# Fountain's pump pressure

1. Apr 16, 2008

### Zacthor

A fountain designed to spray a column of water 11.5 m into the air has a 1.34-cm-diameter nozzle at ground level. The water pump is 2.75 m below the ground. The pipe to the nozzle has a diameter of 2.68 cm. Find the pump pressure necessary if the fountain is to operate as designed. (Assume laminar nonviscous steady-state flow).

I used .5mv^2=mgh to find the velocity of the water coming out of the nozzle. Then found the volume flow rate (Av) and then found the velocity for the water flowing through the pipe. Ive tried just solving for the pressure in the tube by using .5(density of water)v^2. that didnt work so i tried finding the volume of water in the pipe solving for its mass with (volume)(densityofwater) then finding the wieght of the water then solving for the pressure of that on the pump and adding it to what i found before with the velocitys. Can someone PLEASE tell me what im doing wrong?

2. Apr 16, 2008

### alphysicist

Hi Zacthor,

You probably got the expression .5(density of water)v^2 from Bernoulli's equation. That equation accomplishes in one equation what you are splitting up into several steps. Bernoulli's equation is closely related to the conservation of energy in that you must pick two points along the water stream to apply it to.

Obviously one point will have to be at the pump, since that's where you need to find the pressure. But where would you put the other point? You actually have a couple of good alternatives.

(I think once you compare Bernoulli's equation with what you did before you'll find that you mixed the values from several points on one side of the equation and also perhaps left out a known pressure.)

3. Jun 18, 2008

### Anony-mouse

I'm confused on this problem:
v1=velocity at the pump
v2=velocity at the nozzle point on the ground

velocity at the nozzle: 1/2m(v2)^2=mgh..... so v2 = 15.01 m/s
A2 = pi r^2 = pi (0.0134/2)^2 = 1.41 x 10^-4 m
A1 = pi r^2 = pi (0.0268/2)^2 = 5.64 x 10^-4 m
(v2xa2)/a1 = v1 = 3.75 m/s

P1= pressure needed at pump
h = -2.75m
P1 + pgh + 1/2p(v1)^2 = 1/2 p(v2)^2
P1 = 500(15.01^2) + 9800(2.75) - 500(3.75^2) = 133 kPa

I'm confused as to where my math went wrong, since that is not the right answer. Thanks

EDIT: I just realized it has something to do with my Bernoulli's equation, but I can't figure out why I can't make it so that the value is constant on both sides.

Last edited: Jun 18, 2008
4. Jun 18, 2008

### alphysicist

Hi Anony-mouse,

Bernoulli's equation is:

$$P_1 + \rho g h_1 + \frac{1}{2}\rho v_1 ^2 = P_2 + \rho g h_2 + \frac{1}{2}\rho v_2 ^2$$

for any two points along a flowline. It looks like you done almost everything right, but I think you are missing one of those terms in your calculation. Do you see what it is?

5. Jun 18, 2008

### Anony-mouse

Not really. The only two terms I don't have are P2 and pg(h2), but because I have the point at the nozzle, the h2=0. Thus, the only term left is P2, but I don't really understand how that works.

6. Jun 18, 2008

### alphysicist

That's a very important thing that comes up a lot: anytime the fluid flow is touching the air, it is at air pressure.

That's why when you used your first equation in post#3:

1/2m(v2)^2=mgh

you did not need to include the pressure. At those two points (at the nozzle and at the highest point of the water stream) the water flow is in air, and so the fluid is at the same atmospheric pressure at both points. Therefore the pressure cancels from both sides of the equation (and so Bernoulli's equation just gives you the 'regular' energy equation that you used).

7. Jun 18, 2008

### Anony-mouse

Wow, thank you, that helped a lot. Here is the correct answer:

P2 = 101325 Pa
P1 + pgh + 1/2p(v1)^2 = 1/2 p(v2)^2 + P2
P1 = 500(15.01^2) + 9800(2.75) - 500(3.75^2) + 101325 = 234 kPa