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vapor power cycles Rankine reheat regenerative

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Introduction to Vapor Power Cycles

Vapor power cycles form the backbone of thermal power generation, where heat energy is converted into mechanical work using steam as the working fluid. These cycles are fundamental to the operation of steam power plants, which supply a significant portion of electricity worldwide, including in India.

The Rankine cycle is the idealized thermodynamic cycle that models the operation of steam power plants. Understanding this cycle and its enhancements, such as reheat and regenerative cycles, is crucial for designing efficient and reliable power generation systems.

In this section, we will start by exploring the basic Rankine cycle, then progressively introduce improvements like reheating and regeneration that help increase efficiency and reduce operational issues. Each concept will be supported by diagrams, formulas, and practical examples to build a clear and thorough understanding.

Basic Rankine Cycle

The Rankine cycle is a closed loop in which water is converted into steam, expanded to produce work, condensed back to water, and pumped again to repeat the process. It consists of four main components:

  • Boiler: Converts water into high-pressure steam by adding heat.
  • Turbine: Expands the steam to produce mechanical work.
  • Condenser: Condenses the exhaust steam from the turbine back into liquid water.
  • Pump: Pressurizes the condensed water and feeds it back to the boiler.

Each component corresponds to a thermodynamic process, which together form the Rankine cycle:

  1. 1-2: Isentropic expansion in the turbine (steam does work).
  2. 2-3: Constant pressure heat rejection in the condenser (steam condenses).
  3. 3-4: Isentropic compression in the pump (water is pressurized).
  4. 4-1: Constant pressure heat addition in the boiler (water converts to steam).
Boiler Condenser Turbine Pump High pressure steam Expansion Condensed water Pressurized water T-s Diagram 3 4 1 2 Entropy (s) Temperature (T) Heat Addition Heat Rejection

Ideal Rankine Cycle Assumptions:

  • All processes are internally reversible (isentropic expansion and compression).
  • Negligible pressure losses in pipes and components.
  • Steam is dry and saturated or superheated at turbine inlet.
  • Pump work is much smaller than turbine work but included for accuracy.

Reheat Cycle

In practical steam turbines, as steam expands to low pressure, it tends to become wet (contains moisture). Moisture causes blade erosion and reduces turbine efficiency. To overcome this, the reheat cycle is used.

Purpose of Reheating:

  • Increase average temperature of heat addition, improving efficiency.
  • Reduce moisture content at turbine exhaust, protecting blades.

Process Description: Steam expands partially in the high-pressure turbine, then returns to the boiler for reheating at constant pressure to near original temperature. It then expands again in the low-pressure turbine.

Boiler Condenser HP Turbine Pump Reheater LP Turbine T-s Diagram with Reheat 5 1 2 3 4 Entropy (s) Temperature (T) Heat Addition + Reheat Heat Rejection

The reheat cycle increases the average temperature at which heat is added, which improves the thermal efficiency. It also reduces the moisture content at the turbine exhaust, minimizing blade erosion and maintenance costs.

Regenerative Cycle

Regenerative cycles improve the Rankine cycle efficiency by preheating the feedwater before it enters the boiler. This is achieved by extracting steam from intermediate turbine stages and using it to heat the feedwater in devices called feedwater heaters.

Why Regeneration? Preheating feedwater reduces the fuel required to raise its temperature in the boiler, thus saving energy and improving cycle efficiency.

Types of Feedwater Heaters:

  • Open Feedwater Heater: Steam and feedwater mix directly, equalizing pressure and temperature.
  • Closed Feedwater Heater: Heat transfer occurs through tubes; steam and feedwater do not mix.
graph TD    A[Boiler] --> B[HP Turbine]    B --> C[Steam Extraction]    C --> D[Feedwater Heater]    D --> E[Feedwater Pump]    E --> A    B --> F[LP Turbine]    F --> G[Condenser]    G --> H[Condensate Pump]    H --> D

Steam is extracted at an intermediate pressure from the turbine and passed to the feedwater heater, where it transfers heat to the feedwater coming from the condenser. The feedwater then goes to the pump and finally to the boiler at a higher temperature than in the basic Rankine cycle.

This process reduces the heat input required in the boiler, increasing the overall thermal efficiency.

