How do you calculate the pull of this piston?

[vdance]

Homework Statement

An interesting question that came to mind on its own.

Relevant Equations

F=ρghs

Neglect piston gravity, neglect friction.
Assume that the height at which the piston draws water is equal to 5 meters.

When the piston is solid.
F=ρgHs=ρgHs=1000×9.8×5×1=49000 N

When the piston is hollow.
F=ρgHS=ρgHS=1000×9.8×5×2=98000 N

Is the calculation of the tension force in Figure 2 correct?
If incorrect, how is it calculated?

Supplementary questions for information.
If S has no effect, is F（F=ρgHs=ρgHs=1000×9.8×5×1=49000 N）still constant when placed vertically?

 

[haruspex]

S is irrelevant. It makes no difference to the height by which the water is raised by a given movement of the piston.

 

[vdance]

S is irrelevant. It makes no difference to the height by which the water is raised by a given movement of the piston.

Thanks.
If S has no effect, then when placed vertically (Fig. 3)? is F unchanged?

 

[haruspex]

Thanks.
If S has no effect, then when placed vertically (Fig. 3)? is F unchanged?

Yes.
To see why, consider the forces on the piston. What are the forces?

 

[Lnewqban]

View attachment 359947
When the piston is hollow.
F=ρgHS=ρgHS=1000×9.8×5×2=98000 N

Is the calculation of the tension force in Figure 2 correct?
If incorrect, how is it calculated?

Not correct.
The column of fluid is not a cylinder with cross section of 2 m², which would weight twice as much as the actual one.

The force F is of the magnitude needed to lift and hold in static equilibrium the mass of fluid contained in the represented volume.

To be precise, figure 3 represents a slightly greater mass of fluid, considering the fluid occupying the hollow piston (respect to the solid piston of neglected mass).

 

[vdance]

I would like to understand the difference in water pressure on the piston when pumping with a solid piston versus a hollow piston. The master used to say that to calculate the water pressure you need to look at the height between the inlet and outlet, and that the water inside the hollow piston is not flowing, so it is logical to ignore it. In that case, whether the piston is solid or hollow, they are all the same water pressure.

 

[jbriggs444]

I would like to understand the difference in water pressure on the piston when pumping with a solid piston versus a hollow piston.

Where do you want to draw the imaginary lines around your system(s) of interest? A good choice will make calculations easier. A poor choice will make calculations more difficult.

You can draw the line at the mouth of the piston. Pretend that there is a shin sheet of aluminum foil across it. Now your "hollow" piston is a piston with a water filled hollow. The included water makes this piston a bit heavier than a piston with an air-filled hollow. And a bit lighter than a piston with no hollow at all.

As long as nothing flows past the imaginary line where the foil would be, it does not matter whether the foil is present or absent. Except for their weight, a hollow piston and a solid piston behave identically. You would want to consider fluid pressure at the center of the mouth of the piston, not at the back end of the hollow.

Alternately you could draw the line so that it follows the sides of the hollow. Now you have an oddly shaped piston. If you want to compute the force of the fluid on the piston, you will have to integrate your way all around the hollow. That is a lot of work.

 

[vdance]

Where do you want to draw the imaginary lines around your system(s) of interest? A good choice will make calculations easier. A poor choice will make calculations more difficult.

You can draw the line at the mouth of the piston. Pretend that there is a shin sheet of aluminum foil across it. Now your "hollow" piston is a piston with a water filled hollow. The included water makes this piston a bit heavier than a piston with an air-filled hollow. And a bit lighter than a piston with no hollow at all.

As long as nothing flows past the imaginary line where the foil would be, it does not matter whether the foil is present or absent. Except for their weight, a hollow piston and a solid piston behave identically. You would want to consider fluid pressure at the center of the mouth of the piston, not at the back end of the hollow.

Alternately you could draw the line so that it follows the sides of the hollow. Now you have an oddly shaped piston. If you want to compute the force of the fluid on the piston, you will have to integrate your way all around the hollow. That is a lot of work.

Yes, the perspectives shared by the teachers earlier align with this, I've learned a lot, thank you!
I'm trying to figure out a method to extract water while maintaining constant water pressure during the piston's up-and-down motion, which is quite challenging to conceptualize.

 

[haruspex]

I'm trying to figure out a method to extract water while maintaining constant water pressure during the piston's up-and-down motion, which is quite challenging to conceptualize.

Not following. Do you mean you want the water to flow out throughout the cycle? If not, constant pressure where?

 

[vdance]

Not following. Do you mean you want the water to flow out throughout the cycle? If not, constant pressure where?

I would like to find a way to pump water from a rigid tube at a constant pressure, relying on the reciprocating motion of a piston. From the standpoint of the law of conservation of energy, perhaps such a method will never exist, I'm just casually exploring.

