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I have a piston pump with incompressible, inviscid fluid. Let me write out the mathematics of the flow field-

$$ \text{Continuity Equation: } \frac{\partial u}{\partial x}=0 \\ \text{Euler's Equation:} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = -\frac{1}{\rho} \frac{\partial p}{\partial x} $$

Simplifying them: $$ \frac{du}{dt}=\frac{\Delta P}{\rho L} $$

I put in a harmonic displacement $x_p=x_0 \sin{\omega t}$, and correspondingly get $\Delta P=\rho L x_0 \omega^2 \sin{\omega t}$ and $u=x_0 \omega \cos{\omega t}$.

Note at time $t=0$, $u=x_0 \omega$ and $\Delta P=0$. How is this possible? Velocity is there inspite of pressure difference being 0. Can anyone explain my mistake in the mathematics?

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1 Answer 1

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You can imagine this situation using a simpler scenario. The displacement taken by you is simple harmonic. The spring block system for small displacement is also simple harmonic. To start the oscillation in a spring block system you need to provide either kinetic energy to the block at mean position(this is the case described by you, x=0 and v$\neq$0 at t=0) or potential energy to the spring by pulling it to some distance from the mean position(In this case v=0 and x$\neq$0 at t=0). In an ideal non-dissipative environment, this initial energy provided remains in the system and keeps oscillating between KE and PE. In this case too, the velocity at the mean position is non-zero while the force is equal to zero (F=-kx; x=0).

Thus in the situation described by you, some initial energy would have been provided to start the harmonic displacement. This energy keeps oscillating between KE(velocity) and PE(pressure difference). At t=0, ∆P=0 which means all the energy is in the form of KE which is the reason for a non-zero velocity.

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