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Steam power cycles – Rankine cycle

Introduction to Steam Power Cycles and the Rankine Cycle

Steam power cycles form the backbone of many thermal power plants worldwide, converting heat energy into mechanical work and eventually into electrical energy. Among these, the Rankine cycle is the fundamental thermodynamic cycle used in steam power plants. Understanding this cycle requires a solid grasp of the thermodynamic laws and the properties of steam, which govern how energy is transferred and transformed.

At its core, the Rankine cycle describes how water is converted into steam, expanded through a turbine to produce work, condensed back into water, and then pumped back to the boiler to repeat the process. This continuous loop efficiently harnesses thermal energy from fuel combustion or other heat sources.

Before diving into the cycle itself, it is essential to understand the behavior of steam under various pressures and temperatures, which is captured in steam tables and graphical tools like the Mollier diagram. These resources allow engineers to determine key properties such as enthalpy, entropy, and specific volume at different points in the cycle, enabling precise analysis and optimization.

Rankine Cycle Overview

The ideal Rankine cycle consists of four main thermodynamic processes that occur sequentially:

  1. Isentropic Compression in Pump (1 -> 2): Liquid water at low pressure is compressed to high pressure by a pump. Since water is nearly incompressible, this process requires relatively little work.
  2. Constant Pressure Heat Addition in Boiler (2 -> 3): The high-pressure water is heated in the boiler at constant pressure until it becomes superheated steam.
  3. Isentropic Expansion in Turbine (3 -> 4): The high-pressure steam expands through the turbine, producing mechanical work while its pressure and temperature drop.
  4. Constant Pressure Heat Rejection in Condenser (4 -> 1): The low-pressure steam is condensed back into liquid water at constant pressure, releasing heat to the surroundings.

These four processes form a closed loop, continuously converting heat energy into useful work.

T-s Diagram 1 2 3 4 Pump (1->2) Boiler (2->3) Turbine (3->4) Condenser (4->1) P-v Diagram 1 2 3 4 Pump (1->2) Boiler (2->3) Condenser (4->1) Turbine (3->4)

Steam Properties and Steam Tables

Steam, or water vapor, exists in different phases depending on temperature and pressure. To analyze the Rankine cycle accurately, we need to know properties such as pressure (P), temperature (T), specific volume (v), enthalpy (h), and entropy (s) at various points.

Steam tables provide these properties for saturated and superheated steam at different pressures and temperatures. They are essential tools for engineers to find the thermodynamic state of steam without complex calculations.

Steam tables typically include:

  • Saturated liquid properties (water about to vaporize)
  • Saturated vapor properties (steam about to condense)
  • Superheated steam properties (steam at temperature above saturation)

Using these tables, you can find enthalpy and entropy values at each state point of the Rankine cycle, which are crucial for calculating work and heat transfer.

Sample Steam Table Extract at 10 MPa
Property Saturated Liquid (Water) Saturated Vapor (Steam) Superheated Steam (500°C)
Pressure (MPa) 10 10 10
Temperature (°C) 311.0 311.0 500.0
Enthalpy, \(h\) (kJ/kg) 762.81 2796.4 3375.1
Entropy, \(s\) (kJ/kg·K) 2.138 6.507 6.592
Specific Volume, \(v\) (m³/kg) 0.001127 0.0199 0.0281

Thermal Efficiency and Work Output of the Rankine Cycle

The performance of a Rankine cycle is measured by its thermal efficiency, which is the ratio of the net work output to the heat input supplied in the boiler.

Mathematically, thermal efficiency \(\eta\) is expressed as:

Thermal Efficiency of Rankine Cycle

\[\eta = \frac{W_{net}}{Q_{in}} = \frac{(W_{turbine} - W_{pump})}{Q_{in}}\]

Efficiency is net work output divided by heat input

\(W_{net}\) = Net work output (kJ/kg)
\(W_{turbine}\) = Work output from turbine (kJ/kg)
\(W_{pump}\) = Work input to pump (kJ/kg)
\(Q_{in}\) = Heat added in boiler (kJ/kg)

The work done by the turbine and the pump can be calculated using enthalpy differences at the respective state points:

Turbine Work Output

\[W_{turbine} = h_1 - h_2\]

Work done by steam during expansion

\(h_1\) = Enthalpy at turbine inlet (kJ/kg)
\(h_2\) = Enthalpy at turbine outlet (kJ/kg)

