Your town is installing a fountain in the main square. If the water is to rise 20 m (65.6 feet) above the fountain, how much pressure must the water have as it moves slowly toward the nozzle that sprays it up into the air? Assume atmospheric pressure equal to 100,000 Pa.
For water to move out of the nozzle, it must have pressure high enough to over come the atmospheric pressure(100000 pa). If the atmospheric pressure will be high compared to water pressure at the nozzle then the there will be no water fountain.
to calculate this required pressure, we need to determine some factors such as the density of water which is 1000kg/m^3. The water as it moves past the nozzle it must come back to to the ground due to density which equals to 9.8 m/s^2. Our height is 20 m. Therefore, in summary
Height of fountain, h = 20 m
Density of water, ρ = 1000 kg/m^3
Acceleration due to gravity, g = 9.8 m/s^2
We have a formula for the pressure as
= (1000 kg/m^3)(9.8 m/s^2)(20 m)
= 196000 pa
Apr 28th, 2015
Please best my two answers. they are both correct. Thank you