Asking for help with a physical problem that uses a piston to push water

[vxiaoyu18]

Homework Statement

One of my own problems.

Relevant Equations

F=ρgHS

Consult a physical problem that USES piston to absorb water:
Pistons with surface areas S1, S2 and S are placed horizontally at the same height (H).
S1 = S2
S=S1+S2
When F starts moving to the left, how do you calculate its force?
Is F=ρgHS?
(*The friction force doesn't have to be calculated.)

 

[mfb]

The easiest approach is via conservation of energy. Figure out how much water moves from where to where as function of the displacement, that gives you the force.

You can also note that S1, S2, S always have a fixed relative orientation. The water between them doesn't matter, moving that around doesn't change anything either. Only what happens at S1 matters.

Is F=ρgHS?

No.

 

[vxiaoyu18]

The pressure of the water in the tube will drive the piston to the left. If F=ρgHS it's wrong, what's right?

 

[Baluncore]

1. Because S1 and S2 are coupled, the volume of water common to S1, S2 and S is fixed.
2. The area of S2 exposed to atmosphere cancels an equal area of S exposed to the atmosphere.
3. Only the area of S1 remains, moving with S2 and S.
4. So the force F is being applied directly to S1 only.
5. Eliminate redundant pistons S2 and S, along with the common pool.
6. Connect force F directly to S1. Analyse that.

 

[vxiaoyu18]

1. Because S1 and S2 are coupled, the volume of water common to S1, S2 and S is fixed.
2. The area of S2 exposed to atmosphere cancels an equal area of S exposed to the atmosphere.
...

I used to do the same thing, F=ρgHS1, some people didn't agree with my results, so they came here to consult physics experts.

 

[mfb]

F=ρgHS1

That is correct.

 

[Baluncore]

The sign of F will depend on convention. F = - ρ⋅g⋅H⋅S₁

 

[vxiaoyu18]

The sign of F will depend on convention. F = - ρ⋅g⋅H⋅S₁

Thank you. I can see what you mean.
Now I'm wondering, if S1 = S2 = S, is the piston at rest?

 

[vxiaoyu18]

That is correct.

Thank you.

 

[Baluncore]

Now I'm wondering, if S1 = S2 = S, is the piston at rest?

The areas S1, S2 and S of the pistons is fixed by design.
The balance of forces decides acceleration and motion.

 

[Chestermiller]

What will be the force F when the system is at equilibrium, before the piston even begins to move to the left? I get $-\rho g H S_1$

 

[mfb]

Now I'm wondering, if S1 = S2 = S, is the piston at rest?

If S1=S2=S then the area S1 will move slower than S and your force will be different. This setup is very different from what you had before.

 

[Chestermiller]

That is correct.

Except for the sign.

 

[vxiaoyu18]

1，F =-ρgHS₁

 

[vxiaoyu18]

2，F = 0

 

[vxiaoyu18]

3，F =-ρgSH₂

 

[vxiaoyu18]

4，F =-ρgS(H₁+H₂)

 

[vxiaoyu18]

Is 1 minus 4 all right?
I am looking for a structure that can be implemented: S=S₁=S₂, F=-ρgSH; when S=2S₁, F=-2ρgSH.

 

[Chestermiller]

https://www.physicsforums.com/attachments/254780
1、F=ρgHS₁

Your sign is incorrect. F is negative. Haven’t you ever sucked up on a straw?

 

[vxiaoyu18]

Your sign is incorrect. F is negative. Haven’t you ever sucked up on a straw?

I'll correct it. Thanks for your advice.

 

[mfb]

Except for the sign.

Depends on the sign convention, and on the question if F is the force acting on the piston from the water or an external force holding the piston in place.

2，F = 0

No, the force won't be zero in figure 2.

3，F =-ρgSH₂

That is not correct either. The piston S2 moves if S moves.

#4 is correct.

 

[vxiaoyu18]

Depends on the sign convention, and on the question if F is the force acting on the piston from the water or an external force holding the piston in place.

