Thermodynamics and Propulsion

13.4 Aircraft Endurance

If the time spent in the air is of interest and not the distance traveled then one is concerned with endurance. The maximum endurance of an aircraft (or the time aloft) refers to a flight condition that requires the minimum fuel power. Assuming that the overall propulsion system efficiency, $\eta_{overall}$ , defined as the propulsive power over the fuel power, is constant, maximum endurance can be obtained by minimizing the required propulsive power or rate of energy expenditure, $P_R$ , as shown in Figure 13.3.

We can determine the aerodynamic configuration which provides the minimum energy expenditure:

$$D = W\frac{D}{L} = W\left(\frac{C_D}{C_L}\right)$$

so

$$P = W\left(\frac{C_D}{C_L}\right) V,$$

where

$$V = \sqrt{\cfrac{W}{\cfrac{1}{2}\rho S C_L}}$$

Then

$$P = \sqrt{\cfrac{W^3}{\cfrac{1}{2}\rho S}}\left(\frac{C_D}{C_L^{\frac{3}{2}}}\right)$$

So the minimum power required (maximum endurance) occurs when $C_L^{\frac{3}{2}}/C_D$ is a maximum. With a little algebra we can arrive at an expression for the maximum endurance. Setting

$$\frac{d}{d C_L}\left(\cfrac{C_{D_0} + \cfrac{C_L^2}{\pi e AR}}{C_L^{\frac{3}{2}}}\right) = 0$$

we find that

$$C_{L,\textrm{min power}} = \sqrt{3\pi e AR C_{D_0}}$$
and

$$C_{D,\textrm{min power}} = 4 C_{D_0}.$$

$$\left(\frac{C_L}{C_D}\right)_{\textrm{min power}} = \sqrt{\frac{3\pi e AR}{16 C_{D_0}}}$$

and

$$V_{\textrm{min power}} = \sqrt{\cfrac{W}{\cfrac{1}{2}\rho S C_{L,... ...}{\rho^2}\frac{1}{C_{D_0}}\left(\frac{1}{\pi e AR}\right)\right]^{\frac{1}{4}}.$$

Thus the minimum power (maximum endurance) condition occurs at a speed which is $3^{-\frac{1}{4}} = 76\%$ of the minimum drag (maximum range) condition. The corresponding lift-to-drag ratio is 86.6% of the maximum lift-to-drag ratio, shown in Figure 13.5.

Figure 13.5: Relationship between condition for maximum endurance and maximum range.

Continuing

$$D_{\textrm{minimum power}} = W\left[\frac{16}{3}\frac{C_{D_0}}{\pi e AR}\right]^{\frac{1}{2}},$$

which can be substituted into

$$\frac{dW}{dt} = \dot{m}_f g = \frac{-T \dot{m}_f g}{T} = \frac{-T}{\textrm{Isp}} = \frac{-D}{\textrm{Isp}}.$$

Such that, for maximum endurance,

$$\frac{dW}{dt} = \frac{-W}{\textrm{Isp}}\left[\frac{16}{3}\frac{C_{D_0}}{\pi e AR}\right]^{\frac{1}{2}},$$

which can be integrated (assuming constant Isp) to yield

$$t_{\textrm{max}} = \textrm{Isp} \left[\frac{16}{3}\frac{C_{D_0}}{... ...^{-\frac{1}{2}}\ln\left(\frac{W_{\textrm{initial}}}{W_{\textrm{final}}}\right).$$