Finding gauge pressure when the faucet is turned off

AI Thread Summary
The discussion revolves around calculating gauge pressure at a faucet located 5.2 m above the main water line, which has a gauge pressure of 1.70 X 10^5 Pa. To find the gauge pressure at the faucet when turned off, the pressure difference due to height is calculated using the formula P = ρgh, leading to a gauge pressure of approximately 118988 Pa. For the second part of the problem, participants clarify that to determine how high a faucet can be before water stops flowing, the pressure difference must equal the gauge pressure of zero. The key takeaway is that the gauge pressure must be zero for no water to flow, and the calculations should reflect the pressure difference from the main line pressure. Understanding these principles is essential for solving both parts of the homework problem effectively.
tyro008
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Homework Statement


The main water line enters a house on the first floor. The line has a gauge pressure of 1.70 X 10^5 Pa.

(a) A faucet on the second floor, 5.2 m above the first floor, is turned off. What is the gauge pressure at this faucet?
Pa
(b) How high above the water main could a faucet be before no water would flow from it, even if the faucet were open?


Homework Equations


P(gauge)=\rhogh
Pabs = Patm + Pgauge
Pabs = 1.01 X 10^5Pa + \rhogh
density of water = 1000kg/m^3

The Attempt at a Solution


i tried to use both of these equations but it didn't work. do i need to use 1.70 X 10^5Pa when solving the problem?
I tried the first equation and got 51012Pa (1000 X 9.81 X 5.2)
i tried using the second one and got the wrong answer too.
then i tried to find the density at the closed water fountain.. but i wasn't even
sure how to use that information.
which equation should i use to solve a?
i'm sure i can solve b after finding this.
 
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How do you find the difference in pressure between two points in a fluid?
 
i'll just use "p" for rho..

i thought the difference might be pgh, and that pgh would be the gauge pressure.
so i just multiplied density of water by gravity by height.. but that didn't work.
i just don't know if there's supposed to be a different equation, since the faucet is
closed and thus a closed valve (i'm not even sure it's a closed valve..)
 
tyro008 said:
i thought the difference might be pgh
That's it.
, and that pgh would be the gauge pressure.
so i just multiplied density of water by gravity by height.. but that didn't work.
Try it again. Note that ρgh is the change in pressure.
 
ok this time i tried the differences in gauge pressure,
so i subtracted pgh from 1.70 X 10^5Pa, and got 118988Pa..
would this be the right answer?
 
Looks good.
 
thank you!
that really helped a lot :)
 
to find the height, would i use torricelli's theorum and then find V1 (velocity 1)?
 
tyro008 said:
to find the height, would i use torricelli's theorum and then find V1 (velocity 1)?
No. To find the height needed for part (b) you'll do the same sort of thing you did for part (a). In part (a) you were given a height and you found the pressure difference; for part (b) you have the pressure difference (you can figure it out) and need to find the height.

What must the gauge pressure be for no water to flow? What's the pressure difference at that height compared to the pressure at the main level?
 
  • #10
would the gauge pressure have to be zero for no water to flow?
and the pressure difference would be gauge pressure, what I calculated in (a), I think.
 
  • #11
tyro008 said:
would the gauge pressure have to be zero for no water to flow?
Right!
and the pressure difference would be gauge pressure, what I calculated in (a), I think.
No. What you calculated in (a) was the pressure difference for a 5.2 m change in height.

For (b) the pressure difference is between the main line pressure (given) and a gauge pressure of zero.
 
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