Hydostatic pressure at bottom of a cylindrical and conical tanks

  1. Please help me with the following simple and dumb questions:

    First, assumptions:
    - Ignore atmospheric pressure.
    - The liquid in the tanks is water with uniform density = 1g/cc.
    - all tanks are sealed at the bottom and are vertically positioned.
    - cross sections of the tanks are circular.

    1. I have two tanks of same height (H) and same volume (V). One is cylindrical in shape and other is inverted conical. What is the pressure at the bottom of each tank?

    2. I have one tank which has two parts:
    Upper part is cylindrical with area = "A" and height = "H".
    Lower part is an inverted cone with upper surface area = "A" (this is where it meets the cylindrical part) and the lower surface area = "a", and the height of the conical part = "h".
    What is the pressure at the bottom surface of this tank?

    3. I have a cylindrical tank of area "A" and height "H" with a hole in at the center of its circular bottom. The hole area is "a" and it leads to another small cylindrical tank directly below the upper tank. The height of the small lower tank = "h". What is the pressure at the bottom of the small lower tank.

    4. Same question as #3 except that the hole (and the smaller tank that it leads to) at the bottom of the upper tank is not at the center, but to a side.

    Thanks in advance for enlightening me!

    sai_2008
     
    Last edited: Dec 26, 2008
  2. jcsd
  3. stewartcs

    stewartcs 2,284
    Science Advisor

    Chegg
    Is this a homework question? If so, which it sounds like it is, you'll need to show your work/thoughts first.

    HINT: What is the formula for hydrostatic pressure?

    CS
     
  4. Hi StewartCS,

    I anticipated that someone would just say its homework question and direct me to the basics. But looking closer - it is (at least, as I see it) beyond the basic hydrostatic formula thing: P = hdg.

    But I will show below all my work and where I got stuck...

    No, this is not a home work question - It is a home improvement project question :-D.

    I need to design an automatic pump controller system to turn on and off a high power pumps that pump water from an underground sump to the over head tanks which are mixed cylindrical and conical shapes, on top of my 3-floor house. The critical piece for my design to work is to accurately sense the static water pressure from inside the building and turn on one or two or all the three pumps. No, I don't want to install anything on or in the tanks. I got the system to work in my garage as a prototype,
    but to calibrate the gauges with the real thing, my calculations seem to be incorrect..hence these questions.

    Instead of asking the answers to my questions directly, may be I should ask you
    guys to correct where I have gone wrong below.

    Here is my work so far: Let d = density of water, g = acc. due to gravity.

    Q1. I have two tanks of same height (H) and same volume (V). One is cylindrical in shape and other is inverted conical. What is the pressure at the bottom of each tank?

    My work:-
    Cylindrical tank: P = Hgd - independent of the area, no biggie here....
    Conical tank: P = wt/Area = Vdg / ??? <<< this is where the problem is!!
    Ideally, the area at the bottom tip of an inverted cone = 0, so is the pressure
    at the bottom of the tank infinite...? What is the dependence on H in the case
    of conical tank?

    Q2. 2. I have one tank which has two parts:
    Upper part is cylindrical with area = "A" and height = "H".
    Lower part is an inverted cone with upper surface area = "A" (this is where it meets the cylindrical part) and the lower surface area = "a", and the height of the conical part = "h". What is the pressure at the bottom surface of this tank?

    My work:-
    The entire weight of water falls on the lower surface.
    Volume of Cyl section V1 = AH
    Volume of Conical section V2 = (1/3)*h*{A+a+square_root(A*a)}
    So the pressure at the bottom:
    P = total water wt/a = (V1 + V2)*dg/a
    That is, P = [AH + (1/3)*h*{A+a+square_root(A*a)}] * dg/a

    Now, if A/a = f, then
    P = fHgd + (1/3)fhgd + (1/3)hgd + (1/3)hd*square_root(f)

    Because f > 1, the pressure at the bottom of the tank is not just (H+h)dg; it is more than that. My explanation: this is due to the conical nature of the lower part of the tank. If f = 1, then the whole tank is just cylindrical, so P = (H + h)dg, as expected.

    Did I do this correctly? Am I right in saying that due to conical nature of lower part
    of the tank, the pressure at the bottom is more than that it would've been had the
    whole thing was just cylindrical, top to bottom?

    Q3. I have a cylindrical tank of area "A" and height "H" with a hole in at the center of its circular bottom. The hole area is "a" and it leads to another small cylindrical tank directly below the upper tank. The height of the small lower tank = "h". What is the pressure at the bottom of the small lower tank

    My work:-
    The entire wight of the water falls on the lower surface of the lower tank. The
    center of weight of the upper tank portion and the lower tank portion are vertically
    lined up, so the the entire weight just falls thru the tank's central vertical axis on to
    the lower tank's bottom surface.

    So P = total water wt /a = (AH + ah)dg/a = hdg + AHdg/a.
    This is also more than (H+h)gd because a < A !!!

