Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations

[vdance]

Homework Statement

Out of personal interest, I designed a piston-driven water pumping mechanism. However, an extension of this design has left me uncertain about its calculation method.

Relevant Equations

F=ρgsh

Figure 1 Overall Structure Diagram

Figure 2: Top view of the piston when it is cylindrical​

A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.

Figure 3: Modifying the structure to incorporate a fixed internal piston​

When I modify the piston structure by adding a fixed internal piston to achieve the configuration shown in Figure 3, does its tensile force remain F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N?

When the pulling force moves the piston an equal distance to the right, the water intake in Figure 3 is twice that of Figure 2.

 

[haruspex]

Homework Statement: Out of personal interest, I designed a piston-driven water pumping mechanism. However, an extension of this design has left me uncertain about its calculation method.
Relevant Equations: F=ρgsh

View attachment 365450
Figure 1 Overall Structure Diagram

View attachment 365451
Figure 2: Top view of the piston when it is cylindrical​

A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.

View attachment 365452
Figure 3: Modifying the structure to incorporate a fixed internal piston​

When I modify the piston structure by adding a fixed internal piston to achieve the configuration shown in Figure 3, does its tensile force remain F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N?

When the pulling force moves the piston an equal distance to the right, the water intake in Figure 3 is twice that of Figure 2.

The lower portion is not raising any water, only moving it sideways.

 

[vdance]

The lower portion is not raising any water, only moving it sideways.

No, water is indeed drawn upward; the dark green areas in the diagram indicate the increased water volume at the top.

 

[haruspex]

No, water is indeed drawn upward; the dark green areas in the diagram indicate the increased water volume at the top.

Then I misinterpreted your last diagram. I now presume it is a top view. That being so, a given movement of the piston raises twice the water than originally, so the force must be doubled.

 

[vdance]

Then I misinterpreted your last diagram. I now presume it is a top view. That being so, a given movement of the piston raises twice the water than originally, so the force must be doubled.

I also believe the required force is twice the original amount, but the terminal force-bearing area appears to remain constant at s=1 m², h=5m unchanged, with ρ and g also unchanged. Therefore, I am uncertain how to correctly calculate this value.

 

[Steve4Physics]

When I modify the piston structure by adding a fixed internal piston to achieve the configuration shown in Figure 3, does its tensile force remain F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N?

If I understand correctly, 's' is the piston's effective surface area.

By going from the single-piston to the double-piston, you have doubled this area. You need to take this into account.

 

[256bits]

I also believe the required force is twice the original amount, but the terminal force-bearing area appears to remain constant at s=1 m², h=5m unchanged, with ρ and g also unchanged. Therefore, I am uncertain how to correctly calculate this value.

The whole pump casing is moving outward. The second design has double the casing area from the first design. The pulling force is doubled. The work is also doubled. W=Fd.

 

[Steve4Physics]

.. but the terminal force-bearing area appears to remain constant at s=1 m², h=5m unchanged ...

It has doubled - see the 2 areas indicated in yellow:

Minor edit.