Question 5 of 7: Complete Performance Analysis of a Jet Transport

Question 5: Complete Performance Analysis of a Jet Transport (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 transport is analysed through all five parts, so its data are collected once here and the derived polar constants are reused throughout.

Find. Maximum speed at two altitudes; minimum glide angle and the speed at which it occurs; maximum rate of climb at two altitudes; minimum-drag speed at 8000 m with the parasite/induced split; and the absolute ceiling.

Approach. All five parts follow from one drag equation. Writing $q = \tfrac{1}{2}\rho V^2$, the drag in level flight with $L = W$ is

a parasite term rising as $V^2$ and an induced term falling as $1/V^2$. Maximum speed is where this equals the thrust available; minimum drag is where the two terms are equal; the glide angle and the ceiling follow from the minimum-drag condition; and the rate of climb follows from the excess power $(T - D)V/W$.

(a) Maximum speed at sea level and at 5000 m

1. Set thrust equal to drag and clear the fractions. With $T = qSC_{D0} + KW^2/(qS)$, multiplying through by $q$ gives a quadratic in the dynamic pressure, $$S\,C_{D0}\,q^2 - T q + \frac{K W^2}{S} = 0,$$ whose larger root corresponds to the high-speed intersection (the smaller root is the low-speed, induced-drag-limited intersection, which is normally below the stall): $$q = \frac{T + \sqrt{T^2 - 4 C_{D0} K W^2}}{2 S C_{D0}}.$$
2. Sea level. $T = 160\,000\ \text{N}$, $W = 495\,405\ \text{N}$, so $T^2 - 4C_{D0}KW^2 = 2.560\times10^{10} - 1.126\times10^{9} = 2.447\times10^{10}$ and $$q = \frac{160\,000 + 156\,442}{2(120)(0.031)} = 42\,532\ \text{Pa}, \qquad V = \sqrt{\frac{2q}{\rho_0}} = \boxed{263.5\ \text{m}/\text{s} = 949\ \text{km}/\text{h}}$$
3. At 5000 m. $T_{5000} = 288.15 - 0.0065(5000) = 255.65\ \text{K}$ gives $\rho = 1.225(255.65/288.15)^{4.2559} = 0.7361\ \text{kg}/\text{m}^3$, so the available thrust falls to $T = 160(0.7361/1.225) = 96.15\ \text{kN}$. Repeating the quadratic with this thrust and this density, $$V = \boxed{260.8\ \text{m}/\text{s} = 939\ \text{km}/\text{h}}$$

The result deserves comment, because it is the standard trap in this question. Under the assumed lapse the maximum speed barely changes with altitude — it falls by only 1 % over 5000 m — and it certainly does not rise. The reason is visible in the algebra: at high speed the induced term is negligible, so $q_{\max} \approx T/(SC_{D0})$, and since both $T$ and $q$ are proportional to density, $V_{\max}$ is nearly independent of altitude. An answer showing a large increase with altitude has almost certainly forgotten to lapse the thrust. What does increase is the Mach number, from $263.5/340.3 = 0.774$ at sea level to $260.8/320.5 = 0.814$ at 5000 m, and it is that — not thrust — which limits the real aeroplane, since the question's instruction to ignore compressibility has by then become physically untenable.

(b) Minimum glide angle at 1000 m and the corresponding speed

1. Glide angle from the force balance. In a steady unpowered glide the weight component along the flight path balances drag and the component normal to it balances lift, so $\tan\gamma = D/L = C_D/C_L$. The glide angle is therefore minimised when $L/D$ is a maximum — it does not depend on weight or on density at all.
2. Maximum lift-to-drag ratio for a parabolic polar. Minimising $C_D/C_L = (C_{D0} + KC_L^2)/C_L$ gives $C_{L,md} = \sqrt{C_{D0}/K}$ and $$\left(\frac{L}{D}\right)_{\max} = \frac{1}{2\sqrt{C_{D0}K}} = \frac{1}{2\sqrt{0.031 \times 0.037}} = 14.76.$$
3. Minimum glide angle. $$\gamma_{\min} = \arctan\!\left(\frac{1}{14.76}\right) = \boxed{3.87^\circ}$$ which corresponds to 14.76 m of ground covered for every metre of height lost — about 66 km from a cruise altitude of 4500 m.
4. Speed in that glide at 1000 m. The lift coefficient is $C_{L,md} = \sqrt{0.031/0.037} = 0.9153$, and at 1000 m, $T = 281.65\ \text{K}$ so $\rho = 1.1116\ \text{kg}/\text{m}^3$. Since lift supports only the component $W\cos\gamma$, $$V = \sqrt{\frac{2W\cos\gamma}{\rho S\,C_{L,md}}} = \sqrt{\frac{2(495\,405)(0.99772)}{1.1116 \times 120 \times 0.9153}} = \boxed{90.0\ \text{m}/\text{s} = 324\ \text{km}/\text{h}}$$

The corresponding rate of descent is $V\sin\gamma = 90.0(0.06757) = 6.08\ \text{m}/\text{s}$. Note that the $\cos\gamma$ factor changes the speed by only 0.1 % at this shallow angle and is routinely dropped; it is retained here because it costs nothing and keeps the force balance exact.

