# Homework Help: Reducing pressure in a pipe

1. Nov 1, 2007

### Vanessa23

1. The problem statement, all variables and given/known data
The pressure in a section of horizontal pipe with a diameter of 1.80 cm is 144 kPa. Water flows through the pipe at 2.30 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should the diameter of the constricted section be?

2. Relevant equations
p1 + 1/2pv^2 = p2 + 1/2pv^2

A1V1=A2V2

A=pi*r^2

3. The attempt at a solution
convert .0023 m^3 to velocity
.0023/((.018^2)*pi) = velocity is 2.26

144*(10^3) +.5(1000)(2.26)^2=101*(10^3)+.5*(1000)v^2
v=9.54 m/s

(.018^2)*pi*.0023=9.54*A
A=2.45x10^-7 m^2

2.54x10^-7=pi*r^2
r=2.79x10^-4
x2--> diameter=5.59x10^2 cm

2. Nov 4, 2007

### Vanessa23

The answer is supposed to be 1.50 cm. I seem to only get answers 10^-2. Thanks for any help!

3. Jun 18, 2008

### Anony-mouse

I'm having the same problem. I understand that Pascals is in units of N/m^2, but because the velocity is in m^3/s, rather than m/s, how can you use Bernoulli's equation? The units don't work.

4. Jun 18, 2008

### Astronuc

Staff Emeritus
Go back here [ (.018^2)*pi*.0023=9.54*A ] and realize that 0.018 m is the diameter.

Remember the area of the circle is given by $\pi\,r^2\,=\,\pi{d^2}/4$.

Make sure the units are consistent and that one does not mix SI with cgs.

The flow is given in volumetric flow rate Q, which is the product of the cross-sectional area and the mean flow velocity perpendicular to the cross-sectional area.

And realize the OP is Nov-07, so that the HW is long turned in.