How Does Conservation of Momentum Explain Pressure Changes in a Pipe System?

[sarahh]

Our problem is that we measured the rate of change of pressure of a liquid at different length of a pipe, for example, x=0, x=5cm, ... etc, caused by a pump at x=0-15cm=-15cm and got a result that at x=0, dP1/dt1 = -dP2/dt2, where dP1 is the pressure difference over a fixed interval, del t1, and dP2 is the pressure difference over a fixed interval, del t2,
i.e. -----------
- -
- -
del t1 |3 minutes | del t2
(just like a trapezium without the bottom part), and del t1=del t2. Pumping power is decreasing from t=0 to t=4minutes and pumping power =0 when t>4 minutes.
Is it accurate if we try to explain this observation as follows:
Due to conservation of momentum, the rate of momentum-changing force per unit area, dP1/dt, produced by the pump is balanced by an equivalent negative rate of momentum-changing force per unit area, -dP2/dt produced by the system after 3 minutes at x=0.

Thank you very much for your kind assistance.

Sarah

 

[M98Ranger]

Yes, it is accurate to explain this observation using the concept of conservation of momentum. The rate of change of pressure, dP1/dt, represents the rate at which momentum is being added to the system by the pump. However, after 3 minutes, the system has reached a state of equilibrium where the pressure difference, dP2, is equal and opposite to the pressure difference created by the pump. This means that the rate of momentum-changing force per unit area, -dP2/dt, is balancing out the rate of momentum-changing force produced by the pump, dP1/dt. This is in line with the principle of conservation of momentum, which states that in a closed system, the total momentum remains constant. Therefore, your explanation is accurate.