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Author Topic: School boy experiment with water  (Read 15563 times)

vineet_kiran

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School boy experiment with water
« on: May 30, 2012, 07:30:12 AM »
 
School boy experiment with water and tube
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Low-Q

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Re: School boy experiment with water
« Reply #1 on: May 30, 2012, 12:50:29 PM »
Quote from: vineet_kiran on May 30, 2012, 07:30:12 AM

School boy experiment with water and tube
I have tried it and you are wrong in your conclusion. The air bubbles is forced to push the water out of the tube first before air can be pushed out. You will feel a counterforce by the buoyancy of the immersed air column. The weight of the tube will probably be the same as the counterforce, but when you take the tube out of the water to repeate the cycle, you must lift both the tube and the water inside the tube above the water level in the two tanks before water is released.


Vidar
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vineet_kiran

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Re: School boy experiment with water
« Reply #2 on: May 30, 2012, 02:05:47 PM »
@Vidar,
 
1)      The  counter force  on the air column   acts  only on the right side portion of  the tube  ie.,  on the end from which air comes out,   which can be absorbed  by  fixing that portion  of the  tube  to an external  support  attached to container or some other place.  Since tube is flexible, the movement of  left portion of the  tube will not be affected.
 
2)      I have mentioned that  you have to provide a valve (or plug) on the tube  preferably  on the portion which is not immersed in water   to repeat the cycle again.    When  you open the valve the pressure inside the tube will become atmospheric  and then if you lift the tube (left portion only)  keeping the valve open,  water  column will not come along with the tube.   Once you lift the tube  you have to close the valve again and repeat  the cycle. 
  Kindly confirm the results and  tell  me if I am wrong.
 
Regards,
Vineet.K.
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Low-Q

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Re: School boy experiment with water
« Reply #3 on: May 30, 2012, 08:22:20 PM »
No matter how you put it, there is a given amount of force you have to move. If there isn't any force on the "bubble side", you have ALL force on the other side. Try with 4 inch tubes instead so more water can be displaced. Then you can actually feel what forces you're dealing with.


Vidar.
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TinselKoala

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Re: School boy experiment with water
« Reply #4 on: May 30, 2012, 10:18:18 PM »
Vineet, your diagram misrepresents the true situation. Please consider my corrected and annotated version of your diagram which I have attached below.
What your diagram omits is the displaced water. When you push one end of the tube into the water, keeping the other end at the same level as before, you are decreasing the volume _above the waterline_ occupied by the air. It must escape or become pressurized. It pushes the water out of the other end of the tube, RAISING the level of the water in the container by an equivalent volume. The force required to raise this water and produce the bubbles is coming from your hand pushing down the other end of the tube.
If you push both ends down simultaneously of course you will get no bubbling and the level of water in both tubs will rise as you submerge more and more of the tube (and your hands). This lifting up of the weight of the water IS the buoyancy force that tries to push your tube and your hands back up out of the water.

(Actually the level in the tube to the left will also be pressed downward and the level in its tub will rise as well, and should be the same as in the other tub. The difference being that the air can escape in the right tube end but the left tube end is pushed down farther so the air cannot escape there.)
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TinselKoala

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Re: School boy experiment with water
« Reply #5 on: May 30, 2012, 10:42:06 PM »
So, from examining my diagram, we can see that the pressure in the system is set by the depth of the opening in the tube on the right -- that is, by the weight of the water column above the opening, or equivalently the water pressure at that depth. As soon as the pressure in the tube reaches that pressure, the water will be all displaced from the fixed end and the air will bubble out. Push the free end down faster and the air will bubble out faster, but not at any higher pressure. This is why it doesn't seem like your hand is pushing very much... it isn't.

But if you make your tank very deep and put the fixed end of the tube very deep in the tank, then you will feel how much pressure it takes to displace all that water.

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vineet_kiran

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Re: School boy experiment with water
« Reply #6 on: May 31, 2012, 08:17:59 AM »
 
@TinselKoala,
 
You have done an excellent analysis of the experiment with a very nice diagram.   But what I am thinking is totally different  which is about direction and area of application of forces.    I have  given my reply vide attachment showing one more experiment   in which you can totally absorb the buoyancy forces at  any depth.  I request you to kindly go through it and let me have your  comments.
with thanks and regards,
Vineet.K.
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TinselKoala

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Re: School boy experiment with water
« Reply #7 on: May 31, 2012, 09:38:56 PM »
Vineet, once again your assumptions and diagram do not correspond to reality. Now you are apparently failing to consider the stretch of the balloon. When you initially fill the balloon with water and seal it, you have stored energy in the stretch of the balloon itself. This depends on the _volume_ inside the balloon, not the weight. I'm sure you have noticed that filled balloons tend toward a spherical shape if allowed: this is because the sphere has the smallest surface area for a given volume, and the stretchy skin tries to minimize _area_, but the volume is _conserved_.
When you then immerse this balloon in water you are not only _displacing_ the water that is in the tank, raising its level, but you also still have the stored tension in the balloon's skin due to the volume of water inside the balloon, even though its "weight" has been, you believe, neutralized. (It hasn't really, it's just been transferred to the lifted water in the container.) Since the volume in the balloon isn't changed, the skin tension is still there. Now you open the valve to the tube.... and the skin of the balloon pushes the water level up in the tube. The air that now bubbles out is due to the stretched balloon skin.
If you use a limp sack instead of a balloon, what happens? Nothing.
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vineet_kiran

