12.1 Steam Power Plants & Rankine Cycles

Steam power plants generate the vast majority of central-station base-load electricity worldwide by converting chemical or nuclear thermal energy into mechanical shaft work and electrical energy. The fundamental thermodynamic benchmark for vapor power plants is the Rankine cycle, which resolves the practical engineering operational challenges associated with attempting to compress wet vapor mixtures in the theoretical Carnot vapor cycle.

---

1. The Simple Ideal Rankine Cycle

The ideal Rankine cycle executes four internally reversible state changes across four main plant components: the condensate/boiler feed pump, steam generator (boiler), steam turbine, and surface condenser.

State Point Characterization

• State 1: Saturated liquid entering the pump at low condenser pressure ($P_1 = P_{cond}$). Enthalpy $h_1 = h_{f@P1}$, specific volume $v_1 = v_{f@P1}$.
• State 2: Compressed liquid exiting the pump and entering the boiler at high operating pressure ($P_2 = P_{boiler}$). Since liquid water is virtually incompressible, pump work is calculated via $w_p \approx v_1 (P_2 - P_1)$, and $h_2 = h_1 + w_p$.
• State 3: Superheated vapor (or saturated vapor) exiting the boiler drum/superheater at elevated temperature $T_3$ and pressure $P_3$.
• State 4: Wet vapor mixture exiting the steam turbine at condenser pressure ($P_4 = P_1$). Moisture content is determined by entropy equality $s_4 = s_3 = s_{f@P4} + x_4 s_{fg@P4}$, yielding quality $x_4 = (s_3 - s_f) / s_{fg}$ and enthalpy $h_4 = h_f + x_4 h_{fg}$.

Thermal Efficiency Formula

$$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{w_t - w_p}{q_{in}} = \frac{(h_3 - h_4) - (h_2 - h_1)}{h_3 - h_2}$$

Notice that because pump work $w_p$ is very small compared to turbine work $w_t$ (typically $< 1\%$ in subcritical plants), $w_{net} \approx w_t$ is often approximated in preliminary calculations, though exact PRC board problems require inclusion of $w_p$.

---

2. Thermodynamic Property Diagrams: T-s and h-s (Mollier)

Temperature-Entropy (T-s) Diagram

On the $T-s$ plane, constant pressure lines diverge in the superheated vapor region. Heat input $q_{in}$ corresponds to the area under curve 2-3, while heat rejected $q_{out}$ is the area under curve 4-1. The enclosed area represents net work output $w_{net}$. Thermal efficiency represents the ratio of the enclosed area to the area under process 2-3.

Enthalpy-Entropy (h-s / Mollier) Diagram

The Mollier diagram plots specific enthalpy $h$ against specific entropy $s$. Isentropic turbine expansion (3 $\rightarrow$ 4) appears as a vertical straight downward line. Isothermal/isobaric condensation proceeds along sloped constant-pressure lines in the wet mixture region. The vertical length of line 3-4 directly indicates isentropic enthalpy drop and ideal work output $w_{t,ideal} = h_3 - h_4$.

---

3. Rankine Cycle Efficiency Enhancement Strategies

Thermodynamic efficiency increases whenever the average temperature of heat addition $\bar{T}_{in}$ is raised or the average temperature of heat rejection $\bar{T}_{out}$ is lowered.

A. Lowering Condenser Backpressure ($P_{cond}$)

Lowering condenser operating pressure drops the saturation temperature $T_{cond}$, which expands the cycle net work area downward. However, $P_{cond}$ is bounded by ambient cooling water temperatures (typically $5\text{ kPa}$ to $10\text{ kPa}$, corresponding to $33^\circ\text{C}$–$45^\circ\text{C}$). Excessively low pressure increases moisture content at turbine exhaust ($x_4 < 0.88$), causing severe blade erosion from high-velocity liquid droplet impact.

B. Superheating Steam to Higher Temperatures ($T_{boiler}$)

Superheating raises $\bar{T}_{in}$ without changing boiler pressure, increasing both cycle thermal efficiency and turbine exhaust steam quality $x_4$. Modern metallurgical limits cap maximum steam temperatures at approximately $565^\circ\text{C}$ to $620^\circ\text{C}$ to prevent superheater tube creep failure.

C. Increasing Boiler Operating Pressure ($P_{boiler}$)

Raising boiler pressure shifts the boiling process to higher temperatures, elevating $\bar{T}_{in}$. However, higher pressure narrows the vapor dome, moving turbine expansion deeper into the wet mixture region and decreasing moisture quality $x_4$ below safe operational thresholds unless combined with reheating.

---

4. Modified Rankine Cycles: Reheat & Regeneration

The Reheat Rankine Cycle

To take advantage of high boiler pressures while protecting low-pressure turbine blades from wet steam erosion, steam is expanded in two or more turbine stages:

1. Steam expands in the High-Pressure (HP) turbine from state 3 to intermediate pressure $P_{reheat}$ (state 4).
2. Steam is returned to the boiler reheater section and heated at constant pressure $P_{reheat}$ back to elevated temperature $T_5 \approx T_3$.
3. Reheated steam expands in the Low-Pressure (LP) turbine to condenser backpressure (state 6).

$$\text{Total Turbine Work } w_t = (h_3 - h_4) + (h_5 - h_6)$$ $$\text{Total Heat Input } q_{in} = (h_3 - h_2) + (h_5 - h_4)$$ $$\eta_{th,reheat} = \frac{(h_3 - h_4) + (h_5 - h_6) - w_p}{(h_3 - h_2) + (h_5 - h_4)}$$

The Regenerative Rankine Cycle & Feedwater Heaters (FWH)

Regenerative heating preheats cold liquid exiting the condensate pump before it enters the boiler by extracting (bleeding) small fractions of expanding steam from turbine stages.

