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S.1 The Basic Steam CycleHistorically, the first functioning power cycle is the steam cycle, which commonlyis working with water vapor (steam). The Rankine cycle is an ideal case from thecommon steam cycle.Steam power plants constitutes around 80% of the world's electric powergeneration.ABB Steam Turbine ATP-4P1.1 AcknowledgementsAuthor: Samuel Roy, <strong>KTH</strong>, 1999;Author: Catharina Erlich, <strong>KTH</strong>, 2005; updated 2006Reviewer: Catharina Erlich, <strong>KTH</strong>, 2005CompEdu support: Vitali Fedulov, <strong>KTH</strong>, 2005P1.2 Literature (recommended further reading)Elliot. T Chen, K and Swanekamp, R. 1997; "Standard Handbook of Power plantengineering", ISBN 0070194351Granryd E., Ekroth, I; 1991 "Applied Thermodynamics", ISBN 91-7170-067-6Moran M. J., Shapiro H. N., 1998;Fundamentals of engineering thermodynamics: SI versionJohn Wiley & sons ltd, ISBN 0471979600Weston, K, 1992 "Energy Conversion – The EBook",http://www.personal.utulsa.edu/~kenneth-weston/P1.3 PrerequisitesIt is expected that the reader has knowledge about:Basic thermodynamics (at least 160 LU = 4 weeks of fulltime studies),At least one year of studies in an engineering career at university level.1

P1.4 LU and TULearning Units: 3Teaching Units: 1P1.5 ATP 4 LayoutTypical plantlayout for anATP 4condensing/extracting steamturbine.Back pressure or condensing turbine for a wide range of uses, including:• IPP and utility grade power stations and combined cycle plants,• Waste-to-energy plants,• Pulp and paper and other industrial process,• Data manufacturer: ABB,• Electrical output: up to 100 MW,• Turbine outlet: back pressure mode: 3 – 16 bar,condensing mode: 0,03 – 0,25 bar.S.2 Educational ObjectivesAfter this chapter the student should:Know and understand the limitations for the basic steam cycle.Be able to make and analysis of the basic steam cycle through its design, T-s & h-sdiagrams and the thermal efficiency.2

Know the effect of irreversibilities in the pump and steam turbine and be able toanalyze this numerically and illustrative in a T-s or h-s diagram,S.3 Development of a realistic cycle from Carnot cycleThe Carnot-cycle is a theoretical cycle whichcannot be realized. Main reasons are:- Second law of thermodynamics: Allprocesses are irreversible and as a result,entropy always increases.- Working fluid properties and phasechanges.- Limitations in the practical equipmenthandling the working fluid, such aspumps, turbines and heat exchangers.A steam cycle, (or vapor cycle), is a cycle inwhich the working fluid is compressed,vaporized, expanded and condensed;thereafter the cycle repeats.The state of a fluid and then its phase isdefined by its pressure p, volume V andtemperature T.Different phases of a substance:CP means Critical Point, whichfor water is 221 bar and 374ºCOther viewsP3.1 Other views3

Longitudinal section at criticalvolumeProjection of the 3D surfaceS.4 Components of a Basic Steam CycleLiquid water is compressed to a highpressure by a pump.The pressurized water is heated andvaporized in a boiler, which is fuelled bycoal, oil, gas, biomass or nuclear fission.Hot and compressed water vapor hashigh energy content, which is utilized bythe turbine generating work.After expansion in the turbine the vaporenters a condenser, which brings thevapor to liquid form.P4.1 PumpA pump is a device, which increases the pressure of a liquid.Pumps can generally not handle liquid-vapor mixtures, only pure liquids.Compressing a liquid requires very little work compared to compressing a gas.This is because a liquid will not gain much temperature with the increasedpressure, only a few degrees, meanwhile a gas being compressed rapidlyincreases in temperature. The liquid is incompressible, which means thatincreased pressure does not change the volume.From first law of thermodynamics:4

