§ 16

Short circuit at the end of a supply cable

Is the short-circuit current for the end of the cable large enough for the fuse to break?

Answer: The fuse in the transformer substation or cable cabinet must be able to see a fault all the way at the cable end. The longer and thinner the cable, the smaller the short-circuit current becomes, and eventually the fuse cannot break in time. The calculator computes the minimum single-phase short-circuit current at the cable end and compares it with the current the fuse must have to break within the required time.

Formula

Ik1,min = c · U₀ / (Z_før + R_kabel), hvor R_kabel = ρ · L · (1/A_fase + 1/A_nul)

ρ at 80 °C: copper 0.0221 and aluminium 0.0355 Ω·mm²/m (20 °C value × 1.24, which DS/EN 60909-0 uses for minimum short-circuit current)

The fuse breaks in time when Ik1,min is at least as large as the current that the fuse's time/current curve requires for the permitted disconnection time.

Længste kabel = (c · U₀ / I_a − Z_før) / (ρ · (1/A_fase + 1/A_nul))

Symbols and units in the formula
Symbol	Meaning	Unit
z_for	Phase-neutral loop impedance before the cable (stated or measured)	mΩ
rho	Conductor material (resistivity at 80 °C)	Ω·mm²/m
laengde	Cable length	m
a_fase	Phase conductor cross-section	mm²
a_nul	Neutral or PEN conductor's cross-section	mm²
ia	Current that makes the fuse break in time (from its curve)	A
c	Voltage factor c (0.95 for minimum short-circuit current)	—
u0	Voltage phase-neutral	V
=	Minimum single-phase short-circuit current at cable end	A
=	Ratio to the breaker's trip current (must be at least 1)	gange
=	Cable loop resistance (phase + neutral)	Ω
=	Total loop impedance	Ω
=	Longest cable where circuit breaker still trips in time	m

Validity and source

Source: DS/EN 60909-0: spændingsfaktor c_min = 0,95 og ledere ved 80 °C · fagbilag: kortslutningseffekt og valg af kabeldimension

Does not apply when

Estimate: impedance before the cable is added to cable resistance as a figure and cable reactance is left out. With long cables with large cross-sections reactance matters and then the grid company's calculation programme must be used. The allowed disconnection time and the current the fuse should have are in the grid company's sizing rules and the fuse datasheet.

Worked example

The calculator's own default values inserted. The figures are calculated with exactly the formula the tool uses.

Substituted

Phase-neutral loop impedance before the cable (stated or measured)40 mΩ
Conductor material (resistivity at 80 °C)0.04 Ω·mm²/m
Cable length200 m
Phase conductor cross-section150 mm²
Neutral or PEN conductor's cross-section150 mm²
Current that makes the fuse break in time (from its curve)1,100 A
Voltage factor c (0.95 for minimum short-circuit current)0.95
Voltage phase-neutral230 V

Giver

Minimum single-phase short-circuit current at cable end1,623 A
Ratio to the breaker's trip current (must be at least 1)1.48 gange
Cable loop resistance (phase + neutral)0.0947 Ω
Total loop impedance0.1347 Ω
Longest cable where circuit breaker still trips in time335 m

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Sådan bruger du den

Når et forsyningskabel forlænges, et kabelskab flyttes, eller en kunde får en ny stikledning, skal du tjekke, at sikringen stadig kan se en fejl helt ude for enden. Er kablet for langt eller for tyndt, bliver kortslutningsstrømmen så lille, at sikringen ikke bryder i tide. Beregneren regner den mindste fejlstrøm mellem fase og nul.

Step by step

Find sløjfeimpedansen før kablet: fra netselskabets beregning eller målt i kabelskabet, hvor kablet starter.
Vælg ledermaterialet, og indtast længde og tværsnit for fase og nul.
Aflæs på sikringens tid/strøm-kurve, hvilken strøm der skal til for at bryde inden for den tilladte tid.
Er forholdet under 1, bryder sikringen ikke i tide. Vælg et større tværsnit, en mindre sikring eller et kortere kabel.
Se også den længste kabellængde, der kan bruges med de valgte data.

Typical errors

At bruge den trefasede kortslutningsstrøm. En fejl mellem fase og nul giver en mindre strøm, og det er den, sikringen skal se.
At bruge 20 °C-resistiviteten. Ved en fejl er lederen varm, og modstanden større.
At glemme nullederen. Fejlstrømmen løber ud i fasen og tilbage i nullen.
At aflæse sikringens kurve ved den forkerte tid.

Examples from everyday life

Aluminiumskabel på 200 m

Et nyt 4 × 150 mm² aluminiumskabel skal gå 200 m fra stationen til et kabelskab. Sløjfeimpedansen før kablet er 40 mΩ, og sikringens kurve viser, at den skal have 1.100 A for at bryde i tide.

Phase-neutral loop impedance before the cable (stated or measured): 40 mΩ
Conductor material (resistivity at 80 °C): Aluminium (0,0355)
Cable length: 200 m
Phase conductor cross-section: 150 mm²
Neutral or PEN conductor's cross-section: 150 mm²
Current that makes the fuse break in time (from its curve): 1,100 A
Voltage factor c (0.95 for minimum short-circuit current): 0.95
Voltage phase-neutral: 230 V

It means: Den mindste fejlstrøm ved kabelenden er 1,623 A, altså 1.48 times sikringens udløsestrøm. Sikringen bryder i tide. Kablet kunne være op til 335 m langt med samme sikring.

Samme kabel forlænget til 450 m

Kunden ønsker skabet flyttet, så kablet bliver 450 m langt.

Phase-neutral loop impedance before the cable (stated or measured): 40 mΩ
Conductor material (resistivity at 80 °C): Aluminium (0,0355)
Cable length: 450 m
Phase conductor cross-section: 150 mm²
Neutral or PEN conductor's cross-section: 150 mm²
Current that makes the fuse break in time (from its curve): 1,100 A
Voltage factor c (0.95 for minimum short-circuit current): 0.95
Voltage phase-neutral: 230 V

It means: Fejlstrømmen falder til 864 A, kun 0.79 times sikringens udløsestrøm. Den bryder ikke i tide. Vælg en mindre sikring eller et større tværsnit.

Mindre kabel med mindre sikring

Et 4 × 95 mm² aluminiumskabel går 250 m til et skab. Sikringen, der skal bruges, har en kurve, der kræver 700 A.

Phase-neutral loop impedance before the cable (stated or measured): 40 mΩ
Conductor material (resistivity at 80 °C): Aluminium (0,0355)
Cable length: 250 m
Phase conductor cross-section: 95 mm²
Neutral or PEN conductor's cross-section: 95 mm²
Current that makes the fuse break in time (from its curve): 700 A
Voltage factor c (0.95 for minimum short-circuit current): 0.95
Voltage phase-neutral: 230 V

It means: Fejlstrømmen er 963 A, altså 1.38 times udløsestrømmen. Det går, og den største længde er 364 m.

Rule of thumb

Dobbelt kabellængde giver næsten halv fejlstrøm, når kablet er den største del af sløjfen.

How to check the answer

Mål sløjfeimpedansen i kabelskabet efter montering med en sløjfeimpedansmåler. Den skal ligge tæt på den beregnede.