Thermodynamics Fundamentals for Chemical Engineers

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WELCOME TO
THERMODYNAMICS
CHEM 222
Zin-Eddine Dadach
2014-2015

What is thermodynamics?
 A branch of physics that studies the
effects of changes in temperature,
pressure, and volume on physical
systems at the macroscopic scale by
analyzing the collective motion of
their particles using statistics

DEFINITION OF
THERMODYNAMICS
 Thermodynamics (from the Greek
thermos meaning heat and dynamics
meaning power)
 During our course, we will study the
effects of heat and work transfer on
a specified system.

LAWS OF
THERMODYNAMICS
The thermodynamics is the lawyer
of nature and has different laws to
be applied.

THE IMPORTANCE OF THE
LAWS
 The starting point for most
thermodynamic considerations are
the laws of thermodynamics, which
postulate that energy can be
exchanged between physical systems
as heat or work.
 They also postulate the existence of
a quantity named entropy, which can
be defined for any system.

THE SYSTEM
The most important point in
thermodynamics:
DEFINING THE SYSTEM

SYSTEM VS SURROUNDINGS
 In thermodynamics, interactions
between large ensembles of
objects are studied and
categorized.
 Central to this are the concepts
of system and surroundings.

WHAT IS A SYSTEM?
 A system is composed of particles,
whose average motions define its
properties, which in turn are related
to one another through equations of
state.
 In chemical engineering, a system
could be a gas or a liquid flowing in a
pipe or in a tank or reactor

PROPERTIES OF A SYSTEM
 Any system can be defined
by its pressure,
temperature, composition,
or other more complicated
properties like SPECIFIC
entropy , SPECIFIC
enthalpy….

SYSTEM & THERMODYNAMICS
 Thermodynamics describes how systems
respond to changes in their surroundings.
 This can be applied to a wide variety of
topics in science and engineering, such as
engines, phase transitions, chemical
reactions, transport phenomena, and even
black holes.
 The results of thermodynamics are
essential for other fields of physics and for
chemistry, chemical engineering, cell
biology, biomedical engineering, and
materials science to name a few

DEFINING A SYSTEM

WHAT IS A SYSTEM?
 A Thermodynamic System is that
part of the universe that is under
consideration or study.
 A real or imaginary boundary
separates the system from the rest
of the universe, which is referred to
as the environment or surroundings
(sometimes called a reservoir.)

Exchange between Systems &
Surroundings
 A useful classification of
thermodynamic systems is based on
the nature of the boundary and the
flows quantities through it as matter,
energy, work, heat, and entropy.
 A system can be anything, for
example a piston, a fluid in a test
tube, a living organism, or a planet

DIFFERENT KINDS OF
SYSTEMS

WHAT IS AN ISOLATED
SYSTEM?
 Isolated systems are completely
isolated in every way from their
environment.
 They do not exchange heat, work or
matter with their environment.
 An example of an isolated system
would be an insulated rigid
container, such as an insulated gas
cylinder.

WHAT IS A CLOSED SYSTEM?
 Closed systems are able to exchange
energy (heat and work) but not matter
with their environment.
 A greenhouse is an example of a closed
system exchanging heat but not work with
its environment.
 Whether a system exchanges heat, work
or both is usually thought of as a property
of its boundary.
 Figure 1.16 page 10

WHAT IS AN OPEN SYSTEM?
 open systems: exchanging both
energy (heat and work) and matter
with their environment.
 A boundary allowing matter
exchange is called permeable.
 The ocean would be an example of
an open system.

PROPERTIES OF A
SYSTEM

WHAT IS A PROPERTY OF A
SYSTEM ?
 Any characteristic of a system is called a
property.
 Some familiar properties are pressure P,
temperature T, volume V and mass m.
 We can include viscosity, thermal
conductivity, modulus of elasticity,
thermal expansion, ….
 Not all properties are independent
however, some are defined in terms of
others.
 Example : ρ = m/V

WHAT IS AN INTENSIVE
PROPERTY?
 Thermodynamic properties can be divided into two gen
eral classes, intensive and extensive properties.
 An intensive property is independent of the amount
of mass.
 Examples of intensive properties include:
 temperature
 viscosity
 density
 electrical resistivity
 melting point
 boiling point
 color (in solution)
 flammability .

WHAT IS AN EXTENSIVE
PROPERTY?
 An extensive property is a property that
changes when the size of the sample
changes.
 Examples of extensive properties include:
 mass
 volume
 entropy
 energy
 electrical resistance
 texture
 Figure 1-20 page 12

STATE OF A SYSTEM
DEFINING COMPLETELY THE
SYSTEM

STATE OF A SYSTEM
 A system not undergoing any change
 All the properties can be measured
or calculated throughout the entire
system
 This set of properties describe the
state of the system

THERMODYNAMICS &
EQUILIBRIUM
 Thermodynamics deal ONLY with
equilibrium states, that means there
is no driving forces within the
system
 The word Equilibrium implies a state
of balance
 Outside the equilibrium, systems can
exchange heat and work and matter
with the surroundings

ZEROTH LAW OF
THERMODYNAMICS
 It is observed that a higher temperature
object which is in contact with a lower
temperature object will transfer heat to
the lower temperature object.
 The two objects will approach the same
temperature, and in the absence of loss to
other objects, they will then maintain a
constant temperature.
 They are then in thermal equilibrium.
 Thermal equilibrium is the subject of the
Zeroth Law of Thermodynamics.
 Also Figure 1.24 page 14

WHAT IS A THERMODYNAMIC
EQUILIBRIUM?
 If you put a hot cup of coffee or tea in a cold room, heat
will flow from the cup and its contents into the room. But if
you leave it there for two hours and come back, the cup of
coffee or tea will be at the same temperature as the room.
 The coffee and the room will have achieved a state of
thermodynamic equilibrium Thermodynamic
Equilibrium or Thermal equilibrium.
 Thermodynamic equilibrium is when heat ceases to flow
between two systems.
 There is no heat transfer when thermodynamic
equilibrium is reached.

Class Work
 Study examples 1-1 and 1-2 page 8

The different phases of
the system
Remember in our course the
system will be a fluid or solid
SOLID-LIQUID-GAS

DIFFERENT PHASES OF THE
MATTER (THE SYSTEM)
 Matter can exist in four phases (or
states), solid,
liquid,
gas, and
plasma
plus a few other extreme phases, like
critical fluids and degenerate gases

SOLID PHASE
 A solid is matter in which the
molecules are very close together
and cannot move around.
 Examples of solids include rocks,
wood, and ice (frozen water).

LIQUID PHASE
 A liquid is matter in which the
molecules are close together and
move around slowly.
Examples of liquids include
drinking water, alcohol, oil, mercury
at room temperature, and lava
(molten rock).

GAS PHASE
 A gas is matter in which the
molecules are widely separated,
move around freely, and move at
high speeds.
Examples of gases include the
gases we breathe (nitrogen, oxygen,
and others), the helium in balloons,
and steam (water vapor).

This P-T phase diagram of water
shows its phases at various
temperatures and pressures

LET’S ANALYZE THE DIAGRAM
 FROM THE SOLID STATE:
 When a solid is heated (or as pressure
decreases), it will change to a liquid form,
 and when the liquid is heated it will
eventually become a gas.
 FOR THE DIAGRAM:
 ice (frozen water) melts into liquid
water when it is heated or pressure
decreases.
As the water is heated , the water
evaporates and becomes water vapor.

