I've come across another fluid pressure problem I don't understand

[vdance]

You may have some misundestandings - even about the forces in Fig. 1. Try working through this:

Consider the simple Fig. 1 setup. Assume the piston is of negligible weight and is stationary and in equilibrium. Take atmospheric pressure as 101kPa (=101,000 N/m²).

Assume friction, surface tension and adhesion forces can be ignored. There are then three forces acting on the piston:
__________________

1) The force due to the hydrostatic pressure of the water acting on the bottom of the piston.
... ...

Yes, you're right.
My English isn't very good, so I used software to translate. Some parts might not be entirely accurate, but I'll do my best to understand both sides' meanings.

 

[jbriggs444]

What I need to address is the resultant force exerted by water on two pistons fixed together.

Yes. I think that everyone here understands that.

You want the total of:

1. The force of water pressure on the bottom surface of the top piston $F_{3bt}$ plus...
2. The force of water pressure on the top surface of the bottom piston $F_{3tb}$ plus...
3. The force of water pressure on the bottom surface of the bottom piston $F_{3bb}$

This with the understanding that the pressure of the ambient atmosphere on the top surface of the top piston is taken as zero [all pressures are gauge pressures] and that there are no other external vertical forces acting on the two pistons. That is, the rope has been cut.

If we denote the acceleration of the two piston assembly as $a_p$ and the mass of the two piston assembly by $m_p$ then we can write down a formula:$$F_3 = F_{3bt} + F_{3tb} + F_{3bb} = m_p a_p$$if $m_p$ = 0 it follows that $F_3 = 0$.

[Assume that all forces use the same sign convention, e.g. up is positive]

Since Figures 3 and 4 are equivalent, I believe the value of F3 should not be zero.

We need to be clear. In neither figure 3 nor figure 4 is there any rope. So the logic above applies with full force. In both scenarios $F_3 = 0$.

Due to the 1-meter water level difference between the top opening and the M opening, installing a hand scale on the top piston will display a reading.

Do you know how a scale works? A scale works by restraining the motion of a test body and measuring the force required to do this. For instance, one stands still on a bathroom scale.

That is equivalent to re-fastening the rope to the piston assembly and measuring the tension in the rope while the pistons are held motionless by the rope.

Is that the measurement you are interested in?

 

[jbriggs444]

Yes, I am seeking an energy-efficient water pump design that draws water from the top as the piston moves up and down, while maintaining relatively constant gravitational force on the water within the pipe.

Assuming this goal is achieved and devices are installed at both ends of the balancing lever, the initial minimum gravitational force is preset to 100 N. When the water level rises to its peak, the maximum force applied can reach 110 N.

What is the stroke distance? That is, what is the height difference between the piston at the bottom of a stroke and the piston at the top?

From the drawings, it looks like about 0.1 meters. That means that you will be drawing up about 1 meter² times 0.1 meters = 0.1 m³ of water with each stroke. That is 100 liters. Or 100 kg.

You will be effectively drawing this 100 kg of water up over a vertical distance of 4 meters against a gravitational acceleration of 10 m/sec². So this is 4000 J of energy output from your mechanism over one cycle.

You plan to achieve this by applying a net force of 10 N over a stroke of 0.1 m. That is a 1 J of energy input to your mechanism over one cycle.

Unless you can clarify the arrangement, you have described a perpetual motion device. We do not discuss those here. See https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

 

[berkeman]

Thread closed temporarily for Moderation.

 

[berkeman]

Thread is reopened provisionally.

Yes, I am seeking an energy-efficient water pump design that draws water from the top as the piston moves up and down, while maintaining relatively constant gravitational force on the water within the pipe.

@vdance -- what numerical efficiency do you anticipate?

 

[vdance]

That is equivalent to re-fastening the rope to the piston assembly and measuring the tension in the rope while the pistons are held motionless by the rope.

Yes, that's exactly what I'm trying to solve. My English isn't very good, so I use translation software to communicate, but sometimes the software doesn't express things accurately or clearly.

 

[berkeman]

@vdance -- what numerical efficiency do you anticipate?

 

[vdance]

What is the stroke distance? That is, what is the height difference between the piston at the bottom of a stroke and the piston at the top?

I originally only intended to understand how F3 is calculated and did not consider the stroke issue. The piston can be fixed to the crankshaft, and the crankshaft's advertised diameter can be 0.1 meters, 0.5 meters, or 0.9 meters—all are acceptable. However, based on your analysis, this configuration cannot achieve my objective. Therefore, I will no longer pursue further discussion on the stroke issue.

You plan to achieve this by applying a net force of 10 N over a stroke of 0.1 m. That is a 1 J of energy input to your mechanism over one cycle.

