# Non-inertial vs Inertial Rotating Pipe Pressure Contradiction

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!