WHAT IS A SUBCOOLED LIQUID?
 Liquid refrigerant which is cooled below its
saturation temperature.
 NO READY TO VAPORISE
 EXAMPLE:
 Consider a piston cylinder ( Figure 3-6,
page 114) device containing liquid water
at 200C and 1 atm. Under these
conditions, water exist in liquid phase and
is called compressed liquid or
subcooled liquid

WHAT IS A SATURATED LIQUID?
 A Liquid that is about to vaporize
Example:
 If we heat the water liquid at 400C , the liquid
will expand and its specific volume increases
 If we continue to heat the water liquid up to
1000C ( Figure 3.7 page 114)
 At this point the water is still liquid BUT any
additional heat will cause some liquid to vaporize
( Phase change process from liquid to gas)
 WE HAVE A SATURATED LIQUID

WHAT IS A SATURATED VAPOR?
 A Vapor that is about to
condense
 A vapor + one drop of liquid
 Once boiling starts, the temperature
of the liquid stops rising and the
vapor starts to be produced
  we have change of phase from
vapor to liquid

WHAT IS A SUPERHEATED
VAPOR?
 Once the phase change is completed,
we are in a single phase state which
is vapor only
 When a vapor is far from
condensing , it’s called Superheated
vapor
 Figure 3-10 page 115

WHAT IS WET STEAM?
 Mixture of liquid and steam of the same
substance in which both are at saturation
temperature.
 If additional heat is added to the wet
steam at constant pressure, the
temperature remains constant until all
liquid is evaporated (saturated steam);
 it is only at this point that the temperature
increases above the saturation
temperature (superheated steam )

WHAT IS SATURATION
TEMPERATURE?
 The term saturation defines a
condition in which a mixture of vapor
and liquid can exist together at a
given temperature and pressure.
 The temperature at which
vaporization ( Boiling) starts to occur
for a given pressure is called the
saturation temperature or boiling
point

WHAT IS A SATURATION
PRESSURE?
 The pressure at which the vaporization (
boiling) starts to occur for a given
temperature is called the saturation
pressure
 For water at 2120F, the saturation
pressure is 14.7 psia
 For water at 14.7 psia, the saturation
temperature is 2120F.
 The graphical representation of this
relationship between temperature and
pressure at saturated conditions is the
VAPOR PRESSURE CURVE

DRAW A VAPOR PRESSURE
CURVE (Psat vs Tsat) OF WATER
 T ( 0C) :
-10, -5, 0, 5,10,15,20,25,30,40,50,100,
150,
 Saturation Pressure ( kPa)
0.26,0.40,0.61,0.87,1.23,1.71,2.34,3.17,
4.25,7.38,12.35, 101.3, 475.8,

GROUP- WORK
 Draw a T-V diagram for the
temperature change from 100 to
3000C.

CHANGES OF PHASE

Latent heat and sensitive heat
 when your system is heated up or cooled down
without changing phase  the temperature
changes and we have sensitive heat
 When the system is changing phase, the heat is
used by the system for the change of phase by
breaking down the bridges between molecules.
 if we have a pure component, the
temperature stays constant
 if we have a mixture, the temperature will
change during the change of phase

WHAT IS LATENT HEAT?
 The amount of energy needed or
is released during a phase-
change process is called the
latent heat
 For a pure substance, the change
of phase happens in constatnt
temperature

Latent heat of fusion
 The amount of heat absorbed during
melting

Latent heat of vaporization
 A change of state from saturated
liquid to saturated vapor at constant
temperature also requires the input
of energy, called the latent heat of
vaporization.

Magnitudes of latent heat
 The magnitudes of the latent heats
of condensation and vaporization
depend on the temperature or
pressure at which phase change
occurs.
 At 1atm , the latent heat of fusion of
water is 333.7 kJ/kg and the latent
heat of vaporization of water is
2257.1 kJ/kg around 99.80 0C

Wet steam region
 Figure 3.34 page 129: relative
amount of liquid and vapor
 Figure 3.35: Average volume

Specific volume of Sat. Liquid and
Sat. vapor
 In the wet steam region, we have a mixture of
saturated liquid and saturated vapor
 Let’s study saturated steam tables
A-4 & A-5 pages 916-917:
At 450C => PS= 9.5953 kPa ( A-4)
At 20kPa =>TS= 60.060C ( A-5)
At 1000C
=> Specific Volume of Sat.Liq= 0.001043
m3/kg
=> Specific Volume of Sat.Vap= 1.6720
m3/kg

Quality of wet steam
 The quality x of a wet steam is
defined as the ratio of the mass of
vapor over the Total mass
 x= Massvapor/Masstotal
And:
 Masstotal= Mass vapor + Massliquid

Example of the use of Quality
 For the volume:
Vtotal= Vliquid + Vvapor
In table A-4 ( page 916) for 1000C
Vf ( liquid)= 0.001043 m3/kg
Vg(vapor) = 1.6720 m3/kg
 V= (mf .vf) + (mg.vg)
V= ({mt-mg} .vf) + (mg.vg)

Average value of a property of a
mixture
 V=m.Vaverage
 Diving both sides by mt
 Vaverage= (1-x).Vf + x.Vg
 Vaverage= Vf-x.Vf+x.Vg
 = Vf + x. ( Vg-Vf)
 = Vf +x. Vfg
 Can be applied to other properties in
the tables.

CLASS WORK
 Work problems 3-1c to 3-9c page
154
 WORK EXAMPLES 3-1 TO 3-9 PAGES
128-135

Class work
 Complete the tables 3.23 to 3.28
page 155
 Solve : 3-29 to 3-36 pages 155-156

EXCHANGE OF HEAT
AND WORK
SYSTEM AND SURRONDINGS

SYSTEM & SURRONDINGS
 DURING ANY PROCESS, A CLOSED
SYSTEM AND SURRONDINGS EXCHANGE
HEAT AND WORK.
 THE LEVEL OF ENERGY OF THE SYSTEM
WILL BE CHANGED DURING THE CHANGE
OF STATE.
1) Across a pump, the pressure of liquid
will increase ( Energy increases)
2) Across a heat exchanger, the
temperature of the system will change (
energy changes).

WHAT IS ENERGY?
 Energy is defined as: "the ability to
do work.“
 Our bodies transform the energy
stored in the food into energy to do
physical work
 Work means moving something,
lifting something, warming
something, lighting something. All
these are a few of the various types
of work.
 But where does energy come from?

SOURCES OF ENERGY?
 The forms of energy in nature we will look
at include:
Electricity
Biomass Energy - energy from plants
Geothermal Energy
Fossil Fuels - Coal, Oil and Natural Gas
Hydro Power and Ocean Energy
Nuclear Energy
Solar Energy
Wind Energy

Total and specific energy
 Energy can exist in different forms
such as thermal, mechanical, kinetic,
potential, electric, magnetic,
chemical and nuclear
 The total energy E is an extensive
property
 The specific energy is an intensive
property defined as the total energy
of a system by unit mass:
e=E/m

Energy in thermodynamics
 Thermodynamics deals only with the
change of the total energy and give no
information about the absolute value of
the total energy ( we need a reference).
 The macroscopic forms of energy are
those a system posses as a whole such as
kinetic energy and potential energy.
 The microscopic energy are related to
the molecular structure and the sum of all
the microscopic energies is called internal
energy (U)

WHAT IS KINETIC ENERGY?
 Kinetic energy is energy of motion.
The kinetic energy of an object is the
energy it possesses because of its
motion.
 The kinetic energy* of a point mass
m is given by : ( Figure 2.3 page 53)


WHAT IS POTENTIAL ENERGY?
 Gravitational potential energy is
energy an object possesses because
of its position in a gravitational field.
Figure 2.37 page 70

WHAT IS INTERNAL ENERGY?
 Internal energy is defined as the energy
associated with the random, disordered motion of
molecules. Figures 2.5 and 2.6 page 55
 it refers to the invisible microscopic energy on
the atomic and molecular scale

WHAT IS WORK?
 Systems possess energy BUT they
can not posses work or heat
 Energy can be transferred to or from a
closed system in two distinct forms : work
or heat ( Figure 2-11 page 60)
 Work is an energy interaction between a
system and its surroundings ( Figures 2.24-
2.30 pages 65-67)
 Work is an energy transfer
associated with a force acting
through a distance

WHAT IS HEAT?
 Form of energy that is transferred
between a system and its
surroundings by virtue of a
temperature difference ( Figures
2.12 to 2-14 page 60)
 Energy is recognized as heat
transfer only as it crosses the
system boundaries ( Figure 2.13
page 61)

WHAT IS POWER?
 Power is defined as work (or energy
transfer) per unit time:
P = dW / dt
The SI unit of power is watt (w).
 1 w= 1 J/s

CLASS WORK
 Do problems 2-7 to 2-11 page 98.
 Do problems 2-18 to 2-20 page 98.

