Knowledge / Electrical and Electronics Engineering / Power generation: thermal, hydro, nuclear and renewable

Power generation: thermal, hydro, nuclear and renewable

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

In short

Power generation is the conversion of a primary energy source — coal, gas, falling water, nuclear fission, sunlight, wind, biomass — into bulk electrical energy at a generating station. The topic covers plant layout and working, the thermodynamic or fluid-mechanical basis, site selection, performance factors (heat rate, plant load factor, CUF) and the Indian institutional frame. It is examined in GATE EE (power systems), university Power Systems I and Power Plant Engineering papers, and is the standard opening area in interviews at NTPC, NHPC, NPCIL, SJVN, THDC, Tata Power, Adani Power, JSW Energy, Torrent, ReNew, Suzlon, BHEL, state genco/transco boards and PSU recruitment through GATE. It also underpins energy-audit, O&M, planning and EPC roles where coal, water and fuel costs per kWh decide commercial viability.

Explained

India's generation mix and who runs it

Thermal (steam) power plants

Hydroelectric plants

Nuclear power plants

Renewable generation

Performance factors, load curves and choice of plant

Key points

Formulas and examples

Hydro power: P = ρ g Q H_net η — ρ = 1,000 kg/m³, g = 9.81 m/s², Q = discharge (m³/s), H_net = net head (m), η = overall (turbine × generator) efficiency; P in watts.
Specific speed (metric): Ns = N √P / H5/4 — N in rpm, P in kW (per runner), H in m. Pelton ≈ 8-30, Francis ≈ 50-250, Kaplan ≈ 250-850.
Synchronous speed: N = 120 f / P_poles (rpm)
Thermal efficiency and heat rate: η = 3600 / HR(kJ/kWh) = 860 / HR(kcal/kWh); specific coal consumption = HR(kcal/kWh) / GCV(kcal/kg) in kg/kWh
Fission energy: 1 fission ≈ 200 MeV = 200 × 1.602×10⁻¹³ = 3.2×10⁻¹¹ J; P_thermal = (fissions/s) × 3.2×10⁻¹¹ W; P_electrical = η × P_thermal
Wind power: P = ½ ρ A V³ Cp, A = πD²/4 — ρ ≈ 1.225 kg/m³ at sea level, V = hub-height wind speed (m/s), Cp = power coefficient (≤ 16/27)
Solar PV yield: E(kWh/yr) = P_rated(kWp) × H_poa(kWh/m²/yr) × PR / 1 (kW/m²); module area ≈ P_rated / (1,000 × η_module)
Plant factors: Load factor = average load / peak load; PLF or CUF = E_actual / (P_installed × T); Utilisation factor = max demand / installed capacity
Unit cost: cost per kWh = (annual fixed cost + annual running cost) / annual kWh generated

Example 1 — Thermal unit: efficiency, coal and ash. A 500 MW unit has a station heat rate of 2,400 kcal/kWh, burns coal of GCV 3,500 kcal/kg with 35% ash, and runs at 85% PLF.

Example 2 — Hydro station output and energy. Design discharge Q = 30 m³/s, net head 60 m, turbine-generator overall efficiency 88%.

Example 3 — Nuclear fuel consumption. A 220 MWe PHWR has 29% thermal efficiency.

Example 4 — Wind turbine power. Rotor diameter 120 m, hub-height wind 9 m/s, Cp = 0.42, ρ = 1.225 kg/m³.

Example 5 — 1 MWp solar plant. Plane-of-array insolation 1,900 kWh/m²/yr, PR = 0.78, module efficiency 20%, tariff ₹2.60/kWh.

ParameterCoal (supercritical)CCGTHydro (storage)Nuclear (PHWR)Solar PVWind
ConversionRankineBrayton + RankineHydraulicFission + RankinePhotovoltaicAerodynamic
Efficiency38-40%55-60%85-92%29-33%18-23% (module)Cp 0.40-0.48
Typical PLF/CUF65-70%25-55%30-45%75-85%17-22%25-35%
Capital ₹ crore/MW8.5-9.54-58-1215-204-4.56.5-7
Running costHigh (coal)High (gas)NegligibleLowNilNil
Start-up4-8 h (cold)15-30 min2-5 minVery slown/an/a
RoleBase loadMid/peakPeak + reserveBase loadDay energyEnergy
CO₂~0.9 kg/kWh~0.4 kg/kWhNilNilNilNil

Where it is used

Interview questions

Common mistakes

Related topics

Power system operation, SCADA and EMS · Economic dispatch and unit commitment · Smart grid and renewable integration · Distribution systems and substations · Power system stability · Electricity markets and regulation in India

Open in the Knowledge reader →Find jobs →