- #1

Raddy13

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## Homework Statement

This was a problem on my fluids final and I'm trying to figure out if my rationale was correct here. The problem was an open air tank filled with water and at the bottom was a pipe pointing straight down. You had to find the exit velocity at the bottom of the pipe and account for head loss in the pipe.

## Homework Equations

[tex]\frac{p_1}{γ} + z_1 + \frac{V^2}{2g} = \frac{p_2}{γ} + z_2 + \frac{V^2}{2g} + h_L[/tex]

[tex]h_L = f\frac{L}{D}\frac{V^2}{2g}[/tex]

[tex]f = \frac{64}{Re}[/tex]

[tex]Re= \frac{VD}{ν}[/tex]

## The Attempt at a Solution

Crossing out the irrelevant terms in the Bernoulli eqn:

[tex]z_1=\frac{V^2}{2g}+h_L[/tex]

Rewriting head loss equation in terms of V:

[tex]f=\frac{64ν}{VD}[/tex]

[tex]h_L=\frac{64ν}{VD}\frac{L}{D}\frac{V^2}{2g}=\frac{64νV}{2gD^2}=bV[/tex]

Where b is the coefficient of all the variables I had values for (I don't remember the exact numbers now). So going back to Bernoulli:

[tex]z_1=\frac{V^2}{2g}+bV[/tex]

Rewriting as a quadratic equation:

[tex]\frac{V^2}{2g}+bV-z_1=0[/tex]

When I solved the equation, I would up with two possible answers, one positive and one negative. My reasoning at the time was that since the value for gravity I used in the velocity term was positive and pointing down, then the velocity for the flow of the pipe should be as well. But now that I think about it, the z term in negative pointing down (z2, the pipe exit, was 0, while z1 was 2 m), so I'm not sure what the correct answer is.