The rate of change of pressure

[sarahh]

URGENTThe rate of change of pressure

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.
dP1/dt=-dP2/dt, where dP1 is the pressure changes over a fixed interval of time (del t1) and dP2 is the pressure changes over another fixed interval of time (del t2) at x=0, and del t1 = del t2.

Thank you very much for your kind assistance.


Sarah

 

[JohnDubYa]

I have this feeling of deja vu that I cannot quite get over. I have no idea why. How odd.

 

[M98Ranger]

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Based on the information provided, it seems like your explanation is accurate. The conservation of momentum principle states that the rate of change of momentum in a system is equal to the external forces acting on the system. In this case, the pump is creating a positive rate of momentum-changing force per unit area, while the system is producing an equal negative rate. This balance results in a decrease in pumping power over time, until it reaches zero when t>4 minutes. It's important to note that this explanation assumes that there are no other external forces acting on the system, and that the measurements are accurate. Overall, your approach seems reasonable and logically sound.