Solve Bernoulli Problem: Velocity of Water Outlet

[Sall1230]

Homework Statement

" A water filled verticle pipe of cross section area 25 cm^2 is open at the top. If an outlet of cross section area 15 cm^2 is made on it's side at a depth of 10 m from the level of the water at the top, then the velocity of the water exiting from the outlet is?

Homework Equations

" P1 + ρgh1 + 0.5 ρv1^2 = P2 + ρgh2 + 0.5 ρv2^2 .

The Attempt at a Solution

Ok so what made me more confused is that v1 doesn't equal 0 because I was about to use this equation : √2gh

Then I tried to find both of v1 and v2 with the equation of A1 v1 = A2 v2 and I got this " v2 = 1.667 v1" which didn't help much because it is already obvious that the velocity will increase since the area is smaller.

After that I started cancelling some of the equation ( the one in brackets are canceled ) :

(Patm)+ ρgh1 + 0.5 ρv1^2 = (Patm) + (ρgh2) + 0.5 ρv2^2

And I tried to plug in numbers :
1000 • 9.8 • 10 + 0.5 • 1000 v1^2 = 0.5 • 1000 v2^2
98000 + 500 v1^2 = 500 v2^2

And I was left blank. So I'm now thinking whether the cross section area has anything to do with the pressure... But yeah that was my tries .

 

[SteamKing]

Homework Statement

" A water filled verticle pipe of cross section area 25 cm^2 is open at the top. If an outlet of cross section area 15 cm^2 is made on it's side at a depth of 10 m from the level of the water at the top, then the velocity of the water exiting from the outlet is?

Homework Equations

" P1 + ρgh1 + 0.5 ρv1^2 = P2 + ρgh2 + 0.5 ρv2^2 .

The Attempt at a Solution

Ok so what made me more confused is that v1 doesn't equal 0 because I was about to use this equation : √2gh

v1 = 0 only when you are dealing with something like a reservoir, where the change in the volume of the fluid does not produce a large change in the depth of the fluid per unit time. Since the problem here concerns water running out of a pipe, v1 ≠ 0.

Then I tried to find both of v1 and v2 with the equation of A1 v1 = A2 v2 and I got this " v2 = 1.667 v1" which didn't help much because it is already obvious that the velocity will increase since the area is smaller.

This is the continuity equation. It is important not because it solves the problem by itself, but it can be used with the Bernoulli equation to provide a solution.

After that I started cancelling some of the equation ( the one in brackets are canceled ) :

(Patm)+ ρgh1 + 0.5 ρv1^2 = (Patm) + (ρgh2) + 0.5 ρv2^2

And I tried to plug in numbers :
1000 • 9.8 • 10 + 0.5 • 1000 v1^2 = 0.5 • 1000 v2^2
98000 + 500 v1^2 = 500 v2^2

And I was left blank. So I'm now thinking whether the cross section area has anything to do with the pressure... But yeah that was my tries .

Using the continuity equation, you can find a relationship between v1 and v2 based on the area at the open end of the pipe and the hole further down.

By eliminating v1 or v2 (yer cherce), you can solve for the other velocity.