Power Grids Overview
- 3-phase AC, 50 Hz (India / most of world) with a few HVDC links
- High voltage minimizes I²R losses for long distance bulk power transfer
- Frequency: 50 Hz (India/Europe), 60 Hz (USA). Japan has both!
- SIL (Surge Impedance Loading) = V²/Zc determines practical line capacity
Meshed vs Radial
Radial (Distribution)
- Single path source→load
- Easy protection (overcurrent)
- Less reliable; one break = outage
- Lower fault currents
Meshed (Transmission)
- Multiple parallel paths
- Harder protection (distance/differential)
- More reliable (redundancy)
- Higher fault currents
In meshed AC grids, power flow depends on impedances and cannot be controlled locally. To change flow:
- Change generation dispatch
- Change network topology (open/close breakers)
- Use Phase Shifting Transformer or FACTS devices
This is a major motivation for HVDC — DC power flow is fully controllable.
Protection in Meshed AC Systems
- Detect faults reliably
- Trip quickly before equipment/personnel affected (~100 ms CB opening)
- Selectivity: trip only the affected zone
- Backup: if primary protection fails, backup trips
- Single-pole or three-phase tripping
- In AC systems, arc interrupts at current zero crossing
Transformers in the Grid
- Auto-transformers for close voltage ratios (220/400, 132/220 kV) — no electrical isolation, but smaller/cheaper
- Weight ratio vs 2-winding: (1 - N2/N1). Max advantage when voltages are close.
- Paralleled transformers must have same phase shift (e.g., both Yd11)
- Y→Δ transformer: secondary voltages are -30° out of phase with primary
3-Phase Systems & Per-Unit
Three-Phase Fundamentals
- 3 windings displaced by 120° → voltages: va, vb (lag 120°), vc (lag 240°)
- Algebraic sum of balanced 3-phase voltages/currents = 0 at every instant
- Total instantaneous power = constant (no pulsation!)
- Phase sequence a-b-c: determines motor rotation direction
- VLL = √3 · Vphase
- Iline = Iphase
- Neutral point available
- Used for transmission
- VLL = Vphase
- Iline = √3 · Iphase
- No neutral wire
- Triple harmonics circulate in loop
For balanced systems, analyze one phase and multiply power by 3.
Per-Unit System
Transformer p.u. impedance is the same on both sides if voltage bases follow turns ratio. The winding disappears in p.u. equivalent circuits!
Advantages of Per-Unit
- P.u. values are similar for apparatus of same type regardless of size
- Eliminates √3 factors in 3-phase calculations
- Networks with multiple voltage levels solved easily (transformers vanish)
- Convenient for digital computation
Harmonics in 3-Phase
- Fundamental (n=1,7,13...): positive sequence rotation
- 2nd harmonic (n=2,5,8,11...): negative (reverse) sequence rotation
- 3rd harmonic (n=3,6,9...): zero sequence — all in phase, circulate in Δ loops
- Line-to-line voltages have no triple harmonics (they cancel)
Maxwell's Equations & TEM Mode
Maxwell's Equations
- Faraday: curl E ≠ 0. Changing B induces circulating E (voltage!)
- Displacement current: ε0 ∂E/∂t. Changing E creates B even without current.
- Result: electromagnetic waves can propagate through space
Two-Wire Transmission Line (Static)
- Two parallel conductors: +ρ and -ρ line charge density
- E between conductors (radial), B loops around conductors
- No E or B in z-direction → 2D Laplace equation
- Equipotential surfaces are circles (not centered on conductors)
TEM Mode
- TEM = Transverse Electromagnetic — Ez = Bz = 0
- Valid when λ >> conductor spacing (50 Hz: λ = 6000 km!)
- Allows static formulas to be used for time-varying signals
- Requires at least 2 conductors (cannot exist in a hollow waveguide)
Q: How is there a voltage drop V(z+Δz) - V(z) in the z-direction when there's no E field in that direction (TEM)?
