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Transformer Sizing & Specification

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Overview

Sizes and specifies power transformers per AS/NZS 60076 and IEC 60076. Selects the next standard kVA rating for a given load with future growth, estimates or accepts user-specified impedance, losses, and efficiency, calculates voltage regulation, secondary fault current, tap changer voltages, altitude derating, and temperature margins. Supports oil-filled (core and shell type), dry-type VPI, cast resin, and hermetically sealed constructions with all standard cooling designations.

Input Parameters

Load & Sizing

Parameter Type Default Range Description
load_kva Number 500 1 – 100,000 Maximum demand load to be supplied (kVA)
load_power_factor Number 0.85 0.5 – 1.0 Operating power factor of the load
future_growth_factor Number 1.2 1.0 – 2.0 Multiplier for anticipated future load growth
overload_capability Boolean false — If true, the rating is at least load / 0.8, so the present load is at most 80% of nameplate (125% headroom)

Voltage Levels

Parameter Type Default Range Description
primary_voltage_kv Number 11.0 — Primary winding rated voltage, line-to-line (kV)
secondary_voltage_kv Number 0.415 — Secondary winding rated voltage, line-to-line (kV)
phase_config Enum: Three-phase / Single-phase Three-phase — Number of phases
frequency_hz Enum: 50 / 60 50 — System frequency (Hz)

Transformer Construction

Parameter Type Default Range Description
construction_type Enum oil_filled_core — Oil-filled Core, Oil-filled Shell, Dry-type VPI, Cast Resin, Hermetically Sealed
cooling_type Enum ONAN — Cooling designation per AS/NZS 60076: ONAN, ONAF, OFAF, OFWF, ODAF, AN, AF, AA, AFA
vector_group Enum Dyn11 — Winding connection and phase shift: Yy0, Dyn11, Dyn1, Yd11, Dd0, Yy6, Yzn11

Tap Changer

Parameter Type Default Range Description
tap_changer_type Enum DETC — OLTC (On-Load), DETC (De-Energized), Fixed ±2.5%, Fixed ±5%, No taps
tap_range_percent Enum: 2.5 / 5.0 / 7.5 / 10.0 / 15.0 / 20.0 5.0 — Total tap range (±%)
number_of_taps Enum: 1 / 3 / 5 / 7 / 9 / 17 / 19 / 21 / 33 5 — Total number of tap positions
tap_winding Enum: Primary / Secondary / Tertiary Primary — Winding on which the tap changer is fitted

Impedance & Losses

Parameter Type Default Range Description
impedance_percent Number 0 (auto) 0 – 20 Short circuit impedance (%); 0 = auto-calculate from rating and voltage
no_load_loss_w Number 0 (auto) 0 – 100,000 Core losses at rated voltage (W); 0 = estimate from rating
load_loss_w Number 0 (auto) 0 – 500,000 Winding losses at rated current (W); 0 = estimate from rating

Temperature & Insulation

Parameter Type Default Range Description
insulation_class Enum F (155°C) — Temperature class per IEC 60076-2: A (105°C), E (120°C), B (130°C), F (155°C), H (180°C), C (>180°C)
temperature_rise_k Enum: 55 / 60 / 65 / 80 / 100 / 115 65 — Winding temperature rise above ambient (K)
ambient_temperature_c Number 40 −40 – 60 Maximum ambient temperature (°C)

Protection & Instrumentation

Parameter Type Default Range Description
buchholz_relay Boolean true — Gas/oil flow protection (oil-filled only)
temperature_indicators Boolean true — Winding and oil temperature indicators
pressure_relief_device Boolean true — Overpressure protection (oil-filled only)
oil_level_gauge Boolean true — Visual oil level indicator (oil-filled only)
conservator_tank Boolean true — Oil expansion tank (oil-filled only)

Standards & Application

Parameter Type Default Range Description
application Enum Distribution — Distribution, Industrial, Commercial, Utility, Mining, Renewable Energy
installation_location Enum Outdoor — Indoor, Outdoor, Pad-mounted, Pole-mounted, Substation
altitude_m Number 0 0 – 5,000 Installation altitude above sea level (m); derating applied above 1,000 m

Output Results

Output Unit Description
selected_rating_kva kVA Next standard transformer rating that meets the required capacity
utilization_percent % Load as a percentage of selected rating \(\left(\frac{S_{\text{load}}}{S_{\text{rated}}} \times 100\right)\)
sizing_basis text Description of the sizing criteria applied
voltage_ratio — Primary-to-secondary voltage ratio
impedance_percent % Short circuit impedance (user-specified or estimated)
no_load_loss_w W Core losses at rated voltage
load_loss_w W Winding losses at rated current
total_loss_rated_w W Sum of no-load and load losses at rated conditions
efficiency_50_percent % Efficiency at 50% rated load
efficiency_75_percent % Efficiency at 75% rated load
efficiency_100_percent % Efficiency at 100% rated load
voltage_regulation_percent % Voltage regulation at rated load and specified power factor
fault_current_symmetrical_ka kA Three-phase (or single-phase) symmetrical fault current on the secondary
fault_current_peak_ka kA IEC 60909-0 peak short-circuit current \(i_p = \kappa \sqrt{2}\, I_{sc}\)
peak_factor_kappa — IEC 60909-0 peak factor \(\kappa = 1.02 + 0.98\, e^{-3R/X}\)
resistance_percent / reactance_percent % Resistive and reactive parts of \(Z_{\%}\) used for regulation and \(\kappa\)
x_r_ratio / x_r_source — Transformer X/R and whether it came from the nameplate load loss or the typical value 5
tap_info — Tap positions with corresponding voltages, step size, and min/max range
max_winding_temperature_c °C Ambient temperature plus winding temperature rise
temperature_margin_k K Margin between max winding temperature and insulation class limit
altitude_derating_factor — Derating factor for altitude above 1,000 m (1.0 if at or below 1,000 m)
derated_rating_kva kVA Effective rating after altitude derating

Formulas

Rating Selection

\[ S_{\text{req}} = S_{\text{load}} \times k_{\text{growth}} \]

Where \(S_{\text{load}}\) is the maximum demand (kVA) and \(k_{\text{growth}}\) is the future growth factor (default 1.2). The next standard rating \(S_{\text{rated}} \geq S_{\text{req}}\) is selected from the IEC preferred series.

With overload capability:

\[ S_{\text{req}} = \max\!\left( S_{\text{load}} \times k_{\text{growth}},\; \frac{S_{\text{load}}}{0.8} \right) \]

Keeps the present load at no more than 80% of the nameplate rating, i.e. the transformer can carry 125% of the base load before reaching its rating. The capacity check still compares \(S_{\text{load}} \times k_{\text{growth}}\) with the (derated) rating.

Utilization

\[ \text{Utilization} = \frac{S_{\text{load}}}{S_{\text{rated}}} \times 100 \;\; (\%) \]

Impedance Estimation

\[ Z_{\%} = Z_{\text{base}} \times k_{\text{voltage}} \]

Where \(Z_{\text{base}}\) depends on transformer rating (4.5% for <500 kVA, 6.0% for 500–2,500 kVA, 7.5% for 2,500–10,000 kVA, 8.5% for ≥10,000 kVA) and \(k_{\text{voltage}}\) adjusts for maximum winding voltage (1.0 for <11 kV, 1.1 for 11–33 kV, 1.2 for 33–66 kV, 1.3 for ≥66 kV). The result is rounded to the nearest 0.5%, halves rounding up (e.g. 7.5 × 1.1 = 8.25% → 8.5%).

Loss Estimation

No-load losses (core losses):

\[ P_0 = S_{\text{rated}} \times 1000 \times k_{P_0} \quad (\text{W}) \]

Where \(k_{P_0}\) is a percentage factor dependent on rating (1.8% for <100 kVA, 1.2% for 100–500 kVA, 0.8% for 500–2,500 kVA, 0.5% for 2,500–10,000 kVA, 0.4% for ≥10,000 kVA). Dry-type and cast resin transformers apply a 0.85× multiplier.

Load losses (winding losses):

\[ P_k = S_{\text{rated}} \times 1000 \times k_{P_k} \quad (\text{W}) \]

Where \(k_{P_k}\) is a percentage factor dependent on rating (2.5% for <100 kVA, 1.8% for 100–500 kVA, 1.3% for 500–2,500 kVA, 1.0% for 2,500–10,000 kVA, 0.8% for ≥10,000 kVA). Dry-type and cast resin transformers apply a 1.1× multiplier.

Efficiency

\[ \eta = \frac{S_{\text{rated}} \times n \times \cos\phi \times 1000} {S_{\text{rated}} \times n \times \cos\phi \times 1000 \;+\; P_0 + P_k \times n^2} \times 100 \;\; (\%) \]

Where \(n\) is the per-unit loading (e.g., 0.75 for 75% load), \(\cos\phi\) is the load power factor, \(P_0\) is no-load loss (W), and \(P_k\) is load loss at rated current (W). Core losses are constant; winding losses scale with the square of load.

Voltage Regulation

\[ \text{Reg}_{\%} \approx R_{\%} \cos\phi + X_{\%} \sin\phi \]

Where the impedance is split into resistive and reactive components. When the nameplate load loss \(P_k\) is given, the resistive part follows from it (the load loss is the winding I²R loss at rated current):

\[ R_{\%} = \frac{P_k}{S_{\text{rated}} \times 1000} \times 100, \qquad X_{\%} = \sqrt{Z_{\%}^2 - R_{\%}^2} \]

Otherwise a typical X/R ratio of 5 is used:

\[ R_{\%} = \frac{Z_{\%}}{\sqrt{1 + (X/R)^2}}, \qquad X_{\%} = R_{\%} \times (X/R) \]

A nameplate load loss that gives \(R_{\%} \geq Z_{\%}\) is rejected as inconsistent.

Secondary Fault Current

Three-phase symmetrical fault current:

\[ I_{sc} = \frac{S_{\text{rated}} \times 1000}{\sqrt{3} \times V_2 \times 1000} \times \frac{100}{Z_{\%}} \quad (\text{A}) \]

Where \(V_2\) is the secondary voltage (kV) and \(Z_{\%}\) is the transformer impedance. For single-phase, the \(\sqrt{3}\) factor is omitted.

