sem4 termo slide nota dr wan fulll
TRANSCRIPT
Week-1
Thermodynamics(Classical Thermodynamics)
1
“Therme” meaning heat
2
“Dynamics” meaning strength
Thermodynamics
- the science concerned with the causes and effects arising from energy transformations,(e.g heat & work) and the physical properties of substances that are involved in energy transformations.
3
Thermodynamics is based on definitions, concepts, and physical laws.
4
5
Thermodynamics
Statistical
(Microscopic)Classical
(Macroscopic)
Microscopic viewpoints- A collection of atoms within a container, each with a unique velocity
Macroscopic viewpoints (classical)-the properties of system are
assigned to the system as a whole.(does not require knowledge of the behaviour of individual particle)
6
The energy in both cases is the same, E
In the macroscopic description,atomistic concepts are disregarded.
7
Thermodynamics, is fields of physics that describes and correlates the physical properties of macroscopic systems of matter and energy
8
9
Thermodynamics
Processes
properties principles
Thermodynamics Triangle
Definitions:System : the part of the universe that has been chosen to study.
Surroundings: the rest of the universe
Boundary: the surface (can be real or imaginary) dividing the system from the surroundings.
10
11
boundarysystem
surroundings
System can be:Closed: energy can transfer between the system and the surroundings, but not the mass.Open: Mass and energy can transfer between the system and the surrounding.Isolated: Neither mass nor energy can transfer between the system and the surrounding
12
13
(OPEN SYSTEM) or control volume.e.g: pumps, compressors, turbines, valves, heat exchangers.
(CLOSED SYSTEM)or control mass.Mass = constante.g: sealed tanks, piston & cylinder devices.Boundary can move.
(ISOLATED SYSTEM)
Mass=constant
14
System
Surr 1
Work
Surr 4
Surr 3Mass
Surr 2
Heat
Mass
Isolated system Boundary
Heat=0
Work=0
Mass=0
Across
Isolated
boundary
15
16
Property of a System
– Any characteristic of a system in
equilibrium that can be calculated or
measured. The property is independent of
the path used to arrive at the system
condition. Also known as variables of the
system.
Some thermodynamic properties are
pressure P, temperature T, volume V, and
mass m.
17
Two kinds of properties:
1) Intensive properties – independent of the
size of the system, e.g. T, P, r, any mass
independent property.
2) Extensive properties – proportional to the
quantity of material, or those that vary directly
with size of the system. e.g. m,V, E(total), H
(mass dependent property)
18
system
m, V, T
P, r
½ m, ½ V,
T, P, r
½ m, ½ V,
T, P, r
System 1
System 2
pseudointensive properties:
- extensive properties expressed per unit quantity of material.Or extensive properties per unit massEg:specific volume : v = V/m
(m3/kg)
[molar properties : expressed on a unit mole basis]
19
Basic thermodynamics properties
1.Density : mass per unit volume
20
21
)/( 3mkgV
m
volume
massr
Low densiy
Density : mass per unit volume
High density
22
specific gravity or relative density:
the ratio of the density of a substance to the
density of water at a specific temperature.
(usually 4oC, for which rH2O = 1000 kg/m3)
rs = r/rH2O
23
1 2 3 ………..12 ……………….6.022 x 1023
Amount of substance: how much is there?
Avogadro’s
NumberDozen
24
A
FP
Impact
Weight
2. Pressure (P) : force per unit area
F
F
A
25
3. Temperature (T) (thermal potential) :
-a measure of relative hotness or coldness of a material
26
27
4. Volume (V) (mechanical displacement) :
the quantity of space possessed by a material.
28
5. Entropy (S) (thermal displacement) :
the quantity of disorder possessed by a material.
29
6. Internal energy (U) :
the energy of a material which is due to the kinetic and potential energies of its constituent parts (atoms and molecules, usually).
30
Secondary thermodynamic properties
Enthalpy [H] - internal energy plus the pressure-volume product.
Heat capacity [Cp or Cv] (specific heatcp or cv) –-the amount of energy required to increase the temperature of one unit quantity of material by one degree, under specific conditions.
31
State (of System)
Consider a system that is not undergoing any change. The properties can be measured or calculated throughout the entire system. This gives us a set of properties that completely describe the condition or state of the system. At a given state all of the properties are known and have fixed values. If the value of even one properties changes, the state will change to another state.
32
Thermodynamic state, or macrostate,
is specified by stating a complete set of state variables, minimum any two of the three STATE VARIABLES (P, V or T).
33
34
Eg: Water
T = 40oC
v= 0.87 m3/kg
The state of the water is fixed by two properties
The system can change states only through an exchange of internal energy with the environment (unless it's a vacuum).
35
WaterT = 40oCV = 0.87 m3/kg
(State 1)
WaterT = 90oCV = 1.7 m3/kg
(state 2)
State FunctionAny physical quantity having a unique value for a given state of the system.
Examples: Temperature, entropy, enthalpy, pressure, volume,
N.B.: "work" and "heat", although thermodynamic variables, are not state functions.
36
Equation of State :-any equation that relates properties of a system at equilibrium
37
Equilibrium : implies a state of balance
Thermodynamic equilibrium: is achieved when all the
relevant types of equilibrium are satisfied.
iii. Phase equilibrium: the mass of each phase
reaches an equilibrium level and stays there.
38
iv. Chemical equilibrium: chemical composition
does not change with time.
ii. Mechanical equilibrium: no change in
pressure at any point of the system with time.
i.Thermal equilibrium: temperature is the same
throughout the entire system
A Simple Compressible System:
- a system without the effects of electrical, magnetic, gravitational, motion and surface tension.
The State Postulate- The state of a simple compressible system is completely specified by two independent, intensive properties.
e.g. T and v are independent properties.T and P : are independent properties for single phase systemT and P : are dependent properties for multiphase system.
39
Thermodynamic Processes and Cycles.
Process - any succession of events; or any change
from one state to another.
States of a thermodynamic system can be changed by
interacting with its surrounding through work and heat.
When this change occurs in a system, it is said that the
system is undergoing a process.
40
thermodynamic process - is the specific way in which internal energy is exchanged.
Path of the process:a series of states through which a system passes
during a process
41
State 2
State 1 Process path
Property B
P
r
op.
A
Process path
The points (P1,V1) and (P2,V2) characterize the initial and
final equilibrium states.