THERMODYNAMIC
STATES

WHAT IS A THERMODYNAMIC
STATE?
 A thermodynamic state (from
Latin status = to stand) is the state
of a thermodynamic system as
defined by its properties
 A minimum number of parameters
are necessary to specify the state of
the system.

STATE OF A SYSTEM
 The state of a system can be
thought of as an optimal
ensemble of thermodynamic
parameters, namely
temperature, pressure, density,
composition, etc., which
characterize the system, but
neither by its surroundings nor
by its history.

WHAT IS A STATE FUNCTION?
 In thermodynamics, a state
function, is any property of a
system that depends only on the
current state of the system, not
on the way in which the system
got to that state
 Some familiar state functions are
volume, V, pressure, P, and
temperature, T.


STATE FUNCTION
 For example, if we take a state from 00C to
1000C, the change in the temperature is +100oC
whether we go straight up the temperature scale
or we first cool the sytem for a few degrees then
take the system to the final temperature.
 The change in temperature is independent
of the route taken so temperature is a state
function ( FIGURE 1-26 page 15 & figure
1.28 page 16)
 Conversely, if the change in a function is
dependent on the route taken, then the
function is known as a path function

PATH FUNCTION
 A function that depends on the path
 Heat and Work depend on the way
the exchange is performed between
system and surroundings.
 Efficiency depends on the way
 Work and Heat are path functions

Internal energy U &
Enthalpy H

WHAT IS INTERNAL ENERGY?
. Internal energy, represented by U , is
essentially the thermal energy contained in a
system (or particles making up the system)
. A change in internal energy ΔU is due to the
transfer of energy into or out of a system, but
the volume stays constant.
For example, energy transferred into the system,
usually heat (q) and work (w), represents an
increase of internal energy, ΔU, of the system.
Thus,
 ΔU = q + w

Internal Energy
 Internal Energy U of a system does not
tell how energy was transferred.
 It is purely an accounting of energy
content of the system.
 The internal energy, U, is called
a state function.

WHAT IS ENTHALPY?
 Enthalpy (symbolized H, also called heat
content)
 Is the sum of the internal energy of matter and
the product of its volume multiplied by the
pressure
 Enthalpy is defined by the following equation:
 Enthalpy is a quantifiable state function, and the
total enthalpy of a system cannot be measured
directly; the enthalpy change of a system is
measured instead.
 We need a reference. Example : Enthalpy at 00C.

Enthalpy
 The enthalpy of a system includes
the internal energy and work (
external energy).
 Work is performed only if there is a
volume change (Figures 4-1 & 4-2
page 166)
W= P.ΔV

P-V Diagram
 In a P-V diagram, Work is the area
under the process from point 1 to
point 2 ( Figure 4-3 & 4-4 page 167)

Class work
 Work examples 4-1, 4-2, 4-3 and 4-
6 pages 168-176)
 Work problem 4.9 page 202

Sensitive Heat &
Specific heat
Intensive property in heat
transfer

Latent & Sensitive heat
 Latent heat or hidden is the heat for
the change of phase. Heat needed to
destroy the intra-molecular forces.
 Sensitive heat is visible heat due to
the rise of temperature during the
heating of a single phase ( solid,
liquid or gas).

WHAT IS SPECIFIC HEAT?
 The specific heat is an intensive property.
 The amount of sensitive heat per unit mass
required to raise the temperature by one degree
Celsius.
 The relationship between sensitive heat and
temperature change is usually expressed in the
form shown below where c is the specific heat.
 unlike the extensive variable heat capacity, which
depends on the quantity of material, specific heat
is an intensive variable and has units of energy
per mass per degree (or energy per number of
moles per degree).

Specific heat a constant pressure
 A heat capacity by a capital C with a
subscript denoting which variable is held
constant during the temperature change.
Figures 4-17 & 4-18 page 178
 For example, the heat capacity at constant
pressure is defined by:
CP= ( ΔQ/ΔT)p
= (ΔH/ΔT)
P
 where (ΔQ/ ΔT) is the change in heat
with temperature. Here we can have work
 Enthalpy is used

Specific heat at constant volume
At constant volume : no work
CV= ( ΔQ/ΔT)V
= (ΔU/ΔT)v
where (ΔQ/ ΔT) is the change in heat
with temperature.

Class Work
 Work examples 4-7, 4-8 and 4-10
pages 183 -185

PROCESSES

WHAT IS REVERSIBLE
PROCESS?
 A reversible process for a system
is defined as a process that,
once having taken place, can be
reversed, and in so doing leaves
no change in either the system
or surroundings
 In other words, the system and
the surroundings are returned to
their original condition

REVERSIBLE PROCESS
 In reality, there are no truly
reversible process , therefore the
reversible process is a starting point
for thermodynamics calculations.
 One way to make a real process
close to a reversible process is to
carry out the process in a series of
infinitesimal steps

WHAT IS AN IRREVERSIBLE
PROCESS?
 An irreversible process is a
process that can not return both
the system and the surroundings
to their original conditions
 It means if the system is
reversed, the system and the
surroundings will not return to
their original conditions.

EXAMPLE
 For example, an automobile engine
does not give back fuel it took to
drive up a hill as it coasts back down
the hill.
 Main causes of Irreversibility:
Friction, unrestrained expansion,
heat transfer through finite
temperature difference and mixing
two different substances

WHAT IS A CYCLIC PROCESS?
 When a system in a given initial
state goes through a number of
different states ( go through various
processes) and finally returns to its
initial values , the system has
undergone a cyclic process or cycle.
 Therefore, all the properties have
the same value they had at the
beginning.

EXAMPLE
 Steam that circulates
through a closed cooling loop
undergoes a cycle.

WHAT IS AN ISOTHERMAL
PROCESS?
 An isothermal process is a
thermodynamic process in which the
temperature of the system stays constant:
ΔT = 0.
 This typically occurs when a system is in
contact with an outside thermal reservoir
(heat bath), and processes occur slowly
enough to allow the system to continually
adjust to the temperature of the reservoir
through heat exchange

 In an isothermal process, therefore,
all of the heat absorbed by the
system is converted into work.

WHAT IS AN ISOBARIC
PROCESS?
 An isobaric process is a
pressure stays constant; ΔP = 0.
 The heat transferred to the system
does work but also changes the
internal energy of the system.

WHAT IS AN ISOCHORIC
PROCESS?
 An isochoric process, also called an
isometric process, is a
volume stays constant; ΔV = 0.
 This implies that the process does no
pressure-volume work, since such
work is defined by  ΔW = PΔV,

WHAT IS AN ADIABATIC
PROCESS?
 In thermodynamics, an adiabatic process is a
process in which no heat is gained or lost in the
working fluid.
 For example, there are no chemical processes
taking place in the fluid and there is no heat
transfer from the environment.
 The term "adiabatic" describes things that are
impermeable to heat transfer; for example, an
adiabatic boundary is a boundary that is
impermeable to heat transfer and the system is
said to be adiabatically (or thermally) insulated.
 An adiabatic process which is also reversible is
called an isentropic process.

CLASS WORK # 1
 DO EXAMPLES 2-1 to 2-9 in pages
57-69 of the book

THERMODYNAMIC
DIAGRAMS

THE P-V-T SURFACE
 The equilibrium states of a simple,
compressible substance can be specified in
terms of its pressure, volume and
temperature.
 If any two of these state variables is
specified, the third is determined.
 This implies that the states of the
substance can be represented as a
surface in a three dimensional P-V-T
space

PVT SURFACE ( FIGURE 2.26)

WHAT ARE THE REGIONS IN A
PVT SURFACE?
 The solid, liquid and gas (vapor) phases can be
represented by regions on the surface.
 Note that there are regions on the surface which
represent a single phase, and regions which are
combinations of two phases.
 A point lying on a line between a single-phase
and a two-phase region represents a "saturation
state".