Power input = mass flow · (enthalpy outlet — enthalpy inlet )P4.2 BoilerThe boiler is a component containing a furnace and heat exchangers.In the furnace coal, oil, gas, biomass, municipal solid waste or any other fuel iscombusted generating a hot flue gas.This flue gas is heat exchanged with water, so that the pressurized water (from thepump) is preheated in the so-called economizer.Thereafter the hot pressurized liquid water is evaporated to steam in theevaporator. In the basic steam cycle this saturated steam is expanded in theturbine.A special case of boiler is the nuclear reactor, in which the economizer andevaporator are integrated with the reactor. Thus, this boiler has no furnace. Mostnuclear power plants work with saturated steam, meanwhile all other steam plantswork with superheated steam (see next chapter).The heating process in the boiler is separated into 2 steps:preheating and steam generation.P4.3 TurbineA turbine is a work-generating machine consisting of astatic part (stator) and a rotating part (rotor).On the rotor there are angled blades attached, with acertain size and distance, that are catching up the pressureand velocity of the working fluid.The larger the blades and the bigger the distance betweenthe blades, the lower becomes the pressure of the passingsteam, i.e. the smallest blades are found in the turbine inletand the largest in the turbine outlet (opposite direction in acompressor).The high pressure and temperature of vapor causes therotor to rotate, and thus the vapor looses pressure andtemperature while passing through the turbine (= expands).5

The rotational movement (mechanical work) is transferredto electrical energy in a generator.The steam is expanded down to the pressure held in thecondenser = pressure in the turbine outlet. The steamleaving the turbine most often wet, i.e. the steam contains alittle amount of liquid.However to avoid erosion on the turbine blades, the largefraction has to be in vapor form. The fraction is calledsteam quality (see next chapter).P4.4 CondenserA condenser is a heat exchanger that brings thewet steam to liquid form with help of coolingwater from, for example, a lake or a river.A cold-condenser has a pressure below theatmospheric one on the steam side, and thispressure is created with ejectors.The lower the temperature of the cooling water,the lower the pressure that can be created by theejectors and the more the steam is let to beexpanded in the turbine.A turbine working with a cold-condenser iscalled Condensing turbine/Condensing powerplant.There are also hot-condensers, where thecooling media is district heating water orindustrial process water. In this case thepressure on the steam side is higher than theatmospheric pressure.A turbine working with a hot-condenser is calledBackpressure turbine.Condensing is needed so that the pump is ableto increase the pressure of the working fluid.6

S.5 T-s and h-s Diagrams for an Ideal Steam CycleThe ideal steam cycle is also called Rankine Cycle:Process 1-2: Isentropic pressure increase by the pump,Process 2-3: Heat supply in the boiler,Process 3-4: Isentropic expansion in the turbine,Process 4-1 Condensing in the condenserP5.1 IdealIdeal means no irreversibilities through the cycle:• no frictional pressure drops,• no heat transfer with the surroundings,• isentropic (constant entropy, s) compression and expansion.P5.2 Process 1-2: Isentropic pressure increaseThe working fluid is pumped adiabatically and isentropically and the pressureincreases. Pumping requires a power input (for example mechanical or electrical).The POWER input required is the massflow of liquid through the pump times theenthalpy increase:P P( − )= m& ⋅ h 2h 1 [kg/s · kJ/kg = kW ]h 1 is determined from tables or diagram knowing temperature and pressure of theliquid before the pump. However, as a liquid is incompressible (constant volumeand specific heat capacity), the enthalpy can be calculated as:T ref = 273 K (normal conditions).h 1 = c p,liquid · (T – T ref )7

