Cable calculation Star-Delta 1

[St_Isakovic]

Hello,

I am calculating a cable (selecting fuse/protection, calculating voltage drop) for star/delta starter of squirrel-cage motor by IEC standard, IEC 60364-4-43.

When there are two cables in parallel that supply some load, let's say two cables that supply some electrical cabinet from a substation, we take maximum allowed current as two times maximum allowed current of one cable, and take cross section in voltage drop calculation as two times cross section of one cable. Assumption is that load is divided equally between cables, of course cables are of the same type and same length.

In case of star/delta starter currents are not halved in the branches of circuit, they are square root of 3 times smaller than current that passes through protection device. Now, is it correct to do calculation for this case by looking at it like having "1.73 cables in parallel"? I think this will give correct result when sizing cable and proving it is protected from overcurrents.

 

[che12345]

Mr. St_Isakovic (Electrical)(OP)14 Jul 23 06:24
".....In case of star/delta starter currents are not halved in the branches of circuit, they are square root of 3 times smaller than current that passes through protection device. Now, is it correct to do calculation for this case by looking at it like having "1.73 cables in parallel"?..... I".
There was a similar question raised by Mr. zaboli on 11 inst. See Electrical motor & generator & engineering section.
I cut and paste my reply as following for your consideration.
1. For simplicity, take the motor (run/delta) or rated current be I A.
2. A SD starter, when in run/delta condition:
a) the source current (through the main breaker) is I A.
b) the current flow through the main contactor = the delta contactor= (the thermal over load)* = the current through each of the six conductors from the starter to the motor. Note: the thermal over load* is wired in series after the main contactor.
c) The current is [0.58 x I ] ...A
3. The voltage drop of the conductors to the motor and the thermal over load* setting are based on current, see above 2. c).
Che Kuan Yau (Singapore)

 

[St_Isakovic]

Thank you Mr. Che.

Motor in question is 160kW, 279A.
Cables are 4x95mm2, XLPE isolation, main protection is NH2 315A fuse. According to standard it needs to be shown that fuse protects the cable (Ib<In<Iz and I2<1.45Iz conditions from standard). Looking at single cable I can't show that it is protected, but plugging in "1.73 cables in parallel" works, it just feel weird putting that into project. I couldn't find similar calculation anywhere online or in literature.

 

[che12345]

Mr. St_Isakovic (Electrical)(OP)14 Jul 23 11:25
"#1....Motor is 160kW, 279A. Cables are 4x95mm2, XLPE isolation, main protection is NH2 315A fuse. According to standard it needs to be shown that fuse protects the cable (Ib<In<Iz and I2<1.45Iz conditions from standard). #2. Looking at single cable I can't show that it is protected, but plugging in "1.73 cables in parallel" works, it just feel weird putting that into project. I couldn't find similar calculation anywhere online or in literature".
I try to answer your question as following for your consideration. Note: this is pertaining to the selection of the main cable size , which is protected by fuse.
1. Ib = 279 A, In =315 A, Iz = 326 A for (3 or 4 conductors in air with ambient air 30deg C, max operating 90deg C. Attention: the actual value of Iz is depending on the installation method and the ambient temperature.
2. According to the IEC standards:
a) Ib<In<Iz and I2<1.45Iz are valid,
b) I2= 1.45In is valid for circuit breaker complied with IEC 60898,
c) I2= 1.3In is valid for circuit breaker complied with IEC 60947-2,
d) In < 0.9 Iz is valid with fuse complied with IEC 60269-2-1.
3. In case the VD > 4% due to long length, increase 95 to 120 mm2 etc....
4. If two of smaller size conductors in parallel are used, the new Iz is [ Iz x 2], not Iz x 1.7.
Che Kuan Yau (Singapore)

 

[waross]

The answer may be as confused as the question.
When two cables are in parallel, each carries current in the inverse proportion to their respective impedances.
It is good practice and often a code requirement that the cables are identical and of equal lengths.
In that case, each cable will carry one half of the load. (1/2)
But, you throw in star delta starting and it becomes unclear.
If a cable is replaced by two parallel cables, still one half of the total.
But if the load is split, such as the A-B corner of a delta winding, so that one cable feeds the C-A winding and one cable feeds the A-B winding, the cables are no longer in parallel. The cables are now carrying currents at different phase angles. When combined, the total current is less than the sum of the parts.
For example, if each winding draws 10 Amps, each cable will carry 10 Amps.
However, if the cables are joined together at the A-B delta corner so that they are in true parallel, the resultant current will be 17.3 Amps rather than 20 Amps.
So if each phase of a delta winding draws 10 Amps, the current in each phase will be 10A/1.73 = 5.8 A.
So for a delta winding drawing 10 Amps per phase, each phase conductor will carry 10 Amps.
If two cables are used in true parallel, that is joined at each end, then each cable will draw 10A/2 = 5A.
If, however, two cables are used so that each cable feeds one phase winding, such as may be done for series parallel starting, then each cable will carry 10A/1.73 = 5.8A.
As is frequently said here;
"It depends".
It depends here on the actual connections.

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Ohm's law
Not just a good idea;
It's the LAW!