Find the gauge pressure at a second point in the line

Click For Summary

Discussion Overview

The discussion revolves around a problem involving fluid dynamics in a horizontal pipeline, specifically focusing on finding the gauge pressure at a second point in the line given the speed and pressure at the first point. Participants explore the application of Bernoulli's equation and the continuity equation in this context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses difficulty in solving the problem and questions how to find the area at point 1 without knowing the radius.
  • Another participant suggests using Bernoulli's equation instead of the pressure-area relationship, noting that doubling the area results in halving the flow velocity due to the incompressibility of water.
  • A third participant confirms that the pressures at points 1 and 2 are gauge pressures and that the heights at both points are zero.
  • Another participant reiterates the principle of conservation of mass, indicating that the density of water remains constant, which supports the continuity equation.
  • One participant recommends using a combination of the continuity equation and Bernoulli's equation to solve the problem.

Areas of Agreement / Disagreement

Participants generally agree on the relevance of Bernoulli's equation and the continuity principle, but there is no consensus on the specific approach to finding the gauge pressure at the second point, as some participants focus on different aspects of the problem.

Contextual Notes

There are unresolved assumptions regarding the specific values needed for calculations, such as the radius of the pipe, which may affect the application of the equations discussed.

CollectiveRocker
Messages
137
Reaction score
0
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>
 
Physics news on Phys.org
CollectiveRocker said:
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>

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.


http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html
 
So are P1 & P2 = gauge pressure, y1 = 0, y2 = 0.
 
rho * Area * Velocity = constant.. in your case rho 1 = rho 2 since density of water does not change.. so there you go...
 
Yes, a combination of Continuity, and Bernoulli, should get you there.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
13K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K