Question 4 of 7: Complete Performance Analysis of a Jet Aircraft

Question 4: Complete Performance Analysis of a Jet Aircraft (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single jet aircraft is analysed in five different flight conditions, all from the one set of data below and the parabolic drag polar $C_D = C_{D0} + K C_L^2$ with $C_{D0} = 0.027$ and $K = 0.033$.

Find. (a) the parasite and induced drag at maximum sea-level speed; (b) the minimum-drag speed at sea level and the parasite share of the total drag there; (c) the maximum climb angle at $8600\ \text{m}$; (d) the minimum glide angle and the speed at which it occurs at $1500\ \text{m}$; (e) the maximum-range and maximum-endurance speeds at $8000\ \text{m}$.

Approach. Every part is a different reading of the same drag equation $D = q S C_{D0} + K W^2/(qS)$. Maximum speed is where thrust available equals that drag (the larger root of the resulting quadratic in $q$); minimum drag is where its two terms are equal, at $C_L = \sqrt{C_{D0}/K}$; climb and glide angles are the excess-thrust and lift-to-drag readings of the same curve; and range and endurance are two different optima of $C_L$ for a jet. Thrust is lapsed with altitude as $T = T_{SL}(\rho/\rho_0)$.

Part (a) — parasite and induced drag at maximum sea-level speed.

1. Write drag as a function of dynamic pressure at fixed weight. In level flight $L = W$, so $C_L = W/(qS)$ and $$D = q S C_{D0} + \frac{K W^2}{q S}.$$
2. Set thrust available equal to drag required. At maximum speed $T = D$, and multiplying through by $q$ gives a quadratic in the dynamic pressure: $S C_{D0}\,q^2 - T\,q + K W^2/S = 0$, i.e. $1.485\,q^2 - 67\,000\,q + 1.2992\times 10^{7} = 0$.
3. Take the larger root. The two roots are the fast and slow level-flight speeds on full thrust; maximum speed is the larger. $q_{max} = \dfrac{67\,000 + \sqrt{67\,000^2 - 4(1.485)(1.2992\times 10^{7})}}{2(1.485)} = 44\,923\ \text{Pa}$.
4. Convert to a true airspeed. $V_{max} = \sqrt{2q_{max}/\rho_0} = \sqrt{2(44\,923)/1.225}$, so $$\boxed{\ V_{max} = 270.8\ \text{m}/\text{s} = 975\ \text{km}/\text{hr}\ (M = 0.796)\ }$$
5. Split the drag into its two parts at that condition. The parasite term is $D_0 = q_{max} S C_{D0} = 44\,923 \times 55 \times 0.027$ and the induced term is $D_i = K W^2/(q_{max}S) = 1.2992\times 10^{7}\,(55)/(44\,923 \times 55 \times 55)$, giving $$\boxed{\ D_{parasite} = 66\,711\ \text{N}, \qquad D_{induced} = 289\ \text{N}\ }$$

The arithmetic check is that the two must add to the thrust available: $66\,711 + 289 = 67\,000\ \text{N}$ exactly, which confirms that the correct root was taken. The physical message is the extreme asymmetry — at maximum speed the induced drag is only 0.43 per cent of the total, because the aircraft is flying at a lift coefficient of just $C_L = W/(q S) = 0.0596$. Fast flight is essentially a contest between thrust and parasite drag alone.

Part (b) — minimum-drag speed and the parasite share of total drag.

