User:Ssscienccce/sandbox

An actuator disk accelerating a fluid flow from right to left

Inviscid, frictionless fluid

Actuator disc (uniform flow)

Air far above the rotor is at ambient pressure p_a and zero velocity, far downstream it's at wake velocity V_w and at ambient pressure p_a again (we assume inviscid and frictionless fluid, so the wake and the surrounding air don't mix). Directly under the rotor, the velocity is V_d and the pressure is p_d.

The thrust delivered is equal to the change in momentum with time, so (with dotted m meaning the mass flow rate):

$T={\dot {m}}V_{w}$ (why V_w and not V_d? Because at V_d there's an additional pressure component)

The power delivered by the rotor is

$P={\tfrac {1}{2}}{\dot {m}}V_{w}^{2}\quad or\quad P={\tfrac {1}{2}}TV_{w}$

But at the location of the propeller disc, the power is equal to the work done by the thrust:

$P=TV_{d}$

So

${\tfrac {1}{2}}TV_{w}=TV_{d}\quad or\quad V_{d}={\tfrac {1}{2}}V_{w}$, the velocity under the rotor is half of the ultimate wake velocity.

The mass flow rate is:

${\dot {m}}=\rho V_{d}A$

We can write the thrust as:

$T={\dot {m}}V_{w}=\rho V_{d}AV_{w}={\tfrac {1}{2}}\rho AV_{w}^{2}$

This gives us Vw in function of the thrust

$V_{w}={\sqrt {2T \over \rho A}}$

So the power becomes:

$P={\tfrac {1}{2}}TV_{w}=T{\sqrt {T \over 2\rho A}}$

\tfrac12\, \rho\, v^2\, +\, p\, =\, </math>

User:Ssscienccce/miscell

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