A: Use Faraday's law: ∮ E·dl = -d/dt ∫ B·dS
- E field can be expressed as gradient of potential only if curl E = 0
- In the x-y plane, curl E = 0 (no B passing through loops in x-y plane in TEM mode) → E = -∇V works in x-y
- But in any loop in the y-z plane, there IS changing B passing through it
- So there CAN be a voltage drop [V(z+Δz) - V(z)] in the z-direction without having Ez
- The voltage drop comes from Faraday's law (changing magnetic flux), not from an electric field in z
From Fields to Circuit Parameters
If spacing D is doubled:
- C' marginally decreases (C' ∝ 1/ln(D/r), logarithmic)
- L' marginally increases
- Zc = √(L'/C') increases
Bundle Conductors & Corona
- EHV lines use bundled conductors (2, 3, or 4 sub-conductors per phase)
- Purpose: reduce corona (main reason) by lowering surface E-field
- Side effect: also reduces inductance, increases capacitance
- Maximum E field in a triangular bundle is at the outer vertex (superposition of fields from all sub-conductors)
When voltage on a line becomes very high → E field exceeds a critical value → air breaks down → air molecules become ionized → partial discharge around the conductor = corona
- E field is stronger at sharp edges and points (field concentration)
- Bundling creates a smoother, larger effective radius → less field concentration → less corona
- Corona causes power loss, radio interference, and audible noise
Ground Wire (Shield Wire)
- Top wire on transmission tower = ground conductor
- Purpose: prevents phase conductors from direct lightning hit
- Wire has some potential w.r.t. ground (not exactly zero)
- Any metallic object near HV lines acquires a potential due to capacitive coupling
E and B Field Patterns
- B field around two-conductor line (opposite currents): individual loops around each conductor (like two concentric circles, one CW one CCW)
- E field around two-conductor bundle (same potential): radial outward from both conductors (divergent pattern)
- 3-phase line at point P: both magnitude AND direction of E vary with time (superposition of 3 time-varying fields)
Transmission Line Models
Telegrapher's Equations
Boundary Conditions
No reflection. f2=0.
Full reflection. V doubles.
Sign flip. I doubles.
Lossy Line
Sinusoidal Steady State (SSS)
ABCD Matrix
Property: AD - BC = 1. For lossless: cos²(βl) + sin²(βl) = 1.
Cascading ABCD Matrices
Two networks in series: [ABCD]total = [ABCD]1 × [ABCD]2
For parallel networks: special formulas (A0=(A1B2+A2B1)/(B1+B2), etc.)
Equivalent Pi Circuit
Line Length Approximations
| Type | Length | Model | ABCD |
|---|---|---|---|
| Short | <80 km | Series Z only | [[1, Z], [0, 1]] |
| Medium | 80-250 km | Nominal Π | [[1+YZ/2, Z], [Y(1+YZ/4), 1+YZ/2]] |
| Long | >250 km | Exact | cosh/sinh |
OTL vs Cables
| Parameter | Overhead (230 kV) | Cable (230 kV) | Ratio |
|---|---|---|---|
| xl (Ω/km) | 0.488 | 0.339 | ~1.4x |
| bc (μS/km) | 3.371 | 245.6 | ~73x |
| Zc (Ω) | 380 | 37.1 | ~10x |
| SIL (MW) | 140 | 1426 | ~10x |
Cables generate massive reactive power under typical loading (high bc). At 30-40 km, the charging current alone can approach the thermal limit. Need shunt reactors to compensate.
Solution: Compensate with lumped shunt reactors. For long distances, use HVDC cables (no reactive power issue with DC).
| Property | Overhead Lines | Cables |
|---|---|---|
| Zc | 230-400 Ω | 25-50 Ω |
| bc | Baseline | 50-100x more reactive power |
| SIL | Lower | Much higher (cable SIL >> OTL SIL) |
| X/R ratio | 10-12 | 2-9 |
| α, β | Lower | Higher (more lossy/km, dielectric losses) |
| Structure | Open air | Coaxial: 2 conductors close together, dielectric between |
| Max AC length | Hundreds of km | 30-40 km (charging current limit) |
3-Phase Line: Mutual Coupling & Impedance Matrix
Key simplification (balanced system, Ia+Ib+Ic=0):
This is why we can treat a balanced 3-phase system as a single-phase equivalent!
