How do inlet and outlet times relate to speed of sound and flow velocity?

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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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