User:Ssscienccce/sandbox


 * Inviscid, frictionless fluid
 * Actuator disc (uniform flow)


 * Air far above the rotor is at ambient pressure pa and zero velocity, far downstream it's at wake velocity Vw and at ambient pressure pa again (we assume inviscid and frictionless fluid, so the wake and the surrounding air don't mix). Directly under the rotor, the velocity is Vd and the pressure is pd.
 * 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 Vw and not Vd? Because at Vd there's an additional pressure component)


 * The power delivered by the rotor is
 * $$P = \tfrac{1}{2} \dot{m} V_w^2 \quad or \quad  P =  \tfrac{1}{2} T V_w $$
 * But at the location of the propeller disc, the power is equal to the work done by the thrust:
 * $$ P = T V_d $$
 * So
 * $$ \tfrac{1}{2} T V_w = T V_d \quad or \quad V_d = \tfrac{1}{2} V_w $$, 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:
 * $$T = \dot{m} V_w = \rho V_d A V_w = \tfrac12 \rho A V_w^2 $$
 * This gives us Vw in function of the thrust
 * $$V_w = \sqrt{2T \over \rho A} $$
 * So the power becomes:
 * $$ P = \tfrac{1}{2} T V_w = T \sqrt{T \over 2 \rho A}  $$

\tfrac12\, \rho\, v^2\, +\, p\, =\,
 * User:Ssscienccce/miscell


 * Ex@mplet@lk

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