Question 2 of 7: Drag Mechanisms, Level-Flight Equilibrium and Wing Sizing

Question 2: Drag Mechanisms, Level-Flight Equilibrium and Wing Sizing (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.

Part (a) — the four drag terms. Skin friction drag is the streamwise resultant of the shear stress the air exerts on every wetted surface of the aircraft. It exists because the air sticks to the surface, so a boundary layer develops in which the velocity rises from zero at the wall to the local external value, and the wall shear stress is $\tau_w = \mu (\partial u/\partial y)_{y=0}$. Its magnitude depends strongly on whether that boundary layer is laminar or turbulent: a turbulent layer has a much fuller velocity profile, a steeper wall gradient and therefore several times the skin friction of a laminar layer at the same Reynolds number. Skin friction dominates the drag of long, slender, well-streamlined bodies at low speed.

Induced drag — more precisely, drag due to lift — is the price of generating lift with a wing of finite span. The pressure difference between the lower and upper surfaces drives a flow around the tips, which rolls up into trailing vortices. Those vortices induce a downward velocity, the downwash, over the wing itself, which tilts the local relative wind downwards through the induced angle $\alpha_i$. The lift vector, being perpendicular to the local relative wind, is therefore tilted rearwards, and its component along the flight path is the induced drag. For a wing of aspect ratio $AR$ and span efficiency $e$, $C_{D,i} = C_L^2/(\pi\, AR\, e)$ — quadratic in lift coefficient, so it matters most when the aircraft is slow, heavy or manoeuvring, and it vanishes only in the limit of infinite span.

Parasite drag is the collective name for everything that is not drag due to lift: skin friction over the whole airframe, the pressure or form drag associated with boundary-layer displacement and any separation, the drag of undercarriage, aerials, excrescences and cooling flows, and the interference drag generated where components meet. It is very nearly independent of lift coefficient, so it scales with dynamic pressure and rises with the square of speed. In the parabolic polar $C_D = C_{D0} + K C_L^2$ used throughout this paper, $C_{D0}$ is the parasite term and $K C_L^2$ the induced term.

Compressibility drag, or wave drag, appears only when the flow field contains supersonic regions. Above the critical Mach number a pocket of supersonic flow forms over the wing and is closed by a shock wave; the entropy rise across the shock represents an irreversible loss which appears as drag, and the steep adverse pressure gradient the shock imposes on the boundary layer commonly separates it, adding a large pressure drag. The result is the abrupt transonic drag rise addressed in Question 3(f).

Part (b) — thrust required and weight in steady level flight.

Given. The aircraft is in steady, level, unaccelerated flight with the following data.

Find. The thrust required to hold these conditions and the weight of the aircraft.

Approach. Steady level flight means lift balances weight and thrust balances drag, so both answers follow from one dynamic pressure and the two given coefficients.

1. Standard density at 1500 m. $T = 288.15 - 0.0065(1500) = 278.40\ \text{K}$, so $\rho = \rho_0 (T/T_0)^{4.2559} = 1.225 \times (278.40/288.15)^{4.2559} = 1.0581\ \text{kg}/\text{m}^3$.
2. Dynamic pressure. With $V = 320/3.6 = 88.89\ \text{m}/\text{s}$, $q = \tfrac{1}{2}\rho V^2 = 0.5 \times 1.0581 \times 88.89^2 = 4180\ \text{Pa}$.
3. Weight from the lift equation. In level flight $L = W$, and $L = q S C_L = 4180 \times 45 \times 1.05$, so $$\boxed{\ W = 197\,506\ \text{N} \approx 197.5\ \text{kN}\ }$$ which corresponds to a mass of $W/g = 197\,506/9.81 = 20\,133\ \text{kg}$.
4. Thrust from the drag equation. In unaccelerated flight $T = D$, and $D = q S C_D = 4180 \times 45 \times 0.07$, so $$\boxed{\ T = 13\,167\ \text{N} \approx 13.2\ \text{kN}\ }$$

The two answers are consistent with each other in the way the question intends: because both forces share the same $qS$, their ratio is simply the ratio of the coefficients, $W/T = C_L/C_D = 1.05/0.07 = 15.0$. A lift-to-drag ratio of 15 at a relatively high lift coefficient is typical of a subsonic transport in a low-speed, high-lift condition such as a climb-out or a holding pattern, and the thrust required is only about 6.7 per cent of the weight.

Part (c) — wing area required for a specified minimum speed.

Given. Mass $m = 5000\ \text{kg}$ (so $W = 49\,050\ \text{N}$), maximum lift coefficient $C_{L,max} = 1.4$, and a required minimum flight speed of $V_{min} = 220\ \text{m}/\text{s}$ at sea level, where $\rho_0 = 1.225\ \text{kg}/\text{m}^3$.

Find. The wing area $S$ that makes $220\ \text{m}/\text{s}$ the minimum speed.

Approach. The minimum speed is the stalling speed: it is the slowest speed at which the wing can still generate a lift equal to the weight, which happens when the wing is working at $C_{L,max}$. Setting $L = W$ at that condition and solving for $S$ is the whole calculation.

1. Write the stall condition. At the minimum speed the wing is at its maximum lift coefficient and lift still equals weight: $W = \tfrac{1}{2}\rho_0 V_{min}^2 S\, C_{L,max}$.
2. Rearrange for the wing area. $S = \dfrac{2W}{\rho_0 V_{min}^2 C_{L,max}}$.
3. Substitute the given data. $S = \dfrac{2 \times 5000 \times 9.81}{1.225 \times 220^2 \times 1.4} = \dfrac{98\,100}{83\,006}$, hence $$\boxed{\ S = 1.18\ \text{m}^2\ }$$

Whichever datum is used, the structure of the result is the message: required wing area varies inversely with the square of the target minimum speed and inversely with $C_{L,max}$. Halving the acceptable approach speed quadruples the wing you must carry for the rest of the flight, and that is precisely why high-lift devices exist — raising $C_{L,max}$ from 1.4 to 2.6 with flaps and slats delivers the same low-speed capability from a wing 46 per cent smaller, whose lower parasite drag and lower weight pay dividends at cruise.