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

[daiplayer]

Any explanations on how this is calculated ?

Diversity :

By referin to table 1b in the osg, it states : 100% largest motor ?? ...3kw ....???

3000/230 = 13.04 A....is this it !!?? max demand ??


How there on would you make cable selection , mcb selection ?

 

[ezzzekiel]

once design current is established select a protective device (in) rated equal or greater to ib then select a cable rated equal or greater than in

 

[daiplayer]

ok, thankyou. ...How then is volt drop calciulated...eg///length of run = 6mm...13A = ib....cable=2.5mm

 

[smartspark]

Is it going to be run in armoured cable or twin and earth?, is the circuit a radial? A standard ring with 2.5mm t+e will take anything up-to 3kw on a 32A breaker and a radial on a 20A breaker.
Check your table in BS7671 for voltage drop per milliamp per metre, is the cable being clipped direct or going through thermal insulation, blah blah blah.

 

[ezzzekiel]

vd = mva taken from regs x length of circuit x design current / 1000

 

[Sav]

Is it going to be run in armoured cable or twin and earth?, is the circuit a radial? A standard ring with 2.5mm t+e will take anything up-to 3kw on a 32A breaker and a radial on a 20A breaker.
Check your table in BS7671 for voltage drop per milliamp per metre, is the cable being clipped direct or going through thermal insulation, blah blah blah.

I think that an A3 radial 2.5mm cable within a max floor area of 50m2 can take upto 5kW and be protected by a 20A breaker.

I think thats right.......
Sav

 

[smartspark]

I wouldn't, The max I would put on a radial is 3KW, ie: immersion heater 16A or 20A breaker with a 20A double pole switch. Any higher I would go 4mm or 6mm cable 32A breaker. Just to air on the side of safety.

 

[telectrix]

I think that an A3 radial 2.5mm cable within a max floor area of 50m2 can take upto 5kW and be protected by a 20A breaker.

I think thats right.......
Sav

5kw >20A. thats a bit high for a 20A MCB. then there's volt drop to consider. 4mm and 32A minimum in my mind.

 

[daiplayer]

this is the drawing ive been given to design for my final practical exam. A 3 kw motor will be wired in. It will run from the cu to a dp switch via twin and earth in conduit ( 2.5m run) from the dp switch, then run in swa cable to motor , 500mm, then back through it all. Ive have to show how i calculated all the cable design, considering volt drop and diversity. ??? For some reason i come up with : volt drop ( approx 2v) ....for 1.5mm cable, And since the osg says a 1.5mm cable can carry 17.5 A (ref method b) table 6D1. ?????????????? surely im wrong !!!!!!

 

[ezzzekiel]

3kw load at 230v is just over 13a so nearest mcb would be a 16 so although a 1.5mm cable can handle it (going off basic info given) then your design would be correct although not ideal, but do they want exact or for your judgement, now providing all your calcs are correct and you can show how you have designed the circuit then all should be good.

 

[SimpleSimon]

this is the drawing ive been given to design for my final practical exam. A 3 kw motor will be wired in. It will run from the cu to a dp switch via twin and earth in conduit ( 2.5m run) from the dp switch, then run in swa cable to motor , 500mm, then back through it all. Ive have to show how i calculated all the cable design, considering volt drop and diversity. ??? For some reason i come up with : volt drop ( approx 2v) ....for 1.5mm cable, And since the osg says a 1.5mm cable can carry 17.5 A (ref method b) table 6D1. ?????????????? surely im wrong !!!!!!

No your not wrong.
3kW motor.
Diversity= 100%. So 13A
Volt drop= mV/a/m (in osg) x 13 x 2.5m/ 1000.
So 1.5 is ok as a minimum but not the best. You would be better off on 2.5mm but I'm saying that out of instinct so I'm sure some would say you can get away with 1.5mm

 

[smartspark]

I.5mm would be ok if you had a 2inch run of cable from the CU, if the run of cable is two floors away in a 5 bed house, I think you would be in trouble.

