Nuclear reactor output scales with volume but cost scales with surface area. So a larger plant is more efficient.

If the smaller reactors can take advantage of easier fabrication and logistics they might be cheaper, but that is an unknown.

You are talking past the point I made to only focus on raw scaling efficiency, I am talking about capabilities, both technical as well as economic. You are arguing big-o.

How could cost scale with surface area? The mass of a nuclear power plant scales proportionally to volume. The thickness of a pressure vessel or pipe increases as you scale it up, at constant pressure.

The pressure vessel is not a solid mass of steel, it's a hollow container. The thickness of the walls remains the same for the same pressure (it does need to increase slightly because it needs to support the weight of the pressure vessel itself) but the volume increase is cubic.

Cubic volume increase vs. quadratic surface area increase means larger sizes are more effective.

No, that's wrong. For a pressure vessel made of the same material, the same geometry (scaled), and operating at the same safety factor, the thickness of the wall is proportional to the linear dimensions of the vessel.

Here's a simple argument about why that's true:

https://yarchive.net/space/launchers/fuel_tank_scaling_laws....

> Take any plane that slices through the pressure vessel, and consider the problem of trying to prevent the vessel from separating at that plane. The pressure load trying to cause such a separation is proportional to the surface area of the slice. The wall length available to resist it is the outer edge of the slice. The load scales with the square of size, the wall length only linearly... so as the size goes up, the wall thickness must grow as well. Chase it through the math, and wall mass is simply proportional to volume.

It's also a consequence of the virial theorem of mechanics.

Below a certain size you run into code minimums, but nuclear power plants will be well above that size.

Here's a simple analogy: you have a 2 foot kiddie pool in your back yard. The pressure is whatever pressure water is at under a depth of 2 feet. If you have 2 pools, one radius 5 feet and one radius 10 feet they're both at the same pressure. The depth of the water didn't change. Length of the wall doubled (circumference) but the area squared. This is why you can hold back a storm surge with sandbags. The fact that effectively infinite water is behind it doesn't matter.

If you have some gas stored at 100 psi, you need a wall thickness to withstand 100 psi. If you create a sphere with twice the radius you still need a wall thickness to withstand 100 psi. A bit more than that because the tank also needs to support itself, but that's minor relative to its contents. That's why methane storage tanks are big spheres, to minimize surface area: https://i0.wp.com/tmicoatings.com/wp-content/uploads/2019/09...

I'm not sure how applicable rocket motors are to this comparison (maybe you missed that they were talking about rocket motors?). Rocket motors have a big bell and throttle at the end, which is analogous to a big load. And the bigger the rocket combustor the more thrust it needs to support. This is a whole different kind of load, it's not a simple pressure vessel.

In particular:

> The pressure load trying to cause such a separation is proportional to the surface area of the slice.

This is probably taking into account something like greater fuel combustion. Because otherwise it's blatantly wrong. A tank at 100 psi still has 100 psi of pressure whether it's got a volume of 1 cubic meter or 10 cubic meters.

There is no "wall thickness to withstand 100 psi" that is independent of the size of the tank. Stop thinking that that is a real thing. The thickness of the wall needed to withstand 100 psi increases as you make the tank bigger.

Here's a simple online calculator for wall thickness of pressure vessels. Change the radius of the vessel and watch the wall thickness go up in proportion.

https://www.engineersedge.com/calculators/shell_internal_pre...

Efficiency is only one dimension, these reactors can be distributed to the point of use. Efficiency should be measured in its local universe.

You are forgetting about security. One of these things and some explosives could do a lot of damage. Think dirty bombs. Each and every one will need 24x7 security for the duration of it's life. They simply can't be left unattended. Not impossible to do but it adds a lot of cost One of these things falling into the hands of extremists would be very bad.

The sensible way to deploy these would be in groups close to existing infrastructure in places that can be easily secured. For example de-commissioned coal plants or nuclear plants. Security also raises the cost of operation of course. Just because it is small physically does not mean the risk is smaller. So, same level of security as with a big nuclear plant. Same level of cost. But less energy. Co-locating allows you to reduce that cost.

Electricity is pretty easy to transport over 50-100 miles. One big plant servicing a whole metro is as efficient as smaller reactors distributed around. It's only when you have to transport electricity over 300+ miles where more exotic things like HVDC lines become necessary.

I am not against that in general, but your never gone do that seriously with PWRs. There are designs where this does make some sense.

However that is not the backbone of your infrastructure. I makes much more sense to start with one large scale design 500MW+ and deploy that as many times as you can and then you can see the places where that is overkill and you develop smaller reactors for those regions.

But in most countries with ~500MW design you can reach 80-90%+ of the population and since the design would be so well known and so mass produced it might make sense to just build more of those rather then some 50MW reactor that might be only be 25% the cost but 10% the production.