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.

 

[vdance]

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.

However, both structures also appear to be single-piston, and the area subjected to force remains unchanged (referring to the area of the red part), unless you can indicate another area subjected to force.

Figure 4
​

*Note: The red bidirectional arrow indicates the interaction between the water and the acting area S.

 

[vdance]

It has doubled - see the 2 areas indicated in yellow:
View attachment 365460
Minor edit.

In my understanding, the area of force application for a liquid is analogous to the point where the rope is tied to the support frame when swinging on a swing. I believe the area of force application shown in Figure 4 may be the correct one.
I'll need to think about your analysis some more before I can fully grasp it.

 

[vdance]

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.

I also believe the key point lies in correctly understanding the area subjected to force in this diagram.

 

[256bits]

I also believe the key point lies in correctly understanding the area subjected to force in this diagram.

In Fig 4, the water pressure acts upon the casing. The piston in the upper cylinder is not moving but acts as a backstop, so the water does not empty out.

.To see this, make a horizontal cut separating the 2 cylinders, so thar each can move independently, but keep them connected hydraulically with a seal, so they can slide one atop the other without breaking the water column.. A force of 50,000 N will need be applied to each cylinder to counteract the water pressure.

 

[Steve4Physics]

However, both structures also appear to be single-piston, and the area subjected to force remains unchanged (referring to the area of the red part), unless you can indicate another area subjected to force.

View attachment 365477
Figure 4
​

*Note: The red bidirectional arrow indicates the interaction between the water and the acting area S.

The key is to identify all the forces exerted by the water on the various internal surfaces of the double- piston.

Note that the small blue piston (top left) is assumed frictionless: it exerts a force on the water but no force on the double-piston.

Name the various surfaces of the double-piston 1 to 8, as shown. Surfaces 2, 3 and 4 form the right hand-face of the double-piston.

‘f1’ means the force of the water acting on the surface 1, ,etc.

f1 and f8 act in opposite directions: their sum = 0.
f7 and f5 act in opposite directions: their sum = 0.
f3 and f6 act in opposite directions: their sum = 0

f2 and f4 act in the same direction.

The net force of the water on the double-piston is therefore due to combined forces acting on surfaces 2 and 4 (shown yellow).

 

[Lnewqban]

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.

If you are lifting more volume of water with the same displacement of the pulling force; then, you will need to exert a greater value of force.

Work = Force x Distance

 

[vdance]

The key is to identify all the forces exerted by the water on the various internal surfaces of the double- piston.

Note that the small blue piston (top left) is assumed frictionless: it exerts a force on the water but no force on the double-piston.

Name the various surfaces of the double-piston 1 to 8, as shown. Surfaces 2, 3 and 4 form the right hand-face of the double-piston.
View attachment 365482
‘f1’ means the force of the water acting on the surface 1, ,etc.

f1 and f8 act in opposite directions: their sum = 0.
f7 and f5 act in opposite directions: their sum = 0.
f3 and f6 act in opposite directions: their sum = 0

f2 and f4 act in the same direction.

The net force of the water on the double-piston is therefore due to combined forces acting on surfaces 2 and 4 (shown yellow).

Your explanation seems quite easy to understand, and I accept your analysis.

 

[vdance]

Thank you all for your thoughtful analysis.