# Static Pressure Poorly Described by Texts

1. Jun 14, 2006

### Cyrus

I just finished reading the theory of static pressure in relation to the bernoulli equation:

$$P + \frac{1}{2} \rho V^2 + \gamma z = Const$$

Here, we have three terms and a corresponding relationship.

P- is the static or thermodynamic pressure measurment of the fluid. It is the measurment obtained if we move with the flow.

$\frac{1}{2} \rho V^2$ - is the dynamic pressure term in relation to the movement (or kinetic energy) of the flow.

$\gamma z$ is the gravitational potential energy term in relation to the hydrostatic pressure.

Here is the picture (Which I made in word and took forever), and the corresponding text in my book.

The picture:

The problem:

This solution to finding the static pressure cannot be valid. Consider point (1). Clearly, the pressure at point one will have two components. We can write this using a manometer equation and obtain the same results as the text, $p_1 = p(3) + \gamma h_{3-1}$.

Had we chosen a point above or below point (1), the hydrostatic term would have changed. If we move down, the hydrostatic term would become larger. If we move up, the hydrostatic term would become smaller.

This is a direct violation of the condition that static pressure is a constant normal to straight laminar flow.

Therefore, in order to get the true static pressure, we would have to look at point (3) and write a manometer equation. This would give us the static pressure reading as simply the weight of the column of fluid from point (4) to point (3).

In other words, the text should clearly state:

$$p_{static} = \gamma h_{4-3}$$

If the static pressure is to remain constant as you move normal to the laminar flow.

What say you?

Last edited: Jun 14, 2006
2. Jun 14, 2006

### RainmanAero

Hi Cyrus,
Learning about the pressure energy equation can indeed be confusing. However, your confusion between hydrostatic pressure and static pressure comes from an assumption inherent in the problem, and that is the area of the flow channel is constant. We could equally make the assumption that the flow is level, and this would mean that we could ignore the hydrostatic pressure terms at any two points along the flow.

While we could argue that the term "static" might be confusing, the concept of what is being measured when we say "static pressure" is better understood by assuming level flow, with a converging or diverging flow channel. When doing this we define the measurement of static pressure as normal to the flow (since measuring parallel/into the flow we would measure static+dynamic = total pressure for the level flow).

When you have a converging or diverging flow channel, we find that as dynamic pressure goes UP for a converging channel (because velocity must go up to satisfy continuity), then we see that static (normal) pressure must go down to maintain constant total pressure. Alternately, for a diverging flow channel the velocity will go down, and thus the static pressure will increase.

The term "static" pressure comes from a consideration of having no flow velocity (the fluid mass is static). In this case, indeed the static pressure is the same as the hydrostatic pressure due to the weight of the fluid above the measurement point. In your problem, this would not ONLY be the weight of the liquid, but must also include the weight of the atmosphere above that liquid. So you would need to correct your equation to account for the specific weight of the liquid multiplied by its height, as well as the atmospheric pressure above/around the liquid.

However, I do agree that using the term "static pressure" can be confusing, but depending on how you analyze the problem it is not necessarily incorrect. Perhaps a better term for it would be "normal pressure" (as in normal to any flow velocity); however, tradition has given us the term "static pressure" and that is what we typically use.

Take care,
Ray

3. Jun 14, 2006

### Staff: Mentor

Your understanding of how fluids work is correct, but your wording of Bernoulli's principle is not correct. Bernoulli states that static pressure is constant along a streamline. When the streamlines are close together and the weight of the fluid is small compared to the pressure, the way you phrased it is approximately true, but it is not always true. See:
http://wright.nasa.gov/airplane/bern.html

4. Jun 14, 2006

### Cyrus

You stated:

You are correct, I should have stated that I was thinking in terms of the gage pressure and not the absolute pressure.

More importantly,

Based on this definition, the "static" pressure at any point would always be equal to the weight of fluid above it, as read from said point to the top of the manometer.

This would mean that I get different static pressure values if I stay at the same x-location, and only move up or down in the z direction. (Because the column of fluid above the point would now differ)

Isn’t it required that the static pressure along any point vertically of the cross section remain constant?

Thanks,

Cyrus

5. Jun 15, 2006

### FredGarvin

The term "static" is used because you can think of the pressure while moving with the fluid.

Where are you getting the constraint that the pressure has to be constant across the entire cross section? I looked back at the derivation of the Bernoulli equation.

1) The fluid particle along a streamline used to set up F=ma along the streamline includes, on the top faces of the element a $$(p + \delta p_n)\delta s \delta y$$ and a $$(p - \delta p_n)\delta s \delta y$$term on the bottom face. The directions being "s" in the direction of the streamline and "n" as the normal direction to the streamline. The assumption made is that along the streamline, n is constant and the $$\delta n$$ terms go to 0.

2) I'll just quote my book.
Also, there is the same annotation that application of Bernoulli across streamlines leads to errors, which can be large at times depending on the nature of the flow. In general, the Bernoulli constant varies from streamline to streamline.

We measure static pressures at the walls for practical purposes. That static pressure is going to give you a very good average of the flow across the entire duct at that point.

Last edited: Jun 15, 2006
6. Jun 15, 2006

### Cyrus

Once again my friend, you are correct. This makes perfect sense. The reason why I said that the pressure along the direction normal to the flow is because my professor said so. In fact, I told him that it should vary, and he disagreed with me (quite politely, he's a great guy). We discussed this in his office for about 20mins as well.

I will go back to his office and show him that this is only approximately true. I know another professor who is basically considered a god in fluids around here concerning this question if and when I find him in his office. I am sure he will simply state the same thing.

7. Jun 18, 2006

### Cyrus

Turns out the streamlines are constant. Condsider applying the bernoulli equation along a streamline at the center where we define $$z_o =0$$.

In that case, application of the bernoulli equation yields:

$$p_1 + \frac{1}{2} \rho V^2 + \gamma z_1 =p_2 + \frac{1}{2} \rho V^2 + \gamma z_2 = C_{1,2}$$

Since this streamline acts at z=0, it reduces to:

$$p_1 + \frac{1}{2} \rho V^2 =p_2 + \frac{1}{2} \rho V^2= C_{1,2}$$

Where (1) and (2) are any two points along this stream line.

Now pick any other stream line. For this example I will pick a streamline that is above the median line of the pipe. I will choose two points along this streamline, (3) and (4). Applying the bernoulli equation yields:

$$p_3 + \frac{1}{2} \rho V^2 + \gamma z_3 =p_4 + \frac{1}{2} \rho V^2 + \gamma z_4 = C_{3,4}$$

But we can write $$p_3 = p_1 - \gamma z_3$$ and $$p_4 = p_1 - \gamma z_4$$.

This will cause the bernoulli equation on the streamline (3) and (4) to yield:

$$p_1 + \frac{1}{2} \rho V^2 =p_2 + \frac{1}{2} \rho V^2 = C_{3,4}$$

So we conclude that $$C_{1,2} = C_{3,4}$$.

This is not to say that the thermodynamic (static) pressure does not vary in the transverse direction of flow, it does. But the bernoulli constant is the same across flow lines.

Last edited: Jun 18, 2006
8. Jun 18, 2006

### Cyrus

Actually, this also holds true and is reflected on the bernoulli equation across streamlines:

$$p + \int \rho \frac{V^2}{ \Re} + \gamma z = C$$

Where $$\Re$$ is the local radius of curvature. Because the flow is uniform, V is the same. Also, the flow is horizontal and $$\Re$$ is $$\infty$$ so the terms go to zero on both sides.

Again you get:

$$p + \gamma Z = C$$

Last edited: Jun 18, 2006