EL
Evidence Ledger

Claim

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The Energy Losses Due to Air Friction on the Spinning Screws are Acceptable

air frictionpower lossscrews

Evidence

If we assume that the vacuum level inside the evacuated tube, PP, is 5 Pa and the air temperature, TT, is 20°C (293.15 K), the air density inside the tube can be estimated using the ideal gas law:

ρ=PRT=5287.05×293.15=5.94×10−5 kg/m3\rho = \frac{P}{RT} = \frac{5}{287.05 \times 293.15} = 5.94 \times 10^{-5}\ \text{kg/m}^3

where RR is the specific gas constant for dry air, 287.05 J/(kg·K).

The screw flights rotate with angular velocity ω\omega. Based on a maximum flight-tip speed of 530 m/s and a flight-tip radius of 0.5 m, the angular velocity is:

ω=vr=5300.5≈1,060 rad/s\omega = \frac{v}{r} = \frac{530}{0.5} \approx 1{,}060\ \text{rad/s}

We initially modeled the screw flights as flat plates travelling through stationary air, but later learned that this approach overestimates the drag because the rotating screws entrain the small quantity of residual gas surrounding them. We then switched to a different analytical approach based on Taylor–Couette flow. In fluid dynamics, Taylor–Couette flow consists of a viscous fluid confined in the gap between two rotating cylinders. A standard reference is Landau and Lifshitz, Fluid Mechanics, 2nd ed., Section 18, "Flow between rotating cylinders."

For an inner cylinder of radius aa, rotating at angular velocity ω\omega, inside a stationary cylindrical boundary of radius bb, the torque per unit length is

τL=4πμωa2b2b2−a2\frac{\tau}{L} = 4\pi\mu\omega \frac{a^2b^2}{b^2-a^2}

where μ\mu is the dynamic viscosity of the gas. The corresponding power loss per unit length is

PL=ωτL=4πμω2a2b2b2−a2.\frac{P}{L} = \omega\frac{\tau}{L} = 4\pi\mu\omega^2 \frac{a^2b^2}{b^2-a^2}.

For air at room temperature,

μ≈1.81×10−5 Pa⋅s.\mu \approx 1.81\times10^{-5}\ \text{Pa}\cdot\text{s}.

The actual launcher geometry does not consist of concentric cylinders. The evacuated tube has a radius of 4.5 m, while the two screw axes are separated by 6 m and are located 0.5 m above the tube centerline. The distance from either screw axis to the tube center is therefore

d=32+0.52=3.04 m.d = \sqrt{3^2+0.5^2} = 3.04\ \text{m}.

With a screw outer radius of 0.5 m, the minimum distance from the screw axis to the tube wall is approximately

bsmall=4.5−3.04=1.46 m.b_{\text{small}} = 4.5-3.04 = 1.46\ \text{m}.

Because the screw is offset within the much larger vacuum tube, there is no single outer-cylinder radius that exactly represents the actual geometry. We therefore estimate the loss using two concentric-cylinder cases. The first assumes that the screw is enclosed by a relatively small tube with b=1.46b=1.46 m, corresponding to its closest distance to the tube wall. The second assumes a substantially larger surrounding tube with b=3.0b=3.0 m, approximately the distance from the screw axis to the tube centerline. The actual viscous loss is expected to lie somewhere between these two idealized cases.

Using

a=0.5 m,ω=1060 rad/s,μ=1.81×10−5 Pa⋅s,a=0.5\ \text{m}, \qquad \omega=1060\ \text{rad/s}, \qquad \mu=1.81\times10^{-5}\ \text{Pa}\cdot\text{s},

the smaller-tube case gives

PL=4π(1.81×10−5)(1060)2(0.5)2(1.46)2(1.46)2−(0.5)2\frac{P}{L} = 4\pi (1.81\times10^{-5}) (1060)^2 \frac{(0.5)^2(1.46)^2} {(1.46)^2-(0.5)^2} PL≈72 W/m\frac{P}{L} \approx 72\ \text{W/m}

per screw.

For the larger-tube case,

PL=4π(1.81×10−5)(1060)2(0.5)2(3.0)2(3.0)2−(0.5)2\frac{P}{L} = 4\pi (1.81\times10^{-5}) (1060)^2 \frac{(0.5)^2(3.0)^2} {(3.0)^2-(0.5)^2} PL≈66.5 W/m\frac{P}{L} \approx 66.5\ \text{W/m}

per screw.

For two screws, the corresponding power losses are

2(66.5)=133 W/m2(66.5) = 133\ \text{W/m}

and

2(72)=144 W/m.2(72) = 144\ \text{W/m}.

The total length of the launcher sections containing spinning screws is 773 km + 75 km = 848 km. The total aerodynamic power loss is therefore estimated to be between

Ptotal=133(848,000)≈1.13×108 WP_{\text{total}} = 133(848{,}000) \approx 1.13\times10^8\ \text{W}

and

Ptotal=144(848,000)≈1.22×108 W.P_{\text{total}} = 144(848{,}000) \approx 1.22\times10^8\ \text{W}.

This corresponds to a total power loss of approximately

113 to 122 MW.113\text{ to }122\ \text{MW}.

To put this in perspective, the vehicles in a single lane of an 848 km highway, assuming one vehicle passes every 5 seconds, and each vehicle consumes 0.20 kWh/km, would use about 122 MW.

Over a full 14 day launch season, the aerodynamic losses would consume about 38 to 41 GWh, compared with 28.3 GWh transferred to the 56 launched vehicles. So, even at 5 Pa, while the aerodynamic losses may be acceptable, they are still significant. Given that other facilities such as the LIGO gravitational wave observatory achieve much lower internal operating pressures, it may be worth specifying a lower vacuum level for the launch system in an effort to reduce the aerodynamic losses further.

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