tl;dr:
How could the 2018-08-30 Soyuz MS-09 / ISS leak be so slow?
Answer:
By being about 2 millimeters in diameter!
@DavidHammen's comment converts 0.8 mbar/hr to about 0.8 m^/hr air loss rate presumably at standard conditions. Let's see how that's done, how it checks against "a 2mm hole" and what it means if there were no response of any kind (human or make-up air).
He uses the first order relationships
$$ \frac{\dot{p}}{p}=\frac{\dot{m}}{m}=\frac{\dot{V}}{V} = 0.8 \times 10^{-3}/hr$$
where I'm guessing $p$ is the pressure of the remaining air (assuming no make-up air and no change in temperature, which is reasonable considering the air is in intimate contact with so much solid surface area), $m$ is the mass of the remaining air, and $V$ is the equivalent volume of the remaining air if it were at standard conditions.
An ISS pressurized volume above about 938 m^3 (matches values on the internet) times $$0.8 \times 10^{-3}/hr$$ does indeed give about 0.8 m^3/hr!
Now let's see what a 2 mm hole in a thin plate is expected to do. I found two online calculators, although they may have somewhat different assumptions, and the hole has some depth (the wall thickness of the Soyuz at this location) and side roughness, but still we can try.
- http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm#calc
- https://www.tlv.com/global/TI/calculator/air-flow-rate-through-orifice.html
Number 1. gives 2.74E-04 m^3/sec or 1.0 m^3/hr, almost identical to the quoted 0.8 m^3/hr value, and number 2 gives something at least close; about 3 m^3/hr.
So baring any make-up air, that's about a 1% drop in pressure every ten hours. That's enough to be alarming, and of course would probably trigger an alarm in less than ten hours since strikes by meteorites and debris so likely over time that one would expect the ISS to be hyper-vigilant about leaks.
So to answer the question
How could the 2018-08-30 Soyuz MS-09 / ISS leak be so slow?
The answer is
By being about 2 millimeters in diameter!