Steam cycles form the foundation of thermodynamic analysis in many automotive engineering applications, particularly where auxiliary power generation or hybrid system integration is involved. Understanding steam cycles enables automotive engineers to optimize energy conversion processes, improve efficiency, and reduce emissions in systems such as steam-powered vehicles or combined heat and power units. In Kenya, where energy efficiency and sustainable transport solutions are increasingly prioritized, mastery of steam cycle principles is essential for automotive professionals engaged in research, design, and maintenance of advanced propulsion technologies.

3.1 Steam cycles

A steam cycle is a sequence of processes in which water is pumped, heated into steam, expanded to produce work and condensed so that it can be used again. Steam cycles are the basis of steam power plants, factory boiler houses and waste-heat recovery units, and they set the limits of efficiency that real plant can reach. This topic covers the Rankine, Carnot, reheat and regenerative cycles, with the diagrams, equations and worked calculations used to compare them. All worked examples use the same steam conditions, 40 bar and 400 °C at the turbine inlet and 0.1 bar in the condenser, so that the effect of each modification can be seen directly.

3.1.1 Rankine

The Rankine cycle is the standard practical steam cycle and the base against which other steam cycles are compared. Water is pumped to high pressure, heated and vaporised in the boiler, expanded in a turbine or engine to produce work, and condensed back to liquid in the condenser. Example: a factory boiler house that drives a steam turbine for electricity operates on a Rankine cycle, and its output depends on the boiler pressure, the steam temperature and the condenser pressure.

3.1.1.1 Schematic diagram of a steam engine or turbine plant

A schematic diagram shows the main components of a steam plant and the state of the working fluid at each point. The four numbered states used throughout this topic are: 1 turbine inlet, 2 turbine outlet, 3 condenser outlet (saturated liquid) and 4 pump outlet. Example: a technician tracing the pipework of a small steam plant follows the same four stages from the boiler to the turbine, the condenser and the feed pump.

        Heat in (Q_in)
              |
              v
        +-----------+   1    +-----------+
        |  Boiler   |------->|  Turbine  |-----> Work out (W_t)
        +-----------+        +-----------+
              ^                    | 2
              | 4                  v
        +-----------+   3    +-----------+
        |   Pump    |<-------| Condenser |-----> Heat out (Q_out)
        +-----------+        +-----------+
              ^
        Work in (W_p)

The components and their functions are:

• Pump: It increases the pressure of the liquid water, so that it enters the boiler at high pressure.

Automotive engine fuel pump assembly (Source: Wikimedia Commons)

• Boiler: It adds heat to convert the high-pressure liquid into saturated or superheated steam.
• Turbine or engine: It expands the steam and converts thermal energy into mechanical work to drive a load or generator.
• Condenser: It removes heat from the exhaust steam and condenses it back to liquid to complete the cycle.
• Superheater: Where fitted, it raises the steam temperature above saturation before it enters the turbine.

3.1.1.2 Determine Rankine efficiency

The thermal efficiency of a cycle is the net work produced divided by the heat supplied in the boiler. For the Rankine cycle, the net work is the turbine work minus the pump work. Example: an engineer compares a proposed boiler and turbine with an existing one by calculating the efficiency of each from steam table values.

Formula: efficiency = (turbine work − pump work) ÷ heat supplied

Formula: heat supplied = h1 − h4, where h1 is the enthalpy leaving the boiler and h4 is the enthalpy of the water leaving the pump

Data from the steam tables:

• State 1 (40 bar, 400 °C): h1 = 3,214 kJ/kg and s1 = 6.769 kJ/kg·K.
• State 2 (isentropic expansion to 0.1 bar): at 0.1 bar, hf = 191.83 kJ/kg, hfg = 2,392.8 kJ/kg, sf = 0.6493 kJ/kg·K and sfg = 7.5009 kJ/kg·K.
• State 3 (saturated liquid at 0.1 bar): h3 = 191.83 kJ/kg and the specific volume v = 0.00101 m³/kg.

