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

[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 ?

 

[daiplayer]

how the hell do i find a list of cb's/ select them from the OSG ????????? ...or have i just overworked my brain ????????? lol

 

[pushrod]

how the hell do i find a list of cb's/ select them from the OSG ????????? ...or have i just overworked my brain ????????? lol

Off the top of my head i don't think there is a list in the OSG - i always use appendix 3 of BS7671.

edit - oops table 7.1 seems to show them

 

[daiplayer]

that q has really baffled me....Is the MCB ratings differnet when talking about 400v/ triple pole ? if so how do i find out which type to use and therefore do the resulting equations become affected ?

 

[Sparky Ninja]

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
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 ?

Selected reference method was C (No. 57).

In questions where it states within building materials, it refers to direct in masonry.
It's all down to interpretation of reference methods, the syllabus I use have in this example gone with the above. But your opinion of reference method B can also be justified.
It's one of the biggest debates I have with my guys.

 

[Sparky Ninja]

that q has really baffled me....Is the MCB ratings differnet when talking about 400v/ triple pole ? if so how do i find out which type to use and therefore do the resulting equations become affected ?

The question is answered with the exact same approach and formula bar one. The design current (Ib).

With Three phase (400V), we use the following formula:

Ib =

Power (P)
√3xUL​

 

[pushrod]

Selected reference method was C (No. 57).

In questions where it states within building materials, it refers to direct in masonry.
It's all down to interpretation of reference methods, the syllabus I use have in this example gone with the above. But your opinion of reference method B can also be justified.
It's one of the biggest debates I have with my guys.

Cheers for the reply, on reflection 57 is probs a better fit

I suppose we all have to carry our thermal resistivity meter in our kit as well

Thanks for keeping my brain ticking over!

 

[malcolmsanford]

The question is answered with the exact same approach and formula bar one. The design current (Ib).

With Three phase (400V), we use the following formula:



Ib =

Power (P)​

√3xUL​

Not sure what other posters think but Widdler deserves that for this thread, it was good of you to help that lad out the way you did. Top man

 

[pushrod]

Not sure what other posters think but Widdler deserves that for this thread, it was good of you to help that lad out the way you did. Top man

Totally agree
Private tuition is probs £15-£20 per hour!

 

[daiplayer]

Not sure what other posters think but Widdler deserves that for this thread, it was good of you to help that lad out the way you did. Top man

No argument there !....Nice to know there are actually people still out there that will help . Thanks again

 

[Sparky Ninja]

Any progress with question 4?

 

[daiplayer]

ok here i go :

For deign current : 36.1 A

In : 40A
...................

Ok..... here goes...

Ca : 35 oc...... : factor of 0.94....table 4b1 ?
Cg : 0.82........table 4C1 ???

so ...It = 50.8 A ...????

Im lost as to where i go from here....

I can do the volt drop calc....but i cant suss how to find the correct table for the cable selection ?

Thanks

 

[i=p/u]

come on set the sum out clear as ice, peas

 

[telectrix]

how do you get 36.1A for a 3kw motor?

 

[telectrix]

come on set the sum out clear as ice, peas

frozen peas???

 

[i=p/u]

you set sum out for me and paste orig question in too. thank you if you could aint done 3phase before

 

[daiplayer]

how do you get 36.1A for a 3kw motor?

Its a three phase 400 v heater

 

[Sintra]

you set sum out for me and paste orig question in too. thank you if you could aint done 3phase before

The question is in post #49

 

[daiplayer]

edit...the load in q is i three phase (400v ) 25kw heater..

 

[Sintra]

The question is about a 25KW 400V heater.

 

[daiplayer]

you set sum out for me and paste orig question in too. thank you if you could aint done 3phase before

3 phase design current = power / (sqaure root of 3) x Uo (nominal voltage)

 

[daiplayer]

Ib -

Ib = Power / SQUARE ROOT of three x Uo = 36.1 A

In = Greater than Ib = 40A breaker

It : = In / Cg = 40 / 0.82 (table 4c1) ......It = 48.78 A

(these values have changed)

Cable : 4mm cable (Table 4G1A , column 9) ?????????????????

 

[i=p/u]

is that per phase

 

[daiplayer]

is that per phase

no...in three phase systems the ninal voltage between phases is 400v....

in single phase it is 230v

 

[i=p/u]

kk, so would you share that over 3 phases for balancing, i know of initial q. but but but

 

[Sparky Ninja]

Ib -

Ib = Power / SQUARE ROOT of three x Uo = 36.1 A

In = Greater than Ib = 40A breaker

Correct

It : = In / Cg = 40 / 0.82 (table 4c1) ......It = 48.78 A

You've identified Cg, but need to identify Ca. Table 4B1. Mineral, thermoplastic @ 35degrees = 0.93 factor.

 

[daiplayer]

If your talking about load balancing over phases.....then....Yeah...kind of...but that would be in the inititial design of the heater itself......not considered during the heater installation...

Seeing as it a three phase supply...You consider that the load in q has already been balanced by the Maker of it.

Installers may need to consider balancing when installing single phase items...into a premesis that uses/ are provided by three phases...( ie,, three single phase (identical heaters) would be installed on a phase each ?


Correct me on this if im wrong though guys..

 

[daiplayer]

Correct



You've identified Cg, but need to identify Ca. Table 4B1. Mineral, thermoplastic @ 35degrees = 0.93 factor.

Dammit...i had tha to start !!! ....then changed my answer cos i thought it would fall between the bracket ( 30 - 70 degrees ?)

 

[pushrod]

Ib = Power / SQUARE ROOT of three x Uo = 36.1 A

is that per phase?

The 36.1A figure is the design current for each line (L1, L2, L3) it is not a total value that is then shared between 3 cables if that is what you are asking

edit : you have to be a little careful with terminology because the phase current (Ip) only equals the line current (Il) when connected in star. If in delta, as it would most likely be in a motor, then Il = √3* Ip

When designing for cable size it is just the line current that you are interested in.

A complex subject (which i am rapidly forgetting) but just to be aware that phase current and line current are slightly different and are not always the same value.

 

[telectrix]

The question is about a 25KW 400V heater.

sorry, my fault for not reading all the posts. was going from original thread title 3kw motor. DOH!

 

[daiplayer]

The 36.1A figure is the design current for each line (L1, L2, L3) it is not a total value that is then shared between 3 cables if that is what you are asking

edit : you have to be a little careful with terminology because the phase current (Ip) only equals the line current (Il) when connected in star. If in delta, as it would most likely be in a motor, then Il = √3* Ip

When designing for cable size it is just the line current that you are interested in.

A complex subject (which i am rapidly forgetting) but just to be aware that phase current and line current are slightly different and are not always the same value.

Much better explanation than mine...