# Bernoulli's law for water coming out of a faucet

1. Jan 30, 2012

### smeiste

1. The problem statement, all variables and given/known data

Water is supplied to a building through a pipe of radius R = 3.0cm. In the building, a faucet tap of radius r = 1.0cm is located 2.0m above the entering pipe. When the faucet is fully open, it allows us to fill a 25-L bucket in 0.5 minutes. What is gauge pressure in the pipe entering the building? Assume no other faucets are opened during the experiment.

2. Relevant equations

ΔV/Δt = A*v
A1v1 = A2v2
p1 + 1/2ρ(v1^2) = p2 + 1/2ρ(v2^2) (Bernoulli's law)
pgauge = p - patm

3. The attempt at a solution

Using the first two equations I have determined that v1 = 0.3 m/s and v2 = 2.7 m/s. Now I am unsure how to apply Bernoulli's law when I am missing one of the pressures. When I try to use atmospheric pressure (101325 Pa) as one of the pressures, I do not get the correct gauge pressure, which is 23 kPa. I really just need a push in the right direction! Thank you.

2. Jan 30, 2012

I would use atm pressure as the value for the pressure leaving the faucet. Since it hits the air as it exits. Is that what you attempted? Also remember that Bernoulli's equation will give you the absolute pressure. Did you subtract the atmospheric pressure from that to get the gauge pressure?

The velocity as it enters at the bottom is v1=.3 m/s and the velocity as it exits the faucet is v2=2.65 m/s; I get the same thing...

h1=0
h2=2m

Last edited: Jan 30, 2012
3. Jan 30, 2012

Your form of Bernoulli's EQ seems to be missing a gravitational potential term. This is where the 2 meter height comes into play.

4. Jan 30, 2012

Since the pressure coming out of the faucet is just atmospheric you can just solve for the difference P_1-P_2 that's the gauge pressure. That's what I did and I got the 23kPa. Make sure you take the height into account too; as saladsamurai pointed out you left those PE terms out of your equation.

I'm actually preparing for an exam on this stuff myself.

5. Jan 31, 2012

### smeiste

okay, now I see. so patm + ρgh + 1/2ρ(v1^2) = p2 + 1/2ρ(v2^2)? Would that be the formula?

6. Jan 31, 2012

### smeiste

haha yes it is. Thank you! I think that form of the equation is a combination of Bernoulli's law and Pascal's Law

7. Jan 31, 2012

NO! It is not. (Unless you already did some simplifications. Read on just to make sure you understand the equation properly.)

The correct equation is:

$$\frac{v^2}{2} + gz + \frac{p}{\rho} = \text{constant}\qquad(1)$$

This is a statement of the Conservation of Energy. It say that for an incompressible, inviscid flow,the total energy at a point in the fluid (along a stream line) is the same every where. Meaning that the energy at point 1 along a streamline is the same as at point 2 along the streamline. The form of (1) that you would use is:

$$\frac{v_1^2}{2} + gz_1 + \frac{p_1}{\rho}= \frac{v_2^2}{2} + gz_2 + \frac{p_2}{\rho}\qquad(2)$$

where z1 and z2 are the elevation of the fluid at each point. Though you were not given the elevation of each point, you were given the difference which is all that you need since you can group the terms in (2) in such a way that there will be a (z2-z1) term. And you know that z2-z1 = 2 meters.

8. Jan 31, 2012

### smeiste

Okay, thank you for the formula. It's not something that has ever been introduced in class though. I will bring it up with my professor. Thanks!

9. Jan 31, 2012