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
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
Impedance Estimation
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
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
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
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
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
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).
| # | Case | Sreq (kVA) | Rating (kVA) | Z (%) | η75 (%) | Reg (%) | Isc (kA) | ip (kA) | Result |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 500 kVA load, 11 kV / 415 V distribution transformer | 600.0 | 630 | 6.5 | 97.65 | 4.44 | 13.48 | 29.71 | PASS |
| 2 | 1,600 kVA load at 11 kV / 415 V | 1,920.0 | 2,000 | 6.5 | 97.65 | 4.44 | 42.81 | 94.31 | PASS |
| 3 | Required capacity lands exactly on 1,250 kVA | 1,250.0 | 1,250 | 6.5 | 97.65 | 4.44 | 25.64 | 56.49 | PASS |
| 4 | Required capacity just above 1,250 kVA | 1,250.4 | 1,500 | 6.5 | 97.65 | 4.44 | 32.10 | 70.73 | PASS |
| 5 | Overload capability governs the rating | 1,625.0 | 2,000 | 6.5 | 97.65 | 4.44 | 41.03 | 90.39 | PASS |
| 6 | Growth allowance governs over the overload margin | 1,500.0 | 1,500 | 6.5 | 97.65 | 4.44 | 30.77 | 67.79 | PASS |
| 7 | 33 kV / 690 V at 2,500 m with overload capability | 2,250.0 | 2,500 | 9 | 98.45 | 5.44 | 23.24 | 51.21 | PASS |
| 8 | 2,500 kVA at 6.6 kV / 690 V | 2,400.0 | 2,500 | 7.5 | 98.36 | 5.12 | 27.89 | 61.45 | PASS |
| 9 | Typical impedance on a half-percent tie (8.25 % → 8.5 %) | 2,500.0 | 2,500 | 8.5 | 98.36 | 5.81 | 40.92 | 90.15 | PASS |
| 10 | 10 MVA, 33 kV / 11 kV zone substation | 9,600.0 | 10,000 | 10 | 98.68 | 6.83 | 5.25 | 11.56 | PASS |
| 11 | Cast-resin 1,000 kVA indoor transformer | 960.0 | 1,000 | 6.5 | 97.72 | 4.44 | 21.40 | 47.15 | PASS |
| 12 | Nameplate impedance and losses | 900.0 | 1,000 | 5 | 98.89 | 3.08 | 27.82 | 60.38 | PASS |
| 13 | Voltage regulation at unity power factor | 480.0 | 500 | 6.5 | 98.00 | 1.27 | 10.70 | 23.58 | PASS |
| 14 | Voltage regulation at pf 0.7 lagging | 480.0 | 500 | 6.5 | 97.17 | 5.44 | 10.70 | 23.58 | PASS |
| 15 | Altitude derating fails the capacity check | 2,000.0 | 2,000 | 6.5 | 97.65 | 4.44 | 42.81 | 94.31 | FAIL |
| 16 | Altitude exactly 1,000 m — no derating | 1,560.0 | 1,600 | 6.5 | 97.65 | 4.44 | 34.25 | 75.45 | PASS |
| 17 | Derated rating exactly equals the requirement | 1,560.0 | 1,600 | 6.5 | 97.65 | 4.44 | 32.82 | 72.31 | PASS |
| 18 | Class A insulation overheats at 45 °C ambient | 600.0 | 630 | 6.5 | 97.65 | 4.44 | 13.48 | 29.71 | FAIL |
| 19 | Dry-type class B at its temperature limit | 300.0 | 315 | 5 | 96.76 | 3.42 | 8.76 | 19.31 | PASS |
| 20 | Single-phase 180 kVA dry-type with nameplate data | 180.0 | 200 | 4 | 98.41 | 2.89 | 21.74 | 41.18 | PASS |
| 21 | 25 kVA single-phase SWER pole transformer | 24.0 | 25 | 5 | 95.21 | 3.42 | 2.00 | 4.41 | PASS |
| 22 | Low-impedance 2,500 kVA unit — fault level above 50 kA | 2,500.0 | 2,500 | 5 | 98.36 | 3.42 | 69.56 | 153.25 | PASS |
| 23 | 60 Hz, 13.8 kV / 480 V | 1,800.0 | 2,000 | 6.5 | 97.65 | 4.44 | 37.01 | 81.54 | PASS |
| 24 | 5 MVA with ±10 % OLTC, 17 positions | 4,800.0 | 5,000 | 9 | 98.36 | 6.15 | 2.92 | 6.42 | PASS |
| 25 | 315 kVA hermetically sealed, no tap changer | 300.0 | 315 | 5 | 96.65 | 3.42 | 8.76 | 19.31 | PASS |
| 26 | Demand beyond the largest standard rating | 54,000.0 | 50,000 | 11 | 98.68 | 7.52 | 7.95 | 17.52 | FAIL |
| 27 | Nameplate load loss sets a high X/R and peak factor | 9,600.0 | 10,000 | 9.5 | 98.96 | 4.63 | 5.52 | 14.40 | PASS |
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 capacity | Sreq = 500 × 1.2 = 600.0 kVA |
|---|---|
| Rating | next standard rating ≥ 600.0 kVA → 630 kVA; utilisation = 500 / 630 × 100 = 79.4 % |
| Impedance | Typical Z = Zbase × kV = 6 % (630 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 630 × 1000 × 0.8 % = 5,040 W; Pk = 630 × 1000 × 1.3 % = 8,190 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 630 / (√3 × 0.415) = 876.5 A; Isc = 876.5 × 100 / 6.5 = 13.48 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 630 kVA |
| Capacity check | Sload × growth = 600.0 kVA ≤ Sderated → 600.0 vs limit 630.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 630 kVA, Z 6.5 %, Isc 13.48 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 1,600 × 1.2 = 1,920.0 kVA |
|---|---|
| Rating | next standard rating ≥ 1,920.0 kVA → 2,000 kVA; utilisation = 1,600 / 2,000 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 2,000 / (√3 × 0.415) = 2,782.4 A; Isc = 2,782.4 × 100 / 6.5 = 42.81 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA |
| Capacity check | Sload × growth = 1,920.0 kVA ≤ Sderated → 1,920.0 vs limit 2,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 42.81 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 1,000 × 1.25 = 1,250.0 kVA |
|---|---|
| Rating | Sreq equals a standard rating → 1,250 kVA; utilisation = 1,000 / 1,250 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,250 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 1,250 × 1000 × 0.8 % = 10,000 W; Pk = 1,250 × 1000 × 1.3 % = 16,250 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,250 / (√3 × 0.433) = 1,666.7 A; Isc = 1,666.7 × 100 / 6.5 = 25.64 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 1,250 kVA |
| Capacity check | Sload × growth = 1,250.0 kVA ≤ Sderated → 1,250.0 vs limit 1,250.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,250 kVA, Z 6.5 %, Isc 25.64 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 1,042 × 1.2 = 1,250.4 kVA |
|---|---|
| Rating | next standard rating ≥ 1,250.4 kVA → 1,500 kVA; utilisation = 1,042 / 1,500 × 100 = 69.5 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,500 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 1,500 × 1000 × 0.8 % = 12,000 W; Pk = 1,500 × 1000 × 1.3 % = 19,500 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,500 / (√3 × 0.415) = 2,086.8 A; Isc = 2,086.8 × 100 / 6.5 = 32.10 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 1,500 kVA |
| Capacity check | Sload × growth = 1,250.4 kVA ≤ Sderated → 1,250.4 vs limit 1,500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,500 kVA, Z 6.5 %, Isc 32.10 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 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 |
|---|---|
| Rating | next standard rating ≥ 1,625.0 kVA → 2,000 kVA; utilisation = 1,300 / 2,000 × 100 = 65.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 2,000 / (√3 × 0.433) = 2,666.7 A; Isc = 2,666.7 × 100 / 6.5 = 41.03 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA |
| Capacity check | Sload × growth = 1,300.0 kVA ≤ Sderated → 1,300.0 vs limit 2,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 41.03 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 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 |
|---|---|
| Rating | Sreq equals a standard rating → 1,500 kVA; utilisation = 1,000 / 1,500 × 100 = 66.7 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,500 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 1,500 × 1000 × 0.8 % = 12,000 W; Pk = 1,500 × 1000 × 1.3 % = 19,500 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,500 / (√3 × 0.433) = 2,000.1 A; Isc = 2,000.1 × 100 / 6.5 = 30.77 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 1,500 kVA |
| Capacity check | Sload × growth = 1,500.0 kVA ≤ Sderated → 1,500.0 vs limit 1,500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,500 kVA, Z 6.5 %, Isc 30.77 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 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 |
|---|---|
| Rating | next standard rating ≥ 2,250.0 kVA → 2,500 kVA; utilisation = 1,800 / 2,500 × 100 = 72.0 % |
| Impedance | Typical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1.2 (Vmax 33 kV) = 9 % → nearest 0.5 % = 9 % |
| Losses | P0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 9 / √(1 + 52) = 1.7650 %, X = 5 × 1.7650 = 8.8252 % |
| Regulation | sin 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 level | IFL = 2,500 / (√3 × 0.69) = 2,091.8 A; Isc = 2,091.8 × 100 / 9 = 23.24 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | kalt = 1 − (2,500 − 1,000) / 100 × 0.005 = 0.925; Sderated = 2,500 × 0.925 = 2,312.5 kVA |
| Capacity check | Sload × growth = 2,070.0 kVA ≤ Sderated → 2,070.0 vs limit 2,312.5 kVA PASS |
| Winding temperature | T = 45 + 80 = 125 °C ≤ class H limit (margin 55 K) → 125 vs limit 180 °C PASS |
| Result | 2/2 checks pass → PASS; 2,500 kVA, Z 9 %, Isc 23.24 kA; governing: capacity. Altitude 2500.0m requires 7.5% derating |
| Reference | Hand 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 capacity | Sreq = 2,000 × 1.2 = 2,400.0 kVA |
|---|---|
| Rating | next standard rating ≥ 2,400.0 kVA → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1 (Vmax 6.6 kV) = 7.5 % → nearest 0.5 % = 7.5 % |
| Losses | P0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 7.5 / √(1 + 52) = 1.4709 %, X = 5 × 1.4709 = 7.3544 % |
| Regulation | sin 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 level | IFL = 2,500 / (√3 × 0.69) = 2,091.8 A; Isc = 2,091.8 × 100 / 7.5 = 27.89 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA |
| Capacity check | Sload × growth = 2,400.0 kVA ≤ Sderated → 2,400.0 vs limit 2,500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,500 kVA, Z 7.5 %, Isc 27.89 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 2,000 × 1.25 = 2,500.0 kVA |
|---|---|
| Rating | Sreq equals a standard rating → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 7.5 % (2,500 kVA band) × 1.1 (Vmax 11 kV) = 8.25 % → nearest 0.5 % = 8.5 % |
| Losses | P0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 8.5 / √(1 + 52) = 1.6670 %, X = 5 × 1.6670 = 8.3349 % |
| Regulation | sin 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 level | IFL = 2,500 / (√3 × 0.415) = 3,478.0 A; Isc = 3,478.0 × 100 / 8.5 = 40.92 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA |
| Capacity check | Sload × growth = 2,500.0 kVA ≤ Sderated → 2,500.0 vs limit 2,500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,500 kVA, Z 8.5 %, Isc 40.92 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 8,000 × 1.2 = 9,600.0 kVA |
|---|---|
| Rating | next standard rating ≥ 9,600.0 kVA → 10,000 kVA; utilisation = 8,000 / 10,000 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 8.5 % (10,000 kVA band) × 1.2 (Vmax 33 kV) = 10.2 % → nearest 0.5 % = 10 % |
| Losses | P0 = 10,000 × 1000 × 0.4 % = 40,000 W; Pk = 10,000 × 1000 × 0.8 % = 80,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 10 / √(1 + 52) = 1.9612 %, X = 5 × 1.9612 = 9.8058 % |
| Regulation | sin 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 level | IFL = 10,000 / (√3 × 11) = 524.9 A; Isc = 524.9 × 100 / 10 = 5.25 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 10,000 kVA |
| Capacity check | Sload × growth = 9,600.0 kVA ≤ Sderated → 9,600.0 vs limit 10,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 10,000 kVA, Z 10 %, Isc 5.25 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 800 × 1.2 = 960.0 kVA |
|---|---|
| Rating | next standard rating ≥ 960.0 kVA → 1,000 kVA; utilisation = 800 / 1,000 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 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 |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,000 / (√3 × 0.415) = 1,391.2 A; Isc = 1,391.2 × 100 / 6.5 = 21.40 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 1,000 kVA |
| Capacity check | Sload × growth = 960.0 kVA ≤ Sderated → 960.0 vs limit 1,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,000 kVA, Z 6.5 %, Isc 21.40 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 750 × 1.2 = 900.0 kVA |
|---|---|
| Rating | next standard rating ≥ 900.0 kVA → 1,000 kVA; utilisation = 750 / 1,000 × 100 = 75.0 % |
| Impedance | Z = 5 % (nameplate) |
| Losses | P0 = 1,700 W (nameplate); Pk = 10,500 W (nameplate) |
| Efficiency | η = 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 split | R = Pk / S = 10,500 / (1,000 × 1000) × 100 = 1.0500 %, X = √(52 − 1.05002) = 4.8885 % (X/R = 4.656) |
| Regulation | sin 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 level | IFL = 1,000 / (√3 × 0.415) = 1,391.2 A; Isc = 1,391.2 × 100 / 5 = 27.82 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 1,000 kVA |
| Capacity check | Sload × growth = 900.0 kVA ≤ Sderated → 900.0 vs limit 1,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,000 kVA, Z 5 %, Isc 27.82 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 400 × 1.2 = 480.0 kVA |
|---|---|
| Rating | next standard rating ≥ 480.0 kVA → 500 kVA; utilisation = 400 / 500 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (500 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 500 × 1000 × 0.8 % = 4,000 W; Pk = 500 × 1000 × 1.3 % = 6,500 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin φ = √(1 − 12) = 0.0000; Reg = R·cos φ + X·sin φ = 1.2748 × 1 + 6.3738 × 0.0000 = 1.27 % |
| LV fault level | IFL = 500 / (√3 × 0.415) = 695.6 A; Isc = 695.6 × 100 / 6.5 = 10.70 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 500 kVA |
| Capacity check | Sload × growth = 480.0 kVA ≤ Sderated → 480.0 vs limit 500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 500 kVA, Z 6.5 %, Isc 10.70 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 400 × 1.2 = 480.0 kVA |
|---|---|
| Rating | next standard rating ≥ 480.0 kVA → 500 kVA; utilisation = 400 / 500 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (500 kVA band) × 1.1 (Vmax 22 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 500 × 1000 × 0.8 % = 4,000 W; Pk = 500 × 1000 × 1.3 % = 6,500 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 500 / (√3 × 0.415) = 695.6 A; Isc = 695.6 × 100 / 6.5 = 10.70 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 500 kVA |
| Capacity check | Sload × growth = 480.0 kVA ≤ Sderated → 480.0 vs limit 500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 500 kVA, Z 6.5 %, Isc 10.70 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 1,600 × 1.25 = 2,000.0 kVA |
|---|---|
| Rating | Sreq equals a standard rating → 2,000 kVA; utilisation = 1,600 / 2,000 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 2,000 / (√3 × 0.415) = 2,782.4 A; Isc = 2,782.4 × 100 / 6.5 = 42.81 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | kalt = 1 − (3,000 − 1,000) / 100 × 0.005 = 0.9; Sderated = 2,000 × 0.9 = 1,800.0 kVA |
| Capacity check | Sload × growth = 2,000.0 kVA ≤ Sderated → 2,000.0 vs limit 1,800.0 kVA FAIL |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 1/2 checks pass → FAIL; 2,000 kVA, Z 6.5 %, Isc 42.81 kA; governing: capacity. Altitude 3000.0m requires 10.0% derating |
| Reference | Hand 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 capacity | Sreq = 1,300 × 1.2 = 1,560.0 kVA |
|---|---|
| Rating | next standard rating ≥ 1,560.0 kVA → 1,600 kVA; utilisation = 1,300 / 1,600 × 100 = 81.2 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,600 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 1,600 × 1000 × 0.8 % = 12,800 W; Pk = 1,600 × 1000 × 1.3 % = 20,800 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,600 / (√3 × 0.415) = 2,225.9 A; Isc = 2,225.9 × 100 / 6.5 = 34.25 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 1,000 m ≤ 1,000 m → kalt = 1, Sderated = 1,600 kVA |
| Capacity check | Sload × growth = 1,560.0 kVA ≤ Sderated → 1,560.0 vs limit 1,600.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,600 kVA, Z 6.5 %, Isc 34.25 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 1,248 × 1.25 = 1,560.0 kVA |
|---|---|
| Rating | next standard rating ≥ 1,560.0 kVA → 1,600 kVA; utilisation = 1,248 / 1,600 × 100 = 78.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (1,600 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 1,600 × 1000 × 0.8 % = 12,800 W; Pk = 1,600 × 1000 × 1.3 % = 20,800 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 1,600 / (√3 × 0.433) = 2,133.4 A; Isc = 2,133.4 × 100 / 6.5 = 32.82 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | kalt = 1 − (1,500 − 1,000) / 100 × 0.005 = 0.975; Sderated = 1,600 × 0.975 = 1,560.0 kVA |
| Capacity check | Sload × growth = 1,560.0 kVA ≤ Sderated → 1,560.0 vs limit 1,560.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 1,600 kVA, Z 6.5 %, Isc 32.82 kA; governing: capacity. Altitude 1500.0m requires 2.5% derating |
| Reference | Hand 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 capacity | Sreq = 500 × 1.2 = 600.0 kVA |
|---|---|
| Rating | next standard rating ≥ 600.0 kVA → 630 kVA; utilisation = 500 / 630 × 100 = 79.4 % |
| Impedance | Typical Z = Zbase × kV = 6 % (630 kVA band) × 1.1 (Vmax 11 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 630 × 1000 × 0.8 % = 5,040 W; Pk = 630 × 1000 × 1.3 % = 8,190 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 630 / (√3 × 0.415) = 876.5 A; Isc = 876.5 × 100 / 6.5 = 13.48 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 630 kVA |
| Capacity check | Sload × growth = 600.0 kVA ≤ Sderated → 600.0 vs limit 630.0 kVA PASS |
| Winding temperature | T = 45 + 65 = 110 °C ≤ class A limit (margin -5 K) → 110 vs limit 105 °C FAIL |
| Result | 1/2 checks pass → FAIL; 630 kVA, Z 6.5 %, Isc 13.48 kA; governing: capacity. Low temperature margin (-5.0K). Consider higher insulation class. |
| Reference | Hand 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 capacity | Sreq = 250 × 1.2 = 300.0 kVA |
|---|---|
| Rating | next standard rating ≥ 300.0 kVA → 315 kVA; utilisation = 250 / 315 × 100 = 79.4 % |
| Impedance | Typical Z = Zbase × kV = 4.5 % (315 kVA band) × 1.1 (Vmax 11 kV) = 4.95 % → nearest 0.5 % = 5 % |
| Losses | P0 = 315 × 1000 × 1.2 % × 0.85 (dry/cast) = 3,213 W; Pk = 315 × 1000 × 1.8 % × 1.1 (dry/cast) = 6,237 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 % |
| Regulation | sin 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 level | IFL = 315 / (√3 × 0.415) = 438.2 A; Isc = 438.2 × 100 / 5 = 8.76 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 315 kVA |
| Capacity check | Sload × growth = 300.0 kVA ≤ Sderated → 300.0 vs limit 315.0 kVA PASS |
| Winding temperature | T = 50 + 80 = 130 °C ≤ class B limit (margin 0 K) → 130 vs limit 130 °C PASS |
| Result | 2/2 checks pass → PASS; 315 kVA, Z 5 %, Isc 8.76 kA; governing: capacity. Low temperature margin (0.0K). Consider higher insulation class. |
| Reference | Hand 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 capacity | Sreq = 180 × 1 = 180.0 kVA |
|---|---|
| Rating | next standard rating ≥ 180.0 kVA → 200 kVA; utilisation = 180 / 200 × 100 = 90.0 % |
| Impedance | Z = 4 % (nameplate) |
| Losses | P0 = 600 W (nameplate); Pk = 2,800 W (nameplate) |
| Efficiency | η = 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 split | R = Pk / S = 2,800 / (200 × 1000) × 100 = 1.4000 %, X = √(42 − 1.40002) = 3.7470 % (X/R = 2.676) |
| Regulation | sin 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 level | IFL = 200 / 0.23 = 869.6 A (single-phase, no √3); Isc = 869.6 × 100 / 4 = 21.74 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 200 kVA |
| Capacity check | Sload × growth = 180.0 kVA ≤ Sderated → 180.0 vs limit 200.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 200 kVA, Z 4 %, Isc 21.74 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 20 × 1.2 = 24.0 kVA |
|---|---|
| Rating | next standard rating ≥ 24.0 kVA → 25 kVA; utilisation = 20 / 25 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 4.5 % (25 kVA band) × 1.1 (Vmax 12.7 kV) = 4.95 % → nearest 0.5 % = 5 % |
| Losses | P0 = 25 × 1000 × 1.8 % = 450 W; Pk = 25 × 1000 × 2.5 % = 625 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 % |
| Regulation | sin 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 level | IFL = 25 / 0.25 = 100.0 A (single-phase, no √3); Isc = 100.0 × 100 / 5 = 2.00 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 25 kVA |
| Capacity check | Sload × growth = 24.0 kVA ≤ Sderated → 24.0 vs limit 25.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 25 kVA, Z 5 %, Isc 2.00 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 2,000 × 1.25 = 2,500.0 kVA |
|---|---|
| Rating | Sreq equals a standard rating → 2,500 kVA; utilisation = 2,000 / 2,500 × 100 = 80.0 % |
| Impedance | Z = 5 % (nameplate) |
| Losses | P0 = 2,500 × 1000 × 0.5 % = 12,500 W; Pk = 2,500 × 1000 × 1 % = 25,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 % |
| Regulation | sin 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 level | IFL = 2,500 / (√3 × 0.415) = 3,478.0 A; Isc = 3,478.0 × 100 / 5 = 69.56 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,500 kVA |
| Capacity check | Sload × growth = 2,500.0 kVA ≤ Sderated → 2,500.0 vs limit 2,500.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,500 kVA, Z 5 %, Isc 69.56 kA; governing: capacity. High fault current (69.6 kA). Ensure switchgear rated adequately. |
| Reference | Hand 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 capacity | Sreq = 1,500 × 1.2 = 1,800.0 kVA |
|---|---|
| Rating | next standard rating ≥ 1,800.0 kVA → 2,000 kVA; utilisation = 1,500 / 2,000 × 100 = 75.0 % |
| Impedance | Typical Z = Zbase × kV = 6 % (2,000 kVA band) × 1.1 (Vmax 13.8 kV) = 6.6 % → nearest 0.5 % = 6.5 % |
| Losses | P0 = 2,000 × 1000 × 0.8 % = 16,000 W; Pk = 2,000 × 1000 × 1.3 % = 26,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 6.5 / √(1 + 52) = 1.2748 %, X = 5 × 1.2748 = 6.3738 % |
| Regulation | sin 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 level | IFL = 2,000 / (√3 × 0.48) = 2,405.6 A; Isc = 2,405.6 × 100 / 6.5 = 37.01 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 2,000 kVA |
| Capacity check | Sload × growth = 1,800.0 kVA ≤ Sderated → 1,800.0 vs limit 2,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 2,000 kVA, Z 6.5 %, Isc 37.01 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 4,000 × 1.2 = 4,800.0 kVA |
|---|---|
| Rating | next standard rating ≥ 4,800.0 kVA → 5,000 kVA; utilisation = 4,000 / 5,000 × 100 = 80.0 % |
| Impedance | Typical Z = Zbase × kV = 7.5 % (5,000 kVA band) × 1.2 (Vmax 33 kV) = 9 % → nearest 0.5 % = 9 % |
| Losses | P0 = 5,000 × 1000 × 0.5 % = 25,000 W; Pk = 5,000 × 1000 × 1 % = 50,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 9 / √(1 + 52) = 1.7650 %, X = 5 × 1.7650 = 8.8252 % |
| Regulation | sin 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 level | IFL = 5,000 / (√3 × 11) = 262.4 A; Isc = 262.4 × 100 / 9 = 2.92 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Taps | Step = 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 5,000 kVA |
| Capacity check | Sload × growth = 4,800.0 kVA ≤ Sderated → 4,800.0 vs limit 5,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 5,000 kVA, Z 9 %, Isc 2.92 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 250 × 1.2 = 300.0 kVA |
|---|---|
| Rating | next standard rating ≥ 300.0 kVA → 315 kVA; utilisation = 250 / 315 × 100 = 79.4 % |
| Impedance | Typical Z = Zbase × kV = 4.5 % (315 kVA band) × 1.1 (Vmax 11 kV) = 4.95 % → nearest 0.5 % = 5 % |
| Losses | P0 = 315 × 1000 × 1.2 % = 3,780 W; Pk = 315 × 1000 × 1.8 % = 5,670 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 5 / √(1 + 52) = 0.9806 %, X = 5 × 0.9806 = 4.9029 % |
| Regulation | sin 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 level | IFL = 315 / (√3 × 0.415) = 438.2 A; Isc = 438.2 × 100 / 5 = 8.76 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Taps | No tap changer: single position at 11 kV |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 315 kVA |
| Capacity check | Sload × growth = 300.0 kVA ≤ Sderated → 300.0 vs limit 315.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 315 kVA, Z 5 %, Isc 8.76 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 45,000 × 1.2 = 54,000.0 kVA |
|---|---|
| Rating | Sreq exceeds the largest standard rating → capped at 50,000 kVA; utilisation = 45,000 / 50,000 × 100 = 90.0 % |
| Impedance | Typical Z = Zbase × kV = 8.5 % (50,000 kVA band) × 1.3 (Vmax 132 kV) = 11.1 % → nearest 0.5 % = 11 % |
| Losses | P0 = 50,000 × 1000 × 0.4 % = 200,000 W; Pk = 50,000 × 1000 × 0.8 % = 400,000 W |
| Efficiency | η = 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 split | typical X/R = 5: R = 11 / √(1 + 52) = 2.1573 %, X = 5 × 2.1573 = 10.7864 % |
| Regulation | sin 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 level | IFL = 50,000 / (√3 × 33) = 874.8 A; Isc = 874.8 × 100 / 11 = 7.95 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 50,000 kVA |
| Capacity check | Sload × growth = 54,000.0 kVA ≤ Sderated → 54,000.0 vs limit 50,000.0 kVA FAIL |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 1/2 checks pass → FAIL; 50,000 kVA, Z 11 %, Isc 7.95 kA; governing: capacity |
| Reference | Hand 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 capacity | Sreq = 8,000 × 1.2 = 9,600.0 kVA |
|---|---|
| Rating | next standard rating ≥ 9,600.0 kVA → 10,000 kVA; utilisation = 8,000 / 10,000 × 100 = 80.0 % |
| Impedance | Z = 9.5 % (nameplate) |
| Losses | P0 = 10,000 × 1000 × 0.4 % = 40,000 W; Pk = 55,000 W (nameplate) |
| Efficiency | η = 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 split | R = Pk / S = 55,000 / (10,000 × 1000) × 100 = 0.5500 %, X = √(9.52 − 0.55002) = 9.4841 % (X/R = 17.244) |
| Regulation | sin 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 level | IFL = 10,000 / (√3 × 11) = 524.9 A; Isc = 524.9 × 100 / 9.5 = 5.52 kA symmetrical (infinite source) |
| Peak fault current | IEC 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 |
| Altitude | 0 m ≤ 1,000 m → kalt = 1, Sderated = 10,000 kVA |
| Capacity check | Sload × growth = 9,600.0 kVA ≤ Sderated → 9,600.0 vs limit 10,000.0 kVA PASS |
| Winding temperature | T = 40 + 65 = 105 °C ≤ class F limit (margin 50 K) → 105 vs limit 155 °C PASS |
| Result | 2/2 checks pass → PASS; 10,000 kVA, Z 9.5 %, Isc 5.52 kA; governing: capacity |
| Reference | Hand 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 buildercomponents/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 reportcomponents/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 kVASTANDARD_VOLTAGES— dict grouping LV, MV, and HV standard voltagesVECTOR_GROUPS— dict mapping vector group codes to descriptions and common usesCOOLING_TYPES— dict mapping cooling designations to type and forced-cooling flagCONSTRUCTION_TYPES— dict mapping construction types to compatible cooling optionsTAP_CHANGER_TYPES— dict mapping tap changer types to typical step countsINSULATION_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:
- To add new construction types, update
CONSTRUCTION_TYPESincomponent.pyand add compatible cooling options. - To add new vector groups, update
VECTOR_GROUPSwith description and common use. - For manufacturer-specific loss data, create additional lookup
tables in
calculations.pykeyed by manufacturer and efficiency tier. - To refine impedance estimation, add rating-specific Z% tables from standards or manufacturer catalogues.
- 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.