Patent App. (2009 abstract)

[Red]

Why Silver-Plating Is a Spectacularly Bad Idea for FLiBe

Fig. 23 assumes a pristine 92% reflectivity silver-plated sphere as the mirror cavity. Beautiful optics, sure…if you’re operating in a vacuum. But in a real-world BSF reactor filled with molten FLiBe (~750-850°C, highly corrosive, and loaded with fluorine), silver’s tenure would last about as long as an uncoated marshmallow in a blast furnace.

Here’s the breakdown of this fantasy:

• Chemistry Fail #1: Molten FLiBe aggressively attacks most metals. Silver, in particular, reacts violently with fluorine, forming volatile silver fluoride. FLiBe’s full of it. Expect rapid formation of AgF, a compound with all the sticking power of dry ice on Teflon. You’ll get silver vapor, flakes, or a smoldering soup—pick your failure mode.
• Thermal Fail #2: Silver melts and diffuses at 960°C. Any surface “finish” modeled in Fig. 23 would degrade catastrophically under reactor conditions. Even if you plate it with care, it’ll last about as long as an oatmeal patch on a leaky roof.
• Surface Fail #3: Silver doesn’t build a protective oxide. Once corrosion starts, it’s open season. There’s no passivating oxide layer to slow the damage.
• Optical Fail #4: Doesn’t matter how many bounces you simulate—if the mirror coating doesn’t survive contact with FLiBe, your model is toast. Even thin films are fatal—you lose more than reflectivity; you lose function.

You want the mirror to survive in FLiBe? Skip the silver, and start looking at high-Ni alloys, TZM, tungsten, or other refractory metals.

[Fermi]

@Red

You want the mirror to survive in FLiBe? Skip the silver, and start looking at high-Ni alloys, TZM, tungsten, or other refractory metals.

Iridium (Ir) vs Silver (Ag):

Chemistry Pass — Unlike Ag, Ir will neither dissolve nor delaminate. It is nonvolatile, only forming IrF₆ slowly. It has a redox potential lower than gold and higher than W — survivable in moderately oxidizing salt.

Thermal Pass — Melting point: 2446 °C. That’s nearly 2.5× silver. No phase change issues near 850 °C. Stays solid and dense under reactor conditions.

Surface Pass — Iridium (Ir) is one of the most corrosion-resistant metals known. It’s considered chemically inert to fluorides at moderate temperatures (~700–900 °C).

Optical Compromise — Reflectivity at 1050 nm: ~82%. Lower than Ag (92%), but stable across temp.

So yeah — silver’s toast. Iridium’s not flawless, but it survives the salt, the heat, and the optics long enough to matter.

[BS-Fusion]

@Fermi

Optical Compromise — Reflectivity at 1050 nm: ~82%. Lower than Ag (92%), but stable across temp.

Not ideal, but so far, this is the best we have:

[Red]

@BS-Fusion

large cone-shaped pipes that lead into and out from the sphere. These function as low-speed settling chambers, reducing turbulence and making the flow more laminar.

Cone-shaped pipes for “low-speed settling chambers”? Adorable. Let’s pretend molten FLiBe behaves like chamomile tea being poured into fine chine, because nothing say laminar flow like shoving molten, radioactive lava into the belly of a violently burping and benching thermonuclear blast chamber. No, this isn’t herbal tea-it’s dense, corrosive, and eager to strip metal faster than your CFD software can say convergence error.

@BS-Fusion

One way to cope with the stress is to have the ends of the fiberoptic cables fused so that they would form durable silica plugs that could be inserted into the shafts that were drilled to let light in/out of the sphere. The ends of the cables could then be attached to the outside of the sphere using springs that would allow the plugs to function like shock absorbing pistons. This gives the connecting parts some flexure, and minimizes the jarring stress and strain that would otherwise interfere with and inhibit the sphere’s natural resonance.

The thermal gradient (750 K - 300 K) will twist and torque the metal of your sphere like a stress-relief toy at a balloon animal festival. It will not ‘settle’ into harmony-it’ll rattle itself apart like a cheap washing machine on spin cycle.

[Dr.Razor]

@BS-Fusion

An observation worthy of attention, is the fact that, because of spherical geometry, a ray of light inside the sphere has its path confined, bouncing on a single plane that it cannot leave. That plane is determined by the ray’s origin, the first point of reflection, and the sphere’s center. A close examination reveals that if a ray of light passes close to the center it will return after two reflections, revisiting the same approximate location.

After a few months of heat cycling, vibration, and neutron hammering, those mirrored surfaces will reflect like the inside of a Halloween fun house, and the fiber optics beam paths will be alligned like a drunken disco ball.

