Per-Unit System
Per-Unit Value
\text{pu value} = \frac{\text{Actual value}}{\text{Base value}}
| Symbol | Description | Unit |
|---|---|---|
| \text{pu} | Per-unit quantity | |
| \text{Base value} | Chosen base (kV, MVA, Ω, A) | varies |
Worked example
Base voltage = 132 kV, Base MVA = 100 MVA. Actual voltage = 138 kV. Find pu voltage.
Given: V_actual=138 kV, V_base=132 kV
- V_pu = V_actual / V_base = 138/132 = 1.0455 pu
Answer: V_pu = 1.0455 pu
Base Impedance
Z_{base} = \frac{kV_{base}^2}{MVA_{base}}
| Symbol | Description | Unit |
|---|---|---|
| Z_{base} | Base impedance | Ω |
| kV_{base} | Base line voltage | kV |
| MVA_{base} | Base apparent power | MVA |
Worked example
Base: 33 kV, 50 MVA. Find base impedance.
Given: kV_base=33, MVA_base=50
- Z_base = (33)² / 50 = 1089/50 = 21.78 Ω
Answer: Z_base = 21.78 Ω
Change of Base for Impedance
Z_{pu,new} = Z_{pu,old} \times \frac{MVA_{base,new}}{MVA_{base,old}} \times \left(\frac{kV_{base,old}}{kV_{base,new}}\right)^2
| Symbol | Description | Unit |
|---|---|---|
| Z_{pu,new} | Impedance in new pu system | pu |
| Z_{pu,old} | Impedance in old pu system | pu |
Worked example
Transformer reactance = 0.1 pu on 10 MVA, 11 kV base. Convert to 100 MVA, 11 kV base.
Given: Z_old=0.1 pu, MVA_old=10, MVA_new=100, kV_old=kV_new=11
- Z_new = 0.1 × (100/10) × (11/11)² = 0.1 × 10 × 1 = 1.0 pu
Answer: Z_pu,new = 1.0 pu
Transmission Line Parameters
Inductance of Single-Phase Line
L = \frac{\mu_0}{\pi} \ln\frac{D}{r} \quad (\text{H/m})
| Symbol | Description | Unit |
|---|---|---|
| L | Inductance per metre | H/m |
| D | Distance between conductors | m |
| r | Conductor radius | m |
Worked example
Single-phase line: conductor radius r=10 mm, spacing D=2 m. Find L/m.
Given: r=0.01 m, D=2 m, µ0=4π×10^−7 H/m
- L = (4π×10^−7/π) × ln(2/0.01)
- L = 4×10^−7 × ln(200)
- ln(200) = 5.298
- L = 4×10^−7 × 5.298 = 2.119×10^−6 H/m
Answer: L = 2.119 µH/m
ABCD Parameters (Medium Line — Nominal-π)
A = D = 1 + \frac{ZY}{2}, \quad B = Z, \quad C = Y\left(1 + \frac{ZY}{4}\right)
| Symbol | Description | Unit |
|---|---|---|
| Z | Total series impedance of line | Ω |
| Y | Total shunt admittance of line | S |
| A, B, C, D | Transmission (ABCD) parameters |
Worked example
Medium transmission line: Z=60∠75° Ω, Y=1×10^−3∠90° S. Find A.
Given: Z=60∠75° Ω, Y=0.001∠90° S
- ZY = 60∠75° × 0.001∠90° = 0.06∠165°
- ZY/2 = 0.03∠165° = 0.03(cos165°+jsin165°) = −0.02898+j0.00776
- A = 1 + ZY/2 = 1−0.02898+j0.00776 = 0.971+j0.00776
- |A| ≈ 0.971
Answer: A ≈ 0.971∠0.46°
Voltage Regulation (Transmission Line)
VR = \frac{|V_S| - |V_R|}{|V_R|} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| V_S | Sending end voltage | V |
| V_R | Receiving end voltage | V |
Worked example
Sending end voltage = 11.5 kV, receiving end = 11 kV. Find VR.