Saturation Lines
 The line between the liquid and the
liquid-vapor regions is called the
liquid-saturation line and any
point on that line represents a
saturated-liquid state.
 A point on the boundary between the
vapor and the liquid-vapor regions is
called a saturated-vapor state.

P-V-T SURFACE FOR
LIQUID-VAPOR REGION

P-V Diagram
 A process performed at constant pressure is
called an isobaric process. It’s a horizontal
line
 A process performed at constant volume is called
an isochoric process. It’s a vertical line
 A process performed at constant temperature is
called an isothermal process.
 On a p-V diagram, lines of constant temperature
curve from the upper left to the lower right.
 During an adiabatic process no heat is
transferred to the gas, but the temperature,
pressure, and volume of the gas change as
shown by the dashed line.

T-V Diagram

 In this figure :
 The isotherms are ………..
 The isochores are ……….
 The isobars are …………..



WHAT IS THE CRITICAL POINT?
 The two phase region is bounded by the
saturated liquid curve on one side and the
saturated vapor curve on the other.
 Notice that these two curves meet at a point
that corresponds to where one (and only
one) of the isobars becomes horizontal at a
single point.
 This point is called the critical point and
 it corresponds to the highest temperature
and highest pressure for which a vapor and
liquid can coexist.
 The temperature at this point is called the
critical temperature Tc.

 The isobar that passes through this
point (the only isobar that is
horizontal at a single point) is called
the critical pressure Pc.
 The specific volume at this point is
called the critical volume Vc.

WHERE IS THE SUBCOOLED
LIQUID REGION?
 The single phase liquid region, also
known as the subcooled liquid
region is to the left of the two phase
region.
 It is sometimes considered to end at
the critical temperature as shown

WHERE IS THE SUPERHEATED
VAPOR REGION?
 The single phase vapor region, also
known as the superheated vapor
region, is to the right of the two phase
region.
 This region often is (arbitrarily)
considered to end at the critical isobar.

WHAT IS THE SUPERCRITICAL
FLUID REGION?
 You may have noticed that there is a
region on the diagram that we have
not yet discussed.
 The supercritical fluid region is so
called because states in this region
are above both the critical
temperature and critical pressure.


Supercritical Fluid
 Material in this region has properties
somewhat intermediate between what
most people would call a gas and what
most would call a liquid.
 Sometimes this is called the dense gas
region and other times the expanded
liquid region.
 The word fluid covers both gases and
liquids so perhaps supercritical fluid is
the best way to describe this region.

WHAT IS THE CRITICAL STATE?
 Note the critical state where the
saturated-liquid and saturated-vapor
lines meet.
 The state variables of this unique
point are denoted by Pc, vc and Tc.
 If a substance is above the critical
temperature Tc, it cannot condense
into a liquid, no matter how high the
pressure.

Supercritical State
 This merging of the liquid and vapor
states above the critical temperature is a
characteristic of all known substances.
 While a pure vapor state can exist at a
pressure lower than Pc, at pressures
above Pc it is constrained to be a vapor.
 States with pressures above Pc are
described as "supercritical states".

WHERE IS THE SUPERCRITICAL
FLUID REGION?

THE P-T DIAGRAM

 The vapor pressure curve which is
shown in red.
 The vapor pressure curve terminates
at the critical point. Remember that
this point marks the highest
temperature (T = Tc) and highest
pressure (P = Pc) for which vapor and
liquid can coexist.
 The other end of the vapor pressure
curve is marked by the triple point


CLASS WORK
 DO EXAMPLES 3-1 TO 3-9 PAGE 128
BY STUDYING THE DIFFERENT
THERMODYNAMIC CURVES
CORRESPONDING TO THE
PROBLEMS.

THE GAS PHASE

WHAT IS A GAS?
 In a gas phase, the atoms or
molecules constituting the matter
basically move independently, with
no forces keeping them together or
pushing them apart.
 Their only interactions are rare and
random collisions.

THE GAS ATOMS

VAPOR VS GAS ( FIGURE 3.26
page 125)
 Although vapor and gas are frequently
(incorrectly) used interchangeably,
 vapor refers to a gas phase in a state of
equilibrium with identical matter in a liquid or
solid state below its boiling point, or at least
capable of forming solid or liquid at the
temperature of the vapor.
 The term gas refers to a compressible fluid
phase, as in common usage
 GAS EXIST AT TEMPERATURES HIGHER THAN TC
 VAPOR EXIST AT TEMPERATURES LOWER THAN
TC

THE GAS PHASE
 The thermodynamic state of a gas is
characterized by:
 its volume, its temperature, which is determined
by the average velocity or kinetic energy of the
molecules,
 and its pressure, which measures the average
force exerted by the molecules colliding against a
surface.
 These variables are related by the fundamental
gas laws, which state that the pressure in an
ideal gas is proportional to its temperature and
number of molecules, but inversely proportional
to its volume.

IDEAL GAS VS REAL GAS
 An ideal gas or perfect gas is a hypothetical
gas consisting of identical particles of zero
volume, with no intermolecular forces.
 Real gases do not exhibit these exact
properties, although the approximation is often
good enough to describe real gases.
 The approximation breaks down at high
pressures and low temperatures, where the
intermolecular forces play a greater role in
determining the properties of the gas

IDEAL GAS

IDEAL-GAS LAW
 Any equation that relates the pressure, temperature and
specific volume of a substance is called equation of state.
 In 1662, Robert Boyle observed, during his experiments
with a vacuum chamber that the pressure of gases is
inversely proportional to their volume
 In 1802, Guy Lussac, experimentally determined that a low
pressures the specific volume of a gas is proportional to its
temperature
That is Pv = RT -1-
With R being the gas constant, this equation is called
And v is the specific volume by unit mass
IDEAL-GAS EQUATION OF STATE

UNIVERSAL GAS CONSTANT
 In equation -1-, R is different for each gas
( Figure 3.45 page 137)
 ON THE OTHER HAND, The universal gas
constant RU is THE SAME FOR ALL GASES
AND defined as :
RU= R.M -2-
Where M is the molar mass or molecular weight
of the gas.
Values of RU in different units are given in page
1`38 .
For example in SI units RU= 8.314 kJ/kmol.K

IDEAL GAS
 Pv = Ru.T with v is the specific
volume by unit mole OR PV=NRU T
 Pv=RT with v is the specific volume
by unit mass OR PV=mRT

EXAMPLES OF IDEAL GASES?
 At low pressures and high temperatures, the
density of gases decreases and the gas behaves
like a ideal gas ( FIGURE 3.49 page 140)
 Gases like Air, O2, N2, H2 and also heavy gases
like CO2 CAN be treated as ideal gases at low
densities
 However, dense gases such as water vapor
in steam power plant and refrigerant vapors
in refrigerators can not be treated as ideal
gases.

REAL GASES

WHAT IS A REAL GAS?
 If a gas behaves exactly as the ideal
gas laws would predict it to behave
in terms of volume, pressure, moles,
and temperature, then the gas is
said to be an ideal gas.
 If, on the other hand, the gas
deviates from Ideal Gas
behavior, then the gas is said to be
acting like a "real gas".

What causes deviation from
ideal gas behavior?
 Intermolecular forces called Van Der Waals forces
. There are three such types of Van Der
Waals forces:
 London Dispersion Forces which are forces
that exist between molecules as a result of
positive nuclei of one molecule attracting the
electrons of another molecule..
 Dipole-Dipole interactions which are forces
that exist between polar molecules where the
positive end of one molecule attracts the
negative end of another molecule.
 Hydrogen bonding interactions are forces that
exist between molecules that have a hydrogen
atom bonded to a highly electronegative atom
such as Oxygen, Nitrogen, or Flourine.

What is the compressibility factor?
 The term "compressibility" is used in thermodynamics to
describe the deviance in the thermodynamic properties of a
real gas from those expected from an ideal gas.
 The compressibility factor is defined as
•
 In the case of an ideal gas, the compressibility factor Z is
equal to unity, and the familiar ideal gas law is recovered:
•
 Z can, in general, be either greater or less than unity for a
real gas.

