1

S.1 The Basic Steam Cycle

Historically, the first functioning power cycle is the steam cycle, which commonly is working with water vapor (steam). The Rankine cycle is an ideal case from the common steam cycle.

Steam power plants constitutes around 80% of the world's electric power generation.

ABB Steam Turbine ATP-4

P1.1 Acknowledgements

Author: Samuel Roy, KTH, 1999; Author: Catharina Erlich, KTH, 2005; updated 2006 Reviewer: Catharina Erlich, KTH, 2005

CompEdu support: Vitali Fedulov, KTH, 2005

P1.2 Literature (recommended further reading)

Elliot. T Chen, K and Swanekamp, R. 1997; "Standard Handbook of Power plant

engineering", ISBN 0070194351

Granryd E., Ekroth, I; 1991 "Applied Thermodynamics", ISBN 91-7170-067-6

Moran M. J., Shapiro H. N., 1998; Fundamentals of engineering thermodynamics: SI version John Wiley & sons ltd, ISBN 0471979600

Weston, K, 1992 "Energy Conversion – The EBook",

http://www.personal.utulsa.edu/~kenneth-weston/

P1.3 Prerequisites

It 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.

2

P1.4 LU and TU

Learning Units: 3 Teaching Units: 1

P1.5 ATP 4 Layout

Typical plant layout for an ATP 4 condensing /extracting steam turbine.

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 Objectives

After 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-s

diagrams and the thermal efficiency.

3

Know the effect of irreversibilities in the pump and steam turbine and be able to analyze this numerically and illustrative in a T-s or h-s diagram,

S.3 Development of a realistic cycle from Carnot cycle

The Carnot-cycle is a theoretical cycle which cannot be realized. Main reasons are: - Second law of thermodynamics: All

processes are irreversible and as a result, entropy always increases.

- Working fluid properties and phase changes.

- Limitations in the practical equipment handling the working fluid, such as pumps, turbines and heat exchangers.

A steam cycle, (or vapor cycle), is a cycle in

which the working fluid is compressed, vaporized, expanded and condensed; thereafter the cycle repeats.

The state of a fluid and then its phase is defined by its pressure p, volume V and temperature T.

Different phases of a substance:CP means Critical Point, which for water is 221 bar and 374ºC

Other views

P3.1 Other views

4

Longitudinal section at critical volume

Projection of the 3D surface

S.4 Components of a Basic Steam Cycle

Liquid water is compressed to a high pressure by a pump.

The pressurized water is heated and

vaporized in a boiler, which is fuelled by coal, oil, gas, biomass or nuclear fission.

Hot and compressed water vapor has

high energy content, which is utilized by the turbine generating work.

After expansion in the turbine the vapor

enters a condenser, which brings the vapor to liquid form.

P4.1 Pump

A 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 increased pressure, only a few degrees, meanwhile a gas being compressed rapidly increases in temperature. The liquid is incompressible, which means that increased pressure does not change the volume.

From first law of thermodynamics:

5

Power input = mass flow · (enthalpyoutlet — enthalpyinlet)

P4.2 Boiler

The boiler is a component containing a furnace and heat exchangers.

In the furnace coal, oil, gas, biomass, municipal solid waste or any other fuel is combusted generating a hot flue gas.

This flue gas is heat exchanged with water, so that the pressurized water (from the

pump) is preheated in the so-called economizer.

Thereafter the hot pressurized liquid water is evaporated to steam in the evaporator. In the basic steam cycle this saturated steam is expanded in the turbine.

A special case of boiler is the nuclear reactor, in which the economizer and

evaporator are integrated with the reactor. Thus, this boiler has no furnace. Most nuclear power plants work with saturated steam, meanwhile all other steam plants work with superheated steam (see next chapter).

The heating process in the boiler is separated into 2 steps:

preheating and steam generation.

P4.3 Turbine

A turbine is a work-generating machine consisting of a static part (stator) and a rotating part (rotor).

On the rotor there are angled blades attached, with a certain size and distance, that are catching up the pressure and velocity of the working fluid.

The larger the blades and the bigger the distance between the blades, the lower becomes the pressure of the passing steam, i.e. the smallest blades are found in the turbine inlet and the largest in the turbine outlet (opposite direction in a compressor).

The high pressure and temperature of vapor causes the rotor to rotate, and thus the vapor looses pressure and temperature while passing through the turbine (= expands).

6

The rotational movement (mechanical work) is transferred to electrical energy in a generator.

The steam is expanded down to the pressure held in the condenser = pressure in the turbine outlet. The steam leaving the turbine most often wet, i.e. the steam contains a little amount of liquid.

However to avoid erosion on the turbine blades, the large fraction has to be in vapor form. The fraction is called steam quality (see next chapter).

P4.4 Condenser

A condenser is a heat exchanger that brings the wet steam to liquid form with help of cooling water from, for example, a lake or a river.

A cold-condenser has a pressure below the atmospheric one on the steam side, and this pressure is created with ejectors.

