# Homework Help: Differences in Pressure

1. Dec 20, 2009

### yankees26an

1. The problem statement, all variables and given/known data
Water is flowing through a horizontal pipe that has a radius of 6 cm gradually narrowing to a
radius of 4 cm. The volume flow rate is 0.008 m3/s. What is the difference in the water pressurebetween the 6 cm radius region and the 4 cm radius region of the pipe?

2. Relevant equations

P + .5pv^2 + pgy = constant

3. The attempt at a solution

y is 0 so that part cancels

p(density) is 1000 kg/m^3 for water

P1 + .5pv1^2 = P2 + .5pv2^2

Not sure where to go from there

2. Dec 20, 2009

### HallsofIvy

Since water can't "build up" any where in the pipe, the same volume of water must flow through both 6 cm radius and 4 cm radius portions in the pipe at the same time. Since the area of of the 4 cm radius portion is smaller than the 6 cm radius portion, the water must flow faster through the 4 cm portion to do that. That is, it must accelerate. How much faster must it go? What force is necessary to product that acceleration? That force, divided by the area is the pressure difference.

3. Dec 20, 2009

### yankees26an

Any hints?

choice 1: pressure?

If so, how?

Last edited: Dec 20, 2009
4. Dec 21, 2009

### rl.bhat

The volume flow rate Q = A1*v1 = A2*v2.
from the radii of the pipes you can find A1 and A2. From that you can find v1 and v2. Substitute these values in the relevant equation to find P1 - P2.