Vapour power cycle: Rankine cycle

Why it matters

Most of India's electricity, and the process steam used in sugar, paper, chemical and steel plants, comes from steam power plants that run on the Rankine cycle. Many large factories also run captive or cogeneration steam plants. Knowing how boiler pressure, superheat, condenser pressure, reheat and feedwater heating change efficiency explains how real plants are designed and operated.

Key ideas

Why not a Carnot vapour cycle? A Carnot cycle using wet steam would need to compress a wet mixture (difficult for pumps) and would limit the maximum temperature to the saturation temperature. The Rankine cycle fixes this by condensing completely to liquid before pumping and by allowing superheating.

The ideal Rankine cycle (state numbering 1 = pump inlet).

• 1→2 Isentropic compression of saturated liquid in the pump to boiler pressure. Pump work is small because liquid has a tiny specific volume.
• 2→3 Constant-pressure heat addition in the boiler: compressed liquid is heated, evaporated and (usually) superheated.
• 3→4 Isentropic expansion in the turbine to condenser pressure, ending as wet steam.
• 4→1 Constant-pressure heat rejection in the condenser to saturated liquid. Each component is a steady-flow device, so the SFEE gives each work and heat as an enthalpy difference.

Using steam tables. Enthalpies and entropies come from steam tables or the Mollier (h–s) chart, not from ideal-gas formulas. In the wet region, a property y = y_f + x·y_fg, where x is the dryness fraction. Find the turbine exit state from s₄ = s₃; then x₄ = (s₃ − s_f)/s_fg at condenser pressure. Values used in examples are data-book values — always take them from your own steam table.

Back-work ratio. Pump work is typically under 1–2% of turbine work, compared with 40–60% for a gas turbine. This is the big advantage of compressing a liquid. In rough calculations pump work is sometimes neglected, but GATE problems usually expect you to include it.

Ways to raise efficiency (raise the mean temperature of heat addition or lower that of heat rejection).

• Lower condenser pressure: lowers the temperature of heat rejection and increases turbine work, but increases moisture at the turbine exit and needs good vacuum and cooling water.
• Superheat the steam: raises the mean temperature of heat addition and makes the exit steam drier. Limited by metallurgy (about 600 °C for conventional plants).
• Raise boiler pressure: raises the mean temperature of heat addition, but makes the turbine exhaust wetter at a fixed maximum temperature. Exit dryness should generally stay above about 0.88–0.90 to limit blade erosion.
• Reheat: expand partly in a high-pressure turbine, reheat at intermediate pressure, expand again in a low-pressure turbine. Its main purpose is to permit high boiler pressure without excessive exit moisture; efficiency rises modestly.
• Regenerative feedwater heating: bleed some steam from the turbine to preheat the feedwater (open or closed heaters). Heat is then added in the boiler at a higher mean temperature, so efficiency rises. All large plants use several heaters.

Real cycles. Turbine and pump irreversibilities are described by isentropic efficiencies: η_T = (h₃ − h₄)/(h₃ − h₄s) and η_P = (h₂s − h₁)/(h₂ − h₁). Pressure losses in piping and heat losses also reduce performance. Heat rate (heat input per kWh of output) and specific steam consumption (kg of steam per kWh) are plant-level measures.

Formulas

w_P = v_f·(p₂ − p₁) = h₂ − h₁

• Pump work (kJ/kg); v_f specific volume of saturated liquid (m³/kg); p in kPa.

w_T = h₃ − h₄

• Turbine work (kJ/kg).

q_in = h₃ − h₂, q_out = h₄ − h₁

• Boiler and condenser heat (kJ/kg).

η = (w_T − w_P)/q_in = 1 − q_out/q_in

• Cycle thermal efficiency.

x₄ = (s₄ − s_f)/s_fg, h₄ = h_f + x₄·h_fg

• Wet-steam state at turbine exit; s (kJ/kg·K), h (kJ/kg) from steam tables.

