Rankine Cycle and Steam Power Plants
The Rankine cycle on the T-s diagram, pump and turbine work, thermal efficiency, effects of pressure and superheat, reheat and regeneration, and steam rate.
Steam power plants run on the Rankine cycle. It is the practical version of the Carnot cycle for a working fluid that changes phase.
The ideal Rankine cycle
| Process | Device | Energy |
|---|---|---|
| 1 → 2: isentropic compression | Pump | w_p = h₂ − h₁ ≈ v₁(p₂ − p₁) |
| 2 → 3: constant-pressure heating | Boiler | q_in = h₃ − h₂ |
| 3 → 4: isentropic expansion | Turbine | w_t = h₃ − h₄ |
| 4 → 1: constant-pressure cooling | Condenser | q_out = h₄ − h₁ |
Thermal efficiency η = (w_t − w_p)/q_in.
Back work ratio = w_p/w_t is tiny (about 1%), because pumping a liquid needs little work. This is the big advantage over gas cycles.
Specific steam consumption (steam rate) = 3600/(w_t − w_p) kg/kWh (with w in kJ/kg).
Raising efficiency
- Lower condenser pressure: lowers the mean temperature of heat rejection; but the turbine exhaust gets wetter.
- Higher boiler pressure: raises the mean temperature of heat addition; also makes the exhaust wetter.
- Superheating: raises efficiency and dries the exhaust.
- Reheat: expand partly, reheat at lower pressure, expand again. Mainly keeps the exhaust dry (x ≥ 0.88 or so) at high boiler pressure.
- Regeneration: bleed steam to heat the feedwater (open or closed feedwater heaters), raising the mean temperature of heat addition.
Mean temperature of heat addition: T_m = (h₃ − h₂)/(s₃ − s₂). Efficiency ≈ 1 − T_condenser/T_m: anything that raises T_m helps.
Actual cycle
Turbine and pump losses are given by isentropic efficiencies:
- η_t = (h₃ − h₄)/(h₃ − h₄s)
- η_p = (h₂s − h₁)/(h₂ − h₁)
Worked example
Given h₁ = 192 kJ/kg, pump work 10 kJ/kg, h₃ = 3400 kJ/kg and h₄ = 2300 kJ/kg:
- h₂ = 202 kJ/kg; q_in = 3400 − 202 = 3198 kJ/kg.
- w_net = 1100 − 10 = 1090 kJ/kg.
- η = 1090/3198 = 34.1%; steam rate = 3600/1090 = 3.30 kg/kWh.
Practice
Water (v = 0.001 m³/kg) is pumped from 10 kPa to 10 MPa. What is the ideal pump work in kJ/kg? (Give 2 decimal places.)
Show answer and explanation
Answer: 9.99w_p = vΔp = 0.001 × 9990 = 9.99 kJ/kg.
A Rankine cycle has net work 1000 kJ/kg. What is its specific steam consumption in kg/kWh?
Show answer and explanation
Answer: 3.6SSC = 3600/w_net = 3.6 kg/kWh.
The ideal turbine enthalpy drop is 1000 kJ/kg and the actual drop is 850 kJ/kg. What is the isentropic efficiency?
Show answer and explanation
Answer: 0.85η_t = actual/ideal = 850/1000 = 0.85.
The main purpose of reheating in a Rankine cycle is to:
Show answer and explanation
Answer: BReheat keeps the steam dry enough at the last turbine stages; it can also raise efficiency slightly.
Which of these raise the efficiency of a Rankine cycle? (Select all that apply.)
Show answer and explanation
Answer: A, B, CEach of the first three raises the mean temperature of heat addition or lowers that of rejection.