3.1 Steam cycles

A steam cycle is a closed loop in which the same water is boiled, expanded through an engine or turbine to give work, condensed, and pumped back to the boiler. Boiler houses in factories and workshops, and steam power stations, run on such cycles. A technician meets them in boiler tubes, steam pipework and turbine casings, and when judging how pressure and temperature affect the working life of plant and of welded joints. This topic describes four cycles: the Rankine cycle (the basic practical cycle), the Carnot cycle (the ideal against which all cycles are judged), the reheat cycle and the regenerative cycle (two ways of improving the Rankine cycle).

Property values used in the worked examples (specific enthalpy h in kJ/kg, specific entropy s in kJ/kg·K, specific volume v in m³/kg) are from the steam tables based on the IAPWS-IF97 formulation. Printed steam tables differ from them in the last digit, so a student's answer may differ slightly. All pressures are absolute.

Four quantities are used to judge a cycle:

• Thermal efficiency = net work output ÷ heat supplied in the boiler.
• Work ratio = net work output ÷ gross work of the turbine. It is near 1 when the pump work is small.
• Specific steam consumption (SSC) = 3600 ÷ net work per kg, in kg of steam per kWh. A lower figure means a smaller plant for the same power.
• Net work per kg of steam, in kJ/kg. Multiplied by the steam flow in kg/s, it gives the power in kW.

3.1.1 Rankine

The Rankine cycle is the basic cycle of a steam power plant. It has four processes: heating and boiling at constant pressure, expansion in the turbine, condensing at constant pressure, and pumping the liquid back to the boiler pressure. It differs from the Carnot cycle because the steam is fully condensed to liquid before it is pumped, so the pump handles a liquid and needs little work.

Worked example used in this section. A boiler supplies steam at 4 MPa and 400 °C to a turbine. The condenser pressure is 10 kPa (saturation temperature 45.81 °C). The turbine expansion is taken as ideal (constant entropy) and the feed pump as ideal.

3.1.1.1 Schematic diagram of a steam engine or turbine plant

A steam plant has four main units joined in a loop.

        +---------------------+   3    +-----------+
        |  Boiler and         |------->|  Turbine  |-----> shaft work
        |  superheater        |        |  (or      |       to generator
        |                     |        |  engine)  |       or machine
        +---------------------+        +-----------+
                 ^                           |
                 | 2                         | 4
                 |                           v
        +---------------------+   1    +-----------+   cooling water in
        |  Feed pump          |<-------| Condenser |<------------------
        |  (work input)       |        |           |------------------>
        +---------------------+        +-----------+   cooling water out

The states are numbered as follows:

• State 1: saturated liquid leaving the condenser (10 kPa, 45.81 °C).
• State 2: compressed liquid leaving the feed pump, at boiler pressure (4 MPa).
• State 3: steam leaving the superheater and entering the turbine (4 MPa, 400 °C).
• State 4: wet steam leaving the turbine and entering the condenser (10 kPa).

The four processes are: 1 to 2, pumping of the liquid (work input); 2 to 3, heating in the boiler and superheater (heat input); 3 to 4, expansion in the turbine (work output); 4 to 1, condensing (heat rejected to the cooling water). In a steam engine the cylinder and piston take the place of the turbine, and the steps are the same.

3.1.1.2 Determine Rankine efficiency

The efficiency is the net work divided by the heat supplied:

η = [(h₃ − h₄) − (h₂ − h₁)] ÷ (h₃ − h₂)

where h₃ − h₄ is the turbine work and h₂ − h₁ is the pump work. The pump work is small, so the cycle efficiency is often written as η ≈ (h₃ − h₄) ÷ (h₃ − h₁).