Matched Load (R = Zc): Equivalent Circuit View
- When terminated with R = Zc: Vk = Vm ejβl (just a phase shift, no magnitude change)
- Sending and receiving voltages are equal in magnitude
- Split line at midpoint: Vmid = Vk e-jβl/2 = Vm ejβl/2
- Equivalent circuit seen from receiving end: source Vk/cos(βl) in series with jZctan(βl)
Power Flow Through Lossless Line
The equivalent circuit (V/cosβl source + jZctanβl) is a Thevenin equivalent from the receiving end. It lumps all the distributed shunt capacitance into the series element.
- P: Works because Psending = Preceiving (lossless line, active power is conserved)
- Q: Does NOT work because the line itself generates Q (capacitance) and absorbs Q (inductance), so Qs ≠ Qr
For Q, you MUST use the full ABCD matrix:
Notice: Qs = -Qr only when δ = βl (SIL loading). At any other loading, the line has a net Q exchange with the system.
- δ < βl (below SIL): cosδ > cosβl → Qs > 0, Qr < 0 → line generates Q (capacitive dominates)
- δ > βl (above SIL): cosδ < cosβl → Qs < 0, Qr > 0 → line absorbs Q (inductive dominates)
Voltage Profile Concepts
- SIL loading: I²XL = V²BC → flat voltage profile
- Below SIL (light load): line capacitance dominates → voltage rises (Ferranti effect)
- Above SIL (heavy load): line inductance dominates → voltage drops
- At no-load with both ends at 1 p.u., midpoint voltage = 1/cos(βl/2) > 1 p.u.
Active & Reactive Power
Instantaneous Power (Single Phase)
Average power. Does real work.
Peak of oscillating energy exchange. No net work.
Complex Power
Convention (IEC): S = EI* → inductive load absorbs +Q, capacitive load absorbs -Q (generates +Q).
Reactive Power Sources & Sinks
| Component | Q Behavior |
|---|---|
| Inductor / Induction motor | Absorbs Q (Q>0) |
| Capacitor | Generates Q (absorbs Q<0) |
| Over-excited synchronous gen | Generates Q |
| Under-excited synchronous gen | Absorbs Q |
| Lightly loaded OTL | Generates Q (capacitive) |
| Heavily loaded OTL | Absorbs Q (inductive) |
| Cables (always) | Generates Q (high capacitance) |
Power Transfer Through a Line
- P controlled by angle difference δ
- Q controlled by voltage magnitudes
- Voltage drop ΔV ≈ (RP + XQ)/VL (approximate, for small δ)
- For transmission (R<<X): ΔV ≈ XQ/V — voltage depends mainly on Q!
Tellegen's Theorem
Works for ANY circuit: linear, nonlinear, time-varying. At each node: ΣP = 0, ΣQ = 0.
Power System Components
Synchronous Machines
- f = np/60 (n=rpm, p=pole pairs). 50 Hz, 2-pole: 3000 rpm (turbo). 20-pole: 150 rpm (hydro).
- Equivalent circuit: E = Vt + I(RL + jXs)
- Xs = synchronous reactance = Xm + XL (magnetizing + leakage)
- Over-excited: E > V, generates Q (capacitive to system)
- Under-excited: E < V, absorbs Q (inductive to system)
Short-Circuit Ratio (SCR)
- SCR = (field current for rated V on OC) / (field current for rated I on SC)
- SCR ≈ 1/Xs(p.u.) approximately
- Low SCR (0.55) = high Xs = cheaper but less stable
- Design trade-off: economy vs stability
Four-Terminal (ABCD) Networks
Review Questions (Part B) — MCQ Practice
Click any question to reveal the solution. These are from the Moodle quiz bank.
