1

CAMBRIDGE UNIVERSITY ENGINEERING DEPARTMENT PART IB Paper 4: Thermodynamics

Lectures 4, 5, 6

R. J. Miller

4 Working fluids 1

1

NB air-standard cycles (i.e. Joule cycle) using an ideal gas do not approach

Carnot because Q not at constant T. Other working fluids will permit

practical cycles which are closer to the ideal.

4. Working fluids I

4.1 Introduction

Every thermodynamic device (e.g. refrigerators, heat pumps, engines) relies

on a working substance.

The working substance is usually a fluid (gas or a liquid) whose function is

to provide a medium for the transfer of heat and work.

Common working fluids:

1. Air - reciprocating engines, gas turbines.

2. Water - steam engines, steam turbine power plant.

3. HFC s - heat pumps, refrigeration plant.

Examples paper 4, introduction so no questions.

http://www-diva.eng.cam.ac.uk/linkstonotesib.html

2

2

Example of a pure substance:

1. Water H2O, present as liquid, vapour (steam) or as a mixture of liquid and

vapour.

2. Air is not strictly a pure substance, since its composition varies during

liquefaction due to the different boiling points of nitrogen (-195.8°C) and

oxygen (-183.0°C). Nevertheless a substance that is present in only a single

phase may usually be considered to be pure. Gaseous air (79% nitrogen

and 21% oxygen by volume) is pure to the extent that its composition is

constant.

4.2 Reminder

Recall that:

• Properties (e.g. p, v, T) at a particular state have one value.

• A state refers to the condition of a system as described by its properties.

• A pure substance has a homogeneous and invariable composition.

• Phases are solid, liquid vapour.

In this course each working fluid will be considered to be:

1. A pure substance.

2. One, two or more different phases.

3

Two-property rule:

In the absence of external effects (motion, gravity, electromagnetism or

surface tension) the state of a pure substance is fixed by the values of 2

independent properties.

This is known as the state principle and allows us to say p=f(T,v) etc.

For the case of a ideal gas p=f(T,v) can be written as the simple equation

of state above.

4.3 How many properties fixes a state?

We know that not all properties are independent.

Example: For an ideal gas 2 properties set the state.

v

RTp =

3

4

1. Equations of state depend on the detailed atomic and molecular

structure of a substance, and cannot be deduced from the Laws of

Thermodynamics.

2. They must either be obtained by experimental measurement or

deduced from statistical mechanics.

3. It is not normally possible to write an equation of state in closed form,

except for a very limited range of properties. The major exception is

the ideal gas equation of state, pv=RT.

4.4 Some remarks about the state principle

So if p=f(T,v), we should be able to experimentally determine p-v-T surfaces.

5

4.5 Experimentally measured p-v-T surfaces

Liquid-vapour

Surface shows different phases.

1. At high T becomes vapour/gas.

2. As T is dropped becomes liquid.

3. Finally at low T becomes solid.

4. In between two-phase regions.

4

From experiments it is known that the temperature and specific volume can be

regarded as independent and pressure determined as a function of these two

p=p(T,v)

The graph of such a function is a surface, the p-v-T surface.

Terminology:

The saturation state is a state at which phase change starts and ends.

The vapour dome is the region composed of the two-phase liquid-vapour states. The

saturated liquid and vapour lines are the lines bordering the vapour dome.

The critical point is the point where the saturated liquid and vapour lines meet. The

critical temperature of a pure substance is the maximum temperature at which liquid

and vapour phases can coexist.

6

Projecting the p-v-T surface onto the p-v plane

4.6 p-v diagram

Ideal gas

equation

5

7

p-v diagrams normally omit the solid phase, as this has little practical relevance.

1. The saturated liquid line (where the fluid is sometimes described as being

wet saturated).

2. The saturated vapour line (where the fluid is sometimes described as being

dry saturated).

3. The liquid, region, (where the fluid is sometimes described as being

subcooled).