REAL PHASE REGION
 The deviation from ideal gas behavior
tends to become particularly significant
(or, equivalently, the compressibility
factor stays far from unity) :
near the critical point,
 in the case of high pressures
 low temperatures.
In these cases, an alternative equation of
state better suited to the problem must be
utilized to produce accurate results.

COMPRESSIBILITY FACTOR
 The compressibility factor can be
estimated from figure 3-51 page 141 if we
know the reduced temperature and the
reduced pressure of the gas:
 PR= P/PC and TR= T/TC
The compressibility factor Z is almost the
same for all the gases at the same
reduced pressure and reduced
temperature . This is called the principle
of corresponding states

CLASS WORK
Work examples 3-10 to 3-12 page 139
143
Work problems 3.73, 3.74 and 3.77
page 158
Work problems : 3-82; 3-85;3-88 and
3.91 page 159

REAL GASES
EQUATIONS OF STATE

EQUATIONS OF STATE FOR
REAL GASES
 The ideal-gas equation is very
simple, but its range and applicability
are limited.
 Several equations of state have been
proposed in the literature to describe
the behavior of real gases
 We will discuss :
Van der Waals equation
 Beattie- Bridgeman equation

Van der Waals equation of state
 Proposed in 1873
 It has two constants
 a/v2 is a correction of the ideal gas equation
related to the intermolecular forces
 b is a correction of the ideal gas equation
related to the volume occupied by the gas
molecules
RT)bv)(
v
a
P( 2


Determination of a and b
cr
cr
P
TR
a
64
27 22

cr
cr
P
RT
b
8


Beattie- bridgeman equation
of state
 Proposed in 1928
 It has five experimental constants
 Values are given in table 3.4 page 146
232
u
v
A
)Bv)(
vT
c
1(
v
TR
P 
)1(....).......1( 00
v
b
BBand
v
a
AA 

ENERGY OF GASES

SPECIFIC HEATS
vv
T
u
C )(



pp
T
h
C )(




INTERNAL ENERGY OF IDEAL
GAS
 The internal energy of an ideal gas is a
function of the temperature only, that is
 u=u (T)
Figure 2.66 shows the experiment done by
Joule.
du= Cv(T) dT
dTTCu v )(
2
1


Enthalpy of ideal gas
 Definition: h=u+pv and for ideal gas (
Pv=RT), we have the relation h=u+RT
 Therefore h depends also only on
temperature for an ideal gas
 dh = Cp(T)dT
 And

2
1
)( dTTCh p

Specific heat relations of ideal
gases
 h=u+RT
 dh= du +RdT
 Replacing dh by CpdT and du by
CvdT
 WE OBTAIN Cp= Cv +R
 When the specific heats are
given on a molar basis, R is
replaced by Ru

CLASS WORK
 Work examples 3-13 and 3-14 pages
147 and 152
 Work problems 3.85; 3.88, 3.91
3.98, page 159-160

ENERGY TRANSFER
DURING PROCESSES

ENERGY TRANSFER BY HEAT,
WORK AND MASS ( Chapter 3)
 HEAT TRANSFER:
 Energy can cross a boundary of a
system as heat or work
 Heat is defined as the form of
energy that is transferred by virtue
of temperature difference
 A process with no heat transfer is
called adiabatic

Heat Transfer
 Heat can be transferred by
conduction, convection or radiation
 Heat is easy to recognize , its
driving force between the system
and the surroundings being the
temperature difference ( Figure
2.12 Page 60)

ENERGY TRANSFER BY WORK
 If the energy crossing a boundary is
not heat  It must be WORK
 The work done per unit time is
called POWER
 Work can be electrical ,
mechanical work ( moving
boudary, shaft work, spring
work,….)

CONVENTION SIGNS
 Heat and work are directional
quantities
 A complete description of work and
heat need a magnitude and a
direction
 The generally accepted formal sign
convention for heat and work is as
follow:
 Heat amd work In :POSITIVE
 Heat and work OUT : NEGATIVE

Moving Boundary work
( Figure 4.1, page 166)
 It is associated to an expansion or
compression of a gas in a piston-cylinder
device.
 It is also called PdV work
 In Figure 4.3, the work can be defined
as
PdVPAdsFdsWb 

Boundary Work
 The total boundary work done during the
entire process is defined by:
 The area under the process is defined by
( Figure 4.3 page 167):

2
1
.dVPWb

2
1
PdVAAREA

A GAS CAN FOLLOW DIFFERENT
PATHS ( FIGURE 4.4)
 A gas can follow different paths and
since work is a path function 
the area under each path is different
( Figure 4.4 )

Class work
 1) Work examples 4.1 to 4.3 pages
168-171)
 2) Work problems 4.11, 4.12, 4.14
page 202

FIRST LAW OF
THERMODYNAMICS
CONSERVATION OF ENERGY

FIRST LAW OF
THERMODYNAMICS?
 Also known as the conservation of energy
principle, the first law states that:
 energy can never be created not
destroyed, it can only changes form
 Example: ( Figure 2.37 page 70) The
rock at some elevation possesses some
potential energy. As the rock falls, part of
its potential energy is converted into
kinetic energy.

WHAT IS ENERGY?
 Energy can exist in different
forms:
 Internal ( sensible, latent, chemical
and nuclear)
 Kinetic
 Potential
 Electrical
 Magnetic
 Surface tension effects

FOR A SIMPLE COMPRESSIBLE
SYSTEM
 The change of total energy is :
PEKEUE 
)(
)(
2
1
)(
12
2
1
2
2
12
zzmgPE
vvmKE
uumU




FOR A STATIONARY ( CLOSED)
SYSTEM
0 PEKE
UE 

MECHANISMS OF ENERGY
TRANSFER
 Energy can be transferred to or from
a system in three forms:
Heat transfer: Q
 Work : W
 Mass Flow : m

HEAT TRANSFER?
 Caused by a difference of TEMPERATURE
 TO A SYSTEM : Increases the ENERGY of
molecules and thus the INTERNAL ENERGY
 FROM A SYSTEM : Decreases the
INTERNAL ENERGY of the system because
the heat comes from the energy of
molecules

WORK TRANSFER?
 ENERGY interactions not caused by a
temperature difference
 Examples :
 Rising Piston
 Rotating Shaft
 Electrical wire
 Work transfer to a system  Esystem ↑
 Work transfer from a system  Esystem↓

MASS FLOW ?
Because mass carries
energy with it
 Mass enters the system 
Esystem↑
 Mass leaving the system
Esystem↓

NET ENERGY TRANSFER?
 Energies transferred by heat, work and mass
can de added ( FIGURE 2.45 page 74):
Ein-Eout=( Qin-Qout)+( Win-Wout)+(Emass,in-Emass out)
 IN RATE FORM ( KW):
ΔË = Ëin-Ëout
outinsystem EEE 

ENERGY BALANCE FOR
CLOSED SYSTEM
 QNET IN= QIN-QOUT
 WNET OUT= WOUT-WIN
outnetinnet WQE ,, 
outnetinnet WQU ,, 

CLASS WORK
 WORK EXAMPLES 4.5, 4.8,4.9 and
4.10 pages 174-187

CLASS WORK#1
 Work examples 5.1 page 226; 5.3
page 231; 5.6 page 238 and 5.7
page 239.

HOMEWORK #2
 Do problems : 4.5 , 4.6E , 4.18,
4.24, 4.26E, 4.33, 4.36 pages 201-
204.

