2365 Design Project - May need some advice.

[RDB85]

This is Q5:

Determine for one of the circuits supplying the fridges the minimum possible csa of cpc will satisfy the requirements disconnection as ADS under earth fault conditions and the adiabatic equations.

I am going to have a go at the calculation as I think I should be okay doing that. But how do you do:

Actual CPC size using Adiabatic Equation, please?

 

[Richard Burns]

When you do the adiabatic equation you will get a value and you then select the next available cpc size that is greater than that value, e.g. if you get 1.6mm² as the result of the adiabatic equation you would use a 2.5mm² cpc, if the result was 1.4mm² then you could use 1.5mm² cpc.

 

[RDB85]

So I have had a go:

Circuit on 16A

Zs for a 16A Type B breaker is 2.73 ohms

Ze is 0.11 ohms

R1+R2 in the OSG for 16A is 2.30 ohms

Calc 2.30 x 30 divided by 1000 x 1.20 = 0.0828 ohms

Zs = Ze = (R1+R2)
0.11 + 0.0828 = 0.1928 ohms

Adiabatic Equation:

If= Uo/Ze

If= 230v/0.11 ohms = 2090.90A

t = 0.1

S= √(I²×t)/k

S= √(2090²×0.1)/115 - I used table 54.3 pg 197.

= 61.63mm²


If someone could check that I am on the right lines that would be great.

 

[RDB85]

I then did:

k1 divided by k2 x S

115/103 x 16 = 17.86 CSA.

 

[Richard Burns]

The calculation you have done for assessing compliance with the minimum Zs for a type B 16A BSEN60898 circuit breaker shows that the theoretical value is well inside the maximum limit of 2.73Ω at 70°C (normal operating temperature).

Your calculation for the adiabatic is flawed because the time for a B16 MCB to disconnect at a current of 2090.90 A will not be 0.1 seconds.
You have correctly identified that a fault could occur at the origin of the circuit and so the maximum current to be experienced would be the prospective fault current.
(It is worth noting that at lower currents the disconnection time increases and this may mean a greater thermal stress on the conductor, this is especially true for fuses.)
The tables in BS7671 only cover the currents up to the point of the magnetic trip operating for a B16 breaker this would be at worst a fault current of 5 x 16 = 80A. Therefore it might be possible to use 80A and 0.1s in the adiabatic equation and get a csa of 0.22mm² however the actual current experienced is much higher than 80A so you would need to find the manufacturers data for the MCB that would give values for I²t at your calculated fault current.
For 2000A fault current your I²t might be something like 6000 A²s which would give a csa of 0.67mm².


k1/k2 x S is only used where the conductors are not made of copper.

 

[RDB85]

The calculation you have done for assessing compliance with the minimum Zs for a type B 16A BSEN60898 circuit breaker shows that the theoretical value is well inside the maximum limit of 2.73Ω at 70°C (normal operating temperature).

Your calculation for the adiabatic is flawed because the time for a B16 MCB to disconnect at a current of 2090.90 A will not be 0.1 seconds.
You have correctly identified that a fault could occur at the origin of the circuit and so the maximum current to be experienced would be the prospective fault current.
(It is worth noting that at lower currents the disconnection time increases and this may mean a greater thermal stress on the conductor, this is especially true for fuses.)
The tables in BS7671 only cover the currents up to the point of the magnetic trip operating for a B16 breaker this would be at worst a fault current of 5 x 16 = 80A. Therefore it might be possible to use 80A and 0.1s in the adiabatic equation and get a csa of 0.22mm² however the actual current experienced is much higher than 80A so you would need to find the manufacturers data for the MCB that would give values for I²t at your calculated fault current.
For 2000A fault current your I²t might be something like 6000 A²s which would give a csa of 0.67mm².


k1/k2 x S is only used where the conductors are not made of copper.

That's quite a lot to take in, so the First part is okay, its just the adiabatic equation that would need some work. As its a Project I don't think I would need manufacturers data for the MCB that would give values for I²t at your calculated fault current.

I will have another look at it again and see what I can come up with, based on the information you have given me.

 

[Richard Burns]

I would be tempted to answer that the adiabatic at 80A and 0.1s gives 0.22mm², this would mean a 1mm² is possible but the minimum csa specified in table 52.3 for power circuits is 1.5mm² therefore a 1.5mm² cpc is the minimum suitable.

 

[RDB85]

I have had another go:

• Ib 16A
• In 20A
• Iz 26.59A
• CSA: 4mm² Ref Method B 32A Volt Drop 11
• Actual VD: 5.28 Volts
• Max Volt Drop: 11.5V
• Zs: 2.73 ohms
• Ze: 0.11 ohms
• R1+ R2 for 4mm² is 9.22 x 30 x 1.20 divided by 1000 = 0.33ohms
• 0.11 + 0.33192 = 0.44192 ohms

AE

• 230/0.11 = 2090A
• 2090² x 0.4 = 1747240
• square root of 1747240 = 1321/115 = 11.4mm²

So the next size cable would be 16mm²

If that calculation could be checked that would be great.

 

[Richard Burns]

You are certainly working hard on this one. We do not have the full information from the question and scenario to be able to fully calculate the results and doing this would not help you to progress so I am going to try some suggestions that may not match with the scenario but might give you an idea of things to consider in the question.