(c) Maximum rate of climb at sea level and at 8000 m

1. Rate of climb from excess power. For a steady climb at small angle, the rate of climb is the specific excess power, $$R/C = \frac{(T - D)V}{W}.$$ Maximising this with respect to $V$, with $D$ from the drag equation and $T$ independent of speed, gives the condition $$1.5\,\rho S C_{D0}\,x^2 - T x - \frac{2KW^2}{\rho S} = 0, \qquad x = V^2,$$ so the best-climb speed is the positive root $V = \sqrt{x}$.
2. Sea level. With $\rho_0 = 1.225$ and $T = 160\ \text{kN}$ the root is $V = 155.4\ \text{m}/\text{s}$. At that speed $q = 14\,790\ \text{Pa}$, the drag is $D = 60.15\ \text{kN}$, and $$R/C = \frac{(160\,000 - 60\,153)(155.4)}{495\,405} = \boxed{31.3\ \text{m}/\text{s} = 1879\ \text{m}/\text{min}}$$
3. At 8000 m. $T = 236.15\ \text{K}$, $\rho = 0.5252\ \text{kg}/\text{m}^3$, and the thrust available has fallen to $160(0.5252/1.225) = 68.59\ \text{kN}$. The same root-finding gives $V = 164.4\ \text{m}/\text{s}$, $D = 37.07\ \text{kN}$, and $$R/C = \frac{(68\,593 - 37\,073)(164.4)}{495\,405} = \boxed{10.5\ \text{m}/\text{s} = 628\ \text{m}/\text{min}}$$

Two checks confirm the roots are the right ones. The best-climb speed is close to twice the minimum-drag speed found in part (d) — 155.4 m/s against a sea-level minimum-drag speed of 85.8 m/s — which is the expected result for a jet, and it increases slightly with altitude while the climb rate falls by two-thirds. The rate of climb has dropped by 66 % over 8000 m although thrust has dropped by only 57 %, because the drag at the best-climb speed does not fall as fast as the thrust does.

(d) Minimum-drag speed at 8000 m and the parasite/induced split

1. Minimum-drag condition. Differentiating $D = qSC_{D0} + KW^2/(qS)$ with respect to $q$ and setting the result to zero gives $qSC_{D0} = KW^2/(qS)$: the parasite and induced drags are equal at minimum drag. This is equivalent to the condition $C_L = C_{L,md} = \sqrt{C_{D0}/K} = 0.9153$ already used in part (b).
2. Speed at 8000 m. With $\rho = 0.5252\ \text{kg}/\text{m}^3$, $$V_{md} = \sqrt{\frac{2W}{\rho S\,C_{L,md}}} = \sqrt{\frac{2(495\,405)}{0.5252 \times 120 \times 0.9153}} = \boxed{131.1\ \text{m}/\text{s} = 472\ \text{km}/\text{h}}$$
3. The two drag components. At this speed $q = \tfrac{1}{2}(0.5252)(131.1)^2 = 4513\ \text{Pa}$, so $$D_{parasite} = qSC_{D0} = 4513(120)(0.031) = 16.78\ \text{kN}, \qquad D_{induced} = \frac{KW^2}{qS} = \frac{0.037(495\,405)^2}{4513(120)} = 16.78\ \text{kN},$$ which are equal to five significant figures — the arithmetic check that the minimum-drag condition has been applied correctly.
4. Total minimum drag. $$D_{\min} = 2W\sqrt{C_{D0}K} = 2(495\,405)(0.033867) = \boxed{33.56\ \text{kN}}$$ and $W/D_{\min} = 14.76 = (L/D)_{\max}$, consistent with part (b).

It is worth emphasising that $D_{\min}$ contains no density term: the minimum drag of an aeroplane in level flight is the same at every altitude, and only the speed at which it occurs changes. That fact is what makes part (e) a one-line calculation.

(e) Absolute ceiling

1. Ceiling condition. At the absolute ceiling the rate of climb is zero, so the thrust available exactly equals the minimum drag the aircraft can achieve: $$T_0\,\frac{\rho}{\rho_0} = D_{\min} = 2W\sqrt{C_{D0}K}.$$
2. Density at the ceiling. $$\rho_{ceiling} = \rho_0\,\frac{D_{\min}}{T_0} = 1.225 \times \frac{33\,556}{160\,000} = 0.2569\ \text{kg}/\text{m}^3.$$
3. Check which atmospheric layer this lies in — the step that is easy to skip. At the tropopause ($h = 11\,000\ \text{m}$, $T = 216.65\ \text{K}$) the ISA density is $$\rho_{11} = 1.225\,(216.65/288.15)^{4.2559} = 0.3639\ \text{kg}/\text{m}^3.$$ Since $0.2569 < 0.3639$, the ceiling lies above the tropopause and the tropospheric power law must not be used.
4. Invert the isothermal stratospheric relation. In the lower stratosphere the temperature is constant and the density decays exponentially, $\rho = \rho_{11}\exp[-g(h - 11\,000)/(RT_{11})]$, so $$h = 11\,000 + \frac{R\,T_{11}}{g}\ln\!\left(\frac{\rho_{11}}{\rho_{ceiling}}\right) = 11\,000 + 6342\,\ln\!\left(\frac{0.3639}{0.2569}\right)$$ $$h = 11\,000 + 6342(0.3482) = \boxed{13\,208\ \text{m}}$$
5. Speed at the ceiling. The aircraft can fly only at the minimum-drag lift coefficient there, so $$V = \sqrt{\frac{2W}{\rho_{ceiling}S\,C_{L,md}}} = 187.4\ \text{m}/\text{s},$$ corresponding to $M = 187.4/295.1 = 0.635$ at the stratospheric speed of sound.

Had step 3 been skipped and the tropospheric power law applied to $\rho_{ceiling} = 0.2569$, the answer would have come out near 13 620 m — some 410 m too high, with nothing in the arithmetic to signal that anything had gone wrong. Checking the computed density against $\rho_{11}$ before inverting is therefore not a formality; on this class of question it is the difference between a right and a wrong answer.