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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.
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