Calculate water pressure in tube end when tube is compressed

AI Thread Summary
The discussion centers on a DIY project involving a bed supported by a water-filled tube with a liquid pressure sensor. The user seeks to calculate the pressure on the sensor due to a total weight of 1600N distributed across the tube. Concerns are raised about the tube's potential to flatten under pressure and the stability of the setup, suggesting that using pistons might be a better alternative. Additionally, the low compressibility of water is noted, but the elastic properties of the tube material and potential issues like leaks and thermal effects are highlighted as significant factors. Ultimately, the advice leans towards using more robust sensors or alternative designs to ensure reliability.
johboh
Messages
2
Reaction score
0
Hello!

Background
I have a hobby in creating an automated home, controlling lights, reading temperatures, and so on from self build sensors and actuators. For a while I have had force sensing resistors under each leg of my bed so I can messaure my own weight every day, sleeping patterns and so on. This has been working quite good for a while, but the sensors does not stand the constant pressure and fails. I need a new solution.

Setup
Instead of a sensor under each leg, the bed will fully stand on a plastic tube filled with water. The tube will have a liquid pressure sensor in one end and a stop in the other end.

Problem
I need to know the pressure on the sensor so I can buy a suitable sensor for that working range.

Data
  • The tube will have a diameter of about 10mm.
  • The length of the tube will be around 10m to go around and under all legs of the bed.
  • The pressure in total will be around (60+70+30) = 1600N.
  • The bed stands only on the tube and the tube will be filled with water.

Question
In Bar, what will be the pressure applied on the sensor?
 
Last edited:
Physics news on Phys.org
Forget the idea!
You would have equip all the legs not with tubes (which get flattened under pressure), but rather with pistons with well knows area. Other problem is that such suspension for your bed is not stable - if the weight is not distributed centrally, some legs would totally collapse, while other stay lifted.
Better think about more robust tensometers.
 
If the tube is filled with water I think that this will not happen because water have a very low compressibility, right?
 
Water has low compressibility, but the pipe must be ellastic (not quite predictably for plastic/rubber pipes), it suffers from ageing, water loss due to leaks and vapourisation, then you have thermal effects, and lots of others absolutely unpredictable.

Believe me - that is the worst possible tensometer you could design ;)

You can't beat the commercially available ones - just find a more robust one or think about proper shielding. I guess your tensometer got damaged not because of long term pressure (they are designed to withstand it), but because of lateral moves of bed legs, scratching it. Just think about shielding envelope made of hardened steel.

Just try your design to see its flaws - install such a tube without any pressure sensor and watch how it behaves as someone sits on your bed. Or as she rolls from one side to side.
 
Last edited:
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top