4
$\begingroup$

For the last hour I have been trying to understand what $\dfrac{\partial p}{\partial s}ds$ is in Euler's equation but I have a problem. You can see this image:

enter image description here

enter image description here

I know that $p\ dA = \text{Force}$

$γ ds dA = mg$

but what is $\left(p + \dfrac{\partial p}{\partial s} dA\right)$?

$\endgroup$
3
$\begingroup$

$\frac{\partial p}{\partial s}$ is the pressure gradient through the control volume. Multiplying it by $ds$, the length of the control volume, gives the change in pressure from one side of the control volume to the other. It is essentially a first-order Taylor-series expansion of the function $p(s)$. This method is used a lot in fluid dynamics derivations; the most thorough explanation I've seen is in "Computational Fluid Dynamics" by John Anderson.

$\endgroup$
0
$\begingroup$

Whenever a fluid flows in a pipe, it flows due to difference in pressure. Due to this difference in pressure, there is a pressure gradient which is $\dfrac{\partial p}{\partial s}$. $\dfrac{\partial p}{\partial s}\ \text{d}s$ is the change in pressure over the length $\text{d}s$. $p+\dfrac{\partial p}{\partial s}\ \text{d}s$ is the pressure at the exit in terms of pressure function $p$ at inlet + change in pressure over length $\text{d}s$. So, $\left(p+\dfrac{\partial p}{\partial s}*ds\right)dA$ is also the force like $p\cdot \text{d}A$ with pressure $p$ at the exit equals $p+\dfrac{\partial p}{\partial s}\ \text{d}s$.Pressure variation in fluid is given by Tailor's series expansion with respect to length $s$ about $p$. $p(s + \Delta s) = p(s) + \dfrac{\partial p}{\partial s}\ \Delta s + O(\Delta s2)$. $O(\Delta s2)$ is the higher order term which is neglected.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.