Worked example (4 MPa, 400 °C, 10 kPa):

State	Description	h (kJ/kg)	Source
1	Saturated liquid at 10 kPa	191.81	Steam table
2	Compressed liquid at 4 MPa	195.84	h₁ + pump work
3	Steam at 4 MPa, 400 °C (s = 6.7712)	3214.37	Steam table
4	Wet steam at 10 kPa, dryness x = 0.8163	2144.46	h = h_f + x × h_fg

• Pump work: w_p = v₁ × (p₂ − p₁) = 0.001010 × (4000 − 10) = 4.03 kJ/kg, so h₂ = 191.81 + 4.03 = 195.84.
• Dryness after the turbine: states 3 and 4 have the same entropy, 6.7712 kJ/kg·K. At 10 kPa, s_f = 0.6492 and s_g = 8.1489, so x = (6.7712 − 0.6492) ÷ (8.1489 − 0.6492) = 0.8163. Then h₄ = 191.81 + 0.8163 × (2583.89 − 191.81) = 2144.46.
• Turbine work: 3214.37 − 2144.46 = 1069.91 kJ/kg.
• Net work: 1069.91 − 4.03 = 1065.88 kJ/kg.
• Heat supplied: 3214.37 − 195.84 = 3018.53 kJ/kg.
• Efficiency: η = 1065.88 ÷ 3018.53 = 0.3531, or 35.31 %.
• Heat rejected: h₄ − h₁ = 2144.46 − 191.81 = 1952.65 kJ/kg. Check: 3018.53 − 1065.88 = 1952.65.

Three changes raise the Rankine efficiency: a higher boiler pressure, a higher superheat temperature, and a lower condenser pressure. Each limit comes from material strength at the pipe and boiler walls, and from the wetness of the steam leaving the turbine, which erodes blades.

3.1.1.3 T-S and h-s graphs

A temperature-entropy (T-s) graph has entropy on the horizontal axis and temperature on the vertical axis. The saturation dome is drawn first: the liquid line on the left, the vapour line on the right, and the two meeting at the top at the critical point. The area under any heating curve on a T-s diagram is the heat added, and the area enclosed by the cycle is the net work.

Points to plot for the worked example:

State	T (°C)	s (kJ/kg·K)	h (kJ/kg)
1 Saturated liquid at 10 kPa	45.81	0.6492	191.81
2 Compressed liquid at 4 MPa	about 46	0.6492	195.84
2a Saturated liquid at 4 MPa	250.36	2.7967	1087.43
2b Saturated vapour at 4 MPa	250.36	6.0697	2800.90
3 Superheated steam at 4 MPa	400	6.7712	3214.37
4 Wet steam at 10 kPa	45.81	6.7712	2144.46

To draw the T-s diagram: join 1 to 2 with a short almost vertical line (pumping at nearly constant entropy). Join 2 to 2a with a curve that rises along the liquid line (heating the liquid), 2a to 2b with a horizontal line (boiling at 250.36 °C), and 2b to 3 with a rising curve in the superheat region. Join 3 to 4 with a vertical line (constant entropy expansion) down to the 45.81 °C line, then 4 to 1 with a horizontal line to the left (condensing at constant pressure and temperature).

An enthalpy-entropy (h-s) graph has entropy on the horizontal axis and enthalpy on the vertical axis. The same points are used. The ideal expansion 3 to 4 is a vertical line, and its length is the turbine work: 3214.37 − 2144.46 = 1069.91 kJ/kg. The h-s graph is more useful for calculations because the work is read directly as a length.

3.1.1.4 Modified Rankine Cycle

In a reciprocating steam engine the cylinder would have to be very large to let the steam expand all the way down to the condenser pressure, because the steam volume at low pressure is large. To save size and cost, the expansion in the cylinder is stopped at a pressure p₄′ that is higher than the condenser pressure p_c. The steam is then released and its pressure falls to p_c at nearly constant volume, with no further work. This is the modified Rankine cycle.

On a p-v diagram the modified cycle differs from the Rankine cycle in one place. The Rankine cycle keeps expanding along the adiabatic curve from p₄′ down to p_c. The modified cycle drops vertically from p₄′ to p_c, so it loses the thin area at the toe of the diagram.

Worked example data (same boiler as before: 4 MPa, 400 °C, and condenser at 10 kPa). The expansion stops at p₄′ = 100 kPa (1 bar). The entropy stays 6.7712 kJ/kg·K, so at 100 kPa (s_f = 1.3026 and s_g = 7.3588) the dryness is x₄′ = (6.7712 − 1.3026) ÷ (7.3588 − 1.3026) = 0.9030. Then h₄′ = 417.44 + 0.9030 × (2674.95 − 417.44) = 2455.91 kJ/kg, and the specific volume is v₄′ = x × v_g = 0.9030 × 1.6940 = 1.5297 m³/kg.

