2365 Design Project - May need some advice.

[Richard Burns]

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.

Somewhat odd advice perhaps, I must be missing something.
I can only think they mean determine cable size from breaker size and Zs
R1+R2 maximum value is straightforward
Minimum cable size to meet breaker rating would work, and then compare to required R1+R2.
Then with cable selected R1+R2 and Zs is OK
Adiabatic not without more information or using the 80A and 0.1s values.
Not sure where the circuit length comes into it.

 

[RDB85]

So do you think what I have already is fine? I’m concerned about the time more than anything. As I know on page 92 in BS7671 there is a time calculation. When I’ve tried it with those numbers it’s ends up being massive. But I don’t know if I’m doing it wrong or I have the wrong information.

 

[Richard Burns]

I think that if you have put this much effort in and can describe why you are doing these things then you should at least gain a lot of marks for effort, even if it turns out the final figure is in error. But at least you know it will be suitable from table 54.7.
With only the information from BS7671 the highest current you can use reliably in the adiabatic equation is the lowest current that is guaranteed by the circuit breaker standards to disconnect the circuit breaker magnetically. In this case that current is 80A, five times the rating of the type B breaker. However this will give a smaller csa than is required for a higher fault current. Looking at the graph in post 24 you can see that after the magnetic trip operates a large rise in current is required to make a small reduction in the time of disconnection, so the current rises faster than the time drops and consequently larger cpc sizes are required.

 

[RDB85]

I think that if you have put this much effort in and can describe why you are doing these things then you should at least gain a lot of marks for effort, even if it turns out the final figure is in error. But at least you know it will be suitable from table 54.7.
With only the information from BS7671 the highest current you can use reliably in the adiabatic equation is the lowest current that is guaranteed by the circuit breaker standards to disconnect the circuit breaker magnetically. In this case that current is 80A, five times the rating of the type B breaker. However this will give a smaller csa than is required for a higher fault current. Looking at the graph in post 24 you can see that after the magnetic trip operates a large rise in current is required to make a small reduction in the time of disconnection, so the current rises faster than the time drops and consequently larger cpc sizes are required.

Thanks, I will see what feedback I get. I’m onto the last question now which is maximum demand and diversity for all my circuits. I’m just going over Q4 to make sure what I have is correct before attempting this. Thanks again for the help, much appreciated.

 

[RDB85]

Had another go with:

• Calculate Actual R1+R2 = 7.41x30x1.20 / 1000 = 0.26676Ω + 0.11Ω = 0.37676Ω
• 230/0.37 = 621A
• t = k²S²/I² = 115²*2.5²/621² = 0.21s
• 621² x 0.21 = 80984.61
• Square Root of 82993.9 = 284.5779507/115 = 2.475mm²
• Cable Size 2.5mm²

 

[Richard Burns]

There is no problem with the calculation of your expected Zs, this is correct.

For the adiabatic part I think this needs some thought about how you are calculating this value.
You have selected 2.5mm² as a cpc size and put that into the t=k²S²/I² equation and got 0.21s, this is all OK.
However reversing the equation operations to get back the original 2.5mm² is nominally valid but redundant maths.

You have chosen S and then used t=k²S²/I² and then used that t result in the standard form equation S= √(I²t) / k to get back the S you chose in the first place.
I.e. with
S= √(I²t) / k multiply both sides by k
kS = √(I²t) square both sides
k²S² = I²t divide both sides by I²
k²S²/I² = t

this shows both equations are the same so you cannot use the result of one form of the equation, where you have selected one of the values (S), and then use the other form of the equation to get a value for S because it will always be the original value you chose.
Substituting k²S²/I² in place of t in S= √(I²t) / k gives S= √(I²(k²S²/I²)) / k
Multiply by k and square both sides gives k²S² = I²k²S²/I²
The I² cancel each other out to give k²S²=k²S²
divide both sides by k² to get S²=S² and take the square root to get
S=S
which whilst accurate is not worthwhile.

However where you have a value of 0.21s as the maximum time before the cable is damaged from the first part of the calculation you can say that the magnetic trip will operate and will trip in at least 0.01 seconds therefore the cable will not be damaged as 0.01 is less than 0.21s and the cpc will be OK at that csa.

 

[RDB85]

There is no problem with the calculation of your expected Zs, this is correct.

For the adiabatic part I think this needs some thought about how you are calculating this value.
You have selected 2.5mm² as a cpc size and put that into the t=k²S²/I² equation and got 0.21s, this is all OK.
However reversing the equation operations to get back the original 2.5mm² is nominally valid but redundant maths.

You have chosen S and then used t=k²S²/I² and then used that t result in the standard form equation S= √(I²t) / k to get back the S you chose in the first place.
I.e. with
S= √(I²t) / k multiply both sides by k
kS = √(I²t) square both sides
k²S² = I²t divide both sides by I²
k²S²/I² = t

this shows both equations are the same so you cannot use the result of one form of the equation, where you have selected one of the values (S), and then use the other form of the equation to get a value for S because it will always be the original value you chose.
Substituting k²S²/I² in place of t in S= √(I²t) / k gives S= √(I²(k²S²/I²)) / k
Multiply by k and square both sides gives k²S² = I²k²S²/I²
The I² cancel each other out to give k²S²=k²S²
divide both sides by k² to get S²=S² and take the square root to get
S=S
which whilst accurate is not worthwhile.