The thermodynamic process can be indicated by a line
passing through the intermediate equilibrium states the
system passed through.
42
Quasi-static process: when a process proceeds
in such a manner that the system remains
infinitesimally close to an equilibrium state at all
time ………slow-process
43
Prefix iso- : to designate a process for which a particular
property remains constant
Types of Thermodynamic Processes
Reversible: Can happen slowly in either direction.
Irreversible: Involves net increase in entropy (can’t go
backwards).
for a sample of ideal gas: Isothermal: Constant temperature T = 0
44
V1
P1
T
V2
P2
T
T
V
State 1 State 2
Isobaric: Constant pressure P or P = 0
45
V1
P
T1
V2
P
T2
P
V
State 1 State 2
Isochoric: Constant volume V or V= 0
and W = 0
46
V
P1
T1
V
P2
T2
P
V
State 1
State 2
Adiabatic: No heat added or release or Q = 0
47
Q
Q
Q
Q
Cycle:is a sequence of different processes that begins and ends
at the same thermodynamic state.
48
Initial state
final state
V
P
final state
initial state
V
P
Forces and Potentials
49
Solid matter is comprised of interacting atoms and
molecules – how might we model those
interactions?
50
U(x)
xU = ½(kx)
Potential energy curve for SHM
Morse curve – approximation to potential energy variation for two neutral atoms.
51
200 ))](exp(1[ rrEU
52
ro
Eo
U(r)
r
Bottom of interatomic potential curve may be approximated by SHM potential.
53
U(r)
r
SHM
Bottom of interatomic potential curve may be approximated by SHM potential.
Temperature
54
•a concept originates with our sense perceptions.
•a term used to quantify the difference between
warm and cold level of internal energy of a
substance.
•a measure of the average translational kinetic
energy associated with the disordered microscopic
motion of atoms and molecules.
•a physical quantity which gives us an idea of how
hot or cold an object is.
55
temperature
Molecular kinetic energy
Temperature scale
Standard temperature
MAY BE DEFINED IN TERM OF
And experessed in term of
Based on
The temperature of an object depends on
how fast the atoms and molecules which
make up the object can shake, or
oscillate.
56
As an object is cooled, the oscillations of its atoms
and molecules slow down.
With increasing temperature the inter-atomic bond
length increases (centre of oscillations shifts to
larger r).
Hence object expands at higher temperatures.
57
Thermometric Properties : properties
of materials that change with temperature. (may be used to define a temperature)
volumes of most liquids increase with temperature.
length of a metal rod increases as the temperature increases.
58
59
pressure of a constant volume of gas
increases with temperature.
volume of a gas at constant pressure
increases with temperature.
electrical resistance of a piece of wire
changes with temperature.
Thermometer : a device for measuring
temperature.
60
0 K = - 273.15 oCkelvin or K is absolute temperature scale
61
h
heat
pressure
P
T
62
-273.15 temperature
P
63
0K T (K)
P
Heat is internal energy that is
exchanged between two objects due to their difference in temperature.
Heat always transfers, or flows, from a body at higher temperature to one at lower temperature.
64
Thermal Equilibrium
65
A B
UAUB
HOT COLD
Internal energy of A is greater than internal energy of B
UA > UB
heat
UAAUBA
UA: decreasing UB : increasing
to UAA to UBA
UAA ≡ UBA
No energy transfer
Thermal equilibrium is the subject of the Zeroth Law of thermodynamics.
66
hot cold
Heat is transferred from hot object to cold object
Both objects are at the same temperature –
THERMAL EQUILIBRIUM - Zeroth Law of Thermodynamics
Th TcTh > Tc
Te Te
Q
67
The "zeroth law" states that if two systems are at the same
time in thermal equilibrium with a third system, they are in thermal
equilibrium with each other.
B
A C
A is in thermal equilibrium with B; B is in thermal equilibrium with CTherefore A and C are in thermal equilibrium
No heat transferA C
The Zeroth Law implies existence of some universal property of systems in thermal equilibrium.
Temperature is that property of thermal equilibrium that is the same for any two systems in thermal equilibrium.
Temperature being a property of a body and heat being an energy flow to or from a body by virtue of a temperature difference.
The Zeroth law is used to define temperature
68
Let TA is the temperature of ATB is the temperature of BTC is the temperature of C
If TA = TB and TB = TC Then TA = TC
Any two objects in thermal equilibrium have the same temperature
So temperature determines if two objects are in thermal equilibrium.
Any one thermodynamic system can be used as a “thermometer” to define a temperature scale.
69
Internal energy
Energy is often defined as the ability to do work
Two types of energy:Kinetic Energy is energy of motion Potential Energy is energy of position
(technically position in a gradient "field")
70
Potential energy is often thought of as "stored" kinetic energy, meaning that bodies remain stationary in a potential field while held in place by some force, and upon change in this force (breaking the bond between two atoms), potential energy is converted to kinetic form.
71
Many forms of energy:
Nuclear Energy :The potential energy required
to bind nucleons in the nucleus.
Light Energy: The potential energy possessed
by the oscillating electric and magnetic fields that make up electromagnetic radiation.
Chemical Energy: The potential energy stored
in the electrostatic bonding relationships among atoms in a molecule.
72
73
Electrical Energy : The potential energy
involved in initiating and maintaining electron flow.
Mechanical Energy :The energy generated
(or stored) by machines which induces (or results
from) concerted motion processes in a system.
Heat Energy: The kinetic energy associated
with random motion of matter including the
vibratory and rotatory action of molecules
Internal energy is defined as the energy
associated with the random, disordered motion of molecules.
74
Molecular attractive forces are associated with potential energy
Molecular kinetic energy is a part of internal energy
Internal energy involves energy on the microscopic scale.
75
Kinetic : energy due to velocity, Ek = ½ mv2
Potential : energy due to position in a force field.
Internal : Sum of kinetic and potential energies
acting on all molecules in a system.
