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Pancyprian TS Elec. Eng. III ΘΚ study materials
Pancyprian TS Elec. Eng. III ΘΚ study resources

Pancyprian TS Elec. Eng. III ΘΚ: Generation, Transmission and Distribution of Electrical Energy

Pancyprian TS Elec. Eng. III ΘΚ · Work through your exam, one topic at a time.

All guide sections

Generation, Transmission and Distribution of Electrical Energy

This domain covers the power system from the Cyprus generation mix to the consumer terminals. Expect questions on the EAC network structure (thermal stations at Vasilikos, Dhekelia and Moni; 132 kV transmission; 66 kV subtransmission; 11 kV distribution; 400/230 V LV), on losses and voltage control, and on fault-level arithmetic from transformer impedance.

Core rules

  • Generation: synchronous generators with governor (real power, frequency) and AVR (reactive power, voltage); parallel operation requires matching voltage, frequency, phase sequence and phase angle.
  • Transmission at high voltage reduces current and therefore I²R losses for a given power; losses are also cut by reactive compensation, load balancing and reconductoring.
  • Per-unit calculation simplifies multi-voltage-level networks: choose base MVA and base kV, and impedances scale as Z × Sbase/Vbase².
  • Fault level MVA = S/(|Z| per unit); fault current I = fault MVA/(√3 × kV). Motors and pre-fault load add to the calculated value.
  • Transformer vector groups (Dyn11 typical for distribution) affect parallel operation and harmonic paths.
  • Voltage control uses on-load tap changers, capacitor banks and voltage-regulating relays; frequency is a system-wide balance of generation and demand.
  • Renewables (wind, PV) are variable and distributed, so the grid code adds ride-through, reactive capability and curtailment rules; the planned EuroAsia interconnector changes the fault level and inertia picture.

Worked example

A 132/11 kV transformer rated 40 MVA has a per-unit impedance of 0.12. Its fault level is 40/0.12 = 333 MVA, so the 11 kV fault current is 333 × 10⁶/(√3 × 11 × 10³) = 17.5 kA. The 11 kV switchgear must be rated above this, say 20 kA. If a series reactor reduces the fault level to 250 MVA, the current falls to 250 × 10⁶/(√3 × 11 × 10³) = 13.1 kA, which allows lower-rated equipment downstream - the standard reason for installing reactors near large infeeds.

Traps that cost marks

  • Dividing MVA by the percentage impedance but forgetting to convert percent to per unit.
  • Using line-to-line voltage where phase voltage belongs, or vice versa.
  • Assuming a tap changer materially changes the fault level - it changes voltage, not impedance.
  • Ignoring motor contribution to fault current close to large drives.
  • Confusing transmission and distribution voltage levels in the Cyprus system.