THE USES OF THE FIRST
LAW OF
THERMODYNAMICS
1) ENERGY BALANCES
2) DESIGN OF THE EQUIPMENTS
( FIND THEIR DUTY)

WHAT IS ENERGY ( ENTHALPY
OR INTERNAL ENERGY)?
 THE LINK BETWEEN THE SYSTEM
AND THE SURRONDINDS
 CAN BE DEFINED BY THE
PROPERTIES OF THE SYSTEM
FROM TABLES
 CAN BE CALCULATED BY THE
ENERGY BALANCE
out,netin,net WQE 

SOLVING PROBLEM
CASE I
( ENERGY BALANCE)
WE WANT TO DETERMINE ONE
OF THE STATES OF THE SYSTEM
KNOWING THE DUTIES QNET IN
AND WNET OUT

STATIONARY RIGID TANK
SYSTEM
 ΔKE=ΔPE =0
 NO MASS EXCHANGE WITH THE
SURRONDINDS
 NO BOUNDARY WORK
 ΔE=ΔU outnet,innet, WQΔU 

STATIONARY AND CLOSED
PISTON CYLINDER
 ΔKE=ΔPE =0
 NO MASS EXCHANGE WITH THE
SURRONDINDS
 THERE IS BOUNDARY WORK
 WNET OUT SHOULD NOT INCLUDE
BOUNDAY WORK
outnet,innet, WQΔH 

STEADY FLOW SYSTEM
 WE WILL STUDY NEXT
)gz
2
v
h(m)gz
2
v
h(mWQ i
2
i
iie
2
e
ee  

IN CASE I
1) WE SHOULD KNOW THE SPECIFIC
ENERGY OF ONE STATE ( ENTHALPY OR
INTERNAL ENERGY)
2) CALCULATE THE TOTAL ENERGY
KNOWING THE MASS OR MASSFLOW
3) CALCULATE THE ENERGY OF THE
SECOND STATE USING ENERGY BALANCE
( KNOWN DUTIES OF THE EQUIPMENTS)

 CALCULATE THE SPECIFIC ENERGY OF
THE SECOND STATE BY DIVIDING BY THE
MASS ORMASS FLOW
 USING THE CALCULATED SPECIFIC
ENERGY TO DEFINE THE SECOND STATE
AND CALCULATE ALL ITS PROPERTIES

HOW TO FIND u OR h FOR
DIFFERENT SYSTEMS
 SUBCOOLED OR COMPRESSED
LIQUID  WE NEED TWO
VARIABLES FROM P,T,v
 APPROXIMATION FOR COMPRESSED
LIQUIDS:
 v(P,T)= vsat (T)
 u(P,T)=usat (T)
 h(P,T)= hsat (T)+ vsat(T) { P-Psat(T)}

 SATURATED LIQUID: WE NEED
ONE VARIABLE P OR T OR v
 WET STEAM: WE NEED TWO
VARIABLES ( P OR T) AND OTHER
VARIABLE ( EX: QUALITY OR
SPECIFIC VOLUME)

 SATURATED VAPOR: WE NEED ONE
VARIABLE P OR T OR v
 SUPERHEATED VAPOR: WE NEED TWO
VARIABLES BETWEEN P,T AND v
 IDEAL GAS: WE NEED ONLY
TEMPERATURE BECAUSE U AND H
DEPENS ONLY ON TEMPERATURE

CASE 2
( DESIGN PROBLEM )
WE WILL HAVE THE ENERGIES OF
BOTH STATES AND WILL CALCULATE
THE MISSING DUTY OF THE
EQUIPMENT (WORK OR HEAT )
EXCHANGE WITH THE SURRONDINGS

 WE SHOULD DEFINE THE TWO STATES
 FIND THE SPECIFIC ENERGIES AT BOTH STATES
 CONVERT TO TOTAL ENERGIES
 APPLY THE ENERGY BALANCE
 FIND THE DUTY OF THE EQUIPMENT USED TO
EXCHANGE WORK ( TURBINE, COMPRESSORE,
PUMP) OR HEAT ( HEAT EXCHANGER , COOLER.
HEATER, FURNACE)

STEADY FLOW SYSTEMS ?
( FIGURE 5.7 page 225)
 During a steady flow process, no
intensive or extensive property
within the control volume change
with time  V, m and E of the
control volume remains constant
 mcv=CST AND ECV=CST
A large number of engineering devices (
turbines, compressors, nozzles) are classified as
steady flow devices

MASS BALANCE FOR STEADY
FLOW SYSTEMS
 mcv=CST  min=mout
 With multiple inlets and exits: FIGURES
4.23 and 4.24 page 181
  ei mm

ENERGY BALANCE FOR STEADY
FLOW SYSTEMS
ECV=CST  ΔECV=0

Rate of net energy transfer by heat, work
and mass IN is equal to the rate of net
energy transfer by heat, work and mass
OUT
).(0 FLOWSTEADYEEE systemoutin 
outin EE 

ENERGY BALANCE FOR A
GENERAL STEADY FLOW
SYSTEM
)
2
()
2
(
22
i
i
iie
e
ee gz
v
hmgz
v
hmWQ  

SECOND LAW OF
THERMODYNAMICS
DIRECTION OF A PROCESS
AND ENTROPY

THE SECOND LAW OF
THERMODYNAMICS
 In the last chapter, we have studied
the first law of thermodynamics or
the conservation of energy principle.
 Any process can take place only if
the first law is respected
 Is This enough?
 The answer belongs to the second
law of thermodynamics or law of
entropy

EXAMPLE #1
 If we put a hot coffee in a cold room 
the coffee will lose heat to the room ( Fig
6.1 page 284)
 If we consider the opposite process: Can
the coffee get hotter getting heat from the
cold room?
 The answer is NO but the inverse process
still respect the first law of
thermodynamics which is conservation of
energy
 SO WHAT IS WRONG IN THE REVERSE
PROCESS?

RESTRICTION OF THE SECOND
LAW
 It is clear, from this example, that
the first law of thermodynamics
places no restriction on the
direction of the process.
The reverse process violates
the second law of
thermodynamics

EXAMPLE #2
 CONSIDERING A REFRIGERATOR:
THE INLET IS GETTING COLDER AND
THE OUTSIDE AIR IS GETTING
HOTER
 DOES IT VIOLATE THE
SECOND LAW OF
THERMODYNAMICS?

EXAMPLE #3
 From Figure 6.8 page 286, we can observe that
the mechanical energy done by the shaft is first
converted into INTERNAL ENERGY of water
 This energy may leave the water as heat
 However, can we rotate the shaft
by heating the water?
 ANSWEER: LATER

CAN WE TRANSFORM HEAT
INTO WORK?
 To explain this , is often convenient to
have a hypothetical body with a
relatively large thermal energy capacity
that can supply or absorb finite amount of
energy without undergoing any change
in temperature
 Such a body is called THERMAL
ENERGY RESERVOIR ( FIGURE 6.6
page 285)

WHAT IS A THERMAL ENERGY
RESERVOIR?
 A thermal energy reservoir that
supplies energy in the form of heat is
a SOURCE ( FIGURE 6.7)
 A thermal energy reservoir that
absorbs energy in the form of heat is
a SINK (FIGURE 6.7)
 Thermal energy reservoirs are known
as HEAT RESERVOIRS.

THE SECOND LAW ( FIGURE 6.9)
 In figure 6.9 PAGE 286, we can see that
only a part of heat received by a heat
engine is converted into work while the
rest is rejected to a sink
CONCLUSION:
WE CAN NEVER TRANSFORM HEAT
COMPLETELY INTO WORK

FINAL ANSWEER
 WORK CAN BE CONVERTED TO HEAT
DIRECTLY AND COMPLETELY
 HOWEVER, TO TRANSFORM HEAT
INTO WORK REQUIRES THE USE OF
SOME DEVICES CALLED: HEAT
ENGINES

HEAT ENGINES

HEAT ENGINES?
 Heat engines differ considerably from one
another but all can be characterized by
the following STEPS:
 They receive heat from a high
temperature source ( solar, furnace,
nuclear reactor
 They convert part of this heat into
work ( usually in the form of rotating
shaft: turbine)
 They reject the remaining waste heat to a
low temperature sink ( air, rivers, sea,…)
 They operate on a cycle

EXAMPLE: STEAM POWER
PLANT ( figure 6.10) page 287
 Qin= amount of heat supplied to steam in
boiler from a high temperature source (
furnace)
 Qout = amount of heat rejected from
steam in condenser to a low temperature
sink ( atmosphere)
 Wout= amount of work delivered by steam
as it expands in turbine
 Win = amount of work required to
compress water to boiler pressure

NET WORK OUT?
 WE PRODUCE Wnet.out
Wnet, out= Wout- Win ( kJ)
 Since is a cycle  ΔU=0
 Wnet,out= Qin-Qout

