Calculating Inlet and Outlet Times: c=451.8, u=135.5

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Homework Help Overview

The discussion revolves around calculating the times for a wave to travel from a valve to the inlet and outlet in a pipe, given specific values for sound speed and flow velocity. The context involves understanding wave propagation in a fluid medium, specifically methane.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of travel times using the formulae based on the given parameters. Questions arise regarding the directionality of wave travel and the influence of flow velocity on the perceived speed of the wave.

Discussion Status

Participants have shared calculations and interpretations of the wave behavior in the context of the problem. Some have provided clarifications on the mechanics of wave propagation in relation to the flow of methane, while others express curiosity about the underlying assumptions and seek further understanding.

Contextual Notes

There is a mention of the wave starting at the valve, which is positioned in the middle of the pipe, leading to discussions about the distances involved and the effects of flow direction on wave speed. The participants are exploring the implications of these factors on their calculations.

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Homework Statement
Methane is transported along a 4km pipeline at a Mach number of 0.3 and temperature 30◦C. A valve positioned 2km from the inlet begins to close, creating a pressure wave up and down the pipe. Calculate the time it takes for the pressure wave to reach a) the inlet, and b) the outlet of the pipe. [for methane: Cp = 2.2537kJ/kgK, Cv = 1.7354kJ/kgK, R = 0.5182kJ/kgK
Relevant Equations
Methane is transported along a 4km pipeline at a Mach number of 0.3 and temperature 30◦C. A valve positioned 2km from the inlet begins to close, creating a pressure wave up and down the pipe. Calculate the time it takes for the pressure wave to reach a) the inlet, and b) the outlet of the pipe. [for methane: Cp = 2.2537kJ/kgK, Cv = 1.7354kJ/kgK, R = 0.5182kJ/kgK Relevant Equations: The answers given are 6.3s for inlet and 3.4s for outlet, which I was able to obtain by dividing 2000/c-u and 2000/c+u
c=451.8 i calculated and u=135.5...

Could someone explain why 2000m is used instead of length of pipe which is 4km.
Struggling to visualise this in operation. Thanks
The answers given are 6.3s for inlet and 3.4s for outlet, which I was able to obtain by dividing 2000/c-u and 2000/c+u
c=451.8 i calculated and u=135.5...
 
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The wave starts at the valve which is in the middle of the pipe (i.e., 2000m from the inlet and outlet).
 
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so the wave goes from valve to outlet in 3.4s and then hits a wall? and returns to inlet at 6.3s. wish i had a diagram of this
 
Two waves emanate from closing of the valve. One travels toward the inlet and one travels towards the outlet.
 
oh inlet and outlet 2km away from valve which is middle, so why is outlet quicker. is it because methane is flowing in that direction, giving it a push etc. thanks for reply
 
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The sound speed, c, of the methane is identical on both sides of the valve. Imagine if you were traveling with the methane on either side of the valve at velocity u. It would look like the wave was moving towards you with the speed c on both sides. Now go to the lab frame. The apparent velocity of the wave also needs to change to reflect this. You end up with c+u as it goes with the flow and c-u when you go against it.
 
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brilliant explanation.
 

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