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So I’m doing my level 3 design project at the moment, and I’m on the question where you do all the calculations on each circuit, I’ve taken the max earth impedance from the regs, but it’s asking me for the earth fault loop impedance of the circuit.

I think it’s asking me to calculate it, and I know the formula is Zs = Ze + R1 + R2 but I can only get the values for R1 + R2.
 
TL;DR
How do you calculate Zs with only R1 + R2?
So I’m doing my level 3 design project at the moment, and I’m on the question where you do all the calculations on each circuit, I’ve taken the max earth impedance from the regs, but it’s asking me for the earth fault loop impedance of the circuit.

I think it’s asking me to calculate it, and I know the formula is Zs = Ze + R1 + R2 but I can only get the values for R1 + R2.
If its anything like the project I did, only last year, the Ze should be given to you as part of the project handout. If not you might have to give the tutor an email but it should have been included.

Hope it goes well, it's less daunting than it first seems
 
Of course Ze is essential to computing Zs and would normally be something you have measured during an initial investigation or are given as part of the process.

However, if you are designing for a generic case there are a couple of values in the OSG for up to 100A supplies that can be used for this (0.8 ohm TN-S, 0.35 ohm TN-C-S) or in some cases you can ask the DNO about the supply characteristics.

Of course, if it is a TT supply then you almost certainly are relying on RCD protection and the regs have max values for earth rods to be acceptable (don't have them to hand to quote it). In such cases the R1+R2 is unlikely to make any meaningful difference.
 
The handout will show the following.
The existing hotel building forms part of a 400 V three-phase TN-C-S system. Ze and
PFC have been measured and recorded as 0.09 Ω and 5 kA respectively.
A new three-phase distribution circuit to the pool/café block,
Zdb and PFC for the new distribution board within the block will be 0.11 Ω and 3.7 kA respectively.
 
The handout will show the following.
The existing hotel building forms part of a 400 V three-phase TN-C-S system. Ze and
PFC have been measured and recorded as 0.09 Ω and 5 kA respectively.
A new three-phase distribution circuit to the pool/café block,
Zdb and PFC for the new distribution board within the block will be 0.11 Ω and 3.7 kA respectively.
Hi David. I'm currently doing this same design project from home. Can I ask you a question? When working out the design current for the circuits, I'm using the following equation based on it being a 3-phase supply. Watts÷ /3×400×pf = ib.
Is this correct? The pf is for the type of lights I'm using. If I'm doing this wrong ill get all my circuit calcs wrong.
 

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