Cooker calc

[silva.foxx]

Question from a 2400 paper

Cooker @ 12000w. TN-C-S @ Ze= 0.25ohms. Socket on cooker switch. BS1361 fuse. 25 meter run.

PVC t&e clipped direct with 4 other cables. Passes through 100mm length of surrounded insulation. Leaves group and in HG pvc conduit in ambient of 45 degrees.

Ib = 52amps diversity - 10+(30% of remaining 42A)+5 for socket = 28 Amps

In = 30A fuse

Correction factors Cg= 0.65 (4 cables... or should this be 5 to include this cable?)
Ca= 0.81 and Ci= 0.75

Worst 2 = Cg x Ci = 0.507 (or do we use all three?)

30 / 0.507 = 59.17 Amps to be carried

Cable selected is 10mm with tab ccc of 64 Amps - 4mm cpc

4.4mV x 28 x 25 = 3.08 volts so ok as <11.5v

Zs = 0.25 + ((6.44) x (1.2 group x 1.18 ambient)) = 0.477 ohms

adiabatic I get 1.32mm


Can anyone point out where I may have made a mistake, any pointers or if I've got this right, please?

Thanks in advance - s.f

 

[ezzzekiel]

cables in total, 5

 

[Mark.W]

Hi

You need to use 5 cables.

I might be completely wrong but here goes.

I used Installation Method C as it is "Clipped Direct"

Ib = 12000/230V = 52 A , Diversity = 27.65A
In = 30A

Correction Factors
Ca = 0.79 Table 4B1 BS 7671
Cg = 0.73 (5 Cables Reference Method C) Table 4C1 BS 7671
Ci = 0.78 (100mm) Table 52.2 BS 7671

It >= In / Cg Ca Ci
= 30 / 0.73*0.79*0.78
= 30 / 0.449
= 66.69A

Selected Cable size 16mm

2.8mv*27.65A*25m = 1.94V

Zs = 0.25 + (R1+R2)
= 0.25 + (0.15 * 1.2)
= 0.42 ohms

Adiabatic I get 2.99mm^2

Like I said I am probably wrong, been a long time!!!

16mm Cable does seem extreme.

 

[electro]

Hi


A 12KW is usually measure against a voltage of 240v working pressure so gives you a maximum of 50amps and 11.5KW @ 230v these calcs are dependant on manufacture (if not it will be 52Amps). Why are you using diversity for In this is not required, your design current is 50 amps if above is correct so In being greater than or equal to Ib then In= 50A cb not 30A cb.


Regards

 

[silva.foxx]

Ok...thanks so far guys.

Are we saying that I should not be applying diversity as this is just for determining total load of all circuits?

I can see that the fuse should handle total loading of the cooker!

Something that threw me was that part of the install is in HG pvc conduit so should the cable selection be based on this as it has a lower ccc than clipped direct?

Also, what are the correct correction factors? Is the ambient to be factored in with the insulation factor being more detrimental?

Finally... with regards Zs...do we add the grouping and ambient factors again?

Thanks guys.

s.f

 

[Mark.W]

No, when determining CCC you do use diversity. Maybe not on the cb, but I assumed that there was already a 30A fuse in place.

 

[silva.foxx]

Confusing myself big-time now.

12000w/240 = 50A.

Paper states apply diversity in this case. Diversity reads 10A + 30% of f.l (not remaining f.l) + 5A. Therefore Ib now = 30A

Should BS1361 fuse (paper states to use) be 30A or 60A to allow flc?

Correction factors:

Do we use Ib or In for this one? I've used Ib.

Cg = 0.6 as 5 cables bunched
Ci = 0.78 as 100mm surrounded insulation
Ca = 0.81

Paper states use onerous factors as not all apply at same time.

0.468 is bunched and in insualtion. 64A. Do we calculate to ref method C (clipped) or A (in conduit)?

0.81 is in high ambient. 37A. Do we then used ref method A (conduit)?

Both point to 10.0mm with 4mm cpc.

60A fuse obviously doesn't protect the 37A cable. This is whats confusing me. Do we use the 30A fuse to protect a circuit that can use the flc??? I'm not making sense to myself here!

Further thanks for your help.

Regards - s.f

 

[Mark.W]

I'd use a 30A mcb!
Still don't know where you get 0.81 for ambient! I get 0.79
Your OP says that it is clipped direct, so I'd use method C

12000/240 = 50A

Ib after diversity I get 27A, It is 10A + 30% f.l of connected cooking appliances in excess of 10A ( in this case 50 - 10 = 40) + 5A if a socket outlet is connected)
In other words 10A+(30% of 40A)+5A

The formula is In/CgCaCi...not In/CgCi (64A) or In/Ca (37A) all factors are in the same formula.
Using your figures for Correction Factors i still get 79.14A.
30/(0.81*0.60*0.78)
Ca = 0.81
Cg = 0.60
Ci = 0.78

It really would help if I saw the actual question, Can you scan in the paper of write the question as it stands?

 

[Steve D]

Surely diversity shouldn't be applied when you're working out the load of the cable in this type of instance.

 

[Mark.W]

Why wouldn't you use diversity in this instance?