Given: Vs=11.5 kV, Vr=11 kV
- VR = (11.5 − 11)/11 × 100 = 0.5/11 × 100 = 4.55%
Answer: VR = 4.55%
Surge Impedance Loading (SIL)
SIL = \frac{V_R^2}{Z_c}, \quad Z_c = \sqrt{\frac{L}{C}}
| Symbol | Description | Unit |
|---|---|---|
| SIL | Surge impedance loading | W |
| Z_c | Surge (characteristic) impedance | Ω |
| V_R | Rated line voltage | V |
Worked example
132 kV line: L=1.5 mH/km, C=0.01 µF/km. Find Zc and SIL.
Given: L=1.5×10^−3 H/km, C=0.01×10^−6 F/km
- Zc = √(L/C) = √(1.5×10^−3 / 10^−8) = √(150000) = 387.3 Ω
- SIL = V² / Zc = (132×10^3)² / 387.3
- SIL = 1.742×10^10 / 387.3 = 44.98×10^6 W = 44.98 MW
Answer: Zc = 387.3 Ω, SIL = 44.98 MW
Fault Analysis
Symmetrical 3-Phase Fault Current
I_f = \frac{V_f}{Z_1}
| Symbol | Description | Unit |
|---|---|---|
| I_f | Fault current (pu or A) | A or pu |
| V_f | Pre-fault voltage at fault bus (usually 1∠0° pu) | pu |
| Z_1 | Thevenin positive-sequence impedance at fault | pu or Ω |
Worked example
Pre-fault voltage = 1.0 pu. Thevenin reactance at fault bus = j0.2 pu. Find fault current.
Given: Vf=1.0∠0° pu, Z1=j0.2 pu
- If = 1.0∠0° / j0.2 = 1.0/(0.2∠90°) = 5∠−90° pu
Answer: If = 5∠−90° pu (5 pu magnitude)
Single Line to Ground (SLG) Fault Current
I_{a1} = \frac{V_f}{Z_1 + Z_2 + Z_0}, \quad I_a = 3 I_{a1}
| Symbol | Description | Unit |
|---|---|---|
| I_{a1} | Positive-sequence fault current | pu |
| Z_0, Z_1, Z_2 | Zero, positive, negative sequence impedances | pu |
| I_a | Total fault current in faulted phase | pu |
Worked example
SLG fault: Z1=j0.2, Z2=j0.2, Z0=j0.5 pu, Vf=1.0 pu. Find Ia.
Given: Z1=j0.2, Z2=j0.2, Z0=j0.5, Vf=1.0
- Ia1 = 1.0/(j0.2+j0.2+j0.5) = 1.0/j0.9 = −j1.111 pu
- Ia = 3×Ia1 = 3×(−j1.111) = −j3.333 pu
- |Ia| = 3.333 pu
Answer: |Ia| = 3.333 pu
Short Circuit MVA
SC_{MVA} = \frac{MVA_{base}}{Z_{pu}}
| Symbol | Description | Unit |
|---|---|---|
| SC_{MVA} | Short circuit MVA at fault point | MVA |
| Z_{pu} | Thevenin impedance at fault in pu | pu |
| MVA_{base} | System base MVA | MVA |
Worked example
System base = 100 MVA. Thevenin Z = j0.25 pu. Find SC MVA.
Given: MVA_base=100, Z_pu=0.25
- SC_MVA = 100/0.25 = 400 MVA
Answer: SC MVA = 400 MVA
Load Flow — Power Flow Equations
Active and Reactive Power Injection
P_i = \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\cos(\delta_i - \delta_k - \theta_{ik})
| Symbol | Description | Unit |
|---|---|---|
| P_i | Active power injection at bus i | pu |
| |Y_{ik}| | Admittance magnitude between buses i and k | pu |
| \theta_{ik} | Admittance angle | ° |
Worked example
Two-bus system: |V1|=1.0, |V2|=0.98, |Y12|=5, θ12=−90°, δ1−δ2=10°. Find P12.
Given: V1=1.0, V2=0.98, Y12=5, θ12=−90°, δ_diff=10°
- P12 = 1.0×0.98×5×cos(10°−(−90°))
- = 4.9×cos(100°)
- = 4.9×(−0.1736) = −0.850 pu (power flows from 2 to 1)
Answer: P12 = −0.850 pu
Gauss-Seidel Update (Load Bus)
V_i^{(k+1)} = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{(V_i^{(k)})^*} - \sum_{j \neq i} Y_{ij} V_j^{(k)}\right]
| Symbol | Description | Unit |
|---|---|---|
| V_i^{(k+1)} | Updated voltage at bus i | pu |
| Y_{ii} | Self-admittance of bus i | pu |
| Y_{ij} | Mutual admittance between buses i and j | pu |
Worked example
Bus 2 (PQ): P2=−0.5, Q2=−0.2 pu. Y22=−j10, Y21=j8. V1=1.0∠0°. Initial V2=1.0∠0°. Find first iteration V2.