4. The liquid- vapour region.

5. The vapour region, (where the fluid is said to be superheated).

6. The critical point is the point with the maximum temperature at which liquid

and vapour phases can exist.

7. Above the critical point no distinction can be made between liquid and vapour.

8. The triple line is the line along which three phases can exist together.

For engineering purposes this diagram is probably the most useful. The properties of

a substance are shown by plotting the boundaries between the phases together with

lines of constant temperature (isotherms). The diagram divides into regions of solid,

liquid and vapour together with two-phase regions containing solid-vapour or liquid-

vapour mixtures in equilibrium.

It is evident that the temperature and pressure are not independent in two-phase

regions.

The temperature at which a phase change takes place for a given pressure is called the

saturation temperature, and the pressure at which a phase change takes place for a

given temperature is called the saturation pressure.

The solid-vapour and liquid-vapour regions are separated by the triple point line.

They are also separated from the single-phase regions by the saturation lines. The

liquid-vapour region is of greatest engineering interest since it involves a phase

change between fluids.

On the boundary between the liquid-vapour region and the liquid region the substance

is said to be a saturated liquid.

On the boundary between the liquid-vapour region and the vapour region the

substance is said to be a saturated vapour. If the pressure is lowered at constant

temperature or the temperature is raised at constant pressure the vapour is said to be

6

an unsaturated vapour or a superheated vapour. The difference between the

temperature of a superheated vapour and the saturation temperature at the same

pressure is called the degree of superheat.

The saturated liquid line meets the saturated vapour line at the critical point. The

isotherm running through this point is called the critical isotherm. A substance in a

state above the critical isotherm cannot be liquefied by isothermal compression alone

and must first be cooled to below the critical temperature. Thus the critical isotherm

marks a boundary between vapour and gas. If a substance at normal temperatures and

pressures lies above its own critical isotherm it is called a gas, otherwise it is a

vapour. This is not a rigid distinction and no phase change is involved.

8

Examples of phase changes are:

1. Liquid to vapour is called vaporisation.

2. Solid to liquid is called melting.

3. Solid to vapour is called sublimation.

4.7 Phase change

1. During a phase change both phases can coexist.

2. A change of phase is accompanied by absorption or release of energy

resulting from the breaking or reformation of intermolecular bonds.

3. During phase change p and T not independent properties.

4. When a phase change occurs the heat transferred is called latent heat.

5. Latent heat is a function of the p or T.

6. Melting or freezing is known as the latent heat of fusion.

7. Boiling or condensation is known as the latent heat of vaporisation.

Best to see example of phase change.

7

9

Water (H2O): Consider a lump of ice at atmospheric pressure and 270K, i.e.

just below the normal freezing point. The sequence of events during

heating at constant pressure can be traced out on a T-v diagram.

1. Ice at 270K. Heating causes a rise in temperature according to the

specific heat capacity of ice, and there is an increase in specific volume

due to thermal expansion.

2. Ice at 273.15K. Further heating causes no rise in temperature, but

instead a phase change from ice to water occurs at constant temperature.

This temperature is the melting point at atmospheric pressure. For ice,

melting is accompanied by a reduction in specific volume. This is

anomalous, since most substances expand on melting. The heat

supplied per unit mass in going from ice to water is the latent heat of

fusion for water at atmospheric pressure. During the phase change ice

and water coexist in equilibrium.

Example 1: Phase change on a T-v diagram

10

T-v diagram of water

Fill in

8

11

3. Water at 273.15K. The phase change is complete. Further heating

causes the temperature to rise according to the specific heat capacity

of water. The specific volume falls slightly up to 277K (for water), then

rises due to thermal expansion.

4. Water at 373.15K. Further heating causes no further rise in

temperature, but instead a phase change from water to steam occurs

at constant temperature. This temperature is the boiling point at

atmospheric pressure. Boiling (vaporisation) is accompanied by a

large increase in specific volume. The heat supplied per unit mass in

going from water to steam is the latent heat of vaporisation for water

at atmospheric pressure. During the phase change water and steam

coexist in equilibrium.