The specific heat capacity for liquid water is: c p = 4.18 kJ/(kg K) as long as thewater stays in liquid form.h 2 = h 2s (s denotes isentropic) is determined by the constant entropy and thepressure after the pump, either numerically or graphically.Example:Saturated liquid water with a pressure of 0.2 bar is pumped to 60 bar. The mass flowof water is 50 kg/s. Assume that the pump works isentropically. What is the powerinput of the pump?Numerical solution:Saturated liquid water at 0.2 bar (corresponds to the saturation temperature 60.0°C) istabulated in saturation tables* for water: h 1 = 251.4 kJ/kg.It can also be read that the entropy for this state is s 1 = 0.8320 kJ/(kg K).Now increasing the pressure of the saturated liquid , the state will no longer besaturated: it will be a sub-cooled liquid. Proceeding isentropically, the entropy afterthe pump is the same, i.e. s 2 = 0.8320 (kJ/kg K). The pressure is 60 bar and with thevalue of entropy, in a water/steam table* it can be found that the outlet temperature, t 2 ,is t 2 = 60.3°C and with the pressure and temperature that h 2s = 257.5 kJ/kg.Thus:P pump = 50· (257.5 – 251.4) kW = 305 kW* Interactive table for all properties of water/steam and saturation at different pressures andtemperatures is found in S1B2C6 "Water steam thermodynamic properties"P5.3 Process 2-3: Heat supplyThe high-pressure liquid enters a boiler where it is heated at constant pressure byan external heat source. The working fluid is heated from state 2 to state 2’ (liquidsaturation), and then vaporized to become saturated vapor (state 3).The heat needed for vaporization of a liquid is dependent on the pressure. Oftenthis heat of vaporization heat is tabulated for different pressures.A liquid is vaporized at constant temperature (isothermal), going from saturatedliquid state (2') to saturated gaseous state (3).The heating rate to the cycle is (neglecting efficiency of the boiler and thus thelosses):Q&= m&⋅( − )h 3h 2[kg/s · kJ/kg = kW ]The enthalpy h 3 is saturated steam and is found in saturation tables forwater/steam at the pressure in question. The enthalpy h 2 is the state of the waterafter the pump.Example:8

Water of 60 bar pressure, 60.3°C and 50 kg/s is to be brought to gaseous state(=saturated steam). Which heating power is needed?Solution:Enthalpy of water at 60 bar and 60.3°C is: h 2 = 257.5 kJ/kg (from a table).The boiling temperature at 60 bar is 275.6°C (from a saturation table for water/steam)and the enthalpy for saturated vapor is h 3 = 2785.0 kJ/kg.Thus, the heating rate needed is:( 2785.0 − 257.5) kW = 126375kW126. MWQ&= 50⋅≈ 4P5.4 Process 3-4: Isentropic expansionThe vapor expands through a turbine to generate power output. Ideally, thisexpansion is adiabatic and isentropic. The expansion decreases the temperatureand pressure of the vapor.For the basic steam cycle the steam goes from saturated vapor form to a mixtureof saturated vapor and saturated liquid at the pressure determined by thecondenser in the turbine outlet.The power output of the turbine is:P T( − )= m& ⋅ h 3h 4 [kg/s · kJ/kg = kW]For the basic steam cycle, the enthalpy h 3 is saturated vapor and can be found inwater steam tables.For isentropic expansion, h 4 = h 4s (s denotes isentropic) is determined by theconstant entropy and the pressure in the turbine outlet (= condenser pressure),either numerically or graphically.Example:Saturated steam at 60 bar expands isentropically in a turbine to a pressure of 0.2 bar(= condenser pressure). The steam flow is 50 kg/s. Which is the turbine power output?Solution:The enthalpy for steam at 60 bar saturation condition is h 3 = 2785.0 kJ/kg (water/steamtable or h-s/T-s diagram). The entropy for steam at this condition is s 3 = 5.89 kJ/(kg K).The expansion takes place isentropically i.e. s 4 = s 3 = 5.89 kJ/(kg K) down to p 4 = 0.2bar.Solving the enthalpy graphically in an h-s or T-s diagram is easiest. Starting from h 3 =2785 kJ/kg and p = 60 bar, a straight vertical line (=isentropic) is drawn untilintersected with the 0.2 bar pressure line. Alternatively, knowing the entropy and9

pressure of state 4 as well gives the enthalpy in the same diagram. The enthalpy h 4can now be read directly.h 4 = h 4s ≈ 1950 kJ/kgObserve that state 4 is within the saturation region in the diagram, i.e. state 4 is notsaturated liquid or saturated vapor, rather a mixture between these two conditions.The power output is now calculated as:( 2785−1950) kW = 41750kW41. MWP T= 50⋅≈ 810

P5.4.1h-s diagramP5.5 Process 4-1: CondensingThe wet vapor is brought to saturated liquid in the condenser so that the pump isable to start the cycle again.The heat lost in the condenser is thus a little bit less than the vaporization heat (asthe vapor is wet).The heat rejected in the condenser is (numbered subscripts for enthalpyaccording to the diagram):11

