Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Motion In Fluid. Help Me Solve This Question.?
At a certain point in a pipeline the velocity is 1.00ms^-1 and the gauge pressure is 3.0x10^5 Pa. Find the gauge pressure at a second point in the line 20.0m lower than the first, if the cross section at the second point is one-half that at the first. The liquid in the pipe is water.
1 Answer
- ?Lv 71 decade agoFavorite Answer
Again, from Bernouilli's equation along a streamline,
p + 1/2 rho v^2 + rho g h = constant
Because water is incompressible to a high degree, the flow speed at half the cross section is twice the flow speed upstream (I assume you do not mean half the diameter, in that cause the cross section is one-fourth and the speed 4 times as high). So we must have the equality:
3.0 10^5 Pa + 1/2 * 10^3 kg/m^2 * (1.00 m/s)^2 + 10^3 kg/m^3 * 9.81 m/s^2 * 10 m =
p + 1/2 * 10^3 kg/m^3 * (2.00 m/s)^2 + 0
Hence
p = 3.0 * 10^5 Pa + 1/2 10^3 kg/m^3 * ( 1.00 m^2/s^2 - 4.00 m^2/s^2) + 9.81 *10^4 Pa
= 3.97 10^5 Pa