Worked Examples: Airplane Performance

J. Gordon Leishman

86 Worked Examples: Airplane Performance

Many of these worked examples have been fielded as homework problems or exam questions.

Worked Example #1

A propeller-driven airplane has a fuel flow ${\overbigdot{W}_f}$ to the engines of the form

$\frac{dW_f}{dt} = {\overbigdot{W}_f} = c_{b} \left( A V_{\infty}^3 + \frac{B}{V_{\infty}} \right)$

where ${\overbigdot{W}_f}$ is the weight of fuel burned per unit time, ${V_{\infty}}$ is the true airspeed, $c_b$ is the brake specific fuel consumption. The values of $A$ and $B$ depend on the characteristics of the airplane, its weight, and the density altitude at which it is flying, i.e.,

$A = \frac{1}{2} \varrho_{\infty} S C_{D_{0}} \quad \text{and} \quad B = \frac{2 W^2}{\varrho_{\infty} S (\pi \, AR e)}$

Based on this fuel flow equation, show that the airspeed for the best endurance for the airplane will be obtained when the true airspeed is

$V_{\infty} = V_{\text{end}} = \left(\frac{B}{3A}\right)^{1/4}$

and the corresponding airspeed for the best range for the airplane will be obtained when the true airspeed is

$V_{\infty} = V_{\text{range}} = \left(\frac{B}{A}\right)^{1/4}$

Explain the operational significance of flying at the best-endurance and best-range airspeeds, and give an example of a flight profile, or a part of one, when such airspeeds might be specifically used.

The first term in the fuel flow equation is the contribution from the profile/parasitic (non-lifting) drag component, which grows with the cube of the airspeed. The second part is the contribution from the induced drag, which is inversely proportional to airspeed. Starting from

$\frac{dW_f}{dt} = {{\overbigdot{W}_f}} = c_b\left( A V_{\infty}^3 + \frac{B}{V_{\infty}} \right)$

then the lowest fuel burn rate, hence maximum flight endurance, can be determined by finding the point at which ${\overbigdot{W}_f}$ is a minimum. Differentiating the fuel flow equation with respect to ${V_{\infty}}$ gives

${ \frac{d{\overbigdot{W}_f}}{dV_{\infty}} = c_b \left( 3 A V_{\infty}^2 - \frac{B}{V_{\infty}^2} \right) }$

which is zero for a minimum, i.e.,

$3 A V_{\infty}^2 - \frac{B}{V_{\infty}^2} = 0$

and so

$V_{\infty} = V_{\text{end}} = \left( \frac{B}{3A} \right)^{1/4}$

The best range is obtained when the ratio ${\overbigdot{W}_f}/V_{\infty}$ is a minimum. In this case

$\frac{{\overbigdot{W}_f}}{V_{\infty}} = c_b \left( A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \right)$

so that

$\frac{d({\overbigdot{W}_f}/V_{\infty})}{dV_{\infty}} = c_b \left( 2 A V_{\infty} - 2\frac{B}{V_{\infty}^3} \right)$

which is zero for a minimum, i.e.,

$2 A V_{\infty} - \frac{2B}{V_{\infty}^3} = 0$

and so

$V_{\infty} = V_{\text{range}} = \left( \frac{B}{A} \right)^{1/4}$

There are various missions in which flying at the optimal speed for maximum range or endurance is crucial. For example, an airplane’s long-range ferry mission, such as over water, may require it to be flown at or near its optimal airspeed for maximum range, even though this airspeed is typically lower than its best cruise speed. Likewise, flying at an airspeed that maximizes endurance may be necessary for a reconnaissance mission in which the aircraft must remain at or near a particular location for an extended period.

Worked Example #2

Starting from the steady, level-flight flight assumption of ${L = W}$ and ${T = D}$ , and assuming the engine thrust specific fuel consumption TSFC ( $c_t$ ) is constant, then show that for a jet-propelled airplane flying at a constant altitude that the fuel flow to the engines in terms of weight of fuel burned per unit time ${\overbigdot{W}_f}$ can be expressed in the form

${{\overbigdot{W}_f}} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2}$

where you should evaluate $A$ and $B$ in terms of the the wing area, $S$ , wing aspect ratio, $AR$ , flight weight, $W$ , parasitic drag coefficient & ${C_{D_{0}}}$ , Oswald’s efficiency factor, ${e}$ , engine TSFC, $c_t$ .

The standard formula gives the lift $L$ on the airplane.

