Calculating volt drop, cable selection and mcb selection on an 3kw motor

[daiplayer]

ok....when applying diversity...to socket outlets...the osg says : 100% first socket....40% of all other...Is this 13A ?

 

[malcolmsanford]

IMO no. I do like the OSG guide in many ways and it is a good tool, but this is one of the few pieces of advice that I personally don't agree with.

I'm not sure how they arrive at this, as it must be pure speculation what is plugged into a socket outlet final circuit. I think socket outlet diversity is purely down to the designer to decide.

 

[Sparky Ninja]

ok....when applying diversity...to socket outlets...the osg says : 100% first socket....40% of all other...Is this 13A ?

no it's 100% of current demand of the largest circuit + 40% of current demand of every other circuit (I assume you refer to domestic)

For example..

Consider a domestic installation comprising:

i an electric cooker (without socket on panel)
ii a 7.2 kW shower
iii two lighting circuits (10 x GLS fittings on each)
iv two socket circuits

Cooker consists of:
Hob with 4 x 3kW elements
Main oven 2kW
Grill/top oven 2kW

Therefore the total installed capacity = 16kW @ 240V this will result in an appliance demand of 16000/240 = 66A

From there work out the demand.. = 10+(0.3x56) = 26.8A


So..

Cooker = 26.8A
Shower = 30A (7200/240)
Lights 1 = 4.16A (10x100w = 1kW/240)
Lights 2 = 4.16A (10x100w = 1kW/240)
Sockets 1 = 30A
Sockets 2 = 30A


To apply diversity we take 100% of the largest circuit out of the equation and then multiply the remaining tally by 0.4 (40%). We then add the largest circuit back on to the resultant value.

So, in this case the largest circuit is the cooker. Although with demand the rated current 'appears' to be lower than the shower, it is still the largest rating and therefore the largest circuit.

Therefore we take away the cooker for now.... 26.8A
and tally the remaining.. 30A + 4.16A + 4.16A + 30A + 30A = 98.29A total.

98.29 x 0.4 (40%) = 39.33A

39.33A + 26.8A = 66.13A Installation demand.

 

[daiplayer]

ok...that baffled me to no end...lol...but will have a good look at it later...many thanks

 

[daiplayer]

aha...so thats the max demand for the entire installation yeah ? ...ok...got that..but how then do you work out the max demand/ diversity of just a socket circiut ?

 

[malcolmsanford]

What Widdler as done for you is a full design scenario,the type of thing that you would have to do on the 2391-20 course or even as a design engineer and is spot on.

If like me your a rag A**e spark you could use a rule of thumb method similar to what the DNO lads do, which is 0.4 domestic 0.6 commerical 0.8 industrial. What this means is an installation you add up your final circuit protection devices and multiply by the factor, so in domestic it would be 0.4. so if you did that in Widdlers installation you would get over 57 amps about 9 below the proper calcualtion. As you can see this is quite close for a rule of thumb, and both would tell you that in reality that installation would work on a 60amp Fuse as it describes a typical domestic house, but chances are you would fit an 80amp most likely for future proofing.

As I said in my OP there really isn't a tried and trusted method for calculating a socket circuit. This is why the OSG give guidance of socket circuit design by use of areas. If you had a kitchen which needed 8 double sockets you may decide the best design would be a A1 ring or perhaps two A3 radials. Where in the upstairs there are only 7 double sockets in bedrooms you might go for a single A3 radial, it is just a method the designer will judge is best

 

[pushrod]

aha...so thats the max demand for the entire installation yeah ? ...ok...got that..but how then do you work out the max demand/ diversity of just a socket circiut ?

Are you asking about mcb size for sockets circuits???

To a large extent you can't because you don't know what is going to be be plugged in - if you do know then you plan accordingly. There are however general guidelines based on the floor area being served by the circuit - look at table 8A p158 of the OSG.

 

[telectrix]

if your sockets are on a 2.5mm ring final assume a current load of 25A. if a 2.5mm radial assume 15A. that's my rule of thumb guide to calculating.

 

[daiplayer]

ok...thanks guys....it just threw me when trying to calc socket circiuts diversity. How can you calculate each soscket demand when you dont know whats being plugged in ? ... ..in this exam i must show my cable selection and the reasons for doing so.....I assume that saying ' socket are usaully 2.5mm so i'll just do that ' wont be acceptable.....Can i just refer to the osg (as pushrod so kindly pointed out) , table 8A pg 158 ? ..is this acceptable to the examiner ?

Many thanks for your responses All.

 

[Sparky Ninja]

How can you calculate each soscket demand when you dont know whats being plugged in ? ... ..

We don't, we just apply the rated current of the protective device.

Similar to cable selection where we would use In instead of Ib.

in this exam i must show my cable selection and the reasons for doing so.....I assume that saying ' socket are usaully 2.5mm so i'll just do that ' wont be acceptable.....Can i just refer to the osg (as pushrod so kindly pointed out) , table 8A pg 158 ? ..is this acceptable to the examiner ?

Never write done assumptions in an exam. Some exam scenarios are deliberately exaggerated to avoid this.

I can give you a couple of cable selection questions if you like??

 

[daiplayer]

yes please mate. I could do with all the practise possible. Thankyou

 

[Sparky Ninja]

Okies well I'll first off give you a cable selection question. We can go back and forth between this and maximum demand. Repetition is the best form of study.

