# Sign on velocity term in Bernoulli Equation

In summary, the problem is about finding the exit velocity at the bottom of a pipe in an open air tank filled with water, taking into account head loss. The Bernoulli equation and head loss equation are used, and the velocity is found through a quadratic equation. The sign convention for velocity and gravity is discussed, and it is determined that the positive value should be chosen for velocity in this problem.

## 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:
$$z_1=\frac{V^2}{2g}+bV$$
$$\frac{V^2}{2g}+bV-z_1=0$$
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.

## 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:
$$z_1=\frac{V^2}{2g}+bV$$
$$\frac{V^2}{2g}+bV-z_1=0$$
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?

I would expect it to be negative.

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.

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.

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.

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

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.

## 1. What is the Sign on Velocity Term in Bernoulli Equation?

The sign on velocity term in Bernoulli Equation refers to the direction of the flow in relation to the reference point. It is represented by the variable V and can be either positive or negative depending on the direction of the flow.

## 2. What does the Sign on Velocity Term represent in Bernoulli Equation?

The sign on velocity term represents the kinetic energy of the fluid in motion. It is a crucial part of the Bernoulli Equation as it accounts for the change in velocity of the fluid as it moves through different points in the system.

## 3. How does the Sign on Velocity Term affect Bernoulli Equation calculations?

The sign on velocity term plays a significant role in determining the pressure and velocity of a fluid at different points in a system. It is essential to consider the direction and magnitude of this term in order to accurately calculate the pressure and velocity of the fluid.

## 4. Can the Sign on Velocity Term ever be zero in Bernoulli Equation?

Yes, the sign on velocity term can be zero in Bernoulli Equation when the fluid is not in motion or when the flow is parallel to the reference point. In such cases, the term does not contribute to the overall equation and can be ignored.

## 5. Why is the Sign on Velocity Term important in Bernoulli Equation?

The sign on velocity term is important in Bernoulli Equation because it accounts for the kinetic energy of the fluid and helps in accurately predicting the pressure and velocity at different points in a system. It also allows for the analysis and understanding of fluid flow behavior in various applications such as aerodynamics and hydraulics.

Replies
2
Views
1K
Replies
6
Views
1K
Replies
6
Views
7K
Replies
1
Views
2K
Replies
5
Views
2K
Replies
1
Views
2K
Replies
31
Views
2K
Replies
2
Views
2K
Replies
8
Views
7K
Replies
7
Views
2K