Pump Work Input

\[W_{pump} = v (P_2 - P_1)\]

Work required to compress liquid water

v = Specific volume of liquid water (m³/kg)
\(P_2\) = Pressure after pump (Pa)
\(P_1\) = Pressure before pump (Pa)

The heat added in the boiler and heat rejected in the condenser are:

Heat Added in Boiler

\[Q_{in} = h_1 - h_4\]

Heat supplied to steam in boiler

\(h_1\) = Enthalpy of steam at boiler outlet (kJ/kg)
\(h_4\) = Enthalpy of liquid water at boiler inlet (kJ/kg)

Heat Rejected in Condenser

\[Q_{out} = h_2 - h_3\]

Heat removed from steam in condenser

\(h_2\) = Enthalpy of steam at turbine outlet (kJ/kg)
\(h_3\) = Enthalpy of liquid water at condenser outlet (kJ/kg)
Boiler Turbine Condenser Pump Work Out Heat Rejected Heat Added Work In

Worked Examples

Example 1: Calculate Thermal Efficiency of an Ideal Rankine Cycle Medium

Steam enters the turbine at 15 MPa and 600°C and expands isentropically to a condenser pressure of 10 kPa. The condenser pressure is maintained at 10 kPa. Using steam tables, calculate the thermal efficiency of the ideal Rankine cycle.

Step 1: Identify state points and their properties.

State 1 (Turbine inlet): P₁ = 15 MPa, T₁ = 600°C

From steam tables: \(h_1 = 3583.1\) kJ/kg, \(s_1 = 6.6\) kJ/kg·K

State 2 (Turbine outlet): P₂ = 10 kPa, isentropic expansion -> \(s_2 = s_1 = 6.6\)

At 10 kPa, find \(h_2\) corresponding to \(s_2 = 6.6\) (superheated steam).

From steam tables: \(h_2 = 2300.0\) kJ/kg (approximate)

State 3 (Condenser outlet): Saturated liquid at 10 kPa

From steam tables: \(h_3 = 191.8\) kJ/kg

State 4 (Pump outlet): P₄ = 15 MPa, liquid water

Calculate pump work:

Specific volume of liquid water at 10 kPa, \(v = 0.001\) m³/kg

\(W_{pump} = v (P_4 - P_3) = 0.001 \times (15 \times 10^6 - 10 \times 10^3) = 14,990 \text{ Pa·m}^3/\text{kg} = 14.99 \text{ kJ/kg}\)

Enthalpy at pump outlet:

\(h_4 = h_3 + W_{pump} = 191.8 + 14.99 = 206.79 \text{ kJ/kg}\)

Step 2: Calculate turbine work and heat added.

\(W_{turbine} = h_1 - h_2 = 3583.1 - 2300.0 = 1283.1 \text{ kJ/kg}\)

\(Q_{in} = h_1 - h_4 = 3583.1 - 206.79 = 3376.31 \text{ kJ/kg}\)

Step 3: Calculate net work output and thermal efficiency.

\(W_{net} = W_{turbine} - W_{pump} = 1283.1 - 14.99 = 1268.11 \text{ kJ/kg}\)

\(\eta = \frac{W_{net}}{Q_{in}} = \frac{1268.11}{3376.31} = 0.3755 = 37.55\%\)

Answer: The thermal efficiency of the ideal Rankine cycle is approximately 37.55%.

Example 2: Determine Work Output of Turbine and Pump Medium

In an ideal Rankine cycle, steam enters the turbine at 8 MPa and 480°C and expands isentropically to 0.008 MPa. Calculate the turbine work output and pump work input per kg of steam. Use steam tables for enthalpy values.

Step 1: Find enthalpy and entropy at turbine inlet (state 1).

At 8 MPa and 480°C:

\(h_1 = 3316.5\) kJ/kg, \(s_1 = 6.7\) kJ/kg·K

Step 2: Find enthalpy at turbine outlet (state 2) assuming isentropic expansion.

At 0.008 MPa, \(s_2 = s_1 = 6.7\)

From steam tables or interpolation, \(h_2 \approx 2300.0\) kJ/kg

Step 3: Calculate turbine work output:

\(W_{turbine} = h_1 - h_2 = 3316.5 - 2300.0 = 1016.5\) kJ/kg

Step 4: Calculate pump work input.

