- #1
hongiddong
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Homework Statement
A pump forces water at a constant flow rate through a pipe whose cross-sectional area, A, gradually decreases. At the exit point, A has decreased to 1/3 its value at the beginning of the pipe. If y= 60 cm(the distance from point 1 to the exit, where point 1 is where y=0) and the flow speed of the water just after it leaves the pump, which is point 1, is 1m/s what is the gauge pressure at point 1?
I have the solution, but I don't know why a certain part of the solution is the way it is. More specifically, why is P at point 1 - Patm = to the gauge pressure. I know that the gauge pressure= pressure total - pressure at surface.
Homework Equations
Here is the solution.
P1 + 1/2density*v1^2 = Patm + 1/2density*v2^2+density*gravity*y2(60cm)
Then for some reason, which I do not know why P1 is the total pressure and we subtract
P1-Patm = density*gravity*y2+1/2density*(3v1)^2-1/2density*v1^2
The answer for the gauge pressure is 10^4 Pascals.
Thanks Physics Forums!