Specific internal energy : internal energy per unit
mass [Joules/kg]
latent energy : a portion of internal energy
associated with the phase of the system (binding forces between the molecules
chemical energy : is a potential energynuclear energy: is a potential energy
Internal energy is the sum of sensible energy +
latent energy + chemical energy + nuclear energy
76
sensible energy : a portion of internal energy
associated with kinetic energies of the molecules
77
Internal energy for the system with the same temperature
Solid or fluidMolecular gasMono- atomic gas
Potential energy
Vib. and rot. K.E
Trans. K.E
U
U: internal energy
78
Rotation about x-axis(vertical)
Rotation about y-axis (horizontal)
Vibrational motion along the bond
e-
Molecular translation Electron
translation Nuclear spinElectron spin
Total energy of the system
static : energy stored in a system
dynamic or energy interaction:-able to cross the system boundry
79
80
2 forms of energy interaction:
i. heat transfer (or heat): the driving force is
a temperature difference.
ii. work: other than heat interaction.
organized form of energy (work):orderly motion of all molecules in one direction.
disorganized form of energy (heat):random motion of molecules
81
82
Random motion
Oderly motion
Major area of thermodynamics – conversion of
disorganized energy (heat) into organized energy (work).
THE END
83
WEEK 2
1
PURE SUBSTANCES
2
matter
Homogenous?
yesHeterogenous
mixture
no
Separated by
physical means ?yes
Homogenous
mixturePure substance
no
Decomposed by
chemical processes?
compoundelement
yesno
3
Pure substances have an invariable
composition and are composed of either
elements or compounds.
4
Elements - Substances which cannot be
decomposed into simpler substances by
chemical means.
Compounds can be decomposed into two or more elements.
In Thermodynamics we’ll define it as
something that has a fixed chemical
composition throughout.
5
6
Mixtures vs Pure Substances
7
Matter
Mixtures Pure Substances
Mechanical
mixtures
one can
separate
mixtures by
physical means.
Solutions
Completely
mixed.
Suspensions
Example:
homogenized milk.
Particles slightly
clumpled together
but remain in
suspension
Elements
Substances
made of only
1 type of atom
Compounds
Substances that have
more than 1 element
in them (chemically
united)
Eg: NaCl
Physical Versus Chemical Change
8
Physical Changes Chemical Changes
Water freezing
Ice melting
Sugar dissolving in water
Detergent removing
grease from dirty pot
Piece of wood burns in a
fire
A bomb explodes
burning of fuel inside the
engine
Phase
9
A pure substance may exist in more than one phase.
Different phases of a substance have the same chemical composition but different physical structures, such as solid, liquid, and gas.
e.g : water; water + steam
Air : a mixture of nitrogen,oxygen and argon, therefore is not a pure substance. However since air is nonreactive mixture and has a constant chemical composition on macroscopic level, air can be approximated as pure substance.
Phase
10
Solid
11
Molecules are
(a) closely bound, therefore relatively dense
(b) arranged in rigid three-dimensional patterns so that they do not easily deform.
(c) having large attractive forces between atoms or molecules
(d) in constant motion – they vibrate in place
(e) vibrating more at higher temperature
Liquid
12
Molecules are
(a) closely bound, therefore also relatively
dense and unable to expand to fill a space.
(b) no longer rigidly structured so that they are
free to move within a fixed volume.
(c) look a lot like a solid.
Gas
13
Molecules have
(a) virtually no attraction one to another.
(b) not arranged in a fixed pattern. They are far apart.
(c ) high kinetic energy.
heating of water at constant pressure
14
15
16
Saturation Pressure is the
pressure at which the liquid and vapour phases are in equilibrium at a given temperature (point 2 &
4).
Compressed or Subcooled
liquid: (Between states 1
& 2) NOT about to vaporize
(Sub-cooed liquid) e.g., water
at 20oC and 1 atmosphere
Saturation Temperature is
the temperature at which the liquid and vapour phases are in equilibrium at a given pressure (point 2 & 4).
Saturated liquid:
(State 2) about to vapourize e.g., water at 100oC and 1
atmosphere
17
Saturated Liquid-Vapour
Mixture or Quality
region: (Between states 2 &
4) Liquid and vapour exist together in a mixture.
Saturated Vapour:
(State 4) about to condense
e.g., water vapour (steam) at
100oC and 1 atm.
Superheated Vapour:
(Between States 4 & 5)
NOT about to condense
e.g., water vapour (steam) at
>100oC and 1 atm
changing the pressures
18
CRITICAL POINT:
the saturated liquid and saturated vapour states are identical.No saturated mixture exists - the substance changes
directly from the liquid to vapour states.
Above the critical point there is no sharp difference
between liquid and gas!!
19
single phase liquid region
20
single phase vapour region
21
supercritical fluid region
22
T – V diagram
23
P-v diagram
24
P-v Diagram of a Substance that Contracts on Freezing
25
P-v Diagram of a Substance that Expands on Freezing
26
P-T Diagram
27
P-T Diagram
28
At any point in the areas separated by the curves, the pressure and temperature allow only one phase (solid, liquid, or gas) to exist, and changes in temperature and pressure, up to the points on the curves, will not alter this phase.
At any point on the curves, the temperature and pressure allow two phases to exist in equilibrium: solid and liquid, solid and vapor, or liquid and vapor.
Triple point: all three phases of a substance
will exist in equilibrium
P-v-T surfacePutting all the three properties on the same diagram
29
Contracts on Freezing
P-v-T surfacePutting all the three properties on the same diagram
30
Expands on Freezing
ENTHALPY, H
31
another property of pure substances (like internal energy, U)
U is a function ONLY of the temperature of the substance
H = U + PV ; U : internal energy ; PV : work done
(note: PV (N/m2)(m3) = N-m = Joules )
on a per unit mass basis: h = u + Pv (specific enthalpy)
H is a function of BOTH temperature and pressure.
Property Tables
32
The steam table are taken as an example.
Data on the table are;
P pressure
T temperature
v specific volume
u specific internal energy
h specific enthalpy
s specific entropy
subscript fg : difference between the saturated vapour and saturated liquid values
33
e.g :
vf : specific volume of saturated liquid
vg: specific volume of saturated vapour
vfg = vg - vf
example of the table
34
35
using the steam tables:
36
given P and T find the saturation pressure Psat corresponding to T
for P > Psat the point corresponds to compressed liquid
for P < Psat the point corresponds to superheated vapour
for P = Psat the point is in saturation region
but P and T are not independent, needs additional properties to determine state
37
given P and v or T and v first find vf and vg
if vg < v the state is in superheated vapour
if vf v vg the state is in saturation (SLVM)
if vf > v the state is in compressed liquid
38
same procedure for
P and u, T and u, etc.
