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Fluid Mechanics Question

  1. Oct 3, 2004 #1
    I have been going over this problem for at least 3 hours now, without any sign of improvement. I was wondering if you guys could at least point me in the direction of something I missed. Problem: At a certain point in a horizontal pipeline, the water's speed is 2.50 m/s and the gauge pressure is 1.80 * 10^5 Pa. Find the gauge pressure at a second point in the line if the crosssection area at the second point is twice that of the first. Now I realize the critical part of this problem is finding the area, thus Pressure = Force/Area. But how do you find the area at point 1 without knowing the radius? Is there another formula for area which i don't know>
  2. jcsd
  3. Oct 3, 2004 #2


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    This looks like a job for Bernoulli's eq. to me, not pressure = force/area. The trick is that you need to know that when you double the area, you halve the flow velocity. This happens because water is incompressible, so velocity*area = constant. So know you know the pressure and velocity at point 1, and you can figure out the velocity at point 2, thus you can compute the pressure with Bernoulli's eq.

  4. Oct 3, 2004 #3
    So are P1 & P2 = gauge pressure, y1 = 0, y2 = 0.
  5. Oct 4, 2004 #4
    rho * Area * Velocity = constant.. in your case rho 1 = rho 2 since density of water does not change.. so there you go....
  6. Oct 4, 2004 #5


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    Yes, a combination of Continuity, and Bernoulli, should get you there.
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