Formula: dryness fraction x2 = (s1 − sf) ÷ sfg

Substitution: x2 = (6.769 − 0.6493) ÷ 7.5009

Working: 6.1197 ÷ 7.5009 = 0.8159

Result: x2 = 0.816

Formula: h2 = hf + x2 × hfg

Substitution: h2 = 191.83 + 0.8159 × 2,392.8

Working: 0.8159 × 2,392.8 = 1,952.3, so h2 = 191.83 + 1,952.3

Result: h2 = 2,144.1 kJ/kg

Formula: pump work = v × (p4 − p3)

Substitution: pump work = 0.00101 × (4,000 − 10)

Working: 0.00101 × 3,990 = 4.03

Result: pump work = 4.03 kJ/kg, so h4 = 191.83 + 4.03 = 195.86 kJ/kg

Formula: efficiency = [(h1 − h2) − pump work] ÷ (h1 − h4)

Substitution: efficiency = [(3,214 − 2,144.1) − 4.03] ÷ (3,214 − 195.86)

Working: (1,069.9 − 4.03) ÷ 3,018.1 = 1,065.9 ÷ 3,018.1 = 0.3532

Result: efficiency = 35.3 %

Factors that affect the Rankine efficiency:

• Boiler pressure: A higher pressure raises the average temperature at which heat is added, which improves efficiency.
• Turbine inlet temperature: Superheating the steam before expansion increases the work output and efficiency.
• Condenser pressure: A lower condenser pressure increases the work output by allowing greater expansion.
• Pump work: It is small compared with turbine work, but reducing it improves the net output.
• Heat losses: Insulation and good component design reduce heat losses and maintain efficiency.

3.1.1.3 T-S and h-s graphs

Graphs of temperature against entropy (T-s) and of enthalpy against entropy (h-s) show the cycle as a closed shape on the property diagram. On the T-s diagram the area enclosed by the cycle represents the net work, and on the h-s diagram vertical distances represent the work of the turbine. Example: a student plots the four state points from the calculation above on a T-s sheet and joins them to see how the cycle moves through the liquid, wet steam and superheated regions.

The processes and their appearance on the graphs are:

Process	Description	T-s diagram	h-s diagram
4 to 1	Heating in the boiler at constant pressure, ending in superheat	Curve rising along the liquid line, a horizontal evaporation line, then a rising superheat curve	Rising curve, steepest in the superheat region
1 to 2	Isentropic expansion in the turbine	Vertical line downward	Vertical line downward, its length is the turbine work
2 to 3	Constant pressure condensation	Horizontal line to the left inside the dome	Falling line to the saturated liquid state
3 to 4	Isentropic pumping	Very short vertical line upward, almost on the liquid line	Very short vertical line upward

Points to note on the diagrams:

• Saturation dome: The bell-shaped curve separates liquid, wet steam and superheated steam regions.
• Constant pressure lines: In the wet region they are horizontal on a T-s diagram.
• Expansion line: For an ideal turbine it is vertical, because entropy is constant.
• Moisture at exhaust: The point 2 lies inside the dome, and the closer it lies to the vapour line the drier the exhaust steam.
• Pump process: It is nearly invisible on a T-s diagram, because liquid is almost incompressible.

3.1.1.4 Modified Rankine cycle

A steam engine cylinder cannot expand steam all the way to condenser pressure, because the cylinder would have to be impractically large. In the modified Rankine cycle, the steam expands only to an intermediate pressure and is then released, so its pressure drops suddenly to the condenser pressure at constant volume. Example: a small reciprocating steam engine that exhausts steam at 1 bar into a condenser at 0.1 bar follows the modified cycle.

Points that describe the modified cycle:

• Incomplete expansion: Expansion stops at a pressure p2 above the condenser pressure p3.
• Constant volume pressure drop: The pressure falls from p2 to p3 at the volume of the steam at point 2, with no work done.
• Lost work: The part of the expansion between p2 and p3 is not used, so the work is less than in the complete Rankine cycle.
• Cylinder size: Incomplete expansion keeps the cylinder size practical.
• Heat supply: The heat supplied in the boiler is the same as in the Rankine cycle.

Formula: work done per kg = (h1 − h2) + (p2 − p3) × v2 − pump work

Here h2 and v2 are the enthalpy and specific volume at the end of the incomplete expansion, and p is in kPa so that the product is in kJ/kg.

3.1.1.5 Work-done using Rankine equations

The work of the cycle is found from the enthalpy values at the four states. The turbine produces work and the pump absorbs work, and their difference is the net work output. Example: the 40 bar cycle above shows how the work per kilogram is converted into power for a given steam flow.

Equations:

• Turbine work: W_t = h1 − h2, because the expansion is adiabatic.
• Pump work: W_p = h4 − h3, which equals v × (p4 − p3) for an incompressible liquid.
• Net work: W_net = W_t − W_p.
• Heat rejected: Q_out = h2 − h3.
• Energy balance: Q_in − Q_out = W_net.