$L = \frac{1}{2} \varrho_{\infty} \, V_{\infty}^2 \, S \, C_L$

which equals the weight $W$ . The drag $D$ on the airplane is

$D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_D$

where the drag coefficient $C_D$ is given by

$C_D = C_{D_{0}} + C_{D_{i}} = C_{D_{0}} + \frac{{C_L}^2}{\pi \, AR \, e}$

i.e., the sum of non-lifting and lifting (induced) parts. $AR$ is the wing’s aspect ratio, and ${e}$ is Oswald’s efficiency factor. Using the equation for lift (as well as the assumption that $L = W$ ), the lift coefficient is

$C_L = \frac{2 L}{\varrho_{\infty} V_{\infty}^2 S} = \frac{2 W}{\varrho_{\infty} V_{\infty}^2 S}$

Also, because the thrust $T$ equals drag, then

$T = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_{D_{0}} + \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \left( \frac{2 W}{\varrho_{\infty} V_{\infty}^2 S} \right)^2 \left( \frac{1}{\pi \, AR \, e} \right)$

and after rearrangement, then

$T = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_{D_{0}} + \frac{2 W^2}{\varrho_{\infty} V_{\infty}^2 S (\pi \, AR \, e)}$

For a jet engine, the fuel burn rate ${\overbigdot{W}_f}$ will be the product of the thrust and the thrust-specific fuel consumption, i.e.,

$\frac{d W_f}{dt} = {\overbigdot{W}_f} = T c_t$

Substituting for $T$ gives

${\overbigdot{W}_f} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S c_t C_{D_{0}} + \frac{2 c_t W^2}{\varrho_{\infty} V_{\infty}^2 S (\pi \, AR \, e)}$

or

${ {\overbigdot{W}_f} = \left( \frac{1}{2} \varrho_{\infty} S c_t C_{D_{0}} \right) V_{\infty}^2 + \left( \frac{2 c_t W^2}{\varrho_{\infty} S (\pi \, AR \, e)} \right) \frac{1}{V_{\infty}^2} }$

which is of the form

${\overbigdot{W}_f} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2 }$

where

${ A = \frac{1}{2} \varrho_{\infty} S c_t C_{D_{0}} }$

and

$B = \frac{2 c_t W^2}{\varrho_{\infty} S (\pi \, AR \, e)}$

Both constants are based on the stated assumptions of all the values involved for a constant weight $W$ and altitude, i.e., $\varrho_{\infty}$ = constant.

Worked Example #3

Using the fuel flow equation from the previous worked example, i.e.,

$\overbigdot{W}_f = \frac{d W_f}{dt} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2}$

show that the airspeed for the best endurance for a jet airplane will be obtained when

$V_{\infty} = V_{\text{end}} = \left(\frac{B}{A}\right)^{1/4}$

In this problem, we are asked to start from

$\frac{d W_f}{dt} = {\overbigdot{W}_f} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2}$

The lowest fuel burn rate, hence maximum flight endurance, can be determined by differentiating the fuel flow equation with respect to ${V_{\infty}}$ , i.e.,

$\frac{d{\overbigdot{W}_f}}{dV_{\infty}} = 2 A V_{\infty} - \frac{2 B}{V_{\infty}^3}$

which is zero for a minimum, i.e.,

$2 A V_{\infty} - \frac{2B}{V_{\infty}^3} = 0$

and so

$V_{\infty} = V_{\text{end}} = \left( \frac{B}{A} \right)^{1/4}$

where

$A = \frac{1}{2} \varrho_{\infty} S c_t {C_{D_{0}}}$

and

$B = \frac{2 c_t W^2}{\varrho_{\infty} S (\pi \, AR \, e)}$

Therefore, $V_{\text{end}}$ is the airspeed at which the airplane should fly to obtain the minimum fuel flow rate. This speed depends on the airplane’s weight and altitude.

Worked Example #4

An ERAU Cessna 172 Skyhawk airplane is flying along at an estimated in-flight weight of 2,100 lb and an airspeed of 120 knots at 2,000 ft, where the air density is 0.00216 slugs/ft ${^3}$ . The student pilot is flying solo on a long cross-country flight. While flying over Jacksonville Airport on the return to ERAU (125 nautical miles remaining), the pilot observes from the fuel gauge that only 8 gallons of usable AVGAS remain in the tanks. Assume that the airplane’s weight for this analysis is the initial in-flight weight minus half the remaining fuel weight. Assume also that AVGAS fuel weighs 6.0 lb per gallon. The engineering characteristics of the airplane are given below:

Wing span, $b$ = 36 ft
Wing area, $S$ = 174 ft ${^{2}}$
Non-lifting drag coefficient, ${C_{D_{0}}}$ = 0.02
Average propeller efficiency, $\eta_p$ = 0.85
Oswald’s efficiency factor, ${e}$ = 0.81
Engine BSFC, $c_b$ = 0.45 lb hp $^{-1}$ hr $^{-1}$

Calculate the following:
(a) The operating lift coefficient of the wing, $C_L$ .
(b) The induced drag coefficient, ${C_{D_{i}}}$ and the total drag coefficient, $C_D$ .
(c) The lift-to-drag ratio of the airplane.
(d) The propeller thrust and engine power (in hp) required for the airplane to fly.
(e) The fuel flow in gallons per hour. Will the pilot be able to return to ERAU using the remaining fuel?

We assume that the airplane’s weight for this analysis is the initial in-flight weight, denoted $W_0$ , minus half the remaining fuel weight. We have only 8 gallons of AVGAS fuel and are instructed to assume that AVGAS weighs 6.0 lb/gallon, so ${W_f}$ = 48 lb of fuel. Therefore, the weight to perform the analysis is

$W = W_0 - \frac{W_f}{2} = 2,100 - \frac{48.0}{2} = 2,076~\mbox{lb}$

An airspeed of 120 kts is equivalent to 120 $\times$ 1.688 = 202.54 ft/s = ${V_{\infty}}$ .

