# Bernoulli's Equation on water flow

1. Apr 21, 2006

### mb85

Water flows through a horizontal pipe and then out into the atmosphere at a speed v1 = 19 m/s. The diameters of the left and right sections of the pipe are 4.3 cm and 2.1 cm, respectively. (a) What volume of water flows into the atmosphere during a 16 min period? In the left section of the pipe, what are (b) the speed v2, and (c) the gauge pressure?

16 min = 960s

D1 = Area = 3.46x10^-4
D2 = Area = 1.45x10^-3

So for part A.
Rv = Av = (3.46x10^-4)(19)(960) = 6.32 m^3

For part B.
A1v1 = A2v2
(3.46x10^-4)(19) = (1.45x10^-3)v2
V2 = 4.53m/s

Im having problems with the gauge pressure:
Pgauge = Po - Patm
So for Po im using Burnoulli's Equation

P1 - P2 = rho(g)(Y1-Y2) + 1/2 (rho)(V2^2 - V1^2)
I assumed y to be constant and equal = 0 and density of water = 1000
so i had P1 -P2 = 1/2(1000)(19^2 - 4.53^2)
so i got Pgauge = 1.6855 - Patm

But i cant get it right. can somehow help me.

2. Apr 21, 2006

### Andrew Mason

It would help if you could explain the configuration of the pipe or provide a drawing. Which end is the water coming out at 19 m/s? Where is the pressure being measured?

AM