Formula: net work = (h1 − h2) − (h4 − h3)

Substitution: net work = (3,214 − 2,144.1) − (195.86 − 191.83)

Working: 1,069.9 − 4.03 = 1,065.9

Result: net work = 1,065.9 kJ/kg

Formula: work ratio = net work ÷ turbine work

Substitution: work ratio = 1,065.9 ÷ 1,069.9

Working: 1,065.9 ÷ 1,069.9 = 0.996

Result: work ratio = 0.996

Formula: steam consumption = 3,600 ÷ net work, in kg per kWh

Substitution: steam consumption = 3,600 ÷ 1,065.9

Working: 3,600 ÷ 1,065.9 = 3.377

Result: steam consumption = 3.38 kg/kWh

Formula: power = steam flow rate × net work

Substitution: power = 10 kg/s × 1,065.9 kJ/kg

Working: 10 × 1,065.9 = 10,659

Result: power = 10,659 kW, which is about 10.7 MW

3.1.1.6 Efficiency of modified Rankine cycle

The efficiency of the modified cycle is lower than that of the complete cycle, because the final part of the expansion is not converted to work. It is found from the net work of the modified cycle and the same heat supply. Example: for the 40 bar, 400 °C steam, the engine is assumed to expand only to 1 bar before exhausting into a condenser at 0.1 bar.

Data at 1 bar: hf = 417.5 kJ/kg, hfg = 2,257.5 kJ/kg, sf = 1.3028 kJ/kg·K, sfg = 6.0561 kJ/kg·K and vg = 1.694 m³/kg.

Formula: dryness fraction x2 = (s1 − sf) ÷ sfg

Substitution: x2 = (6.769 − 1.3028) ÷ 6.0561

Working: 5.4662 ÷ 6.0561 = 0.9026

Result: x2 = 0.903

Formula: h2 = hf + x2 × hfg

Substitution: h2 = 417.5 + 0.9026 × 2,257.5

Working: 0.9026 × 2,257.5 = 2,037.6, so h2 = 417.5 + 2,037.6

Result: h2 = 2,455.1 kJ/kg

Formula: v2 = x2 × vg

Substitution: v2 = 0.9026 × 1.694

Working: 0.9026 × 1.694 = 1.529

Result: v2 = 1.529 m³/kg

Formula: work done = (h1 − h2) + (p2 − p3) × v2 − pump work

Substitution: work done = (3,214 − 2,455.1) + (100 − 10) × 1.529 − 4.03

Working: 758.9 + 137.6 − 4.03 = 892.5

Result: net work = 892.5 kJ/kg

Formula: efficiency = net work ÷ (h1 − h4)

Substitution: efficiency = 892.5 ÷ 3,018.1

Working: 892.5 ÷ 3,018.1 = 0.2957

Result: efficiency = 29.6 %

3.1.1.7 Theoretical loss of work per kg of steam due to incomplete expansion

The theoretical loss of work is the difference between the work that complete expansion would give and the work that the modified cycle gives. It is the work that would be produced if the steam were expanded from p2 to the condenser pressure but is lost when the pressure is released instead. Example: for the engine above, the loss shows how much work is given up by using a cylinder of practical size.

Formula: loss of work = (h2 − h3') − (p2 − p3) × v2, where h3' is the enthalpy after complete isentropic expansion to the condenser pressure

Substitution: loss of work = (2,455.1 − 2,144.1) − (100 − 10) × 1.529

Working: 311.0 − 137.6 = 173.4

Result: loss of work = 173.4 kJ/kg

Check: the complete cycle gives 1,069.9 kJ/kg of turbine work and the modified cycle gives 896.5 kJ/kg, and 1,069.9 − 896.5 = 173.4 kJ/kg.

3.1.1.8 Loss in Rankine efficiency due to restricted expansion of steam

Restricted expansion reduces the efficiency, because the same heat is supplied while less work is obtained. The loss in efficiency is found by comparing the two cycles. Example: using the efficiencies calculated above, the loss for expansion limited to 1 bar is found directly.

Formula: loss in efficiency = efficiency of complete cycle − efficiency of modified cycle

Substitution: loss in efficiency = 35.3 − 29.6

Working: 35.3 − 29.6 = 5.7

Result: loss = 5.7 percentage points, which is about 16 % of the Rankine efficiency

Points on restricted expansion:

• Heat rejected: The exhaust steam leaves with a higher enthalpy, so more heat is rejected in the condenser.
• Work lost: The shaded area on the p-v diagram between p2 and p3 is not converted to work.
• Compromise: The cylinder is limited in size, so a small loss is accepted.
• Multi-stage engines: Compound engines expand the steam in several cylinders, which reduces the loss.
• Turbines: A turbine can expand steam almost to condenser pressure, which avoids most of this loss.