39
example:
Find the volume, the internal energy, and the enthalpy of 2 kg water at the following condition;
a. T = 300oC, P=2 MPa
b. T = 300oC, P= 20 MPa
c. T = 300oC, P= 8.59 MPa
a. T = 300oC, P=2 MPa, mass = 2 kg, to find V, H, U
See table A4: at T=300oC,
Psat = 8.588 Mpa.
So P < Psat, this implies
that the point corresponds
to Superheated water
(vapour), so see Table A6
See Pressure P=2.00MPa
See at Temperature
T=300oC
Read: v = 0.12551 m3/kg
v= V∕m or V = mv
V= 2 kg (0.12551)m3/kg
V= 0.25102 m3
V≈0.25 m3.
40
Read: u = 2773.2 kJ/kg
u= U∕m or U = mu
V= 2 kg (2773.2)kJ/kg
V= 5546.4 kJ
Read h=3024.2 kJ/kg
(H=6,048.4 kJ)
b. T = 300oC, P= 20 MPa
41
See table A4: at T=300oC,
Psat = 8.588 Mpa.
So P › Psat, this implies
that the point corresponds
to Compressed Liquid, so
see Table A7
See Pressure P=20MPa
See at Temperature
T=300oC
Read: v = 0.0013611
m3/kg
v= V∕m or V = mv
V= 2 kg (0.0013611)
m3/kg
V= 0.0027222 m3
V≈0.003 m3. Read: u = 1307.2 kJ/kg
u= U∕m or U = mu
V= 2 kg (1307.2)kJ/kg
V= 2,614.4 kJ
Read h=1334.4 kJ/kg
(H=2668.8 kJ)
c. T = 300oC, P= 8.581 MPa
42
See table A4: at T=300oC,
Psat = 8.588 Mpa.
So P = Psat, this implies
that the point corresponds
to SLVM, so see Table A4
See Pressure P=8.581MPa
See at Temperature
T=300oC
Have to consider X:
Quality Factor
Saturated Liquid – Vapour Mixture(SLVM)
43
U kJ/kg
T(oC)
P
QUALITY ( symbol x ) : the ratio of mass of vapour to the total mass of the mixture.
44
mass
massx
m
m msaturated vapour
total
g
f g
= =+
X = mg /mt
since mt = mf + mg
mt : total mass of the mixture.X = 0 : the mixture is all saturated liquid
X = 1 : the mixture is all saturated vapour
Properties at point P (in the diagram above)
-a mixture of liquid & vapour
45
46
Vapour
liquid
Let say x = 0.4
40% Vapour
60% liquid
vg
vf
vav
mf : mass of sat.
liquid
mg : mass of sat.
vapour
mt : total mass
mt = mf + mg
total volume V = Vf + Vg
but V = mvso mtvav = mfvf + mgvg
mtvav = (mt – mg )vf + mgvg
vav = [(mt/mt – (mg/mt) ]vf + (mg/mt)vg
vav = [1 – X ]vf + X vg
or vav = vf – x vf + x vg
vav = vf + x [vg - vf ]
** vav = vf + Xvfg
47
specific properties of saturated mixture
uav = uf + x·ufg
vav = vf + x·vfg
hav = hf + x·hfg
sav = sf + x·sfg
x = 0 : uav = uf : all liquid
x = 1 : uav = uf + ufg = uf + (ug – uf) =ug
: all vapour
0 < x < 1 : liquid + vapour48
e.g:
Ans:
i) At the given temperature, determine the sat. pressure (from the steam table A4)
T = 90oC , P = 70.18 kPa
Properties of Saturated Water -Temperature Table (SI)
49
Vapour
2 kg
Water
8 kg
Temperature of the mixture is 90oC.
Find
i) pressure in the tank (look at table)
ii) quality of the mixture (X)
iii) volume of the tank (V=mt vav
Vav=vf+Xvfg )
T(oC) P(kPa) S.Vol., v (m3/kg) Int.Enr, u(kJ/kg) Enthl., h(kJ/kg)
vf vg uf ug hf hg
85 57.868 0.001032 2.8261 355.96 2487.8 356.02 2651.4
90 70.183 0.001036 2.3593 376.97 2494.0 377.04 2659.6
95 84.609 0.001040 1.9808 398.00 2500.1 398.09 2667.6
ii) X = mg / mt = mg / (mf + mg )
= 2 kg / ( 2 + 8 ) kg = 0.2
at 90oC, vf = 0.001036 m3/kg
vg = 2.3593 m3/kg
(from the steam table)
vav = vf + xvfg
= 0.001036 + (0.2)(2.3593-0.001036)
= 0.473 m3/kg
V = mvav = 10 kg (0.473 m3 / kg)
= 4.73 m3
50
51
v m3/kg
ToC
90o
P=70.18 kPa
T = 90oC , mg = 2 kg
mf = 8 kg
Quality is used to find other properties of
saturated liquid-vapor mixtures (SLVM):
uavg = uf + x ufg and, x = (uavg - uf )/ufg
havg = hf + x hfg and, x = (havg - hf )/hfg
ALWAYS check: hf ≤ havg ≤ hg,
52
53
another e.g.
GIVEN: 0.05 kg of water at 25oC in a container of 1.0 m3 volume.
FIND: the phase description, pressure, and quality (if appropriate).
54
Answer: 1.Find the proper table: Water + given T, use Table, Look up
vf = 0.001003 m3/kg , vg = 43.340 m3/kg
2.Calculate v (= vavg) = V/m = 1.0 m3/0.05 kg = 20 m3/kg
3.Since vf < vavg < vg the phase is saturated liquid-vapor
mixture (SLVM).
4.Since this is a saturated state: P = P sat@25C = 3.1698 kPa
5.Finally, x = (vavg - vf)/(vg - vf) = (20 - 0.001003)/(43.34 -
0.001003) = 0.461 or ( 46.1% )
55
SUPERHEATED VAPOUR:NOT about to condense.
56
e.g:Water at P = 0.5 MPa, h = 2890 kJ/kg. Find T.
Answer:
1.Check Table (given P) finding hg = 2748.7
kJ/kg
2.since h > hg, this is a superheated vapor !