5. Steam at 373.15K. The phase change is complete. Further heating

causes the temperature to rise according to the specific heat capacity

for steam at constant pressure. The steam is said to be superheated.

6. Steam at 400K. Further heating causes no further phase changes

and the equation of state approximates to the ideal gas equation of

state.

12

p-T diagram is known as a phase diagram since all three phases are

separated by lines.

4.8 p-T diagram

The vapourisation line ends at the critical point as not distinction can be made

between vapour or liquid.

p and T linked and constant during phase change.

9

The same information can be displayed on a p-T diagram by plotting the pressure

against temperature during a change of phase. Curves of sublimation, fusion and

vaporisation appear, separating regions of constant phase. The triple point appears at

the junction of the three curves. The vaporisation curve ends abruptly at the critical

point.

Water (H2O) is a good example: Consider a system consisting of a unit mass of ice at

a temperature below the triple point temperature. Let us begin where the system is at

state a shown in the figure below. This point has a pressure greater than the triple

point pressure. Suppose the system is slowly heated while maintaining the pressure

constant and uniform throughout. The temperature increases with heating until point

b. At this point the ice is a saturated solid. Additional heat at a fixed pressure results

in the formation of liquid without a change in temperature. For most substance the

specific volume increases during melting, but with water the opposite occurs.

Next consider a system initially at a’ where the pressure is less than the triple point

pressure. In this case as it is heated it passes through the two-phase solid-vapour

region into the vapour region along a’-b’-c’. The case of vaporisation by heating is

shown by line a’’-b’’-c’’.

13

4.9 Expansion and contraction on freezing

Water

(expands on freezing)

Most other substances

(contract on freezing)

10

14

4.10 Non-ideal gases

The conditions for a gas to be ideal are:

a) Molecules are rigid, and no momentum or time is lost during collisions.

b) Volume occupied by the molecules negligible compared to total volume.

c) The forces between adjacent molecules are negligible.

Condition (a) is satisfied for all gases, or p drops with time.

Condition (b) is satisfied at low pressures, since the molecules are far apart.

Condition (c) is also satisfied at low p and high T, when the molecules are

moving faster and spend less time in close proximity to each other.

If conditions (b) and (c) are not satisfied equation of state must be modified.

15

Clearly Z=1 for an ideal gas. In general Z depends on the

gas, and also on the pressure and temperature. A charts

for Z for various gases is given in the notes.

RT

vp

p

=→

lim0

is called the universal gas constant.R

As the pressure is raised two molecule and then three molecule

interaction becomes more significant and Z diverges from 1

11

At the pressures and temperatures that usually occur in engineering applications many

substances are found to lie in a state which is to the right of the critical isotherm (the

constant temperature line which runs through the critical point). In this section we

will restrict ourselves to two regions in this area: (1) the region where the temperature

is in excess of twice absolute critical temperature; and (2) the region of low pressure

where the pressure is one atmosphere or less. The so-called ‘permanent’ gases

oxygen, hydrogen, nitrogen etc., all lay in the first region at ordinary working

temperatures (the critical temperature of the specified gases being 154.8k, 33.99k and

126.2k). Water vapour at atmosphere pressure lies in the second and is in fact super-

heated steam at low pressure. The behaviour of substances in these two regions

approximate very closely to those of a ideal gas. An ideal gas is a particularly useful

concept as its properties are related in a very simple way.

From kinetic theory, the conditions for a gas to be ideal are

a) The molecules are rigid, and no momentum or time is lost during collisions.

b) The volume occupied by the molecules is negligible compared to the total volume.

c) The forces between adjacent molecules are negligible.

Condition (a) is satisfied for all gases, or else the pressure of a gas contained in a

constant volume closed vessel would fall with time.

Condition (b) is satisfied at low pressures, since the molecules are far apart. It is less

accurate at high pressures when the molecules are closer together.