3.1.1.9 Enthalpy-entropy chart

The enthalpy-entropy chart, also called a Mollier chart, is a graph with entropy on the horizontal axis and enthalpy on the vertical axis. It shows the properties of steam as lines, so that the work of expansion can be read directly instead of being calculated. Example: an engineer reads the exhaust condition of a turbine from the chart by starting at the inlet point and moving straight down to the condenser pressure line.

Features of the chart:

• Saturation line: It separates the wet region below from the superheated region above.
• Constant pressure lines: They run across the chart, rising to the right, and are straight lines inside the wet region.
• Constant temperature lines: They are the same as constant pressure lines in the wet region, and curve apart in the superheated region.
• Constant dryness lines: They show the fraction of vapour in wet steam.
• Isentropic expansion: It is a vertical line, and its length is the enthalpy drop, which is the work in an ideal turbine.

Using the chart for the 40 bar cycle:

1. Locate the point at 40 bar and 400 °C, which reads h = 3,214 kJ/kg and s = 6.769 kJ/kg·K.
2. Draw a vertical line down to the 0.1 bar pressure line.
3. Read the enthalpy at the end of the line, which gives about 2,144 kJ/kg, with a dryness fraction of about 0.82.
4. Subtract to find the work, 3,214 − 2,144 = 1,070 kJ/kg.

3.1.2 Carnot

The Carnot cycle is an ideal reversible cycle that operates between two fixed temperatures and gives the greatest efficiency possible between them. It consists of two isothermal and two adiabatic processes: isothermal heat addition at the high temperature, adiabatic expansion, isothermal heat rejection at the low temperature and adiabatic compression. It is not built in practice, but it is the standard against which real steam cycles are judged. Example: an engineer compares the efficiency of a boiler plant with the Carnot value for the same temperatures to see how much improvement is theoretically possible.

3.1.2.1 Draw schematic diagrams of Carnot engine

A Carnot engine takes heat from a hot source, converts part of it to work and rejects the rest to a cold sink. The schematic shows only the energy flows, without describing the equipment. Example: a student draws the engine as a box between two reservoirs, with the heat and work arrows labelled.

        Hot source at T_H
               |
               |  Q_in
               v
        +--------------+
        | Carnot engine|-------> Net work W
        +--------------+
               |
               |  Q_out
               v
        Cold sink at T_L

Points on the schematic:

• Heat input: Q_in is taken from the hot source at the constant temperature T_H.
• Work output: W = Q_in − Q_out.
• Heat rejection: Q_out is delivered to the cold sink at the constant temperature T_L.
• Absolute temperatures: T_H and T_L must be in kelvin.
• Reversibility: The engine has no friction and no heat transfer across a temperature difference.

3.1.2.2 Carnot Cycle with Steam as Working Substance

When steam is the working fluid, the isothermal processes occur inside the saturation dome, because boiling and condensing take place at constant temperature and pressure. The cycle is a rectangle on the T-s diagram. Example: a Carnot cycle for steam between 40 bar and 0.1 bar uses saturated steam at 523.5 K in the boiler and condenses partly at 319 K.

The four processes are:

• 1 to 2, isothermal heat addition: Saturated liquid is evaporated in the boiler at constant temperature and pressure.
• 2 to 3, isentropic expansion: The steam expands in the turbine, and its temperature falls to T_L.
• 3 to 4, isothermal heat rejection: The wet steam is partly condensed in the condenser at constant temperature.
• 4 to 1, isentropic compression: The wet steam is compressed back to saturated liquid.
• Practical difficulty: The compressor would have to handle a mixture of liquid and vapour, which is impractical, and the cycle cannot use superheat without leaving constant temperature heat addition.