(a) The operating lift coefficient of the wing, $C_L$ , is

$C_L = \frac{2W}{\varrho \, V^2 S} = \frac{2 \times 2,076}{0.00216 \times 202.54^2 \times 174.0} = 0.269$

(b) To find the induced drag coefficient, we need the aspect ratio of the wing, $AR$ , i.e.,

$AR = \frac{b^2}{S} = \frac{(36.0)^2}{174.0} = 7.45$

The induced drag coefficient ${C_{D_{i}}}$ will be

$C_{D_{i}} = \frac{{C_L}^2}{\pi \, AR \, e} = \frac{0.269^2}{\pi \times 7.45 \times 0.81} = 0.00382$

The total drag coefficient, $C_D$ , is

$C_D = C_{D_0} + C_{D_i} = C_{D_0} + \frac{{C_L}^2}{\pi \, A\! R \, e} = 0.02 + 0.00382 = 0.02382$

(c) The lift-to-drag ratio of the airplane at the given conditions of flight is

$\frac{L}{D} = \frac{C_L}{C_D} = \frac{0.269}{0.02382} = 11.3$

(d) The thrust $T$ from the propeller required for the airplane to fly is

${ T = \frac{W}{C_L/C_D} = \frac{2,076}{11.3} = 183.7~\mbox{lb} }$

The power required for flight $P_{\rm req}$ will be

${P_{\rm req} = \frac{ T \, V_{\infty} }{\eta_p} = \frac{183.7 \times 202.54}{0.85 \times 550} = 79.6~\mbox{hp} }$

where the conversion factor 550 is used to convert base USC units to standard units of horsepower (hp).

(e) The fuel flow now follows directly, i.e.,

$\frac{d W_f}{dt} = {\overbigdot{W}_f} = P_{\rm req} \, c_b = 79.6 \times 0.45 = 35.8~\mbox{lb hr$^{-1}$}$

so the airplane is using about 6 gallons per hour. The time to burn off all of the available fuel is

$\mbox{Time} = T_i = \frac{48.0}{35.8} = 1.341~\mbox{hrs}$

So, the potential approximate range of the airplane is

$\mbox{Range} = R = 1.341 \times 120 \approx 160~\mbox{nautical~miles}$

In conclusion, the airplane has enough fuel to cover the 125 nautical miles remaining to return to ERAU under these simplified assumptions. However, in practice, FAA fuel-reserve requirements must also be considered. For airplane operations under VFR, the required reserve is 30 minutes of flight at normal cruising speed during the day and 45 minutes at night. Under IFR, the fuel planning requirement includes the flight to the destination, then to an alternate airport if required, and then 45 minutes at normal cruising speed. Therefore, although the simplified calculation suggests the airplane could cover the remaining 125 nautical miles to return to ERAU, the student pilot would not have an adequate reserve and should land to refuel before completing the final leg back to ERAU.

Worked Example #5

Consider a small jet-powered airplane with an initial in-flight weight of 21,000 lb flying in unaccelerated level flight at an airspeed of 250 knots where the air density is 0.0015 slugs/ft ${^3}$ . The airplane weighs 850 lb of usable fuel remaining in its tanks. The airplane’s wingspan is 48.0 ft, and the wing panels have a trapezoidal shape with a root chord $c_r$ = 9 ft and a tip chord $c_{\rm tip}$ = 4 ft. The other characteristics of the airplane are: Non-lifting drag coefficient, ${C_{D_{0}}}$ = 0.02; Oswald’s efficiency factor, ${e}$ = 0.81; Engine TSFC = $c_{\rm TSFC}$ = 0.5 lb lb^-1 hr^-1. Assume for the following analysis that the airplane’s weight from fuel burning is the initial weight minus half the remaining fuel weight. Calculate the following:

(a) The operating lift coefficient of the wing, $C_L$ .
(b) The induced drag coefficient, ${C_{D_{i}}}$ and the total drag coefficient, $C_D$ .
(c) The lift-to-drag ratio of the airplane in the given flight conditions.
(d) The thrust $T$ required from the engines for the airplane to fly.
(e) The approximate maximum potential remaining flight range of the airplane.
(f) The approximate flight time to reach the remaining flight range.

We need initial information, including the wing area and aspect ratio, as well as the weight at which to run the calculations. Calculating the area of the wing gives

$S = b\frac{(c_r + c_{\rm tip})}{2} = 48.0 \left( \frac{9.0 + 4.0}{2} \right) = 312.0~\mbox{ft}^2$

The aspect ratio can be calculated using

$AR = \frac{b^2}{S} = \frac{(48.0)^2}{312.0} = 7.38$

We are told to use the weight of the airplane at a point which is its initial weight minus half the remaining fuel weight, so

$W = W_0 - \frac{W_f}{2} = 21,000 - \frac{850.0}{2} = 20,575~\mbox{lb}$

(a) An airspeed of 250 kts is equal to 422 ft/s. The operating lift coefficient of the wing, $C_L$ , is

$C_L = \frac{2W}{\varrho_{\infty} \, V_{\infty}^2 S} = \frac{2 (20,575)}{0.0015 (422.0)^2 312.0} = 0.494$