At point Y (outer vertex), E fields from all 3 sub-conductors point roughly in the same outward direction, so they add up (constructive superposition) → maximum E.
At point X (center), fields from all 3 cancel out. At W (between two conductors), fields partially cancel.
Why 10-15? At 400 kV (EHV), conductors are thick aluminium with very low resistance. But the spacing between phases is large (several metres), which gives significant inductance.
From Table 3.2a (Weedy): 400 kV, 4×258mm² ACSR:
Compare with distribution (11 kV): thinner wires, closer spacing → X/R ≈ 1-2
Rule of thumb: Higher voltage → thicker conductor (lower R) → higher X/R
Method: Use Zc = √(xl/bc) to check which option is consistent.
Check Option 1: R=20.8, xl=24, bc=256.375 μS
Check Option 2: R=6.4, xl=73.6, bc=800 μS
Both options give Zc ≈ 300 Ω, but only Option 2 has a realistic X/R ratio for a 400 kV line. This is why you need to know typical X/R values (Q6)!
Magnitude is constant (|Vs| = |Vr|) but phase shift = βl, which depends on length.
Logarithmic dependence → only a small (marginal) decrease.
Step 1: Find the electrical length of the line:
Step 2: At no-load, Ir = 0. Use ABCD from sending end to midpoint (half the line, length l/2):
Step 3: By symmetry (both ends at same voltage, no load), the midpoint is where I = 0 and V is maximum. So at midpoint Imid = 0:
Why does voltage rise? At no-load, no current flows at the ends, but the line's distributed capacitance still generates reactive power (charging current). This Q has nowhere to go → it pushes the voltage up in the middle. This is the Ferranti effect.
Shape: Voltage profile is a bowl (concave up) — 1.0 p.u. at both ends, peaks at 1.035 at midpoint.
This is the line from Review Q4 & Q5 (600 km, 400 kV). βl = 1.05×10-3 × 600 = 0.63 rad ≈ 36°. Also: Zc = √(L'/C') = √(10-3/11.1×10-9) ≈ 300 Ω.
Open circuit: Ir = 0. From ABCD: Vs = cos(βl) · Vr
|Vr| = |Vs| / cos(0.39) = 1.0 / 0.9247 = 1.082 p.u.
Q16 had 400 km with both ends connected at 1 p.u. → midpoint rose to 1.035 (Ferranti). Here the line is open at Vr, so the full rise appears at the receiving end.
At SIL (P = V²/Zc, δ = βl): distributed capacitive Q and inductive Q exactly cancel → flat voltage profile, zero net Q.
Below SIL: capacitance dominates → line generates Q → Ferranti-like voltage rise.
Above SIL: inductance dominates → line absorbs Q → voltage sags.
Parallel lines: series impedance per unit length halves (z' = z/2), shunt admittance per unit length doubles (y' = 2y).
Contrast with Q18 (series): series connection leaves Zc unchanged. Parallel halves it.
VLL is the line-to-line voltage. This 3-phase formula works because: 3 × (VLL/√3)²/Zc = VLL²/Zc. At SIL, voltage profile is flat and Q exchange is zero.
Logarithmic dependence: marginal increase. Compare with Q15 (capacitance): C' = πε0/ln(D/r) → marginally decreases when D doubles. The two effects are complementary: more separation → more flux linkage (L↑), weaker electric coupling (C↓). Note: Zc = √(L'/C') increases as D increases.
Short line: shunt capacitance C' is negligible → A = D = 1, B = Ztotal = R + jX, C = 0.
As line length increases: use nominal π (medium line, add C as lumped shunt halves at each end). For >300 km: use exact distributed-parameter ABCD with cosh/sinh.
This is a cable — very high bc (capacitance) compared to OTL gives a much lower Zc. Compare with OTL (Q10): Zc ≈ 230–400 Ω vs cable ≈ 30–50 Ω.
Formula: for a lossless line, Zc = √(L'/C') = √(ωL'/ωC') = √(xl/bc) where xl and bc are total for the line.
