Lower altitude = higher endurance?

At first glance this didn’t make sense to me. Air Force pilots will be the first to tell you that you don’t hug the earth en route to maximize your time on station, it is harder slogging in thicker air. At higher altitudes you have lower air pressure and colder air temperatures, low pressure is bad for thrust but low temperatures are more favorable. How does this all shake out? And what are the differences between jet- and prop-driven aircraft?

My initial hunch was that for jets, higher altitude requires more speed for level flight and for every gram of fuel burned, you would cover more unit of range in more rarefied air, but endurance is really about turbine efficiency. Higher altitudes decrease the mass flow through the engine, but engine cycle performance is mostly correlated with the ratio of maximum temperature achievable in the turbine $T_{max}$ to ambient temperatures. $T_{max}$ is fixed by engine material properties so the main way to increase efficiency with altitude is to decrease your ambient temperature. With a standard day, the beginning of the isotherm is around 36,089 feet where temperatures remains close to -70 F for about 30k more feet. At 70kft, the outside temperature actually begins to increase. Pressure decreases at a rate that is not subjected to the same complex interplay between the sun’s incoming radiation and the earths outbound radiation, which means any altitude above 36kft should really not give you a performance advantage.

However, modern high endurance unmanned platforms are propeller-driven aircraft. While I haven’t done to math to show how much more efficient these are when compared to jets, I want to explore particularly how the efficiency of propeller-driven aircraft is affected by altitude.

All aircraft share common interactions with the air, so I had to start with the basics of the airfoil. The lift coefficient is traditionally defined as the ratio of force is that pulling an aircraft up over resistive forces. Dynamic pressure (the difference between the stagnation and static pressures) is simply the pressure due to the motion of air particles. When applied against a given area, it is the force that must be overcome by lift for upward motion. If $L$ is the the lift force, and $q$ is the dynamic pressure applied across a planform area $A$ for an air density of $\rho$, by rearranging the lift equation, we have,

$$ C_L = \frac{L}{q_{\infty}\,A}=\frac{2\,L}{\rho\,v^2 A}. $$

Solving for velocity gives us our first insight on how altitude is going to impact the performance of any airfoil:

$$ v = \sqrt{\frac{2\,L}{A}}\,\sqrt{\frac{1}{\rho}}. $$

since, lower altitudes require lower velocity to generate the same lift force. But how does lower velocity equate to more endurance?

Using climatic data from MIL-STD-210C under the conservative assumption of a high density which occurs 20{aaa01f1184b23bc5204459599a780c2efd1a71f819cd2b338cab4b7a2f8e97d4} of the time we have a basic 1/x relationship with density decreasing dramatically with altitude.

From, this alone, we can then plot the velocity needed to keep $C_L$ constant.

To understand the impact this has on endurance, we have to look at brake specific fuel consumption (BSFC). This is a measure of fuel efficiency within a shaft reciprocating engine. It simply the rate of fuel consumption divided by the power produced and can also be considered a as power-specific fuel consumption. If we consume a given number of grams per second of fuel at rate, $r$ for a given power, $P$,

$$ \text{BFSC} = \frac{r}{P} = \frac{r}{\tau \, \omega}, $$

with $\tau$ as engine torque $NÂ·m$ and ($\omega$) as engine speed (rad/s).

$BFSC$ varies from complex interactions with engine speed and pressure:

To understand how this affects endurance, let’s just consider how much fuel we have divided by the rate we use it. Because fuel consumption for piston engines is proportional to power output, we use the average value method to predict the endurance of a propeller-driven aircraft. This is very simply looking at endurance as the fuel weight you have divided by the rate that fuel is spent, all the way to zero fuel. If $W_f$ is the weight of fuel available to be burned:

$$ E = \frac{\Delta W_f}{\dot W_f} = \frac{\Delta W_f}{(\text{BSFC}/\eta_{prop})\,D_{avg}\,V_{\infty}}, $$

where $\eta_{prop}$ is the propeller efficiency factor and $V_{\infty}$ is the speed of the air in the free-stream. The speed for maximum endurance needs to consider flight at the minimum power available,

and at this power, we should be able to maximize our endurance at the appropriate velocity. For increased accuracy, let’s consider that fuel is burned continuously,

$$ E = \int_{W_2}^{W_1} \frac{\eta_{prop}}{\text{BSFC}} \frac{dW}{D\,V_{\infty}}. $$

If $C_L$ is constant and $W=L$, then we can substitute $V_{\infty} = \sqrt{2W/(\rho\,A\,f\,C_L)}$,

$$ E = \frac{\eta_{prop}\,C_L^{3/2}}{\text{BSFC}\,C_D} \sqrt{\frac{\rho,A}{2}} \int_{W_2}^{W_1} \frac{dW}{W^{3/2}} = \frac{\eta_{prop}\,C_L^{3/2}}{\text{BSFC}\,C_D} \sqrt{2 \rho A} \left( W_2^{-1/2} – W_1^{1/2} \right)$$

This tells us that if we want maximum endurance, we want high propeller efficiency, low BSFC, high density (both *low altitude* and temperature) high amount (weight) of fuel available, and a maximum value of the ratio of $C_L^{3/2}/C_D$. Naturally, the higher density is going to directly assist, but the coefficients of lift over drag is maximized when we minimize the power output required,

$$ P_R = V_{\infty} \, D = V_{\infty} , \frac{W}{C_L / C_D} = \sqrt{\frac{2\,W^3}{\rho S}}\frac{1}{C_L^{3/2}/C_D} = \frac{\text{constant}\,C_D}{C_L^{3/2}} $$

So we want to minimize $P_R$. For an assumed drag polar, the condition for minimizing $C_D/C_L^{3/2}$ is found by expressing the ratio in terms of the drag polar, taking the derivative, and setting it equal to zero:

$$ 3 C_{D_0}=k C_L^2 $$

This has the interesting result that the velocity which results in minimum power required has an induced drag three times higher than parasitic drag.

In all, it seems that platform endurance is a function of a number of design parameters and the air density, so in general the higher the air density, the higher the endurance. Please comment to improve my thoughts on this.

## Leave a Reply