The lower the temperature of the cooling water, the lower the pressure that can be created by the ejectors and the more the steam is let to be expanded in the turbine.

A turbine working with a cold-condenser is called Condensing turbine/Condensing power plant.

There are also hot-condensers, where the cooling media is district heating water or industrial process water. In this case the pressure on the steam side is higher than the atmospheric pressure.

A turbine working with a hot-condenser is called Backpressure turbine.

Condensing is needed so that the pump is able to increase the pressure of the working fluid.

7

S.5 T-s and h-s Diagrams for an Ideal Steam Cycle

The 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 condenser

P5.1 Ideal

Ideal 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 increase

The working fluid is pumped adiabatically and isentropically and the pressure increases. Pumping requires a power input (for example mechanical or electrical).

The POWER input required is the massflow of liquid through the pump times the enthalpy increase:

( )12 hhmPP −⋅= & [kg/s · kJ/kg = kW ]

h1 is determined from tables or diagram knowing temperature and pressure of the

liquid before the pump. However, as a liquid is incompressible (constant volume and specific heat capacity), the enthalpy can be calculated as:

h1 = cp,liquid · (T – Tref)

Tref = 273 K (normal conditions).

8

The specific heat capacity for liquid water is: cp = 4.18 kJ/(kg K) as long as the water stays in liquid form.

h2 = h2s (s denotes isentropic) is determined by the constant entropy and the

pressure 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 flow of water is 50 kg/s. Assume that the pump works isentropically. What is the power input of the pump? Numerical solution: Saturated liquid water at 0.2 bar (corresponds to the saturation temperature 60.0°C) is tabulated in saturation tables* for water: h1 = 251.4 kJ/kg. It can also be read that the entropy for this state is s1 = 0.8320 kJ/(kg K). Now increasing the pressure of the saturated liquid , the state will no longer be saturated: it will be a sub-cooled liquid. Proceeding isentropically, the entropy after the pump is the same, i.e. s2 = 0.8320 (kJ/kg K). The pressure is 60 bar and with the value of entropy, in a water/steam table* it can be found that the outlet temperature, t2, is t2 = 60.3°C and with the pressure and temperature that h2s = 257.5 kJ/kg. Thus:

Ppump = 50· (257.5 – 251.4) kW = 305 kW

* Interactive table for all properties of water/steam and saturation at different pressures and temperatures is found in S1B2C6 "Water steam thermodynamic properties"

P5.3 Process 2-3: Heat supply

The high-pressure liquid enters a boiler where it is heated at constant pressure by an external heat source. The working fluid is heated from state 2 to state 2’ (liquid saturation), and then vaporized to become saturated vapor (state 3).

The heat needed for vaporization of a liquid is dependent on the pressure. Often this heat of vaporization heat is tabulated for different pressures.

A liquid is vaporized at constant temperature (isothermal), going from saturated liquid state (2') to saturated gaseous state (3).

The heating rate to the cycle is (neglecting efficiency of the boiler and thus the losses):

( )23 hhmQ −⋅= && [kg/s · kJ/kg = kW ]

The enthalpy h3 is saturated steam and is found in saturation tables for

water/steam at the pressure in question. The enthalpy h2 is the state of the water after the pump.

Example:

9

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: h2 = 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 h3 = 2785.0 kJ/kg. Thus, the heating rate needed is:

( ) MWkWkWQ 4.1261263755.2570.278550 ≈=−⋅=&

P5.4 Process 3-4: Isentropic expansion

The vapor expands through a turbine to generate power output. Ideally, this expansion is adiabatic and isentropic. The expansion decreases the temperature and pressure of the vapor.

For the basic steam cycle the steam goes from saturated vapor form to a mixture

of saturated vapor and saturated liquid at the pressure determined by the condenser in the turbine outlet.

The power output of the turbine is:

( )43 hhmPT −⋅= & [kg/s · kJ/kg = kW]

For the basic steam cycle, the enthalpy h3 is saturated vapor and can be found in

water steam tables.

For isentropic expansion, h4 = h4s (s denotes isentropic) is determined by the constant 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 h3 = 2785.0 kJ/kg (water/steam table or h-s/T-s diagram). The entropy for steam at this condition is s3 = 5.89 kJ/(kg K). The expansion takes place isentropically i.e. s4 = s3 = 5.89 kJ/(kg K) down to p4 = 0.2 bar. Solving the enthalpy graphically in an h-s or T-s diagram is easiest. Starting from h3 = 2785 kJ/kg and p = 60 bar, a straight vertical line (=isentropic) is drawn until intersected with the 0.2 bar pressure line. Alternatively, knowing the entropy and

10

pressure of state 4 as well gives the enthalpy in the same diagram. The enthalpy h4 can now be read directly. h4 = h4s ≈ 1950 kJ/kg Observe that state 4 is within the saturation region in the diagram, i.e. state 4 is not saturated liquid or saturated vapor, rather a mixture between these two conditions. The power output is now calculated as:

( ) MWkWkWPT 8.41417501950278550 ≈=−⋅=

11

P5.4.1 h-s diagram

P5.5 Process 4-1: Condensing

The wet vapor is brought to saturated liquid in the condenser so that the pump is able to start the cycle again.

