Disk loading

For an actuator disk of area ${\displaystyle A}$, with uniform induced velocity ${\displaystyle v}$ at the rotor disk, and with ${\displaystyle \rho }$ as the density of air, the mass flow rate ${\displaystyle {\dot {m}}}$ through the disk area is:

${\displaystyle {\dot {m}}=\rho \,A\,v.}$

By conservation of mass, the mass flow rate is constant across the slipstream both upstream and downstream of the disk (regardless of velocity). Since the flow far upstream of a helicopter in a level hover is at rest, the starting velocity, momentum, and energy are zero. If the homogeneous slipstream far downstream of the disk has velocity ${\displaystyle w}$, by conservation of momentum the total thrust ${\displaystyle T}$ developed over the disk is equal to the rate of change of momentum, which assuming zero starting velocity is:

${\displaystyle T={\dot {m}}\,w.}$

By conservation of energy, the work done by the rotor must equal the energy change in the slipstream:

${\displaystyle T\,v={\tfrac {1}{2}}\,{\dot {m}}\,{w^{2}}.}$

Substituting for ${\displaystyle T}$ and eliminating terms, we get:

${\displaystyle v={\tfrac {1}{2}}\,w.}$

So the velocity of the slipstream far downstream of the disk is twice the velocity at the disk, which is the same result as for an elliptically loaded wing predicted by lifting-line theory.[9]

To compute the disk loading using Bernoulli's principle, we assume the pressure in the slipstream far downstream is equal to the starting pressure ${\displaystyle p_{0}}$, which is equal to the atmospheric pressure. From the starting point to the disk we have:

${\displaystyle p_{0}=\,p_{1}+\ {\tfrac {1}{2}}\,\rho \,v^{2}.}$

Between the disk and the distant wake, we have:

${\displaystyle p_{2}+\ {\tfrac {1}{2}}\,\rho \,v^{2}=\,p_{0}+\ {\tfrac {1}{2}}\,\rho \,w^{2}.}$

Combining equations, the disk loading ${\displaystyle T/\,A}$ is:

${\displaystyle {\frac {T}{A}}=p_{2}-\,p_{1}={\tfrac {1}{2}}\,\rho \,w^{2}}$

The total pressure in the distant wake is:

${\displaystyle p_{0}+{\tfrac {1}{2}}\,\rho \,w^{2}=\,p_{0}+{\frac {T}{A}}.}$

So the pressure change across the disk is equal to the disk loading. Above the disk the pressure change is:

${\displaystyle p_{0}-{\tfrac {1}{2}}\,\rho \,v^{2}=\,p_{0}-\,{\tfrac {1}{4}}{\frac {T}{A}}.}$

Below the disk, the pressure change is:

${\displaystyle p_{0}+{\tfrac {3}{2}}\,\rho \,v^{2}=\,p_{0}+\,{\tfrac {3}{4}}{\frac {T}{A}}.}$

The pressure along the slipstream is always falling downstream, except for the positive pressure jump across the disk.[9]

From the momentum theory, thrust is:

${\displaystyle T={\dot {m}}\,w={\dot {m}}\,(2v)=2\rho \,A\,v^{2}.}$

The induced velocity is:

${\displaystyle v={\sqrt {{\frac {T}{A}}\cdot {\frac {1}{2\rho }}}}.}$

Where ${\displaystyle T/A}$ is the disk loading as before, and the power ${\displaystyle P}$ required in hover (in the ideal case) is:

${\displaystyle P=Tv=T{\sqrt {{\frac {T}{A}}\cdot {\frac {1}{2\rho }}}}.}$

Therefore, the induced velocity can be expressed as:

${\displaystyle v={\frac {P}{T}}=\left[{\frac {T}{P}}\right]^{-1}.}$

So, the induced velocity is inversely proportional to the power loading ${\displaystyle T/P}$.[10]