17.1 Hydroelectric, Thermal & Combined Cycle Power Plants

17.1 Hydroelectric, Thermal & Combined Cycle Power Plants

Power generation systems form the core of utility infrastructure and represent a vital major area on the PRC Registered Electrical Engineer (REE) Licensure Examination. Electrical power engineers must master the fundamental mechanical-to-electrical energy conversion principles, thermodynamic cycles, component functions, and mathematical formulations governing hydroelectric stations, thermal fossil-fueled plants, gas turbines, and combined cycle facilities.

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1. Hydroelectric Power Generation Systems

Hydroelectric power plants convert the potential energy of elevated water into kinetic energy, then into rotational mechanical energy via hydraulic turbines, and finally into electrical energy via synchronous generators.

Fundamental Power Output Equation

Consider water flowing from a reservoir through a penstock at volumetric flow rate $Q$ (cubic meters per second, $\text{m}^3/\text{s}$) under effective net head $H$ (meters, $\text{m}$). The theoretical mechanical power available in the water stream is:

$P_{\text{theoretical}} = \rho \cdot g \cdot Q \cdot H \quad (\text{Watts, W})$

where:

• $\rho$ = density of water ($1000\ \text{kg/m}^3$ at standard conditions)
• $g$ = acceleration due to gravity ($9.81\ \text{m/s}^2$)
• $Q$ = volumetric flow rate ($\text{m}^3/\text{s}$)
• $H$ = net effective hydraulic head ($\text{m}$) = Gross Head ($H_g$) minus friction head losses ($h_f$)

Accounting for turbine hydraulic efficiency $\eta_t$ and generator electromechanical efficiency $\eta_g$, the net output electrical power $P_e$ in kilowatts (kW) or megawatts (MW) is:

$P_e = \frac{1000 \times 9.81 \times Q \times H \times \eta_t \times \eta_g}{1000} = 9.81 \cdot Q \cdot H \cdot \eta_{\text{overall}} \quad (\text{kW})$

$P_e = 0.00981 \cdot Q \cdot H \cdot \eta_{\text{overall}} \quad (\text{MW})$

where $\eta_{\text{overall}} = \eta_t \times \eta_g$.

Penstock Head Losses (Manning & Darcy-Weisbach Equations)

The head loss $h_f$ due to friction inside a cylindrical penstock of length $L$ (m) and diameter $D$ (m) with water velocity $v = \frac{4 Q}{\pi D^2}$ is evaluated using the Darcy-Weisbach equation:

$h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2 g}$

where $f$ is the dimensionless Darcy friction factor.

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2. Classification of Hydraulic Turbines

Hydraulic turbines are classified into impulse turbines and reaction turbines based on the action of water on the runner blades, or by operating head and specific speed $N_s$.

Specific Speed ($N_s$)

The specific speed of a hydraulic turbine is defined as the rotational speed (in rpm) at which a geometrically similar turbine would run if it were scaled to produce 1 metric horsepower ($0.7355\ \text{kW}$) under a unit head of 1 meter:

$N_s = \frac{N \sqrt{P}}{H^{5/4}}$

where:

• $N$ = actual rotational speed in rpm
• $P$ = turbine output power in metric HP or kW (standard metric HP in classical PRC board formulas)
• $H$ = effective net head in meters

Hydraulic Turbine Types & Operational Parameters

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3. Essential Hydroelectric Auxiliary Components

                     HYDROELECTRIC STATION SCHEMATIC & SURGE TANK

  Reservoir                                                     Surge Tank
   ~~~~~~~~~                                                        || 
   |       |                                                        || 
   | Water |========= Intake ======\                                 || 
   | Level |                      \\                                || 
   ~~~~~~~~~                       \\ Penstock                      || 
                                    \\==============================||=======\ 
                                                                            \\ Valve / Governor
                                                                             \\   +----------+
                                                                              \\--| Turbine  |
                                                                                  +----------+
                                                                                       | Draft Tube
                                                                                       v Tailrace

Surge Tanks

A surge tank is an open standpipe or storage reservoir connected to the conduit between the high-pressure penstock and the turbine casing.

• Water Hammer Relief: When electrical load drops suddenly, the governor rapidly closes the turbine inlet valves. The immense momentum of water in the long penstock generates extreme hydraulic shock waves (water hammer). The surge tank allows water to rise freely, absorbing the excess pressure surge and preventing penstock rupture.
• Storage Buffer during Load Acceptance: When load increases suddenly, the surge tank supplies immediate additional water to the turbine while the main water column in the penstock accelerates.