  • The Rankine cycle is the fundamental steam power cycle with four main components and processes.
  • Reheat cycles improve efficiency and reduce moisture by reheating steam between turbine stages.
  • Regenerative cycles use feedwater heaters to recover heat from turbine exhaust steam, increasing efficiency.
  • Understanding T-s diagrams helps visualize thermodynamic processes and state changes.

Formula Bank

Formula Bank

Thermal Efficiency of Rankine Cycle
\[ \eta = \frac{W_{net}}{Q_{in}} = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4} \]
where: \(h_1\) = enthalpy at turbine inlet, \(h_2\) = enthalpy at turbine outlet, \(h_3\) = enthalpy at condenser outlet, \(h_4\) = enthalpy at pump outlet
Work done by Turbine
\[ W_{turbine} = m \times (h_1 - h_2) \]
where: \(m\) = mass flow rate of steam, \(h_1\) = enthalpy at turbine inlet, \(h_2\) = enthalpy at turbine outlet
Work done by Pump
\[ W_{pump} = m \times (h_4 - h_3) \]
where: \(m\) = mass flow rate of steam, \(h_3\) = enthalpy at condenser outlet, \(h_4\) = enthalpy at pump outlet
Heat Added in Boiler
\[ Q_{in} = m \times (h_1 - h_4) \]
where: \(m\) = mass flow rate of steam, \(h_1\) = enthalpy at boiler outlet, \(h_4\) = enthalpy at pump outlet
Back Work Ratio
\[ BWR = \frac{W_{pump}}{W_{turbine}} \]
where: \(W_{pump}\) = pump work, \(W_{turbine}\) = turbine work
Specific Steam Consumption (SSC)
\[ SSC = \frac{3600 \times m}{W_{net}} \]
where: \(m\) = mass flow rate of steam (kg/s), \(W_{net}\) = net power output (kW)

Worked Examples

Example 1: Thermal Efficiency Calculation of Basic Rankine Cycle Easy
Calculate the thermal efficiency of an ideal Rankine cycle with the following data:
  • Boiler pressure = 15 MPa
  • Condenser pressure = 10 kPa
  • Steam temperature at turbine inlet = 500°C

Step 1: Identify state points and find enthalpy values using steam tables.

At turbine inlet (state 1): \(P_1 = 15\, \text{MPa}\), \(T_1 = 500^\circ C\)

From superheated steam tables, \(h_1 = 3375\, \text{kJ/kg}\), \(s_1 = 6.6\, \text{kJ/kg·K}\)

At condenser pressure (state 2), assuming isentropic expansion, \(s_2 = s_1 = 6.6\)

From steam tables at 10 kPa, find \(h_2\) corresponding to \(s=6.6\), approximately \(h_2 = 2100\, \text{kJ/kg}\)

At condenser outlet (state 3): saturated liquid at 10 kPa, \(h_3 = 191\, \text{kJ/kg}\)

At pump outlet (state 4): pressure raised to 15 MPa, pump work is small, approximate \(h_4 = h_3 + v (P_4 - P_3)\)

Specific volume of liquid at 10 kPa, \(v \approx 0.001\, \text{m}^3/\text{kg}\)

Pressure difference \(= 15 \times 10^3 - 10 = 14990\, \text{kPa}\)

Pump work: \(W_{pump} = v \times \Delta P = 0.001 \times 14990 = 14.99\, \text{kJ/kg}\)

So, \(h_4 = 191 + 15 = 206\, \text{kJ/kg}\) (approximate)

Step 2: Calculate turbine work and pump work per kg of steam.

\(W_{turbine} = h_1 - h_2 = 3375 - 2100 = 1275\, \text{kJ/kg}\)

\(W_{pump} = h_4 - h_3 = 206 - 191 = 15\, \text{kJ/kg}\)

Step 3: Calculate heat added in boiler.

\(Q_{in} = h_1 - h_4 = 3375 - 206 = 3169\, \text{kJ/kg}\)

Step 4: Calculate thermal efficiency.

\[ \eta = \frac{W_{turbine} - W_{pump}}{Q_{in}} = \frac{1275 - 15}{3169} = \frac{1260}{3169} = 0.397 \approx 39.7\% \]

Answer: The thermal efficiency of the Rankine cycle is approximately 39.7%.