The heat lost in the condenser is thus a little bit less than the vaporization heat (as

the vapor is wet).

The heat rejected in the condenser is (numbered subscripts for enthalpy according to the diagram):

12

( )14 hhmQcondenser −⋅= && [kg/s · kJ/kg = kW ] Example: Wet steam with the enthalpy 1950 kJ/kg enters the condenser at a pressure of 0.2 bars. 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 thus h1 = 251.3 kJ/kg (from saturated steam/water tables). The heat rejected is thus:

( ) MWkWkWQcondenser 4.89849353.251195050 ≈=−⋅=&

S.6 Irreversibilities in the Pump and the Turbine

T-s and h-s diagrams with losses in the pump and the turbine

The isentropic efficiency for a process is describing the irreversibility compared to the 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 the

efficiency the larger the irreversibility for the process, i.e. larger increase of entropy.

Definitions of isentropic efficiencies for a pump and for a turbine.

P6.1 Pump

Isentropic efficiency for a pump is:

13

12

12hhhh s

P −−

=η which gives that p

PlP WWη1

isentropicrea ⋅=

Remember! For a work-requiring device, for example compressor or pump, the

real amount of work that has to be supplied is larger than in the ideal case and entropy increases during the compression.

P6.2 Turbine

Isentropic efficiency for a turbine is:

sT hh

hh

43

43−−

=η which gives that TTT WW η⋅= isentropicreal

Remember! For a work-giving device, such as a turbine, the real amount of

work that can be obtained is less than in the ideal case and entropy increases during the expansion

Example: A turbine is working with 60 bar saturated steam at 50 kg/s that expands down to 0.2 bar. The isentropic efficiency is 90% for the turbine. What is the power output for the turbine? Solution: Proceed first as in the ideal case. Determine the enthalpy of steam in the turbine inlet (60bar, saturated steam): h3 = 2785.0 kJ/kg. Isentropic expansion gives from state 3 to 4 gives h4s ≈ 1950 kJ/kg. Now the real enthalpy in 4 can be calculated from the definition of the isentropic efficiency. Solving for h4:

h4 = h3 - ηST⋅(h3 – h4s) = 2785 – 0.90⋅(2785 – 1950) kJ/kg = 2033.5 kJ/kg The power output is:

PT = 50⋅(2785.0 – 2033.5) kW = 37575 kW ≈ 37.6 MW Observe 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.

14

S.7 Cycle Net Work and Thermal Efficiency

The net work output is:

PT WWW −=

However, pump work << turbine work,

which means that the pump work can be neglected.

Net thermal efficiency is the ratio

between the net work output and the energy in the fuel supplied:

Fuel

pT

QWW )(−

=η

Example of efficiency calculation

P7.1 Pump work << turbine work

In a steam cycle, the pump work can be neglected, as the turbine power output is very large in comparison to the pump work.

In the calculation example earlier in this chapter; the pump work was 350 kW,

meanwhile the turbine output for the same steam was 37600 kW with irreversibility included. Thus:

Pump work << turbine work

P7.2 Fuel supplied

A heat balance on the water/steam side of a boiler gives the heat absorbed in the steam cycle (equals QB).

However, a boiler-efficiency indicates which is the fuel input, i.e. how much

energy must be supplied with the fuel.

In all boilers there are always losses while transferring the heat from the fuel to the water/steam. For example, the flue gases may need to pass cleaning equipment and for this need a relative high temperature, thus all heat content cannot be transferred to the steam.

The boiler efficiency is:

Fuel

BB Q

Q=η

15

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 saturated steam (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 h2 = 257.5 kJ/kg. Enthalpy of saturated steam at 60bar is h3 = 2785 kJ/kg. The heat transferred to the steam cycle Q = 50*(2785 – 257.5) = 126 375 kW The fuel input in the boiler is:

kWkWQQB

BFuel 148680

85.0126375

===η

Thus observe that the worse is the boiler efficiency, the more fuel that has to be supplied.

P7.3 Example of efficiency calculation

Example: A steam turbine in a basic steam cycle gives 37 575 kW with a fuel input in the boiler of 148 680 kW. What is the efficiency? Solution: Neglecting the pump work, the net efficiency of the steam cycle is:

%3.25253.014868037575

→==η

This says that in this basic steam cycle, only 25.3 % of the energy available in the fuel becomes useful work.

S.8 Summary

The basic steam cycle consists of a pump, boiler, turbine and condenser.

In enthalpy-entropy or temperature-entropy diagrams, enthalpies for the different states of the working fluid can be obtained in order to calculate power output and cycle efficiency.

16

The prevailing irreversibilities take place in the turbine and pump and are assessed by the isentropic efficiency of the expansion and compression.

In real processes, the entropy during expansion and compression always increases.

In a steam cycle, pump work can be neglected as pump work << turbine work.