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3.1 · Analyze a simple ideal Rankine cycle

Learn to analyze a simple ideal rankine cycle through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Vapour Power Cycles

State-by-state balances for a water-vapour power cycle

The working fluid is water circulating in a closed loop. Model the pump, boiler, turbine, and condenser as steady-flow control volumes. State 1 is the condenser exit and pump inlet; state 2 is the pump exit; state 3 is the boiler exit and turbine inlet; state 4 is the turbine exit. The ideal model assumes an incompressible-liquid estimate for pump work, isentropic pump and turbine processes, constant-pressure heat addition and rejection, and negligible kinetic- and potential-energy changes. These are assumptions, not property data. Use only values supplied in a problem or obtained from a named water-property source.

What you will learn

  • Identify the four states and ideal processes in a simple Rankine cycle.
  • Apply steady-flow energy balances to its pump, boiler, turbine, and condenser.
  • Calculate heat transfer, work, and thermal efficiency from supplied property data.
  • Check units, signs, the cycle energy balance, and physical plausibility.

1. Define the loop and energy convention

The process order is pump from state 1 to 2, boiler from 2 to 3, turbine from 3 to 4, and condenser from 4 to 1. With no leaks or branches, the same mass flow passes through each component. Apply a steady-flow energy balance to each component.
For a steady-flow component with negligible kinetic- and potential-energy changes, heat transfer into the device minus work transfer out equals outlet enthalpy minus inlet enthalpy.
The pump consumes work and the turbine produces work. Because the working fluid returns to its initial state after a complete cycle, the cycle’s net energy change is zero. Therefore net work output equals boiler heat input less condenser heat rejection. This gives an independent check on component calculations.
q - w = h_{out} - h_{in}
  • State 1: condenser exit and pump inlet; state 2: pump exit.
  • State 3: boiler exit and turbine inlet; state 4: turbine exit.
  • The pump requires work input; the turbine supplies work output.

2. Determine states from the ideal-process assumptions

At state 1, water is liquid at the condenser pressure. For the ideal pump estimate, multiply the liquid specific volume at state 1 by the pressure rise. Pressure in kilopascals multiplied by specific volume in cubic metres per kilogram gives energy per unit mass in kilojoules per kilogram.
Heat is added in the boiler at approximately constant pressure from state 2 to state 3. Determine state 3 from the stated boiler-exit condition and property data. For an ideal turbine, inlet and exit entropy are equal. Combine that relation with the turbine-exit pressure and use supplied data to determine state 4.
The condenser rejects heat at approximately constant pressure and returns the fluid to state 1. A pressure–specific-volume or temperature–entropy plot can show process order, but a schematic plot does not provide numerical coordinates or enthalpies.
w_{p,in} \simeq v_1(p_2 - p_1), h_2 \simeq h_1 + w_{p,in}, s_4 = s_3
  • The incompressible-liquid pump estimate uses state-1 specific volume.
  • Equal turbine inlet and exit entropy follows from the ideal isentropic-turbine assumption.
  • Keep model assumptions separate from supplied or looked-up property values.

3. Calculate heat, work, and thermal efficiency

For the boiler and condenser, neglecting work and kinetic- and potential-energy changes, heat transfer per unit mass follows from the enthalpy change. For the turbine and pump, neglecting heat transfer, work transfer per unit mass follows from the enthalpy change. Using positive magnitudes, the boiler adds heat, the condenser rejects heat, the turbine delivers work, and the pump consumes work.
Find net cycle work output by subtracting pump work input from turbine work output. Thermal efficiency is net work output divided by boiler heat input. Keep specific transfers in kilojoules per kilogram.
Check the complete-cycle balance independently: boiler heat input less condenser heat rejection should equal net work output.
Thermal efficiency is given by ηth=wnetqin\eta_{\mathrm{th}} = \frac{w_{\mathrm{net}}}{q_{\mathrm{in}}}.
  • Calculate turbine output and pump input separately before finding net work.
  • Thermal efficiency uses boiler heat input, not net heat transfer, as its denominator.
  • Use the complete-cycle energy balance as an independent check.

4. Use mixture properties only when the state supports them

Before looking up a property, list what is known at each state: pressure, a specified temperature or phase condition, and any supplied enthalpy or entropy. Apply the ideal-process relations to connect states, then obtain needed values from supplied data.
If state 4 is established as a saturated liquid–vapour mixture, quality is the vapour mass fraction. The quality relation uses saturated-liquid enthalpy and enthalpy of vaporization at that pressure. A value between zero and one is consistent with a saturated mixture.
x=h−hfhfgx = \frac{h - h_f}{h_{fg}}
  • A schematic plot is not a substitute for water-property data.
  • Quality applies only to a saturated liquid–vapour mixture.
  • Check pressure units before using the pump-work estimate.

Worked example

Calculate ideal-cycle efficiency

A water Rankine cycle operates between 10 kPa and 8.00 MPa. Supplied properties are h1=191.8 kJ/kg and v1=0.00101 m^3/kg at state 1, and h3=3330 kJ/kg and h4=2100 kJ/kg at states 3 and 4. The supplied state-4 enthalpy is the result of an ideal-turbine property lookup. Assume an ideal pump, an isentropic turbine, constant-pressure boiler and condenser, and negligible kinetic- and potential-energy changes. Find pump work input, boiler heat input, turbine work output, net work, and thermal efficiency.
Ideal Rankine cycle, schematic P–v plot
Ideal Rankine cycle, schematic P–v plotSpecific volume, vPressure, PPumpBoilerTurbineCondenser11223344Schematic · not to scale

Water is the working fluid. State numbers show process order. This P–v plot is schematic and not to scale.

