Recent content by NexPhil

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    Roman Inverted Siphons and Fluid Dynamics

    Thank you again for helping. Here's what I have done so far: I converted all the pressure units from atmospheres to Pascals and redo the problem using the Bernoulli's equation. 1 atm = 101,325 Pascals I canceled the whole terms of the (pv^2)/2 from the equation and substituted the known...
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    Roman Inverted Siphons and Fluid Dynamics

    I reread the problem again and decided to use a different approach using the pressure formula since I'm only given the height, but know density of water and acceleration of gravity. pressure = density of water * acceleration of gravity * height pressure = 1000*9.81*10 pressure = 9810 Pa 9810 Pa...
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    Roman Inverted Siphons and Fluid Dynamics

    (1 atmosphere)+(1000 kg/m3)/2 + (1000 kg/m3 (9.81 m/s2)(10 meters) = P2 + (1000 kg/m3)/2 + (1000 kg/m3)(9.81 m/s2)(10 meters)I actually wanted to know if the bernoulli's equation is appropriate for this problem or not. I am only given the height difference and asked to find the pressure at the...
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    Roman Inverted Siphons and Fluid Dynamics

    Homework Statement "Inverted siphons were used by the Romans to cross some valleys. If the height difference between the top of the valley and the bottom is 10 meters, the pressure in the pipe at the bottom of the valley will be ____. Homework Equations Bernoulli's Equation P1+(pv^2)/2 + pgh1...
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