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Water fountain in eye of hurricane?
When the atmospheric pressure is 1 atm, a water fountain ejects a stream of water that rises to a height of 5m. There is a 1-cm-radius pipe that leads from a pressurized tank to the opening that ejects the water. What would happen if the fountain were operating when the eye of a hurricane passes through? Assume that the atmospheric pressure in the eye is 0.877 atm and the tank's pressure remains the same.
Please describe how you came to your conclusion.
1 Answer
- ?Lv 79 years agoFavorite Answer
Energy conservation implies
p - p_atm = rho g h
Comparison of the two cases thus gives
(h2 - h1) = (p1 - p2) / ( rho g)
= 0.877 * 1.013 * 10^5 N/m^2 / ( 1000 kg/m^3 * 9.81 m/s^2)
= 9.06 m
So the fountain in that case rises to 14.1 m
