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The rate of change of pressure

by sarahh
Tags: pressure, rate, urgentthe
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sarahh
#1
Jul20-04, 04:12 AM
P: 5
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
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JohnDubYa
#2
Jul20-04, 04:35 AM
P: 1,322
I have this feeling of deja vu that I cannot quite get over. I have no idea why. How odd.


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