The things around short circuit (SC) calculations are fairly complex as @Antonio51 already had mentioned. However they boil down to some basic principles. Physics are involved and the parameters vary greatly so there is most likely no standardised method which can mitigate the collection of parameters.
My source is:
„Berechnung von Kurzschlussströmen in Hoch- und Niederspannungsanlagen“ written by „Dr.-Ing. Martin Gerlach“, third edition, published by „VEB Verlag Technik Berlin“ in 1955.
This Book refers to VDE0670/XII.47
Question No. 1:
Basic principle is divided in the collection of all impedances between the short circuit and power sources and the calculations depending on the type of short circuit.
Accepted simplifications:
- With some other exceptions only the devices designed as power sources (generators) are considered as sources. Motors will act as generators due to persisting excitation and the momentum, but all asynchronous/synchronous motors are uncared-for except they have a size similar to the transformer they are fed by.
- The load on the network will increase the load on the generators and upstream components during SC. However it will lower the SC power at the place of the SC. So load can be neglected.
Impedances:
- Generator's (3-phase) initial SC reactance per phase(!): $$ X_{st} = \frac{U^2_n}{p \cdot N} [\Omega] $$ with \$ U_n = \$ nom. voltage of generator, i.e. line voltage, \$ p= \$ ratio of initial AC-SC vs. nominal current, \$ N= \$ nominal apparent power in MVA. p in big powerstations running with \$ n=3000 \frac{1}{min}\$ or \$ n=1500 \frac{1}{min}\$ is typically 8. For slower machines p goes down to 4. Asynchronous generators can be assumed with a p between 4 and 5.
- Transformer's (3-phase) initial SC reactance per phase: $$ X_t \approx \frac{U^2_n \cdot u_k}{N \cdot 100}[\Omega] $$ with \$u_k\$ being the so called short circuit voltage of the transformer in percent (!). This is a value available from every transformer's datasheet.
- Transformer's (3-phase) resistance per phase: $$R_t \approx \frac{U^2_n \cdot u_r}{N \cdot 100}[\Omega] $$. \$u_r\$ is typically not given. So instead: $$R_t = \frac{N_w}{1000}\cdot\frac{U^2_n}{N^2}[\Omega]$$ with \$N_w\$ being the ohmic losses (from datasheet)
- overhead lines and cables have no iron cores, their impedances depend on their line to line and line to ground distances. Mostly you will refer to tables, because calculations will be fairly complex.
Except for networks with many relevant loops you can safely add the reactances and resistances of the network between source and short circuit separately. However in many cases you have parallel circuits and then calculations get tricky. There are simplifications, but there is no general rule to do so, because the deviation by simplification varies greatly.
Having calculated or estimated a series reactance and resistance the initial SC AC current for 3-phase is: $$ I_{sw}=\frac{1.1 \cdot U}{\sqrt{3}\cdot\sqrt{X^2 + R^2}} [kA]$$ The initial SC AC current doesn't differ greatly for 1, 2 or 3-phase short circuits.
The initial SC current including the DC current portion is $$ I_s= x \cdot \sqrt{2}\cdot I_{sw} $$ with \$ x \$ depending on the ratio between R and X. e.g. $$ x=1.8 ; \frac{R}{X}=0 $$ $$ x=1.45 ; \frac{R}{X}=0.2$$ $$ x=1.2 ; \frac{R}{X}=0.5$$ $$ x=1.07 ; \frac{R}{X}=1.0$$ $$ x=1.05 ; \frac{R}{X}=1.2$$
To calculate the reactances and resistances over networks with different voltages you have to select a line voltage for calculations at your whim and transform all impedances to that level. I.e. if you have selected 20 kV for calculations the impedances at a 10 kV side of an attached transformer have to be multiplied with the square of the voltage ratios of the transformer. Ratio is 20/10=2. Multiply the reactances and resistances of the 10 kV side with 4. THen you can simply add up the impedances. The resulting SC currents have to be transformed back to the voltage level at the place of the SC subsequently.
Simplifications
However there is some reason to simplify calculations.
It depends on the position and purpose of the equipment to rate.
We can distinguish two cases.
The complex one is a SC happening in a part of the network where it has direct contact to a network loop. This means there are two or more significantl, different paths to a busbar or to a single power source. This usually happens at voltage levels above 50 kV, because the levels below are usually distribution grids and not transportation grids from their intended main purpose.
These SCs are very difficult to calculate. And to control those currents theres usually very special equipment needed.
The simple one is happening at a lower voltage level and there's only a single path to the next higher level of the grid. This includes parallel lines connecting the same busbar at each endpoint. For this case the underlying power grid can be assumed a "hard grid" except for the "last mile" to the SC.
You may do some example calculations if you like. You will find that a hard net with zero impedance and constant voltage output will produce a SC current roughly 5% bigger than a fully calculated grid. That's because the higher levels of a power grid are designed to appear that way.
So exactly for the rating of elements at lower grid level this is a good method. You will get additional 5% safety margin from the "real soft grid" vs. a hard grid.
Question 2:
I don't know which type of equipment you are referring to. This might differ slightly on the purpose of the equipment. But we can rule some things out. The steady state case in many cases doesn't bring any useful information about the robustness of some device. The most interesting question is whether a device will survive the initial SC including DC curent. For a circuit breaker this is the only interesting value because it must be able to separate the connection under worst conditions to protect lines, generators, all sorts of devices and people.
So I think in most cases the testing will assume some standardized network impedance and then go vor the maximum current possible with that.