Example 2: Impact of Reheat on Work Output Medium
A Rankine cycle operates with a boiler pressure of 15 MPa and condenser pressure of 10 kPa. Steam enters the turbine at 500°C. A single reheat is introduced at 4 MPa to 500°C. Calculate the increase in net work output and thermal efficiency compared to the basic cycle.

Step 1: Identify state points and enthalpy values from steam tables.

State 1: Turbine inlet at 15 MPa, 500°C, \(h_1 = 3375\, \text{kJ/kg}\), \(s_1=6.6\)

State 2: After expansion to 4 MPa (isentropic), find \(h_2\) at \(s=6.6\), approx. \(h_2=3200\, \text{kJ/kg}\)

State 3: After reheating at 4 MPa to 500°C, \(h_3=3375\, \text{kJ/kg}\), \(s_3=6.6\)

State 4: After expansion to 10 kPa (isentropic), find \(h_4\) at \(s=6.6\), approx. \(h_4=2100\, \text{kJ/kg}\)

State 5: Condenser outlet saturated liquid at 10 kPa, \(h_5=191\, \text{kJ/kg}\)

State 6: Pump outlet at 15 MPa, \(h_6 \approx 206\, \text{kJ/kg}\)

Step 2: Calculate turbine work.

High-pressure turbine work: \(W_{t1} = h_1 - h_2 = 3375 - 3200 = 175\, \text{kJ/kg}\)

Low-pressure turbine work: \(W_{t2} = h_3 - h_4 = 3375 - 2100 = 1275\, \text{kJ/kg}\)

Total turbine work: \(W_{t} = 175 + 1275 = 1450\, \text{kJ/kg}\)

Step 3: Calculate pump work.

\(W_{pump} = h_6 - h_5 = 206 - 191 = 15\, \text{kJ/kg}\)

Step 4: Calculate heat added.

Heat added in boiler: \(Q_{boiler} = h_1 - h_6 = 3375 - 206 = 3169\, \text{kJ/kg}\)

Heat added in reheater: \(Q_{reheat} = h_3 - h_2 = 3375 - 3200 = 175\, \text{kJ/kg}\)

Total heat added: \(Q_{in} = 3169 + 175 = 3344\, \text{kJ/kg}\)

Step 5: Calculate thermal efficiency.

\[ \eta = \frac{W_t - W_{pump}}{Q_{in}} = \frac{1450 - 15}{3344} = \frac{1435}{3344} = 0.429 \approx 42.9\% \]

Step 6: Compare with basic cycle efficiency (39.7%).

Efficiency improvement = 42.9% - 39.7% = 3.2%

Work output increase = 1450 - 1260 = 190 kJ/kg

Answer: Reheat increases net work output by 190 kJ/kg and improves thermal efficiency by 3.2%.

Example 3: Regenerative Cycle with Open Feedwater Heater Hard
In a regenerative Rankine cycle, steam is extracted at 1 MPa to heat the feedwater in an open feedwater heater. The boiler pressure is 15 MPa, condenser pressure is 10 kPa, and turbine inlet temperature is 500°C. Calculate the thermal efficiency improvement if the extraction steam enthalpy is 2800 kJ/kg and feedwater enthalpy before heating is 500 kJ/kg.

Step 1: Understand the process.

Steam is extracted at 1 MPa with enthalpy \(h_{ex} = 2800\, \text{kJ/kg}\) to heat feedwater from \(h_{fw} = 500\, \text{kJ/kg}\) to a higher enthalpy in the feedwater heater.

Step 2: Calculate mass fraction of steam extracted (\(y\)) assuming steady state and energy balance.

Let total steam mass flow = 1 kg/s.

Energy balance on feedwater heater:

\[ y \times h_{ex} + (1 - y) \times h_{fw} = h_{fw,heated} \]

Assuming feedwater is heated to \(h_{fw,heated} = 1500\, \text{kJ/kg}\) (typical value at 1 MPa saturated liquid),

\[ y \times 2800 + (1 - y) \times 500 = 1500 \]

\[ 2800y + 500 - 500y = 1500 \]

\[ 2300y = 1000 \implies y = \frac{1000}{2300} \approx 0.435 \]

Step 3: Calculate heat saved in boiler.