3.1.1.5 Work-done using Rankine equations

The work of the ideal Rankine cycle comes from enthalpy differences, per kg of steam:

• Turbine work: w_t = h₃ − h₄
• Pump work: w_p = h₂ − h₁ ≈ v₁ × (p₂ − p₁)
• Net work: w_net = w_t − w_p
• Heat supplied: q_in = h₃ − h₂, made of three parts: the economiser or liquid heating (h₂ to saturated liquid), boiling (the latent heat h_fg) and superheating
• Heat rejected: q_out = h₄ − h₁
• Energy balance: q_in − q_out = w_net

For the worked example the heat supplied is divided as 1087.43 − 195.84 = 891.59 kJ/kg for heating the liquid, 2800.90 − 1087.43 = 1713.47 kJ/kg for boiling and 3214.37 − 2800.90 = 413.47 kJ/kg for superheating. The total is 3018.53 kJ/kg.

Performance figures for the worked example:

• Net work: 1065.88 kJ/kg
• Work ratio: 1065.88 ÷ 1069.91 = 0.996
• Specific steam consumption: 3600 ÷ 1065.88 = 3.377 kg/kWh
• Power from a steam flow of 5 kg/s: 5 × 1065.88 = 5329 kW

For the modified cycle, the work per kg between steam admission and exhaust is the expansion work plus the work represented by the pressure fall at constant volume. When the pump work is neglected, it is:

w_mod = (h₃ − h₄′) + (p₄′ − p_c) × v₄′

with pressure in kPa and v in m³/kg, which gives kJ/kg. The first term is the adiabatic expansion down to p₄′, and the second term is the work gained from the steam that is still at p₄′ when it is exhausted against the lower condenser pressure.

For the worked example: w_mod = (3214.37 − 2455.91) + (100 − 10) × 1.5297 = 758.46 + 137.67 = 896.13 kJ/kg.

3.1.1.6 Efficiency of Modified Rankine Cycle

The heat supplied is still taken from the feed water at condenser temperature (state 1) to the superheated steam, so the efficiency of the modified cycle, with pump work neglected, is:

η_mod = [(h₃ − h₄′) + (p₄′ − p_c) × v₄′] ÷ (h₃ − h₁)

For the worked example, the heat supplied is 3214.37 − 191.81 = 3022.56 kJ/kg, so η_mod = 896.13 ÷ 3022.56 = 0.2965, or 29.65 %.

The Rankine cycle with complete expansion, on the same basis (pump work neglected), has a net work of 1069.91 kJ/kg and an efficiency of 1069.91 ÷ 3022.56 = 0.3540, or 35.40 %. The modified cycle is therefore less efficient, which is the price paid for the smaller cylinder.

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

The loss of work is the difference between the work of the complete expansion and the work of the modified cycle:

loss = w_Rankine − w_mod = (h₄′ − h₄) − (p₄′ − p_c) × v₄′

where h₄ is the enthalpy at the end of complete expansion to p_c (state 4 of the Rankine cycle) and h₄′ is the enthalpy at the end of the shortened expansion.

For the worked example: loss = (2455.91 − 2144.46) − 137.67 = 311.45 − 137.67 = 173.78 kJ/kg. Check: 1069.91 − 896.13 = 173.78 kJ/kg. This is the work that the toe of the p-v diagram would have produced, and it is 16.2 % of the work of the complete expansion.

3.1.1.8 Loss in Rankine efficiency due to restricted expansion of steam

The loss in efficiency is the loss of work divided by the heat supplied:

loss in efficiency = loss of work ÷ (h₃ − h₁) = η_Rankine − η_mod

For the worked example: 173.78 ÷ 3022.56 = 0.0575, so the efficiency falls by 5.75 percentage points, from 35.40 % to 29.65 %. The steam consumption rises from 3600 ÷ 1069.91 = 3.365 to 3600 ÷ 896.13 = 4.017 kg/kWh.

The earlier the expansion is stopped (the higher p₄′), the larger the loss, and the loss falls to zero when p₄′ reaches the condenser pressure. Engine designers choose the cut-off point as a compromise between cylinder size and this loss.

