The drag polar of a propeller driven aircraft is $C_{D}=0.039+0.071 C_L^2$. The lift curve slope of the wing ($C_{L_\alpha}$) is 0.09/deg and zero lift angle ($\alpha_{L=0})= -1.5^o$. Stall angle of the wing is $15^o$.

I am trying to find out the aerodynamic efficiency (L/D). Also if it even can fly with such specifications? If it can, what could be the angle of attack?

  • $\begingroup$ How are you finding the Design of Fixed Wing UAV course ? $\endgroup$ – Subhrajit Ghosh Oct 2 '19 at 17:02

At cruise condition, a propeller propulsor can be modeled as: $P=TV$, with constant output power as a function of airspeed. Your objective is to find the cruise condition such that the output power is minimized.

Assuming you know apriori what altitude you want to cruise, then density $\rho$ is determined. At level flight:

$$T=\frac{P}{V}=(C_{D_0}+KC_L^2)\frac{1}{2}\rho V^2S$$

At cruise, weight is equal to lift. At minimum power, we have $\frac{\partial P}{\partial V}=0$:

$$\frac{\partial P}{\partial V}=\frac{3}{2}C_{D_0}\rho V^2S-\frac{KW^2}{\frac{1}{2}\rho SV^2}=0$$

Do some algebra, and we have:


Therefore, the minimum power condition is not equal to maximum L/D (as an aside, it is equivalent to the condition for maximum rate of climb, which is pretty intuitive).

Now you can find the corresponding $C_L$ and verify whether it's within the valid range of the lift curve. Generally, we want at least 1.3G margin to stall warning or stall buffet. So a 1.4G margin to $C_{L_{max}}$ might not be a bad place to start.

  • $\begingroup$ To determine lift coefficient we have the stall angle and zero lift angle. At minimum power condition we calculate at (L/D) maximum. I am finding it hard to go with the answer. Can you create the equations as such. $\endgroup$ – Shishir Maurya Oct 3 '19 at 4:52
  • $\begingroup$ @ShishirMaurya See modified answer. Remember, your OP is asking about minimum power condition of a propeller propulsed aircraft, not max L/D. $\endgroup$ – JZYL Oct 3 '19 at 6:04
  • $\begingroup$ That really helped. Thanks. $\endgroup$ – Shishir Maurya Oct 3 '19 at 7:42

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