# How can a system of hoses each defined in their own non-inertial reference frame transmit energy information?

Rotating pipes <- image

Since energy is frame dependent, I can't find a relationship that makes energy invariant at the hose connection points in respect to each local frame in the general case. If the hose topology is purely in series, you can derive an origin energy term that links the flow energies between two connected hoses such that the flow energy is invariant at the connection point. I think the only way to solve this issue is solve the energy equation for each hose in an inertial frame and the flow energies then always become uniquely defined for each hose.

Below is the flow energy term for a position in a pipe that has frame $L$. You can see it's comprised of a specific internal energy term $e$, a gravitation energy term $g ^L h$, a local kinetic energy term $\dfrac{^L \bar V_{p/o^{'}}^2}{2}$, an euler energy term $^F \bar a_{o^{'}/o} \bar * r_{p/o^{'}}$, a centrifugal energy term $\dfrac{{(^F \Omega^L \wedge \bar r_{p/o^{'}})}^2}{2}$.

$^L \kappa = e + g ^L h + \dfrac{^L \bar V_{p/o^{'}}^2}{2} + ^F \bar a_{o^{'}/o} \bar * r_{p/o^{'}} - \dfrac{{(^F \Omega^L \wedge \bar r_{p/o^{'}})}^2}{2} + \dfrac{p}{\rho} + ^L q_{origin}$

The origin energy term is normally set to zero, however, it could take on other values based on say time. If we equate the left and right frames flow energy you can get the below relationship. This guarantees that flow energy that passes from the left pipe to the right pipe have the same value invariant of the two frames. This type of relationship doesn't hold for pipes in parallel and when loops are formed due to mainly the fictitious force energy terms. As I mentioned above, since energy is frame dependent the only real solution in the general case appears to be solving the energy equations in the global fixed frame $F$.

$^{L_{2}} q_{origin} = ^{L_{1}} q_{origin} + ...$

Rotating Pipes

• Not clear what you are asking. Can you provide a diagram to illustrate? Some background context to your difficulty would also be helpful. – sammy gerbil Jul 27 '18 at 18:04