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Power plant engineering

Mechanical Engineering (B.E./B.Tech) · Energy · free study notes for exams, placements and interviews

In short

Power plant engineering is the study of how bulk electricity is generated — the thermodynamic cycle, the equipment that realises it (boiler, turbine, generator, condenser, feed system), and the economics of running it against a varying load. India's generation mix is dominated by coal-fired steam plants, with hydro, nuclear, gas/combined-cycle and diesel sets supplying base, peak and captive needs. It is a core paper in Mechanical and Electrical Engineering, carries steady marks in GATE (ME Thermodynamics/Applied Thermal) and is the main technical subject in NTPC, BHEL, NPCIL, NHPC, Tata Power, Adani Power, JSW Energy, Thermax and L&T Power interviews. The same material supports energy-audit and performance-engineering work under the Energy Conservation Act and BEE's PAT scheme.

Explained

Classification of power plants and the Indian context

Steam power plant: the four circuits and the Rankine cycle

Boilers, fuel, combustion and draught

Turbine, condenser, cooling and feed system

Gas turbine, combined cycle, cogeneration and diesel plants

Hydroelectric and nuclear plants

Economics of generation, load curve and tariffs

Environment, emissions and efficiency regulation

Key points

Formulas and examples

Rankine efficiency η = [(h₁ − h₂) − (h₄ − h₃)] / (h₁ − h₄); h in kJ/kg, states 1 turbine inlet, 2 turbine exhaust, 3 condensate, 4 boiler inlet. Pump work ≈ v_f(p_boiler − p_cond), v_f in m³/kg, p in kPa
Turbine isentropic efficiency η_T = (h₁ − h₂,actual) / (h₁ − h₂,ideal)
Specific steam consumption SSC = 3600 / W_net (kg/kWh), W_net in kJ/kg
Heat rate HR = 3600/η kJ/kWh = 860/η kcal/kWh; specific fuel consumption = HR / GCV (kg/kWh)
Boiler efficiency η_b = m_s(h_s − h_fw) / (m_f × GCV); m in kg/h, GCV in kJ/kg
Equivalent evaporation (from and at 100 °C) E = m_s(h_s − h_fw) / 2257 kg/h; factor of evaporation = (h_s − h_fw)/2257
Chimney draught h_w = 353 H [1/T_a − (m+1)/(m·T_g)] mm of water; H in m, T in K, m = kg air per kg fuel
Condenser heat balance m_s(h₂ − h_f) = m_w c_p (t_out − t_in); c_p = 4.187 kJ/kg·K
Hydro output P = ρ g Q H η; ρ=1000 kg/m³, g=9.81 m/s², Q in m³/s, H = net head in m
Combined cycle η = η_gt + η_st − η_gt·η_st
Load factor = average load / max demand; Plant capacity factor = kWh generated / (capacity × hours); Plant use factor = kWh generated / (capacity × hours actually run); Demand factor = max demand / connected load; Diversity factor = Σ individual max demands / system max demand
Cost per unit = (annual fixed charges + annual running cost) / annual units sent out
Nuclear: energy per fission of U-235 ≈ 200 MeV = 3.2 × 10⁻¹¹ J, so ~1 g of U-235 ≈ 1 MW-day of thermal energy

1. Ideal Rankine cycle. Steam enters the turbine at 40 bar, 400 °C (h₁ = 3,214 kJ/kg, s₁ = 6.769 kJ/kg·K); condenser at 0.05 bar (h_f = 137.8, h_fg = 2,423.7, s_f = 0.4764, s_fg = 7.9187, v_f = 0.001005).

2. Boiler efficiency and equivalent evaporation. A boiler makes 8,000 kg/h of dry saturated steam at 20 bar (h_g = 2,799.5 kJ/kg) from feedwater at 60 °C (h = 251.1 kJ/kg), burning 1,000 kg/h of coal of GCV 25,000 kJ/kg.

3. A 500 MW Indian coal unit. Station heat rate 2,450 kcal/kWh, coal GCV 3,500 kcal/kg, PLF 70%, auxiliary consumption 8%, coal carbon 38%.

4. Load curve and cost of generation. A station with 120 MW installed meets 40 MW for 10 h, 60 MW for 8 h and 100 MW for 6 h each day.

5. Hydro and combined cycle. (a) A run-of-river plant passes Q = 50 m³/s at a net head of 60 m with 88% overall efficiency: P = 1,000 × 9.81 × 50 × 60 × 0.88 = 25.9 MW; at a 45% capacity factor, annual energy = 25.9 × 8,760 × 0.45 ≈ 102 million kWh. (b) A CCGT with gas-turbine efficiency 38% and bottoming steam cycle efficiency 30%: η = 0.38 + 0.30 − 0.114 = 56.6%, heat rate = 3600/0.566 = 6,360 kJ/kWh. For 200 MW net on gas of CV 48,000 kJ/kg, fuel input = 200/0.566 = 353.4 MW → 353,400/48,000 = 7.36 kg/s = 26.5 tonnes/hour.

Chimney draught check: H = 40 m, T_a = 300 K, T_g = 600 K, m = 18: h_w = 353 × 40 × [1/300 − 19/(18×600)] = 14,120 × 0.0015741 = 22.2 mm of water — far too little for a utility boiler, which is why fans are used.

Plant typeCapital costRunning costStart-upTypical efficiencyDuty
Coal (subcritical/supercritical)HighModerate4–8 h cold33–42%Base load
Nuclear (PHWR)Very highVery lowDays30–33%Base load
Combined cycle gasModerateHigh (fuel)1–2 h55–60%Intermediate
Open-cycle gas turbineLowVery high10–20 min30–40%Peak
Diesel engine setLowVery high1–2 min35–45%Standby/captive
Hydro (storage)HighNegligible1–5 min85–92%Peak/base

Where it is used

Interview questions

Common mistakes

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