Peak short-circuit current (IEC 60909-0):

\[ \kappa = 1.02 + 0.98\, e^{-3R/X}, \qquad i_p = \kappa \sqrt{2}\, I_{sc} \]

\(R/X\) is the transformer's own ratio from the impedance split above (nameplate \(P_k\), or the typical X/R of 5, giving \(\kappa = 1.558\) and \(i_p = 2.20 \times I_{sc}\)). With an infinite source the transformer is the whole fault path, so its R/X sets the DC offset. Use \(i_p\) for switchgear making (peak withstand) ratings and \(I_{sc}\) for breaking ratings.

Tap Voltage Calculation

\[ V_{\text{tap}} = V_{\text{nom}} \times \left(1 + \frac{\Delta_{\%}}{100}\right) \]

Where the percentage offset from nominal for each tap position is:

\[ \Delta_{\%} = (t - t_{\text{nom}}) \times \frac{2 \times \text{Range}_{\%}}{N - 1} \]

\(t\) is the tap number (1 to \(N\)), \(t_{\text{nom}}\) is the centre tap, and \(\text{Range}_{\%}\) is the ±% range.

Altitude Derating

\[ k_{\text{alt}} = 1 - \frac{(h - 1000)}{100} \times 0.005 \]

Applied when altitude \(h > 1000\) m. Reduces the effective rating by 0.5% for every 100 m above 1,000 m.

\[ S_{\text{derated}} = S_{\text{rated}} \times k_{\text{alt}} \]

The standard rating is selected before derating. If \(S_{\text{load}} \times k_{\text{growth}} > S_{\text{derated}}\) the capacity check fails and the next larger rating is needed.

Temperature Margin

\[ T_{\text{max}} = T_{\text{ambient}} + \Delta T_{\text{rise}} \] \[ \text{Margin} = T_{\text{insulation\;limit}} - T_{\text{max}} \]

A warning is issued if the margin falls below 10 K.

Worked Examples

The 27 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_transformer; worked examples for every calculation).

#CaseSreq (kVA)Rating (kVA)Z (%)η75 (%)Reg (%)Isc (kA)ip (kA)Result
1500 kVA load, 11 kV / 415 V distribution transformer600.06306.597.654.4413.4829.71PASS
21,600 kVA load at 11 kV / 415 V1,920.02,0006.597.654.4442.8194.31PASS
3Required capacity lands exactly on 1,250 kVA1,250.01,2506.597.654.4425.6456.49PASS
4Required capacity just above 1,250 kVA1,250.41,5006.597.654.4432.1070.73PASS
5Overload capability governs the rating1,625.02,0006.597.654.4441.0390.39PASS
6Growth allowance governs over the overload margin1,500.01,5006.597.654.4430.7767.79PASS
733 kV / 690 V at 2,500 m with overload capability2,250.02,500998.455.4423.2451.21PASS
82,500 kVA at 6.6 kV / 690 V2,400.02,5007.598.365.1227.8961.45PASS
9Typical impedance on a half-percent tie (8.25 % → 8.5 %)2,500.02,5008.598.365.8140.9290.15PASS
1010 MVA, 33 kV / 11 kV zone substation9,600.010,0001098.686.835.2511.56PASS
11Cast-resin 1,000 kVA indoor transformer960.01,0006.597.724.4421.4047.15PASS
12Nameplate impedance and losses900.01,000598.893.0827.8260.38PASS
13Voltage regulation at unity power factor480.05006.598.001.2710.7023.58PASS
14Voltage regulation at pf 0.7 lagging480.05006.597.175.4410.7023.58PASS
15Altitude derating fails the capacity check2,000.02,0006.597.654.4442.8194.31FAIL
16Altitude exactly 1,000 m — no derating1,560.01,6006.597.654.4434.2575.45PASS
17Derated rating exactly equals the requirement1,560.01,6006.597.654.4432.8272.31PASS
18Class A insulation overheats at 45 °C ambient600.06306.597.654.4413.4829.71FAIL
19Dry-type class B at its temperature limit300.0315596.763.428.7619.31PASS
20Single-phase 180 kVA dry-type with nameplate data180.0200498.412.8921.7441.18PASS
2125 kVA single-phase SWER pole transformer24.025595.213.422.004.41PASS
22Low-impedance 2,500 kVA unit — fault level above 50 kA2,500.02,500598.363.4269.56153.25PASS
2360 Hz, 13.8 kV / 480 V1,800.02,0006.597.654.4437.0181.54PASS
245 MVA with ±10 % OLTC, 17 positions4,800.05,000998.366.152.926.42PASS
25315 kVA hermetically sealed, no tap changer300.0315596.653.428.7619.31PASS
26Demand beyond the largest standard rating54,000.050,0001198.687.527.9517.52FAIL
27Nameplate load loss sets a high X/R and peak factor9,600.010,0009.598.964.635.5214.40PASS

Example 1 — 500 kVA load, 11 kV / 415 V distribution transformer

The default 20 % growth takes a 500 kVA demand to 600 kVA and the next standard rating, 630 kVA, with typical %Z and losses.

Given: Maximum demand 500 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 500 × 1.2 = 600.0 kVA
Ratingnext standard rating ≥ 600.0 kVA → 630 kVA; utilisation = 500 / 630 × 100 = 79.4 %
ImpedanceTypical Z = Zbase × kV = 6 % (630 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 630 × 1000 × 0.8 % = 5,040 W; Pk = 630 × 1000 × 1.3 % = 8,190 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 267,750 / (267,750 + 5,040 + 8,190 × 0.52 = 7,088) = 97.42 %; η75 = 401,625 / (401,625 + 5,040 + 8,190 × 0.752 = 9,647) = 97.65 %; η100 = 535,500 / (535,500 + 5,040 + 8,190 × 12 = 13,230) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 630 / (√3 × 0.415) = 876.5 A; Isc = 876.5 × 100 / 6.5 = 13.48 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 13.48 = 29.71 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 630 kVA
Capacity checkSload × growth = 600.0 kVA ≤ Sderated → 600.0 vs limit 630.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 630 kVA, Z 6.5 %, Isc 13.48 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 500 × 1.2 = 600.0 kVA; next standard rating → 630 kVA; utilisation = 500/630 = 79.4 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 5,040 W, Pk = 8,190 W; η75 = 401,625 / (401,625 + 5,040 + 8,190 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 630/(√3 × 0.415) = 876.5 A, Isc = 876.5 × 100/6.5 = 13.48 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 13.48 = 29.71 kA; capacity 600.0 ≤ 630.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-500kva-11kv-415v-distribution)

Example 2 — 1,600 kVA load at 11 kV / 415 V

A larger LV switchboard supply: 1,920 kVA required selects 2,000 kVA, giving a 42.8 kA transformer-limited LV fault level.

Given: Maximum demand 1,600 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,600 × 1.2 = 1,920.0 kVA
Ratingnext standard rating ≥ 1,920.0 kVA → 2,000 kVA; utilisation = 1,600 / 2,000 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 850,000 / (850,000 + 16,000 + 26,000 × 0.52 = 22,500) = 97.42 %; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.752 = 30,625) = 97.65 %; η100 = 1,700,000 / (1,700,000 + 16,000 + 26,000 × 12 = 42,000) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 2,000 / (√3 × 0.415) = 2,782.4 A; Isc = 2,782.4 × 100 / 6.5 = 42.81 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 42.81 = 94.31 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA
Capacity checkSload × growth = 1,920.0 kVA ≤ Sderated → 1,920.0 vs limit 2,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 42.81 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,600 × 1.2 = 1,920.0 kVA; next standard rating → 2000 kVA; utilisation = 1,600/2000 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 16,000 W, Pk = 26,000 W; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 2000/(√3 × 0.415) = 2,782.4 A, Isc = 2,782.4 × 100/6.5 = 42.81 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 42.81 = 94.31 kA; capacity 1,920.0 ≤ 2,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1600kva-11kv-415v)

Example 3 — Required capacity lands exactly on 1,250 kVA

1,000 kVA × 1.25 growth = 1,250 kVA exactly; an equal standard rating is accepted rather than stepping to 1,500 kVA.

Given: Maximum demand 1,000 kVA at pf 0.85 lagging, growth factor 1.25, overload capability not required; 22 kV / 0.433 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,000 × 1.25 = 1,250.0 kVA
RatingSreq equals a standard rating → 1,250 kVA; utilisation = 1,000 / 1,250 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,250 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,250 × 1000 × 0.8 % = 10,000 W; Pk = 1,250 × 1000 × 1.3 % = 16,250 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 531,250 / (531,250 + 10,000 + 16,250 × 0.52 = 14,062) = 97.42 %; η75 = 796,875 / (796,875 + 10,000 + 16,250 × 0.752 = 19,141) = 97.65 %; η100 = 1,062,500 / (1,062,500 + 10,000 + 16,250 × 12 = 26,250) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,250 / (√3 × 0.433) = 1,666.7 A; Isc = 1,666.7 × 100 / 6.5 = 25.64 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 25.64 = 56.49 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 1,250 kVA
Capacity checkSload × growth = 1,250.0 kVA ≤ Sderated → 1,250.0 vs limit 1,250.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,250 kVA, Z 6.5 %, Isc 25.64 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,000 × 1.25 = 1,250.0 kVA; next standard rating → 1250 kVA; utilisation = 1,000/1250 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 10,000 W, Pk = 16,250 W; η75 = 796,875 / (796,875 + 10,000 + 16,250 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1250/(√3 × 0.433) = 1,666.7 A, Isc = 1,666.7 × 100/6.5 = 25.64 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 25.64 = 56.49 kA; capacity 1,250.0 ≤ 1,250.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1000kva-22kv-433v-exact-1250)

Example 4 — Required capacity just above 1,250 kVA

1,042 kVA × 1.2 = 1,250.4 kVA, 0.4 kVA over a standard rating, so the next size up (1,500 kVA) is selected.

Given: Maximum demand 1,042 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,042 × 1.2 = 1,250.4 kVA
Ratingnext standard rating ≥ 1,250.4 kVA → 1,500 kVA; utilisation = 1,042 / 1,500 × 100 = 69.5 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,500 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,500 × 1000 × 0.8 % = 12,000 W; Pk = 1,500 × 1000 × 1.3 % = 19,500 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 637,500 / (637,500 + 12,000 + 19,500 × 0.52 = 16,875) = 97.42 %; η75 = 956,250 / (956,250 + 12,000 + 19,500 × 0.752 = 22,969) = 97.65 %; η100 = 1,275,000 / (1,275,000 + 12,000 + 19,500 × 12 = 31,500) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,500 / (√3 × 0.415) = 2,086.8 A; Isc = 2,086.8 × 100 / 6.5 = 32.10 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 32.10 = 70.73 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 1,500 kVA
Capacity checkSload × growth = 1,250.4 kVA ≤ Sderated → 1,250.4 vs limit 1,500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,500 kVA, Z 6.5 %, Isc 32.10 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,042 × 1.2 = 1,250.4 kVA; next standard rating → 1500 kVA; utilisation = 1,042/1500 = 69.5 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 12,000 W, Pk = 19,500 W; η75 = 956,250 / (956,250 + 12,000 + 19,500 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1500/(√3 × 0.415) = 2,086.8 A, Isc = 2,086.8 × 100/6.5 = 32.10 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 32.10 = 70.73 kA; capacity 1,250.4 ≤ 1,500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1042kva-11kv-415v-just-above-1250)

Example 5 — Overload capability governs the rating

With no growth allowance, the 80 % loading limit (1,300 / 0.8 = 1,625 kVA) governs and selects 2,000 kVA instead of 1,500 kVA.

Given: Maximum demand 1,300 kVA at pf 0.85 lagging, growth factor 1, overload capability required; 11 kV / 0.433 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,300 × 1 = 1,300.0 kVA; overload (≤ 80 % loading): 1,300 / 0.8 = 1,625.0 kVA → Sreq = max(1,300.0, 1,625.0) = 1,625.0 kVA
Ratingnext standard rating ≥ 1,625.0 kVA → 2,000 kVA; utilisation = 1,300 / 2,000 × 100 = 65.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 850,000 / (850,000 + 16,000 + 26,000 × 0.52 = 22,500) = 97.42 %; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.752 = 30,625) = 97.65 %; η100 = 1,700,000 / (1,700,000 + 16,000 + 26,000 × 12 = 42,000) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 2,000 / (√3 × 0.433) = 2,666.7 A; Isc = 2,666.7 × 100 / 6.5 = 41.03 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 41.03 = 90.39 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA
Capacity checkSload × growth = 1,300.0 kVA ≤ Sderated → 1,300.0 vs limit 2,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 41.03 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,300 × 1 = 1,300.0 kVA, overload 1,300/0.8 = 1,625.0 kVA → max = 1,625.0 kVA; next standard rating → 2000 kVA; utilisation = 1,300/2000 = 65.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 16,000 W, Pk = 26,000 W; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 2000/(√3 × 0.433) = 2,666.7 A, Isc = 2,666.7 × 100/6.5 = 41.03 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 41.03 = 90.39 kA; capacity 1,300.0 ≤ 2,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1300kva-overload-governs)

Example 6 — Growth allowance governs over the overload margin

With 50 % growth (1,500 kVA) the overload term (1,250 kVA) is smaller, so growth sets the rating at exactly 1,500 kVA.

Given: Maximum demand 1,000 kVA at pf 0.85 lagging, growth factor 1.5, overload capability required; 22 kV / 0.433 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,000 × 1.5 = 1,500.0 kVA; overload (≤ 80 % loading): 1,000 / 0.8 = 1,250.0 kVA → Sreq = max(1,500.0, 1,250.0) = 1,500.0 kVA
RatingSreq equals a standard rating → 1,500 kVA; utilisation = 1,000 / 1,500 × 100 = 66.7 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,500 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,500 × 1000 × 0.8 % = 12,000 W; Pk = 1,500 × 1000 × 1.3 % = 19,500 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 637,500 / (637,500 + 12,000 + 19,500 × 0.52 = 16,875) = 97.42 %; η75 = 956,250 / (956,250 + 12,000 + 19,500 × 0.752 = 22,969) = 97.65 %; η100 = 1,275,000 / (1,275,000 + 12,000 + 19,500 × 12 = 31,500) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,500 / (√3 × 0.433) = 2,000.1 A; Isc = 2,000.1 × 100 / 6.5 = 30.77 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 30.77 = 67.79 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 1,500 kVA
Capacity checkSload × growth = 1,500.0 kVA ≤ Sderated → 1,500.0 vs limit 1,500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,500 kVA, Z 6.5 %, Isc 30.77 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,000 × 1.5 = 1,500.0 kVA, overload 1,000/0.8 = 1,250.0 kVA → max = 1,500.0 kVA; next standard rating → 1500 kVA; utilisation = 1,000/1500 = 66.7 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 12,000 W, Pk = 19,500 W; η75 = 956,250 / (956,250 + 12,000 + 19,500 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1500/(√3 × 0.433) = 2,000.1 A, Isc = 2,000.1 × 100/6.5 = 30.77 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 30.77 = 67.79 kA; capacity 1,500.0 ≤ 1,500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1000kva-growth-governs-overload)

Example 7 — 33 kV / 690 V at 2,500 m with overload capability

Overload margin sets 2,500 kVA, altitude derates it by 7.5 % to 2,312.5 kVA, and 2,500 kVA at 33 kV takes the 7.5 % × 1.2 typical impedance.

Given: Maximum demand 1,800 kVA at pf 0.9 lagging, growth factor 1.15, overload capability required; 33 kV / 0.69 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class H, 80 K rise, ambient 45 °C, altitude 2,500 m.

Required capacitySreq = 1,800 × 1.15 = 2,070.0 kVA; overload (≤ 80 % loading): 1,800 / 0.8 = 2,250.0 kVA → Sreq = max(2,070.0, 2,250.0) = 2,250.0 kVA
Ratingnext standard rating ≥ 2,250.0 kVA → 2,500 kVA; utilisation = 1,800 / 2,500 × 100 = 72.0 %
ImpedanceTypical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1.2 (Vmax 33 kV) = 9 % → nearest 0.5 % = 9 %
LossesP0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 1,125,000 / (1,125,000 + 12,500 + 25,000 × 0.52 = 18,750) = 98.36 %; η75 = 1,687,500 / (1,687,500 + 12,500 + 25,000 × 0.752 = 26,562) = 98.45 %; η100 = 2,250,000 / (2,250,000 + 12,500 + 25,000 × 12 = 37,500) = 98.36 %
Impedance splittypical X/R = 5: R = 9 / √(1 + 52) = 1.7650 %, X = 5 × 1.7650 = 8.8252 %
Regulationsin φ = √(1 − 0.92) = 0.4359; Reg = R·cos φ + X·sin φ = 1.7650 × 0.9 + 8.8252 × 0.4359 = 5.44 %
LV fault levelIFL = 2,500 / (√3 × 0.69) = 2,091.8 A; Isc = 2,091.8 × 100 / 9 = 23.24 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 23.24 = 51.21 kA peak
Altitudekalt = 1 − (2,500 − 1,000) / 100 × 0.005 = 0.925; Sderated = 2,500 × 0.925 = 2,312.5 kVA
Capacity checkSload × growth = 2,070.0 kVA ≤ Sderated → 2,070.0 vs limit 2,312.5 kVA PASS
Winding temperatureT = 45 + 80 = 125 °C ≤ class H limit (margin 55 K) → 125 vs limit 180 °C PASS
Result2/2 checks pass → PASS; 2,500 kVA, Z 9 %, Isc 23.24 kA; governing: capacity. Altitude 2500.0m requires 7.5% derating
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,800 × 1.15 = 2,070.0 kVA, overload 1,800/0.8 = 2,250.0 kVA → max = 2,250.0 kVA; next standard rating → 2500 kVA; utilisation = 1,800/2500 = 72.0 %; Z = 7.5 × 1.2 = 9 → 9 % (nearest 0.5); P0 = 12,500 W, Pk = 25,000 W; η75 = 1,687,500 / (1,687,500 + 12,500 + 25,000 × 0.5625) = 98.45 %; R = 9/√26 = 1.7650 %, X = 5 × R = 8.8252 % (typical X/R 5); Reg = 1.7650 × 0.9 + 8.8252 × 0.4359 = 5.44 %; I_FL = 2500/(√3 × 0.69) = 2,091.8 A, Isc = 2,091.8 × 100/9 = 23.24 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 23.24 = 51.21 kA; k_alt = 1 − (2500 − 1000)/100 × 0.005 = 0.925, derated = 2500 × 0.925 = 2,312.5 kVA; capacity 2,070.0 ≤ 2,312.5 kVA → PASS; T = 45 + 80 = 125 °C vs class H 180 °C → PASS (validation case c0024-1800kva-33kv-690v-altitude-overload)

Example 8 — 2,500 kVA at 6.6 kV / 690 V

Below 11 kV the voltage factor is 1.0, so the 2,500 kVA unit takes the 7.5 % base impedance unchanged.

Given: Maximum demand 2,000 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 6.6 kV / 0.69 kV, three-phase, 50 Hz, oil-filled core-type ONAF, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 2,000 × 1.2 = 2,400.0 kVA
Ratingnext standard rating ≥ 2,400.0 kVA → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1 (Vmax 6.6 kV) = 7.5 % → nearest 0.5 % = 7.5 %
LossesP0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 1,062,500 / (1,062,500 + 12,500 + 25,000 × 0.52 = 18,750) = 98.27 %; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.752 = 26,562) = 98.36 %; η100 = 2,125,000 / (2,125,000 + 12,500 + 25,000 × 12 = 37,500) = 98.27 %
Impedance splittypical X/R = 5: R = 7.5 / √(1 + 52) = 1.4709 %, X = 5 × 1.4709 = 7.3544 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.4709 × 0.85 + 7.3544 × 0.5268 = 5.12 %
LV fault levelIFL = 2,500 / (√3 × 0.69) = 2,091.8 A; Isc = 2,091.8 × 100 / 7.5 = 27.89 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 27.89 = 61.45 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA
Capacity checkSload × growth = 2,400.0 kVA ≤ Sderated → 2,400.0 vs limit 2,500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,500 kVA, Z 7.5 %, Isc 27.89 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 2,000 × 1.2 = 2,400.0 kVA; next standard rating → 2500 kVA; utilisation = 2,000/2500 = 80.0 %; Z = 7.5 × 1.0 = 7.5 → 7.5 % (nearest 0.5); P0 = 12,500 W, Pk = 25,000 W; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.5625) = 98.36 %; R = 7.5/√26 = 1.4709 %, X = 5 × R = 7.3544 % (typical X/R 5); Reg = 1.4709 × 0.85 + 7.3544 × 0.5268 = 5.12 %; I_FL = 2500/(√3 × 0.69) = 2,091.8 A, Isc = 2,091.8 × 100/7.5 = 27.89 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 27.89 = 61.45 kA; capacity 2,400.0 ≤ 2,500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-2500kva-6.6kv-690v)

Example 9 — Typical impedance on a half-percent tie (8.25 % → 8.5 %)

2,000 kVA × 1.25 = 2,500 kVA at 11 kV: 7.5 % × 1.1 = 8.25 % sits exactly between 8.0 and 8.5 %; rounding half up gives 8.5 % and about 40.9 kA.

Given: Maximum demand 2,000 kVA at pf 0.85 lagging, growth factor 1.25, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 2,000 × 1.25 = 2,500.0 kVA
RatingSreq equals a standard rating → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1.1 (Vmax 11 kV) = 8.25 % → nearest 0.5 % = 8.5 %
LossesP0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 1,062,500 / (1,062,500 + 12,500 + 25,000 × 0.52 = 18,750) = 98.27 %; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.752 = 26,562) = 98.36 %; η100 = 2,125,000 / (2,125,000 + 12,500 + 25,000 × 12 = 37,500) = 98.27 %
Impedance splittypical X/R = 5: R = 8.5 / √(1 + 52) = 1.6670 %, X = 5 × 1.6670 = 8.3349 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.6670 × 0.85 + 8.3349 × 0.5268 = 5.81 %
LV fault levelIFL = 2,500 / (√3 × 0.415) = 3,478.0 A; Isc = 3,478.0 × 100 / 8.5 = 40.92 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 40.92 = 90.15 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA
Capacity checkSload × growth = 2,500.0 kVA ≤ Sderated → 2,500.0 vs limit 2,500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,500 kVA, Z 8.5 %, Isc 40.92 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 2,000 × 1.25 = 2,500.0 kVA; next standard rating → 2500 kVA; utilisation = 2,000/2500 = 80.0 %; Z = 7.5 × 1.1 = 8.25 → 8.5 % (nearest 0.5); P0 = 12,500 W, Pk = 25,000 W; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.5625) = 98.36 %; R = 8.5/√26 = 1.6670 %, X = 5 × R = 8.3349 % (typical X/R 5); Reg = 1.6670 × 0.85 + 8.3349 × 0.5268 = 5.81 %; I_FL = 2500/(√3 × 0.415) = 3,478.0 A, Isc = 3,478.0 × 100/8.5 = 40.92 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 40.92 = 90.15 kA; capacity 2,500.0 ≤ 2,500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-2500kva-11kv-415v-impedance-tie)

Example 10 — 10 MVA, 33 kV / 11 kV zone substation

8 MVA × 1.2 = 9.6 MVA selects 10 MVA, entering the ≥10 MVA band (8.5 % base impedance, 0.4 % / 0.8 % losses).

Given: Maximum demand 8,000 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 33 kV / 11 kV, three-phase, 50 Hz, oil-filled shell-type ONAF, Dyn1, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 8,000 × 1.2 = 9,600.0 kVA
Ratingnext standard rating ≥ 9,600.0 kVA → 10,000 kVA; utilisation = 8,000 / 10,000 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 8.5 % (10,000 kVA band) × 1.2 (Vmax 33 kV) = 10.2 % → nearest 0.5 % = 10 %
LossesP0 = 10,000 × 1000 × 0.4 % = 40,000 W; Pk = 10,000 × 1000 × 0.8 % = 80,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 4,250,000 / (4,250,000 + 40,000 + 80,000 × 0.52 = 60,000) = 98.61 %; η75 = 6,375,000 / (6,375,000 + 40,000 + 80,000 × 0.752 = 85,000) = 98.68 %; η100 = 8,500,000 / (8,500,000 + 40,000 + 80,000 × 12 = 120,000) = 98.61 %
Impedance splittypical X/R = 5: R = 10 / √(1 + 52) = 1.9612 %, X = 5 × 1.9612 = 9.8058 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.9612 × 0.85 + 9.8058 × 0.5268 = 6.83 %
LV fault levelIFL = 10,000 / (√3 × 11) = 524.9 A; Isc = 524.9 × 100 / 10 = 5.25 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 5.25 = 11.56 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 10,000 kVA
Capacity checkSload × growth = 9,600.0 kVA ≤ Sderated → 9,600.0 vs limit 10,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 10,000 kVA, Z 10 %, Isc 5.25 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 8,000 × 1.2 = 9,600.0 kVA; next standard rating → 10000 kVA; utilisation = 8,000/10000 = 80.0 %; Z = 8.5 × 1.2 = 10.2 → 10 % (nearest 0.5); P0 = 40,000 W, Pk = 80,000 W; η75 = 6,375,000 / (6,375,000 + 40,000 + 80,000 × 0.5625) = 98.68 %; R = 10/√26 = 1.9612 %, X = 5 × R = 9.8058 % (typical X/R 5); Reg = 1.9612 × 0.85 + 9.8058 × 0.5268 = 6.83 %; I_FL = 10000/(√3 × 11) = 524.9 A, Isc = 524.9 × 100/10 = 5.25 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 5.25 = 11.56 kA; capacity 9,600.0 ≤ 10,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-10mva-33kv-11kv-zone-substation)

Example 11 — Cast-resin 1,000 kVA indoor transformer

Cast-resin construction applies 0.85 × to the typical no-load loss and 1.1 × to the typical load loss.

Given: Maximum demand 800 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, cast-resin AA, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 800 × 1.2 = 960.0 kVA
Ratingnext standard rating ≥ 960.0 kVA → 1,000 kVA; utilisation = 800 / 1,000 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,000 × 1000 × 0.8 % × 0.85 (dry/cast) = 6,800 W; Pk = 1,000 × 1000 × 1.3 % × 1.1 (dry/cast) = 14,300 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 425,000 / (425,000 + 6,800 + 14,300 × 0.52 = 10,375) = 97.62 %; η75 = 637,500 / (637,500 + 6,800 + 14,300 × 0.752 = 14,844) = 97.72 %; η100 = 850,000 / (850,000 + 6,800 + 14,300 × 12 = 21,100) = 97.58 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,000 / (√3 × 0.415) = 1,391.2 A; Isc = 1,391.2 × 100 / 6.5 = 21.40 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 21.40 = 47.15 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 1,000 kVA
Capacity checkSload × growth = 960.0 kVA ≤ Sderated → 960.0 vs limit 1,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,000 kVA, Z 6.5 %, Isc 21.40 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 800 × 1.2 = 960.0 kVA; next standard rating → 1000 kVA; utilisation = 800/1000 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 6,800 W, Pk = 14,300 W; η75 = 637,500 / (637,500 + 6,800 + 14,300 × 0.5625) = 97.72 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1000/(√3 × 0.415) = 1,391.2 A, Isc = 1,391.2 × 100/6.5 = 21.40 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 21.40 = 47.15 kA; capacity 960.0 ≤ 1,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1000kva-cast-resin)

Example 12 — Nameplate impedance and losses

Manufacturer test data (Z 5 %, P0 1.7 kW, Pk 10.5 kW) replaces the typical values in efficiency, regulation and fault level.

Given: Maximum demand 750 kVA at pf 0.9 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; nameplate Z 5 %, P0 1,700 W, Pk 10,500 W; insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 750 × 1.2 = 900.0 kVA
Ratingnext standard rating ≥ 900.0 kVA → 1,000 kVA; utilisation = 750 / 1,000 × 100 = 75.0 %
ImpedanceZ = 5 % (nameplate)
LossesP0 = 1,700 W (nameplate); Pk = 10,500 W (nameplate)
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 450,000 / (450,000 + 1,700 + 10,500 × 0.52 = 4,325) = 99.05 %; η75 = 675,000 / (675,000 + 1,700 + 10,500 × 0.752 = 7,606) = 98.89 %; η100 = 900,000 / (900,000 + 1,700 + 10,500 × 12 = 12,200) = 98.66 %
Impedance splitR = Pk / S = 10,500 / (1,000 × 1000) × 100 = 1.0500 %, X = √(52 − 1.05002) = 4.8885 % (X/R = 4.656)
Regulationsin φ = √(1 − 0.92) = 0.4359; Reg = R·cos φ + X·sin φ = 1.0500 × 0.9 + 4.8885 × 0.4359 = 3.08 %
LV fault levelIFL = 1,000 / (√3 × 0.415) = 1,391.2 A; Isc = 1,391.2 × 100 / 5 = 27.82 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2148 = 1.5345; ip = κ·√2·Isc = 1.5345 × √2 × 27.82 = 60.38 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 1,000 kVA
Capacity checkSload × growth = 900.0 kVA ≤ Sderated → 900.0 vs limit 1,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,000 kVA, Z 5 %, Isc 27.82 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 750 × 1.2 = 900.0 kVA; next standard rating → 1000 kVA; utilisation = 750/1000 = 75.0 %; Z = 5 % (nameplate); P0 = 1,700 W, Pk = 10,500 W; η75 = 675,000 / (675,000 + 1,700 + 10,500 × 0.5625) = 98.89 %; R = 10,500/(1000 × 1000) × 100 = 1.0500 %, X = √(5² − 1.0500²) = 4.8885 % (X/R 4.656); Reg = 1.0500 × 0.9 + 4.8885 × 0.4359 = 3.08 %; I_FL = 1000/(√3 × 0.415) = 1,391.2 A, Isc = 1,391.2 × 100/5 = 27.82 kA; κ = 1.02 + 0.98·e^(−3 × 0.2148) = 1.5345 (IEC 60909-0), i_p = 1.5345 × √2 × 27.82 = 60.38 kA; capacity 900.0 ≤ 1,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1000kva-nameplate-losses)

Example 13 — Voltage regulation at unity power factor

At pf 1.0 the reactive term vanishes and the regulation equals the resistive component R % only.

Given: Maximum demand 400 kVA at pf 1 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 400 × 1.2 = 480.0 kVA
Ratingnext standard rating ≥ 480.0 kVA → 500 kVA; utilisation = 400 / 500 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (500 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 500 × 1000 × 0.8 % = 4,000 W; Pk = 500 × 1000 × 1.3 % = 6,500 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 250,000 / (250,000 + 4,000 + 6,500 × 0.52 = 5,625) = 97.80 %; η75 = 375,000 / (375,000 + 4,000 + 6,500 × 0.752 = 7,656) = 98.00 %; η100 = 500,000 / (500,000 + 4,000 + 6,500 × 12 = 10,500) = 97.94 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 12) = 0.0000; Reg = R·cos φ + X·sin φ = 1.2748 × 1 + 6.3738 × 0.0000 = 1.27 %
LV fault levelIFL = 500 / (√3 × 0.415) = 695.6 A; Isc = 695.6 × 100 / 6.5 = 10.70 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 10.70 = 23.58 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 500 kVA
Capacity checkSload × growth = 480.0 kVA ≤ Sderated → 480.0 vs limit 500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 500 kVA, Z 6.5 %, Isc 10.70 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 400 × 1.2 = 480.0 kVA; next standard rating → 500 kVA; utilisation = 400/500 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 4,000 W, Pk = 6,500 W; η75 = 375,000 / (375,000 + 4,000 + 6,500 × 0.5625) = 98.00 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 1 + 6.3738 × 0.0000 = 1.27 %; I_FL = 500/(√3 × 0.415) = 695.6 A, Isc = 695.6 × 100/6.5 = 10.70 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 10.70 = 23.58 kA; capacity 480.0 ≤ 500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-500kva-unity-pf)

Example 14 — Voltage regulation at pf 0.7 lagging

A poor 0.7 pf puts most of the impedance drop in phase with the load current, roughly 5.5 % regulation for a 6.5 % unit.

Given: Maximum demand 400 kVA at pf 0.7 lagging, growth factor 1.2, overload capability not required; 22 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 400 × 1.2 = 480.0 kVA
Ratingnext standard rating ≥ 480.0 kVA → 500 kVA; utilisation = 400 / 500 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (500 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 500 × 1000 × 0.8 % = 4,000 W; Pk = 500 × 1000 × 1.3 % = 6,500 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 175,000 / (175,000 + 4,000 + 6,500 × 0.52 = 5,625) = 96.89 %; η75 = 262,500 / (262,500 + 4,000 + 6,500 × 0.752 = 7,656) = 97.17 %; η100 = 350,000 / (350,000 + 4,000 + 6,500 × 12 = 10,500) = 97.09 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.72) = 0.7141; Reg = R·cos φ + X·sin φ = 1.2748 × 0.7 + 6.3738 × 0.7141 = 5.44 %
LV fault levelIFL = 500 / (√3 × 0.415) = 695.6 A; Isc = 695.6 × 100 / 6.5 = 10.70 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 10.70 = 23.58 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 500 kVA
Capacity checkSload × growth = 480.0 kVA ≤ Sderated → 480.0 vs limit 500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 500 kVA, Z 6.5 %, Isc 10.70 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 400 × 1.2 = 480.0 kVA; next standard rating → 500 kVA; utilisation = 400/500 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 4,000 W, Pk = 6,500 W; η75 = 262,500 / (262,500 + 4,000 + 6,500 × 0.5625) = 97.17 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.7 + 6.3738 × 0.7141 = 5.44 %; I_FL = 500/(√3 × 0.415) = 695.6 A, Isc = 695.6 × 100/6.5 = 10.70 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 10.70 = 23.58 kA; capacity 480.0 ≤ 500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-500kva-pf-0.7)

Example 15 — Altitude derating fails the capacity check

2,000 kVA selected at sea-level rating derates 10 % at 3,000 m to 1,800 kVA, below the 2,000 kVA required — capacity FAIL.

Given: Maximum demand 1,600 kVA at pf 0.85 lagging, growth factor 1.25, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 3,000 m.

Required capacitySreq = 1,600 × 1.25 = 2,000.0 kVA
RatingSreq equals a standard rating → 2,000 kVA; utilisation = 1,600 / 2,000 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 850,000 / (850,000 + 16,000 + 26,000 × 0.52 = 22,500) = 97.42 %; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.752 = 30,625) = 97.65 %; η100 = 1,700,000 / (1,700,000 + 16,000 + 26,000 × 12 = 42,000) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 2,000 / (√3 × 0.415) = 2,782.4 A; Isc = 2,782.4 × 100 / 6.5 = 42.81 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 42.81 = 94.31 kA peak
Altitudekalt = 1 − (3,000 − 1,000) / 100 × 0.005 = 0.9; Sderated = 2,000 × 0.9 = 1,800.0 kVA
Capacity checkSload × growth = 2,000.0 kVA ≤ Sderated → 2,000.0 vs limit 1,800.0 kVA FAIL
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result1/2 checks pass → FAIL; 2,000 kVA, Z 6.5 %, Isc 42.81 kA; governing: capacity. Altitude 3000.0m requires 10.0% derating
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,600 × 1.25 = 2,000.0 kVA; next standard rating → 2000 kVA; utilisation = 1,600/2000 = 80.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 16,000 W, Pk = 26,000 W; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 2000/(√3 × 0.415) = 2,782.4 A, Isc = 2,782.4 × 100/6.5 = 42.81 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 42.81 = 94.31 kA; k_alt = 1 − (3000 − 1000)/100 × 0.005 = 0.9, derated = 2000 × 0.9 = 1,800.0 kVA; capacity 2,000.0 ≤ 1,800.0 kVA → FAIL; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-2000kva-altitude-3000m-fail)

Example 16 — Altitude exactly 1,000 m — no derating

Derating starts only above 1,000 m, so the factor is exactly 1.0 at the boundary.

Given: Maximum demand 1,300 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 1,000 m.

Required capacitySreq = 1,300 × 1.2 = 1,560.0 kVA
Ratingnext standard rating ≥ 1,560.0 kVA → 1,600 kVA; utilisation = 1,300 / 1,600 × 100 = 81.2 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,600 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,600 × 1000 × 0.8 % = 12,800 W; Pk = 1,600 × 1000 × 1.3 % = 20,800 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 680,000 / (680,000 + 12,800 + 20,800 × 0.52 = 18,000) = 97.42 %; η75 = 1,020,000 / (1,020,000 + 12,800 + 20,800 × 0.752 = 24,500) = 97.65 %; η100 = 1,360,000 / (1,360,000 + 12,800 + 20,800 × 12 = 33,600) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,600 / (√3 × 0.415) = 2,225.9 A; Isc = 2,225.9 × 100 / 6.5 = 34.25 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 34.25 = 75.45 kA peak
Altitude1,000 m ≤ 1,000 m → kalt = 1, Sderated = 1,600 kVA
Capacity checkSload × growth = 1,560.0 kVA ≤ Sderated → 1,560.0 vs limit 1,600.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,600 kVA, Z 6.5 %, Isc 34.25 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,300 × 1.2 = 1,560.0 kVA; next standard rating → 1600 kVA; utilisation = 1,300/1600 = 81.2 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 12,800 W, Pk = 20,800 W; η75 = 1,020,000 / (1,020,000 + 12,800 + 20,800 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1600/(√3 × 0.415) = 2,225.9 A, Isc = 2,225.9 × 100/6.5 = 34.25 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 34.25 = 75.45 kA; capacity 1,560.0 ≤ 1,600.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1600kva-altitude-1000m-no-derating)

Example 17 — Derated rating exactly equals the requirement

At 1,500 m the 1,600 kVA rating derates 2.5 % to 1,560 kVA, exactly the 1,248 × 1.25 = 1,560 kVA required — PASS at the limit.

Given: Maximum demand 1,248 kVA at pf 0.85 lagging, growth factor 1.25, overload capability not required; 11 kV / 0.433 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 1,500 m.

Required capacitySreq = 1,248 × 1.25 = 1,560.0 kVA
Ratingnext standard rating ≥ 1,560.0 kVA → 1,600 kVA; utilisation = 1,248 / 1,600 × 100 = 78.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (1,600 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 1,600 × 1000 × 0.8 % = 12,800 W; Pk = 1,600 × 1000 × 1.3 % = 20,800 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 680,000 / (680,000 + 12,800 + 20,800 × 0.52 = 18,000) = 97.42 %; η75 = 1,020,000 / (1,020,000 + 12,800 + 20,800 × 0.752 = 24,500) = 97.65 %; η100 = 1,360,000 / (1,360,000 + 12,800 + 20,800 × 12 = 33,600) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 1,600 / (√3 × 0.433) = 2,133.4 A; Isc = 2,133.4 × 100 / 6.5 = 32.82 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 32.82 = 72.31 kA peak
Altitudekalt = 1 − (1,500 − 1,000) / 100 × 0.005 = 0.975; Sderated = 1,600 × 0.975 = 1,560.0 kVA
Capacity checkSload × growth = 1,560.0 kVA ≤ Sderated → 1,560.0 vs limit 1,560.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 1,600 kVA, Z 6.5 %, Isc 32.82 kA; governing: capacity. Altitude 1500.0m requires 2.5% derating
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,248 × 1.25 = 1,560.0 kVA; next standard rating → 1600 kVA; utilisation = 1,248/1600 = 78.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 12,800 W, Pk = 20,800 W; η75 = 1,020,000 / (1,020,000 + 12,800 + 20,800 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 1600/(√3 × 0.433) = 2,133.4 A, Isc = 2,133.4 × 100/6.5 = 32.82 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 32.82 = 72.31 kA; k_alt = 1 − (1500 − 1000)/100 × 0.005 = 0.975, derated = 1600 × 0.975 = 1,560.0 kVA; capacity 1,560.0 ≤ 1,560.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-1600kva-altitude-1500m-boundary)

Example 18 — Class A insulation overheats at 45 °C ambient

45 °C ambient + 65 K rise = 110 °C exceeds the 105 °C class A limit — winding temperature FAIL.

Given: Maximum demand 500 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class A, 65 K rise, ambient 45 °C, altitude 0 m.

Required capacitySreq = 500 × 1.2 = 600.0 kVA
Ratingnext standard rating ≥ 600.0 kVA → 630 kVA; utilisation = 500 / 630 × 100 = 79.4 %
ImpedanceTypical Z = Zbase × kV = 6 % (630 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 630 × 1000 × 0.8 % = 5,040 W; Pk = 630 × 1000 × 1.3 % = 8,190 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 267,750 / (267,750 + 5,040 + 8,190 × 0.52 = 7,088) = 97.42 %; η75 = 401,625 / (401,625 + 5,040 + 8,190 × 0.752 = 9,647) = 97.65 %; η100 = 535,500 / (535,500 + 5,040 + 8,190 × 12 = 13,230) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 630 / (√3 × 0.415) = 876.5 A; Isc = 876.5 × 100 / 6.5 = 13.48 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 13.48 = 29.71 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 630 kVA
Capacity checkSload × growth = 600.0 kVA ≤ Sderated → 600.0 vs limit 630.0 kVA PASS
Winding temperatureT = 45 + 65 = 110 °C ≤ class A limit (margin -5 K) → 110 vs limit 105 °C FAIL
Result1/2 checks pass → FAIL; 630 kVA, Z 6.5 %, Isc 13.48 kA; governing: capacity. Low temperature margin (-5.0K). Consider higher insulation class.
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 500 × 1.2 = 600.0 kVA; next standard rating → 630 kVA; utilisation = 500/630 = 79.4 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 5,040 W, Pk = 8,190 W; η75 = 401,625 / (401,625 + 5,040 + 8,190 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 630/(√3 × 0.415) = 876.5 A, Isc = 876.5 × 100/6.5 = 13.48 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 13.48 = 29.71 kA; capacity 600.0 ≤ 630.0 kVA → PASS; T = 45 + 65 = 110 °C vs class A 105 °C → FAIL (validation case c0024-class-a-winding-temperature-fail)

Example 19 — Dry-type class B at its temperature limit

50 °C ambient + 80 K rise = 130 °C equals the class B limit: PASS with zero margin (the calc warns below 10 K).

Given: Maximum demand 250 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, dry-type VPI AN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class B, 80 K rise, ambient 50 °C, altitude 0 m.

Required capacitySreq = 250 × 1.2 = 300.0 kVA
Ratingnext standard rating ≥ 300.0 kVA → 315 kVA; utilisation = 250 / 315 × 100 = 79.4 %
ImpedanceTypical Z = Zbase × kV = 4.5 % (315 kVA band) × 1.1 (Vmax 11 kV) = 4.95 % → nearest 0.5 % = 5 %
LossesP0 = 315 × 1000 × 1.2 % × 0.85 (dry/cast) = 3,213 W; Pk = 315 × 1000 × 1.8 % × 1.1 (dry/cast) = 6,237 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 133,875 / (133,875 + 3,213 + 6,237 × 0.52 = 4,772) = 96.56 %; η75 = 200,812 / (200,812 + 3,213 + 6,237 × 0.752 = 6,721) = 96.76 %; η100 = 267,750 / (267,750 + 3,213 + 6,237 × 12 = 9,450) = 96.59 %
Impedance splittypical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %
LV fault levelIFL = 315 / (√3 × 0.415) = 438.2 A; Isc = 438.2 × 100 / 5 = 8.76 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 8.76 = 19.31 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 315 kVA
Capacity checkSload × growth = 300.0 kVA ≤ Sderated → 300.0 vs limit 315.0 kVA PASS
Winding temperatureT = 50 + 80 = 130 °C ≤ class B limit (margin 0 K) → 130 vs limit 130 °C PASS
Result2/2 checks pass → PASS; 315 kVA, Z 5 %, Isc 8.76 kA; governing: capacity. Low temperature margin (0.0K). Consider higher insulation class.
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 250 × 1.2 = 300.0 kVA; next standard rating → 315 kVA; utilisation = 250/315 = 79.4 %; Z = 4.5 × 1.1 = 4.95 → 5 % (nearest 0.5); P0 = 3,213 W, Pk = 6,237 W; η75 = 200,812 / (200,812 + 3,213 + 6,237 × 0.5625) = 96.76 %; R = 5/√26 = 0.9806 %, X = 5 × R = 4.9029 % (typical X/R 5); Reg = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %; I_FL = 315/(√3 × 0.415) = 438.2 A, Isc = 438.2 × 100/5 = 8.76 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 8.76 = 19.31 kA; capacity 300.0 ≤ 315.0 kVA → PASS; T = 50 + 80 = 130 °C vs class B 130 °C → PASS (validation case c0024-dry-class-b-temperature-boundary)

Example 20 — Single-phase 180 kVA dry-type with nameplate data

Single-phase full-load current is S / V (no √3); user losses and impedance replace the typical values.

Given: Maximum demand 180 kVA at pf 0.9 lagging, growth factor 1, overload capability not required; 0.415 kV / 0.23 kV, single-phase, 50 Hz, dry-type VPI AN, Dyn11, off-circuit tap changer ±5 % in 5 positions; nameplate Z 4 %, P0 600 W, Pk 2,800 W; insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 180 × 1 = 180.0 kVA
Ratingnext standard rating ≥ 180.0 kVA → 200 kVA; utilisation = 180 / 200 × 100 = 90.0 %
ImpedanceZ = 4 % (nameplate)
LossesP0 = 600 W (nameplate); Pk = 2,800 W (nameplate)
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 90,000 / (90,000 + 600 + 2,800 × 0.52 = 1,300) = 98.58 %; η75 = 135,000 / (135,000 + 600 + 2,800 × 0.752 = 2,175) = 98.41 %; η100 = 180,000 / (180,000 + 600 + 2,800 × 12 = 3,400) = 98.15 %
Impedance splitR = Pk / S = 2,800 / (200 × 1000) × 100 = 1.4000 %, X = √(42 − 1.40002) = 3.7470 % (X/R = 2.676)
Regulationsin φ = √(1 − 0.92) = 0.4359; Reg = R·cos φ + X·sin φ = 1.4000 × 0.9 + 3.7470 × 0.4359 = 2.89 %
LV fault levelIFL = 200 / 0.23 = 869.6 A (single-phase, no √3); Isc = 869.6 × 100 / 4 = 21.74 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.3736 = 1.3395; ip = κ·√2·Isc = 1.3395 × √2 × 21.74 = 41.18 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 200 kVA
Capacity checkSload × growth = 180.0 kVA ≤ Sderated → 180.0 vs limit 200.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 200 kVA, Z 4 %, Isc 21.74 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 180 × 1 = 180.0 kVA; next standard rating → 200 kVA; utilisation = 180/200 = 90.0 %; Z = 4 % (nameplate); P0 = 600 W, Pk = 2,800 W; η75 = 135,000 / (135,000 + 600 + 2,800 × 0.5625) = 98.41 %; R = 2,800/(200 × 1000) × 100 = 1.4000 %, X = √(4² − 1.4000²) = 3.7470 % (X/R 2.676); Reg = 1.4000 × 0.9 + 3.7470 × 0.4359 = 2.89 %; I_FL = 200/0.23 = 869.6 A (single-phase), Isc = 869.6 × 100/4 = 21.74 kA; κ = 1.02 + 0.98·e^(−3 × 0.3736) = 1.3395 (IEC 60909-0), i_p = 1.3395 × √2 × 21.74 = 41.18 kA; capacity 180.0 ≤ 200.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-180kva-1ph-415v-230v-dry-type)

Example 21 — 25 kVA single-phase SWER pole transformer

A 12.7 kV / 250 V single-phase unit: below 100 kVA the 1.8 % / 2.5 % loss band and 4.5 % × 1.1 impedance apply.

Given: Maximum demand 20 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 12.7 kV / 0.25 kV, single-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 20 × 1.2 = 24.0 kVA
Ratingnext standard rating ≥ 24.0 kVA → 25 kVA; utilisation = 20 / 25 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 4.5 % (25 kVA band) × 1.1 (Vmax 12.7 kV) = 4.95 % → nearest 0.5 % = 5 %
LossesP0 = 25 × 1000 × 1.8 % = 450 W; Pk = 25 × 1000 × 2.5 % = 625 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 10,625 / (10,625 + 450 + 625 × 0.52 = 606) = 94.60 %; η75 = 15,938 / (15,938 + 450 + 625 × 0.752 = 802) = 95.21 %; η100 = 21,250 / (21,250 + 450 + 625 × 12 = 1,075) = 95.18 %
Impedance splittypical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %
LV fault levelIFL = 25 / 0.25 = 100.0 A (single-phase, no √3); Isc = 100.0 × 100 / 5 = 2.00 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 2.00 = 4.41 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 25 kVA
Capacity checkSload × growth = 24.0 kVA ≤ Sderated → 24.0 vs limit 25.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 25 kVA, Z 5 %, Isc 2.00 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 20 × 1.2 = 24.0 kVA; next standard rating → 25 kVA; utilisation = 20/25 = 80.0 %; Z = 4.5 × 1.1 = 4.95 → 5 % (nearest 0.5); P0 = 450 W, Pk = 625 W; η75 = 15,938 / (15,938 + 450 + 625 × 0.5625) = 95.21 %; R = 5/√26 = 0.9806 %, X = 5 × R = 4.9029 % (typical X/R 5); Reg = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %; I_FL = 25/0.25 = 100.0 A (single-phase), Isc = 100.0 × 100/5 = 2.00 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 2.00 = 4.41 kA; capacity 24.0 ≤ 25.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-25kva-1ph-swer-pole)

Example 22 — Low-impedance 2,500 kVA unit — fault level above 50 kA

A 5 % impedance 2,500 kVA transformer gives about 70 kA at 415 V, which the calc flags for switchgear rating.

Given: Maximum demand 2,000 kVA at pf 0.85 lagging, growth factor 1.25, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; nameplate Z 5 %; insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 2,000 × 1.25 = 2,500.0 kVA
RatingSreq equals a standard rating → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 %
ImpedanceZ = 5 % (nameplate)
LossesP0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 1,062,500 / (1,062,500 + 12,500 + 25,000 × 0.52 = 18,750) = 98.27 %; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.752 = 26,562) = 98.36 %; η100 = 2,125,000 / (2,125,000 + 12,500 + 25,000 × 12 = 37,500) = 98.27 %
Impedance splittypical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %
LV fault levelIFL = 2,500 / (√3 × 0.415) = 3,478.0 A; Isc = 3,478.0 × 100 / 5 = 69.56 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 69.56 = 153.25 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA
Capacity checkSload × growth = 2,500.0 kVA ≤ Sderated → 2,500.0 vs limit 2,500.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,500 kVA, Z 5 %, Isc 69.56 kA; governing: capacity. High fault current (69.6 kA). Ensure switchgear rated adequately.
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 2,000 × 1.25 = 2,500.0 kVA; next standard rating → 2500 kVA; utilisation = 2,000/2500 = 80.0 %; Z = 5 % (nameplate); P0 = 12,500 W, Pk = 25,000 W; η75 = 1,593,750 / (1,593,750 + 12,500 + 25,000 × 0.5625) = 98.36 %; R = 5/√26 = 0.9806 %, X = 5 × R = 4.9029 % (typical X/R 5); Reg = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %; I_FL = 2500/(√3 × 0.415) = 3,478.0 A, Isc = 3,478.0 × 100/5 = 69.56 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 69.56 = 153.25 kA; capacity 2,500.0 ≤ 2,500.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-2500kva-415v-high-fault-warning)

Example 23 — 60 Hz, 13.8 kV / 480 V

A North American voltage pair: frequency does not enter the rating, impedance or fault arithmetic.

Given: Maximum demand 1,500 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 13.8 kV / 0.48 kV, three-phase, 60 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 1,500 × 1.2 = 1,800.0 kVA
Ratingnext standard rating ≥ 1,800.0 kVA → 2,000 kVA; utilisation = 1,500 / 2,000 × 100 = 75.0 %
ImpedanceTypical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 13.8 kV) = 6.6 % → nearest 0.5 % = 6.5 %
LossesP0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 850,000 / (850,000 + 16,000 + 26,000 × 0.52 = 22,500) = 97.42 %; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.752 = 30,625) = 97.65 %; η100 = 1,700,000 / (1,700,000 + 16,000 + 26,000 × 12 = 42,000) = 97.59 %
Impedance splittypical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %
LV fault levelIFL = 2,000 / (√3 × 0.48) = 2,405.6 A; Isc = 2,405.6 × 100 / 6.5 = 37.01 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 37.01 = 81.54 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA
Capacity checkSload × growth = 1,800.0 kVA ≤ Sderated → 1,800.0 vs limit 2,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 37.01 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 1,500 × 1.2 = 1,800.0 kVA; next standard rating → 2000 kVA; utilisation = 1,500/2000 = 75.0 %; Z = 6.0 × 1.1 = 6.6 → 6.5 % (nearest 0.5); P0 = 16,000 W, Pk = 26,000 W; η75 = 1,275,000 / (1,275,000 + 16,000 + 26,000 × 0.5625) = 97.65 %; R = 6.5/√26 = 1.2748 %, X = 5 × R = 6.3738 % (typical X/R 5); Reg = 1.2748 × 0.85 + 6.3738 × 0.5268 = 4.44 %; I_FL = 2000/(√3 × 0.48) = 2,405.6 A, Isc = 2,405.6 × 100/6.5 = 37.01 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 37.01 = 81.54 kA; capacity 1,800.0 ≤ 2,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-2000kva-60hz-13.8kv-480v)

Example 24 — 5 MVA with ±10 % OLTC, 17 positions

Tap voltages: step = 2 × 10 / (17 − 1) = 1.25 %, tap 1 at 29.7 kV, centre tap 9 at 33 kV, tap 17 at 36.3 kV.

Given: Maximum demand 4,000 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 33 kV / 11 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, on-load tap changer ±10 % in 17 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 4,000 × 1.2 = 4,800.0 kVA
Ratingnext standard rating ≥ 4,800.0 kVA → 5,000 kVA; utilisation = 4,000 / 5,000 × 100 = 80.0 %
ImpedanceTypical Z = Zbase × kV = 7.5 % (5,000 kVA band) × 1.2 (Vmax 33 kV) = 9 % → nearest 0.5 % = 9 %
LossesP0 = 5,000 × 1000 × 0.5 % = 25,000 W; Pk = 5,000 × 1000 × 1 % = 50,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 2,125,000 / (2,125,000 + 25,000 + 50,000 × 0.52 = 37,500) = 98.27 %; η75 = 3,187,500 / (3,187,500 + 25,000 + 50,000 × 0.752 = 53,125) = 98.36 %; η100 = 4,250,000 / (4,250,000 + 25,000 + 50,000 × 12 = 75,000) = 98.27 %
Impedance splittypical X/R = 5: R = 9 / √(1 + 52) = 1.7650 %, X = 5 × 1.7650 = 8.8252 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 1.7650 × 0.85 + 8.8252 × 0.5268 = 6.15 %
LV fault levelIFL = 5,000 / (√3 × 11) = 262.4 A; Isc = 262.4 × 100 / 9 = 2.92 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 2.92 = 6.42 kA peak
TapsStep = 2 × 10 / (17 − 1) = 1.25 %; nominal tap 9; tap 1 = 33 × (1 − 10 %) = 29.700 kV, tap 17 = 33 × (1 + 10 %) = 36.300 kV
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 5,000 kVA
Capacity checkSload × growth = 4,800.0 kVA ≤ Sderated → 4,800.0 vs limit 5,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 5,000 kVA, Z 9 %, Isc 2.92 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 4,000 × 1.2 = 4,800.0 kVA; next standard rating → 5000 kVA; utilisation = 4,000/5000 = 80.0 %; Z = 7.5 × 1.2 = 9 → 9 % (nearest 0.5); P0 = 25,000 W, Pk = 50,000 W; η75 = 3,187,500 / (3,187,500 + 25,000 + 50,000 × 0.5625) = 98.36 %; R = 9/√26 = 1.7650 %, X = 5 × R = 8.8252 % (typical X/R 5); Reg = 1.7650 × 0.85 + 8.8252 × 0.5268 = 6.15 %; I_FL = 5000/(√3 × 11) = 262.4 A, Isc = 262.4 × 100/9 = 2.92 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 2.92 = 6.42 kA; capacity 4,800.0 ≤ 5,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-5mva-33kv-11kv-oltc-taps)

Example 25 — 315 kVA hermetically sealed, no tap changer

Without taps the only tap voltage is the nominal primary; 315 kVA sits in the 100–500 kVA loss band (1.2 % / 1.8 %).

Given: Maximum demand 250 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 11 kV / 0.415 kV, three-phase, 50 Hz, hermetically sealed ONAN, Dyn11, no tap changer; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 250 × 1.2 = 300.0 kVA
Ratingnext standard rating ≥ 300.0 kVA → 315 kVA; utilisation = 250 / 315 × 100 = 79.4 %
ImpedanceTypical Z = Zbase × kV = 4.5 % (315 kVA band) × 1.1 (Vmax 11 kV) = 4.95 % → nearest 0.5 % = 5 %
LossesP0 = 315 × 1000 × 1.2 % = 3,780 W; Pk = 315 × 1000 × 1.8 % = 5,670 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 133,875 / (133,875 + 3,780 + 5,670 × 0.52 = 5,198) = 96.26 %; η75 = 200,812 / (200,812 + 3,780 + 5,670 × 0.752 = 6,969) = 96.65 %; η100 = 267,750 / (267,750 + 3,780 + 5,670 × 12 = 9,450) = 96.59 %
Impedance splittypical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %
LV fault levelIFL = 315 / (√3 × 0.415) = 438.2 A; Isc = 438.2 × 100 / 5 = 8.76 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 8.76 = 19.31 kA peak
TapsNo tap changer: single position at 11 kV
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 315 kVA
Capacity checkSload × growth = 300.0 kVA ≤ Sderated → 300.0 vs limit 315.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 315 kVA, Z 5 %, Isc 8.76 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 250 × 1.2 = 300.0 kVA; next standard rating → 315 kVA; utilisation = 250/315 = 79.4 %; Z = 4.5 × 1.1 = 4.95 → 5 % (nearest 0.5); P0 = 3,780 W, Pk = 5,670 W; η75 = 200,812 / (200,812 + 3,780 + 5,670 × 0.5625) = 96.65 %; R = 5/√26 = 0.9806 %, X = 5 × R = 4.9029 % (typical X/R 5); Reg = 0.9806 × 0.85 + 4.9029 × 0.5268 = 3.42 %; I_FL = 315/(√3 × 0.415) = 438.2 A, Isc = 438.2 × 100/5 = 8.76 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 8.76 = 19.31 kA; capacity 300.0 ≤ 315.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-315kva-hermetic-no-taps)

Example 26 — Demand beyond the largest standard rating

45 MVA × 1.2 = 54 MVA exceeds the 50 MVA top of the series: the calc returns 50 MVA and the capacity check FAILs.

Given: Maximum demand 45,000 kVA at pf 0.85 lagging, growth factor 1.2, overload capability not required; 132 kV / 33 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; no nameplate data (typical impedance and losses); insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 45,000 × 1.2 = 54,000.0 kVA
RatingSreq exceeds the largest standard rating → capped at 50,000 kVA; utilisation = 45,000 / 50,000 × 100 = 90.0 %
ImpedanceTypical Z = Zbase × kV = 8.5 % (50,000 kVA band) × 1.3 (Vmax 132 kV) = 11.1 % → nearest 0.5 % = 11 %
LossesP0 = 50,000 × 1000 × 0.4 % = 200,000 W; Pk = 50,000 × 1000 × 0.8 % = 400,000 W
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 21,250,000 / (21,250,000 + 200,000 + 400,000 × 0.52 = 300,000) = 98.61 %; η75 = 31,875,000 / (31,875,000 + 200,000 + 400,000 × 0.752 = 425,000) = 98.68 %; η100 = 42,500,000 / (42,500,000 + 200,000 + 400,000 × 12 = 600,000) = 98.61 %
Impedance splittypical X/R = 5: R = 11 / √(1 + 52) = 2.1573 %, X = 5 × 2.1573 = 10.7864 %
Regulationsin φ = √(1 − 0.852) = 0.5268; Reg = R·cos φ + X·sin φ = 2.1573 × 0.85 + 10.7864 × 0.5268 = 7.52 %
LV fault levelIFL = 50,000 / (√3 × 33) = 874.8 A; Isc = 874.8 × 100 / 11 = 7.95 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.2000 = 1.5578; ip = κ·√2·Isc = 1.5578 × √2 × 7.95 = 17.52 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 50,000 kVA
Capacity checkSload × growth = 54,000.0 kVA ≤ Sderated → 54,000.0 vs limit 50,000.0 kVA FAIL
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result1/2 checks pass → FAIL; 50,000 kVA, Z 11 %, Isc 7.95 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 45,000 × 1.2 = 54,000.0 kVA; next standard rating → 50000 kVA; utilisation = 45,000/50000 = 90.0 %; Z = 8.5 × 1.3 = 11.1 → 11 % (nearest 0.5); P0 = 200,000 W, Pk = 400,000 W; η75 = 31,875,000 / (31,875,000 + 200,000 + 400,000 × 0.5625) = 98.68 %; R = 11/√26 = 2.1573 %, X = 5 × R = 10.7864 % (typical X/R 5); Reg = 2.1573 × 0.85 + 10.7864 × 0.5268 = 7.52 %; I_FL = 50000/(√3 × 33) = 874.8 A, Isc = 874.8 × 100/11 = 7.95 kA; κ = 1.02 + 0.98·e^(−3 × 0.2000) = 1.5578 (IEC 60909-0), i_p = 1.5578 × √2 × 7.95 = 17.52 kA; capacity 54,000.0 ≤ 50,000.0 kVA → FAIL; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-50mva-cap-132kv-fail)

Example 27 — Nameplate load loss sets a high X/R and peak factor

A 10 MVA unit with Pk 55 kW has R = 0.55 % against Z 9.5 %, so X/R ≈ 17: κ rises to about 1.84 and the peak is 2.6 × I_k, well above the 2.2 × that the typical X/R of 5 would give.