Unless you can clarify the arrangement, you have described a perpetual motion device. We do not discuss those here. See https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

What I'm seeking here is an energy-efficient pumping mechanism. The core challenge lies in extracting water from the top while maintaining the gravitational stability of the water body. A single structure cannot possibly lift water that high with just 10N of force. This structure must be installed in pairs on a crankshaft to achieve the potential for 10N pumping.
We all know perpetual motion machines are impossible—this is an indisputable conclusion.
Our discussions are grounded in physical analysis and calculations, exploring the feasibility of certain technologies. Should any approach violate the principles of physics, we will halt it immediately.
Even if I discovered this technology and achieved a 10-Newton pumping effect using a crankshaft mechanism, such a device still could not be called a perpetual motion machine. This is because a perpetual motion machine must operate without relying on external energy input. In this pumping device, the gravitational force acting on the water is always influenced by Earth's gravity. If placed in the space environment outside Earth's weak gravitational field, the water would become suspended, causing the device to fail. This violates the core requirement of perpetual motion machines: “operation without external energy input.”
Recently, I conceived another model and believe I have found a feasible solution to achieve “water extraction from the top while maintaining stable hydrostatic pressure conditions.” If you are interested, we can discuss this privately. I am not yet ready to disclose this technology publicly.

 

[vdance]

Thread is reopened provisionally.

@vdance -- what numerical efficiency do you anticipate?

The effect I aim to achieve is to continuously extract water from the top using a force exceeding 10N, leveraging the action of a 180° phase crankshaft. This is precisely why I have persistently formulated hypotheses and calculated them based on physical formulas. Such an outcome is feasible and does not violate forum regulations or the laws of physics.

 

[berkeman]

The effect I aim to achieve is to continuously extract water from the top using a force exceeding 10N, leveraging the action of a 180° phase crankshaft. This is precisely why I have persistently formulated hypotheses and calculated them based on physical formulas. Such an outcome is feasible and does not violate forum regulations or the laws of physics.

You did not answer my question; short leash.

What efficiency are you shooting for? 80%, 85%? You will need to include the weight of your pistons and piston wall friction in order to get closer to the real efficiency. Please post a number in your next reply, or this thread will be permanently closed as a PMM discussion on your part.

 

[vdance]

You did not answer my question; short leash.

What efficiency are you shooting for? 80%, 85%? You will need to include the weight of your pistons and piston wall friction in order to get closer to the real efficiency. Please post a number in your next reply, or this thread will be permanently closed as a PMM discussion on your part.

I have not yet considered the overall efficiency of the system, but according to the law of conservation of energy, a specific amount of energy must correspond to an equal amount of work. Therefore, taking friction resistance into account, to drive the crankshaft rotation and extract 10 newtons of water, the externally applied force must exceed 10 newtons. Currently, I am solely focused on determining the correct physical calculation method for this single component. This is analogous to the practice in solving general physics problems where the mass and friction forces of irrelevant components are ignored to simplify calculations. I am reviewing the solution from reply #32, but the content generated by the translation software is quite chaotic, making it difficult to fully grasp its meaning at this stage.

 

[vdance]

Yes. I think that everyone here understands that.

You want the total of:

1. The force of water pressure on the bottom surface of the top piston $F_{3bt}$ plus...
2. The force of water pressure on the top surface of the bottom piston $F_{3tb}$ plus...
3. The force of water pressure on the bottom surface of the bottom piston $F_{3bb}$

This with the understanding that the pressure of the ambient atmosphere on the top surface of the top piston is taken as zero [all pressures are gauge pressures] and that there are no other external vertical forces acting on the two pistons. That is, the rope has been cut.

If we denote the acceleration of the two piston assembly as $a_p$ and the mass of the two piston assembly by $m_p$ then we can write down a formula:$$F_3 = F_{3bt} + F_{3tb} + F_{3bb} = m_p a_p$$if $m_p$ = 0 it follows that $F_3 = 0$.

[Assume that all forces use the same sign convention, e.g. up is positive]

We need to be clear. In neither figure 3 nor figure 4 is there any rope. So the logic above applies with full force. In both scenarios $F_3 = 0$.

Do you know how a scale works? A scale works by restraining the motion of a test body and measuring the force required to do this. For instance, one stands still on a bathroom scale.

That is equivalent to re-fastening the rope to the piston assembly and measuring the tension in the rope while the pistons are held motionless by the rope.

Is that the measurement you are interested in?

Yes, you're absolutely right. With the help of AI, I understand what you mean now.
Thank you!

 

[vdance]

These structural designs represent hypothetical solutions I developed while pursuing the goal of “extracting water from the top while maintaining the water's gravitational stability.” I present them here for reference. I believe their calculation methods are fundamentally similar—understand one problem, and you can solve the others.

​

With everyone's help, I've now figured out the calculation method and achieved my goal using other models, so there's no need to continue discussing this issue.
Thank you all!

 

[berkeman]

We still have not figured out if you are trying to construct a PMM, but whatever. Either way this thread is now done.