Calculating short-circuit current in a Low Voltage (LV) distribution system with both a

transformer and a diesel generator (DG) set requires different approaches for each source.
Here’s how you can do it:

1. Short-Circuit Current Contribution from the Transformer

The transformer's short-circuit current can be calculated using the following formula:

Isc=IflZpuI_{sc} = \frac{I_{fl}}{Z_{pu}}

Where:

 IscI_{sc} = Short-circuit current (A)

 IflI_{fl} = Full-load current of the transformer (A), given by:

Ifl=Srated×1033×VLLI_{fl} = \frac{S_{rated} \times 10^3}{\sqrt{3} \times

V_{LL}}

 ZpuZ_{pu} = Per-unit impedance of the transformer (typically given as a percentage

in datasheets, e.g., 5% = 0.05 pu)
 SratedS_{rated} = Transformer rated power in kVA
 VLLV_{LL} = Line-to-line voltage (V)

2. Short-Circuit Current Contribution from the DG Set

A DG set has a different behavior compared to a transformer, as it has a limited short-circuit

current due to its internal reactance and the governor response.

Isc=IflZDGI_{sc} = \frac{I_{fl}}{Z_{DG}}

Where:

 ZDGZ_{DG} = Total reactance of the DG set (includes subtransient reactance Xd′

′X''_d)
 Xd′′X''_d (per unit) is typically 0.1 to 0.2 for synchronous generators.
 IflI_{fl} = Full-load current of the generator, calculated similarly to the transformer:

Ifl=SDG×1033×VLLI_{fl} = \frac{S_{DG} \times 10^3}{\sqrt{3} \times V_{LL}}

Approximate DG short-circuit current:

Isc=IflXd′′I_{sc} = \frac{I_{fl}}{X''_d}
where Xd′′X''_d is the subtransient reactance in per unit.

3. Combined Short-Circuit Contribution

Since both the transformer and DG set contribute to the short-circuit current, the total fault
current at the LV bus is:

Isctotal=IscTransformer+IscDGI_{sc}^{total} = I_{sc}^{Transformer} + I_{sc}^{DG}

where IscTransformerI_{sc}^{Transformer} and IscDGI_{sc}^{DG} are computed

separately using the formulas above.

4. Example Calculation

Given:

 Transformer: 1000 kVA, 5% impedance, 415V, 3-phase

 DG Set: 800 kVA, Xd′′=0.15X''_d = 0.15, 415V, 3-phase

Step 1: Full-load Currents

IflTransformer=1000×1033×415=1391AI_{fl}^{Transformer} = \frac{1000 \times 10^3}{\

sqrt{3} \times 415} = 1391 A IflDG=800×1033×415=1113AI_{fl}^{DG} = \frac{800 \times
10^3}{\sqrt{3} \times 415} = 1113 A

Step 2: Short-Circuit Currents

For the transformer:

IscTransformer=13910.05=27,820AI_{sc}^{Transformer} = \frac{1391}{0.05} = 27,820 A

For the DG:

IscDG=11130.15=7,420AI_{sc}^{DG} = \frac{1113}{0.15} = 7,420 A

Step 3: Total Short-Circuit Current

IscTotal=27,820+7,420=35,240AI_{sc}^{Total} = 27,820 + 7,420 = 35,240 A

Key Takeaways

 Transformers contribute a high short-circuit current due to their low impedance.

 DG sets have a limited fault current based on their subtransient reactance.
 The total fault current is the sum of individual contributions.
 Protection settings should consider both contributions for accurate relay coordination.
Would you like a more detailed breakdown or a specific case study? 🚀
To perform a short-circuit calculation for the system shown in the diagram, we need to
consider contributions from:

1. Two 11kV/400V Dry-Type Transformers

2. Two Diesel Generators (DG Set S & DG Set W)

Step-by-Step Short-Circuit Calculation

1. Assumptions & Given Data Required

Before performing the calculation, we need the following information (or assumptions if not
available):

 Transformers:
o Rating: STS_{T} (in kVA)
o Impedance: ZTZ_{T} (in %)
o Secondary Voltage: VLL=400VV_{LL} = 400V
 DG Sets:
o Rating: SDGS_{DG} (in kVA)
o Subtransient reactance Xd′′X''_d (typically 0.1 to 0.2 pu)
o Secondary Voltage: VLL=400VV_{LL} = 400V

2. Short-Circuit Contribution from Transformer

For each transformer:

IscTransformer=IflTransformerZpuI_{sc}^{Transformer} = \frac{I_{fl}^{Transformer}}
{Z_{pu}}

 Full-load current of transformer:

IflTransformer=ST×1033×VLLI_{fl}^{Transformer} = \frac{S_{T} \times 10^3}{\sqrt{3} \

times V_{LL}}

 Short-circuit current:

IscTransformer=IflTransformerZT/100I_{sc}^{Transformer} = \frac{I_{fl}^{Transformer}}
{Z_{T}/100}

3. Short-Circuit Contribution from DG Set

For each DG:

IscDG=IflDGXd′′I_{sc}^{DG} = \frac{I_{fl}^{DG}}{X''_d}
  Full-load current of DG:

IflDG=SDG×1033×VLLI_{fl}^{DG} = \frac{S_{DG} \times 10^3}{\sqrt{3} \times

V_{LL}}

4. Total Short-Circuit Current

Since the transformers and DGs operate in parallel, the total short-circuit current at the Main
LT Panel is:

IscTotal=IscTransformer1+IscTransformer2+IscDG1+IscDG2I_{sc}^{Total} =
I_{sc}^{Transformer1} + I_{sc}^{Transformer2} + I_{sc}^{DG1} + I_{sc}^{DG2}

Next Steps

1. Provide Transformer & DG Ratings – If you have the actual ratings, I can compute
exact values.
2. Confirm Impedance Values – Transformer impedance (%Z) and DG subtransient
reactance (Xd′′X''_d) are needed.
3. Specify Fault Location – Do you need the short-circuit at the Main LT Panel, UPS
Panel, or another point?

Let me know these details, and I'll finalize the calculations for you! 🚀
To calculate the short-circuit current at the Main LT Panel, let’s define the parameters
based on common assumptions. If you have actual values, please provide them to refine the
calculation.

1. Assumed System Data

Transformers (11kV/400V Dry Type)

 Rating: ST=2000S_T = 2000 kVA (each)

 Impedance: ZT=5%Z_T = 5\%
 Voltage: VLL=400VV_{LL} = 400V

Diesel Generators (DG Set S & W)

 Rating: SDG=1250S_{DG} = 1250 kVA (each)

 Subtransient Reactance: Xd′′=0.15X''_d = 0.15 pu
 Voltage: VLL=400VV_{LL} = 400V

2. Full-Load Current Calculation

For each transformer:

IflTransformer=2000×1033×400I_{fl}^{Transformer} = \frac{2000 \times 10^3}{\sqrt{3} \

times 400} IflTransformer=2887AI_{fl}^{Transformer} = 2887 A

For each DG:

IflDG=1250×1033×400I_{fl}^{DG} = \frac{1250 \times 10^3}{\sqrt{3} \times 400}

IflDG=1805AI_{fl}^{DG} = 1805 A

3. Short-Circuit Current Calculation

Transformers Contribution

IscTransformer=IflTransformerZT/100I_{sc}^{Transformer} = \frac{I_{fl}^{Transformer}}
{Z_T / 100} IscTransformer=28870.05=57,740AI_{sc}^{Transformer} = \frac{2887}{0.05}
= 57,740 A

For two transformers:

IscTotalTransformers=2×57,740=115,480AI_{sc}^{Total Transformers} = 2 \times 57,740 =

115,480 A
Diesel Generators Contribution

IscDG=IflDGXd′′I_{sc}^{DG} = \frac{I_{fl}^{DG}}{X''_d}
IscDG=18050.15=12,033AI_{sc}^{DG} = \frac{1805}{0.15} = 12,033 A

For two DGs:

IscTotalDGs=2×12,033=24,066AI_{sc}^{Total DGs} = 2 \times 12,033 = 24,066 A

4. Total Short-Circuit Current at the Main LT Panel

Since the transformers and DGs contribute in parallel:

IscTotal=IscTotalTransformers+IscTotalDGsI_{sc}^{Total} = I_{sc}^{Total Transformers}

+ I_{sc}^{Total DGs} IscTotal=115,480+24,066=139,546AI_{sc}^{Total} = 115,480 +
24,066 = 139,546 A

5. Conclusion

 The estimated short-circuit current at the Main LT Panel is ~139 kA.

 This is a very high value and must be considered when selecting circuit breakers
(ACB, MCCB) and protection relays.

Let me know if you need further refinements based on actual system data! 🚀
Yes! We can calculate short-circuit current at the loads and sub-boards by following
these steps:

1. Approach for Load & Sub-Board Short-Circuit

Calculation
The short-circuit current at a load or sub-board depends on:

1. The main short-circuit current (ISC) at the Main LT Panel (already calculated:
~139 kA).
2. Cable impedance between the Main LT Panel and the sub-board/load.
3. Any additional transformers or protective devices in between.

To find the short-circuit current at a sub-board, we apply:

IscSub=IscMain1+(ZCableZSource)I_{sc}^{Sub} = \frac{I_{sc}^{Main}}{1 + \left(\

frac{Z_{Cable}}{Z_{Source}}\right)}

Where:

 IscSubI_{sc}^{Sub} = Short-circuit current at the sub-board

 IscMainI_{sc}^{Main} = Short-circuit current at the Main LT Panel (already
calculated)
 ZCableZ_{Cable} = Impedance of the cable between the Main LT Panel and the sub-
board
 ZSourceZ_{Source} = Impedance of the source (transformer + DG set)

2. Cable Impedance Calculation

Cable impedance (ZCableZ_{Cable}) depends on:

 Cable length (L)

 Cable type & size (cross-section area)
 Resistance (RcR_c) and reactance (XcX_c) per meter from cable manufacturer
data

Total cable impedance:

ZCable=Rc+jXcZ_{Cable} = R_c + j X_c

Impedance per phase for a typical 400V LV cable (example: 3-core, 240mm² copper):
ZCable=(0.000125+j0.00009)×LΩ/mZ_{Cable} = (0.000125 + j0.00009) \times L \quad \
Omega/m

For L = 50m,

ZCable=(0.000125+j0.00009)×50Z_{Cable} = (0.000125 + j0.00009) \times 50

ZCable=(0.00625+j0.0045)ΩZ_{Cable} = (0.00625 + j0.0045) \quad \Omega

3. Short-Circuit Current at a Sub-Board

Using the main short-circuit current IscMainI_{sc}^{Main} = 139 kA and assuming the
source impedance is small compared to cable impedance:

IscSub=1390001+(0.00750.05)I_{sc}^{Sub} = \frac{139000}{1 + \left(\frac{0.0075}{0.05}\

right)} IscSub=1390001.15=120,870AI_{sc}^{Sub} = \frac{139000}{1.15} = 120,870 A

For a longer cable (100m), the short-circuit current decreases further.

4. Short-Circuit Current at Load Panels

For a load panel further down the network (e.g., UPS Panel or Emergency Panel):

 Use the same formula with additional cable impedance.

 The more cable length and smaller cable size, the more impedance → lower short-
circuit current.

For example, if the cable from the sub-board to a load is 30m, the additional impedance
reduces the short-circuit current further.

5. Key Takeaways
 Closer to the Main LT Panel → Higher Short-Circuit Current (~139 kA)
 Farther from the LT Panel (longer cable) → Lower Short-Circuit Current
 Smaller cables = higher impedance = lower short-circuit current
 Breakers & protection settings should be selected based on these values

Would you like me to calculate for a specific panel (e.g., UPS Panel, Tunnel Load)? 🚀
Short-Circuit Current Calculation for Jet Fan Load in Tunnel

To calculate the short-circuit current at the Jet Fan Load in the Tunnel, we need to
consider:

1. Main LT Panel Short-Circuit Current – already calculated as 139 kA.

2. Cable Impedance from the Main LT Panel to the Jet Fan.
3. Additional Distribution Panels/Sub-Boards in between (if applicable).
4. Jet Fan Motor Characteristics – Typically, jet fans are large induction motors,
which may contribute some fault current due to their internal reactance.

1. Assumed Data for Calculation

If exact system data isn't available, we assume:

 Cable from Main LT Panel to Jet Fan Load:

o Type: 3-core, 240 mm² Copper
o Length: 100m (assumption, adjust if needed)
o Resistance: Rc=0.000125 Ω/mR_c = 0.000125 \, \Omega/m
o Reactance: Xc=0.00009 Ω/mX_c = 0.00009 \, \Omega/m
 Jet Fan Load:
o Power: 110 kW
o Voltage: 400V
o Motor Locked-Rotor Current (LRC): 6 times full-load current
o Motor Impedance ZMZ_{M}: Assumed as 0.2 pu

2. Cable Impedance Calculation

Total cable impedance for 100m run:

ZCable=(Rc+jXc)×LZ_{Cable} = (R_c + j X_c) \times L

ZCable=(0.000125+j0.00009)×100Z_{Cable} = (0.000125 + j0.00009) \times 100
ZCable=(0.0125+j0.009) ΩZ_{Cable} = (0.0125 + j0.009) \, \Omega

3. Full-Load Current of Jet Fan Motor

The full-load current is given by:

Ifl=P×1033×VI_{fl} = \frac{P \times 10^3}{\sqrt{3} \times V} Ifl=110×1033×400I_{fl} = \

frac{110 \times 10^3}{\sqrt{3} \times 400} Ifl=159AI_{fl} = 159 A
4. Jet Fan Contribution to Fault Current
Since induction motors contribute to short-circuit current for a short time, their
contribution is typically 4-6 times full-load current.

IscMotor=6×Ifl=6×159=954AI_{sc}^{Motor} = 6 \times I_{fl} = 6 \times 159 = 954 A

This contribution decays quickly after a few cycles.

5. Short-Circuit Current at Jet Fan Load

Using:

IscLoad=IscMain1+(ZCableZSource)I_{sc}^{Load} = \frac{I_{sc}^{Main}}{1 + \left(\

frac{Z_{Cable}}{Z_{Source}}\right)}

Assuming source impedance ZSourceZ_{Source} ≈ 0.05 Ω:

IscLoad=1390001+(0.0150.05)I_{sc}^{Load} = \frac{139000}{1 + \left(\frac{0.015}{0.05}\

right)} IscLoad=1390001.3=106,923AI_{sc}^{Load} = \frac{139000}{1.3} = 106,923 A

Adding motor contribution:

IscTotal=106,923+954=107,877AI_{sc}^{Total} = 106,923 + 954 = 107,877 A

6. Conclusion
 Short-Circuit Current at the Jet Fan Load = ~108 kA.
 Protection devices (MCCB, ACB) should be rated accordingly.
 Motor contribution (~954 A) is small but important in the first few cycles.

Would you like me to adjust this based on different cable size, length, or motor power? 🚀