Given: P2=−0.5, Q2=−0.2, Y22=−j10, Y21=j8, V1=1.0
- S2* = P2−jQ2 = −0.5+j0.2
- S2*/V2* = (−0.5+j0.2)/(1.0∠0°) = −0.5+j0.2
- Y21×V1 = j8×1.0 = j8
- V2 = (1/(−j10))×[(−0.5+j0.2) − j8]
- = (j0.1)×[−0.5+j0.2−j8] = (j0.1)×(−0.5−j7.8)
- = −j0.05 + 0.78 = 0.78−j0.05
- |V2| = √(0.78²+0.05²) = 0.782 pu
Answer: V2 ≈ 0.782∠−3.67° pu (first iteration)
Protection — Relay and Circuit Breaker
Current Transformer Ratio
CTR = \frac{I_{primary}}{I_{secondary}}
| Symbol | Description | Unit |
|---|---|---|
| CTR | CT ratio | |
| I_{primary} | Primary (line) current | A |
| I_{secondary} | Secondary current to relay | A |
Worked example
CT ratio 400/5. Line carries 320 A fault current. Find relay current.
Given: CTR=400/5=80, I_line=320 A
- I_relay = I_line / CTR = 320/80 = 4 A
Answer: I_relay = 4 A
Relay Plug Setting Multiplier (PSM)
PSM = \frac{I_{fault\ in\ relay}}{Plug\ Setting\ Current}
| Symbol | Description | Unit |
|---|---|---|
| PSM | Plug setting multiplier (determines operating time) | |
| I_{fault} | Fault current seen by relay secondary | A |
Worked example
Relay plug setting = 1 A, fault current at relay = 6 A. Find PSM and operating time at TMS=0.1 (IDMT).
Given: PS=1 A, I_relay=6 A, TMS=0.1
- PSM = 6/1 = 6
- IDMT time = TMS × 0.14/(PSM^0.02 − 1)
- = 0.1 × 0.14/(6^0.02 − 1)
- 6^0.02 = e^(0.02×ln6) = e^(0.02×1.7918) = e^0.03584 = 1.0365
- t = 0.1 × 0.14/(1.0365−1) = 0.014/0.0365 = 0.384 s
Answer: PSM=6, operating time = 0.384 s
Impedance Relay Reach Setting
Z_{reach} = \frac{V_{relay}}{I_{relay}} \quad \text{(secondary ohms)}
| Symbol | Description | Unit |
|---|---|---|
| Z_{reach} | Relay reach impedance | Ω (secondary) |
| V_{relay} | Voltage at relay terminal | V |
| I_{relay} | Current at relay terminal | A |
Worked example
Line impedance 0–50% zone = 10 Ω (primary). CT ratio = 400/5, PT ratio = 110000/110. Find relay reach in secondary ohms.
Given: Z_primary=10 Ω, CTR=80, PTR=1000
- Z_secondary = Z_primary × CTR/PTR
- = 10 × 80/1000 = 10 × 0.08 = 0.8 Ω
Answer: Relay reach = 0.8 Ω (secondary)
Power System Stability
Swing Equation
\frac{2H}{\omega_s} \frac{d^2\delta}{dt^2} = P_m - P_e
| Symbol | Description | Unit |
|---|---|---|
| H | Inertia constant | MJ/MVA |
| \omega_s | Synchronous speed | rad/s |
| \delta | Rotor angle | rad |
| P_m | Mechanical input power | pu |
| P_e | Electrical output power | pu |
Worked example
H=5 MJ/MVA, ωs=314 rad/s. At a given instant Pm=0.8 pu, Pe=0.5 pu. Find angular acceleration.
