2365 L3 Design Project

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Why are you using this cable?

You know that unless the accessory termination points are rated to 90dec C you have to de-rate the cable.


This is what was on the design spec. It stated that this cable was used for the dishwasher. 

 
Possibly, that depends on whether those lights would be utilised 100% of the time that the main pool lights were.

Who are you studying with?


Okay. I have checked the question and we don't have to do diversity and maximum demand till the next question. https://ufile.io/0ztu58hi

I am doing my course at Trade Skills 4 U. 

 
TS4U should have covered all this in class for you.


They only briefly went over it. I think I am okay doing it. For example, I know the Dishwasher is rated at 32A, so that would be my Design Current. I just intend on working through the questions. Hopefully, it will be okay, just finding the information in the regs and OSG. 

 
ok, but that's unfair on you, design, is really important.


I agree, these places only want your money. I think they essentially want you to just do all the work. Or they will just send me some screens and say the information can be found in there. 

 
I will give the Light a go as I think I have all the information. Would that be correct in saying that 32A for the Dishwasher is the Design Current? As that is what the piece of equipment is drawing. 

As the Calculation, I have been shown is 

Power divided by Voltage to get Amps. But as I am already given the Amps I would assume that its correct? 

 
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A dishwasher will be heating the water most of the time during the wash cycle so the maximum load would be drawn at most times when it is in operation.

Therefore the design current is 32A.

Since you have been given that figure it should not be necessary to use any other value.

In other circumstances where you only have the power (in watts) then you would use P/V = I, using the design voltage of 230V.

 
A dishwasher will be heating the water most of the time during the wash cycle so the maximum load would be drawn at most times when it is in operation.

Therefore the design current is 32A.

Since you have been given that figure it should not be necessary to use any other value.

In other circumstances where you only have the power (in watts) then you would use P/V = I, using the design voltage of 230V.


Thanks very much. That makes sense. I will see how I get on with the questions. I will probably have some questions. 

 
I am just working through the calculations and I am unsure on the last point:

Earth Fault Loop Impedance:

In BS7671 the calculation is

Zs =UoxCmin divided by Ia = Amps

Would the Ia be the Design Current?

Uo=230V Cmin=0.95 Zs

As I have given that a go and got 72.35A

But the Zs for 6A 60898 is 7.28 ohms

 
If you look at appendix 3 of BS7671 the tables beside the graphs give the Ia, minimum disconnection current to use.

For a 6A type B it is 30A as given on page 370.

 
If you look at appendix 3 of BS7671 the tables beside the graphs give the Ia, minimum disconnection current to use.

For a 6A type B it is 30A as given on page 370.


Thanks so much. I am just getting used to BS7671. 

 
I have done my Lighting Circuits. I am pretty happy with them, but I may need someone just to check them if that is okay? I am now working on the Power for the Pool.

I have two Circuits:
 

Circuit 1: 32A 5 core 90 degrees SWA cable clipped direct. I should be okay with that.

Sockets. Would you recommend a Ring or Radial, as I have the CCU in the Pump Room.

I will upload the diagram, so you have an idea.

https://ufile.io/bi8u6jco

 
I've decided that I will include a Ring Circuit in the Kitchen Area. So it will be 32A. As they are General Sockets. 

In the Changing Rooms, there are Sockets I was thinking of running them as a Radial so 16A. Would this be correct? I have attached the drawings above. 

 
Feel free to post a synopsis of the lighting and we can provide feedback on the design.

I am not sure how much design for loading you need to include in this question.  A cafe can be a very heavy load and might need two circuits on its own in order to ensure availability of power to a commercial enterprise so if one circuit has a fault then the cafe might be able to continue operations.

Similarly you might not want a fault on the open area of the cafe to interfere with cooking and heating in the kitchen area.  Though I may be going into too much consideration for this question.

Generally a 32A ring is a good circuit to run as it has a lot of advantages but they can still be overloaded by lots of heating electrics.  I would want to know more about the specific loads in the kitchen.

A 16A or 20A radial would be OK in changing rooms so long as lots of 1.8kW hairdryers and 3kW hand dryers are not in use.

 
So they are happy with what I have submitted, I am now moving on to Q5 which I will try and have a go.  I will let you know if I need any assistance if that is okay? 

 
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.

So one of the Fridge circuits in the design said that it was 16A. I know to look in BS7671 that the Zs for a 16A Type B breaker is 2.73 ohms.

Ze is 0.11 ohms

Using Table I1 in the OSG how would I work out the R1 +R2  value?

 
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.

So one of the Fridge circuits in the design said that it was 16A. I know to look in BS7671 that the Zs for a 16A Type B breaker is 2.73 ohms.

Ze is 0.11 ohms

Using Table I1 in the OSG 

16mm2 Line Conductor and 16mm2 Protective Conductor

R1=R2 is 2.30

To get the actual R1=R2 value is the calculation:

2.30 x Length: 30M then I would need to add in the correction factor of 1.20/ as the Ambient Temp in the Kitchen is 35 degrees C. Then divide by 1000 to get the Actual R1+ R2 Value. 

The cable is PVC surface conduit with 70 degrees C thermoplastic single core non-sheathed cable. 

If that calculation could be checked. How then do you work out 

Actual CPC size using Adiabatic Equation please?

 
You are correct with reading the R1+R2 value from table I1: 2.30 mΩ/m for a 16mm² line and cpc.

Tabulated value x length gives the R1+R2 at 20°C in milliohms.

To get the R1+R2 value at maximum normal conductor operating temperature (70°C) multiply by 1.2.

To get the value at 35°C multiply by (1+((35-20)*0.004)) = 1.06

The adiabatic equation is easier if you have the energy let through (I²t)  for the protective device as the BS7671 graphs do not go up to the sort of fault levels normally seen.

 
You are correct with reading the R1+R2 value from table I1: 2.30 mΩ/m for a 16mm² line and cpc.

Tabulated value x length gives the R1+R2 at 20°C in milliohms.

To get the R1+R2 value at maximum normal conductor operating temperature (70°C) multiply by 1.2.

To get the value at 35°C multiply by (1+((35-20)*0.004)) = 1.06

The adiabatic equation is easier if you have the energy let through (I²t)  for the protective device as the BS7671 graphs do not go up to the sort of fault levels normally seen.




So I have looked at the spec and there are 4 Fridges. 2 16A circuits so 1 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

 
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