Draft Tubes

A draft tube is an airtight diverging conduit connecting the runner outlet of a reaction turbine (Francis or Kaplan) to the tailrace channel.

• Kinetic Energy Recovery: Water leaves the runner with significant kinetic energy ($\frac{v_2^2}{2g}$). The expanding cross-sectional area of the draft tube decreases water velocity ($v_3 < v_2$), converting velocity head into pressure head in accordance with Bernoulli's theorem: $\Delta h_{\text{recovered}} = \frac{v_2^2 - v_3^2}{2g} - h_{f,\text{draft}}$
• Head Preservation: Allows the turbine runner to be installed above the tailrace water level for inspection and maintenance without sacrificing effective net head.

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4. Thermal Power Plants (Steam / Rankine Cycle)

Fossil-fueled thermal power stations convert chemical energy in fuel (coal, heavy fuel oil, or natural gas) into thermal energy via combustion, generating high-pressure, high-temperature steam that expands in a steam turbine to drive a synchronous generator.

The Standard Ideal Rankine Cycle

The fundamental thermodynamic cycle governing steam power plants comprises four main processes:

                         STEAM POWER PLANT RANKINE CYCLE

               +--------------+     High-Temp Steam     +-----------------+
               | Boiler /     |========================>| Steam Turbine   |----
               | Superheater  |                         +-----------------+   |
               +--------------+                                  |            |
                      ^                                          | Exhaust    | Work Output
                      | Feedwater                                v Steam      v (Generator)
               +--------------+     Low-Temp Water      +-----------------+
               | Feed Pump    |<========================| Condenser       |
               +--------------+                         +-----------------+

1. Process 1--2 (Isentropic Expansion): High-pressure superheated steam expands through high-pressure (HP), intermediate-pressure (IP), and low-pressure (LP) turbine stages, performing mechanical shaft work $W_t = h_1 - h_2$.
2. Process 2--3 (Isobaric Condensation): Wet exhaust steam passes into the condenser, rejecting heat $Q_{\text{out}} = h_2 - h_3$ to cooling water, condensing into saturated liquid water.
3. Process 3--4 (Isentropic Compression): Boiler feed pumps (BFP) elevate liquid water pressure to boiler drum pressure, requiring pump work input $W_p = h_4 - h_3 = v_f (P_4 - P_3)$.
4. Process 4--1 (Isobaric Heat Addition): Water passes through economizers, steam drum, boiler tubes, and superheaters, absorbing heat $Q_{\text{in}} = h_1 - h_4$ from combustion flue gases.

Rankine Cycle Efficiency Equations

$\text{Thermal Efficiency: } \eta_{\text{Rankine}} = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{W_t - W_p}{Q_{\text{in}}} = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}$

Since feed pump work $W_p$ is typically small ($< 1%\text{--}2%$ of $W_t$), the approximate thermodynamic efficiency is:

$\eta_{\text{Rankine}} \approx \frac{h_1 - h_2}{h_1 - h_{f4}}$

Station Overall Thermal Efficiency ($\eta_{\text{overall}}$)

Overall station efficiency factors in thermal, mechanical, electrical, and auxiliary energy conversion losses:

$\eta_{\text{overall}} = \eta_{\text{boiler}} \times \eta_{\text{thermal}} \times \eta_{\text{generator}} \times \eta_{\text{auxiliary}}$

Typical coal-fired power station overall efficiencies range from $32%$ to $40%$ (up to $45%$ for ultra-supercritical steam boilers operating above $22.1\ \text{MPa}$ and $600^\circ\text{C}$).

Efficiency Enhancement Auxiliaries

• Superheater: Elevates steam temperature above saturation level at constant pressure. Eliminates moisture droplets during expansion, protecting turbine blades from erosion while increasing Carnot cycle efficiency.
• Reheater: Returns partially expanded steam from the HP turbine back to the boiler to reheat it to high temperature before sending it into IP/LP turbine stages. Increases work output and steam dryness at turbine exit.
• Economizer: A heat exchanger located in the boiler flue gas duct that preheats boiler feed water using residual heat from exhaust gases before entering the steam drum, reducing fuel consumption by $5%\text{--}10%$.
• Air Preheater (APH): Preheats incoming combustion air using low-temperature flue gas exiting the economizer. Improves boiler combustion efficiency and furnace temperature.