Formula: heat added = hfg at the boiler pressure

Substitution: heat added = hfg at 40 bar, which is 1,714.1 kJ/kg

Formula: efficiency = 1 − T_L ÷ T_H

Substitution: efficiency = 1 − 319 ÷ 523.5

Working: 319 ÷ 523.5 = 0.6094, so 1 − 0.6094 = 0.3906

Result: efficiency = 39.1 %

Formula: net work = efficiency × heat added

Substitution: net work = 0.3906 × 1,714.1

Working: 0.3906 × 1,714.1 = 669.5

Result: net work = 669.5 kJ/kg, and heat rejected = 1,714.1 − 669.5 = 1,044.6 kJ/kg

3.1.2.3 Performance Criteria for Carnot Cycle

The performance of a Carnot cycle is judged by its thermal efficiency, which depends only on the temperatures of the source and sink. Other criteria, such as work ratio and steam consumption, show how practical the cycle is. Example: the efficiency formula shows an engineer that raising the boiler temperature or lowering the condenser temperature improves the efficiency.

Formula: Carnot efficiency = (T_hot − T_cold) ÷ T_hot

Points on the performance:

• Dependence on temperatures: Increasing the temperature difference improves the efficiency, so engineers seek higher boiler temperatures and lower condenser temperatures within material limits.
• No other dependence: The efficiency does not depend on the working fluid.
• Upper limit: No real engine working between the same temperatures can have a higher efficiency.
• Work ratio: In the steam Carnot cycle, the compression of wet steam takes a large share of the turbine work, so the work ratio is much lower than for a Rankine cycle, in which it is 0.996.
• Steam consumption: The 669.5 kJ/kg of net work in the example gives 3,600 ÷ 669.5 = 5.38 kg/kWh, which is higher than the 3.38 kg/kWh of the Rankine cycle, because less work is obtained from each kilogram.

Limitations of the Carnot cycle in practical applications

Despite its theoretical importance, the Carnot cycle is impractical for real plant for these reasons:

• Idealised processes: Isothermal and adiabatic processes cannot be achieved exactly.
• Slow operation: Isothermal heat transfer needs slow processes, which are not suited to vehicles and fast-running machines.
• Material limits: The high temperatures that would raise efficiency exceed the limits of common materials.
• Complexity: Reversible processes would add mechanical complexity and cost.
• No ideal components: Real turbines, pumps and boilers cannot operate at ideal efficiency.

Role of the Carnot cycle as a benchmark

The Carnot efficiency is used as a yardstick. By comparing the actual efficiency of a cycle with the Carnot value for the same temperatures, engineers see how much room for improvement remains in component design, operating conditions and system layout.

3.1.3 Reheat

The reheat cycle improves the Rankine cycle by returning the steam to the boiler after partial expansion, heating it again and expanding it further. This raises the average temperature at which heat is added and reduces the moisture content of the steam at the end of expansion. Example: a power plant with a high-pressure and a low-pressure turbine passes the steam through a reheater between the two turbines.

3.1.3.1 T-S diagram of reheat steam cycle

The T-s diagram of the reheat cycle is the Rankine cycle with an extra heating line in the middle of the expansion. The steam expands in the high-pressure turbine from state 1 to state 2, is heated again at constant pressure from 2 to 3, and then expands in the low-pressure turbine from 3 to 4. Example: the 40 bar cycle with reheat at 5 bar returns the steam to 400 °C before the final expansion.

Boiler -> HP turbine -> Reheater -> LP turbine -> Condenser -> Pump
   1          2            3            4             5          6

Points on the diagram:

• 1 to 2: Isentropic expansion in the high-pressure turbine, a vertical line down to the reheat pressure.
• 2 to 3: Constant pressure reheating, a rising line, usually back to the original steam temperature.
• 3 to 4: Isentropic expansion in the low-pressure turbine, a second vertical line, ending at a drier state than a simple Rankine cycle would give.
• 4 to 5: Condensation at constant pressure to saturated liquid.
• 5 to 6: Pumping the liquid back to boiler pressure.

3.1.3.2 Determine work-done using reheat equations

The work of the reheat cycle is the sum of the work of the two turbines minus the pump work. Example: for steam at 40 bar and 400 °C, reheated at 5 bar to 400 °C and expanded to 0.1 bar, the work per kilogram is found as follows.

Data from the steam tables:

• State 1 (40 bar, 400 °C): h1 = 3,214 kJ/kg and s1 = 6.769 kJ/kg·K.
• State 2 (expansion to 5 bar): hf = 640.1, hfg = 2,108.5, sf = 1.8604 and sfg = 4.9603.
• State 3 (5 bar, 400 °C): h3 = 3,272 kJ/kg and s3 = 7.794 kJ/kg·K.
• State 4 (expansion to 0.1 bar): hf = 191.83, hfg = 2,392.8, sf = 0.6493 and sfg = 7.5009.

