Register to reply 
The rate of change of pressure 
Share this thread: 
#1
Jul2004, 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=015cm=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 momentumchanging force per unit area, dP1/dt, produced by the pump is balanced by an equivalent negative rate of momentumchanging 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 


#2
Jul2004, 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.



Register to reply 
Related Discussions  
Related Rate finding the Rate of Change of an Angle  Calculus & Beyond Homework  5  
Measuring rate of reaction via rate of temperature change  Chemistry  3  
Rate of change of pressure  gases  Classical Physics  9  
Rate of change of pressure  General Physics  5  
The rate of change of pressure (Urgent, please help)  Introductory Physics Homework  0 