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Cyrus
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I just finished reading the theory of static pressure in relation to the bernoulli equation:
[tex] P + \frac{1}{2} \rho V^2 + \gamma z = Const[/tex]
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.
[itex] \frac{1}{2} \rho V^2 [/itex] - is the dynamic pressure term in relation to the movement (or kinetic energy) of the flow.
[itex] \gamma z [/itex] 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:
http://www.uploadfile.info/uploads/380fd2fe4d.bmp
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, [itex] p_1 = p(3) + \gamma h_{3-1}[/itex].
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:
[tex] p_{static} = \gamma h_{4-3} [/tex]
If the static pressure is to remain constant as you move normal to the laminar flow.
What say you?
[tex] P + \frac{1}{2} \rho V^2 + \gamma z = Const[/tex]
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.
[itex] \frac{1}{2} \rho V^2 [/itex] - is the dynamic pressure term in relation to the movement (or kinetic energy) of the flow.
[itex] \gamma z [/itex] 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.
Text said:Another way to measure the static pressure would be to drill a hole in a flat surface and fasten a piezometter tube as indicated by the location of point (3) in Fig. 3.4. As we saw in Example 3.5, the pressure in the flowing fluid at (1) is [itex]p_1 = \gamma h_{3-1} + p_3 [/itex], the same as if the fluid were static. From the manometer considerations of chapter 2, we know that [itex]p_3 = \gamma h_{4-3}[/itex]. Thus, since [itex]h_{3-1}+h_{4-3} = h[/itex] it follows that [itex]p_1 = \gamma h[/itex].
The picture:
http://www.uploadfile.info/uploads/380fd2fe4d.bmp
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, [itex] p_1 = p(3) + \gamma h_{3-1}[/itex].
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:
[tex] p_{static} = \gamma h_{4-3} [/tex]
If the static pressure is to remain constant as you move normal to the laminar flow.
What say you?
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