• Ib 16A - 16A is a lot of current for fridges which might generally be about 500W each, though I do not know the specification, so you would be running seven fridges off this circuit to get to 16A. If this is correct no problem.
• If the design current, Ib, is 16A then In should be >= Ib so the In could be 16A however 20A is also acceptable.
• In 20A - Taking In as 20A and looking in, say, the table for 70°C singles, and looking at reference method B for single phase and dividing 20A by 0.94 for 35°C (this is a high ambient temperature) and 0.80 for 2 circuits grouped does give Iz as 26.59A and require a 4mm² cable.
• Iz 26.59A - correct
• CSA: 4mm² Ref Method B 32A Volt Drop 11 - correct
• Actual VD: 5.28 Volts - correct
• Max Volt Drop: 11.5V
• Zs: 2.73 ohms
• Ze: 0.11 ohms
• R1+ R2 for 4mm² is 9.22 x 30 x 1.20 divided by 1000 = 0.33ohms
• 0.11 + 0.33192 = 0.44192 ohms - correct and compliant

AE

• 230/0.11 = 2090A
• 2090² x 0.4 = 1747240
• square root of 1747240 = 1321/115 = 11.4mm²

If you have a fault current flowing of 2090A then the circuit breaker will trip faster than the 0.4 s you have used.
You must use the actual time taken to trip at that fault current in order to get the right answer. However the graphs of trip time will not give you the trip time at that current.
If you look at the graph below for trip time for Eaton type B MCBs you can see that I/In (2090/20=104.5) is off the right hand side of the graph. As a reality check you could use the fastest trip time at 60 I/In (1200A) and use 0.0075s in the adiabatic and get 1.57mm² and from this be able to realise that the csa would be lower than this if the time is faster.
You could also calculate the fault current at the end of the circuit would be (230/0.44192=520A) and 520/20=26 which is on the graph and gives a trip time of 0.01s which then gives a csa using the adiabatic equation of 0.45mm² and you know that the csa should be greater than this for a higher current.
This gives you a range from 1.57 to 0.45 mm² for the minimum cpc size.
With the information you have given it is not possible to use the adiabatic equation to calculate the minimum csa directly.

 

[RDB85]

Thanks, Richard. Its hard work. It's proving difficult to work out the adiabatic equation. I will scan the question in, so you have all the information and the information on the fridges as detailed in the specification.

 

[RDB85]

Here is the Q5:

Determine, for one of the circuits supplying the fridges, the minimum possible cross-sectional area of cpc which will satisfy the requirements disconnection as ADS under earth fault conditions and the

adiabatic equation as Regulation §43.1.3 in BS 7671.

HINT: A two part question. You first need to establish the max (R1+R2) for the circuit based on the Zs for a 16A protective device. From that you can then select a suitable Line & CPC combination for the circuit. The selected size then needs to be confirmed using the adiabatic equation.

Design Spec:

Four Fridges in Kitchen Area, Supplied by 2 x 16A circuits.

 

[RDB85]

Another go:

• Circuit 16A
• Ib 16A
• In 16A
• Zs for a 16A Type B is 2.73Ω
• Ze is 0.11Ω
• Zs - Ze = R1+R2 = 2.62Ω
• Table I1 CSA 2.5mm²
• In 21.27A
• Cable Size: 2.5mm² Ref Method B Volt Drop: 18
• Actual Volt Drop: 8.64v
• Maximum VD: 11.5v
• Calculate Actual R1+R2 = 7.41x30x1.20 / 1000 = 0.26676Ω + 0.11Ω = 0.37676Ω
• UoxCmin / Zs = 80A looking in BS7671 16A Current is 80A

Adiabatic equation.

• 230/0.11 = 2090A
• I then modified to get an exact time: t = k²S²/I² = 115²*2.5²/2090² = 0.019s
• 2090² x 0.019 = 82993.9
• Square Root of 82993.9 = 288.0866189/115 = 2.505mm²
• Cable Size 2.5mm²

 

[Richard Burns]

First part all OK, much better choices even though the other choices were possible, this is in keeping with the information in the question that states the protective device is 16A.
For a type B MCB the highest current required to ensure magnetic trip is 5 times rating, 5*16 = 80A. With the 80A and Uo (230V) and Cmin (0.95) gives (230*0.95)/80 = 2.73Ω i.e. maximum Zs for that breaker.

In the second part you have done a tautological equation.
Effectively you have said A+B=C. Then C-A=B
What you can say from using the rearranged adiabatic equation is that the maximum period of time that the 2.5 mm² cable can carry 2090 A without damage is 0.019 s, because most 16A circuit breakers will actually trip faster than that at 2090 A the cpc is probably OK at 2.5mm².
However it does still rely on using manufacturers graphs to find if the trip time gets that fast and most manufacturers graphs do tend to stop at about 0.01s

 

[RDB85]

Thanks Richard. Do you think that would be satisfactory for answering the question? As your not given no manufacturer data. But then I suppose you can’t really guess the time as Type B breakers have a trip time of 0.1 - 04 seconds. I did try it with 0.1 and 0.4 but the cable sizes were really high.

So I’m not sure to go with that or try another way if there is one.

 

[RDB85]

So I have asked them for some guidance:

• Calc CPC size using selected Circuit Breakers Zs.
• Then work out R1+R2
• Select cable from Table I1 to meet minimum R1+R2 value.
• Once the cable is selected, work out actual R1+R2 and calculate actual circuit Zs - including correction factors 1.20
• Calc to confirm actual CPC size using Adiabatic Equation.
• Circuit length needs to be greater than 15m.