Hypothetical Bernoulli situation with Pascals Law(Efflux torcellis result)

[jlyu002@ucr.e]

Homework Statement

:

There is tank with a certain depth with water that is exposed to air. There is a hole at the bottom of the tank. Using Bernoulli's equation, we can find the efflux speed. density of the water * gravity * depth = 1/2density of water * velocity ^2.

This will give us the velocity of the water at the bottom of the tank which is V= Squareroot of (2*gravity*depth).

My question is now, what would be the velocity of the water at the top as it is draining. I believe we can use pascals law where Flow(area times velocity)= Flow2(area2 times velocity). Since the area of the top of the water is bigger, the velocity of the water at the top draining would be smaller than the efflux velocity?

Do I have the correct interpretation? [/B]

 

[SteamKing]

Yes. The flow out of the tank at the bottom of the tank must equal the rate of change of the volume at the top of the tank, according to the continuity relation. As long as the area of the hole at the bottom is less than the area of the free surface at the top of the tank, then the time rate of change in water depth measured at the free surface must be less than the velocity of the water squirting out the bottom of the tank.

 

[jlyu002@ucr.e]

Sweet! Thank you good sir!