F is the force acting on the piston from the water.

 

[vxiaoyu18]

No, the force won't be zero in figure 2.

If it's not equal to 0, how do you calculate it?

 

[vxiaoyu18]

That is not correct either. The piston S2 moves if S moves.

Figure 3. Suppose I add a tension meter and pull S piston at rest, then how to calculate the tension meter data?

 

[Baluncore]

A homework forum is not the place to rediscover a technical solutions to an unspecified problem using trial and error techniques. Would it not be simpler to first understand hydraulic cylinders, specify the fluid transfer requirement, then design a minimum solution.

The field of “hydraulic regeneration” is not magic. It is well understood and has been used by industry for some time.
https://www.hydraulicspneumatics.co...84332/book-2-chapter-17-regeneration-circuits

 

[vxiaoyu18]

I didn't have a deep understanding of this problem, so I assumed some examples to calculate and see what the result would be. Only by mastering the correct method of physical calculation can I develop and utilize this technology efficiently.

 

[Chestermiller]

Depends on the sign convention, and on the question if F is the force acting on the piston from the water or an external force holding the piston in place.

The figure seems to clearly show that F is the external force oriented to the left on the outside face of the piston. What am I missing?

 

[vxiaoyu18]

You're not missing anything. The source of motivation is the pressure of the water.

 

[vxiaoyu18]

One of my engineer friends told me before: To calculate the water pressure, we must pay attention to the position of the inlet and outlet. Only when there is flow can there be water pressure. The height between the inlet and outlet is the height of water pressure. I used this method to analyze figure 1. Water enters from S and comes out from the bottom of S₁ with a height of H, so F=-ρgHS. At the same time, since S and S₂ move synchronously, the water between S and S₂ keeps a fixed volume in the pipe and does not change, so F=0. Finally I worked out that F=-ρgHS.

 

[vxiaoyu18]

Figure 1, if there is no fixed shafts between S₁ and S₂, from S and S₂ water flows out from the bottom of the S₁, then S and S₁ at the bottom of the nozzle pressure between F₁ = -ρgHS, the S₂ and S₁ at the bottom of the nozzle pressure is F=-ρgHS₂, so the existence of fixed shafts brought inside environment and complicated changes, the key problem is the pipe of the synchronization between S and S₂ in the sports meet will not cause F resistance.

 

[vxiaoyu18]

As far as I am concerned, the inflow and outflow of water refers to the outflow of water from the tube to the place in contact with air, S and S₁ meet this condition, while S and S₂ do not meet this condition, because the water at the mouth of S₂ has no contact with air.

 

[vxiaoyu18]

I think I've figured out how to do it, and I can do it very quickly with PASCAL's law. 1. F=-ρgSH; 2. F=-（1/2）ρgSH; 3. F=-（1/2）ρgS（H₁+H₂）; 4. F =-ρgS(H₁+H₂).

 

[vxiaoyu18]

2. The area of S₂ exposed to atmosphere cancels an equal area of S exposed to the atmosphere.

I can't understand what you said here. I'm not sure if it's because I misunderstood the translation.
If S₂ and S are equal to the air contact surface, how to understand figure 2? Does it also result in F =-ρgHS₁?

 

[vxiaoyu18]

Except for the sign.

You told me what the answer was, but you didn't tell me how it was calculated. There are several possible ways to compute this answer, and I don't know which one is the most accurate and can be applied to other graphs, so I assumed a few graphs and tried to learn how to compute it from more calculations.

 

[haruspex]

I can't understand what you said here. I'm not sure if it's because I misunderstood the translation.
If S₂ and S are equal to the air contact surface, how to understand figure 2? Does it also result in F =-ρgHS₁?

@Baluncore is saying to consider the area S as made of two areas S₁' and S₂' side by side, the first of area S₁ and the second of area S₂. The pressure on S₂' is the same as that on S₂, and they have the same area, so the forces on them are equal and opposite, so cancel.