Pressure change in pipe due to sudden closure of valve

AI Thread Summary
The discussion focuses on the dynamics of pressure changes in a pipe due to the sudden closure of a valve, leading to the propagation of a pressure wave. It emphasizes the importance of using a moving frame of reference to accurately describe the velocities of water particles, as simply equating the mass flow rates with positive and negative velocities is incorrect. The conservation of mass is applied to relate the positions and velocities of water particles before and after the valve closure. Key equations are derived to express the relationships between the initial and final states of the water flow. Understanding these principles is crucial for solving the problem of pressure change in the pipe.
phantomvommand
Messages
287
Reaction score
39
Homework Statement
Please see the attached photo
Relevant Equations
Conservation of Mass
Impulse forumla
Screenshot 2021-07-30 at 2.24.09 AM.png

Water is flowing in the pipe with velocity v0. Upon sudden closure of the valve at T, a pressure wave travels in the -ve x direction with speed c. The task is to find ##\alpha##, where ##\Delta P = \rho_0 c (\Delta v) \alpha##.
The 1st step is to set up an equation using conservation of mass. (picture is below)
I do not understand why we need to use a moving frame c to find the speed of the water particles. Why is simply stating ##\rho_0 v_0 = \rho_1 v_1## incorrect?

Screenshot 2021-07-30 at 2.26.05 AM.png
 
Physics news on Phys.org
phantomvommand said:
Why is simply stating ρ0v0=ρ1v1 incorrect?
One obvious reason is that v0 is positive and v1 is negative, so it cannot be true.

Consider positions x, y, x', y' in the pipe, where,
at time t, the wave front is at x
at time t+dt, the wave front is at y'
at time t+dt, the water that had been at x has moved to y
the water that is at y' at t+dt had been at y at time t
Draw a picture or two.
So the water that was between x and y at time t is between x' and y' at t+dt.
By mass conservation, ##(x-y)\rho_0=(x'-y')\rho_1##.
Also ##x'-x=v_1 \delta t##, ##y'-y=v_0 \delta t## and ##y'-x=c \delta t##.

I leave the working to you.
 
  • Like
Likes Lnewqban and phantomvommand
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top