  1. Estimate pump work input
    The pressure rise is 7990 kPa. Multiplying by the supplied liquid specific volume gives kJ/kg because 1 kPa m^3 is 1 kJ. w_{p,in} \simeq (0.00101)(7990) = 8.07\ kJ/kg, h_2 \simeq 191.8 + 8.07 = 199.87\ kJ/kg
  2. Find heat input and turbine work
    Boiler heat input is the enthalpy increase across the boiler. Turbine work output is the enthalpy decrease across the turbine. Both are positive magnitudes. q_{in} = 3330 - 199.87 = 3130.13\ kJ/kg, w_{t,out} = 3330 - 2100 = 1230\ kJ/kg
  3. Calculate net work and efficiency
    Subtract pump work input from turbine work output, then divide net work output by boiler heat input.
    wnet=1230−8.07=1221.93 kJ/kg,ηth=1221.933130.13=0.390w_{\text{net}} = 1230 - 8.07 = 1221.93\ \text{kJ/kg}, \quad \eta_{\text{th}} = \frac{1221.93}{3130.13} = 0.390
Answer: Pump work input is 8.07 kJ/kg; boiler heat input is 3130.13 kJ/kg; turbine work output is 1230 kJ/kg; net work output is 1221.93 kJ/kg; thermal efficiency is 39.0%.
Check: Condenser heat rejection is 1908.2 kJ/kg. The heat difference is 1221.93 kJ/kg, matching net work.

Worked example

Find turbine-exit quality

At the turbine-exit pressure, supplied saturated-water data are hf=191.8 kJ/kg and hfg=2392 kJ/kg. The supplied ideal-turbine exit enthalpy is h4=2050 kJ/kg. Find exit quality.
Turbine
TurbineTurbine34w_outControl-volume schematic

Single turbine component schematic.

  1. Calculate quality
    Use the vapour mass fraction formula for a saturated mixture.
    x4=2050−191.82392=0.777x_4 = \frac{2050-191.8}{2392} = 0.777
Answer: Turbine-exit quality is 0.777.
Check: Quality is between zero and one, consistent with a saturated mixture.

Worked example

Convert to plant power

Boiler heat-input rate is 18.0 MW. Specific heat input is 3130.13 kJ/kg and specific net work is 1221.93 kJ/kg. Find mass flow rate and net power.
Power Plant
Power PlantPlantinoutQ_inW_netControl-volume schematic

Overall power plant balance.

  1. Find mass flow rate
    Convert MW to kW.
    m˙=180003130.13=5.75 kg/s\dot m = \frac{18000}{3130.13} = 5.75\ \text{kg/s}
  2. Find net power
    Multiply mass flow by specific work.
    W˙net=(5.75)(1221.93)=7026 kW\dot W_{\text{net}} = (5.75)(1221.93) = 7026\ \text{kW}
Answer: Mass flow rate is 5.75 kg/s and power is 7.03 MW.
Check: Net power is less than boiler input.

Common mistakes and how to avoid them

Treating turbine work output as cycle net work.
Correction: Subtract pump work input from turbine work output.
Guessing steam properties from a schematic plot.
Correction: Use supplied data or a named property source.
Setting turbine inlet and exit entropy equal without stating the assumption.
Correction: The equality follows from the ideal isentropic-turbine assumption.
Mixing pressure units in the pump-work estimate.
Correction: Using kPa with m^3/kg yields kJ/kg.

Lesson summary

  • Label states 1–4 as condenser exit, pump exit, boiler exit, and turbine exit.
  • Use the incompressible-liquid pump estimate and isentropic-turbine assumption.
  • Calculate heat and work from enthalpy changes.
  • Calculate thermal efficiency as net work output divided by boiler heat input.
  • Check that net work equals heat input less heat rejection.

Check your understanding

Question 1

How is Rankine-cycle net work per kilogram calculated?
  1. Turbine work plus pump work
  2. Turbine work minus pump work
  3. Pump work minus boiler heat
  4. Condenser heat minus turbine work
Show answer and explanation
Turbine work minus pump work
Net work is turbine output minus pump input.

Question 2

An ideal turbine inlet entropy is 6.80 kJ/(kg·K). What is the exit entropy?
  1. 0 kJ/(kg·K)
  2. 6.80 kJ/(kg·K)
  3. Greater than 6.80
  4. Less than 6.80
Show answer and explanation
6.80 kJ/(kg·K)
Isentropic means entropy is constant.

Question 3

If boiler heat input is 2400 kJ/kg and condenser heat rejection is 1500 kJ/kg, what is net work?
  1. 900 kJ/kg
  2. 1500 kJ/kg
  3. 2400 kJ/kg
  4. 3900 kJ/kg
Show answer and explanation
900 kJ/kg
Net work is input minus rejection.

Key terms

Rankine cycle
A vapour power-cycle model with a pump, boiler, turbine, and condenser.
Specific enthalpy
Enthalpy per unit mass.
Quality
The vapour mass fraction in a saturated mixture.
Thermal efficiency
Net work output divided by boiler heat input.

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