η_T = (h₃ − h₄)/(h₃ − h₄s)

• Isentropic efficiency of turbine.

SSC = 3600/w_net

• Specific steam consumption (kg/kWh), w_net in kJ/kg.

ṁ = P/w_net

• Steam mass flow (kg/s) for net power P (kW).

Worked examples

Steam-table data used below (take from your data book; given here as data): at 10 kPa, h_f = 191.81 kJ/kg, h_fg = 2392.1 kJ/kg, s_f = 0.6492 kJ/kg·K, s_fg = 7.4996 kJ/kg·K, v_f = 0.00101 m³/kg. At 3 MPa and 350 °C, h = 3116.1 kJ/kg, s = 6.7450 kJ/kg·K.

Example 1 (standard): ideal Rankine cycle. Given: steam at 3 MPa, 350 °C enters the turbine; condenser pressure 10 kPa. Find the dryness at the turbine exit, the net work and the efficiency.

1. Pump: w_P = v_f·(p₂ − p₁) = 0.00101 × (3000 − 10) = 3.02 kJ/kg; h₂ = 191.81 + 3.02 = 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 = h₃ − h₄ = 3116.1 − 2136.1 = 980.0 kJ/kg
4. w_net = 980.0 − 3.02 = 976.9 kJ/kg
5. q_in = h₃ − h₂ = 3116.1 − 194.83 = 2921.3 kJ/kg
6. η = 976.9/2921.3 = 0.334 Answer: x₄ ≈ 0.813; w_net ≈ 977 kJ/kg; η ≈ 33.4% (back-work ratio only about 0.3%)

Example 2 (GATE level): real turbine and plant size. Given: the same cycle, but the turbine isentropic efficiency is 85% (pump ideal). The plant must deliver 10 MW net. Find the actual exit dryness, efficiency and steam flow.

1. w_T,actual = 0.85 × 980.0 = 833.0 kJ/kg
2. h₄ = 3116.1 − 833.0 = 2283.1 kJ/kg; x₄ = (2283.1 − 191.81)/2392.1 = 0.874
3. w_net = 833.0 − 3.02 = 829.9 kJ/kg
4. η = 829.9/2921.3 = 0.284
5. ṁ = 10 000/829.9 = 12.05 kg/s Answer: x₄ ≈ 0.874; η ≈ 28.4%; ṁ ≈ 12.0 kg/s — note the irreversibility makes the exhaust drier but cuts efficiency.

Common mistakes

• Forgetting pump work, or computing it with p in bar instead of kPa (v·Δp gives kJ/kg only with kPa).
• Taking h₄ from saturated vapour instead of finding x₄ from s₃ = s₄.
• Using h_g instead of h_f at the pump inlet (it is saturated liquid).
• Applying the isentropic efficiency the wrong way round for a turbine and a pump.
• Thinking reheat mainly raises efficiency; its main job is to keep the turbine exhaust dry at high boiler pressure.
• Using ideal-gas relations for steam.

For GATE PI

Expect Rankine-cycle numericals where enthalpies are supplied (turbine work, pump work, efficiency, steam flow for a given power), dryness at turbine exit from entropy values, and the effect of turbine efficiency. Conceptual questions ask about the effect of boiler pressure, superheat, condenser pressure, reheat and regeneration on efficiency and exit dryness. Practise reading saturated tables quickly and using y = y_f + x·y_fg.

Quick check

1. Why is pump work in a Rankine cycle so small?
2. What is the main purpose of reheating?
3. Turbine work 1000 kJ/kg, pump work 10 kJ/kg, heat input 3000 kJ/kg. Efficiency?
4. Does lowering condenser pressure make the turbine exhaust wetter or drier?

Answers: 1. Liquid has a very small specific volume, so ∫v dp is small 2. To allow high boiler pressure without excessive moisture at the turbine exit 3. 990/3000 = 33% 4. Wetter