With R=0: ΔV = XQ/V. If Q < 0 (capacitive load injects reactive power) → ΔV < 0 → Vr > Vs.
This is consistent with Ferranti: distributed capacitance of a lightly loaded line acts as a Q source (Q < 0 from the line's perspective) → voltage rises at receiving end.
Rule: Reactive power injection raises voltage; absorption lowers it. (ΔV ≈ XQ/V for inductive lines)
Nominal-π model: shunt Y/2 at each end, full Z in series. Applying KVL/KCL:
Nominal-T model (shunt Y in middle, Z/2 each side) gives same A=D but swapped B and C. In practice, π is preferred for medium lines. Reducing to Y=0 recovers the short-line A=D=1, B=Z, C=0.
Z = total series impedance of line, Y = total shunt admittance. For a lossless line (R=G=0): √(ZY) = jβl, so cosh(jβl) = cos(βl) and sinh(jβl) = j sin(βl). The medium-line expression 1+ZY/2 is just the first two terms of the cosh series.
Regulation measures how much the receiving-end voltage rises when load is removed, relative to full-load voltage:
For a capacitive / leading-p.f. load VR can exceed VS, giving negative regulation. Short line with R=0 and purely capacitive load: VR(NL) < VR(FL) → VR < 0.
Zc depends only on line geometry (roughly constant for a given conductor type), so SIL ∝ V². This is why upgrading from 275 kV to 400 kV raises the natural load by ≈ (400/275)² ≈ 2.1×. It also explains why high-voltage transmission is economically advantageous for bulk power transfer.
Bundling several subconductors increases the effective conductor radius req:
- Inductance: L' = (μ0/2π) ln(Deq/req) — larger req → smaller L'
- Capacitance: C' = 2πε0/ln(Deq/req) — larger req → larger C'
From Table 3.2a: 400 kV, 1×258 mm² → Zc=296 Ω, SIL=540 MW; 4×258 mm² → Zc=258 Ω, SIL=620 MW. Bundling also reduces electric-field gradients, suppressing corona.
Review Questions (5A) — Problem Strategies
Given: Lossless line with ABCD = [[cosβl, jZcsinβl], [j(1/Zc)sinβl, cosβl]]
Part (a): Inverse ABCD
Part (b): Open circuit (Ir = 0)
Part (c): Terminated by R
Vr = R·Ir. Substitute into ABCD. |Vr| depends on R/Zc ratio.
Part (d,e): Shunt L or C for |Vr| = |Vs|
Add shunt admittance Ycomp at receiving end. Set |Vr| = |Vs| and solve for Ycomp.
Setup: Vs = V∠δ, Vr = V∠0. Express V(d), I(d) using ABCD from receiving end.
At SIL: voltage profile is flat. Below SIL: midpoint voltage > V (Ferranti).
Method: Cascade ABCD: [ABCD]total = [ABCD]cable × [ABCD]OTL
OTL terminated at Zc,OTL → Vr = Zc,OTL·Ir. Cable and OTL have very different Zc values (37 vs 380 Ω).
Key insight: there will be significant reactive power exchange at the cable-OTL junction due to impedance mismatch.
Setup: Both ends at 400 kV, β = 0.0013 rad/km. θ = βl = 0.78 rad ≈ 45°
Method: Split into two 300 km halves. Each half has ABCD with βl/2.
At midpoint: KCL with shunt susceptance Bc. Set Vmid = 400 kV.
Series capacitor at midpoint: Cancels part of inductive reactance.
With 50% series compensation, Xeff halves, so Pmax doubles.
Similar to Q3, but with shunt reactive element Qc at cable-OTL junction to maintain Vjunction = 230 kV.
Method:
- Find V,I at junction from OTL side (ABCD with load condition)
- Find V,I at junction from cable side (ABCD with source condition)
- Apply KCL at junction: Icable = IOTL + Icomp
- Determine if compensator is L or C based on sign of Qc
Textbook Worked Examples & Practice
These are directly relevant to the review questions. Click to expand solutions.