3.1.1.9 Enthalpy-entropy chart

The enthalpy-entropy chart, also called the Mollier chart, has entropy on the horizontal axis and enthalpy on the vertical axis. It shows the saturation line, lines of constant pressure, lines of constant temperature in the superheat region and lines of constant dryness fraction in the wet region. Only the part of the chart above and around the saturation line is used for steam turbines.

To find the turbine work on the chart:

1. Locate the inlet state 3 where the 4 MPa pressure line meets the 400 °C line. Read h₃ = 3214 kJ/kg.
2. Draw a vertical line from state 3 downwards, because the ideal expansion has constant entropy.
3. Stop where the vertical line meets the 10 kPa pressure line (state 4). Read h₄ = 2144 kJ/kg, and the dryness fraction x = 0.816 from the lines of constant dryness.
4. The turbine work is the vertical distance between the two points: 3214 − 2144 = 1070 kJ/kg.

For a reheat cycle, the vertical line stops at the reheat pressure, the chart is followed along that pressure line to the reheat temperature, and a new vertical line is drawn to the condenser pressure. Values read from a chart are approximate to about 10 kJ/kg, so tables are used when more exact figures are needed.

3.1.2 Carnot

The Carnot cycle is the most efficient cycle that can work between a given hot source temperature T_H and a given cold sink temperature T_L. Its efficiency depends only on the two temperatures, measured in kelvin (K = °C + 273.15):

η_Carnot = 1 − T_L ÷ T_H

It is not built as a steam plant, but it is the standard against which every real cycle is judged. For example, a plant with steam at 400 °C and condensing at 45.81 °C could not exceed 1 − 318.96 ÷ 673.15 = 0.526, or 52.6 %. The Rankine cycle on these limits gives only 35.3 %.

3.1.2.1 Draw schematic diagrams of Carnot engine

A Carnot engine takes heat Q_H from a hot source, converts part of it to work W, and rejects the rest, Q_L, to a cold sink.

         Hot source at T_H
                |
                | Q_H (heat in)
                v
          +-----------+
          |  Carnot   |------> W (net work out)
          |  engine   |
          +-----------+
                |
                | Q_L (heat rejected)
                v
         Cold sink at T_L

W = Q_H − Q_L and η = W ÷ Q_H = 1 − Q_L ÷ Q_H = 1 − T_L ÷ T_H.

The engine works in a cycle of four reversible processes: heat added at constant temperature, expansion with no heat transfer (adiabatic), heat rejected at constant temperature, and compression with no heat transfer.

With steam as the working substance, the same four processes appear as a plant diagram.

        +--------+   2    +---------+   3    +-----------+
        | Boiler |------->| Turbine |------->| Condenser |
        |(boils) |        +---------+        | (partial) |
        +--------+                           +-----------+
            ^                                      |
            | 1                                    | 4
            |        +--------------------+        |
            +--------|  Compressor (wet   |<-------+
                     |  steam to liquid)  |
                     +--------------------+

3.1.2.2 Carnot Cycle with Steam as Working Substance

In the wet region of the steam, constant pressure also means constant temperature, so the isothermal processes are easy: boiling in the boiler and condensing in the condenser. The four processes are:

1. 1 to 2: heat added at constant pressure and temperature, boiling saturated liquid to saturated vapour in the boiler.
2. 2 to 3: isentropic expansion of the saturated vapour in the turbine down to the condenser pressure.
3. 3 to 4: heat rejected at constant pressure and temperature as the wet steam is partly condensed.
4. 4 to 1: isentropic compression of the wet steam back to saturated liquid at the boiler pressure.

Worked example. The boiler is at 2 MPa (saturation temperature 212.38 °C) and the condenser at 10 kPa (45.81 °C).

State	Description	h (kJ/kg)	s (kJ/kg·K)
1	Saturated liquid at 2 MPa	908.62	2.4470
2	Saturated vapour at 2 MPa	2798.38	6.3392
3	Wet steam at 10 kPa, x = 0.7587	2006.66	6.3392
4	Wet steam at 10 kPa, x = 0.2397	765.24	2.4470

The dryness fractions come from the entropy: x₃ = (6.3392 − 0.6492) ÷ (8.1489 − 0.6492) = 0.7587, and x₄ = (2.4470 − 0.6492) ÷ (8.1489 − 0.6492) = 0.2397. The enthalpies are h = 191.81 + x × 2392.08.