However where you have a value of 0.21s as the maximum time before the cable is damaged from the first part of the calculation you can say that the magnetic trip will operate and will trip in at least 0.01 seconds therefore the cable will not be damaged as 0.01 is less than 0.21s and the cpc will be OK at that csa.

I think if I show the time calculation, and then say that the trip time for a 16A Type B breaker is.... then state what you have mentioned that the cpc will be okay at 2.5mm²

 

[Richard Burns]

I think if I show the time calculation, and then say that the trip time for a 16A Type B breaker is.... then state what you have mentioned that the cpc will be okay at 2.5mm²

Yes that should cover it, I did make a typographical error in that the BS7671 graphs only go to 0.1 not 0.01 seconds however the same effect is there as the time you have is 0.21s.

 

[RDB85]

Would anyone mind checking my final circuit for both lighting and power, before I attempt the question on maximum demand and diversity. If I send them via pm as to not clog up the forum posts?

 

[westward10]

Hello RDB85, it is probably best to keep it on the open forum as opposed to PMs so other members can stay on track and join in. The length of thread is not an issue.

 

[Nigel]

Interesting thread but can I ask if they went through any of this during the actual course?

 

[RDB85]

Interesting thread but can I ask if they went through any of this during the actual course?

Briefly, but I’m just asking others to check what I’ve done.

 

[RDB85]

Here is Question 4. Below are some of the circuits that have already been defined in the specification:

Pool Lighting: PVC Conduit 90°C thermosetting singles

1. Circuit 1 - Main Pool Lights
2. Circuit 2 - Main Pool Lights
3. Circuit 3 - Changing Rooms, Pump Room and Plunge Pool

Cafe Lighting: PVC Conduit 70°C thermosetting singles

1. Circuit 1 - Cafe Lights
2. Circuit 2 - Cafe Lights
3. Circuit 3 - Kitchen Lights

Pool Power PVC Conduit 90°C thermosetting singles

1. Circuits 1 - Changing Room Sockets
2. Circuit 2 - Pump Room Equipment ( 5-core 90°C SWA Clipped Direct)

Cafe Power: PVC Conduit 70°C thermosetting singles

1. Circuit 1 - Dining Room Sockets
2. Circuit 2 - Kitchen Fridges - 1 ( 2x 16A Circuits)
3. Circuit 3 - Kitchen Fridges - 1 ( 2x 16A Circuits)
4. Circuit 4 - General Kitchen Sockets
5. Circuit 5 - Dishwasher ( 3-core 70°C SWA Clipped Direct)
6. Circuit 6 - Food Holding Equipment ( 5-core 70°C SWA Clipped Direct)
7. Circuit 7 - Fire Alarm 13A unswitched fuse spur connection unit will be required in the pool/cafe block to supply a repeater panel. To be installed by a specialist contractor.
8. Circuit 8 - Vending Machines ( 2x 230v 1.2kW each)

We had some notes:

• Circuits must be grouped together.
• Pool - 3 Max circuits grouped together
• Cafe - 2 Max circuits grouped together
• Kitchen 4 Max circuits grouped together

Temps:

• Pool - 40°C
• Cafe - 25°C
• Kitchen - 35°C

 

[RDB85]

Main Pool Lights - Circuit 1

• Ib = 1.51A
• In = 6A
• Installation Method = 4B
• RF = Ca 0.91 Cg = 0.80
• Iz = 8.24A
• csa = 1mm² 13.5A VD 44
• Actual VD = 1.99v
• Max Vd = 6.9v
• Max Disconnection Time: 0.4 seconds
• Earth Fault Loop Impedance = 7.28Ω 
• Maximum Earth Fault Loop Impedance
• Ze-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
• Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω

 

[RDB85]

Made a Mistake:

• csa = 1mm² 17A VD 46
• Actual VD = 2.08v

Main Pool Lights - Circuit 2

• Ib = 1.51A
• In = 6A Type B breaker
• Installation Method = 4B
• RF = Ca 0.91 Cg = 0.80
• Iz = 8.24A
• csa = 1mm² 17A VD 46
• Actual VD = 2.08v
• Max Vd = 6.9v
• Max Disconnection Time: 0.4 seconds
• Earth Fault Loop Impedance = 7.28Ω
• Maximum Earth Fault Loop Impedance
• Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
• Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω

Changing Rooms, Pump Room, Plunge Pool Lights

• Ib = 0.45A
• In = 6A Type B breaker
• Installation Method = 4B
• RF = Ca 0.91 Cg = 0.80
• Iz = 8.24A
• csa = 1mm² 17A VD 46
• Actual VD = 2.08v
• Max Vd = 6.9v
• Max Disconnection Time: 0.4 seconds
• Earth Fault Loop Impedance = 7.28Ω
• Maximum Earth Fault Loop Impedance
• Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
• Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω