Your town is installing a fountain in the main square. If the water is to rise 26 m (85.3 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.
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To determine the pressure required for the water to rise 26 m above the fountain, we can use the concept of hydrostatic pressure. Hydrostatic pressure is the pressure exerted by a fluid due to the weight of the column of the fluid above it.
The pressure at a certain depth in a fluid is given by the equation:
P = ρgh
Where:
- P is the pressure
- ρ is the density of the fluid
- g is the acceleration due to gravity
- h is the height or depth of the fluid
In this case, the water is rising above the fountain, so the depth is 26 m. We need to find the pressure at that depth.
The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.8 m/s². Substituting these values into the equation, we get:
P = (1000 kg/m³)(9.8 m/s²)(26 m)
P ≈ 254,800 Pa
However, this pressure is the absolute pressure and includes the atmospheric pressure. To find the pressure exerted specifically by the water, we need to subtract the atmospheric pressure from this value.
Given that atmospheric pressure is 100,000 Pa, we can subtract this from the absolute pressure:
Pressure exerted by water = 254,800 Pa - 100,000 Pa
Pressure exerted by water ≈ 154,800 Pa
Therefore, the water must have a pressure of approximately 154,800 Pa as it moves toward the nozzle.
373000
it has to gain sufficent force to rise which equals the weight of the height of the water.
We will calculate in gauge pressure...
pressure=force/area= density*area*height*g/area= density*g*height
height = 26m
density= 1.04E6kg/m^3
g= 9.8N/kg
so calcualate pressure, it ought to be about three atmosphers, or 3E5Pa