 

[haruspex]

I would like to find a way to pump water from a rigid tube at a constant pressure, relying on the reciprocating motion of a piston. From the standpoint of the law of conservation of energy, perhaps such a method will never exist, I'm just casually exploring.

You could link two back to back so that when one piston is on the pull stroke the other is on the push.

 

[vdance]

You could link two back to back so that when one piston is on the pull stroke the other is on the push.

It looks simple, but it's hard to implement, I've been trying to design it for a week now, but haven't found a way to do it yet.

 

[haruspex]

It looks simple, but it's hard to implement, I've been trying to design it for a week now, but haven't found a way to do it yet.

Please post an attempt.

 

[vdance]

Please post an attempt.

In this structure, I want to know the magnitude and direction of F.
Is it F=ρgH(S-s)=1000×9.8×5×(2-1)=49000 N
or F=ρgH(S-S)=1000×9.8×5×(2-2)=0 N

 

[haruspex]

View attachment 360044
In this structure, I want to know the magnitude and direction of F.
Is it F=ρgH(S-s)=1000×9.8×5×(2-1)=49000 N
or F=ρgH(S-S)=1000×9.8×5×(2-2)=0 N

I do not see how that would work. The space between the pistons is constant volume, so why would any water rise?

 

[vdance]

I do not see how that would work. The space between the pistons is constant volume, so why would any water rise?

Aha, I think I understand now. I shouldn't calculate the water pressure, but rather the weight of the water. This way, my problem can be solved. Thank you!

 

[jbriggs444]

View attachment 360044
In this structure, I want to know the magnitude and direction of F.

I see a one way valve.

If this is a question about hydrostatics then @haruspex is making the relevant observation. As long as the pistons are moving slowly and steadily, the one way valve is under a constant pressure difference. No air is ever allowed in past the one way valve. No water can ever leak out against the unchanging pressure difference.

If the pistons are accelerated back and forth then we can identify dynamic pressure differences due to that intermittent acceleration. While the pistons are being accelerated rightward or leftward by some external force, a pressure gradient will exist in the fluid between the two pistons. This pressure gradient will be consistent with the acceleration of the fluid. The gradient will be given by $\Delta = \rho a s$ where $\rho$ is the fluid density, $a$ is the left-right acceleration and $s$ is the left-right displacement across which a pressure differential is to be measured. [I've not carefully chosen a sign convention for that formula. If you decide to use it, pay some attention to that].

If the back and forth acceleration is vigorous enough, one may find that the left-right pressure difference is enough so that the gauge pressure at the one way valve intermittently goes positive, allowing fluid to escape past the valve.

 

[vdance]

I see a one way valve.

If this is a question about hydrostatics then @haruspex is making the relevant observation. As long as the pistons are moving slowly and steadily, the one way valve is under a constant pressure difference. No air is ever allowed in past the one way valve. No water can ever leak out against the unchanging pressure difference.

If the pistons are accelerated back and forth then we can identify dynamic pressure differences due to that intermittent acceleration. While the pistons are being accelerated rightward or leftward by some external force, a pressure gradient will exist in the fluid between the two pistons. This pressure gradient will be consistent with the acceleration of the fluid. The gradient will be given by $\Delta = \rho a s$ where $\rho$ is the fluid density, $a$ is the left-right acceleration and $s$ is the left-right displacement across which a pressure differential is to be measured. [I've not carefully chosen a sign convention for that formula. If you decide to use it, pay some attention to that].

If the back and forth acceleration is vigorous enough, one may find that the left-right pressure difference is enough so that the gauge pressure at the one way valve intermittently goes positive, allowing fluid to escape past the valve.

This question is the part I broke out of the graphic on the right above. The left end of the piston on the left side of the picture above has a different cross-sectional area than the right end, and at atmospheric pressure the piston must be moving to the right. In the right graphic I modified the right piston to be of equal size and added check valves to the ports, in this case I don't understand if the piston is stationary or moving to the right.Therefore, I need guidance from all teachers.

 

[vdance]

The piston absorbs water and is analyzed at rest, ignoring piston gravity and friction.
Calculate the traction of the piston separately.
The first set of calculations:
F₁=ρgHS=1000*10*5*1=50000 N
F₂=ρg(H+h)S=1000*10*(4+1)*1=50000 N
F₃=ρgHs=1000*10*0.5*1=5000 N
Second set of calculations:
F₁=ρgV=ρgHS=1000*10*5*1=50000 N
F₂=ρg(hS+Hs)=1000*10*(1*1+4*0.1)=14000 N
F₃=ρgV=ρgHS=1000*10*5*1=50000 N
Which group's calculations are correct?
The AI is telling me it's the first set of results.