Condition (c) is also satisfied at low pressures and also at high temperatures, when the

molecules are moving faster and so spend less time in close proximity to each other.

It is less accurate at high pressures and/or low temperatures.

As pressure is raised and or temperature is dropped, conditions (b) and (c) are not

strictly true, and the behaviour of the real gas differs from that of an ideal gas.

Let a gas be confined in a piston and the entire assembly be maintained at a constant

temperature. The piston can be moved to various positions so that a series of

equilibrium states at constant temperature can be obtained. The ratio p v /T, where v

is the volume per mole, of the gas in the cylinder is plotted against pressure below.

12

The lines for several temperatures are plotted. When the ratios are extrapolated to

zero pressure, precisely the same limiting value is obtained for each curve.

RT

vp

p

=→

lim0

Where R denotes the common limit for all temperatures. If this was repeated for all

gases at all temperatures the limit would be R in all cases. R is called the universal

gas constant. Its value as determined experimentally is

KkmolkJR ⋅= /314.8

The compressibility factor, Z, of a gas is denoted as

TR

vpZ = or

RT

pvZ =

As pressure tends to zero Z tends to 1 and the ideal gas equation is obtained. As the

pressure is raised two molecule and then three molecule interaction becomes more

significant and Z diverges from 1. The behaviour of all gases are qualifiedly similar.

The figure below shows Z plotted against reduced pressure, PR , for a number of

reduced temperatures, TR , for a range of substances. Where

c

RP

PP = and

c

RT

TT =

And Tc is the critical temperature and Pc is the critical pressure.

13

Generalised compressibility chart for various gases

The graph can be used to show that the behaviour of all real gases are the same but in

practice measured p-v-T surfaces should be used for accuracy.

Another equation for describing the behaviour of a real gas is van der Waals equation

of state.

( ) RTbvv

ap =−

+

2

where a and b are constants. This equations is determined by allowing for the

attractive force between molecules, a/v2 term, and the volume occupied by them, b.

14

CAMBRIDGE UNIVERSITY ENGINEERING DEPARTMENT PART IB Paper 4: Thermodynamics

R. J. Miller

5 Working fluids II

1

5. Working fluids II

5.1 Introduction

1. Thermodynamic property data can be retrieved in various ways,

including tables, graphs, equations and computer software.

2. The emphasis of the present lecture is the use of tables and graphs.

3. These are commonly available for substances of engineering interest.

4. The lecture explains how the tables and graphs in the CUED

Thermofluids Data Book should be used.

Examples paper 4, questions 1, 2.

http://www-diva.eng.cam.ac.uk/linkstonotesib.html

15

2

5.2 Dryness fraction

gf mmm +=

ggff vmvmV +=

( ) ( ) fgffgfgf

gf

ggffxvvvvxvxvvx

mm

vmvm

m

Vv +=−+=+−=

+

+== 1

1. A substance in the two-phase liquid-vapour region consists of both liquid

and vapour coexisting in equilibrium.

2. x is called the dryness fraction of the mixture.

3. Each 1 kg of mixture contains x kg of vapour and (1-x) kg of liquid.

The volume of the mixture is then

Hence the specific volume of the mixture is

Thus the dryness fraction provides a measure of how far the phase

change has proceeded.

(f=fluid,liquid, g=gas)

3

( ) fgfgf xvvxvvxm

Vv +=+−== 1

vf vg

vfg

x=1

x=0

The above `lever rule' applies to all intensive properties (v, u, h, s, etc.).

16

4

5.3 Tabulated properties

1. There are many ways in which to present thermodynamic data for

substances (tables, graphs, computer programs).

2. CUED Thermofluid’s data book provides a good (and useful!) example.

3. The main tables for this course are concerned with water and steam, since

H2O remains a major working fluid for many engineering applications.

4. The Thermofluids Data Book 2008 Edition was issued to you this year and

this is the version which will be used.

