Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Choosing cables to load - 3-phase engine.

Status
Not open for further replies.

zitrex2

Newbie level 5
Joined
Sep 18, 2014
Messages
9
Helped
1
Reputation
2
Reaction score
1
Trophy points
3
Activity points
66
Hi everyone,
I need to wire up 3-phase motor.
5,5kW | 400V | 11,3A | direct delta connection | cos(phi)=0,86 |
Cable length must be 25m. | Tmax=60*C | Voltage drop max. 3%
From calculation, I see that 2,5mm^2 would be ok, but I have doubts. Can someone give me advice? Thank You.
 
Last edited:

Hi everyone,
I need to wire up 3-phase motor.
5,5kW | 400V | 11,3A | direct delta connection | cos(phi)=0,86 |
Cable length must be 25m. | Tmax=60*C | Voltage drop max. 3%
From calculation, I see that 2,5mm^2 would be ok, but I have doubts. Can someone give me advice? Thank You.

Show us your calculations and we'll review them. But it is essenially quite simple:
1) For the absolute voltage drop and the plate's current rating, calculate the wire's allowable ohmic value.
2) Divide that value by your length, and you'll have an ohms/meter requriement.
3) Go to a Copper Wire Table, and select the gage that meets that requirement. Give yourself a little margin.

BTW, I like your avatar of the jumping transmission tower!
 

My H07RN-F cable have 60*C rating. R(20*C)=7,98Ω/km -> 25m cable have R=0,2Ω (20*C) and R=0,23Ω (60*C) Ur=I*R=2,6V -> Voltage drop = 1,13% < 3%. This is in theory, if I'm right ofcourse. I'm concerned about inrush current which should be like 5-8 times bigger than normal current. I don't this electronics way of calculating wire across section would work in my problem.

Inrush I*5=56.5A -> Ur=13V -> Voltage drop = 5.65%

PS: Thanks, this is how voltage spikes and drops are made.

Calc. update
 
Last edited:

Looks like you didn't actually calculate. 2.5 sqmm is suffcient but not necessary.

- - - Updated - - -

Voltage drop is calculated for rated (steady state) current, not inrush. The relative voltage drop has to related to phase rather than phase-to-phase voltage.

An additional requiremet is that apart from acceptable voltage drop, the short circuit current must be high enough to trip the overcurrent protection means. But that's usually no point with short cables up to 100 m length.
 
  • Like
Reactions: zitrex2

    zitrex2

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top