2365 L3 Design Project

[Pops1]

Adiabatic Equation:

If= Uo/Zs

If= 230v/0.11 ohms =  2090.90A

t = 0.4

S= √(I²×t)/k

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

= 61.63mm²

 

[Pops1]

I then did: 

k1 divided by k2 x S

115/103 x 16 = 17.86 CSA.

 

[Sharpend]

Where does it state cable size is 16mm? 

 

[Pops1]

Where does it state cable size is 16mm? 

It doesn't but If the Ib is 16A then I would use 20A as the In, then selecting the rating factors and working out Iz to be 26.59Amps

Then looking for a suitable cable 4mm at 32A VD 11

I am not sure if I have done it correctly. that's why I was asking about the calculations. 

 

[Pops1]

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. 

 

[poni]

its been some time(no books to hand and using your calculation)

Zs = Ze +(R1+R2)

If= Uo/Zs

0.44192 ohms

should you not be using the Zs (0.44) and not Ze (0.11)? this should bring it down to about 2.8mm ( use 4mm) its just that 16mm2 looks a little big for a 20A circuit 

is this the Max Zs for the circuit ?

Zs: 2.73 ohms

 

[Sharpend]

As pointed out by Poni, you have used Ze instead of Zs in your calc. 

 

[Ardet R]

The first bit is OK with some overkill on the currents and cables but compliant and workable.

For the second why have you used 0.4s as the trip time with a fault current of 2090A?

Your cpc size would usually be expected to be smaller than your line conductor csa when using the adiabatic equation.

 

[Pops1]

The first bit is OK with some overkill on the currents and cables but compliant and workable.

For the second why have you used 0.4s as the trip time with a fault current of 2090A?

Your cpc size would usually be expected to be smaller than your line conductor csa when using the adiabatic equation.

That the part I am getting confused on. I will do the calculations again based on the above. Thank you. 

 

[Pops1]

Question:

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. 

 

[Ardet R]

You are told it is a 16A circuit, you are not given the design current, if you specify a type B 16A BSEN60898 then maximum Zs @70°C is 2.73Ω

Ze is 0.11Ω, therefore R1+R2 must be less than 2.62Ω @70°C.

From the installation method (singles in conduit) and assuming maximum current as you do not have the design current, the cable minimum cable csa is 2.5mm².

If you choose 2.5mm² for the cpc size the resistance R1+R2 is much lower than 2.62Ω so Zs will be compliant.

and then calculate using the rearranged adiabatic equation from 434.5.2

t = k²S²/I² = 115²*2.5²/2090² = 0.019s  if the circuit breaker will disconnect in that time then the cpc is OK.

You can also check at the end of the circuit to find 0.3s which is also OK.

 

[Pops1]

You are told it is a 16A circuit, you are not given the design current, if you specify a type B 16A BSEN60898 then maximum Zs @70°C is 2.73Ω

Ze is 0.11Ω, therefore R1+R2 must be less than 2.62Ω @70°C.

From the installation method (singles in conduit) and assuming maximum current as you do not have the design current, the cable minimum cable csa is 2.5mm².

If you choose 2.5mm² for the cpc size the resistance R1+R2 is much lower than 2.62Ω so Zs will be compliant.

and then calculate using the rearranged adiabatic equation from 434.5.2

t = k²S²/I² = 115²*2.5²/2090² = 0.019s  if the circuit breaker will disconnect in that time then the cpc is OK.

You can also check at the end of the circuit to find 0.3s which is also OK.

Thank you, that's helped. Would you say the Design Current to be 16A, then your In could be 20A? I will have another go at the calculation, just so I understand how you have done it. Thanks once again. 

 

[poni]

can you use the conduit as the cpc?

 

[Pops1]

So I had a 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

 

[Ardet R]

From table 54.7 a 2.5mm² conductor is OK, so however you calculate the adiabatic equation it should come out as suitable.

You can't put 2.5mm² into an equation, square it multiply it and divide it and then reverse the process to get the same answer you put into the equation as it is meaningless. e.g. 2.5 *3 /10 = 0.75...  0.75 *10 / 3 = 2.5

As this is a project you should need to do some research for it and perhaps get I²t values for your breakers.

In essence the adiabatic equation is fairly pointless without these values unless you are checking very low fault currents.

 

[Pops1]

From table 54.7 a 2.5mm² conductor is OK, so however you calculate the adiabatic equation it should come out as suitable.

You can't put 2.5mm² into an equation, square it multiply it and divide it and then reverse the process to get the same answer you put into the equation as it is meaningless. e.g. 2.5 *3 /10 = 0.75...  0.75 *10 / 3 = 2.5

As this is a project you should need to do some research for it and perhaps get I²t values for your breakers.

In essence the adiabatic equation is fairly pointless without these values unless you are checking very low fault currents.

Okay thanks. We have not been told to do any research on breakers, and we have not given any manufacturers data, in the specification. The question does not ask for it either. I have asked for some more help. They have come back with:

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.
  

 

[Ardet R]

So you have done all that except the fifth one to resolve the adiabatic equation for confirming the cpc size.

I also missed your post of "Thank you, that's helped. Would you say the Design Current to be 16A, then your In could be 20A? I will have another go at the calculation, just so I understand how you have done it. Thanks once again."

The question states the circuit is 16A this would mean a circuit with a 16A circuit protective device. in the Hint section it also states "the max (R1+R2) for the circuit based on the Zs for a 16A protective device" Therefore your In must be 16A and based on that your Ib is less than or equal to 16A (in all likelihood your Ib for two fridges is going to be less than 7A based on commercial fridge specifications).

If you calculate your modified adiabatic equation using the fault current at the end of the circuit using I = V/R =V/Zs = 230/0.3767=611A and with S as 2.5 and k as 115 then you get 0.22s which is larger than the maximum possible trip time of 0.1s and so is compliant.

This does not address faults earlier in the circuit, however.

Locked to prevent hi-jacking (again)