3.From Table A6 at P = 0.5 MPa, find two rows
that bracket the given h
T= 200 oC h= 2855.8 kJ/kg
T= 250oC h= 2961.0 kJ/kg
h of 2890 kJ/kg is between these values
make interpolation57
58
COMPRESSED LIQUID:all liquid, NOT about to vaporize
Characteristics:
at a given P, T < TSAT or
at a given T, P > PSAT or
at either a given P or T, v < vf or h < hf or u < uf
approximate compressed liquid behaviour, evaluated at the
given TEMPERATURE (don’t use the pressure)
59
60
Note:-given two independent intensive properties and need to find other properties.
• Before the quantitative properties can be found, must find qualitatively where the point is on the phase diagram – saturated, superheated, subcooled or mixture …etc
61
62
• Algorithm:
– Given T and P:
• Compare to Tsat or Psat.
Decide whether subcooled, saturated, or
superheated.
– Given T or P and one other property (v, u, h, s)
• Look up vf or uf or hf or sf and vg or ug or hg or sg
from table (sat.)
• If v, u, h, or s < sat values (vf , uf,….): subcooled.
• If v, u, h, or s > sat values (vg, ug,….): superheated.
• If vf or uf or hf or sf <v, u, h, or s < vg or ug or hg or sg
: two-phase (SLVM) -- find the quality x.
Properties of Compressed Liquids
• Generally no tabular data for compressed
liquids
• Since liquids are close to being
incompressible, their properties change
little with pressure.
– We know this is true from our everyday
experience with specific volume (density).
– It also holds for u, h, s …
• To find thermodynamic properties for
compressed liquids, we use the properties
of saturated liquids at the same
temperature.63
64
65
66
Example:1. Complete the following steam table. Locate each point on a PT and a PV diagram.
T oC P, kPa v m3/kg Phase
Description
50 ? 4.16 ?
? 200 ? Saturated
vapour
250 400 ? ?
110 600 ? ?
67
T oC P, kPa v m
3/kg Phase
Description
50 12.352 4.16 SLVM
120.21 200 0.89 Saturated
vapour
250 400 0.53 Superheated
vapour
110 600 0.001 Compressed
liquid
At Saturated Liquid & saturated Vapour states:
hfg
• Latent heat of vapourization
• Enthalpy of vapourization
hfg
• represents
• The amount of Energy needed to Vapourize a unit Mass of Saturated Liquid at a given Temperature and Pressure
hfg
• At critical point hfg= 0
• h as T , P
68
2. A piston-cylinder device is used to vaporize 10 kg saturated liquid to saturated vapour at 101 kPa. Determine the volume change and the amount of energy added.[given the Latent heat of vaporization of water at 101
kPa is LV = 2257.0 kJ/kg )
69
Assumptions:
-The pressure is kept as a constant during the process
Volume Change Vg - Vf
(1)Volume change
The total volume of the saturated liquid
Vf = vfm
The total volume when all the water becomes saturated
vapour.
Vg = vgm
The volume difference is:
Vdifference = vgm - vfm (vg & vf from table)
= 10 (1.673 - 0.001) = 16.72 m3
(2) Amount of energy added
Latent heat of vaporization of water at 101 kPa is LV =
2257.0 kJ/kg. It is the same amount of energy needed to
change 1 kg saturated water to saturated vapour at 100
kPa.
Q = m Lv = 10kg (2257.0)kJ/kg = 22,570 kJ
70
THE END
71
72
Example:1. Complete the following steam table. Locate each point on a PT and a PV diagram.
QUIZ
73
1. For a pure subcooled liquidA. the pressure will be lower than the vapour pressure at
the same temperature.
B. the quality will be zero.
C. the temperature will be lower than the boiling
temperature at the same pressure.
2. In the superheated vapour region of a pure substance
A. quality is greater than 1.0
B. the temperature is lower than the boiling temperature
at the same pressure.
C. the pressure is lower than the vapor pressure at the
same temperature.
74
3. Specific properties
A. apply at a point.
B. include temperature and pressure.
C. are intensive properties.
4. Quality
A. is defined by ML/(ML + MV)
B. is defined by v = xvf + (1-x)vg
C. is 1 for a superheated vapour.
D. is the fraction of mass that is vapour.
E. is only defined in the two phase region.
75
9. In the piston-cylinder experiment, volume increased in
the two phase region as heat was added at constant
pressure because gases and liquids both expand as they
are heated.
True or false
10. At a given pressure, the melting temperature and the
boiling temperature can not be equal.
True or false
76
5. The vapour pressure of a pure substance
A. depends on the relative amounts of liquid and vapour
present.
B. is defined between the triple point and critical point.
C. is the pressure exerted by a pure substance in the two
phase (liquid-vapor) region.
6. Which of the following is an extensive property?
A. Temperature
B. Pressure
C. Volume
D. Density
E. None of the above
77
7. The critical point of a pure substance:
A. represents the highest temperature and pressure for
which liquid and vapour can coexist.
B. is the point at which the saturated liquid and
saturated vapour curves meet.
C. is where vapor pressure has its largest possible value.
8. The normal boiling temperature of water is lower in
Tibet than in Malaysia.
True or false
78
9. In the piston-cylinder experiment, volume
increased in the two phase region as heat was
added at constant pressure because gases and
liquids both expand as they are heated.
True or false
10. At a given pressure, the melting temperature
and the boiling temperature can not be equal.
True or false
79
Week 3EQUATION OF STATE
Any equation that relates the pressure, temperature, and specific
volume of a substance
1
• The simplest is the equation of state for substance in the gas phase, best-known as the ideal-gas equation of state.
Gas & Vapour
• Vapour : a gas closes to the state of condensation
• Gas : the vapour phase when it is above the critical temperature (always)
2
3
% of error involved in assuming steam to be an ideal gas & the region where steam can be treated as an ideal gas with less than 1% error.
A gas will fill its container completely and
does not exhibit a free surface.
A liquid will take the shape of its container
but exhibits a free surface.
4
Gases
• very low density that must be enclosed to keep together.
• no definite shape.
• no definite volume, but they completely fill a container. The volume of the container is the volume of the gas in it.
5
gasgas
containergas
PT
VV
,
A gas exerts a pressure on all sides of the container that holds it. Gas can be compressed by pressures greater than the pressure the gas on its container .
6
• particles (atoms or molecules) are not attached to each other.
• When a gas molecule hits another one, they bounce off each other.
7
Ideal gas
• An ideal gas is a special case of a pure substance in the vapour phase and follow the ideal gas law, which was originally derived from the experimentally measured Charles’ law and Boyle’s law.