1. Differentiate the drag with respect to dynamic pressure. Setting $dD/dq = S C_{D0} - K W^2/(q^2 S) = 0$ gives $q_{md} = (W/S)\sqrt{K/C_{D0}}$, which is exactly the condition that the two drag terms are equal.
2. Express it as a lift coefficient. Since $C_L = W/(qS)$, the condition becomes $C_{L,md} = \sqrt{C_{D0}/K} = \sqrt{0.027/0.033} = 0.9045$. Note that this is a pure property of the polar and does not depend on altitude or weight.
3. Convert to a speed at sea level. $V_{md} = \sqrt{\dfrac{2W}{\rho_0 S\, C_{L,md}}} = \sqrt{\dfrac{2(147\,150)}{1.225 \times 55 \times 0.9045}}$, so $$\boxed{\ V_{md} = 69.5\ \text{m}/\text{s} = 250\ \text{km}/\text{hr}\ }$$
4. Evaluate the ratio of parasite to total drag there. Because the two terms are equal at this point by construction, $D_{min} = 2 q_{md} S C_{D0} = 2W\sqrt{C_{D0}K} = 8785\ \text{N}$ and $$\boxed{\ \frac{D_{parasite}}{D_{total}} = \frac{1}{2} = 0.500\ }$$

The value one-half is not a coincidence of these numbers: for any parabolic polar the minimum-drag point is defined by the parasite and induced contributions being equal, so the parasite share is exactly 50 per cent whatever $C_{D0}$, $K$, $W$ or altitude may be. This condition is simultaneously the maximum lift-to-drag ratio, $(L/D)_{max} = 1/(2\sqrt{C_{D0}K}) = 16.75$, and it reappears in parts (c), (d) and (e) as the governing point for the best climb angle, the flattest glide and the best jet endurance.

Part (c) — maximum angle of climb at 8600 m.

1. Resolve the forces along the flight path in a steady climb. With the climb angle $\gamma$ measured from the horizontal, $T = D + W\sin\gamma$, so $\sin\gamma = (T - D)/W$. The climb angle is greatest where the excess thrust $T - D$ is greatest, which for constant thrust means where the drag is least.
2. Substitute the minimum drag. Using $D_{min} = 2W\sqrt{C_{D0}K}$ from part (b), $$\sin\gamma_{max} = \frac{T}{W} - 2\sqrt{C_{D0}K}.$$
3. Lapse the thrust to 8600 m. The standard temperature is $T = 288.15 - 0.0065(8600) = 232.25\ \text{K}$, so $\rho/\rho_0 = (232.25/288.15)^{4.2559} = 0.3994$ and $\rho = 0.4892\ \text{kg}/\text{m}^3$. On the fixed-geometry assumption $T = 67\,000 \times 0.3994 = 26\,758\ \text{N}$.
4. Evaluate the climb angle. $\sin\gamma_{max} = 26\,758/147\,150 - 2\sqrt{0.027 \times 0.033} = 0.18183 - 0.05970 = 0.12213$, hence $$\boxed{\ \gamma_{max} = 7.02^\circ\ }$$

The speed at which this angle is achieved is the minimum-drag speed at that altitude, $V = \sqrt{2W/(\rho S C_{L,md})} = 110.0\ \text{m}/\text{s}$, and the corresponding rate of climb is $V\sin\gamma = 13.4\ \text{m}/\text{s}$. It is worth noticing how much of the sea-level capability has been eaten by altitude: at sea level the same formula would give $\sin\gamma_{max} = 0.4553 - 0.0597 = 0.3956$, a climb angle of $23.3^\circ$. Thrust falls with density while the drag term $2\sqrt{C_{D0}K}$ does not change at all, so the excess thrust is squeezed from both ends as the aircraft climbs; the absolute ceiling is the altitude at which $T/W$ has fallen to $0.0597$ and the climb angle reaches zero.

Part (d) — minimum glide angle and the speed at which it occurs at 1500 m.