 

[pushrod]

Not asked to consider power factor for the 3kW motor? It will increase Ib

 

[malcolmsanford]

this is the drawing ive been given to design for my final practical exam. A 3 kw motor will be wired in. It will run from the cu to a dp switch via twin and earth in conduit ( 2.5m run) from the dp switch, then run in swa cable to motor , 500mm, then back through it all. Ive have to show how i calculated all the cable design, considering volt drop and diversity. ??? For some reason i come up with : volt drop ( approx 2v) ....for 1.5mm cable, And since the osg says a 1.5mm cable can carry 17.5 A (ref method b) table 6D1. ?????????????? surely im wrong !!!!!!

Are you sure that is right. I would have expected the question if it's a design one to want you to calculate the CPC sizing as well by using the adiabatic equation in reg 543.1.3.

So would it not be singles in that conduit?

 

[daiplayer]

yes, it is all run in singles. And in 20mm pvc conduit.

 

[daiplayer]

no...all inforamtion i have given is so far my own calcs. No pf is given

Not asked to consider power factor for the 3kW motor? It will increase Ib

 

[malcolmsanford]

yes, it is all run in singles. And in 20mm pvc conduit.

Then your going to have to work out your CPC by the equation.

 

[lewlec]

Not asked to consider power factor for the 3kW motor? It will increase Ib

hi

i agree with pushrod

i would imagine you need more info to go on, single or three phase? i assume its single phase therefore there is more to consider

i think its I = P/V X EFF X PF......... But not 100% though.

fitted many single phase 3kw 4pole motors and the running current is usually about 18amps

hope this helps
regards
gary

 

[i=p/u]

get onsite guide, appendix 6, easy once you shown once or twice, a rule of thumb on cpc is upto 16mm on cables all the same, but can be significantly lower if you use adibiatic equation

 

[ian.settle1]

I think that an A3 radial 2.5mm cable within a max floor area of 50m2 can take upto 5kW and be protected by a 20A breaker.

I think thats right.......
Sav

Thats for sockets I think you will find not for a 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.

 

[Sparky Ninja]

another..

3.

An upstairs lighting circuit is wired in thermoplastic insulated and sheathed twin and CPC cable. The cable is buried within the fabric of the building. It is run with insulation surrounding it for a distance of 0.5m, although its total length is 30m. There are 8 x 230V GLS lights on the circuit, and the circuit is protected by a 5A BS 3036 fuse. If the ambient temperature is 30 degrees C and there is no grouping, determine an appropriate cable size.

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

I have some maximum demand questions ready and waiting too..

 

[daiplayer]

many many many thanks for taking the time to teach a bit mate. My tutors are known for being ...er...****e...

Hence the reason for seekin help.

Any more q's would be much apprieciated ( around all the subjects) ....Thanks again

 

[daiplayer]

now thats thrown me....Do i consider the cable to be in insulation ?
Do i need to confirm the protective device rating if it is given ? (check it maybe ? )

 

[daiplayer]

Ib = 3.47A
=1mm cable
=fuse 5A

volt drop : 44x3.47x30/100 = 4.58v ?

 

[Sparky Ninja]

Spot on

Hopefully you would have had a correction rating of 0.5 for Ci, giving you a current-carrying capacity of 7A = 1.0mm / 15Amp

I had 4.62V, but that's because I rounded some of my figures up.

 

[daiplayer]

...success....!!! good tutor you see ;-)

 

[Sparky Ninja]

now thats thrown me....Do i consider the cable to be in insulation ?
Do i need to confirm the protective device rating if it is given ? (check it maybe ? )

Yes for the length described, and always as it could be provided incorrect to mislead you.

 

[daiplayer]

sneaky sneaky...lol

 

[Sparky Ninja]

I want to give you one last question on cable selection (to allow for 400V) then we can do some maximum demand/diversity and adiabatics (sizing of protective conductor)

4.

A 25kW, 400 V three-phase heater is wired in thermoplastic sheathed MICC cable fixed to horizontal perforated cable tray. The circuit is run with two other circuits and the maximum permitted volt drop is 6 V. The length of run is 28m and the heater is protected by a BS EN 60898 Type B triple pole circuit-breaker. The ambient temperature is 35 degrees C.

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

 

[pushrod]

Spot on




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



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

Much respect to you Widdler for putting so much time and effort in. One very small point which doesn't affect the outcome at all. The question said

buried within the building material of the installation

So wouldn't that mean ref method B [42 from table 4A2?] not ref method C, so table 4D2A column 4 = 1.5mm² carries 16.5A

or am i wrong ?