• Heat added: q_H = h₂ − h₁ = 2798.38 − 908.62 = 1889.76 kJ/kg.
• Turbine work: h₂ − h₃ = 2798.38 − 2006.66 = 791.72 kJ/kg.
• Compressor work: h₁ − h₄ = 908.62 − 765.24 = 143.38 kJ/kg.
• Net work: 791.72 − 143.38 = 648.34 kJ/kg.
• Efficiency: 648.34 ÷ 1889.76 = 0.3431, or 34.31 %. This equals 1 − 318.96 ÷ 485.53 = 0.3431.

The efficiency from the energy values equals the temperature formula, which confirms the calculation.

3.1.2.3 Performance Criteria for Carnot Cycle

The three criteria are applied to the Carnot cycle and compared with the Rankine cycle that has the same boiler and condenser conditions (saturated vapour at 2 MPa, condensing at 10 kPa, Rankine net work 789.71 kJ/kg and heat supplied 2604.56 kJ/kg).

Criterion	Carnot	Rankine
Thermal efficiency	34.31 %	30.32 %
Net work (kJ/kg)	648.34	789.71
Work ratio	648.34 ÷ 791.72 = 0.819	789.71 ÷ 791.72 = 0.997
Specific steam consumption (kg/kWh)	3600 ÷ 648.34 = 5.55	3600 ÷ 789.71 = 4.56

The Carnot cycle has the higher efficiency, as it must, but it is not used in practice for steam for three reasons:

• The compressor must compress a mixture of liquid and vapour, which is difficult and wears the machine. It also takes a large part (18 %) of the turbine work, which lowers the work ratio.
• The wet steam leaving the turbine has dryness only 0.759, so about 24 % of it is liquid droplets that erode the blades.
• The net work per kg of steam is smaller, so the steam consumption is higher and a larger plant is needed for the same power.

The Rankine cycle avoids these problems by condensing fully and pumping a liquid, and it can use superheat to raise the temperature at which heat is added.

3.1.3 Reheat

In the reheat cycle the steam is expanded in a high-pressure turbine, taken back to the boiler to be reheated at constant pressure, and then expanded in a low-pressure turbine. The purpose is to reduce the wetness of the steam at the end of expansion, which protects the blades, and to raise the efficiency a little.

The worked example continues with the earlier plant: 4 MPa, 400 °C, condenser 10 kPa. The steam leaves the high-pressure turbine at 1 MPa, is reheated to 400 °C at 1 MPa, and is expanded to 10 kPa.

3.1.3.1 T-S diagram of reheat steam cycle

The T-s diagram of the reheat cycle is the Rankine diagram with an extra loop. Points are numbered 1 (saturated liquid), 2 (pump outlet), 3 (high-pressure turbine inlet), 4 (high-pressure turbine exit, the reheat pressure), 5 (low-pressure turbine inlet after reheat) and 6 (condenser inlet).

State	p	T (°C)	s (kJ/kg·K)	h (kJ/kg)
1	10 kPa	45.81	0.6492	191.81
2	4 MPa	about 46	0.6492	195.84
3	4 MPa	400	6.7712	3214.37
4	1 MPa	215.3	6.7712	2864.66
5	1 MPa	400	7.4668	3264.39
6	10 kPa	45.81	7.4668	2366.32 (x = 0.9090)

To draw it: 1 to 2 is the pump (almost vertical). 2 to 3 is the heating, boiling and superheating at 4 MPa. 3 to 4 is a vertical line down to the 1 MPa pressure line. 4 to 5 follows the 1 MPa line to the right and up, as the steam is reheated to 400 °C. 5 to 6 is a vertical line down to the condenser temperature. 6 to 1 is the horizontal line of condensing.

State 6 lies further to the right than state 4 of the plain Rankine cycle, because the reheated steam has a higher entropy (7.4668 instead of 6.7712). The steam therefore leaves the turbine much drier: x = 0.909 instead of 0.816.