8
Boyle’s law PV = C at constant T
9
10
Charles’ law V/T = C at constant P
Boyle’s law, Charles’ law and Avogadro’s hypothesis give the ideal gas law
P = R (T/v) or Pv = RT• P : pressure (absolute)
• T : temperature (absolute)
• v : specific volume
• R : characteristic gas constant
11
• R is different for each gas R = RU /M• unit : kJ/kg.K or kPa.m3/kg.K or kJ/kmol.K• M : molar mass (molecular weight) of the gas• The mass of one kmol (or kilogram-mole) of a substance in
kilograms.• e.g : molar mass of nitrogen is 28• -it means : the mass of 1 kmol of nitrogen is 28 kg.• -the mass of the system,
m = M N or N = m/M• where N : the mole number• (e.g : volume per unit mass, v, m3/kg
volume per unit mole, , m3/kmol)
12
• RU: universal gas constant
(same value for all gases)
RU = 8.314 kJ/kmol.K, or kPa/m3/kmol.K
• Several forms of the ideal gas equation
PV = m R T (V = mv)
PV = NRUT (mR = (MN)R = NRU)
P v = RUT (V = N v )
13
Note : V = mvmR = (MN)R = N RU
V = NV : volume (m3)
v : specific volume (m3/kg): molar specific volume (m3/kmol)v
v
14
what is PV ?
PV = (pressure).(volume)
= 1 (kg/(m.s2))(m3)
= 1 ( kg.m2/s2)
= 1 joule
= 1 J
a measure of energy !!!
15
Compressibility factor Z (a measure of deviation from ideal gas behaviour)
• there is no real substance that satisfies the ideal gas definition for the entire range of state
• the ideal gas is a concept rather than a reality
• the deviation of a real substance to the ideal gas behaviour is given by compressibility factor, Z
Z = Pv/RT or Pv = ZRT• for ideal gases , Z = 1
16
The further away Z is from
unity, the more the gas deviates
from ideal gas behaviour.
17
IDEAL
GAS
Z = 1
REAL GAS
1< Z 0R Z < 1
Molar mass and gas constant properties.
Substance Formula Molar
Mass,M
kg/kmol
Gas constant, R
kJ/kg.K
Air - 28.97 0.2870
Argon Ar 39.948 0.2081
Carbon dioxide CO2 28.011 0.1889
Chlorine Cl2 70.906 0.1173
Helium He 4.003 2.0769
Nitrogen N2 28.013 0.2968
Oxygen O2 31.999 0.2598
Water H2O 18.015 0.4615
18
Example 1:Determine the mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25oC.
Answer:
from the table above :
the gas constant of air R = 0.287 kPa.m3/kg.K
absolute temperature of the room
T = 25 oC + 273 = 198 K
volume of the room , V = 4x5x6 m3 = 120 m3
mass of the air in the room
19
20
m =(100 kPa)(120 m3)
(0.287kPa. m3/kg.K)(298 K)= 140.3 kg
PV = mRT ; therefore m = PV/RT
Example 2:
Using the ideal gas equation of state find the specific volume of steam at 400oC and at pressure,i) 0.01 MPa ii) 0.1 MPa iii) 20.0 MPa
Answer:
• For steam R = 0.4615 kJ/kg.K ( from table above)
• i)v = RT/P = 0.4615x(400+273) / 10 = 31.066 m3/kg
• ii)v = RT/P = 0.4615x(400+273) / 100= 3.1066 m3/kg
• ii) v = RT/P = 0.4615x(400+273) / 20000 = 0.015533 m3/kg
21
The combine gas law formula
• The same amount of the same gas is given at two different sets of conditions [m1=m2].
22
State 1P1, V1, T1, m1
State 2P2, V2, T2, m2=m1
RT
vP
1
11 at state 1 :
at state 2 : R
T
vP
2
22
• therefore or
2
22
1
11
T
vP
T
vP
2
22
1
11
T
VP
T
VP
Other Equations Of State
• -The ideal gas equation of state is a simple algebraic equation.
• -the range of applicability is quite limited
• -accurate for low pressures and/or high temperatures
• - for improving the accuracy and range of applicability in describing the behaviour of actual system, many equation have been proposed.
23
(i) Van der Waals Equation
comparing this equation with the ideal gas equation, we can see
that,
• P is replaced with (P+a/v2) – taking care of attraction forces by other molecules when a molecule strikes the container wall.
• V is replaced with (v-b) – the free volume the molecules can move around is the volume that are not occupied by other molecules.
24
RTb))(vv
a(P
2
Constants of Van der Waals equation of state
25
Substance
a
m6kPa/kmol2b
m3/kmol
Carbon dioxide 365.4 0.04280
Helium 3.46 0.02371
Hydrogen 24.96 0.02668
Nitrogen 232.4 0.03864
Oxygen 138.1 0.03184
Water 552.6 0.03042
(ii) Beattie-Bridgeman Equation
26
)v
b(1BB
)v
a(1AA
v
AB)v)(
vT
c(1
v
RTP
O
O
23
Specific Heats, internal energy and enthalpy
tell us about energy storage capability of a substance.
27
1 kg
IRON
1 kg
WATER
20oC 30oC20oC 30oC
4.5 kJ41.8 kJ
Specific Heats, internal energy and enthalpy
• All matter has a temperature, T associated with it
• T -- a direct measure of the motion of the molecules:
• -greater the motion the higher the temperature
• -Motion requires energy--more energy ®higher T
• -this energy is supplied by heat
• QUESTION: by how much will the temperature of an object increase or decrease by the gain or loss of heat energy?
• The greater the heat capacity of an object, the more heat
• energy is required to raise the temperature of the object
28
29
Specific Heat Capacity : the energy required to raise the temperature of a unit mass of a substance by one degree.
• (The relationship does not apply if a phase change is encountered)
30
m= 1 kg, T = 1oC
specific heat =5 kJ/kg.oC
5 kJ
Q = m c DT(heat added) = (mass) x (specific heat) x
(change in temperature)
Two kinds of specific heat
• specific heat at constant volume, cV
-the energy required to
raise the temperature of
the unit mass of a
substance by one degree
as the volume is
maintained constant.
• specific heat at constant pressure, cP
-the energy required to
raise the temperature of
the unit mass of a
substance by one degree
as the pressure is
maintained constant.