At condenser pressure 0.008 MPa, saturated liquid water specific volume \(v = 0.001\) m³/kg

Pressure rise in pump: \(P_2 - P_1 = 8 \times 10^6 - 0.008 \times 10^6 = 7.992 \times 10^6\) Pa

\(W_{pump} = v (P_2 - P_1) = 0.001 \times 7.992 \times 10^6 = 7.992\) kJ/kg

Answer: Turbine work output is 1016.5 kJ/kg and pump work input is 7.992 kJ/kg.

Example 3: Effect of Reheat on Cycle Efficiency Hard

A Rankine cycle operates with steam entering the turbine at 15 MPa and 600°C. After expanding to 3 MPa, steam is reheated to 500°C and then expanded to 10 kPa. Calculate the thermal efficiency improvement due to reheat compared to the simple Rankine cycle.

Step 1: Identify state points for reheat cycle:

  • State 1: Turbine inlet at 15 MPa, 600°C
  • State 2: After first expansion to 3 MPa (isentropic)
  • State 3: After reheating at 3 MPa to 500°C
  • State 4: After second expansion to 10 kPa (isentropic)
  • State 5: Condenser outlet at 10 kPa saturated liquid
  • State 6: Pump outlet at 15 MPa

Step 2: Use steam tables to find enthalpies and entropies:

  • \(h_1 = 3583.1\) kJ/kg, \(s_1 = 6.6\) kJ/kg·K
  • At 3 MPa, find \(h_2\) after isentropic expansion from 15 MPa:
  • \(s_2 = s_1 = 6.6\), \(h_2 \approx 3120.0\) kJ/kg

  • Reheat to 500°C at 3 MPa:
  • \(h_3 = 3410.0\) kJ/kg, \(s_3 = 7.0\) kJ/kg·K

  • Second expansion to 10 kPa isentropic:
  • \(s_4 = s_3 = 7.0\), \(h_4 \approx 2400.0\) kJ/kg

  • Condensate at 10 kPa saturated liquid:
  • \(h_5 = 191.8\) kJ/kg

  • Pump work (approximate):
  • \(W_{pump} = v (P_6 - P_5) = 0.001 \times (15 \times 10^6 - 10 \times 10^3) = 14.99\) kJ/kg

    \(h_6 = h_5 + W_{pump} = 191.8 + 14.99 = 206.79\) kJ/kg

Step 3: Calculate turbine work output:

First stage turbine work: \(W_{t1} = h_1 - h_2 = 3583.1 - 3120.0 = 463.1\) kJ/kg

Second stage turbine work: \(W_{t2} = h_3 - h_4 = 3410.0 - 2400.0 = 1010.0\) kJ/kg

Total turbine work: \(W_{turbine} = 463.1 + 1010.0 = 1473.1\) kJ/kg

Step 4: Calculate heat added:

Heat added in boiler: \(Q_{in,boiler} = h_1 - h_6 = 3583.1 - 206.79 = 3376.31\) kJ/kg

Heat added in reheater: \(Q_{in,reheat} = h_3 - h_2 = 3410.0 - 3120.0 = 290.0\) kJ/kg

Total heat added: \(Q_{in} = 3376.31 + 290.0 = 3666.31\) kJ/kg

Step 5: Calculate net work and efficiency:

Net work: \(W_{net} = W_{turbine} - W_{pump} = 1473.1 - 14.99 = 1458.11\) kJ/kg

Thermal efficiency: \(\eta = \frac{1458.11}{3666.31} = 0.3976 = 39.76\%\)

Step 6: Compare with simple Rankine cycle efficiency (from Example 1, 37.55%)

Efficiency improvement due to reheat: \(39.76\% - 37.55\% = 2.21\%\)

Answer: Reheating improves the thermal efficiency by approximately 2.21%.

Example 4: Using Mollier Diagram to Find Steam Properties Easy

Using the Mollier diagram (h-s chart), find the enthalpy and entropy of steam at 5 MPa and 400°C for Rankine cycle analysis.

Step 1: Locate the pressure line corresponding to 5 MPa on the Mollier diagram.

Step 2: Move along the 5 MPa pressure line to the temperature curve of 400°C.

Step 3: Read the enthalpy (h) and entropy (s) values from the chart at this intersection.

From the Mollier diagram, approximate values are:

  • Enthalpy, \(h \approx 3200\) kJ/kg
  • Entropy, \(s \approx 6.8\) kJ/kg·K

Answer: At 5 MPa and 400°C, steam has approximately enthalpy 3200 kJ/kg and entropy 6.8 kJ/kg·K.