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Re: School boy experiment with water
« Reply #8 on: June 01, 2012, 07:50:45 AM »
@TinselKoala,
 
1)      Stretching  of balloon requires  force which is provided by  weight of the water and also pressure of the water but not volume of water.    If you attach  a lesser length tube to the balloon,  the stretch in the balloon pushes the water up and water spills  out  bringing the balloon to its normal un stretched  condition but when you fix a lengthy tube (vertically),  water will  rise  to some height  corresponding to the force and pressure offered by the stretch of the balloon and stops at that height holding the balloon in stretched condition.
 
2)      When you immerse the water balloon with lengthy vertical tube attached to it  in water,  the balloon experiences pressure / force from  outside water which allows it to push  more water into the vertical tube.    The deeper the balloon pushed into the water,  it experiences more pressure and hence pushes more water into the tube.  (it works just like  a pressure indicator, indicating pressure at depth).
 
3)      When  water balloon is  immersed  in water  it loses its weight totally  with respect to external measuring balance (ie.,  if you fix this water balloon to a spring balance  and then immerse it in water,  the spring balance shows zero weight) because density of water in balloon and outside water  is same.  Rising of water level in the  container is immaterial  as it happens  when you immerse any solid material in water.   But this solid material immersed in water  loses its weight with respect to external spring balance depending on upthrust  of water as per the laws of Archimedes.
 
4)      Because the water balloon loses its weight with respect external force,  it can be easily moved inside the water applying a very little force.   When you  push the balloon deep in water, it displaces more water into the tube which in turn displaces air  present in upper portion of the tube to the exit  and  air comes out in the form of bubbles with huge force.   
 
5)      The energy required to move the balloon inside the water container is negligible  whereas air bubbles come out of the other container  with huge force.
 
6)      By  providing a valve or plug on the tube you can repeat the cycle moving balloon up and down with very little input energy and get comparatively high energy output in the form  of  air bubbles emerging out.
 
Vineet.K.
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TinselKoala

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Re: School boy experiment with water
« Reply #9 on: June 01, 2012, 07:55:24 PM »
Vineet, have you actually done this experiment yourself? 

Your description, if I am understanding you correctly, is not the way objects behave in reality.

When you fill a balloon with water it is primarily the volume that stretches the balloon, not the weight of the water. You can prove this to yourself very easily.

Fill a balloon with red-colored water to the extent required for your experiment,  and tie a knot in the end. No Air !!

Put this balloon into a bucket of water. Does the balloon shrink or expand? Its weight has "decreased".... but it still encloses the same volume and its skin still has the same overall tension. If you punch a hole in the submerged balloon,  now it will leak out the red water until the balloon's skin is relaxed again.

It is the volume inside the balloon that determines its tension. When  you have a balloon filled with water hanging in air, some of the tension in the skin is caused by the weight of the water but to get the effect you are describing you obviously have to put enough water in the balloon to stretch it substantially.

And don't forget: for any gravity or buoyancy effect to be useful, it has to be repeated in a cycle. If your first cycle depends on stored energy-- as yours does -- and the energy must be replaced for the cycle to repeat.... then you've got a problem. Who is filling up that balloon in the first place, working against its skin tension? Who lifted the water to do that with? What did it cost?

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vineet_kiran

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Re: School boy experiment with water
« Reply #10 on: June 02, 2012, 07:13:16 AM »
 
@TinselKoala,
 
1) I  have done that experiment completely and after confirming I have posted that expriment here.    I think you have just filled the balloon with water and dropped it into water container.   With that you cannot make out anything.   Fix a lengthy vertical tube into it and repeat the sequence which I have said in my earlier reply.   
2) I am not making use of the energy stored in first  place while filling  the water  but by  moving the water balloon up and down inside the water container.   The energy stored in first place stretching the skin doesnot come into picture here.     The stretch of the skin varies  ( gets compressed or expands)  depending on the depth to which water balloon is pushed inside the water container. (only if you fix a vertical tube to the balloon)
I cannot explain anything more than this.    I am retiring from the argument.
Thanks anyway for the nice discussion.
 
Regards,
Vineet.K.
 
 
 
 
 
 
 
 
 
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Ghost

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Re: School boy experiment with water
« Reply #11 on: June 02, 2012, 11:07:42 AM »
so how can we close the loop here? how would you feed the output back to the input? how would we make this self running?
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Ghost

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Re: School boy experiment with water
« Reply #12 on: June 26, 2012, 07:41:36 AM »
vineet_kiran,
Any news on this experiment?
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Newton II

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Re: School boy experiment with water
« Reply #13 on: January 07, 2013, 05:17:18 AM »

This seems to be a interesting experiment.  I tried experiment several times each time with a different diameter tube and also at deeper depths.  I didnot experience much input force in any of those trials.


Since parameters on both sides of the tube are identical,  you will not experience any force even if you use larger diameter tube and go to a higher depth.  I think you are making use of potential difference existing at two different heights of water. (mgH2 - mgH1 ?)


 
 

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