1. Page 12 provides basic data for perfect gases.

2. Page 13 deals with semi-perfect gases. At low p and over the range

of T listed (200K to 3000K) the gases listed are semi-perfect. This

means cp and cv varies and the h can be looked up.

5.3.1 Tables for perfect and semi-perfect gases

5

5.3.2 Tables for water and steam

1. Page 16 contains data for the triple point and the critical point for

2. Page 17-18 covers the liquid-vapour region for H2O as function of

temperature.

vf, uf, hf, sf

vg, ug, hg, sg

17

6

3. Page 19-22 covers the liquid-vapour region for H2O as function of

pressure.

vf, uf, hf, sf

vg, ug, hg, sg

p

7

4. Page 23-26 covers date outside the the liquid-vapour region for H2O

as function of temperature (horizontally) and pressure (vertically).

Liquid

Vapour

p

T

18

8

5.3.3 Transport properties

Page 27 contains transport properties for H2O from the triple point to

the critical point, vf, vg, cpf, cpg, thermal conductivity, dynamic viscosity

and Prandtl number.

Page 28 contains (cp, thermal conductivity, dynamic viscosity and

Prandtl number) data for steam at atmospheric pressure.

Page 28 also contains corresponding data for air at atmospheric

pressure.

Page 29 contains corresponding data for carbon dioxide at atmospheric

pressure.

Page 29 also contains corresponding data for hydrogen at atmospheric

pressure.

Why they are called transport properties is not clear.

The tables tend to contain properties useful in flow problems (such as pipe flow).

9

5.3.4 Properties of gases and liquids at sea level and conditions

Page 31 contains data on air, carbon dioxide, hydrogen and helium

under sea level conditions. Density, dynamic viscosity, kinematic

viscosity, speed of sound and thermal conductivity are tabulated.

Page 31 also contains data on mercury and water at sea level.

Density, dynamic viscosity, kinematic viscosity and thermal

conductivity are tabulated.

5.3.5 The international Standard Atmosphere

Page 31 gives data for the international standard atmosphere for sea

level conditions (mostly repeat of air in table above).

Page 32 for conditions at altitude. Temperature, pressure, density and

kinematic viscosity are tabulated.

e.g. how p, T, ρ vary with altitude up to 30,000m.

19

10

5.3.6 Data for Refrigerant HFC-134a (CH2FCF3)

Page 37 contains a table of

saturation properties for refrigerant

HFC-134a. The format is similar to

the saturation tables for H2O,

except that data h and s at

superheated conditions (at 20K and

at 40K above the saturation

temperature) is included in the

table.

11

Example 1:

Find the h of H2O at 17.5MPa and 455°C? The phase is not known.

First check the saturation tables for H2O. A pressure of 17.5MPa = 175 bar

appears on page 22.

Tsaturation = 354.67°C.

The given T is higher, therefore the H2O is in the vapour phase.

Now go to page 23 for the enthalpy of H2O.

The answer lies between tabulated values so interpolation is necessary.

20

12

( )1

12

121 xx

xx

yyyy −

−

−+=

( ) 8.3109150175150200

9.31577.30619.3157)450( =−

−

−+=h

( ) 2.3196150175150200

6.32368.31556.3236)475( =−

−

−+=h

( ) 1.3127450455450475

8.31092.31968.3109 =−

−−

+=h

then interpolating again in temperature gives the required answer

kJ/kg

The general formula for linear interpolation is

Choosing to interpolate in pressure first (is this choice important?) gives

kJ/kg

kJ/kg

13

Example 2:

ggff vmvmV +=

gf mmm +=

gf

gf

ggffxvvx

mm

vmvm

m

Vv +−=

+

+== )1(

843.0001043.06941.1

001043.07.0/1=

−

−=

−

−=

fg

f

vv

vvx

( ) 23204.2257843.05.4171 =×+=+=+−= fgfgf xhhxhhxh

A rigid vessel of volume 1m3 contains 0.7kg of H2O at a pressure of 1 bar.

Calculate the fraction by mass of vapour and the specific enthalpy of the

mixture.