Problem: 11 kV generator → 11/132 kV transformer (50 MVA, X=10%) → 132 kV line (j100 Ω) → 132/33 kV transformer (50 MVA, X=12%) → 50 MW load at 0.8 p.f. lag. Vload = 30 kV. Find generator terminal voltage.
Key takeaway: Transformers vanish in p.u. circuit! Just add all reactances in series. Vbase must follow turns ratio.
Problem: 275 kV line, 160 km, R=0.034 Ω/km, X=0.32 Ω/km. Load: 600 MW, 300 MVAr. Compare accurate and approximate methods.
Lesson: Approximate formula ΔV ≈ XQ/V works well for short lines / small angles. For heavily loaded long lines (δ > 15°), use the accurate formula.
Problem: 3.3 kV OTL, 1.6 km, horizontal formation, conductor spacing 762 mm, diameter 3.5 mm, R=0.41 Ω/km. Find inductance. Then find Vload for 1 MW at 0.8 p.f. lag.
Problem: 400 kV, 150 km, quad conductors. R=0.017 Ω/km, XL=0.27 Ω/km, bc=10.58 μS/km. Load: 1800 MW at 0.9 lag. Find Vs using all 3 models.
Conclusion: Short-line model gives 530 kV (too high!), while medium and long models agree at 518 kV. For 150 km at 400 kV, the medium line model is adequate. Short-line model overpredicts Vs because it ignores shunt capacitance (which generates Q and supports voltage).
Note: 518 kV is too high for a 400 kV system (max ≈ 440 kV). Would need series capacitors or shunt compensation.
Problem: 60 MW, 75 MVA, 11.8 kV generator. SCR = 0.63 (so Xs = 1/0.63 = 1.59 p.u.). Find excitation E and load angle δ at full load (0.8 p.f. lag) and at unity p.f.
Key point: At unity p.f. the load angle is larger (50° vs 33°) because the machine needs less excitation → less margin to the 90° stability limit. Over-excited machines (0.8 lag) are more stable!
Relevant Textbook Practice Problems
Multiply the matrices: [ABCD]total = [ABCD]1 × [ABCD]2
This is the exact same technique needed for Q5A-Q3 (cable + OTL cascade)!
The approximate method works well for 16 km but starts to diverge at 80 km (1% error).
Power delivered to R: P = I²R where I depends on the impedance seen.
The load impedance Zload = jX2 || R. For maximum power transfer to R, take dP/dR = 0.
This is analogous to impedance matching. Similar concept to Q5A-Q5 where series capacitor changes effective X to maximize power transfer.
Quick Concept Checks (from Weedy Ch 3)
Typical values:
- 132 kV: Zc=150Ω, SIL=50 MW
- 275 kV: Zc=315Ω, SIL=240 MW
- 400 kV: Zc=295Ω, SIL=540 MW
- Cable Zc ≈ 1/10th of OTL
- Cable XL/R ≈ 2-9 (vs 10-15 for OTL)
- Cable C = 2πε0εr/ln(R/r)
- εr ≈ 2.3 (XLPE), so C is much higher than OTL
- Charging current: Ic = VωC per km
- At 30-40 km, Ic ≈ thermal rating!
For unequal spacing d12, d23, d31:
Conductors are rotated through all 3 positions along the line to equalize inductance. In practice, done at substations.
- Y-Y or Δ-Δ: No phase shift
- Y-Δ: Positive sequence voltages advanced by +30° on delta side
- Δ-Y: Positive sequence voltages delayed by -30° on Y side
- Transformers with different phase shifts CANNOT be paralleled
- Typical impedance: 10-20% for large transformers (>20 MVA)
- 10% reactance means short-circuit current = 10x rated
Master Formula Sheet
Transmission Line Core Formulas
Power Formulas
Per-Unit System
Key Numbers to Remember
You've got this!
Focus on ABCD matrices, P-Q flow, and the Part B MCQs.
EE238 Power Engineering II • IIT Bombay • Built from Weedy Ch 2-3, AMK slides, & review questions