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5. Gas Turbines (Brayton Cycle) & Combined Cycle Systems (CCGT)

Open-Cycle Gas Turbines (Brayton Cycle)

Gas turbine units operate on the thermodynamic Brayton Cycle, consisting of:

1. Isentropic compression of ambient air in a multi-stage axial compressor.
2. Isobaric combustion of fuel (natural gas or distillate diesel) in a combustion chamber.
3. Isentropic expansion of high-temperature combustion gases ($1100^\circ\text{C}\text{--}1400^\circ\text{C}$) through the gas turbine.

Ideal Brayton cycle thermal efficiency depends strictly on the pressure ratio $r_p = \frac{P_2}{P_1}$:

$\eta_{\text{Brayton}} = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}}$

where $\gamma = \frac{C_p}{C_v} \approx 1.4$ for air.

Operating Characteristic: Open-cycle gas turbines feature low capital cost and extremely fast start-up capability ($5\text{--}15$ minutes to full load), making them ideal for peaking duty, but their efficiency is modest ($28%\text{--}38%$).

Combined Cycle Gas Turbine (CCGT) Systems

CCGT power plants combine the high-temperature open Brayton cycle gas turbine with a low-temperature Rankine steam cycle using a Heat Recovery Steam Generator (HRSG).

                  COMBINED CYCLE POWER PLANT (CCGT) TOPOLOGY

     Air --> [Compressor] --> [Combustor] --> [Gas Turbine] --> Generator (2/3 Power)
                                                    |
                                             Hot Exhaust Gas (~600°C)
                                                    v
                                           +-----------------+
                                           |   HRSG Unit     |
                                           +-----------------+
                                                    |
                                               Superheated Steam
                                                    v
                                            [Steam Turbine]  --> Generator (1/3 Power)
                                                    |
                                               [Condenser]

CCGT Thermal Efficiency

Let $\eta_1$ be the thermal efficiency of the gas turbine topping cycle and $\eta_2$ be the thermal efficiency of the steam bottoming cycle. The combined cycle overall efficiency $\eta_{\text{cc}}$ is:

$\eta_{\text{cc}} = \eta_1 + \eta_2 (1 - \eta_1)$

Modern CCGT power plants achieve overall thermal efficiencies between $55%$ and $63%$, representing the most efficient fossil-fuel generation technology available.

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Solved Board Exam Examples

Example 1: Hydroelectric Power & Flow Calculation

Problem: A hydroelectric power station operates under an effective net head of $120\ \text{m}$ with a water discharge rate of $25\ \text{m}^3/\text{s}$. The hydraulic turbine efficiency is $90%$ and the synchronous generator efficiency is $95%$. Calculate: (a) Total overall plant efficiency, and (b) Net electrical power output delivered to the grid in MW.

Solution:

1. Compute overall plant efficiency: $\eta_{\text{overall}} = \eta_t \times \eta_g = 0.90 \times 0.95 = 0.855 \quad (85.5\%)$
2. Calculate electric power output using $P_e = 9.81 \cdot Q \cdot H \cdot \eta_{\text{overall}}$: $P_e = 9.81 \times 25\ \text{m}^3/\text{s} \times 120\ \text{m} \times 0.855 = 25,162.05\ \text{kW}$
3. Convert to megawatts: $P_e = \frac{25,162.05}{1000} = 25.16\ \text{MW}$

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Example 2: Thermal Steam Station Coal Consumption & Efficiency

Problem: A $100\ \text{MW}$ coal-fired steam power station burns coal with a higher heating value (HHV) of $28,000\ \text{kJ/kg}$ at a rate of $45\ \text{metric tons}$ per hour. Determine: (a) Total thermal heat input rate in MW, (b) Station overall thermal efficiency, and (c) Station heat rate in $\text{kJ/kWh}$.

Solution:

1. Convert coal consumption rate to kg/s: $\dot{m}_{\text{fuel}} = \frac{45,000\ \text{kg/h}}{3600\ \text{s/h}} = 12.5\ \text{kg/s}$
2. Compute total thermal heat input rate $Q_{\text{in}}$: $Q_{\text{in}} = 12.5\ \text{kg/s} \times 28,000\ \text{kJ/kg} = 350,000\ \text{kJ/s} = 350,000\ \text{kW} = 350\ \text{MW}_{\text{th}}$
3. Calculate overall plant thermal efficiency: $\eta_{\text{overall}} = \frac{P_{\text{electrical}}}{Q_{\text{in}}} \times 100\% = \frac{100\ \text{MW}_{\text{el}}}{350\ \text{MW}_{\text{th}}} \times 100\% = 28.57\%$
4. Calculate station heat rate: $\text{Heat Rate} = \frac{3600}{\eta_{\text{overall}}} = \frac{3600}{0.285714} = 12,600\ \text{kJ/kWh}$