- #36

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**(10)**( from Reynolds Transport Theorem ) but its known as the "Momentum Equation" in fluid mechanics ( It is Newtons Second Law for fluid systems ). Which is basically how I had started this whole thing before I led everyone on the wild goose chase!

$$ \sum F = \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} +\int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A $$

I'm not sure why there are different symbols for the velocity ## \boldsymbol v, \boldsymbol V ## , but that is how it is in the textbook.

Anyhow the control volume is the pipe between the tanks

$$ \sum F = P_1 A - P_2 A -\tau \pi D l $$

Momentum Accumulation in the Control Volume:

Incompressible flow with constant cross sectional area ##A ## implies that the velocity is not varying along the length of the pipe and can come outside of the integral along with the density:

$$ \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} = \frac{d}{dt} \rho v A \int_{cv} dl = \rho A l \frac{dv}{dt}$$

The second integral is the Net Efflux of Momentum across the control surface, and because of my assumptions of uniform velocity distribution, incompressibility, and constant cross sectional area it is identically 0:

$$ \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A = 0 $$

That implies we have:

$$ \sum F = P_1 A - P_2 A -\tau \pi D l = \rho A l \frac{dv}{dt} $$

We've moved ## P_1, P_2 ## from the tank surface to the inlet and outlet of the pipe it follows that:

$$ P_1 = \rho g z = \rho g \frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) $$

$$ P_2 = \rho g h $$

$$ v_p = \frac{A_2}{A_p} \dot h $$

That means that finally, and I hope to the satisfaction of all involved in the hunt for oscillation, exponential decay we arrive at:

$$ \rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h $$

I'm not going to bother to simplify it more at this time.

It's a Second Order,Linear ODE of the Non-Homogeneous variety.