Sign on velocity term in Bernoulli Equation

[Raddy13]

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

$$\frac{p_1}{γ} + z_1 + \frac{V^2}{2g} = \frac{p_2}{γ} + z_2 + \frac{V^2}{2g} + h_L$$
$$h_L = f\frac{L}{D}\frac{V^2}{2g}$$
$$f = \frac{64}{Re}$$
$$Re= \frac{VD}{ν}$$

The Attempt at a Solution

Crossing out the irrelevant terms in the Bernoulli eqn:
$$z_1=\frac{V^2}{2g}+h_L$$
Rewriting head loss equation in terms of V:
$$f=\frac{64ν}{VD}$$
$$h_L=\frac{64ν}{VD}\frac{L}{D}\frac{V^2}{2g}=\frac{64νV}{2gD^2}=bV$$
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:
$$z_1=\frac{V^2}{2g}+bV$$
Rewriting as a quadratic equation:
$$\frac{V^2}{2g}+bV-z_1=0$$
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.

 

[Chestermiller]

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

$$\frac{p_1}{γ} + z_1 + \frac{V^2}{2g} = \frac{p_2}{γ} + z_2 + \frac{V^2}{2g} + h_L$$
$$h_L = f\frac{L}{D}\frac{V^2}{2g}$$
$$f = \frac{64}{Re}$$
$$Re= \frac{VD}{ν}$$

The Attempt at a Solution

Crossing out the irrelevant terms in the Bernoulli eqn:
$$z_1=\frac{V^2}{2g}+h_L$$
Rewriting head loss equation in terms of V:
$$f=\frac{64ν}{VD}$$
$$h_L=\frac{64ν}{VD}\frac{L}{D}\frac{V^2}{2g}=\frac{64νV}{2gD^2}=bV$$
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:
$$z_1=\frac{V^2}{2g}+bV$$
Rewriting as a quadratic equation:
$$\frac{V^2}{2g}+bV-z_1=0$$
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.

Suppose that the kinetic energy term wasn't in there. Would you expect V to be positive or negative if there is a fluid column of height z1-z2 driving the flow?

 

[Raddy13]

I would expect it to be negative.

 

[Chestermiller]

I would expect it to be negative.

Is that negative upward or negative downward? That is, would you expect the velocity vector to be pointed upwards or downwards? And, in your equation, is V supposed to be multiplying an upward pointing unit vector or a downward pointing unit vector.

 

[Raddy13]

I would expect the sign convention to be be positive = up, negative = down. That's the convention for the elevation term, but it's not the convention for the gravity term, that's why I'm unsure.

 

[Chestermiller]

I would expect the sign convention to be be positive = up, negative = down. That's the convention for the elevation term, but it's not the convention for the gravity term, that's why I'm unsure.

In your equation, since z1 is positive, V should be pointed downward. Liquid flows from the higher elevation to lower elevation.

 

[Raddy13]

So I should have chosen the negative value from the quadratic solutions?

 

[Chestermiller]

So I should have chosen the negative value from the quadratic solutions?

No, the positive value. Velocity in this problem is taken as positive downward.