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

[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.