Editing Blade element theory (section)

=== Aerodynamic forces on a blade element ===
Consider the element at [[radius]] ''r,'' shown in Fig. 1, which has the [[infinitesimal]] length ''dr'' and the width ''b.'' The motion of the element in an aircraft propeller in flight is along a helical path determined by the forward velocity ''V'' of the aircraft and the [[Tangent|tangential]] velocity 2πrn of the element in the plane of the propeller disc, where ''n'' represents the revolutions per unit time. The velocity of the element with respect to the air ''Vr'' is then the resultant of the forward and tangential velocities, as shown in Fig. 2. Call the [[angle]] between the direction of motion of the element and the [[plane of rotation]] ''Φ,'' and the blade angle ''β.'' The angle of attack α of the element relative to the air is then <math>\alpha=\beta-\phi</math>''.''  

Applying ordinary airfoil coefficients, the lift force on the element is: 

:<math>dL = \frac{1}{2}V_r^2 C_L b \, dr.</math>

Let ''γ'' be the angle between the lift component and the resultant force, or <math display="inline">\gamma=\arctan\frac{D}{L}</math>. Then the total resultant air force on the element is: <math display="block">dR=\frac{\frac{1}{2}V_r^2 C_L b \, dr}{\cos\gamma}.</math>

The thrust of the element is the component of the resultant force in the direction of the propeller axis (Fig. 2), or 
<math display="block">\begin{align}
dT & = dR\cos(\phi+\gamma) \\
& = \frac{\frac{1}{2}V_r^2C_L b\cos(\phi+\gamma)}{\cos\gamma} dr,
\end{align}</math>
and since <math display="inline">V_r=\frac{V}{\sin\phi}</math>
<math display="block">dT = \frac{\frac{1}{2}V^2 C_L b\cos(\phi+\gamma)}{\sin^2\phi\cos\gamma} dr.</math>

For convenience let 
<math display="block">K=\frac{C_Lb}{\sin^2\phi\cos\gamma}</math> 
and
<math display="block">T_c=K\cos(\phi+\gamma).</math> 

Then 
<math display="block">dT = \frac{1}{2}\rho V^2 T_c \, dr,</math>
and the total thrust for the propeller (of B blades) is: 
<math display="block">T=\frac{1}{2}\rho V^2B\int_{0}^{R} T_c \, dr.</math>

Referring again to Fig. 2, the tangential or torque force is  

:<math>dF=dR\sin(\phi+\gamma),</math> 

and the torque on the element is 

:<math>dQ = r \, dR \, \sin(\phi+\gamma),</math>

which, if <math display="inline">Q_c=Kr\sin(\phi+\gamma)</math>, can be written 

:<math>dQ=\frac{1}{2}\rho V^2 Q_c dr.</math> 

The expression for the torque of the whole propeller is therefore 

:<math>Q = \frac{1}{2}\rho V^2 B\int_{0}^{R} Q_c dr.</math> 

The [[horsepower]] absorbed by the propeller, or the torque horsepower, is 

:<math>QHP=\frac{2\pi nQ}{550}</math>

and the efficiency is

:<math>\eta=\frac{THP}{QHP}=\frac{TV}{2\pi nQ}.</math>