

    What if I could lower the pressure on the surface of the barrel and lower the exhaust end, wouldn't that mean less work done, since the force (pressure) is lower while displacement is the same? But according to the potential energy calculation, the difference in potential energy is bigger, so isn't there a contradiction?
     
  9. Dec 7, 2011 #8

    Delphi51

    User Avatar
    Homework Helper

    Okay, I think I have resolved my difficulty. Air pressure pushes the water up to the top of the siphon. The question doesn't give the height of the inverted U tube, so the work cannot be calculated. This work is returned when the water falls down the other side, so no net work is done by air pressure.
     
  10. Dec 7, 2011 #9
    It obviously takes work to move the water to the top of the barrel. Since most people have siphoned I suspect the person who wrote the problem envisioned a garden hose stuck over the side of the barrel into the water so the height of the tube above the barrel is pretty small. Obviously it would have clarified the problem to include a diagram or some sort of better description. The total work done on or by the system during the process wasn't asked for, just the work for moving the water to the top of the barrel. So mgh for the .30m.
     
  11. Dec 8, 2011 #10
    I don't understand why the amount that the potential energy of the water decreases shows how much work was done by the gas? Gas does the work, potential energy of the water decreases, why do they have to be equal? If the gas was doing work to increase the potential energy of the water, then they would have to be at least equal or the work done has to be higher, as you cannot create energy.

    What I am thinking is that the decrease in potential energy of the water shows how much work the water itself has done and this has to be subtracted from the total work needed to be done to pump the water out.

    Anyway, ignoring the siphon effect, just pumping a 0.6 m tall water column out of a 0.6 m tall and 1 m diameter container requires this much work done by the gas:

    [tex]A=PV[/tex]
    [tex]A=(0,3*1000*10+10^5)*0,5^2*π*0,6=48537,6 J[/tex]

    It's not just 1410J, or do you think it is just 1410J in this case also? The center of mass also changes 0,3 m and the mass of the pumped out water is the same.

    I am thinking that if from 48537,6 I subtract the amount by which the potential energy of the water decreases (for which I would need the height of the other end, where the water goes out), I would get the answer. What do you think?
     
  12. Dec 8, 2011 #11
    The work done by the gas increases the potential energy of the water by the amount mgh where h is the height the center of mass of the water raised moves, it does not decrease it. The water has to be higher than its initial position to get out of the barrel. The water then falls outside the barrel and loses the potential energy but that was not the question being asked. But think if you used a small scoop to get the water out of the barrel. You'd have to lift the water to the top edge of the barrel and then dump it. But you would have to do the weight of the water (calculated using density and volume of the water) times the height you have to lift it to get the work you have to do to take it out scoop by scoop.

    The siphon saves you the trouble because you only have to lift a tiny amount of water enough, to fill the very thin tube up and over to below the level where it is on the inside and for a very small tube, this is quite negligible. Once the tube is filled and the end is lower than the initial part, air pressure keeps it going. So you see something has to do the work to increase the potential energy of the water enough to get it over the edge of the barrel before it can then be removed from the barrel. If you aren't going to use a scoop or a bucket, a siphon will do almost all of it for you but there is still work needed to get the water up to the top of barrel. There are just multiple ways to do this work.
     
  13. Dec 8, 2011 #12
    But doesn't the air also have to do work against air pressure, or more specifically, against the reduced pressure at the top of the siphon tube? Doesn't it mean that the air has to exert a pressure equal to the air pressure at the top of the tube plus the water column to the top of the tube?
     
  14. Dec 8, 2011 #13
    The tube is full of water from one end to the other I assume you realize. One end is submerged, the other is hanging down to the ground. The air pressure pushes down on the surface of the water in the barrel causing more water to flow upward as the water falls down the tube since it's falling farther than its rising so this isn't some perpetual motion type thing. It's just an energy exchange and the only part of this system you're concerned with is the water's center of gravity moving from where it is, to the top of the barrel. What happens from there doesn't matter. And of course it's happening a little at a time but you don't have to consider anything but the whole batch of water being moved by any process from where it is to the top of the barrel.
     
  15. Dec 9, 2011 #14
    Why don't I have to be concerned about anything else? I'm not sure why do I have to be concerned about the potential energy of the water. I have to calculate the work done by the gas, and do you agree that work done by a gas is pressure times volume?
     
  16. Dec 9, 2011 #15

    Delphi51

    User Avatar
    Homework Helper

    PV WOULD work if you actually knew the pressure difference between the water surface in the barrel and at the top of the siphon. (The flow is caused by the reduced pressure at the top of the siphon due to the weight of the water column outside the barrel.) Certainly not atmospheric pressure times volume - that would give the same answer regardless of how high the water was lifted. Go with mgh and guess at the height of the siphon top which wasn't given. Use the average height through which the water was lifted.

    After doing that, set that work equal to (ΔP)V and calculate the ΔP. Then calculate W = (ΔP)V just to show you know about that, too. Then have a chat with the prof or marker so they know you know some physics and this question needs improvement (the height of the top of the siphon needs to be given and some mention of the partial vacuum). Become very familiar with the principle of operation before your meeting; a nice explanation of the siphon is given here: http://www.newton.dep.anl.gov/askasci/phy00/phy00137.htm

    Incidentally, did you give us the exact wording of the question?
    It is quite surprising that a question like this has not appeared in these pages before. Or anywhere else on the web. Often a really new question needs a bit of fine tuning after first use.
     
    Last edited: Dec 9, 2011
  17. Dec 10, 2011 #16
    This isn't actually my homework, just a problem that I had to solve at a time and I think the answer which was later given to me was incorrect, for which I am seeking comments here. It wasn't originally in English, I translated it, but I included all the information which was originally given.

    Ear pressure is the pressure equal to a column of water 10 meters high. The siphons other end can only be 0,5 lower than the center of mass of the moved water, and that is how much it can reduce the pressure. This is why I think the given answer is incorrect by at least an order of magnitude.

    What I am thinking, is that by calculating E=mgh I am only calculating the work needed to push the water up in a vacuum, but there is also air pressure which I have to overcome. If I am a diver underwater, then at 10 m underwater the pressure is 2 atmospheres, but if the earth had no atmosphere, the pressure would only be 1 atmosphere when I'm 10 m underwater. Do you see what I mean?
     
  18. Dec 10, 2011 #17

    Delphi51

    User Avatar
    Homework Helper

    Yes, I'm a former diver myself!
    I don't see what you mean by "air pressure which I have to overcome", though.
    Do you just mean the pressure at the top of the siphon is greater than zero?

    The difference in pressure between the water and the top of the siphon is what overcomes gravity and pushes the water up. The (ΔP)V work done by the pressure difference in pushing the volume of water is equal to the potential energy gained by the lifted water.
     
  19. Dec 10, 2011 #18
    yup it is

    You have to look at what the question is asking because you're getting hung up on details that don't matter. It asks you to determine the work needed to get the water out of the tank. The fact that it is being done by a siphon doesn't really matter in answering the question. Siphon hoses are quite small in diameter and the tiny distance it must move the water above the edge of the barrel is negligible so all you need is the mgh with h being the height you have to move the center of mass of the water just to the top of the barrel, not to the teeeeny bit higher that it will actually move.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work done by a siphon.
  1. Work done and thermo? (Replies: 4)

Loading...