WHAT IS THERMAL
EFFICIENCY?
 In general, we define efficiency as the
amount of energy you give divided by the
amount of energy you spent for:
 The fraction of the heat input that is
converted into net work is a measure
of the performance of the heat engine
and is called thermal efficiency

CLASS WORK #1
 From figure 6.10 page 287 show that
in
out
in
out,net
th
Q
Q
1
Q
W


Class work #2
 Work examples 6.1 and 6.2 page
290
 Work problems 6.20-6.23 page 321
 Work problems 6.18 ,6.19,6.24,6.28

REFRIGERATORS &
HEAT PUMPS

REFRIGERATORS ( FIG 6.19
PAGE 292)
 In order to transfer heat from a cold space (
inside refrigerator) to hot space ( kitchen)
refrigerator must receive work.
 As shown in figure 6.20 page 292 , the
performance of a refrigerator ( COPR) is the ratio
between the desired output over the required
input
COPR= QL/Wnet,in

HEAT PUMPS ( FIGURE 6.21
PAGE 293)
 Work like the refrigerator but the objective
is different
 Now we want to heat up an apartment
during cold weather
 Therefore : form figure 6.21 page 293, we
have
COPHP = QH/Wnet in

Class work #4
 Work examples 6.3 and 6.4 page
295
 Work problems 6.39, 6.42, 6-50 AND
6.52 pages 322-323

CARNOT HEAT
ENGINE
THE CARNOT CYCLE

REVERSIBLE PROCESS
Reversible versus Irreversible process?
 Reversible process is defined as a process
that can be reversed without leaving any
trace on the surroundings.
 Both the system and the surroundings are
returned to their initial states at the end of
the reverse process
 This is possible only if the net heat and
net work exchange is zero for the
combined ( original and reverse process)

IRREVERSIBLE PROCESS
 All real processes are Irreversible
and the Reversible process is a
theoretical process used as a limit
for the corresponding irreversible
process
 Irreversibility is caused by
FRICTION

THE CARNOT CYCLE
 WE HAVE ALREADY MENTIONNED
THAT HEAT ENGINES ARE CYCLIC
DEVICES
 REVERSIBLE CYCLES CAN NOT EXIST
BECAUSE OF THE FRICTION.
 THE CARNOT CYCLE IS A
THEORITICAL REVERSIBLE
CYCLE PROPOSED IN 1824 BY SADI
CARNOT ( A FRENCH ENGINEER)

THE CARNOT CYCLE :
FIGURE 6.37 page 304
 THE CARNOT CYCLE IS COMPOSED
OF:
 Reversible Isothermal Expansion
 Reversible Adiabatic Expansion
 Reversible Isothermal Compression
 Reversible Adiabatic Compression

CARNOT CYCLE

CARNOT CYCLE
 FIGURE 6.38 ( PAGE 305) SHOWS THE
CARNOT CYCLE
 PROCESS 1->2 EXPANSION AT
CONSTANT TH:
The gas is in contact with a source at TH.
 As the gas expands , TH tends↓ by dT
 However, the heat input QH from the
source TH, will increase again T to TH in a
reversible way
 AT THE END OF THE PROCESS , THE GAS
HAS ABSORBED QH

CARNOT CYCLE
 PROCESS 2->3:ADIABATIC
EXPANSION WHERE
TEMPERATURE DROPS FROM TH
TO TL:
 At State 2, the reservoir is removed
and replaced by an insulation
 The gas continues to expand slowly
doing work on the surroundings until
T TL

CARNOT CYCLE
 PROCESS 3->4: ISOTHERMAL
COMPRESSION AT TL
 At state 3, the insulation is removed and
the gas is in contact with a sink at
temperature TL.
 The piston is pushed inward by an
external force doing work on the gas
 As the gas compresses , T ↑by dT but the
sink will absorb this elevation of
temperature
 At the end of the process 3->4, the gas
has rejected QL

CARNOT CYCLE
 PROCESS 4->1: ADIABATIC
COMPRESSION WHERE
TEMPERATURE RISES FROM TL TO
TH:
 The sink is removed and the
insulation is put back and the gas is
compressed in a reversible manner
 The gas returns to its initial state
which completes the cycle

EFFICIENCY OF CARNOT HEAT
ENGINE
 The second law of thermodynamics states that no heat
engine can have efficiency of 100%
 The maximum efficiency is for reversible cycle as Carnot
heat engine. CARNOT EFFICIENCY IS EQUAL TO:
 Temperature in Kelvin.
H
L
H
L
REVTHCARNOT
T
T
Q
Q
 11,

CLASS WORK
 WORK EXAMPLE 6.5 PAGE 310
 Work problems 6.80 and 6.81 page
325

THE CARNOT REFRIGERATOR
 Operates on the reverse Carnot cycle
 The Coefficient of performance of any
refrigerator ( reversible or irreversible):
1
1


L
H
R
Q
Q
COP

THE CARNOT HEAT PUMP
 Operates as a reverse Carnot cycle
 The Coefficient of performance of any
heat pump ( reversible or irreversible):
H
L
HP
Q
Q
COP


1
1

COP FOR REVERSIBLE
PROCESS
 REVERSIBLE REFRIGERATOR:
 REVERSIBLE HEAT PUMP
1
1
,


L
H
REVR
T
T
COP
H
L
REVHP
T
T
COP


1
1
,

CLASS WORK
 WORK EXAMPLES : 6.6 AND 6.7
page 314
 WORK PROBLEMS: 6.94-6.97pages
326-327

ENTROPY?

WHAT IS ENTROPY?
 MEASURES DESORDER
 AS A SYSTEM BECOMES MORE
DESORDERED  THE POSITION OF
MOLECULES BECOME LESS
PREDICTABLE  THE ENTROPY
INCREASES
 USEFUL PROPERTY FOR SECOND
LAW
 EXTENSIVE PROPERTY

CHANGE OF ENTROPY?
 The entropy is a state function
 The entropy change of a process from
state 1 to state 2 can be determined by
the relation:
rev
T
Q
SSS int,
2
1
12 )(



SPECIAL CASE
 For an internally reversible isothermal
heat transfer process, the change of
entropy reduces to :
 WORK EXAMPLE 6-1 page 305
0T
Q
S 

PROPERTY DIAGRAMS
INVOLVING ENTROPY
 In the second law, is very useful to use
the T-S diagram or H-S diagrams
 From the previous equations , we can
write 

2
1
int, .dSTQ rev

T-S DIAGRAM
 In figure 6-16 page 314, we can
calculate Qint,rev as the area of the
curve under the states 1 and 2
 Isentropic process is a vertical line (
Figure 6-17 page 315)
 T-S diagram of water is in appendix
in figure A-9

T-S DIAGRAM OF A CARNOT
CYCLE?
 WORK EXAMPLE 6-6 PAGE 315

CARNOT VAPOR
CYCLE
Chapter 10 page 565

CARNOT VAPOR CYCLE
 CARNOT CYCLE is the most efficient
cycle operating between two
temperatures
 IS CARNOT CYCLE EFFICIENT FOR
VAPOR POWER PLANT?

