Safety Valve/Pressure System question

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Homework Help Overview

The discussion revolves around a pressure system involving a valve with a specific discharge hole diameter and a spring with defined characteristics. Participants are tasked with determining the pressure at which the valve will open, utilizing relevant equations related to force and area.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of pressure and force equations, questioning the use of the spring's compressed length versus its free length. There is also consideration of whether additional information provided in a diagram is necessary for solving the problem.

Discussion Status

Some participants have provided insights into the correct application of Hook's Law and the relevance of the spring's length. Others have raised questions about the necessity of certain data in the problem setup, indicating a productive exploration of the topic.

Contextual Notes

Participants note potential extraneous information in the problem, such as dimensions of the tank and wall thickness, which may not be directly relevant to the solution.

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Homework Statement


A valve for a pressure system has a discharge hole whose diameter is 50mm. The spring has a free length of 250mm (if you stretched the spring out), a coiled length (measured top to bottom while spring is coiled) of 207mm and a spring constant of 120kN/m. At what pressure will the valve open?


Homework Equations


Not 100% but I think
a) p = F/A
b) F = kx

The Attempt at a Solution


Area of discharge hole = ∏r2 = 1.963x10-3m2
Force exerted by spring to keep hole closed = length of spring x spring constant = .207m x 120kN/m = 24.84kN
Therefore valve open pressure = 24.84kN/1.963x10-3m2 = 12.65 kPa.

How does that look? There is a diagram with some other info attached but I'm not sure it's relevant.

Thanks!
 

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atay5510 said:

Homework Statement


A valve for a pressure system has a discharge hole whose diameter is 50mm. The spring has a free length of 250mm (if you stretched the spring out), a coiled length (measured top to bottom while spring is coiled) of 207mm and a spring constant of 120kN/m. At what pressure will the valve open?


Homework Equations


Not 100% but I think
a) p = F/A
b) F = kx

The Attempt at a Solution


Area of discharge hole = ∏r2 = 1.963x10-3m2
Force exerted by spring to keep hole closed = length of spring x spring constant = .207m x 120kN/m = 24.84kN
Therefore valve open pressure = 24.84kN/1.963x10-3m2 = 12.65 kPa.

How does that look? There is a diagram with some other info attached but I'm not sure it's relevant.

Thanks!

Hook's Law requires the change in length of the spring from its equilibrium point (relaxed condition). You've used the compressed length only. What if the spring were compressed to nearly zero length? Would the force then be nearly zero? :wink:
 
Aha! So the resistance force due to the spring is F = -kx where x = 250mm - 207mm and k is as given.
I feel like they are giving me a little more info in the picture than is necessary. Specifically the 400mm diameter of the tank (not the discharge hole) and the 25mm thickness of the walls. My solution doesn't involve these numbers. Is that correct?
Thanks for the help :)
 
atay5510 said:
Aha! So the resistance force due to the spring is F = -kx where x = 250mm - 207mm and k is as given.
I feel like they are giving me a little more info in the picture than is necessary. Specifically the 400mm diameter of the tank (not the discharge hole) and the 25mm thickness of the walls. My solution doesn't involve these numbers. Is that correct?
Thanks for the help :)

Yes, sometimes you'll find extraneous information given, particularly on diagrams. This may be because: It's placed there to confuse you; To get you used to picking the required data out of actual engineering diagrams; Simply because the diagram originally came from some other question entirely. Pick one or more :smile:
 

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