 

[haruspex]

View attachment 360166
The piston absorbs water and is analyzed at rest, ignoring piston gravity and friction.
Calculate the traction of the piston separately.
The first set of calculations:
F₁=ρgHS=1000*10*5*1=50000 N
F₂=ρg(H+h)S=1000*10*(4+1)*1=50000 N
F₃=ρgHs=1000*10*0.5*1=5000 N
Second set of calculations:
F₁=ρgV=ρgHS=1000*10*5*1=50000 N
F₂=ρg(hS+Hs)=1000*10*(1*1+4*0.1)=14000 N
F₃=ρgV=ρgHS=1000*10*5*1=50000 N
Which group's calculations are correct?
The AI is telling me it's the first set of results.

And what do you think, and why?
Forum rules request that you start a new thread for a new question and show an attempt.

 

[vdance]

And what do you think, and why?
Forum rules request that you start a new thread for a new question and show an attempt.

I believe this question shares the same fundamental principle as the previous inquiry, representing an extension of the same physical concept, which is why they are consolidated into a single discussion thread. My view is: Calculating the blocking force based on the piston’s cross-sectional area is valid. In such cases, even though two pools have identical water depths of 1 meter but drastically different internal cross-sectional areas (1 m² vs. 10,000 m², implying a colossal volume disparity), the force required to block their drainage pipes remains equal as long as the pipe diameters are the same. However, a fascinating paradox emerges: If a standalone 1 kg mass of water is freely suspended, I could effortlessly lift it with one hand. Yet, if this same kilogram of water is spread into a thin layer spanning several square meters, held by a piston above and connected to a pool below via a narrow tube with an inner diameter of 1 cm, I would be completely unable to lift this mere 1 kg of water. Alternatively, this line of reasoning might be flawed, and a more accurate approach would involve calculating pressure gradients across varying inner diameters before summing them—in which case, I could still lift the 1 kg of water.

 

[haruspex]

I believe this question shares the same fundamental principle as the previous inquiry, representing an extension of the same physical concept,

Sure, but it is still a new question, so new thread please.

My view is: Calculating the blocking force based on the piston’s cross-sectional area is valid. In such cases, even though two pools have identical water depths of 1 meter but drastically different internal cross-sectional areas (1 m² vs. 10,000 m², implying a colossal volume disparity), the force required to block their drainage pipes remains equal as long as the pipe diameters are the same.

Right.

However, a fascinating paradox emerges: If a standalone 1 kg mass of water is freely suspended, I could effortlessly lift it with one hand. Yet, if this same kilogram of water is spread into a thin layer spanning several square meters, held by a piston above and connected to a pool below via a narrow tube with an inner diameter of 1 cm, I would be completely unable to lift this mere 1 kg of water.

If you were to lift the piston at 10cm/s, how fast would the water in the thin tube rise?

 

[Lnewqban]

Yet, if this same kilogram of water is spread into a thin layer spanning several square meters, held by a piston above and connected to a pool below via a narrow tube with an inner diameter of 1 cm, I would be completely unable to lift this mere 1 kg of water.

Why not?
You may be cofusing force and work.
This hydraulic pistons business is not different from a lever and work: what you gain in reduced force, you lose in increased distance you need to apply that lesser force, and vice-verse.
Please, continue on in a new thread, as required above.

 

[haruspex]

Why not?

Because @vdance has not correctly described the difference between the two scenarios. It's not that the reservoir at the base is spread out wide, it's that the 1kg of water has the form of a tall thin tube, meaning that there is a large pressure difference between top and bottom.

 

[vdance]

If you were to lift the piston at 10cm/s, how fast would the water in the thin tube rise?

Assuming the cross-sectional area of the piston in the upper section is S=10m², and the inner diameter of the lower section is 1cm (corresponding to s=7.85×10⁻⁵m²), when the piston moves upward at V=10cm/s to draw water, the continuity equation S⋅V=s⋅v yields a flow velocity v≈12738.85m/s in the lower section. This result violates physical limits (e.g., the speed of sound is ~343m/s).

 

[jbriggs444]

Assuming the cross-sectional area of the piston in the upper section is S=10m², and the inner diameter of the lower section is 1cm (corresponding to s=7.85×10⁻⁵m²), when the piston moves upward at V=10cm/s to draw water, the continuity equation S⋅V=s⋅v yields a flow velocity v≈12738.85m/s in the lower section. This result violates physical limits (e.g., the speed of sound is ~343m/s).

The speed of sound in air is not a physical limit on the speed of water in a pipe.

It is not clear to me what point you are trying to make. Auto mechanics routinely make good use of hydraulic jacks that involve force ratios much smaller than 120 thousand to one. If you want to make a point about force multiplication using pistons, that is, perhaps, the sort of device that you ought to consider.

Also, normally one makes use of pressure rather than suction. Much less worry about cavitation.

 

[haruspex]

a flow velocity v≈12738.85m/s in the lower section

So do you now see why you would not be able to lift it?

 

[vdance]

So do you now see why you would not be able to lift it?

The power of the individual is too small to fight the atmospheric pressure of nature. Haha!
I appreciate your help.