(b) The induced drag coefficient is

$C_{D_i} = \frac{{C_L}^2}{\pi \, AR \, e} = \frac{0.494^2}{\pi \times 7.38 \times 0.81} = 0.0130$

The total drag coefficient, $C_D$ , is

$C_D = C_{D_0} + C_{D_i} = C_{D_0} + \frac{{C_L}^2}{\pi \, AR \, e} = 0.02 + 0.0130 = 0.033$

(c) The lift-to-drag ratio of the airplane at the given conditions of flight is

$\frac{L}{D} = \frac{C_L}{C_D} = \frac{0.494}{0.033} = 14.97$

(d) The thrust $T$ required for the airplane to fly is

${ T = \frac{W}{C_L/C_D} = \frac{20,575}{14.97} = 1,374.42~\mbox{lb} }$

(e) The fuel flow rate is

$\overbigdot{W}_f = c_{\rm TSFC} \, T = 0.5 \times 1,374.42 = 687.2~\mbox{lb hr$^{-1}$}$

The time to burn off all of the available fuel is

$\mbox{Time} = T_i = \frac{W_f}{c_{\rm TSFC} \, T} = \frac{850.0}{0.5 \times 1,374.42} = 1.24~\mbox{hours}$

So, the potential approximate range of the airplane at this airspeed is

$\mbox{Range} = R = 1.24 \times 250.0 = 309~\mbox{nautical~miles}$

(f) The approximate flight time to reach this remaining range is the same fuel-limited time calculated above, i.e.,

$T_i = 1.24~\mbox{hours}$

Worked Example #6

Consider an airplane with a wing of lifting planform area $S =$ 60 m ${^{2}}$ , an aspect ratio $AR =$ 12, and Oswald’s efficiency factor $e =$ 0.90. The wing has a non-lifting profile drag coefficient of 0.01. The remainder of the airplane has a non-lifting drag coefficient of 0.03. All force coefficients are based on wing area $S$ . The mass of the airplane is 16,000 kg. If the airplane is flying at a density altitude of 10,000 ft and its true airspeed is 253 kts, then calculate (a) The lift force produced by the wing; (b) The lift coefficient of the wing; (c) The drag force on the wing; (d) The lift-to-drag ratio of the wing; (e) The total drag force on the airplane; (f) The lift-to-drag ratio of the airplane.

(a) With the assumption that the airplane is flying along in steady, unaccelerated flight, the lift force produced by the wing will equal the weight of the airplane, i.e.,

$L = W = M \, g = 16,000 \times 9.81 = 156,960~\mbox{N}$

(b) The lift coefficient of the wing is given by

$C_L = \frac{L}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S}$

To find $C_L$ , we need the density of the air at 10,000 ft and the true airspeed in units of m/s. The density can be found from the ISA, assuming standard atmospheric conditions, so

$\varrho_{\infty} = 0.9048~\mbox{kg/m${^{3}}$}$

and converting from nautical miles per hour (kts) to m/s gives

$V_{\infty} = 130.154~\mbox{m s$^{-1}$}$

Inserting the numbers gives

$C_L = \frac{L}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} = \frac{156,960}{\frac{1}{2} \times 0.9048 \times 130.154^2 \times 60.0} = 0.341$

(c) The drag force on the wing will be given by

$D_{\rm wing} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_{D_{\rm wing}}$

where $C_{D_{\rm wing}}$ is the drag coefficient of the wing, which will be given by

$C_{D_{\rm wing}} = 0.01 + \frac{{C_L}^2}{\pi \, AR \, e}$

where the second part is the induced drag, i.e., drag due to lift. Inserting the numbers gives

$C_{D_{\rm wing}} = 0.01 + \frac{0.341^2}{\pi \times 12 \times 0.9} = 0.01 + 0.00343 = 0.01343$

Therefore, the drag force on the wing is

$D_{\rm wing} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_{D_{\rm wing}} = 0.5 \times 0.9048 \times 130.154^2 \times 60.0 \times 0.01343 = 6,175.4~\mbox{N}$

(d) Now that the lift and drag on the wing are known, the lift-to-drag ratio of the wing is

$\frac{L}{D_{\rm wing}} = \frac{156,960}{6,175.4} = 25.41$

(e) The total drag force on the airplane is

$D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D$

where $C_D$ is the net drag coefficient of the airplane. Because all drag coefficients are defined using the wing area as the reference area, then

$C_D = 0.03 + C_{D_{\rm wing}}$

where 0.03 is the non-lifting drag coefficient of the remainder of the airplane, and $C_{D_{\rm wing}}$ includes the wing profile drag and induced drag already calculated. Therefore,

$C_D = 0.03 + 0.01343 = 0.04343$

The total drag force on the airplane is then

$D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D = 0.5 \times 0.9048 \times 130.154^2 \times 60.0 \times 0.04343 = 19,970~\mbox{N}$

(f) Now that the lift and drag are known, the lift-to-drag ratio of the entire airplane is

$\frac{L}{D} = \frac{156,960}{19,970} = 7.86$

Worked Example #7

The goal is to estimate the endurance and range of a general-aviation airplane similar to the Cessna 182 Skylane. The parameters describing this propeller-driven airplane are listed in the table below.

