Solving Hydraulic Systems with Pascal's Law - Joshua

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Discussion Overview

The discussion revolves around the application of Pascal's Law in designing a hydraulic system to lift water from a basement to an upper level using a hydraulic ram concept. Participants explore the necessary calculations and formulas to determine the required weight to generate sufficient pressure for the system.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Joshua expresses uncertainty about applying Pascal's Law to his hydraulic system design and seeks the correct formulas without wanting a direct solution.
  • One participant explains that pressure is transmitted equally in a liquid and provides the relationship between force and area (P1 = P2, F1/A1 = F2/A2) to help Joshua understand the calculations needed.
  • Joshua specifies a requirement for 80 psi output from a 0.5" pipe and asks for clarification on the necessary calculations for F1.
  • Another participant confirms that to achieve 80 psi, the inlet pipe size is needed to calculate A1, suggesting that F1 can be derived from the formula F1 = 80A1.
  • Joshua calculates A1 and finds that he would need 36,173 lbs on a 452 square inch plate to achieve the desired pressure, expressing concern over the excessive weight required.
  • A later reply agrees with Joshua's calculation but notes that he may be using Pascal's Law incorrectly, as it typically involves using a smaller force on a smaller area to lift a larger weight on a larger area.
  • Joshua expresses disappointment and seeks alternative methods to achieve his goal without using electricity or other energy sources.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the feasibility of Joshua's hydraulic system idea, with some agreeing on the calculations while others suggest that the approach may not be practical.

Contextual Notes

There are unresolved assumptions regarding the specific dimensions and configurations of the hydraulic system, as well as the implications of using Pascal's Law in this context.

Barstowrat
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I think the solution I need lies within Pascal's Law, but I'm not quite sure I'm doing it right. I'm not looking for the answer to the problem, that's no fun, but I at least need the right series of formulas.

I have a storage vessel in my basement with water in it. My plan is to basically make a hydraulic ram (in theory), by adding a sealed plate with heavy weights, pressurizing it into a smaller pipe, thus supplying water to the upstairs. How do I go about figuring out what weight it is that I need for this to supply ample pressure? Or is this a completely worthless idea, meaning that it would require the weight a school bus?

My biggest problems is that all I read about Pascal's Law is the exact opposite of what I want and the formulas never seem quite right... probably because all of science exists in metric, and I learned SAE. Again, not looking for a solution to the problem, but need the way to figure it out.

Thank you,
Joshua
 
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For Pascal's Law, the pressure is transmitted equally throughout a liquid. Take the case here:

variation_depth.jpg



Pressure at 1 = Pressure at 2

i.e. P1= P2 but you know that P = F/A

i.e. F1/A1= F2/A2

Based on your piping you will have the values for A1 and A2. F2 would be the weight of what you want to lift. F1 would be how much force you would need to apply.

You can simplify the formula to make F1 the subject to get the required force.


EDIT: I am also not sure if what you are asking is called "Pascal's Law" though.
 
Last edited:
I need F2 to be @80 psi out of .5" pipe. What formula would I use to figure out how much weight its required for F1? Am I missing a step or am I just dense? LOL
 
Barstowrat said:
I need F2 to be @80 psi out of .5" pipe. What formula would I use to figure out how much weight its required for F1? Am I missing a step or am I just dense? LOL


If you have P2 = 80 psi, then you would need to know the inlet pipe size (to get A1).

Then just F1 = 80A1 which give F1 in lbs.
 
A1=452.16sq" • 80psi = 36,172.8lbs

Is that right? Seems a little excessive.
 
Barstowrat said:
A1=452.16sq" • 80psi = 36,172.8lbs

Is that right? Seems a little excessive.

Your calculation is correct.. You would need 36,173 lbs on a 452 square inch plate to pressurize the fluid to 80 PSI. The reason it seems excessive is you're using Pascal's Law backwards. Usually you use a small force on a small surface area to lift a heavier weight on a large surface area.
 
Looks like I will have to abandon that idea. I really hoped it was going to work. Don't suppose anyone has a better idea to do this without using energy (electricity,etc)?
 

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