Formula: x2 = (s1 − sf) ÷ sfg

Substitution: x2 = (6.769 − 1.8604) ÷ 4.9603

Working: 4.9086 ÷ 4.9603 = 0.9896, and h2 = 640.1 + 0.9896 × 2,108.5 = 640.1 + 2,086.6 = 2,726.7

Result: x2 = 0.990 and h2 = 2,726.7 kJ/kg

Formula: x4 = (s3 − sf) ÷ sfg

Substitution: x4 = (7.794 − 0.6493) ÷ 7.5009

Working: 7.1447 ÷ 7.5009 = 0.9525, and h4 = 191.83 + 0.9525 × 2,392.8 = 191.83 + 2,279.2 = 2,471.0

Result: x4 = 0.953 and h4 = 2,471.0 kJ/kg

Formula: net work = (h1 − h2) + (h3 − h4) − pump work

Substitution: net work = (3,214 − 2,726.7) + (3,272 − 2,471.0) − 4.03

Working: 487.3 + 801.0 − 4.03 = 1,284.3

Result: net work = 1,284.3 kJ/kg, compared with 1,065.9 kJ/kg for the simple Rankine cycle

3.1.3.3 Determine efficiency of reheat cycles

The efficiency of the reheat cycle is the net work divided by the total heat supplied, which includes the heat added in the boiler and in the reheater. Example: using the work found above, the efficiency is compared with the 35.3 % of the simple cycle.

Formula: heat supplied = (h1 − h6) + (h3 − h2), where h6 is the enthalpy after the pump, 195.86 kJ/kg

Substitution: heat supplied = (3,214 − 195.86) + (3,272 − 2,726.7)

Working: 3,018.1 + 545.3 = 3,563.4

Result: heat supplied = 3,563.4 kJ/kg

Formula: efficiency = net work ÷ heat supplied

Substitution: efficiency = 1,284.3 ÷ 3,563.4

Working: 1,284.3 ÷ 3,563.4 = 0.3604

Result: efficiency = 36.0 %

Benefits of the reheat cycle

• Increased efficiency: Reheating raises the average steam temperature during expansion, which improves efficiency, here from 35.3 % to 36.0 %.
• Reduced blade erosion: The exhaust dryness rises from 0.816 to 0.953, which reduces wear on the last turbine blades.
• Higher work output: The extra work per kilogram is 1,284.3 − 1,065.9 = 218.4 kJ/kg.
• Improved reliability: Drier steam gives less mechanical stress on the blades and a longer life.
• Better control: The two turbine stages allow more flexible operation over varying loads.

Design considerations for reheat cycles

• Reheater placement: It must be positioned to recover heat without excessive complexity.
• Material selection: Components must withstand the high temperature of the reheated steam.
• Control systems: Steam flows and temperatures must be controlled precisely.
• Space and weight: Compact designs are needed where space is limited, as on a vehicle.
• Maintenance access: Added complexity makes easy maintenance access important.

3.1.4 Regenerative

The regenerative cycle improves efficiency by using steam extracted from the turbine to heat the boiler feed water before it enters the boiler. The feed water enters the boiler at a higher temperature, so less heat has to be supplied by the fuel, and the heat is added at a higher average temperature. Example: a boiler house that preheats its feed water with turbine steam burns less fuel for the same output.

3.1.4.1 Ideal regenerative cycle diagram

In the ideal regenerative cycle, the feed water is heated gradually by the steam as it expands, through an infinite number of heaters, so that heat is transferred with no temperature difference. The feed water is then heated reversibly along a path parallel to the turbine expansion line. Example: on a T-s diagram, the heating line of the liquid is parallel to the expansion line of the turbine, and the area under it equals the heat given up by the steam.

Points on the ideal cycle:

• Heat transfer: The steam gives up its heat to the water at the same temperature, so the process is reversible.
• Heat addition: All the heat from the fuel is supplied at the boiler saturation temperature, which is the highest temperature of the cycle.
• Heat rejection: Heat is rejected at the condenser temperature only.
• Efficiency: It equals the Carnot efficiency between the two temperatures, which is 39.1 % for the 40 bar and 0.1 bar example.
• Practical limit: An infinite number of heaters cannot be built, so real plants use one to several heaters.

3.1.4.2 Regenerative Cycle with Single Feed Water Heater

In a cycle with a single feed water heater, a fraction of the steam is extracted from the turbine at an intermediate pressure and used to heat the feed water. The heater may be of the open type, in which the steam and water mix, or the closed type, in which heat passes through tube walls. Example: for the 40 bar and 400 °C steam, a fraction m is extracted at 5 bar into an open heater, and the rest expands to 0.1 bar.