Wing span

$b$

35.8 ft

Wing area

$S$

174 ft ${^{2}}$

Wing aspect ratio

$AR$

7.37

Gross takeoff weight

$W_{\rm GTOW}$

2,950 lb

Fuel capacity (tankage)

$V_f$

65 U.S. gals AVGAS

Engine rated power

$P_{\rm bhp}$

230 hp @ MSL ISA

Engine BSFC

$c_b$

0.45 lb hp $^{-1}$ hr $^{-1}$

Parasitic drag coefficient

${C_{D_{0}}}$

0.025

Oswald’s efficiency factor

${e}$

0.8

Average propeller efficiency

$\eta_p$

0.8

Based on the provided information, several key performance characteristics of the airplane can be determined, including the power requirements for flight. It can be assumed that the weight of the airplane is the gross takeoff weight, so $L = W =W_{\rm \scriptsize GTOW}$ and so

$L = \frac{1}{2} \varrho_{\infty} \, V_{\infty}^2 \, S \, C_L =W_{\rm \scriptsize GTOW} = W$

Rearranging for the lift coefficient gives

${ C_L = \frac{W}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} }$

Also, the drag coefficient for the airplane (profile drag plus induced drag) is

$C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi \, AR \, e}$

and so the total drag is

$D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S\left(C_{D_{0}} + \frac{{C_L}^2}{\pi \, AR \, e}\right)$

For level flight, the brake power required is

$P_{\rm req} = \frac{D V_{\infty}}{\eta_p}$

remembering to account for propeller efficiency. The power curve is shown in the figure below for a pressure altitude of 5,000 ft, where $\varrho_{\infty}$ = 0.002048 slugs/ft ${^3}$ .

Finally, assuming a constant BSFC gives the fuel flow rate as

$\frac{dW_f}{dt} = {\overbigdot{W}_f} = c_b P_{\rm req}$

which is shown graphically in the figure below.

The airplane’s endurance and range can be estimated using the appropriate Breguet formulas. This airplane’s fuel capacity is 65 gallons of AVGAS, and at 6.01 lb per gallon, the maximum fuel weight it could carry is 390.65 lb. This is about 13% of the gross weight, so the assumption of constant weight over the flight time is reasonable for evaluating the maximum values of $C_L/C_D$ and $C_L^{\, 3/2}/C_D$ . While these values and their corresponding airspeeds vary with weight and altitude, a representative weight for analysis is the in-flight weight, defined as the Gross Takeoff Weight (GTOW) minus half the total fuel weight.

Not all of this fuel would be usable, however, because some would be needed for startup, taxiing, takeoff, and climb, as well as for descent and landing. FAA fuel-reserve requirements must also be considered. For airplane operations under VFR, the required reserve is 30 minutes of flight at normal cruising speed during the day and 45 minutes at night. Under IFR, the fuel planning requirement includes a flight to the destination, then to an alternate airport if required, and then 45 minutes at normal cruising speed. If an average fuel allowance of 120 lb is assumed for non-cruise operations and reserve fuel, this leaves about 270 lb of fuel for the range and endurance estimates. Therefore, for this example, $W_0$ = 2,950 – 60 = 2,890 lb and $W_1$ = 2,890 – 270 = 2,620 lb.

The estimated maximum endurance can be found using

$E = \frac{ \eta_p}{c} \left(\frac{C_L^{\, 3/2}}{C_D}\right) \sqrt{2 \varrho_{\infty} S} \left( \frac{1}{\sqrt{W_1}} - \frac{1}{\sqrt{W_0} } \right)$

From the graph below, the maximum value of $C_L^{\, 3/2}/C_D$ is 12.79 at 81 mph or 118.8 ft/s.

Notice that the BSFC value must be converted into appropriate engineering units so $c_b$ = 0.45/550/3600 = 2.27 $\times 10^{-7}$ in units of (lb) (lb-ft s $^{-1}$ ) $^{-1}$ (s) $^{-1}$ . Substituting the appropriate values gives

$\begin{eqnarray*} E & = & \frac{0.8 \times 12.79}{2.27 \times 10^{-7}} \sqrt{2 \times 0.002048 \times 174} \left( \frac{1}{\sqrt{2,620}} - \frac{1}{\sqrt{2,890} } \right) \\[10pt] & = & 35,582 \mbox{s} = 593~\mbox{mins} = 9.9~\mbox{hrs} \nonumber \end{eqnarray*}$

If we loiter, the airplane at an airspeed of 81 mph would remain airborne for nearly 10 hours.

The estimated maximum range can be found using

$R = \frac{\eta_p}{c} \left( \frac{C_L}{C_D}\right) \ln \left( \frac{W_0}{W_1}\right)$

The maximum value of $C_L/C_D$ is 13.6 at 106 mph or 155 ft/s. Substituting the appropriate values gives

$\begin{eqnarray*} R & = & \frac{0.8 \times 13.6}{2.27 \times 10^{-7}} \ln \left( \frac{2,890}{2,620} \right) \\[10pt] & = & 3.73\times 10^6~\mbox{ft} = 707~\mbox{~miles} \nonumber \end{eqnarray*}$

which would only be obtained if the airplane were flown at or near 106 mph. Of course, this is rather slow compared to the airplane’s top speed of around 170 mph; however, at that airspeed, the range would be reduced by nearly half.