Example 5: Calculate Pump Work Using Specific Volume Approximation Easy

Calculate the work done by the pump to raise the pressure of water from 10 kPa to 8 MPa. Use the specific volume of liquid water as 0.001 m³/kg.

Step 1: Write down the given data:

  • Initial pressure, \(P_1 = 10\) kPa = \(10 \times 10^3\) Pa
  • Final pressure, \(P_2 = 8\) MPa = \(8 \times 10^6\) Pa
  • Specific volume, \(v = 0.001\) m³/kg

Step 2: Calculate pump work using:

\[ W_{pump} = v (P_2 - P_1) \]

\[ W_{pump} = 0.001 \times (8 \times 10^6 - 10 \times 10^3) = 0.001 \times 7,990,000 = 7,990 \text{ Pa·m}^3/\text{kg} \]

Since 1 Pa·m³ = 1 J,

\(W_{pump} = 7,990\) J/kg = 7.99 kJ/kg

Answer: The pump work required is approximately 7.99 kJ/kg.

Formula Bank

Thermal Efficiency of Rankine Cycle
\[ \eta = \frac{W_{net}}{Q_{in}} = \frac{(W_{turbine} - W_{pump})}{Q_{in}} \]
where: \(W_{net}\) = Net work output (kJ/kg), \(W_{turbine}\) = Turbine work output (kJ/kg), \(W_{pump}\) = Pump work input (kJ/kg), \(Q_{in}\) = Heat added in boiler (kJ/kg)
Turbine Work Output
\[ W_{turbine} = h_1 - h_2 \]
where: \(h_1\) = Enthalpy at turbine inlet (kJ/kg), \(h_2\) = Enthalpy at turbine outlet (kJ/kg)
Pump Work Input
\[ W_{pump} = v (P_2 - P_1) \]
where: \(v\) = Specific volume of liquid water (m³/kg), \(P_2\) = Pressure after pump (Pa), \(P_1\) = Pressure before pump (Pa)
Heat Added in Boiler
\[ Q_{in} = h_1 - h_4 \]
where: \(h_1\) = Enthalpy of steam at boiler outlet (kJ/kg), \(h_4\) = Enthalpy of liquid water at boiler inlet (kJ/kg)
Heat Rejected in Condenser
\[ Q_{out} = h_2 - h_3 \]
where: \(h_2\) = Enthalpy of steam at turbine outlet (kJ/kg), \(h_3\) = Enthalpy of liquid water at condenser outlet (kJ/kg)

Tips & Tricks

Tip: Always check steam quality before turbine inlet to avoid wet steam issues.

When to use: When analyzing turbine inlet conditions to ensure realistic efficiency calculations.

Tip: Use specific volume of saturated liquid water (~0.001 m³/kg) for pump work approximation to save time.

When to use: During pump work calculations where high precision is not required.

Tip: Remember that pump work is much smaller than turbine work; neglecting it can simplify quick estimates.

When to use: For quick efficiency approximations in entrance exam problems.

Tip: Use Mollier diagram for quick property estimation when steam tables are not allowed or time is limited.

When to use: During exams or practical situations where rapid property lookup is needed.

Tip: Draw the T-s diagram first to visualize processes and avoid confusion in state point calculations.

When to use: At the start of any Rankine cycle problem solving.

Common Mistakes to Avoid

❌ Confusing enthalpy values at turbine inlet and outlet.
✓ Carefully identify state points and cross-check with steam tables or diagrams.
Why: Because enthalpy decreases during expansion, mixing up values leads to incorrect work output.
❌ Neglecting pump work completely in all calculations.
✓ Include pump work in thermal efficiency calculations unless explicitly stated negligible.
Why: Pump work, though small, affects net work and efficiency; ignoring it can cause errors.
❌ Using saturated steam properties instead of superheated steam where applicable.
✓ Check steam conditions carefully and use correct tables or diagrams for superheated steam.
Why: Steam properties differ significantly between saturated and superheated states, affecting results.
❌ Mixing units, especially pressure in bar and Pa or enthalpy in kJ/kg and J/kg.
✓ Consistently use metric units and convert where necessary before calculations.
Why: Unit inconsistency leads to calculation errors and wrong answers.
❌ Forgetting that heat rejection occurs at constant pressure in the condenser.
✓ Apply constant pressure assumption when calculating heat rejected and enthalpy changes.
Why: Ignoring this leads to incorrect enthalpy differences and efficiency values.
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