3.1.3.2 Determine work-done using reheat equations

The work comes from both turbines, and the pump is subtracted:

• High-pressure turbine work: w_HP = h₃ − h₄ = 3214.37 − 2864.66 = 349.71 kJ/kg
• Low-pressure turbine work: w_LP = h₅ − h₆ = 3264.39 − 2366.32 = 898.07 kJ/kg
• Total turbine work: 349.71 + 898.07 = 1247.78 kJ/kg
• Pump work: 4.03 kJ/kg
• Net work: 1247.78 − 4.03 = 1243.75 kJ/kg

The dryness x₆ comes from the entropy 7.4668 at 10 kPa: x = (7.4668 − 0.6492) ÷ 7.4997 = 0.9090, so h₆ = 191.81 + 0.9090 × 2392.08 = 2366.32. The enthalpy h₄ is read at 1 MPa and the entropy 6.7712, where the steam is still slightly superheated at 215.3 °C.

The net work is 1243.75 − 1065.88 = 177.87 kJ/kg more than in the plain Rankine cycle.

3.1.3.3 Determine efficiency of reheat cycles

The heat is supplied in two places, the boiler and the reheater:

• Heat in the boiler: q₁ = h₃ − h₂ = 3214.37 − 195.84 = 3018.53 kJ/kg
• Heat in the reheater: q₂ = h₅ − h₄ = 3264.39 − 2864.66 = 399.73 kJ/kg
• Total heat supplied: 3018.53 + 399.73 = 3418.26 kJ/kg

η_reheat = w_net ÷ (q₁ + q₂) = 1243.75 ÷ 3418.26 = 0.3639, or 36.39 %.

The plain Rankine cycle gave 35.31 %, so reheating raises the efficiency by 1.08 percentage points. The reason is that the extra heat of 399.73 kJ/kg gives an extra work of 177.87 kJ/kg, an efficiency of 44.5 %, which is higher than the 35.3 % of the main cycle. The steam consumption falls from 3.377 to 3600 ÷ 1243.75 = 2.894 kg/kWh.

Benefits of reheat:

• The steam leaves the turbine drier (x = 0.909 compared with 0.816), so there is less blade erosion and a longer turbine life.
• The work per kg of steam is larger, so the same power needs a smaller steam flow.
• The thermal efficiency rises, especially in large plants with high boiler pressure.

The cost is a more complex boiler with extra pipework and a reheater. Welded joints in the hot reheat pipe work at high temperature and need careful inspection.

3.1.4 Regenerative

In the regenerative cycle, some steam is taken out (bled) from the turbine part-way through the expansion and used to heat the feed water before it enters the boiler. The feed water then needs less heat from the fuel, and the steam that was bled has already done some work in the turbine and gives up its condensing heat to the feed water instead of to the cooling water.

3.1.4.1 Ideal regenerative cycle diagram

In the ideal regenerative cycle the feed water is heated by the steam while the steam expands, through the turbine casing itself, so that at each point the water and the steam are at nearly the same temperature. The heat transfer is then reversible. The feed water goes into the boiler at the saturation temperature, and all of the heat from the fuel is added at the boiler temperature T_H. The heat rejected is at T_L. For these conditions the efficiency equals that of the Carnot cycle:

η_ideal = 1 − T_L ÷ T_H

For saturated steam at 4 MPa (250.36 °C) and a condenser at 10 kPa (45.81 °C): η = 1 − 318.96 ÷ 523.51 = 0.3907, or 39.07 %.

   Feed water in (cool) ---> [ heated along the turbine casing ] ---> Boiler
                                      ^
                                      | heat flows from the expanding
                                      | steam to the water, in the
                                      | opposite direction
   Steam in (hot) ---> [ turbine, steam expands and cools ] ---> Condenser

The ideal cycle cannot be built, because the steam would become very wet inside the turbine as it gives up heat to the water, and the heat flow through the casing is impossible to arrange. It shows the aim of regeneration: to add heat to the cycle only at high temperature. Real plants approach it by bleeding steam at one or more points.

3.1.4.2 Regenerative Cycle with Single Feed Water Heater

In a practical regenerative plant, a fraction m of the steam mass is bled from the turbine at an intermediate pressure and mixed with the feed water in a feed water heater. In the open (direct-contact) heater, the bled steam mixes with the pump discharge and leaves as saturated liquid at the bleed pressure. A second pump then lifts this liquid to the boiler pressure.

