Rankine cycle and its modifications

The ideal Rankine cycle, pump and turbine work from steam tables, effects of boiler pressure, condenser pressure and superheat, reheat, regeneration with open and closed feedwater heaters, and real-cycle losses.

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Why it matters

The Rankine cycle is the ideal cycle of steam power plants, which generate much of the world's electricity, and of the steam and organic-fluid bottoming cycles in combined-cycle gas-turbine plants. For an aerospace engineer it is the reference vapour cycle: it shows how phase change makes compression almost free, and its modifications — superheat, reheat, regeneration — mirror those of the Brayton cycle.

Key ideas

Ideal Rankine cycle (steady flow, per kg of water):

  • 1→2 isentropic compression of saturated liquid in the pump (condenser pressure to boiler pressure)
  • 2→3 constant-pressure heat addition in the boiler (compressed liquid → saturated or superheated vapour)
  • 3→4 isentropic expansion in the turbine
  • 4→1 constant-pressure heat rejection in the condenser to saturated liquid

Why not Carnot? A Carnot vapour cycle would need to compress a wet mixture (impractical) and limits the maximum temperature to the saturation temperature. Rankine fully condenses the steam so that a small pump handles liquid, and allows superheat.

Pump work is tiny. Liquid is nearly incompressible, so w_p = v·(p_high − p_low) — a few kJ/kg against turbine work of about 1000 kJ/kg. The back-work ratio is around 1%, compared with 40–60% for gas turbines.

Raising efficiency — all three raise the average temperature of heat addition or lower that of heat rejection:

  • Lower condenser pressure: lowers T of heat rejection (a condenser at 10 kPa runs near 46 °C). Limited by cooling-water temperature; increases moisture at turbine exit.
  • Superheat: raises average heat-addition temperature and reduces exit moisture. Limited by metallurgy (about 600–620 °C).
  • Higher boiler pressure: raises the saturation temperature at which boiling occurs, but at fixed turbine inlet temperature it increases exit moisture. Exit quality below about 0.88–0.90 erodes the last-stage blades.

Reheat. Expand in a high-pressure turbine, return the steam to the boiler and reheat it, then expand in a low-pressure turbine. Its main purpose is to allow high boiler pressure without excessive exit moisture; efficiency also rises slightly when the reheat temperature is high.

Regeneration (feedwater heating). Steam is bled from the turbine to preheat the feedwater, so heat is added at a higher average temperature. Open (direct-contact) heaters mix bleed steam and feedwater to saturated liquid at the bleed pressure and need a pump after each heater; they also de-aerate. Closed heaters are shell-and-tube exchangers; streams do not mix, so the feedwater can stay at boiler pressure, but heat transfer is less complete. Large plants use several of each.

Real cycle. Turbine and pump irreversibilities (isentropic efficiencies), pressure drops in the boiler and piping, and heat losses all reduce efficiency.

Formulas

w_p = v₁·(p₂ − p₁) (ideal pump, v ≈ v_f at condenser pressure) h₂ = h₁ + w_p w_T = h₃ − h₄; q_in = h₃ − h₂; q_out = h₄ − h₁ η = (w_T − w_p)/q_in = 1 − q_out/q_in x₄ = (s₃ − s_f)/s_fg (isentropic expansion into the wet region, at condenser pressure) h₄ = h_f + x₄·h_fg η_T = (h₃ − h₄)/(h₃ − h₄s) Open feedwater heater: y·h_bleed + (1 − y)·h_in = h_out (y = bled fraction)

  • h: kJ/kg; s: kJ/(kg·K); v: m³/kg; p: kPa (kPa·m³/kg = kJ/kg)
  • subscripts f, g, fg: saturated liquid, vapour, difference; x: quality
  • Steam-table values below are standard data, quoted as given; use your own data book in exams.

Worked examples

Example 1 (standard). An ideal Rankine cycle: steam enters the turbine at 3 MPa, 350 °C and the condenser is at 10 kPa. Find the net work, efficiency and turbine-exit quality. Data: at 3 MPa, 350 °C, h = 3116.1 kJ/kg, s = 6.7450 kJ/(kg·K). At 10 kPa: v_f = 0.00101 m³/kg, h_f = 191.81, h_fg = 2392.1 kJ/kg, s_f = 0.6492, s_fg = 7.4996 kJ/(kg·K).

