does it slow down when it gets nearer the hole ?

jrhartley

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Sep 10, 2008
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dont worry only a pool question- I located a hole but the speed of the water leaving the pool seems to have slowed as it gets nearer to the hole, Im not a scientist so could anyone confirm that the pressure of water would make water run out faster when the water level is high and slower when there is less water in pool
 
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A pool develops a leak 2 meters below the surface of the water. The pool is 3 meters above the ground level and the hole leaking water is 0.5cm in diameter. How far from the pool will the water hit the ground?

use P1 + (1/2)(ρ)(v1^2) + ρgy = P3 + (1/2)(ρ)(v2^2) + ρgy

I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.

I assumed the tank was open topped and therefore P1 = P2.

We're left with:
ρgy = (1/2)(ρ)(v2^2)

so I canceled the ρ, leaving:

(9.8 m/s^2)(20 m) = (1/2)(v2^2)

so v = sqrt(98) at an angle of 0 degrees, right?

-------

yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground

if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 2 meters away from the pool
 

Conchman

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Jul 3, 2002
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I have a hole in my pool, but I pay a fixed monthly fee for water. Therefore, instead of tearing up my pool to find the hole, I just add water every day. Does anybody know of a company that maybe specializes on finding holes in pools? My leak must be in the pipes between the skimmers and the pump house, which are buried several feet underground and are about 30 feet in length, so this will not be an easy project.
 
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engineerfg

Guest
The selfish capitalist in me wonders, if it's not costing you anything, why are you worried?

What makes you suspect the leak is in the pipes? how big is the pool and what's it made of?

Can you for example, drain the pool, and fill it for just 50 cm. leave it for a day. and see if the water level decreases dramatically? if not, then increase by 50 cm per day to zero in on the point of loss?
 

MikeFisher

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Feb 28, 2006
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The selfish capitalist in me wonders, if it's not costing you anything, why are you worried?

What makes you suspect the leak is in the pipes? how big is the pool and what's it made of?

Can you for example, drain the pool, and fill it for just 50 cm. leave it for a day. and see if the water level decreases dramatically? if not, then increase by 50 cm per day to zero in on the point of loss?

Heck,
My Pool doesn't loose/leak water "Significantly" every Day of the Year, but on some Days it is very significant,
so much that I really need to walk several Meters closer to it's edge before I can get my Feet Wet!

Why???

Hey,
I needed Years to figure out who and where is steeling My Pool water on so many Days,
while on other Days of the Year it is just "Overflooding" without Me leaving any waterhose open,
but at least I figured it out, finally.

They name da Thingy "Tidal Changes",
on good days I step off the Veranda and get my feet wet right away,
on Bad Days I have to walk further down the path before I can touch the good stuff.

I just still couldn't find da lil Hole in da darn Atlantic Pool here in the Backyard, but I don't give up the Search.

Mike
 
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Celt202

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May 22, 2004
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A pool develops a leak 2 meters below the surface of the water. The pool is 3 meters above the ground level and the hole leaking water is 0.5cm in diameter. How far from the pool will the water hit the ground?

use P1 + (1/2)(ρ)(v1^2) + ρgy = P3 + (1/2)(ρ)(v2^2) + ρgy

I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.

I assumed the tank was open topped and therefore P1 = P2.

We're left with:
ρgy = (1/2)(ρ)(v2^2)

so I canceled the ρ, leaving:

(9.8 m/s^2)(20 m) = (1/2)(v2^2)

so v = sqrt(98) at an angle of 0 degrees, right?

-------

yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground

if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 2 meters away from the pool

If a vira lata pees in the pool will it change the velocity? :p
 

belmont

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Oct 9, 2009
1,536
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A pool develops a leak 2 meters below the surface of the water. The pool is 3 meters above the ground level and the hole leaking water is 0.5cm in diameter. How far from the pool will the water hit the ground?

use P1 + (1/2)(ρ)(v1^2) + ρgy = P3 + (1/2)(ρ)(v2^2) + ρgy

I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.

I assumed the tank was open topped and therefore P1 = P2.

We're left with:
ρgy = (1/2)(ρ)(v2^2)

so I canceled the ρ, leaving:

(9.8 m/s^2)(20 m) = (1/2)(v2^2)

so v = sqrt(98) at an angle of 0 degrees, right?

-------

yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground

if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 2 meters away from the pool

So that's why my 3 year-old son misses the toilet when he stands right next to it? I'll have to calulate how far away he should stand.