Pressure gradient along a streamline using CV analysis

AI Thread Summary
The discussion focuses on deriving the pressure gradient along a streamline using control volume (CV) analysis. The key equation to prove is \(\partial p/\partial x=-\rho u\partial u/\partial x\). The user attempts to apply the CV method but arrives at an incorrect factor of 2 in their final expression, yielding \(\partial p/\partial x=-2\rho u\partial u/\partial x\). They seek clarification on where their reasoning may have gone wrong, particularly in the manipulation of the equations and terms involved in the analysis. The conversation highlights the importance of careful review of mathematical expressions in fluid dynamics problems.
Bohr1227
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Homework Statement


Use a CV analysis to show that an element of fluid along a streamline gives
\[\partial p/\partial x=-\rho u\partial u/\partial x\]

Homework Equations


\[\sum F=\oint_{CS}^{ } \rho \overrightarrow{V}(\overrightarrow{V_{rel}}\cdot \overrightarrow{n})\]

The Attempt at a Solution


Using a CS with following conditions on left side: u and p.
Following conditions on right side of the element: \[u+\partial u/\partial x\Delta x] and \[p+\partial p/\partial x\Delta x]

Just looking at it per unit width inside the paper: (Neglecting small factors)
\[(p-(p+\partial p/\partial x\Delta x))\Delta y=\rho ((u+\frac{\partial u}{\partial x}\Delta x)^{2}-u^{2})\Delta y=\rho (u^{2}-u^{2}+2u\frac{\partial u}{\partial x}\Delta x))\Delta y\]
This gives:
\[\frac{\partial p}{\partial x}=-2\rho u\frac{\partial u}{\partial x}\]

I get the answer with a factor 2, which I am not supposed to get. What do I do wrong?
 
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Bohr1227 said:

Homework Statement


Use a CV analysis to show that an element of fluid along a streamline gives
\[\partial p/\partial x=-\rho u\partial u/\partial x\]

Homework Equations


\[\sum F=\oint_{CS}^{ } \rho \overrightarrow{V}(\overrightarrow{V_{rel}}\cdot \overrightarrow{n})\]

The Attempt at a Solution


Using a CS with following conditions on left side: u and p.
Following conditions on right side of the element: \[u+\partial u/\partial x\Delta x] and \[p+\partial p/\partial x\Delta x]

Just looking at it per unit width inside the paper: (Neglecting small factors)
\[(p-(p+\partial p/\partial x\Delta x))\Delta y=\rho ((u+\frac{\partial u}{\partial x}\Delta x)^{2}-u^{2})\Delta y=\rho (u^{2}-u^{2}+2u\frac{\partial u}{\partial x}\Delta x))\Delta y\]
This gives:
\[\frac{\partial p}{\partial x}=-2\rho u\frac{\partial u}{\partial x}\]

I get the answer with a factor 2, which I am not supposed to get. What do I do wrong?
Did you review this before you posted it to make sure that the LaTex displayed properly?
 
Sorry about the first post, I didn't review the latex, and apparently it is not possible to edit it...so here is the edited version:

1. Homework Statement

Use a CV analysis to show that an element of fluid along a streamline gives
\frac{\partial p}{\partial x}=-\rho u\frac{\partial u}{\partial x}

Homework Equations


\sum F=\oint_{CS}^{ } \rho \overrightarrow{V}(\overrightarrow{V_{rel}}\cdot \overrightarrow{n})

The Attempt at a Solution


Using a CS with following conditions on left side: u and p.
Following conditions on right side of the element: u+\frac{\partial u}{\partial x}\Delta x and p+\frac{\partial p}{\partial x}\Delta x

Just looking at it per unit width inside the paper: (Neglecting small factors because \Delta x\rightarrow 0)
(p-(p+\partial p/\partial x\Delta x))\Delta y=\rho ((u+\frac{\partial u}{\partial x}\Delta x)^{2}-u^{2})\Delta y=\rho (u^{2}-u^{2}+2u\frac{\partial u}{\partial x}\Delta x))\Delta y
This gives:
\frac{\partial p}{\partial x}=-2\rho u\frac{\partial u}{\partial x}I get the answer with a factor 2, which I am not supposed to get. What do I do wrong?
 
Chestermiller said:
Did you review this before you posted it to make sure that the LaTex displayed properly?
You are right, I did not review it. I'm sorry, but it is fixed now.

Thank you!
 
Bohr1227 said:
Sorry about the first post, I didn't review the latex, and apparently it is not possible to edit it...so here is the edited version:

1. Homework Statement

Use a CV analysis to show that an element of fluid along a streamline gives
\frac{\partial p}{\partial x}=-\rho u\frac{\partial u}{\partial x}

Homework Equations


\sum F=\oint_{CS}^{ } \rho \overrightarrow{V}(\overrightarrow{V_{rel}}\cdot \overrightarrow{n})

The Attempt at a Solution


Using a CS with following conditions on left side: u and p.
Following conditions on right side of the element: u+\frac{\partial u}{\partial x}\Delta x and p+\frac{\partial p}{\partial x}\Delta x

Just looking at it per unit width inside the paper: (Neglecting small factors because \Delta x\rightarrow 0)
(p-(p+\partial p/\partial x\Delta x))\Delta y=\rho ((u+\frac{\partial u}{\partial x}\Delta x)^{2}-u^{2})\Delta y=\rho (u^{2}-u^{2}+2u\frac{\partial u}{\partial x}\Delta x))\Delta y
This gives:
\frac{\partial p}{\partial x}=-2\rho u\frac{\partial u}{\partial x}I get the answer with a factor 2, which I am not supposed to get. What do I do wrong?
I don't see how you get the right hand side of this equation:
(p-(p+\partial p/\partial x\Delta x))\Delta y=\rho ((u+\frac{\partial u}{\partial x}\Delta x)^{2}-u^{2})\Delta y=\rho (u^{2}-u^{2}+2u\frac{\partial u}{\partial x}\Delta x))\Delta y
 
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