Given: Maximum demand 8,000 kVA at pf 0.9 lagging, growth factor 1.2, overload capability not required; 33 kV / 11 kV, three-phase, 50 Hz, oil-filled core-type ONAN, Dyn11, off-circuit tap changer ±5 % in 5 positions; nameplate Z 9.5 %, Pk 55,000 W; insulation class F, 65 K rise, ambient 40 °C, altitude 0 m.

Required capacitySreq = 8,000 × 1.2 = 9,600.0 kVA
Ratingnext standard rating ≥ 9,600.0 kVA → 10,000 kVA; utilisation = 8,000 / 10,000 × 100 = 80.0 %
ImpedanceZ = 9.5 % (nameplate)
LossesP0 = 10,000 × 1000 × 0.4 % = 40,000 W; Pk = 55,000 W (nameplate)
Efficiencyη = S·n·cos φ / (S·n·cos φ + P0 + Pk·n2): η50 = 4,500,000 / (4,500,000 + 40,000 + 55,000 × 0.52 = 53,750) = 98.82 %; η75 = 6,750,000 / (6,750,000 + 40,000 + 55,000 × 0.752 = 70,938) = 98.96 %; η100 = 9,000,000 / (9,000,000 + 40,000 + 55,000 × 12 = 95,000) = 98.96 %
Impedance splitR = Pk / S = 55,000 / (10,000 × 1000) × 100 = 0.5500 %, X = √(9.52 − 0.55002) = 9.4841 % (X/R = 17.244)
Regulationsin φ = √(1 − 0.92) = 0.4359; Reg = R·cos φ + X·sin φ = 0.5500 × 0.9 + 9.4841 × 0.4359 = 4.63 %
LV fault levelIFL = 10,000 / (√3 × 11) = 524.9 A; Isc = 524.9 × 100 / 9.5 = 5.52 kA symmetrical (infinite source)
Peak fault currentIEC 60909-0: κ = 1.02 + 0.98·e−3R/X = 1.02 + 0.98·e−3 × 0.0580 = 1.8435; ip = κ·√2·Isc = 1.8435 × √2 × 5.52 = 14.40 kA peak
Altitude0 m ≤ 1,000 m → kalt = 1, Sderated = 10,000 kVA
Capacity checkSload × growth = 9,600.0 kVA ≤ Sderated → 9,600.0 vs limit 10,000.0 kVA PASS
Winding temperatureT = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS
Result2/2 checks pass → PASS; 10,000 kVA, Z 9.5 %, Isc 5.52 kA; governing: capacity
ReferenceHand calculation (IEC 60076-1 rating selection, c0024 help-page formulas): S_req = 8,000 × 1.2 = 9,600.0 kVA; next standard rating → 10000 kVA; utilisation = 8,000/10000 = 80.0 %; Z = 9.5 % (nameplate); P0 = 40,000 W, Pk = 55,000 W; η75 = 6,750,000 / (6,750,000 + 40,000 + 55,000 × 0.5625) = 98.96 %; R = 55,000/(10000 × 1000) × 100 = 0.5500 %, X = √(9.5² − 0.5500²) = 9.4841 % (X/R 17.244); Reg = 0.5500 × 0.9 + 9.4841 × 0.4359 = 4.63 %; I_FL = 10000/(√3 × 11) = 524.9 A, Isc = 524.9 × 100/9.5 = 5.52 kA; κ = 1.02 + 0.98·e^(−3 × 0.0580) = 1.8435 (IEC 60909-0), i_p = 1.8435 × √2 × 5.52 = 14.40 kA; capacity 9,600.0 ≤ 10,000.0 kVA → PASS; T = 40 + 65 = 105 °C vs class F 155 °C → PASS (validation case c0024-10mva-nameplate-high-x-r-peak)

Standards & References

  • AS 60076.1 — Power transformers — Part 1: General
  • AS 60076.2 — Power transformers — Part 2: Temperature rise for liquid-immersed transformers
  • AS 60076.3 — Power transformers — Part 3: Insulation levels, dielectric tests and external clearances in air
  • AS 60076.5 — Power transformers — Part 5: Ability to withstand short circuit
  • IEC 60076 (series) — International standard counterparts to AS 60076
  • IEC 60076-7 — Loading guide for mineral-oil-immersed power transformers
  • IEC 60909-0 — Short-circuit currents in three-phase a.c. systems — peak factor κ and peak current ip
  • IEEE C57.12.00 — Standard General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
  • IEEE C57.91 — Guide for Loading Mineral-Oil-Immersed Transformers and Step-Voltage Regulators
  • AS/NZS 2374 — Power transformers (older Australian standard, now largely replaced by AS 60076)

Tips & Limitations

Tips
  • The most common distribution vector group in Australia and New Zealand is Dyn11 (delta primary, star secondary with neutral, −30° phase shift).
  • A growth factor of 1.2 (20%) is typical for commercial installations; for mining and industrial, 1.3–1.5 may be more appropriate.
  • If manufacturer test data for impedance and losses is available, enter it directly rather than relying on estimated values — estimates are for preliminary sizing only.
  • Transformers are most efficient when core losses equal winding losses, which typically occurs at 50–70% loading. Avoid both oversizing (poor efficiency at light load) and undersizing (overheating risk).
  • For parallel operation, transformers must have the same vector group, voltage ratio, and similar impedance values (within 10%).
  • OLTC (On-Load Tap Changer) is typically specified when voltage regulation under varying load is critical; DETC is adequate for most distribution applications.
Limitations
  • Impedance estimates are approximate — actual impedance depends on detailed winding geometry. Use manufacturer data for final design.
  • Loss values are estimated — the calculator uses typical percentages by rating class. Actual losses vary significantly between manufacturers and efficiency tiers.
  • Fault current calculation assumes infinite bus — the source impedance upstream of the transformer is not considered. Actual fault levels will be lower.
  • Typical X/R without a nameplate load loss — when \(P_k\) is not given, X/R = 5 is used for regulation and the peak factor. Actual X/R ratios vary by transformer size and design (typically 3–12, higher for large units); enter the nameplate load loss to use the real ratio.
  • Altitude derating is linear — the 0.5% per 100 m above 1,000 m is a simplified rule. Detailed derating per IEC 60076-2 considers both convection and dielectric effects.
  • No inrush current calculation — magnetizing inrush current (typically 8–12× rated) is not modelled.
  • Dry-type correction factors are approximate — the 0.85× no-load and 1.1× load loss adjustments are generic; actual ratios depend on core material and winding design.

Developer Guide

Architecture

File layout:

  • components/c0024_transformer_sizing/component.py — Component registration, parameter schema, SVG diagram generator, summary card builder
  • components/c0024_transformer_sizing/calculations.py — Pure math functions: rating selection, impedance/loss estimation, efficiency, regulation, fault current, tap voltages, altitude derating (no Django dependencies)
  • components/c0024_transformer_sizing/equations.py — LaTeX equation builder for the calculation report
  • components/c0024_transformer_sizing/excel_generator.py — Excel export generation

The calculator class is TransformerSizingCalculator, extending ElectricalCalculator. It delegates all computation to calculate_transformer_specification(data) in calculations.py.

Key data structures:

  • STANDARD_RATINGS_KVA — list of 37 IEC preferred transformer ratings from 5 to 50,000 kVA
  • STANDARD_VOLTAGES — dict grouping LV, MV, and HV standard voltages
  • VECTOR_GROUPS — dict mapping vector group codes to descriptions and common uses
  • COOLING_TYPES — dict mapping cooling designations to type and forced-cooling flag
  • CONSTRUCTION_TYPES — dict mapping construction types to compatible cooling options
  • TAP_CHANGER_TYPES — dict mapping tap changer types to typical step counts
  • INSULATION_CLASS_TEMPS — dict mapping insulation class labels to temperature limits (°C)

Code Paths

Key functions in calculations.py:

Function Purpose
select_transformer_rating(load_kva, growth_factor, overload_required) Select next standard kVA rating from the IEC preferred series
estimate_impedance_percent(rating_kva, primary_kv, secondary_kv) Estimate short circuit impedance based on rating and voltage class
estimate_no_load_loss(rating_kva, primary_kv, construction) Estimate core losses as a percentage of rating
estimate_load_loss(rating_kva, construction) Estimate winding losses as a percentage of rating
calculate_efficiency(rating_kva, no_load_loss_w, load_loss_w, load_percent, power_factor) Compute efficiency at a given load percentage and power factor
split_impedance(impedance_percent, rating_kva, load_loss_w) Split %Z into R% and X% (from the load loss, or the typical X/R of 5)
calculate_voltage_regulation(resistance_percent, reactance_percent, power_factor) Approximate voltage regulation R·cosφ + X·sinφ
calculate_short_circuit_current(rating_kva, secondary_kv, impedance_percent, phase_config, x_r_ratio) Symmetrical rms and IEC 60909-0 peak fault current on the secondary side, and κ
calculate_tap_voltages(nominal_kv, tap_range_percent, num_taps) Generate tap position table with voltages and percentage offsets
calculate_transformer_specification(data) Main orchestrator: runs all steps and returns the full results dict

Key functions in component.py:

Function Purpose
build_summary_cards(results) Build the four summary cards (Rating, Voltage, Performance, Construction)
generate_transformer_diagram_svg(parameters, results) Generate the SVG schematic of the transformer
register_component(registry) Register the component with the calc registry

Testing

Running tests:

pytest calcs/components/c0024_transformer_sizing/ -v
pytest calcs/test_c0024* -v

How to extend:

  1. To add new construction types, update CONSTRUCTION_TYPES in component.py and add compatible cooling options.
  2. To add new vector groups, update VECTOR_GROUPS with description and common use.
  3. For manufacturer-specific loss data, create additional lookup tables in calculations.py keyed by manufacturer and efficiency tier.
  4. To refine impedance estimation, add rating-specific Z% tables from standards or manufacturer catalogues.
  5. Write tests covering edge cases: minimum rating (5 kVA), maximum rating (50,000 kVA), high altitude, low temperature margin, step-up configuration (primary < secondary), and single-phase fault current.