Types of feed water heater:

• Open feed water heater: Steam and feed water mix directly, so the water leaves as saturated liquid at the extraction pressure and a second pump is needed.
• Closed feed water heater: Heat passes through tubes without mixing, so the pressures of the two fluids can differ, and the condensed steam is drained back.

The numbered states for the open heater cycle are: 1 turbine inlet, 2 extraction point at 5 bar, 3 condenser inlet at 0.1 bar, 4 after pump 1, 5 heater outlet (saturated liquid at 5 bar), 6 after pump 2 (boiler inlet).

Formula: mass fraction extracted m = (h5 − h4) ÷ (h2 − h4)

Data: h2 = 2,726.7 kJ/kg, h5 = 640.1 kJ/kg, pump 1 work = 0.00101 × (500 − 10) = 0.495 kJ/kg, so h4 = 191.83 + 0.495 = 192.33 kJ/kg

Substitution: m = (640.1 − 192.33) ÷ (2,726.7 − 192.33)

Working: 447.77 ÷ 2,534.37 = 0.1767

Result: m = 0.177 kg of steam extracted per kg of steam entering the turbine

3.1.4.3 Regenerative cycle with single feed water heater diagram

The diagram shows the flow of steam and water through the plant, with the extraction line leading to the heater. Example: a technician follows the extraction pipe from the turbine to the open heater and then to the second pump.

 Boiler --1--> Turbine --(m)--> Open feed water heater
                  |                  ^        |
               (1-m) 3               | 4      | 5
                  v                  |        v
              Condenser --> Pump 1 --+      Pump 2 --6--> Boiler

Points on the diagram:

• Extraction: A fraction m of the steam leaves the turbine at 5 bar and goes to the heater.
• Remaining steam: The fraction 1 − m continues to expand to the condenser pressure.
• Pump 1: It raises the condensate from condenser pressure to heater pressure.
• Open heater: The extracted steam and condensate mix and leave as saturated liquid.
• Pump 2: It raises the heated water to boiler pressure.

3.1.4.4 Determine work-done by Regenerative cycle

The work of the regenerative cycle is the turbine work, with the reduced flow after extraction, minus the work of both pumps. Example: the cycle above is calculated per kilogram of steam entering the turbine.

Formula: turbine work = (h1 − h2) + (1 − m) × (h2 − h3)

Substitution: turbine work = (3,214 − 2,726.7) + 0.8233 × (2,726.7 − 2,144.1)

Working: 487.3 + 0.8233 × 582.6 = 487.3 + 479.7 = 967.0

Result: turbine work = 967.0 kJ/kg

Formula: pump work = (1 − m) × W_p1 + W_p2

Substitution: pump work = 0.8233 × 0.495 + 0.001093 × (4,000 − 500)

Working: 0.408 + 3.83 = 4.24

Result: pump work = 4.24 kJ/kg

Formula: net work = turbine work − pump work

Substitution: net work = 967.0 − 4.24

Working: 967.0 − 4.24 = 962.7

Result: net work = 962.7 kJ/kg

3.1.4.5 Determine Regenerative cycle efficiency

The efficiency is the net work divided by the heat supplied in the boiler, where the feed water now enters at a higher enthalpy. Example: the efficiency of the single-heater cycle is compared with the simple Rankine cycle.

Formula: h6 = h5 + W_p2 = 640.1 + 3.83 = 643.9 kJ/kg

Formula: heat supplied = h1 − h6

Substitution: heat supplied = 3,214 − 643.9

Working: 3,214 − 643.9 = 2,570.1

Result: heat supplied = 2,570.1 kJ/kg

Formula: efficiency = net work ÷ heat supplied

Substitution: efficiency = 962.7 ÷ 2,570.1

Working: 962.7 ÷ 2,570.1 = 0.3746

Result: efficiency = 37.5 %, compared with 35.3 % for the simple Rankine cycle

3.1.4.6 Regenerative Cycle with Two Feed Water Heaters and its efficiency

With two open heaters, steam is extracted at two pressures, so the feed water is heated in two stages and the efficiency rises further. Example: steam is extracted at 10 bar and at 1 bar, with the rest expanding to 0.1 bar.

