An ideal Rankine cycle uses water as a working fluid

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An ideal Rankine cycle uses water as a working fluid. The water circulates at a rate of 80 kg/s. The boiler pressure is 6 MPa, and the condenser pressure is 10 kPa. The water enters the turbine at 600°C and leaves the condenser as a saturated liquid. Assume that heat is transferred to the working fluid in the boiler from a reservoir at 1400 K and that the fluid in the condenser rejects heat to the surroundings at 25°C. (a) Calculate the rate of entropy in the cycle. Is this power cycle viable? (b) Calculate the irreversibility of the cycle by calculating (w_carnot_cycle – w_actual_cycle).

The Correct Answer and Explanation is:

To analyze the ideal Rankine cycle described, we’ll proceed step-by-step. Given:

  • Mass flow rate of water, ṁ = 80 kg/s
  • Boiler pressure = 6 MPa
  • Turbine inlet temperature = 600°C
  • Condenser pressure = 10 kPa
  • Water leaves condenser as saturated liquid
  • Boiler heat source temperature = 1400 K
  • Surroundings (condenser heat sink) temperature = 25°C = 298 K

(a) Entropy Generation (ΔS_gen) in the Cycle

Steps in the ideal Rankine cycle:

  1. Boiler (3→4): Water enters as compressed liquid and exits as superheated vapor (600°C, 6 MPa).
  2. Turbine (4→1): Isentropic expansion to 10 kPa.
  3. Condenser (1→2): Heat rejection at constant pressure.
  4. Pump (2→3): Isentropic compression from 10 kPa to 6 MPa.

From steam tables:

  • At state 4 (600°C, 6 MPa):
    • Enthalpy, h₄ ≈ 3642 kJ/kg
    • Entropy, s₄ ≈ 7.022 kJ/kg·K
  • At state 1 (10 kPa, saturated mixture after isentropic expansion):
    • Use isentropic expansion: s₁ = s₄ = 7.022 kJ/kg·K
    • At 10 kPa, s_f = 0.6492, s_fg = 7.5010 → x = (s₁ – s_f)/s_fg ≈ (7.022 – 0.6492)/7.501 ≈ 0.85
    • h₁ = h_f + x·h_fg = 191.81 + 0.85×2392.82 ≈ 2224.2 kJ/kg
  • At state 2: Saturated liquid at 10 kPa → h₂ = 191.81 kJ/kg
  • At state 3: Compressed liquid at 6 MPa → h₃ ≈ h₂ + v·ΔP ≈ 191.81 + 0.00101×(6000–10) ≈ 197.87 kJ/kg

Heat and Work:

  • Turbine work = h₄ – h₁ = 3642 – 2224.2 = 1417.8 kJ/kg
  • Pump work = h₃ – h₂ = 197.87 – 191.81 = 6.06 kJ/kg
  • Net work = 1417.8 – 6.06 = 1411.74 kJ/kg
  • Boiler heat input = h₄ – h₃ = 3642 – 197.87 = 3444.13 kJ/kg

Now, entropy generation in the cycle:

  • Ideal Rankine cycle is internally reversible → entropy is generated due to heat transfer with finite ΔT at source/sink.
  • Entropy transferred from heat source: S˙in=Q˙inTH=80×3444.131400≈196.6 kW/K\dot{S}_{in} = \frac{\dot{Q}_{in}}{T_H} = \frac{80 \times 3444.13}{1400} ≈ 196.6 \ \text{kW/K}S˙in​=TH​Q˙​in​​=140080×3444.13​≈196.6 kW/K
  • Entropy rejected to sink: Q˙out=h1−h2=2224.2–191.81=2032.39 kJ/kg⇒S˙out=80×2032.39298≈545.5 kW/K\dot{Q}_{out} = h₁ – h₂ = 2224.2 – 191.81 = 2032.39 \ \text{kJ/kg} \Rightarrow \dot{S}_{out} = \frac{80 \times 2032.39}{298} ≈ 545.5 \ \text{kW/K}Q˙​out​=h1​−h2​=2224.2–191.81=2032.39 kJ/kg⇒S˙out​=29880×2032.39​≈545.5 kW/K
  • Net entropy generation: \Delta S_{gen} = \dot{S}_{out} – \dot{S}_{in} = 545.5 – 196.6 = **348.9 \ \text{kW/K**

Conclusion: Since ΔS_gen > 0, the cycle is irreversible. However, it is viable because it’s within the bounds of the second law (entropy increases).

(b) Irreversibility (I):

Carnot efficiency:ηCarnot=1−TLTH=1−2981400≈0.7871\eta_{Carnot} = 1 – \frac{T_L}{T_H} = 1 – \frac{298}{1400} ≈ 0.7871ηCarnot​=1−TH​TL​​=1−1400298​≈0.7871

  • Carnot work = η_Carnot × q_in = 0.7871 × 3444.13 ≈ 2711.2 kJ/kg
  • Irreversibility per kg = w_Carnot – w_actual = 2711.2 – 1411.74 = 1299.5 kJ/kg
  • Total irreversibility rate: I=m˙×(wCarnot−wactual)=80×1299.5≈∗∗103,960kW∗∗I = \dot{m} \times (w_{Carnot} – w_{actual}) = 80 \times 1299.5 ≈ **103,960 kW**I=m˙×(wCarnot​−wactual​)=80×1299.5≈∗∗103,960kW∗∗

Summary Explanation

The Rankine cycle is a fundamental thermodynamic cycle widely used in steam power plants. In this case, an ideal Rankine cycle operates with water as the working fluid, expanding through a turbine and condensing at low pressure. The turbine inlet conditions are high-pressure superheated steam (600°C at 6 MPa), and the condenser pressure is maintained at 10 kPa.

To evaluate entropy generation, we recognize that although the internal processes of an ideal Rankine cycle are reversible, entropy is still generated due to the heat transfer between finite temperature reservoirs. Specifically, heat is added in the boiler from a reservoir at 1400 K and rejected in the condenser to surroundings at 298 K (25°C). The entropy generation arises due to these thermal interactions.

By calculating the rate of heat addition and rejection, and relating them to their respective reservoir temperatures, we determine the entropy flow into and out of the system. The net entropy generation is positive (348.9 kW/K), confirming the second law of thermodynamics: real processes increase the entropy of the universe. Thus, the cycle is thermodynamically viable, though not perfectly efficient.

To assess irreversibility, we compare the actual work output of the Rankine cycle to that of an ideal Carnot cycle operating between the same thermal reservoirs. The Carnot cycle provides a theoretical upper limit on efficiency. The difference in work outputs—1299.5 kJ/kg—quantifies the irreversibility caused by real-world constraints. For the given mass flow rate, the total irreversibility is about 103.96 MW. This loss represents the energy that could not be converted into useful work due to the entropy increase and other inefficiencies. Understanding and minimizing this irreversibility is central to improving thermal system design.

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