    Now the kicker and my dilemma: as h tends to 0, what happens to P? It becomes AHdg/a!! This is also more than Hdg, since A/a > 1!!!! Where did I go wrong?
    And what happens as "a" tends to 0? Aaaarrrrghhh...!!!

    4. Same question as #3 except that the hole (and the smaller tank that it leads to) at the bottom of the upper tank is not at the center, but to a side.

    My work:- I know I am could be on the wrong track on #3, so I cannot proceed beyond this point. But I think there is no difference where the hole in the bottom of the tank is!!
    The pressure is exactly the same as #3.


    Thanks again. Sorry for not providing my calculations earlier ( now I realize it was too much to ask others to solve the whole thing for me...)

    Thanks!
    sai_2008
     
  5. russ_watters

    Staff: Mentor

    Several misunderstandings:
    Either you haven't had calculus yet or you had it and misunderstood this part. Division by zero does not produce an infinite result, it produces no result. It is undefined. You can't apply that equation to the point of the cone.
    No, it doesn't (the other questions are the same question and the same wrong assumption). Some is supported by the sides of the cone (I assume you are using a cut-off cone with a flat, horizontal piece at the bottom represented by your "a").

    So it appears you are trying to derive the hydrostatic pressure equation - or, rather, derive a new one, but you're making some errors. Why not just use the equation as is? I suppose you think you found an error or think it doesn't apply to these situations, but it does.
     
  6. Well, not sure if this has anything to do with calculus, but I agree /0 means undefined.

    Well, just applying the equation to estimate the pressure lead me is giving errors in my calibration of the system, my pressure gauges values did not agree. So I thought there is something more to that hdg equation that I know, which is what I was after.

    Clearly, I don't understand hydrostatics...OK, I agree.
    So pl. help me out to figure this out...correctly, not saying just apply the equation, the
    only derivations of the equation I remember was using simple tanks.

    Now that you said "sides of the cone" support some of the weight...
    Suppose the tank with the cutoff cone base of area a, as you rightly said, filled with water
    were to sit on a weighing scale. What will the scale read? If you just keep the empty tank
    weight aside for now, (or calibrate the scale to zero with an empty tank on it), the
    scale will read (the entire water weight) right?
    What is the pressure on the surface of the scale exerted by the tank?
    (entire water weight) / a - no?

    So just on the other side of the wall (inside the tank) the pressure, per the equation
    is (H+h)gd. But on the side that touches the scale the pressure is
    (entire water weight) / a. Is this correct?
     
  7. The static condition is simply based on the density of the liquid and the height of the liquid....Think of an ocean and a nearby bay...at 100 feet depth in each, the pressure is the same...the shape and volume of the surrounding waters is irrelevant....

    or think instead of a single length of vertical pipe inserted through all your different shapes...the pressure at the bottom of the pipe is the same as just oustide despite all the different shapes..if it were not identical, water would flow from one to the other until the pressure were equal...

    The location of the holes and connections in your situation also does not matter...center of tank versus side, for example, is irrelevant....

    If you have a fixed column of water, say six feet high, it has a fixed pressure at the bottom regardless of the shapes and volumesthe water may occupy...and the pressure on the bottom of the tank is the same as the pressure on all sides at the bottom....
     

  8. Hi Naty,
    I agree with every bit of what you have said. It makes sense in general to me -
    if I dive into a river or an ocean, at 100 meters, I will have identical pressure effects.

    But academically, I never saw a proof of P = hgd other than using a
    tank of uniform cross section. Could you pl. point me to a link or
    something that proves it more generally, from fluid statics area?

    Thanks
     
  9. russ_watters

    Staff: Mentor

    Ok....
    You haven't told us anything about what you are trying to do, but if you want to calibrate a pressure gage by calculation, using the hydrostatic equation is the proper way. No, there isn't anything more to it than that.

    Just make sure you use the correct density for water (this isn't salt water, is it?) and differentiate between gage and absolute pressure.
    The pressure of the water on the bottom of the tank does not equal the pressure of the scale on the bottom of the tank. Since some of the water is supported vertically by the sides of the tank, the extra force is carried through the sides of the tank. It must be rigid or it will collapse.
    Yes.

    Good example.
     
  10. stewartcs

    stewartcs 2,284
    Science Advisor


  11. Thanks to all of you guys - it was re-assuring that I did learn this fundamental correctly in high school after all...and there is nothing much more complicated than that. The links are fantastic. I dont have my physics textbook anymore...

    Since russ_watters asked...this is what I am trying to do...
    I am trying to sense the static water pressure from an ordinary water tap in the water
    distribution system of my house. I idea is to know if the overhead tanks are almost empty
    or full, so that I can switch on and off a system of pumps.
    The pressure sensor generates a current change signal which I sample and hold
    and then drive a circuit that in turn drives a relay circuit.

    I suspect, now, that what is throwing my sensors is, first they are too sensitive,
    and when water falls into the tank, there is a pressure wave that travels through out the water lines and there is a momentary jump that latches a high pressure reading...

    I believe you guys...:D there nothing more to it than the basic hdg thing, but I was getting paranoid...

    thanks again.
     
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