Remember that these latter results are only estimates of maximum range and endurance, but they are typically within 10% of the actual values demonstrated in flight.

Worked Example #8

It is desired to estimate the endurance and range of a jet-powered airplane in the form of a small jet similar to the Cessna Citation. The parameters describing this airplane are listed in the table below.

Wing span

$b$

53.3 ft

Wing area

$S$

318 ft ${^{2}}$

Wing aspect ratio

$AR$

8.93

Gross takeoff weight

$W_{\rm GTOW}$

19,815 lb

Fuel capacity (tankage)

$V_f$

1,119 U.S. gals JET-A

Engine rated thrust, per engine

${T_A}$

3,650 lb @ MSL ISA

Engine TSFC

$c_t$

0.6 lb lb $^{-1}$ hr $^{-1}$

Parasitic drag coefficient

${C_{D_{0}}}$

0.02

Oswald’s efficiency factor

${e}$

0.81

We can assume for the following calculations that the airplane’s weight is the gross takeoff weight, so $L = W = W_{\rm \scriptsize GTOW}$ , although this assumption will tend to overpredict the fuel burn. Alternatively, one can use a representative in-flight weight, such as $W_{\rm GTOW}$ minus half the usable fuel weight, which is usually a better approach for estimating performance over a finite flight segment. The drag coefficient for the airplane is

$C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi \, AR \, e}$

assuming no transonic wave drag. For level flight, then, the total thrust required is

$T_{\rm req} = D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S\left(C_{D_{0}} + \frac{{C_L}^2}{\pi \, AR \, e}\right)$

where

$C_L = \frac{W}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S}$

Substituting the known values for this exemplar jet airplane gives the thrust required curve shown below for a pressure altitude of 15,000 ft, where $\varrho_{\infty}$ = 0.0014963 slugs/ft ${^3}$ .

Because this is a twin-engine airplane, each engine must produce half of the total required thrust. Assuming a constant TSFC gives the total fuel flow rate as

$\frac{dW_f}{dt} = {\overbigdot{W}_f} = c_t \, T_{\rm req}$

which is shown in the figure below. Notice that the fuel flow mimics the thrust requirements. At higher airspeeds, fuel flow increases rapidly, indicating that for any airplane, there is a high fuel cost associated with flying fast.

We can estimate the endurance and range of this jet airplane using the appropriate Breguet formulas. We should always use the proper formulas for jet airplanes and distinguish them from propeller airplanes.

The fuel capacity of this jet airplane is 1,119 gallons of JET-A, and at 6.8 pounds per gallon, the maximum fuel weight would be 7,609.2 pounds, which is approximately 38% of the airplane’s gross weight. This is why a more accurate estimate of endurance and range would be obtained by using a representative in-flight weight, such as $W_{\rm GTOW}$ minus half the total fuel weight, when evaluating the aerodynamic ratios. Not all of this fuel will be usable, and allowances are required for the reasons discussed above. We can proceed by assuming that $W_0$ = 19,815 – 1,000 = 18,815 lb, where 1,000 lb is the allowance for fuel not used in cruise, and $W_1$ = 18,815 – 5,600 = 13,215 lb. The endurance is evaluated by using

$E = \frac{1}{c_t} \left( \frac{C_L}{C_D} \right) \ln \left(\frac{W_0}{W_1} \right)$

According to the results below, the airplane’s best $C_L/C_D$ ratio of 16.85 occurs at an airspeed of 208 kts (351 ft/s). The required ratios can also be calculated using formulas for $C_L$ and $C_D$ .

Substituting the numbers and remembering to convert $c_t$ into units of s $^{-1}$ by dividing by 3,600 gives

$E = \left(\frac{1}{0.6/3,600} \right) \left( 16.85 \right) \ln \left(\frac{18,815}{13,215} \right) = 9.9~\mbox{hrs}$

where the final result has been converted back from seconds to hours. The range is evaluated by using

$R = \frac{2}{c_t} \sqrt{ \frac{2}{\varrho_{\infty} S}} \left(\frac{C_L^{1/2}}{C_D} \right) \left( W_0^{1/2} - W_1^{1/2} \right)$

The best ${C_L^{1/2}/C_D}$ ratio of 23.4 occurs at an airspeed of 274 kts, or 462 ft/s. Substituting the actual numbers gives

$\begin{eqnarray*} R & = & \frac{2 \times 3,600}{0.6} \sqrt{ \frac{2}{0.0014963 \times 318}} \left( 23.4 \right) \left(18,815^{1/2} - 13,215^{1/2} \right) \\[10pt] & = & 2,422~\mbox{statute miles} \end{eqnarray*}$

which appears reasonable for this class of jet-powered aircraft. Of course, these results would depend on the payload, which may need to be traded for fuel. Note: For all aircraft, a maximum useful load can be carried, which is the sum of the payload and the fuel load. More payload, such as passengers or baggage, usually permits a lower fuel load.