1.

A 3kW heating load is supplied from a 230V, 50Hz supply. The circuit is wired in thermoplastic insulated and sheathed twin and CPC cable protected by a BS 3036 fuse and is situated 15m away from the distribution board. It is run with one other circuit and is buried within the building material of the installation. The temperature within the installation can be assumed to be 25 degrees C. Determine:

i) the design current
ii) the fuse rating
iii) the smallest possible cable size
iv) the voltage drop within the cable.

 

[daiplayer]

ok this is my findings :

Ib = 3000/230 = 13.04 A
Table 2A osg....pg100 bs3036 fuses : 15A ( In greater than Ib ) ?

Table 6E1 pg 131 osg, 1.5MM cable can carry 14A .....Correct cable selection ?????

Now heres where i get confused.....if this is correct : by apllying grouping factors...It (greater than or equal to) : In/Ca x Cg x Ci x Cc

SO... 15/1.03 x0.8 x 0.725 = 25.1 A ?????

Deos this mean i must therefore use 4mm cable ( table 6E1) ...??

And therefore use 4mm for volt drop ??? which should be done first ??

However for 1.5mm cable ....29x13.04 x30 / 1000 = 11.34 v

????????????????????? lost ??????????????/ lol

 

[Sparky Ninja]

ok this is my findings :

Ib = 3000/230 = 13.04 A
Table 2A osg....pg100 bs3036 fuses : 15A ( In greater than Ib ) ?

Spot on

Table 6E1 pg 131 osg, 1.5MM cable can carry 14A .....Correct cable selection ?????

Selecting cable too soon.. will waste time in your exam

Now heres where i get confused.....if this is correct : by apllying grouping factors...It (greater than or equal to) : In/Ca x Cg x Ci x Cc

SO... 15/1.03 x0.8 x 0.725 = 25.1 A ?????

Deos this mean i must therefore use 4mm cable ( table 6E1) ...??

You've selected the correct factors etc, so if you put these workings down you would get a good quantity of the mark. However one thing you need to consider is the potential of overload.
The question reads 'heating load', not 'load' or 'sockets'. Because of this we can assume that the heater is of a fixed rating and so overload won't occur.
This therefore means that instead of using (In) divided by the correction factors, we could instead use (Ib).
It also means that the Cc factor of 0.725 (due to the BS3036 fuse) can also be omitted from the formula.

So the formula can be put down as:

It= Ib/(CaxCg)

= 13A/(0.80x1.03) = 15.7A

So from table 4D2A Column 6 we can pick a tabulated current-carrying capacity = 1.5mm carries 19.5A

And therefore use 4mm for volt drop ??? which should be done first ??

However for 1.5mm cable ....29x13.04x30 / 1000 = 11.34 v

????????????????????? lost ??????????????/ lol

Length = 15m, not 30.

Vd = (29x13x15)/1000 = 5.7V

 

[Sparky Ninja]

If you want to go through anything some more let me know, or if you want another Q then that's not a problem.

 

[daiplayer]

thanks for the reply....Gettin the hang of it now i think. Wasnt aware of the heater load scenario though. LOL. Yeah any q's...(if u have the time would be great mate. Thankyou very much,

I regards to the earlier q....i take it that because the heater is 15m away from the supply....it not 30m ( line out 15m...neutral back) ?

 

[Sparky Ninja]

thanks for the reply....Gettin the hang of it now i think. Wasnt aware of the heater load scenario though. LOL. Yeah any q's...(if u have the time would be great mate. Thankyou very much,

I regards to the earlier q....i take it that because the heater is 15m away from the supply....it not 30m ( line out 15m...neutral back) ?

Line out, neutral back??

Is that what you have been taught, or is that your assumption?

 

[Sparky Ninja]

Another question..

2.

A 7.6kW 230 V single-phase load is fed from a distribution board 25m away. The load is fed via thermoplastic singles and the cable is run on its own in steel conduit on the surface, and is protected by a BS EN 60898 Type B circuit-breaker. The cable is run in an area where the ambient temperature is 35 degrees C. Determine the size of cable required.

i) Design current
ii) Rating of protective device
iii) Current-carrying capacity of the cable
iv) Volt drop within the cable.

 

[daiplayer]

Ib : 33.04 A

40 A mcb ??

6mm cable ??

volt drop : 7.3 x 33.04 x 25 / 1000 = 6.03 V ....( for power : 5% of supply) 230/100 x 5 : 11.5V ....6.03 V accepted. ?

 

[Sparky Ninja]

Ib & In correct.

Cable should be 10.0 mm

(It = In/Ca) = 40/0.94 = 42.6A

It = 42.6A, Table 4D1A Column 4, 10.0 mm carries 57A (6.0 mm carries 41A, just shy)

Vd = (4.4 x 33 x 25) / 1000 = 3.63 V

This one does the opposite to the prior question, and it's exactly that that has caught you out. The first question had a 'heater', we could assume it wouldn't overload. However this question just says single phase load. We cannot determine if this load is subject to overload or not, and because of this we have to use (In) instead of (Ib) when working out the (It).

That is the only place you slipped up on that question, so you would only lose 1 mark out of the 6

I suggest that the first thing you do as you read the question is ask yourself if you can determine weather the load could be subject to overloading or not. Then carry on as you are.