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Let's assume you have a pipe rotating steadily in a horizontal plane. (1) denotes the inlet of the pipe and (2) denotes the exit. A pivot point exists at (1) which the pipe rotates around.

$e:$ specific internal energy, $P:$ static pressure, $r:$ radius component, $u:$ fluid local speed, $\rho:$ density, $l:$ pipe length, $\dot{\theta} :$ angular speed,

The velocity in respect to the global frame for an arbitrary point in the flow where the unit vectors are attached to the pipe.

$^F \bar{V}_{p/o^{'}} = u\hat e_{r} + r\dot{\theta} \hat e_{\theta}$

If we apply the conservation of energy in respect to the global frame for the pipe, I obtained the below equation. Note, all rotation effects are only built into the kinetic energy term.

$e_{1} + \dfrac{u^{2}_{1}}{2} + \dfrac{P_{1}}{\rho} = e_{2} + \dfrac{u^{2}_{2} + l^2 \dot{\theta}^2}{2} + \dfrac{P_{2}}{\rho}$

Assume $e_{1} -e_{2} \approx 0$

Below is the incorrect result, stating the static pressure decreases as the fluid moves from the pivot point to the end of the pipe.

$ P_{2} = P_{1} - \dfrac{\rho l^2 \dot{\theta}^2}{2} $

If you apply the conservation of energy in respect to the local frame for the pipe, I obtained the below equation. Rotation effects come solely from a centrifugal energy term.

$e_{1} + \dfrac{u^{2}_{1}}{2} + \dfrac{P_{1}}{\rho} = e_{2} + \dfrac{u^{2}_{2} - l^2 \dot{\theta}^2}{2} + \dfrac{P_{2}}{\rho}$

Assume $e_{1} -e_{2} \approx 0$

Below is the correct result, stating the static pressure increases as fluid moves from the pivot point to the end of the pipe.

$ P_{2} = P_{1} + \dfrac{\rho l^2 \dot{\theta}^2}{2} $

I would expect the same answer. What am I missing?

Thanks!

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1 Answer 1

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If you write the integral energy balance in the global frame for the pipe, you must include power done on the pipe from the pipe wall. Note, this power is zero in respect to the local frame.

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