  1. Pump: w_p = v_f·(p₂ − p₁) = 0.00101 × (3000 − 10) = 3.02 kJ/kg; h₂ = 194.83 kJ/kg
  2. Turbine exit: x₄ = (6.7450 − 0.6492)/7.4996 = 0.8128; h₄ = 191.81 + 0.8128 × 2392.1 = 2136.1 kJ/kg
  3. w_T = 3116.1 − 2136.1 = 980.0 kJ/kg; q_in = 3116.1 − 194.83 = 2921.3 kJ/kg
  4. w_net = 980.0 − 3.0 = 977.0 kJ/kg; η = 977.0/2921.3 = 0.334

Answer: w_net = 977 kJ/kg, η = 33.4%, x₄ = 0.813

Example 2 (GATE level). Steam enters the HP turbine at 15 MPa, 600 °C (h = 3583.1, s = 6.6796) and expands isentropically to 4 MPa, where h = 3155.0 kJ/kg. It is reheated to 600 °C at 4 MPa (h = 3674.9, s = 7.3706) and expands isentropically in the LP turbine to 10 kPa (data as in Example 1). Find the efficiency and exit quality, and compare with no reheat (which gives η = 43.0% and x = 0.804).

  1. Pump: w_p = 0.00101 × (15 000 − 10) = 15.14 kJ/kg; h_pump-exit = 206.95 kJ/kg
  2. LP exit: x = (7.3706 − 0.6492)/7.4996 = 0.8962; h = 191.81 + 0.8962 × 2392.1 = 2335.7 kJ/kg
  3. w_T = (3583.1 − 3155.0) + (3674.9 − 2335.7) = 428.1 + 1339.2 = 1767.3 kJ/kg
  4. q_in = (3583.1 − 206.95) + (3674.9 − 3155.0) = 3376.2 + 519.9 = 3896.1 kJ/kg
  5. η = (1767.3 − 15.1)/3896.1 = 0.450

Answer: η = 45.0% (up from 43.0%), exit quality 0.896 (up from 0.804) — reheat mainly fixes the moisture problem.

Common mistakes

  • Forgetting the pump work, or calculating it with v of steam instead of liquid.
  • Using h_g at condenser pressure for the turbine exit instead of finding x from s.
  • Saying higher boiler pressure always helps; at fixed turbine inlet temperature it increases moisture.
  • Mixing kPa and MPa in v·Δp.
  • In feedwater heaters, forgetting that only (1 − y) of the steam passes through the low-pressure turbine and the condenser.

For GATE AE

Expect efficiency and work calculations for simple and reheat Rankine cycles with table values supplied, the effect of boiler pressure, condenser pressure and superheat on efficiency and exit quality, and one-mark questions on open versus closed feedwater heaters and why the pump work is small. Practise reading s at the turbine inlet and getting x at the exit quickly.

Quick check

  1. Turbine work 1500 kJ, pump work 100 kJ, boiler heat 4000 kJ: find η.
  2. Why is pump work small compared with turbine work?
  3. What is the main reason for reheat?
  4. Does lowering condenser pressure raise or lower efficiency?
  5. Which feedwater heater mixes bleed steam directly with feedwater?

Answers: 1. 35%. 2. The pump handles liquid with a very small specific volume. 3. To limit moisture at the turbine exit when the boiler pressure is high. 4. Raises it. 5. The open (direct-contact) heater.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Rankine cycle and why is it important in thermodynamics?Concept

    The Rankine cycle is a thermodynamic cycle used to convert heat into work, commonly used in power generation. It consists of four main processes: isentropic expansion in a turbine, isobaric heat rejection in a condenser, isentropic compression in a pump, and isobaric heat addition in a boiler. It is important because it forms the basis for steam power plants, which are a major source of electricity worldwide.

  2. 2.Explain the modifications that can be made to the basic Rankine cycle to improve its efficiency.Concept

    Modifications to the Rankine cycle to improve efficiency include superheating, reheating, and regeneration. Superheating involves heating the steam beyond its saturation temperature before it enters the turbine, increasing the work output. Reheating involves expanding the steam in multiple stages with reheating in between, reducing moisture content at the turbine exit. Regeneration involves using steam to preheat the feedwater before it enters the boiler, reducing the heat required from external sources.

  3. 3.Why is superheating used in the Rankine cycle?Application

    Superheating is used in the Rankine cycle to increase the thermal efficiency of the cycle. By increasing the temperature of the steam above its saturation point before it enters the turbine, the average temperature at which heat is added to the cycle is increased. This leads to a higher thermal efficiency according to the Carnot principle. Additionally, superheating reduces the moisture content of the steam at the turbine exit, which helps prevent turbine blade erosion.