&Q condenser= m&⋅( − )h 4h 1[kg/s · kJ/kg = kW ]Example:Wet steam with the enthalpy 1950 kJ/kg enters the condenser at a pressure of 0.2bars. How much heat is rejected in the condenser if the steam flow is 50kg/s?Solution:State 1, i.e. after the condenser (or before the pump) is saturated liquid water thush 1 = 251.3 kJ/kg (from saturated steam/water tables).The heat rejected is thus:( 1950−251.3) kW = 84935kW89. MWQ&condenser= 50⋅≈ 4S.6 Irreversibilities in the Pump and the TurbineT-s and h-s diagrams with losses in the pump and the turbineThe isentropic efficiency for a process is describing the irreversibility compared tothe ideal process, i.e. the isentropic process (constant entropy).For an isentropic process, the efficiency is 1.For an irreversible process, the isentropic efficiency is < 1. The lower theefficiency the larger the irreversibility for the process, i.e. larger increase ofentropy.Definitions of isentropic efficiencies for a pump and for a turbine.P6.1 PumpIsentropic efficiency for a pump is:12

h2s − h1η P = which gives thath2− h1W P = WrealPisentropic⋅1ηpRemember! For a work-requiring device, for example compressor or pump, thereal amount of work that has to be supplied is larger than in the ideal case andentropy increases during the compression.P6.2 TurbineIsentropic efficiency for a turbine is:h− h3 4η T =which gives thath3− h4sWTreal= W T isentropicRemember! For a work-giving device, such as a turbine, the real amount ofwork that can be obtained is less than in the ideal case and entropy increasesduring the expansionExample:A turbine is working with 60 bar saturated steam at 50 kg/s that expands down to 0.2bar. The isentropic efficiency is 90% for the turbine. What is the power output for theturbine?Solution:Proceed first as in the ideal case. Determine the enthalpy of steam in the turbine inlet(60bar, saturated steam): h 3 = 2785.0 kJ/kg.Isentropic expansion gives from state 3 to 4 gives h 4s ≈ 1950 kJ/kg.Now the real enthalpy in 4 can be calculated from the definition of the isentropicefficiency. Solving for h 4 :h 4 = h 3 - η ST ⋅(h 3 – h 4s ) = 2785 – 0.90⋅(2785 – 1950) kJ/kg = 2033.5 kJ/kgThe power output is:P T = 50⋅(2785.0 – 2033.5) kW = 37575 kW ≈ 37.6 MWObserve that for a turbine, the real outlet enthalpy is larger than the isentropic,resulting in less power output for the real case than for the ideal.⋅ηT13

S.7 Cycle Net Work and Thermal EfficiencyThe net work output is:W = W T −W PHowever, pump work

More details about boiler efficiency can be found in the Combustion shelf.Example:In a boiler, water of 60 bar pressure and 60.3°C is heated and vaporised to saturatedsteam (under constant pressure) with a flow of 50 kg/s. The boiler efficiency is 85%.What is the fuel input in kW?Solution:Enthalpy of water at 60bar and 60.3°C is h 2 = 257.5 kJ/kg.Enthalpy of saturated steam at 60bar is h 3 = 2785 kJ/kg.The heat transferred to the steam cycleQ = 50*(2785 – 257.5) = 126 375 kWThe fuel input in the boiler is:QB126375QFuel = = kW = 148680kWη 0.85BThus observe that the worse is the boiler efficiency, the more fuel that has to besupplied.P7.3 Example of efficiency calculationExample:A steam turbine in a basic steam cycle gives 37 575 kW with a fuel input in the boilerof 148 680 kW. What is the efficiency?Solution:Neglecting the pump work, the net efficiency of the steam cycle is:η =37575148680= 0.253 → 25.3%This says that in this basic steam cycle, only 25.3 % of the energy available in the fuelbecomes useful work.S.8 SummaryThe basic steam cycle consists of a pump, boiler, turbine and condenser.In enthalpy-entropy or temperature-entropy diagrams, enthalpies for the differentstates of the working fluid can be obtained in order to calculate power output andcycle efficiency.15

The prevailing irreversibilities take place in the turbine and pump and areassessed by the isentropic efficiency of the expansion and compression.In real processes, the entropy during expansion and compression alwaysincreases.In a steam cycle, pump work can be neglected as pump work