 

[silva.foxx]

I'd use a 30A mcb!
Still don't know where you get 0.81 for ambient! I get 0.79
Your OP says that it is clipped direct, so I'd use method C

12000/240 = 50A

Ib after diversity I get 27A, It is 10A + 30% f.l of connected cooking appliances in excess of 10A ( in this case 50 - 10 = 40) + 5A if a socket outlet is connected)
In other words 10A+(30% of 40A)+5A

The formula is In/CgCaCi...not In/CgCi (64A) or In/Ca (37A) all factors are in the same formula.
Using your figures for Correction Factors i still get 79.14A.
30/(0.81*0.60*0.78)
Ca = 0.81
Cg = 0.60
Ci = 0.78

It really would help if I saw the actual question, Can you scan in the paper of write the question as it stands?

It's not actually an exam paper, sorry if I led you to believe so, it's the 2400 worksheet from

http://acorninspections.com/city2400/Design%20Erection%20and%20Verification%20Workbook.doc

EXAMPLE.

You are to design the following circuit by assessing the minimum acceptable csa of cable that would comply with all aspects of the Regulations.

A 12Kw cooker is to be fitted in domestic premises served by a TN-C-S type earthing system, having a value of Ze of 0.25 ohms. The cooker switch is to be fitted with a socket outlet. The distribution board is located at mains intake and has spare capacity for additional BS1361 fuses. The PVC twin with earth cable length is 25mtrs from distribution board to the cooker. The cable is grouped and clipped direct with 4 other cables for part of its route and all pass through thermal insulation which totally surrounds them for a distance of 100mm. The cooker cable then leaves the group and is routed through the roof space where it is installed in HG Pvc conduit and subject to an ambient temperature of 45 degrees C.

STEP 1.

Find the design current Ib. In this case you will need to apply diversity!


Ib =

STEP 2.

Now select In. The question implies that it must be a BS1361 device. Note Ib In

In =


STEP 3.

Because overload protection is required then the following conditions must be met;

Ib ≤ In ≤ Iz


and In
It ≥ Applicable correction factors

Now note that the cable is subject to various correction factors but not all at the same time!!

So, work out which set of conditions will be the most onerous.

Where the cable is grouped and is totally enclosed in thermal insulation, or where it is in conduit and subject to a high ambient temperature;

STEP 4.

Now select a cable from the relevant column in TABLE 4D2A which has a current carrying capacity greater than or equal to the highest value obtained in STEP 3.


The cable size is;

STEP 5



Now use TABLE 4D2B to check that the voltage drop does not exceed 4% of the nominal supply voltage of 230V.



STEP 6


Check that the circuit will be of sufficiently low impedance that when it is added to Ze, the maximum values of total earth loop impedance in TABLE 4B1 are not exceeded. Remember to correct for the cables operating temperature which is 70 degrees C.

STEP 7


When we have selected our circuit on the basis of overload protection and we have used one of the protective devices listed in 433-02-02 then it is assumed that the cable will comply with the thermal requirements of 434-03-03.
However, just as an exercise, lets check the cable for thermal compliance!



Note: above is wrt 16th Ed but trying to be up to date.

You are correct with the ambient 0.79. I guestimated from the OSG as it only goes to 40 deg in there but have since located the proper table in the BRB.

As you state Ca, Cg and Ci are all in the same formula but the paper states they don't all apply at the same time so use the most onerous ones, which I take to be grouping and insualtion.

Where does the 'leaves group and run in HG pvc conduit in ambient of 45 degrees' apply?
I wondered if this was to replace the actual install method from clipped direct.

I thought I explained it ok in the beginning but I've confused everyone now!

Thanks for the continued interest and help.

 

[Mark.W]

Hi

Step 1: Ib = 12000/240 = 50A, diversity 10A + (30% of 40A) + 5A = 27A
So Ib = 27A

Step 2: In = 30A

Step3: I'd say in conduit and at an ambient of 45deg. I would not use any Cg or Ci figures as the worst factor compensates for them, this is done when there are combination of factors not all affected at the same time.

Ca = 0.79
Cg = 1
Ci = 1
30/0.79*1*1 = 38A

Step 4: Cable selected is 6mm with 4mm cpc using Ref Method B as the conduit is in the roof space not in an insulated wall! Therefore CCC of 41A.

Step 5: 7.3mV*27*25 = 4.93V = 2.05%

Step 6: Maximum Zs for 30A BS1361 is 1.15 ohms
R1 = 0.08
R2 = 0.12
ZS = 0.25 + ((0.08*0.12)*1.2) = 0.49 ohms

Step 7: csa = 2.74mm so the cpc complies.

 

[aaronstuart]

Hi,
Page 160 of the OSG. 8.4......'A 30 or 32A circuit is usually appropriate for household or similar cookers of rating up to 15kW.

Thoughts

 

[Mark.W]

That sounds about right aaronstuart, that's why In is 30A

 

[silva.foxx]

Hi,
Page 160 of the OSG. 8.4......'A 30 or 32A circuit is usually appropriate for household or similar cookers of rating up to 15kW.

Thoughts

Are you deliberately trying to spoil all this fun?

notabrightspark ; thanks for your continued interest. Calc looks good.

Regards
s.f