31
• cp > cv
because at constant pressure P, the system is allowed to expand, and the energy for this expansion work must also be supplied to the system.
32
Suppose that a body absorbs an amount of heat Q and its temperature consequently rises by T.
33
v constantm=1 kgT = 1oC
cV=3.12 kJ/kg.oC
P constantm= 1 kgT = 1oC
cP=5.2kJ/kg.oC
3.12kJ5.2 kJ
The usual definition of the heat capacity of
the body is
C =QT
• Let say the body absorbs a very small amount of heat, so that its temperature only rises slightly, and its specific heat remains approximately constant
• For a constant volume process, all the heat absorbed will be converted into internal energy ( since dV = 0, no work done by or onto the system)
• ∂Q = du
• for an ideal gas the internal energy is a function of temperature only.
34
dT
Qc
(T)dTcdu v
• For a constant pressure process, the heat absorbed will increase the internal energy and doing some work.
• From the definition of enthalpy,
• h = u + Pv
• dh = du + Pdv (P is constant)
by the definition of specific heat at constant pressure,
• we can write
• The enthalpy of an ideal gas is a function of temperature only.
35
dT
dhcP
(T)dTcdh P
Change in internal energy or enthalpy between state 1 and 2,
• can be simplified by taking average value of specific heat, then we can assume constant specific heat.
36
2
1
V12 (T)dTcuuΔu
2
1
P12 (T)dTchhΔh
)T(TcdTcuuΔu 12(av)V
2
1
(av)V12
)T(TcdTchhΔh 12(av)P
2
1
(av)P12
note : three ways to determine the internal energy and enthalpy changes of ideal gases:
• -by using the tabulated data
• -by using the cV and cP and performing the integration
• -by using average specific heat
37
Specific-heat Relation
• From the ideal gas law, Pv = RT
• And h = u + Pv
• h = u + RT
• differentiating dh = du + RdT
• replacing dh and du with specific heat relations
• cPdT = cVdT + RdT
• or cP = cV + R•
• The specific heat ratio is given the symbol κ
38
V
P
c
c
Summary: Ideal Gas• The gas consists of molecules that are in random
motion and obey the laws of mechanics.
• - The total number of molecules is large, but the volume of the molecules is a negligibly small fraction of the volume occupied by the gas.
• - No appreciable forces act on the molecules except during the collisions.
• An Ideal Gas (perfect gas)is one which obeys Boyle's Law and Charles' Law exactly.
• PV = mRT ; PV =NRUT
• R = 8.314 J K-1 mol-1 if P is in kPa, V is in L, T is in K
39
40
41
42
43
What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?
Given universal gas constant is 8.314 J/K.mol
44
45
What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20 oC?
Given universal gas constant is 8.314 J/K.mol and molecular mass of hydrogen is 2.016g/mol.
46
47
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 30C. How many moles of argon gas are in the cylinder?
Given universal gas constant is 8.314 J/K.mol
48
49
Given molecular mass of He is 4.003g/mol and universal gas constant is 8.314 J/K.mol
50
• A vessel of volume 0.2 m3 contains nitrogen at 1.013 bar and 15oC. If 0.2 kg of nitrogen is now pumped into the vessel, calculate the new pressure when the vessel has returned to its initial temperature. The molar mass of nitrogen is 28 kg/kmol, and RU = 8.3145 kJ/kmol.K
51
Solution 1
52
KkgNmM
RR U ./95.296
28
5.8314
kgx
xx
RT
VPm 237.0
28895.296
2.010013.1 5
1
111
kPaxx
V
RTmP 187
2.0
28895.296437.0
2
222
From
From P1v1 = m1RT1 --- initial state
m2 = m1 + 0.2 = 0.237 + 0.2 = 0.437 kg
final state P2v2 = m2RT2
but v2 = v1 & T2 = T1
53
1 Pa = 1 N/m2
1 atm = 101.325 kPa
= 1.01325 bars
= 760 mm Hg at 0 oC
1 mm Hg = 0.1333 kPa
1 kPa = 103 Pa = 10-3 Mpa
•Thank you!
54
ENERGY ANALYSIS OF CLOSED SYSTEMS
1
Week 4
(Heat & Work Interaction)
• Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors.
• Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems.
• Develop the general energy balance applied to closed systems.
• Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.
2
Objectives
• Heat, Q, energy caused by DT (temp difference)
• Work W, energy caused by DX (physical motion)
• Systems DO NOT have HEAT (Q) or WORK (W)
3
HEAT AND WORK INTERACTION
• HEAT and WORK cross the system’s boundaries,
• which changes the energy of the system,
• which then changes the properties (can be measured)
4
Qin
Qout
Win
Wout
5
• Heat Interaction: the form of energy that is transferred between two systems (or system and its surroundings) by virtue of a temperature difference.
• Symbol : Q : heat (or heat transfer) (kJ)
• q = Q /m heat per unit mass (kJ/kg)
• : Heat transfer rate (kJ/s = kW)
• An object does not possess "heat"
6
Qdt
Q
• Work Interaction: if the energy crossing the boundary of a closed system is not heat, it must be work. Definition: Energy transfer accomplished by a force moving an object some distance.
• Symbol: W unit: kJ
• w = W/m (kJ/kg)
• Power (kJ/s or kW)
7
Wdt
W
8
** Formal sign convention
• OR by using subscript in and out
• Qin : heat is transferred into the system
• Qout : heat lost by the system
• Win : work done onto the system
• Wout: work done by the system
• If the direction is not known, it can be assumed in or out. After solving – positive result indicates the assumed direction is right, otherwise the other way round.
9
Similarities1. Both are boundary phenomena.
2. System possesses energy, not heat or work.
3. Both are associated with a process, not a
state.
4. Both are path function (their magnitudes
depend on the path followed during a
process as well as the end states).
10
11
Note: the final states are indistinguishable – by doing
work or heating
• -Path functions (heat, work) have inexact differentials –
symbol δ
e.g. δQ, δW
(not DW, DQ)
• -Point functions (properties) have exact differentials –
symbol d
• e.g. dV, dT
12
Work
• Mechanical forms of work
• W = F s
• The work done by a constant force F on a body displaced a distance s
• Moving boundary work (Wb)
• P: constant
• V: constant
13
MOVING BOUNDARY WORK
• Moving boundary work
(P dV work): The expansion and compression work in a piston-cylinder device.