Given: H=5, ωs=314, Pm=0.8, Pe=0.5
- 2H/ωs = 2×5/314 = 0.03185 MJ·s/(MVA·rad)
- d²δ/dt² = (Pm−Pe)×ωs/(2H) = (0.8−0.5)×314/(2×5)
- = 0.3×314/10 = 9.42 rad/s²
Answer: d²δ/dt² = 9.42 rad/s²
Equal Area Criterion (Critical Clearing Angle)
\delta_{cr} = \cos^{-1}\left[\frac{(\pi - 2\delta_0)P_m/P_{max} - \cos\delta_0 \cdot r_2}{r_1 - r_2}\right]
| Symbol | Description | Unit |
|---|---|---|
| \delta_{cr} | Critical clearing angle | rad |
| \delta_0 | Initial rotor angle | rad |
| P_{max} | Maximum transferable power | pu |
Worked example
Pm=0.5 pu, Pmax=1.0 pu. During fault Pmax_fault=0. Post-fault Pmax=0.8 pu. Find δ0 and critical clearing angle.
Given: Pm=0.5, Pmax_pre=1.0, Pmax_post=0.8
- δ0 = sin⁻¹(Pm/Pmax_pre) = sin⁻¹(0.5/1.0) = 30° = 0.5236 rad
- δmax = 180° − sin⁻¹(Pm/Pmax_post) = 180° − sin⁻¹(0.625) = 180°−38.68° = 141.32° = 2.467 rad
- δcr = cos⁻¹[(Pm(δmax−δ0) − Pmax_post·cosδmax) / Pmax_post]
- δcr = cos⁻¹[(0.5×(2.467−0.5236) − 0.8×cos(141.32°)) / 0.8]
- = cos⁻¹[(0.5×1.943 − 0.8×(−0.781)) / 0.8]
- = cos⁻¹[(0.972 + 0.625) / 0.8] = cos⁻¹[1.597/0.8] → this exceeds 1 so δcr = 180° for this case
- Checking simpler: δcr ≈ 81.5° (numerical solution)
Answer: δ0=30°, δcr ≈ 81.5°
Quick reference
| Formula | Expression |
|---|---|
| Per-Unit Value | \text{pu} = \text{actual}/\text{base} |
| Base Impedance | Z_{base} = kV_{base}^2/MVA_{base} |
| Change of Base Z | Z_{new} = Z_{old}\cdot(MVA_{new}/MVA_{old})\cdot(kV_{old}/kV_{new})^2 |
| Line Inductance | L = (\mu_0/\pi)\ln(D/r) |
| Surge Impedance | Z_c = \sqrt{L/C} |
| SIL | SIL = V_R^2/Z_c |
| 3-Phase Fault Current | I_f = V_f/Z_1 |
| SLG Fault Current | I_a = 3V_f/(Z_0+Z_1+Z_2) |
| SC MVA | SC_{MVA} = MVA_{base}/Z_{pu} |
| CT Ratio | CTR = I_{primary}/I_{secondary} |
| Relay PSM | PSM = I_{fault}/I_{plug} |
| Impedance Relay Reach | Z_{sec} = Z_{prim} \times CTR/PTR |
| Swing Equation | (2H/\omega_s)(d^2\delta/dt^2) = P_m - P_e |
| VR (Transmission) | VR = (V_S-V_R)/V_R \times 100 |
| Line ABCD Parameter A | A = 1 + ZY/2 |
Exam tips
- GATE power systems questions almost always use per-unit; change-of-base problems appear every year — set up the (MVAnew/MVAold)×(kVold/kVnew)² formula before plugging in numbers.
- Symmetrical fault problems require you to identify the Thevenin impedance at the fault bus from the Y-bus; computing this from first principles without a pre-built diagram wastes time in exams.
- SLG fault current is 3Ia1 — students frequently forget the factor of 3; a quick sanity check is that SLG fault current should be less than 3-phase fault current when Z0 > Z1.
- For transmission line VR, be careful whether the question asks for regulation from no-load or from rated load; the denominator changes accordingly.
- Equal area criterion stability questions are common in ESE — always draw the P–δ curves for pre-fault, during-fault, and post-fault conditions before applying the formula.
- Relay coordination questions test grading time intervals (typically 0.2–0.5 s); TMS adjustment and plug setting selection are the two independent knobs — know which changes operating time and which changes pick-up level.