  4. 4.What happens if the condenser pressure in a Rankine cycle is increased?Application

    If the condenser pressure in a Rankine cycle is increased, the thermal efficiency of the cycle decreases. This is because the increase in pressure raises the temperature at which heat is rejected, reducing the temperature difference between the heat source and sink. A smaller temperature difference results in lower efficiency according to the Carnot principle. Additionally, higher condenser pressure can lead to higher back pressure on the turbine, reducing the net work output.

  5. 5.How does reheating improve the efficiency of the Rankine cycle?Application

    Reheating improves the efficiency of the Rankine cycle by reducing the moisture content of the steam at the turbine exit and increasing the average temperature at which heat is added. In a reheating process, steam is expanded in a high-pressure turbine, reheated in the boiler, and then expanded in a low-pressure turbine. This process allows for more work to be extracted from the steam and reduces the risk of turbine blade erosion due to moisture.

  6. 6.What is the purpose of regeneration in the Rankine cycle?Application

    The purpose of regeneration in the Rankine cycle is to increase the cycle's thermal efficiency by preheating the feedwater before it enters the boiler. This is achieved by extracting some steam from the turbine and using it to heat the feedwater in a feedwater heater. By doing so, the amount of heat required from the external source is reduced, leading to improved efficiency. Regeneration also helps in reducing the thermal stresses in the boiler.

  7. 7.Calculate the thermal efficiency of an ideal Rankine cycle with a boiler pressure of 15 MPa, condenser pressure of 10 kPa, and turbine inlet temperature of 600°C.Numerical

    From steam tables: at 15 MPa, 600 °C, h₃ = 3583.1 kJ/kg and s₃ = 6.6796 kJ/(kg·K); at 10 kPa, h_f = 191.81, h_fg = 2392.1, s_f = 0.6492, s_fg = 7.4996, v_f = 0.00101. Isentropic expansion gives x₄ = (6.6796 − 0.6492)/7.4996 = 0.804 and h₄ = 2115.3 kJ/kg, so w_T = 1467.8 kJ/kg; pump work = 0.00101 × 14 990 = 15.1 kJ/kg and q_in = 3583.1 − 206.9 = 3376.2 kJ/kg. η = (1467.8 − 15.1)/3376.2 ≈ 0.430, i.e. about 43%, but the exit quality of 0.80 is too wet in practice, which is why such plants use reheat.

  8. 8.What is the effect of increasing the boiler pressure on the Rankine cycle efficiency?Application

    Increasing the boiler pressure in the Rankine cycle generally increases the thermal efficiency. This is because higher boiler pressure increases the average temperature at which heat is added to the cycle, which according to the Carnot principle, improves efficiency. However, there are practical limits due to material strength and cost considerations, as higher pressures require more robust and expensive equipment.

  9. 9.Explain how the use of a closed feedwater heater differs from an open feedwater heater in the Rankine cycle.Concept

    An open (direct-contact) heater mixes bleed steam with feedwater so that the mixture leaves as saturated liquid at the bleed pressure; it is simple, gives the best possible heat transfer and also de-aerates the water, but every open heater needs its own pump to raise the water to the next pressure. A closed heater is a shell-and-tube exchanger: the streams do not mix, so the feedwater can stay at boiler pressure and only one feed pump is needed, but the feedwater leaves somewhat below the bleed-steam saturation temperature and the condensed bleed must be drained back or pumped forward. Closed heaters are more expensive; large plants use several closed heaters plus one open heater as a de-aerator.

  10. 10.Steam enters a turbine at 10 MPa and 500°C and expands to a condenser pressure of 5 kPa. If the turbine isentropic efficiency is 85%, calculate the actual work output per kg of steam.Numerical

    From steam tables: at 10 MPa, 500 °C, h₁ = 3375.1 kJ/kg and s₁ = 6.5995 kJ/(kg·K); at 5 kPa, s_f = 0.4762, s_fg = 7.9176, h_f = 137.75, h_fg = 2423.0. Isentropic exit quality x₂s = (6.5995 − 0.4762)/7.9176 = 0.773, so h₂s = 137.75 + 0.773 × 2423.0 = 2011.7 kJ/kg and the isentropic work is 1363.4 kJ/kg. Actual work = 0.85 × 1363.4 ≈ 1159 kJ/kg, and the actual exit enthalpy is 3375.1 − 1159 ≈ 2216 kJ/kg (drier than the isentropic exit).

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