The total volume

and the total mass is

Thus the specific volume of the mixture is

so that the dryness fraction x is

The specific enthalpy is then

kJ/kg.

21

14

5.4 Graphical representation

5.4.1 T-s diagram Constant p

1. Most commonly sketched

graph for cycles.

2. Good for comparing to

Carnot cycle.

15

5.4.2 h-s diagram

1. Easiest graph to solve

cycle problems on.

2. Much quicker than

using tables but you

need to be able to use

both.

Dryness fraction 0.85

Constant p

Constant T

22

16

5.4.3 p-h diagram

1. Used for refrigerant calculations.

2. Not as easy to use. Sometimes have to used tables.

23

CAMBRIDGE UNIVERSITY ENGINEERING DEPARTMENT PART IB Paper 4: Thermodynamics

R. J. Miller

6 Power generation I

1

6. Power generation I

6.1 Ideal Carnot cycle?

Qin

W inWout

Qout

Problems:

1. Significant practical

problems with developing

pumps that operate in the

two-phase region.

2. The heat is transferred

from hot combustion

products that are at

constant pressure.

1-2 Compressor/pump, 2-3 Boiler, 3-4 turbine, 4-1 condenser.

Can we design an ‘ideal’ power plant using steam as the working fluid?

Examples paper 4, questions 3, 4, 5.

24

We mentioned repeatedly that the Carnot cycle is the most efficient cycle operating

between two specified temperature limits. Thus it is natural to look at the Carnot

cycle first as a prospective ideal cycle for vapour power plants. If we could, we

would certainly adopt it as the ideal cycle. However, in practice there are several

impracticalities:

1. In the heat addition (2-3) the temperature is limited to the critical point

value, which is 374°C for water. This limits the maximum temperature of

the heat addition and thus the Carnot efficiency.

2. The expansion process (3-4) is in the two-phase region. The impingement

of liquid droplets onto the turbine blades causes erosion.

3. The compression process (1-2) is also in the two-phase region. It is not

possible to design a pump which handles two-phase flow.

2

6.2 Rankine cycle

Advantages:

1. Takes advantage of heat

transfer at lower

temperature.

2. Pump in single phase

regions.

Problem:

1. Turbine in two phase

region results in high

blade erosion.

25

3

6.3 Superheated Rankine cycle

Boiler heats into superheated region.

Advantages:

1. Turbine operates in less of the two

phase region.

2. Average temperature of heat input

rises and so efficiency rises.

4

6.4 Practical implementation

1

2

3

4

Boiler

PumpCondenser

Turbine

26

5

Boiler from

burning coal, oil,

gas biomassCooling tower to ‘waste’ the

low temperature heat

rejection

Electricity generation

6

1. Cyclic steam-based power plants are the World's biggest man-made

power source.

2. The steam turbine was introduced by Sir Charles Parsons in the

1880’s.

3. All steam cycles are based on the Rankine cycle which is a true

thermodynamic cycle.

4. All plant uses heat to generate 50 to 2000 MW electricity from:

1. combustion of fossil fuels (oil, coal, gas).

2. exhaust of gas turbine.

3. nuclear reactions.

6.5 Background

27

7

0

50

100

150

200

250

300

350

400

450

Year 1980 1990 2000 2003 2004

TWh

Net imports

Other fuels

Hydro

Nuclear

Gas

Oil

Coal

UK Fuel consumption for Electricity Production

>90% of electricity comes from large gas turbines and steam turbines

8

6.6 Calculations

12 hhm

Wp −=&

&

23 hhm

Qin −=&

&

43 hhm

Wt −=&

&

41 hhm

Qout −=&

&

Pump 1-2.

Boiler 2-3.

Turbine 3-4.

Condenser 4-1.

in

pt

ThermalQ

WW

&

&& −=η

28

9

Example 6.1:

A steam power plant operates between a boiler pressure of 20MPa and a

condenser pressure of 0.004MPa. The turbine entry temperature is

550°C. Calculate the net work output per unit mass of steam and the

cycle efficiency.