• Open Feedwater Heater (OFWH): Direct-contact mixing chamber where bled steam fraction $y$ mixes directly with incoming condensate $(1-y)$. Exiting liquid leaves as saturated liquid at FWH operating pressure.
• Closed Feedwater Heater (CFWH): Shell-and-tube heat exchanger where extracted steam condenses on tube surfaces transferring heat to subcooled feedwater flowing inside tubes without direct fluid mixing.

For an Open Feedwater Heater operating at intermediate extraction state $h_{bleed}$, entering condensate $h_{cond,in}$, and exiting saturated liquid $h_{fwh,out}$:

$$y \cdot h_{bleed} + (1 - y) h_{cond,in} = h_{fwh,out} \implies y = \frac{h_{fwh,out} - h_{cond,in}}{h_{bleed} - h_{cond,in}}$$

---

5. Supercritical Power Plants & Plant Heat Rate

Supercritical Steam Power Plants

When boiler operating pressure exceeds the thermodynamic critical pressure of water ($P_c = 22.06\text{ MPa}$ or $3200\text{ psia}$, $T_c = 373.95^\circ\text{C}$), liquid water transforms continuously into superheated vapor without phase-change boiling. Supercritical (SC) and Ultra-Supercritical (USC) plants operate at pressures from $24.1\text{ MPa}$ to $30+\text{ MPa}$ and steam temperatures up to $600^\circ\text{C}$–$620^\circ\text{C}$, elevating plant thermal efficiency to $42\%$–$48\%$.

Station Heat Rate Metrics

Heat Rate ($HR$) quantifies the total fuel heat input required in $\text{kJ}$ (or $\text{Btu}$) to produce one kilowatt-hour ($\text{kWh}$) of net electrical output. Heat rate is inversely proportional to thermal efficiency:

$$\text{Station Heat Rate (kJ/kWh)} = \frac{\dot{Q}_{in} (\text{kW})}{\dot{W}_{net} (\text{kW})} \times 3600 = \frac{3600}{\eta_{th}}$$ $$\text{Station Heat Rate (Btu/kWh)} = \frac{3412.14}{\eta_{th}}$$

---

6. Worked Step-by-Step Steam Power Plant Problem

Problem Statement: A steam power plant operates on an ideal simple Rankine cycle between a boiler pressure of $4.0\text{ MPa}$ ($4000\text{ kPa}$) with steam temperature $400^\circ\text{C}$, and a condenser pressure of $10\text{ kPa}$. Calculate:

1. Specific pump work $w_p$
2. Enthalpy at turbine exit $h_4$
3. Specific turbine work $w_t$
4. Cycle thermal efficiency $\eta_{th}$
5. Station heat rate in $\text{kJ/kWh}$

Step-by-Step Solution

Step 1: State 1 (Condenser Exit / Pump Inlet) At $P_1 = 10\text{ kPa}$ (saturated liquid):

• $h_1 = h_{f@10kPa} = 191.83\text{ kJ/kg}$
• $v_1 = v_{f@10kPa} = 0.001010\text{ m}^3\text{/kg}$
• $s_1 = s_{f@10kPa} = 0.6492\text{ kJ/kg}\cdot\text{K}$
• $s_{fg@10kPa} = 7.5010\text{ kJ/kg}\cdot\text{K}$, $h_{fg@10kPa} = 2392.1\text{ kJ/kg}$

Step 2: State 2 (Pump Outlet / Boiler Inlet)

$$w_p = v_1 (P_2 - P_1) = 0.001010\text{ m}^3\text{/kg} \times (4000 - 10)\text{ kPa} = 4.03\text{ kJ/kg}$$ $$h_2 = h_1 + w_p = 191.83 + 4.03 = 195.86\text{ kJ/kg}$$

Step 3: State 3 (Boiler Outlet / Turbine Inlet) From superheated steam tables at $P_3 = 4.0\text{ MPa}$ and $T_3 = 400^\circ\text{C}$:

• $h_3 = 3213.6\text{ kJ/kg}$
• $s_3 = 6.7690\text{ kJ/kg}\cdot\text{K}$

Step 4: State 4 (Turbine Exit / Condenser Inlet) Isentropic expansion $s_4 = s_3 = 6.7690\text{ kJ/kg}\cdot\text{K}$ at $P_4 = 10\text{ kPa}$:

$$x_4 = \frac{s_4 - s_{f@10kPa}}{s_{fg@10kPa}} = \frac{6.7690 - 0.6492}{7.5010} = \frac{6.1198}{7.5010} = 0.8158\text{ (81.58\% steam quality)}$$ $$h_4 = h_f + x_4 h_{fg} = 191.83 + (0.8158 \times 2392.1) = 191.83 + 1951.48 = 2143.31\text{ kJ/kg}$$

Step 5: Turbine Work, Net Work, and Heat Input

$$w_t = h_3 - h_4 = 3213.60 - 2143.31 = 1070.29\text{ kJ/kg}$$ $$w_{net} = w_t - w_p = 1070.29 - 4.03 = 1066.26\text{ kJ/kg}$$ $$q_{in} = h_3 - h_2 = 3213.60 - 195.86 = 3017.74\text{ kJ/kg}$$

Step 6: Thermal Efficiency and Heat Rate

$$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{1066.26}{3017.74} = 0.35334 = 35.33\%$$ $$\text{Heat Rate} = \frac{3600}{\eta_{th}} = \frac{3600}{0.35334} = 10,188.5\text{ kJ/kWh}$$