I think you may be underestimating the complexity of both weather and climate modelling.
In my area thats 4490887(m^2) ×0,2(m^3 water) = 898 177.4 ^ m3 max rain water. Now all i want is a model of what areas would be reached on a stream if it received this much water.
I know that some here are grumpy because there houses may loose value like a sinking ship very soon. But knowing is better then not knowing. Always.
No fluid-simulations needed, just a basic Topographic map fill the pond.
If you dumped that amount of rain water in one second, you'd have a Tsunami. Dump it over a 24 hour period, it might flow out smoothly out of a little stream. Dump it 10% in an hour, 50% in the next hour, 10% the next hour, 30% over the next 12? Um, that's harder.
Many insurance companies will be happy to write you a policy for something high risk but they will absolutely charge you more to compensate. It's all a numbers and probability game for them.
You can get a rough approximation, and that's been done at great effort for the entire US at the FEMA website:
https://msc.fema.gov/portal/home.
It's not a simple problem to solve. Some of the caveats are discussed in this Practical Engineering video:
https://www.youtube.com/watch?v=EACkiMRT0pc
Volume of outflow through a stream is equal to rainfall times watershed area. Seems intuitive, right? Except what about water that gets absorbed into the ground, or captured in a temporary holding pond, slowed by vegetation, or accelerated in a storm drain...I think Grady put it well when he pointed out that the silly Imperial units of outflow in cubic feet per second, rainfall in inches per hour, and watershed area in acres multiply together to 1.008, which is off by about 1% but close enough that civil engineers just throw away the constant entirely because it's such an imprecise approximation that the error is irrelevant.