A number of people have viewed this post and some may not be familiar with this problem so I will try to give a simplified explanation.
Let's start with a example of a simple system - a battery with a length of cable connected to it that has a resistance of 0.5 ohm for the line and 0.5 ohm for the neutral.
Assume that the battery has an open circuit emf of E = 10 volts. I don't want to get into the use of the terms emf (electro-motive-force) and voltage so just assume that emf is the voltage on the battery terminals when it is not supplying any current to a load.
Batteries are made from materials that have resistance. These are used for the battery plates and the electrolyte. This means that a battery has resistance inside it and we call this internal resistance.
This internal resistance forms part of the system and it will modify the maximum current that the system can deliver.
Now if we give this internal resistance (r) a value of r = 0.1 ohm the battery would deliver a current of
I = E/r = 10/0.1 = 100A
if we put a short circuit across its terminals.
If we now connect our cable and put a short circuit at the far end - the circuit now has a total resistance of
r + R(line) + R(neutral) = 0.1 + 0.5 + 0.5 = 1.1 ohms
The short circuit current at the far end of the cable is
I = 10/1.1 = 9.1A
So the cable has considerably reduced the fault current level at its far end.
When we come to power transformers we have a similar situation. The internal resistance is now the impedance of the transformer windings, and an allowance for the impedance of the high voltage supply network that it is connected to.
The 40kA fault level is at the terminals of the transformer - to work out the prospective fault current at the far end of the 400 yards of cable that is in this installation - you would need to know the impedance (mainly resistance) of the cable. You will find that it will be much less than 40kA.
Now all this said what you are trying to ascertain actually requires a different approach - you do need to calculate the fault current at the mcb - but then you need to realize that it is energy dissipating as heat that actually damages things such as the mcb you are concerned about - not simply current. It may surprise you to told that in many cases it is the lowest value of fault current (usually due to earth faults) that will potential cause the most damage - because they flow for longer.
Take a look at the second paragraph of regulation 434.5.1 - this is the regulation you are trying to apply and you will see that it is written in terms of let through energy - not fault current.
Let through energy is generally represented as I²t A/s - this is called the joule integral (yes we are getting into calculus
).
To calculate an actual value you would need to know the equation for I in your system. In a.c networks at steady state this could be a sine function (hence sine wave). Now a short circuit is not a steady state especially during the first 0.1 seconds or 5 cycles of the supply. It gets even worse at 0.01 seconds because this is the first half cycle of the supply. Imagine you have just woken a tiger and it's response is to immediately lash out
.
Calculating things becomes hard because we do not have a nice simple equation for I when the system is reacting in this way. So the manufacturer has to measure it in a lab and publish the results. That is the data you need to assess this situation.
You need the fuse let through energy at the fault level available at the mcb. You also need the withstand energy of the mcb. Now some manufacturers package this data in nice simple tables which can sometimes indicate that a particular fuse will protect a particular mcb.