Worked Example #9

A small jet airplane weighs 10,000 lb, has a wing area of 200 ft ${^{2}}$ , and a drag polar given by $C_D = 0.02 + 0.05 C_L^2$ . Because of a fuel leak, the airplane runs out of fuel, and the engines stop at an altitude ${h}$ = 20,000 ft. Estimate the best glide range from this altitude.

The drag polar is given by

${ C_D = 0.02 + 0.05 C_L^ {\,2} }$

Therefore,

$\frac{C_D}{C_L} = \frac{0.02}{C_L} + 0.05 C_L$

Differentiating with respect to $C_L$ gives

$\frac{d}{dC_L}\left(\frac{C_D}{C_L}\right) = -\frac{0.02}{C_L^{\,2}} + 0.05$

which will be zero for a maximum or minimum, i.e.,

$-\frac{0.02}{C_L^{\,2}} + 0.05 = 0$

or

$C_L = \sqrt{ \frac{0.02}{0.05}} = 0.632$

The drag coefficient at this lift coefficient is

$C_D = 0.02 + 0.05 \times 0.632^2 = 0.04$

Therefore, the lift-to-drag ratio is

$\frac{C_L}{C_D} = \frac{0.632}{0.04} = 15.81$

The gliding distance is given by

$R = h \left( \frac{C_L}{C_D} \right) = 20,000 \times 15.81 = 316,227~\mbox{ft} \approx 52~\mbox{nautical~miles}$

Worked Example #10 – Flight performance of a drone

A reconnaissance drone aircraft, as shown in the figure below, is cruising in trim in steady flight at an in-flight weight of 3,100 lb at a Mach number of 0.30 at an altitude of 10,000 ft. The aircraft’s drag polar is given by:

$C_D = 0.025 + 0.020 \, C_L^2$

The wing has an area of 122.5 ft² and an aspect ratio of 19.6. Assume the following ambient atmospheric conditions: $a = 1,077.0 \, \text{ft/s}$ and $\sigma = \varrho / \varrho_0 = 0.74$ .

Explain the balance of forces and moments acting on the airplane.
Determine the operating lift coefficient of the wing.
Determine the induced drag and total drag coefficients.
Calculate Oswald’s efficiency factor.
Calculate the total drag force and the lift-to-drag ratio.
Calculate the brake power required for flight if the propeller efficiency is 0.8.
If the BSFC of its piston engine is 0.5 lb hp $^{-1}$ hr $^{-1}$ , approximately how much fuel will be burned in 15 minutes of flying time? Assume that the net change in the aircraft’s weight from burning fuel is negligible.

1. The figure below explains the balance of forces, where lift equals weight and thrust equals drag. All the moments are balanced in trim, so the net moments about all axes are zero.

2. The lift equation is given by:

$L = \frac{1}{2} \varrho_\infty V_\infty^2 S C_L = W$

Because $L = W$ , the operating lift coefficient is

$C_L = \frac{2W}{\varrho_\infty V_\infty^2 S}$

The density of air is

$\varrho_\infty = 0.74 \varrho_0 = 0.74 \times 0.002378 = 0.00176 \, \text{slug/ft$^3$}$

The airspeed is

$V_\infty = a_\infty M_\infty = 1,077 \times 0.3 = 323.1 \, \text{ft/s}$

Substituting all the values gives the lift coefficient as

$C_L = \frac{2 \times 3,100}{0.00176 \times 323.1^2 \times 122.5} = 0.276$

3. The induced drag coefficient is

$C_{D_i} = \frac{C_L^2}{\pi AR \, e} = K C_L^2$

In this case $K = 0.020$ , so

$C_{D_i} = 0.020 \times 0.276^2 = 0.00152$

The total drag coefficient is

$C_D = C_{D_0} + C_{D_i} = 0.025 + 0.00152 = 0.02652$

4. The induced drag $K$ factor is

$K = \frac{1}{\pi AR \, e}$

Rearranging for ${e}$ gives

$e = \frac{1}{\pi AR \, K} = \frac{1}{\pi \times 19.6 \times 0.02} = 0.812$

5. The total drag force is given by

$D = \frac{1}{2} \varrho_\infty V_\infty^2 S C_D$

Substituting the values gives

$D = 0.5 \times 0.00176 \times 323.1^2 \times 122.5 \times 0.02652 = 298.31 \, \text{lb}$

Therefore, the lift-to-drag ratio is

$\frac{L}{D} = \frac{W}{D} = \frac{3,100}{298.31} = 10.39$

6. The brake power (in hp) required for flight is

$P_{\text{bhp}} = \frac{T V_\infty}{550 \eta_p} = \frac{D V_\infty}{550 \eta_p}$

Substituting values gives

$P_{\text{bhp}} = \frac{298.31 \times 323.1}{550 \times 0.8} = 219.16 \, \text{hp}$

7. The fuel flow rate is

$\overbigdot{W}_f = \text{BSFC} \bigcdot P_{\text{bhp}}$

Substituting gives

$\overbigdot{W}_f = 0.5 \times 219.16 = 109.58~\mbox{lb hr$^{-1}$}$

For 15 minutes ( $t = 0.25 \, \text{hr}$ ), then

$W_f = 0.25 \times 109.58 = 27.4 \, \text{lb}$

Worked Example #11

The Nemeth Parasol was an early aircraft designed for short takeoff and landing capabilities. It used a circular wing, which the inventors claimed would operate at very low airspeeds and act as a parachute during landing. In early flight demonstrations, it was reported that the aircraft descended “almost vertically with a gentle landing.” The diameter of the circular wing was 15 ft, and the aircraft had a total weight of 1,000 lb.