1. Resolve the forces in an unpowered glide. With the thrust zero and the glide path at angle $\gamma$ below the horizontal, the component of weight along the path balances the drag and the component normal to it balances the lift: $D = W\sin\gamma$ and $L = W\cos\gamma$. Dividing, $\tan\gamma = D/L = C_D/C_L$.
2. Minimise the glide angle. The shallowest glide is therefore the maximum lift-to-drag ratio, which from part (b) occurs at $C_L = C_{L,md} = 0.9045$: $$\tan\gamma_{min} = \frac{1}{(L/D)_{max}} = 2\sqrt{C_{D0}K} = 0.05970,$$ so $$\boxed{\ \gamma_{min} = 3.42^\circ,\qquad (L/D)_{max} = 16.75\ }$$
3. Find the speed at which it occurs at 1500 m. The glide angle itself is independent of altitude and of weight, but the speed is not. With $\rho = 1.0581\ \text{kg}/\text{m}^3$ at $1500\ \text{m}$ and $L = W\cos\gamma_{min}$, $V = \sqrt{\dfrac{2W\cos\gamma_{min}}{\rho S\, C_{L,md}}} = \sqrt{\dfrac{2(147\,150)(0.99822)}{1.0581 \times 55 \times 0.9045}}$, giving $$\boxed{\ V = 74.7\ \text{m}/\text{s} = 269\ \text{km}/\text{hr}\ }$$

The $\cos\gamma_{min}$ factor changes the speed by less than 0.1 per cent here, so the shallow-glide approximation $V \approx \sqrt{2W/(\rho S C_L)}$ is entirely adequate; it is quoted in full only because it is the exact statement. A glide ratio of 16.75 means the aircraft covers $16.75\ \text{m}$ horizontally for every metre it descends, so from $1500\ \text{m}$ it could reach an airfield $25.1\ \text{km}$ away in still air. Note that flying faster or slower than $74.7\ \text{m}/\text{s}$ steepens the glide in both directions, and that a heavier aircraft glides just as far — it simply does so faster, since $V$ scales as $\sqrt{W}$ while $\gamma_{min}$ does not change.

Part (e) — maximum-range and maximum-endurance speeds at 8000 m.

1. Identify the correct optima for a jet. A jet burns fuel in proportion to the thrust it produces, so endurance — time aloft per unit fuel — is maximised where the thrust required, and hence the drag, is least, i.e. at $(L/D)_{max}$. Range — distance per unit fuel — is maximised where $V/D$ is greatest, equivalently where $C_L^{1/2}/C_D$ is a maximum.
2. Evaluate the two lift coefficients. Maximising $C_L^{1/2}/(C_{D0}+KC_L^2)$ gives $C_{D0} = 3K C_L^2$, so $C_{L,range} = \sqrt{C_{D0}/(3K)} = \sqrt{0.027/0.099} = 0.5222$, while $C_{L,end} = \sqrt{C_{D0}/K} = 0.9045$ as before.
3. Standard density at 8000 m. $T = 288.15 - 0.0065(8000) = 236.15\ \text{K}$, so $\rho = 1.225(236.15/288.15)^{4.2559} = 0.5252\ \text{kg}/\text{m}^3$.
4. Convert both to speeds. Using $V = \sqrt{2W/(\rho S C_L)}$ with $2W = 294\,300\ \text{N}$ and $\rho S = 28.88\ \text{kg}/\text{m}$, $$\boxed{\ V_{endurance} = 106.1\ \text{m}/\text{s},\qquad V_{range} = 139.7\ \text{m}/\text{s}\ }$$

The ratio of the two is a useful check that the right optimum was taken: because the lift coefficients differ by a factor $\sqrt{3}$, the speeds must differ by $3^{1/4} = 1.316$, and indeed $139.7/106.1 = 1.316$. The best-range speed is always the faster of the two for a jet, which is the opposite of the propeller aircraft result, where range is flown at $(L/D)_{max}$ and endurance slower still. In practice a jet is cruised a little faster than the theoretical best-range speed, on the flat part of the specific-air-range curve, because the time saved is worth more than the small fuel penalty; and both speeds here, at $M = 0.345$ and $M = 0.453$, are comfortably below the compressibility rise, so the low-speed polar used throughout remains valid.