Data:

• Extraction at 10 bar: h_a = 2,865.2 kJ/kg, found at s = 6.769 kJ/kg·K from the superheat table.
• Extraction at 1 bar: h_b = 2,455.1 kJ/kg.
• Condenser inlet at 0.1 bar: h_c = 2,144.1 kJ/kg.
• Saturated liquid: 762.5 kJ/kg at 10 bar and 417.5 kJ/kg at 1 bar.
• Pump outputs: after pump 1, h = 191.92 kJ/kg; after pump 2, h = 417.5 + 0.94 = 418.4 kJ/kg; after pump 3, h = 762.5 + 3.38 = 765.9 kJ/kg.

Formula: high-pressure heater, m1 = (762.5 − 418.4) ÷ (2,865.2 − 418.4)

Substitution and working: 344.1 ÷ 2,446.8 = 0.1406

Result: m1 = 0.141

Formula: low-pressure heater, m2 × (h_b − 191.92) = (1 − m1) × (417.5 − 191.92)

Substitution: m2 × 2,263.18 = 0.8594 × 225.58

Working: 0.8594 × 225.58 = 193.86, so m2 = 193.86 ÷ 2,263.18 = 0.0857

Result: m2 = 0.086

Formula: turbine work = (h1 − h_a) + (1 − m1)(h_a − h_b) + (1 − m1 − m2)(h_b − h_c)

Substitution: turbine work = (3,214 − 2,865.2) + 0.8594 × 410.1 + 0.7737 × 311.0

Working: 348.8 + 352.4 + 240.6 = 941.8

Result: turbine work = 941.8 kJ/kg

Formula: pump work = 0.7737 × 0.09 + 0.8594 × 0.94 + 3.38

Working: 0.07 + 0.81 + 3.38 = 4.26

Result: pump work = 4.26 kJ/kg, so net work = 941.8 − 4.26 = 937.5 kJ/kg

Formula: efficiency = net work ÷ (h1 − h6)

Substitution: efficiency = 937.5 ÷ (3,214 − 765.9)

Working: 937.5 ÷ 2,448.1 = 0.3830

Result: efficiency = 38.3 %

The effect of the number of heaters on the efficiency of the same plant is:

Cycle	Efficiency
Simple Rankine	35.3 %
Reheat	36.0 %
One feed water heater	37.5 %
Two feed water heaters	38.3 %
Ideal regenerative (Carnot limit)	39.1 %

Advantages of regenerative cycles

• Improved thermal efficiency: Preheating the feed water reduces fuel consumption and operating cost.
• Reduced emissions: Lower fuel use means lower greenhouse gas emissions.
• Longer boiler life: Warmer feed water reduces thermal shock on the boiler parts.
• Smoother operation: Preheated feed water stabilises boiler temperature.
• Energy recovery: More of the energy in the steam is used before it reaches the condenser.

Challenges and practical implementation

• Complexity: Extra heaters, pumps and pipes increase maintenance.
• Space and weight: Limited space and weight in vehicles restrict heater size.
• Cost: Higher initial cost may discourage adoption.
• Control: Steam extraction must be controlled accurately.
• Practical measures: Compact heat exchangers, automatic controls, lightweight corrosion-resistant materials and modular units help to overcome these problems.

Application of steam cycles in automotive engineering

Steam cycles are used in steam-powered vehicles and in waste-heat recovery systems that capture exhaust heat to produce steam. For example, a hybrid vehicle may use a Rankine cycle unit to turn engine exhaust heat into extra mechanical power, which reduces fuel consumption and emissions. Engineers adapt the cycle parameters to the size, weight and operating conditions of the vehicle.

Chapter Summary

This chapter explored various steam cycles, beginning with the Rankine cycle, where its schematic design, efficiency calculations, and graphical representations such as T-S and h-s diagrams were examined. Modifications to the Rankine cycle were discussed, including the impact on work output and efficiency, as well as theoretical work losses due to incomplete steam expansion and their effect on overall performance. The Carnot cycle was introduced with its schematic and steam-based working substance, highlighting its performance criteria as an idealized thermodynamic cycle. Reheat cycles were analyzed through T-S diagrams, equations for work done, and efficiency determination, showing how reheating improves cycle performance. The regenerative cycle was studied with diagrams illustrating single and multiple feed water heaters, and methods to calculate work done and efficiency enhancements were presented. In the section on thermodynamic steam turbines, key characteristics and classifications were outlined, followed by detailed discussions on the pressure and velocity of steam in impulse turbines. The chapter concluded with an examination of velocity triangles for moving blades and the calculation of power produced by impulse turbines, providing a comprehensive understanding of steam turbine operation within thermodynamic systems.