Calculate the aspect ratio of the circular wing and discuss the implications on induced drag. Why do aircraft typically avoid using wings with such low aspect ratios?
Assuming the aircraft stalls at a maximum lift coefficient of $C_{L_{\rm max}}$ = 1.8, estimate the stall speed of the aircraft at sea level standard conditions. Is this value reasonable?
Calculate the terminal velocity in feet per second at which the aircraft will “parachute” to a landing, assuming purely vertical motion. Hint: The drag coefficient of a circular disk with the flow perpendicular to the disk is 1.4.
Compare your answers from parts (2) and (3). Would the transition from horizontal forward flight to vertical descent happen smoothly or abruptly? What flight characteristics would help ensure a “gentle” landing?
You may decide on the outcome from part (c) that a more comfortable vertical landing speed of 12 ft/s would be appropriate. What would the diameter of the circular wing need to be for this scenario?
What do you think of the inventor’s original claims?

The aspect ratio, $AR$ , of a wing is defined as
$AR = \frac{b^2}{S}$

where $b$ is the wing span and $S$ is the planform area. For a circular wing of diameter ${d}$ = 15 ft, the span $b = d$ and the area is

$S = \frac{\pi \, d^2}{4} = \frac{\pi \times \, 15^2}{4} = 176.7~\text{ft}^2$

so that

$AR= \frac{d^2}{\pi \, d^2/4} = \frac{4}{\pi} \approx 1.273$

A low aspect ratio leads to high induced drag because the wing generates strong trailing vortices and inefficient lift distribution. Aircraft typically avoid low aspect ratios to minimize induced drag and improve aerodynamic efficiency, particularly during cruise.
The stall speed, $V_{\rm stall}$ , can be found from the lift equation at maximum lift, i.e.,
$W = \frac{1}{2} \varrho_{\infty} V_{\rm stall}^2 \, S \, C_{L_{\rm max}}$

Solving for $V_{\rm stall}$ gives

$V_{\rm stall} = \sqrt{ \frac{2 W}{\varrho_{\infty} \, S \, C_{L_{\rm max}}} }$

The air density at sea level standard conditions is $\varrho_{\infty}$ = 0.002378 slug/ft $^{3}$ , so that

$V_{\rm stall} = \sqrt{ \frac{2 \times 1000}{0.002377 \times 176.71 \times 1.8} } = 51.4~\text{ft/s} \approx 35.0~\text{mph}$

This stall speed is very low and reasonable for an aircraft designed for short takeoff and landing.
In a vertical descent, the weight of the aircraft is balanced by drag, i.e.,
$W = \frac{1}{2} \varrho_{\infty} V_{\rm term}^2 S C_D$

Solving for $V_{\rm term}$ gives

$V_{\rm term} = \sqrt{ \frac{2W}{\varrho_{\infty} S C_D} }$

where $C_D$ = 1.4. Therefore,

$V_{\rm term} = \sqrt{ \frac{2 \times 1000}{0.002377 \times 176.71 \times 1.4} } = 58.3~\text{ft/s}$
From parts (2) and (3), $V_{\rm stall} \approx$ 51.4 ft/s and $V_{\rm term} \approx$ 58.3 ft/s. The terminal velocity is slightly higher than the stall speed, suggesting that once the aircraft stalls, it would transition into a somewhat faster vertical descent. The transition would likely be abrupt without careful control.
Given a desired terminal velocity $V_{\rm term}$ = 12 ft/s, solve for required circular wing area, $S$ , using
$S = \frac{2W}{\varrho_{\infty} V_{\rm term}^2 C_D}$

Substituting values gives

$S = \frac{2 \times 1,000}{0.002377 \times (12)^2 \times 1.4} = 4,169~\text{ft}^2$

The corresponding diameter is found from

$S = \frac{\pi d^2}{4} \quad \text{so that} \quad d = \sqrt{ \frac{4S}{\pi} }$

and substituting values gives

$d = \sqrt{ \frac{4 \times 4169}{\pi} } = 72.9~\text{ft}$

Therefore, the diameter of the circular wing would need to be approximately 73 ft to achieve a vertical landing speed of 12 ft/s.
The inventor’s claims are only partially credible. The aircraft likely exhibited a low forward stall speed and substantial drag during descent, but the calculated vertical terminal speed of 58.3 ft/s is still too high to be considered a gentle vertical landing. Therefore, the claim that the aircraft could descend almost vertically and land gently is not supported by this simplified calculation. Achieving a much lower vertical landing speed would require a much larger circular wing, as shown in part (5), or some additional aerodynamic or piloting mechanism to arrest the descent before touchdown.