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Distribution systems and substations

Electrical and Electronics Engineering (B.E./B.Tech) · Power systems · free study notes for exams, placements and interviews

In short

A distribution system is the last stage of the power system: it takes power from the transmission or sub-transmission network at a substation, steps it down in stages (220/132/66 kV → 33/11 kV → 433 V) and delivers it to industrial, commercial and domestic consumers. A substation is the node where this stepping down, switching, protection, metering and sometimes reactive compensation happens. The topic is examined in university Power Systems papers (distributor voltage-drop problems, Kelvin's law, bus-bar schemes, substation earthing) and appears in GATE through voltage-regulation and distributor loading numericals. It is the single most employable part of power systems in India: every DISCOM (BESCOM, MSEDCL, TPDDL, Adani, CESC), every EPC contractor (L&T, KEC, Kalpataru, Sterlite) and every consultancy hires for distribution planning, substation design and O&M.

Explained

Structure of the distribution network

Classification of distribution systems

DC and AC distributor calculations (the examined core)

Substations: classification and equipment

Substation earthing, layout and safety

Losses, economics and the Indian regulatory frame

Key points

Formulas and examples

Voltage drop in a DC section: ΔV = I·R, where I = section current (A), R = section resistance (Ω) = ρL/A for go-and-return as applicable
Uniformly loaded distributor, fed at one end: ΔV = i·r·l²/2 = (I_total · R_total)/2 — i = A/m, r = Ω/m, l = length (m)
Uniformly loaded distributor, fed at both ends (equal voltage): ΔV_max = i·r·l²/8 (at the mid-point)
AC distributor, drop per section (pf referred to receiving-end voltage): ΔV = I(R cos φ + X sin φ) — R, X in Ω; φ = load angle
Percentage voltage regulation = (V_send − V_receive)/V_receive × 100
Line loss (three-phase): P_loss = 3 I² R = (P² + Q²)R / V² = P² R /(V² cos²φ) — V = line voltage (V), R = per-phase resistance (Ω)
Capacitor kVAR for pf correction: Q_c = P (tan φ₁ − tan φ₂) — P in kW, φ₁ initial, φ₂ target angle
Kelvin's law: most economical area A when annual capital cost (P₁ + P₂A) differential equals annual energy loss cost (P₃/A); optimum at A = √(P₃/P₂)
AT&C loss (%) = [1 − (units billed/units input) × (revenue collected/revenue billed)] × 100
Transformer loading: kVA = √3 × V_L × I_L / 1000; % loading = load kVA / rated kVA × 100
Load factor = average load / maximum demand; Diversity factor = Σ individual max demands / max demand of the group

Example 1 - DC distributor fed at one end. A 2-wire DC distributor AD, 400 m long, is fed at A at 240 V. Loads: 50 A at 100 m (B), 40 A at 250 m (C), 30 A at 400 m (D). Resistance of the distributor is 0.0002 Ω per metre (go and return combined).

Example 2 - distributor fed at both ends. A 2-wire distributor AB is 300 m long, fed at A at 250 V and B at 250 V, resistance 0.0005 Ω/m (go + return). Loads: 60 A at 80 m from A (C) and 100 A at 200 m from A (D).

Example 3 - uniformly loaded distributor. A 400 m, 2-wire distributor is uniformly loaded at 0.5 A per metre, resistance 0.0003 Ω per metre (go + return), fed at 230 V.

Example 4 - AC distributor, single-phase. A single-phase distributor fed at A at 240 V supplies load C = 100 A at 0.8 pf lagging and load D = 60 A at 0.9 pf lagging (power factors referred to one common reference - the supply voltage at A, taken as reference). Section impedances: A→C = (0.1 + j0.15) Ω, C→D = (0.08 + j0.12) Ω.

Example 5 - distribution transformer loading and loss saving from pf correction. An 11 kV/433 V, 250 kVA distribution transformer feeds 180 kW at 0.75 pf lagging through an LT line of 0.05 Ω per phase at 415 V.

SchemeReliabilityCostProtectionTypical Indian use
RadialLow - one fault kills the feederLowestSimple overcurrentRural 11 kV, all LT networks
Ring main (open ring)Medium-high - restorable by switchingMediumOvercurrent + RMU sectionalisingUrban UG networks
Interconnected/meshHighestHighestDirectional, distanceCity cores, industrial estates
FeatureAIS (air-insulated)GIS (SF₆)
SpaceLargeAbout 10% of AIS
CostLower3-5× higher
Environment exposureAffected by dust, salt, pollutionSealed, pollution-proof
MaintenanceFrequent, accessibleMinimal, but specialised
ConcernLand, clearancesSF₆ global-warming potential, leak detection

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Related topics

Load flow and fault analysis · Power system protection: relays and circuit breakers · Underground cables · Power quality and harmonics · Smart grid and renewable integration · Electricity markets and regulation in India

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