How does the Boeing TTBW reduce fuel consumption?

Question

Compared to a traditional airliner wing, the Boeing Transonic Truss-Braced Wing (TTBW) showcases a much longer wingspan and an associated higher aspect ratio. Intuitively, this seems to reduce fuel consumption by lowering induced drag by involving more air in the creation of a given lift value.

However, if a given airliner is flying at an example 35,000ft and Mach 0.78, the aircraft will usually be flying faster than its L/D max AOA. This implies that the induced drag is already less than the parasite drag values, and the aircraft could save a bit of fuel by slowing down. It flies faster than the ideal L/D max AOA for economic reasons rather than for fuel savings.

Now if you fly the same flight profile with a larger span TTBW wing, it would seem that the AOA would be even lower than on the previous example, thus resulting in no fuel savings.

In order to realize the potential fuel savings, would a TTBW wing be required to fly either slower or higher than a typical modern airliner configuration in order to keep the AOA values in an efficient range? Or are there also significant parasite drag savings associated with the short chord lengths of the TTBW that would negate this need to go slower / higher?

Answer 1

We have a question here concerning the Boeing TTBW study from the time it was called Sugar VOLT. In my answer I speculated that the cruise Mach number of that design is 0.75 or lower.

You are correct with your suspicion that the TTBW needs to fly slower than current airliners. Since the best L/D is reached when induced and friction drag are about equal (actually, depending on engine characteristics, induced drag should be a bit less than friction drag; see below for proof), the higher aspect ratio will drive this optimum to higher lift coefficients and, consequently, lower speeds. The gains from flying higher drop off once the airplane climbs into the stratosphere since air temperature is constant above the tropopause and even rises in the upper stratosphere. Also, flying higher needs a heavier pressure vessel to keep cabin pressure comfortable. Therefore, a slower cruise speed will be the best way to lower fuel consumption.

Proof along the lines of this answer:

We use the quadratic drag equation to model the airplane and model thrust $T$ over speed $v$ as $T \varpropto v^{n_v}$ with $n_v$ a negative number for propeller and turbofan engines, and positive for ramjets.

We start from the equilibrium in steady flight:

$$T_0\cdot v^{n_v} = c_D\cdot\frac{\rho}{2}\cdot v^2\cdot S$$

$T_0$ is the reference thrust at a specific speed and depends only on fuel flow. For best range we should optimize the ratio of fuel flow over speed:

$$\frac{T_0}{v}\cdot v^{n_v} = c_D\cdot\frac{\rho}{2}\cdot v\cdot S$$ $$\Leftrightarrow\frac{T_0}{v} = c_D\cdot\frac{\rho}{2}\cdot v^{1-n_v}\cdot S$$

Next, we approximate the drag coefficient with the quadratic polar $\left(c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)$:

$$\frac{T_0}{v} = c_{D0}\cdot\frac{\rho}{2}\cdot S \cdot v^{1-n_v} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon} \cdot\frac{\rho}{2}\cdot S \cdot v^{1-n_v}$$

and express the lift coefficient in terms of speed $\left(c_L = \frac{2\cdot m\cdot g}{\rho \cdot v^2 \cdot S} \right)$

$$\frac{T_0}{v} = c_{D0}\cdot\frac{\rho}{2}\cdot S \cdot v^{1-n_v} + \frac{2\cdot (m\cdot g)^2}{\pi\cdot AR\cdot\epsilon \cdot\rho \cdot S} \cdot v^{-3-n_v}$$

Now we can differentiate the right part of the equation with respect to $v$ and set the expression to zero to find its optimum:

$$0 = (1-n_v)\cdot c_{D0}\cdot\frac{\rho}{2}\cdot S \cdot v^{-n_v} - (3+n_v)\cdot\frac{2\cdot (m\cdot g)^2}{\pi\cdot AR\cdot\epsilon \cdot\rho \cdot S} \cdot v^{-4-n_v}$$

Things will look better once we re-introduce the lift coefficient:

$$0 = (1-n_v)\cdot c_{D0} - (3+n_v)\cdot\frac{c_L^2}{\pi\cdot AR \cdot \epsilon}$$

$$\Leftrightarrow c_L = \sqrt{\frac{1-n_v}{3+n_v}\cdot\pi\cdot AR\cdot\epsilon\cdot c_{D0}}$$

This by itself is not yet helpful, but if we look at the ratio of the drag components at specific values of $n_v$, the answer becomes clearer:

$$c_{Di} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon} = \frac{1-n_v}{3+n_v} \cdot c_{D0}$$

Propeller aircraft ($n_v$ = -1): $c_{Di} = c_{D0}$ at the speed for lowest drag,

Turbofan aircraft ($n_v$ = -0.5): $c_{Di} = \frac{3}{5}\cdot c_{D0}$ at the speed for lowest drag,

Turbojet aircraft ($n_v$ = 0): $c_{Di} = \frac{1}{3}\cdot c_{D0}$ at the speed for lowest drag.

At high speed Mach effects cannot be neglected. Here the figure of merit is maximum lift coefficient times Mach squared. A typical value for a modern wing would be 0.4, and when the zero-lift drag coefficient is known, the optimum lift coefficient can be determined which in turn allows to find the maximum cruise Mach number.

Let's assume that $c_{D0}$ is 0.014, given the high Reynolds number and the wing bracing. Aspect ratio is about 20, so the lift coefficient at a $n_v$ of -0.6 and $\epsilon$ = 0.9 is 0.73. This would limit the buffet-free Mach number of the unswept wing to 0.74. Even with rather optimistic assumptions it is hard to justify a cruise Mach number of 0.75 and above for the TTBW.

Nomenclature:
$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent, as in $T = T_0\cdot v^{n_v} $
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing
$\epsilon \:\:\:\:\:$ the wing's Oswald factor
$c_{D0} \:$ zero-lift drag coefficient
$c_{Di} \:\:$ induced drag coefficient