Question

A fountain sends water to a height of 150 meters. What is the difference between the pressure of the water system and the atmospheric pressure?

Answer 1

How can we find the pressure with only one piece of data, the height of the fountain? We must make some crazy assumptions.

Bernoulli equation

Lets call point 1 the water before it leaves the fountain jet, and point 2 is the peak of the fountain water.

(P1/gamma) + (V1^2/2g) + z1 = (P2/gamma) + (V2^2/2g) + z2

z2 = 150 m
z1 = 0 m
V2 = o m/s at the peak

So we get

(P1/gamma) + (V1^2/2g) = (P2/gamma) + z2

There is no way to solve this because there is one equation and two unknowns. There may be another equation that relates the pressure and velocity of the water at point one, but I don't know one.

I believe that they want you to call V1 = 0 m/s before it leaves the nozzle. Of course the water is moving in the pipe before it exits the nozzle, but maybe they are trying to simplify the problem.

If you use V1 = 0 m/s:

(P1/gamma) = (P2/gamma) + z2

(P1/gamma) - (P2/gamma) = z2

P1 - P2 = z2 * gamma

and gamma = density times gravity

P1 - P2 = 150 m * 1000 kg/m^3 * 9.81 m/s^2

P1 - P2 = 1471.5 kPa

Notice that this is similar to the static water pressure at a certain depth, P = density * gravity * depth.