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.
p1 + 0.5ρv12 + ρgh1 = p2 + 0.5ρv22 + ρgh2
A1v1 = A2v2
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.