Bernoulli's Equation on water flow

[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 I am 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 can't get it right. can somehow help me.

 

[Andrew Mason]

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?
...

But i can't get it right. can somehow help me.

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

 

[kyle8921]

Sure, let's work through it together. First, let's define our variables:
P1 = pressure at the left section of the pipe
P2 = pressure at the right section of the pipe
ρ = density of water (1000 kg/m^3)
g = acceleration due to gravity (9.8 m/s^2)
Y1 = height of the water at the left section of the pipe (we can assume this to be 0)
Y2 = height of the water at the right section of the pipe (also 0 in this case)
V1 = velocity of water at the left section of the pipe (19 m/s)
V2 = velocity of water at the right section of the pipe (4.53 m/s)

Now, let's plug these values into Bernoulli's equation:
P1 + 1/2ρV1^2 + ρgy1 = P2 + 1/2ρV2^2 + ρgy2

Since Y1 and Y2 are both 0, they cancel out and we are left with:
P1 + 1/2ρV1^2 = P2 + 1/2ρV2^2

We can rearrange this equation to solve for P2:
P2 = P1 + 1/2ρV1^2 - 1/2ρV2^2

Now, let's plug in our values:
P2 = (1.6855 x 10^5 Pa) + (1/2 x 1000 kg/m^3 x (19 m/s)^2) - (1/2 x 1000 kg/m^3 x (4.53 m/s)^2)

P2 = 1.6855 x 10^5 Pa + 1/2 x 1000 kg/m^3 x 361 m^2/s^2 - 1/2 x 1000 kg/m^3 x 20.52 m^2/s^2

P2 = 1.6855 x 10^5 Pa + 180500 Pa - 10260 Pa

P2 = 1.875 x 10^5 Pa

Now, to find the gauge pressure, we need to subtract the atmospheric pressure (Patm) from P2:
Pgauge = P2 - Patm

So, if we assume atmospheric