Sensitive Equipment Protection Calculator – IEEE, IEC

Protecting sensitive electrical equipment from faults and transient disturbances is critical in modern power systems. Accurate calculations ensure reliability and longevity of devices following IEEE and IEC standards.

This article explores the Sensitive Equipment Protection Calculator, detailing formulas, tables, and real-world applications aligned with IEEE and IEC guidelines. Learn how to optimize protection settings effectively.

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Calculate maximum allowable fault current for a 5 MVA transformer with 11 kV rating.
Determine relay pickup setting for a sensitive motor protection at 400 V, 50 Hz.
Compute coordination time interval between primary and backup relays for a 33 kV feeder.
Estimate transient recovery voltage (TRV) for a 132 kV circuit breaker interrupting a 2000 A fault current.

Common Values for Sensitive Equipment Protection – IEEE and IEC Standards

Parameter	Typical Values	Units	Reference Standard
Rated Voltage (Transformer)	11, 33, 66, 132	kV	IEC 60076-1
Rated Current (Transformer)	50, 1000, 2000	A	IEEE C57.12.00
Fault Current (Symmetrical)	5,000 – 50,000	A	IEEE Std 242-2001
Relay Pickup Current	1.2 – 1.5 × Load Current	A	IEC 60255-1
Time Dial Setting (TDS)	0.05 – 1.0	Unitless	IEEE C37.112
Transient Recovery Voltage (TRV)	1.0 – 2.5 × Rated Voltage	p.u. (per unit)	IEC 62271-100
Coordination Time Interval (CTI)	0.2 – 0.5	seconds	IEEE Std C37.2
Maximum Allowable Voltage Drop	5%	%	IEC 61000-4-30

Fundamental Formulas for Sensitive Equipment Protection Calculations

1. Fault Current Calculation

The maximum fault current at the equipment terminals is essential for setting protective devices.

Ifault = Vrated / (Zsource + Zequipment)

I_fault: Fault current (A)
V_rated: Rated voltage of the equipment (V)
Z_source: Source impedance (Ω)
Z_equipment: Equipment impedance (Ω)

Typical source impedance values depend on system configuration and are often provided by utility companies or calculated from short-circuit MVA.

2. Relay Pickup Setting

Relay pickup current must be set above normal load current but below fault current to ensure selectivity.

Ipickup = K × Iload

I_pickup: Relay pickup current (A)
K: Safety factor (typically 1.2 to 1.5)
I_load: Full load current of the equipment (A)

3. Time Dial Setting (TDS) for Inverse Time Overcurrent Relays

The operating time of the relay is inversely proportional to the magnitude of the fault current.

t = TDS × (0.14 / ((I / Ipickup)0.02 – 1))

t: Operating time of the relay (seconds)
TDS: Time dial setting (unitless)
I: Fault current magnitude (A)
I_pickup: Relay pickup current (A)

4. Coordination Time Interval (CTI)

CTI ensures proper coordination between primary and backup protective devices.

CTI = tbackup – tprimary

CTI: Coordination time interval (seconds)
t_backup: Operating time of backup relay (seconds)
t_primary: Operating time of primary relay (seconds)

5. Transient Recovery Voltage (TRV)

TRV is the voltage that appears across the terminals of a circuit breaker after current interruption.

TRV = k × Vrated

TRV: Transient recovery voltage (V)
k: Factor depending on system and fault type (1.0 to 2.5 p.u.)
V_rated: Rated voltage of the equipment (V)

Detailed Real-World Examples of Sensitive Equipment Protection Calculations

Example 1: Transformer Fault Current and Relay Setting Calculation

A 5 MVA, 11 kV transformer is connected to a power system with a short-circuit MVA of 100 MVA. Calculate the maximum fault current at the transformer terminals and determine the relay pickup current if the full load current is 262 A. Use a safety factor of 1.3.

Step 1: Calculate Transformer Full Load Current

Given:

Transformer rating, S = 5 MVA = 5,000,000 VA
Rated voltage, V = 11 kV = 11,000 V

Full load current, I_load = S / (√3 × V)

Iload = 5,000,000 / (1.732 × 11,000) ≈ 262 A

Step 2: Calculate Source Impedance

Short-circuit MVA, S_sc = 100 MVA

Source impedance, Z_source = V² / S_sc

Zsource = (11,000)2 / 100,000,000 = 1.21 Ω

Step 3: Calculate Fault Current

Assuming transformer impedance is negligible compared to source impedance:

Ifault = V / Zsource = 11,000 / 1.21 ≈ 9,090 A

Step 4: Determine Relay Pickup Current

Using safety factor K = 1.3:

Ipickup = 1.3 × 262 = 340.6 A

The relay should be set to trip at approximately 341 A to protect the transformer effectively.

Example 2: Coordination Time Interval Between Primary and Backup Relays

For a 33 kV feeder, the primary relay has a pickup current of 500 A and a time dial setting (TDS) of 0.2. The backup relay has a pickup current of 1000 A and a TDS of 0.5. Calculate the coordination time interval if a fault current of 2000 A occurs.

Step 1: Calculate Operating Time of Primary Relay

Using the inverse time formula:

tprimary = 0.2 × (0.14 / ((2000 / 500)0.02 – 1))

Calculate the exponent:

2000 / 500 = 4
4^0.02 ≈ 1.029
1.029 – 1 = 0.029

Therefore:

tprimary = 0.2 × (0.14 / 0.029) ≈ 0.2 × 4.83 = 0.966 seconds

Step 2: Calculate Operating Time of Backup Relay

tbackup = 0.5 × (0.14 / ((2000 / 1000)0.02 – 1))