[BS-Fusion]

@Red

nothing say laminar flow like shoving molten, radioactive lava into the belly of a violently burping and benching thermonuclear blast chamber.

Thanks Red, I love your negativity. But this issue was previously addressed elsewhere, where assumptions where listed and the conclusion was posted:

@BS-Fusion

Flow Regime: Laminar vs. Turbulent

The Reynolds number (Re) determines whether flow is laminar or turbulent:

Re= \frac{\rho v D}{\mu}

Where:

• ρ = 2000 kg/m³
• v = 0.08 m/s (fluid velocity)
• D = 1 m (characteristic length, e.g., pipe diameter)
• μ = 3 Pa·s (dynamic viscosity)

Result:

Re= \frac{2000 \cdot 0.08 \cdot 1}{3} ≈ 53

This is well below the turbulence threshold (Re > 4000), indicating laminar flow.

That example assumed thick, corn-syrupy FLiBe. A thinner mixture with water-like viscosity can be produced by adding LiF or increasing the LiF/BeF2 ratio. Because low-viscosity mixtures can circulate more efficiently through heat-exchangers, they are more desirable. Regardless, with two additional control-parameters (flow velocity & pipe diameter), adjusting the Reynolds number to maintain laminar flow should be easy.

In a typical power plant, the coolant circulates 1 to 3 meters per second. In my example (above), the coolant flows 0.08 m/s. That, arbitrary chosen flow rate, leaves lots of room for increased capacity. However, an increased flow rate may not be necessary, because the thermal output is already substantial:

W= \frac{2414\space J}{kg\space K}*350\space K*\frac{2000 \space kg}{m^3}*\pi\space 0.5^2m^2*\frac{0.08\space m}{s}

= 106,000,000 \space Watts.

[Fermi]

@Banana

Chemical explosives convert nearly all their energy into hot, dense gases that push outward, generating intense blast pressure. In contrast, DT fusion releases about 80% of its energy as high-energy neutrons that penetrate deep into the surrounding blanket, where the energy becomes heat—not pressure. Only a tiny fraction contributes to a shock wave. So despite having the same total energy as ~1000 sticks of dynamite, a small DT fusion burst produces nowhere near the same physical destruction.

To be clear, that “tiny fraction” still equates to 204 sticks of dynamite (1/5 of 1020)—not a firecracker.

Instead of hand-waving, let’s do the math. The maximum pressure this reactor can withstand (before rupture) is governed by thin-walled pressure vessel theory:

Given:

• r_i = 2.50 meters, the sphere’s inside radius
• r_o = 2.75 meters, the sphere’s outside radius
• t = 0.25 meters, the walls thickness
• \sigma = 1207 MPa, yield strength of Alloy 4907
• P = internal pressure (to be solved)

If the sphere is cut in half, force equilibrium requires that

\pi (r_o^2-r_i^2)\sigma = \pi r_i^2P

Solving for P:

P = \frac {(r_o^2-r_i^2)}{r_i^2}\sigma = \text {253 MPa}

For Americans, 253 MPa is ~36700 psi.

Because pressure is energy-density, we can calculate the amount of energy (E=PV) this represents, assuming that this pressure is uniformly distributed throughout the sphere: E = P\frac {4}{3}\pi r^3 = 17 GigaJoules! So, to burst the sphere, it would require 700x more energy than in the dynamite (204 MJ).

Conclusion: containing fusion yields of 1.02 GJ (equivelent to 3 mg of DT) pose no threat to this hypothetical BSF reactor.

[Banana]

@Red

The thermal gradient (750 K - 300 K) will twist and torque the metal of your sphere like a stress-relief toy at a balloon animal festival. It will not ‘settle’ into harmony-it’ll rattle itself apart like a cheap washing machine on spin cycle.

Nonsense! The sphere below has a 25 cm thick metal wall with a \frac {5}{25} \frac {\text {K}}{\text {cm}} thermal gradient, untroubled by any of your rediculous make-believe problems.

Here, a zirconia insulator is located at the tip of each standoff rod. Each tip supports one free-standing piezoelectric stack actuator, so the attachment is secure, pre-stressed, shielded from excessive heat, and individually ballasted with a heavy stabilizing weight.

The metal standoff rods do not experience any thermal stress because they are free to expand and contract radially during tempeature changes. The key lies in the absence of mechanical constraint. Thermal stress arises only when expansion or contraction is restricted. If you heat a rod lying on a frictionless table, it just gets longer—no drama, no stress.