Heat input without regeneration (basic cycle):

\(Q_{in,basic} = h_1 - h_4 = 3375 - 500 = 2875\, \text{kJ/kg}\)

Heat input with regeneration:

\(Q_{in,regen} = (1 - y)(h_1 - h_{fw,heated}) = (1 - 0.435)(3375 - 1500) = 0.565 \times 1875 = 1059\, \text{kJ/kg}\)

Step 4: Calculate thermal efficiency improvement.

Assuming turbine and pump work remain similar, net work output is approximately constant.

Efficiency basic: \(\eta_{basic} = \frac{W_{net}}{Q_{in,basic}}\)

Efficiency regenerative: \(\eta_{regen} = \frac{W_{net}}{Q_{in,regen}}\)

Since \(Q_{in,regen} < Q_{in,basic}\), efficiency increases.

Efficiency improvement ratio:

\[ \frac{\eta_{regen}}{\eta_{basic}} = \frac{Q_{in,basic}}{Q_{in,regen}} = \frac{2875}{1059} \approx 2.71 \]

This suggests a significant efficiency gain, but actual improvement is less due to assumptions.

Answer: Regeneration reduces heat input by approximately 63%, leading to a substantial increase in thermal efficiency.

Example 4: Back Work Ratio Calculation Easy
A Rankine cycle has turbine work output of 100 MW and pump work input of 5 MW. Calculate the back work ratio.

Step 1: Use the formula for back work ratio:

\[ BWR = \frac{W_{pump}}{W_{turbine}} = \frac{5}{100} = 0.05 \]

Answer: The back work ratio is 0.05 or 5%, indicating pump work is 5% of turbine work.

Example 5: Specific Steam Consumption for a Power Plant Medium
A steam power plant produces 50 MW net power with a steam flow rate of 120 kg/s. Calculate the specific steam consumption (SSC) in kg/kWh.

Step 1: Use the SSC formula:

\[ SSC = \frac{3600 \times m}{W_{net}} \]

Where \(m = 120\, \text{kg/s}\), \(W_{net} = 50,000\, \text{kW}\)

Step 2: Calculate SSC:

\[ SSC = \frac{3600 \times 120}{50,000} = \frac{432,000}{50,000} = 8.64\, \text{kg/kWh} \]

Answer: The specific steam consumption is 8.64 kg/kWh.

Tips & Tricks

Tip: Always sketch the T-s diagram before solving problems.

When to use: To visualize thermodynamic processes and avoid confusion in enthalpy and entropy values.

Tip: Use steam tables or Mollier charts efficiently by focusing on key states.

When to use: To quickly find enthalpy and entropy values during calculations.

Tip: Remember that pump work is usually much smaller than turbine work and can sometimes be approximated.

When to use: To simplify calculations under time constraints.

Tip: For regenerative cycles, identify extraction points clearly to avoid mixing states.

When to use: When solving problems involving feedwater heaters.

Tip: Use consistent units (SI) throughout calculations to prevent errors.

When to use: Always, especially in entrance exams where unit errors cause loss of marks.

Common Mistakes to Avoid

❌ Confusing enthalpy values at turbine inlet and outlet.
✓ Carefully identify state points on T-s diagram and cross-check with steam tables.
Why: Turbine inlet is high pressure and temperature, outlet is lower; mixing these leads to wrong work calculations.
❌ Ignoring pump work in efficiency calculations.
✓ Include pump work for accurate net work and efficiency.
Why: Though small, neglecting pump work leads to overestimation of efficiency.
❌ Incorrectly assuming reheating raises steam temperature above initial boiler temperature.
✓ Reheat temperature is usually lower or equal to initial steam temperature.
Why: Steam cannot be reheated beyond initial superheated temperature in practical cycles.
❌ Mixing open and closed feedwater heater concepts.
✓ Understand the difference: open heaters mix steam and feedwater, closed heaters transfer heat without mixing.
Why: Misapplication leads to incorrect mass and energy balance.
❌ Using inconsistent units (e.g., pressure in bar and enthalpy in kJ/kg without conversion).
✓ Always convert all units to SI (MPa, kJ/kg, etc.) before calculations.
Why: Unit inconsistency causes calculation errors and incorrect answers.
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