# Find the gauge pressure at a second point in the line

1. Oct 3, 2004

### CollectiveRocker

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. Oct 3, 2004

### pervect

Staff Emeritus
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

3. Oct 3, 2004

### CollectiveRocker

So are P1 & P2 = gauge pressure, y1 = 0, y2 = 0.

4. Oct 4, 2004

### Gopi Prashanth

rho * Area * Velocity = constant.. in your case rho 1 = rho 2 since density of water does not change.. so there you go....

5. Oct 4, 2004

### Gokul43201

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