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
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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
 
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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.
 
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