# Height of fluid and Bernoullis equation

1. May 27, 2013

1. The problem statement, all variables and given/known data

An ideal liquid flows horizontally through a pipe of cross sectional area 3cm2 with a velocity 1ms-1. The pipe narrows to a cross sectional area 1cm2. 2 vertical pipes are connected to the pipe, one in either region.
Calculate the height difference of the liquid between the 2 vertical pipes.

2. Relevant equations

p1 + 0.5ρv12 + ρgh1 = p2 + 0.5ρv22 + ρgh2

A1v1 = A2v2

3. The attempt at a solution

I think i'm a bit confused on what my variables represent in bernoullis equation:
Using the continuity equation gives v2 = 3ms-1.
Then I said that for the pipes p1 = p2 (because they're both open to the atmosphere), putting everything into bernoullis equation and rearranging gives
h1 - h2 = 0.4 m
The book says the answer is 4x10-3 m.

Am I not allowed to consider the heights of the fluid in the vertical pipe because I calculated the speeds of the fluid in the horizontal pipe? I can't think of any other way to do the problem.

2. May 27, 2013

Oh sorry nevermind, I got it.
Turns out the velocity is initially 0.1 ms-1, then doing what I said gives the answer.

:)

3. May 27, 2013

### SteamKing

Staff Emeritus
IMO, the book's answer is wrong.

I see that the OP had missed a decimal in the initial velocity. Book answer is OK.

Last edited: May 27, 2013