10

∫=−=−2

112 vdphhwx

( ) ( ) 1.201041020001004.0 36

12 =×−×=−= ppvwp

5.1411.33962323 −=−= hhq

1−−−−2: Feed pump: State 1 is sat. liquid at 4kPa = 0.04 bar. Tables: h1 = 121.4kJ/kg.

State 2 is liquid with s2=s1, so that h2 could be found by interpolating the

tables. This is tedious and may be inaccurate. Instead use.

Water is virtually incompressible, so a good approximation is v=vf1 = constant.

Then h2 = h1 + wp = 141.5kJ/kg.

2−−−−3: Boiler: State 3 is superheated steam at 20MPa = 200 bar and 550°C.

Using Tables or Enthalpy-Entropy diagram have h3 = 3396.1kJ/kg.

Thus the heat added in the boiler is

= 3254.6kJ/kg.

29

11

3-4: Turbine: State 4 is in the liquid-vapour region.

The turbine is isentropic, so that s4 = s3. Use s4 to get x4.

From Tables have s3 = 6.339kJ/kgK. Then from Tables at p4 = 0.04 bar

gives sf = 0.422kJ/kgK and sg = 8.473kJ/kgK.

Again from Tables at p4 = 0.04 bar have hf = 121.4kJ/kg and hfg =

2432.3kJ/kg

73.04

4 =−

−=

fg

f

ss

ssx

Alternatively, use the Enthalpy-Entropy diagram - have h4 and x4 directly.

kgkJxhhh fgf /0.18972.243273.04.1214 =×+=+=

kgkJhhwt /1.14990.18971.339643 =−=−=

The cycle efficiency is given by

===6.3254

0.1479

23q

wnetη 45.4%

12

6.7 Notes on Rankine cycle

1. The work input required to compress the liquid water is very much

less than the work output of the turbine. The effect of component

irreversibility is likely to be small on cycle.

2. Operating pressures are very high in the boiler and very low in the

condenser. This requires high structural integrity from the enormous

lengths of pipework involved. In the boiler the pipes are subjected to

high temperatures and corrosive flue gases.

3. The temperature at which heat is rejected only just above ambient.

This compensates for the relatively low maximum temperature, which

is much lower than that achieved in a gas turbine or a Diesel engine.

4. The specific volume increases greatly through the turbine such that

the volume flow rate at turbine exit is very high. This means that the

turbine exit area must be very large. The final stage of a large

turbine has blades of about 4m diameter.

5. The steam leaving the turbine typically has a dryness fraction of

about 0.9, with the water suspended in the stream as a fog of

droplets of diameter about 1 micron. These are too small to cause

damage, but may deposit on blades and accumulate into larger

droplets which are thrown off and may cause erosion.

30

13

6.8 Irreversible Rankine cycle

Turbine irreversibilities.

sxideal

xrealT

hh

hh

W

W

43

43

−−

==&

&

η

Example 6.2:

If boiler exit state (p3, T3), turbine efficiency ηT and exit pressure p4

are known then calculate the dryness fraction at turbine exit, x4.

14

1. Ideal isentropic turbine.

Draw vertical line down from 3

until touch exit p4.

2. Use efficiency to get real

turbine enthalpy change.

sxideal

xrealT

hh

hh

W

W

43

43

−−

==&

&

η

3. Get dryness fraction

at 4. Use real turbine

enthalpy change and turbine

exit pressure p4. Read off

the dryness fraction x4.

3

4s4

Constant pConstant p

Constant x

31

15

6.9 Future trends: Burning biomass

Drax Power Station in North Yorkshire has generating capacity of 3,960

megawatts and is the highest of any power station in the United Kingdom and

Western Europe, providing about 7% of the United Kingdom's electricity

supply.

Over the last 5 years, Drax has developed the capability to co-fire renewable

biomass materials with coal and has set itself the target to produce 10% of its

output from co-firing.