The fraction bled is fixed by an energy balance on the heater. For 1 kg of steam entering the turbine, m kg is bled and (1 − m) kg passes on to the condenser:

m × h_bleed + (1 − m) × h_A = h_f(bleed)

so that m = (h_f(bleed) − h_A) ÷ (h_bleed − h_A), where h_A is the enthalpy of the liquid leaving the first pump.

Benefits of regeneration:

• The mean temperature at which heat is added is higher, so the efficiency rises.
• Less steam passes to the condenser, so less heat is rejected to the cooling water, and the low-pressure end of the turbine and the condenser can be smaller.
• The feed water entering the boiler is hot, so the thermal stresses in the boiler drum and tubes are lower.
• An open heater can also work as a deaerator, which removes dissolved air that would corrode boiler tubes.

3.1.4.3 Regenerative cycle with single feed water heater diagram

   +--------+  3   +---------+  4, (1 - m) kg
   | Boiler |----->| Turbine |------------------> Condenser (state 1)
   +--------+      +---------+                         |
       ^                |                              | (1 - m) kg
       | 7              | b: m kg bled at 0.5 MPa      v
   +--------+           v                          +--------+
   | Pump 2 |<--6-- Open feed water <------ 2 -----| Pump 1 |
   +--------+          heater (mixing)             +--------+

States for the worked example (boiler 4 MPa and 400 °C, bleed at 0.5 MPa, condenser 10 kPa):

State	Description	h (kJ/kg)
1	Saturated liquid at 10 kPa	191.81
2	After pump 1, at 0.5 MPa	192.307
3	Turbine inlet, 4 MPa, 400 °C	3214.37
b	Bled steam at 0.5 MPa (x = 0.9900)	2727.12
4	End of expansion at 10 kPa (x = 0.8163)	2144.46
6	Saturated liquid at 0.5 MPa (151.84 °C)	640.19
7	After pump 2, at 4 MPa	644.009

Pump work: w_p1 = 0.001010 × (500 − 10) = 0.495 kJ/kg and w_p2 = 0.001093 × (4000 − 500) = 3.824 kJ/kg.

3.1.4.4 Determine work-done by Regenerative cycle

The mass fraction bled is m = (640.19 − 192.307) ÷ (2727.12 − 192.307) = 447.883 ÷ 2534.813 = 0.1767.

The work per kg of steam entering the turbine:

• Turbine work, first section (all 1 kg): h₃ − h_b = 3214.37 − 2727.12 = 487.26 kJ/kg
• Turbine work, second section (1 − m = 0.8233 kg): 0.8233 × (2727.12 − 2144.46) = 0.8233 × 582.66 = 479.71 kJ/kg
• Total turbine work: 487.26 + 479.71 = 966.97 kJ/kg
• Pump work: (1 − m) × w_p1 + w_p2 = 0.8233 × 0.495 + 3.824 = 4.232 kJ/kg
• Net work: 966.97 − 4.23 = 962.74 kJ/kg

The net work per kg is smaller than the 1065.88 kJ/kg of the plain cycle because part of the steam leaves the turbine early, but less heat is needed from the boiler.

3.1.4.5 Determine Regenerative cycle efficiency

The heat supplied in the boiler is the difference between the turbine inlet and the feed water entering the boiler at state 7:

q_in = h₃ − h₇ = 3214.37 − 644.01 = 2570.36 kJ/kg

η = w_net ÷ q_in = 962.74 ÷ 2570.36 = 0.3746, or 37.46 %

This is 2.15 percentage points above the plain Rankine cycle at 35.31 %. Check by the heat rejected: (1 − m) × (h₄ − h₁) = 0.8233 × 1952.65 = 1607.62 kJ/kg, and 2570.36 − 962.74 = 1607.62 kJ/kg, which agree.

The steam consumption is 3600 ÷ 962.74 = 3.739 kg/kWh, higher than the Rankine cycle, because less work is obtained from each kg of steam.

3.1.4.6 Regenerative Cycle with Two Feed Water Heaters and its efficiency

A second bleed increases the benefit. The worked example uses two open heaters, bleeding at 1 MPa and 0.2 MPa, with three pumps (one after the condenser, one after each heater).