CARNOT VAPOR CYCLE
 FIGURES 10.1 a and b PAGE 566 show two
Carnot vapor cycles:
Figure 10.1 a:
 Process 1-2 : The fluid is heated reversibly and
isothermally in a boiler
 Process 2-3: The fluid is expanded isentropically
in a turbine
 Process 3-4 : The fluid is condensed reversibly
and isothermally in a condenser
 Process 4-1:The fluid is compressed to the
initial state

IMPRACTICALITIES OF THE
CARNOT VAPOR CYCLE
 A) In Figure 10.1 A , Processes 1-2 and 3-4 (
Boiler and condenser) use two phases- region for
the water , This can be easily done in practice
by maintaining a constant pressure in the boiler
and condenser , however the maximum
temperature of the source should remain below
the critical temperature of water ( 3740C)
 This will limit the thermal efficiency
 B) To increase the maximum temperature of
water above the critical temperature, we need to
heat the steam water at constant temperature
 not easy in practice

PROCESS 23
 B) The turbine will have to handle
steam with poor quality ( steam with
high moisture)
 The droplets of water will
cause erosion in the turbine

PROCESS 4 1
 C) In the isentropic compression, the
pump will handle 2 phases fluid ???
 cavitation of the pump

Figure 10.1 b page 566
 This Carnot process involves
isentropic compression at very high
pressures and isothermal heat
transfer at variable pressures
 Not possible in practice

RANKINE VAPOR
CYCLE
IDEAL CYCLE FOR STEAM
POWER PLANT

RANKINE VAPOR CYCLE
 The difficulties encountered in the
CARNOT vapor cycle can be
eliminated using the RANKINE vapor
cycle
 This can be done by;
SUPERHEATING the steam in boiler
CONDENSE all the steam in
condenser
 FIGURE 10.2 PAGE 567

RANKINE VAPOR CYCLE
 IS THE IDEAL CYCLE FOR VAPOR POWER
CYCLES: DIAGRAM IN PAGE 567
 PROCESSES OF THE CYCLE:
 3-4:Isentropic EXPANSION (turbine)
 4-1:Isobaric heat rejection (
condenser)
 1-2:Isentropic COMPRESSION (pump)
 2-3:Isobaric heat addition ( boiler)

WORKS INVOLVED IN RANKINE
CYCLE
Work output of the cycle (Steam
turbine) W1 Work input to the cycle
(Pump) W2
are respectively:
 W1 = m (h1-h2)
 W2 = m (h4-h3)
where m is the mass flow of the cycle.

HEATS INVOLVED IN RANKINE
CYCLE
 Heat supplied to the cycle (boiler),
Q1
 Heat rejected from the cycle
(condenser), Q2 are respectively:
 Q1 = m (h1-h4)
 Q2 = m (h2-h3)

NET WORK OUTPUT
 The net work output of the cycle is:
W = W1 - W2

THERMAL EFFICIENCY
 The thermal efficiency of a Rankine
cycle is:
IN
OUT
IN
NET
RANKINE
Q
Q
Q
W
 1

Class Work
 Work Example 10-1 page 569

SOLUTION OF EXAMPLE 10-1
 A) Draw the process
 B) Draw the T-S Diagram of the
process
 C) Put the data on the process

 STATE 1:
P1= 75 kPa and saturated liquid
 Table A-5: h1=hf= 384.39 kJ/kg
v1=vf= 0.001037
m3/kg

 STATE 2:
P2= 3MPa and s2=s1
wpump = v1 ( P2-P1) = 0.001037 ( 3000-
75)
= 3.03 kJ/kg
h2= h1+wpump= 384.39+3.03= 387.42 kJ/kg

 STATE 3: ( superheated vapor)
P3= 3Mpa and T3= 3500C
 Table A-6 : h3= 3115.3 kJ/kg
s3= 6.7428
kJ/kg.K

 STATE 4: ( MIXTURE)
P4= 75 kPa and s4=s3
WHAT IS THE PERCENTAGE OF VAPOR?
x4= ( 6.7428- 1.213)/ 6.2434 = 0.8857
fg
f
s
ss
x

 4
4

 h4?
h4= hf + x4.hfg
= 384.39 + 0.8857x 2278.6
= 2402.6 kJ/kg

Heats in and out ?
 Qin= h3 - h2= 3115.3- 387.42=
2727.9 kJ/kg
 Qout= h4-h1= 2402.6-384.39=
2018.2 kJ/kg

THERMAL EFFICIENCY
%26
9.2727
2.2018
11 
in
out
Q
Q


Class work
 Solve problems 10.15 , 10.18, 10.19
page 605

RANKINE VAPOR CYCLE
 THE IDEAL CYCLE FOR VAPOR POWER
CYCLES
 PROCESSES OF THE CYCLE:
 1-2:Isentropic EXPANSION (turbine)
 2-3:Isobaric heat rejection (
condenser)
 3-4:Isentropic COMPRESSION (pump)
 4-1:Isobaric heat addition ( boiler)

DEVIATION FROM RANKINE
CYCLE
 FIRST Cause of deviation:
A) Fluid friction causes pressure drop in the boiler,
condenser and the piping:
 RESULTS:
1)The steam leaves the boiler at some LOWER pressure
2) The pressure at the inlet of turbine is somewhat LOWER
3) The pressure drop in the condenser is usually very small
SEE FIGURE 10-4 A PAGE 572 AND COMPARE ACTUAL CYCLE
TO IDEAL RANKINE CYCLE

Solution for pressure drop in the
system
TO COMPENSATE the problem of
pressure drop in the system: THE
PUMP SHOULD WORK AT HIGHER
PRESSURE THAN THE IDEAL CYCLE.

Second cause of deviation from
Rankine cycle
B) The other cause of irreversibility is
the heat loss from the steam to the
surroundings
 As a result : MORE HEAT NEEDS TO
BE TRANSFERRED TO THE STEAM

IRREVERSIBILITY OCCURING IN
THE PUMP AND THE TURBINE
 AS A RESULT OF
IRREVERSIBILITY ON THE IDEAL
RANKINE CYCLE: Figure 10-4b
page 572
 A pump requires a greater work
input and the turbine produces a
smaller work output

ISENTROPIC EFFICIENCIES
 FROM FIGURE 10-4b page 573
 A) ISENTROPIC EFFICIENCY OF THE PUMP
:
 B) ISENTROPIC EFFICIENCY OF THE
TURBINE
12
12
hh
hh
w
w
a
s
a
s
PUMP



s
a
s
a
TURBINE
hh
hh
w
w
43
43




CLASS WORK
 WORK EXAMPLE 10-2 PAGE 573
 WORK PROBLEM 10.19 PAGE 605

HOW CAN WE INCREASE THE
EFFICIENCY OF RANKINE
CYCLE
 Steam power plant are responsible
for the production of most electric
power in the world
 Even small increase in thermal
efficiency can mean large savings in
fuel consumption

The basic concept
 The basic concept is to: η= 1-(TL/TH)
 Increase the average temperature at
which heat is transferred to the working
fluid in the boiler
 Decrease the average temperature at
which heat is rejected from the working
fluid to the condenser
 BUT DO NOT FORGET: ONLY FOR CARNOT
CYCLE , WE HAVE η= 1-(TL/TH)

The actions
 A) Lowering the condenser pressure 
this will lower the TLOW,av
 Figure 10.6 page 574
 B) Superheating the steam to higher
temperatures  Increases Thigh, av
 Figure 10.7 page 575
 C) Increasing the boiler pressure 
Increases Thigh,av  Figure 10.8 page 575
 Increase efficiency but increases the
moisture at the turbine

IDEAL REHEAT RANKINE CYCLE
 How we can take advantage of
increasing the efficiencies at higher
boiler pressures without facing the
problem of excessive moisture at
the final stage of the turbine?
 TWO POSSIBILITIES

POSSIBILITY #1
 SUPERHEAT THE STEAM TO VERY
HIGH TEMPERATURES BEFORE IT
ENTERS THE TURBINE
 COULD BE UNSAFE FOR THE METAL
OF THE TURBINE

POSSIBILITY #2
 EXPAND THE STEAM IN THE
TURBINE IN TWO STAGES AND
REHEAT IT BETWEEN THE TWO
STAGES
REHEAT IS THE PRACTICAL
SOLUTION FOR THE HIGH MOISTURE
IN THE TURBINE T
 FIGURE 10-11 PAGE 579

CLASS WORK
 Work example 10.4 page 580
 Work problems 10.35, 10.39 pages
607-608

HOMEWORK
 TO PREPARE YOUR FINAL EXAM, TRY
TO SOLVE THE PROBLEMS 4.28,
4.31, 4.38, 4.44, 5.50, 5.52, 5.58,
6.17, 6.18,6.42, 6.44, 6.52, 6.82,
6.83, 6.95,6.96, 6.99, 9.17, 9.18 ,
10.15, 10.20,10.22,10.35
 GOOD LUCK

Thermodynamics Fundamentals for Chemical Engineers

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