State	Description	h (kJ/kg)
1	Saturated liquid at 10 kPa	191.81
A	After pump 1, at 0.2 MPa (w_p = 0.19)	192.00
f2	Saturated liquid at 0.2 MPa (120.2 °C)	504.68
B	After pump 2, at 1 MPa (w_p = 0.85)	505.53
f1	Saturated liquid at 1 MPa (179.9 °C)	762.68
C	After pump 3, at 4 MPa (w_p = 3.38)	766.06
3	Turbine inlet, 4 MPa, 400 °C	3214.37
b1	Bled steam at 1 MPa (215.3 °C, slightly superheated)	2864.66
b2	Bled steam at 0.2 MPa (x = 0.9365)	2566.34
4	End of expansion at 10 kPa (x = 0.8163)	2144.46

The mass fractions come from an energy balance on each heater. For the high-pressure heater (1 MPa):

m₁ × 2864.66 + (1 − m₁) × 505.53 = 762.68, so m₁ = (762.68 − 505.53) ÷ (2864.66 − 505.53) = 257.15 ÷ 2359.13 = 0.1090.

For the low-pressure heater (0.2 MPa), with (1 − m₁) kg flowing through it:

m₂ × 2566.34 + (1 − m₁ − m₂) × 192.00 = (1 − m₁) × 504.68, so m₂ = (1 − m₁) × (504.68 − 192.00) ÷ (2566.34 − 192.00) = 0.8910 × 312.68 ÷ 2374.34 = 0.1173.

The mass reaching the condenser is 1 − 0.1090 − 0.1173 = 0.7737 kg.

• Turbine work: (3214.37 − 2864.66) + 0.8910 × (2864.66 − 2566.34) + 0.7737 × (2566.34 − 2144.46) = 349.71 + 265.81 + 326.39 = 941.91 kJ/kg
• Pump work: 0.7737 × 0.19 + 0.8910 × 0.85 + 3.38 = 0.15 + 0.76 + 3.38 = 4.29 kJ/kg
• Net work: 941.91 − 4.29 = 937.62 kJ/kg
• Heat supplied: 3214.37 − 766.06 = 2448.31 kJ/kg
• Efficiency: 937.62 ÷ 2448.31 = 0.3830, or 38.30 %

The fractions are carried with more digits than shown, so a hand calculation with the rounded values may differ in the last digit.

Summary of the cycles in the worked example, all with steam at 4 MPa and 400 °C and the condenser at 10 kPa:

Cycle	Efficiency	Net work (kJ/kg)	SSC (kg/kWh)
Rankine	35.31 %	1065.88	3.377
Reheat (1 MPa, 400 °C)	36.39 %	1243.75	2.894
Regenerative, one heater (0.5 MPa)	37.46 %	962.74	3.739
Regenerative, two heaters (1 MPa and 0.2 MPa)	38.30 %	937.62	3.84

Each added heater gives a smaller gain, so large power stations use several heaters, while a small plant uses one or two. Closed heaters, in which the bled steam heats the water through tube walls and the condensate is drained back, work on the same principle and are used when the water and steam must be kept apart.

Chapter Summary

This chapter explored various steam cycles, beginning with the Rankine cycle, where the schematic of a steam engine or turbine plant was examined alongside methods to determine its efficiency using T-S and h-s graphs. Modifications to the Rankine cycle were discussed to improve performance, including calculations of work done and efficiency, as well as analysis of theoretical losses due to incomplete and restricted expansion of steam illustrated on enthalpy-entropy charts. The Carnot cycle was introduced with schematic diagrams and performance criteria, highlighting its role as an idealized thermodynamic cycle using steam as the working substance. Reheat cycles were presented with T-S diagrams and equations to evaluate work done and efficiency improvements. The regenerative cycle was analyzed through idealized diagrams featuring single and multiple feed water heaters, with detailed methods to compute work output and efficiency gains. The chapter concluded with an examination of steam turbines, focusing on their characteristics, classification, and the dynamics of pressure and velocity within impulse turbines. Velocity triangles for moving blades were developed, culminating in the calculation of power produced by impulse turbines.