A gas does a differential amount of work Wb as it forces the piston to move by a differential amount ds.
14
• Quasi-equilibrium process: A process during which the system remains nearly in equilibrium at all times.
Wb is positive for expansionWb is negative for compression
The work associated with a moving
boundary is called boundary work.
15
• The area under the process curve on a P-V diagram represents the boundary work.
16
• The boundary work done during a process
depends on the path followed as well as the
end states.
17
• The net work done during a cycle is the difference between the work done by the system and the work done on the system
18
Moving boundary work (Wb)P: constant
19
P = f(V)
20
• Total boundary work,
• Wb = δW1 + δW2 + δW3 +…..+ δWN
• For a large N
but
• the boundary work is equal in magnitude to the area under process curve.
21
Constant P, Wb = PDV
unit : Wb = kPa-m3 = kJConstant V, Wb = 0
22
23
W= (P x dV)
Path (A) WA = W12 + W23 = (0) + ( 4 x (4-2)) = 8 unit of work
Path (B) WB = W14 + W43= (2 x (4 – 2)) + (0) = 4 unit of work
• The work done in changing states is dependent on the way the transformation takes place
• Q and W are "path functions":
• the amount of Q and W depends on the process path (NOT on the initial & final states).
24
25
26
V: constant
27
V: constant
28
P: constant
29
Polytropic, Isothermal, and Isobaric processes
30
Polytropic and for ideal gas
Polytropic process:
C, n (polytropic exponent) constants
31
• When n = 1 (isothermal process)
Constant pressure process.
What is the boundary work for a
constant-volume process?
32
Schematic and P-V diagram for a polytropic process.
33
Wb in an IDEAL GAS
Constant Volume process: dV = 0 ; Wb = 0
Constant Pressure process
• Wb = P(V2 – V1) = P (mRT2/P – mRT1/P)
= mR(T2 – T1), for n = 1
Constant Temperature process (isothermal) and more ?
• for n = 1
34
FLOW WORK
• Mass flow rate :
the amount of mass flowing through a cross section per unit time
• For flow in tube = VmA unit : kg/s
where : density of fluid
Vm : mean fluid velocity normal to A
A : cross-sectional area normal to flow direction
• Volume flow rate: : the amount of the fluid flowing through a cross-section per unit time
=VmA unit : m3/s
35
Control Volume
36
F = PA
The work done in pushing the fluid element across the
boundary is called the flow work
Flow work is the energy needed to push a fluid into or out of a control volume, and it is equal to PV.
37
The total energy (per unit mass),
total energy = flow energy + internal energy + kinetic energy +
potential energy
Note: for non-flow process, Pv = 0.
• Total energy of the flowing fluid,
38
The rate of energy being transported
ENERGY BALANCE FOR CLOSED SYSTEMS
• Energy balance for any system undergoing any process
• Energy balance in the rate form
39
-The total quantities are related to the quantities per unit time is
- Energy balance per unit mass basis
- Energy balance in differential form
- Energy balance for a cycle
40
41
Energy balance when sign convention is used (i.e., heat
input and work output are positive; heat output and work
input are negative).
42
For a cycle DE = 0, thus Q = W.
43
Various forms of the first-law relation for closed systems when sign convention is used.
The first law cannot be proven mathematically, but no
process in nature is known to have violated the first law, and
this should be taken as sufficient proof.
44
HWU b DD
For a constant-pressure expansion
or compression process:
An example of constant-pressure process
45
Three ways of calculating Du.
THANK YOU FOR YOUR ATTENTION!!
46
47
Energy balance for a constant-pressure expansion or compression process
• General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process.
(Q is to the system and W is from the system.)
48
For a constant-pressure expansion
or compression process:
HWU b DD
49
Three ways of calculating Du and Dh
1. By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available.
2. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate.
3. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large.
50
• An example of constant-pressure process
Week 5
1
First Law of ThermodynamicsConservation of energy principle:
Energy can be neither created nor destroyed;
it can only change forms.
Conservation of energy the total amount of energy in an isolated system remains constant
2
The 1st law the application of the conservation of energy principle to heat and thermodynamic processes;
e.g
-the kinetic energy of a moving car is converted into
heat energy at the brakes and tire surfaces.
-when chemical energy is released in burning and is
converted into light and heat energy.
-an object can be lifted by heating a rubber band.
Heat is converted into gravitational potential energy
3
NOTE: Process changes in the properties of that system.
When a very small part of a process occurs in a system, and is accompanied by a correspondingly very small change in the properties of the system
-A large (finite) change in the quantity X DX
-A small (infinitesimal) change in X dX
4
Q: -denotes heat, it has units of energy (it is an
energy interaction): Joules (J) or kJ (kilo-Joules).
-energy in transit due to the thermal interaction (temperature difference between the system and
its surroundings).
W: -denotes work interaction, it has units of energy (it is also an energy interaction): Joules (J) or
kilo-Joules (kJ).
-energy in transit due to mechanical interaction between the system and its surroundings
E: -Energy contained by the system (stored energy),
property of the system.
How much work can we possibly get out of heat?
5
6
E: represents the change in energy experienced by the system. If the system experience a change from state 1 to state 2, then
In many cases in thermodynamic analysis
7
Energy balance
8
9
10
11
12
13
14
b. The third process is adiabatic, means Q = 0
DE3 = Q3 –W3 = 32 kJ
= 0 - W3 = 32 kJ
W3 = -DE3 = -32 kJ
15
16
Process a: W’ = 0, why? DV = 0
1st law: Q’ – W’ = DU
Q’ – 0 = DU
Q’ = DU
or DU = Q’
Process b: no heat Q” = 0,
work of stirrer = W”s
1st law: Q” – W” = DU
0 – (-W”s) = DU
or DU = W”s
17
18
c
19
Let say u = f(T,v)
Integrating
20
21
22
c
23
24
25
P2 =?
T2 = 300K
V2 = 0.05 m3 +0.03 m3
= 0.08 m3
means Q = 0
means T2 = 300 K= T1
26
Answer
Hence final state: T2 = 300 K, P2 = 112.5 kPa , V2= 0.08 m3
k
or W = 0
therefore
hence
27
V2 = 0.05 m3 +0.03 m3
= 0.08 m3
Examples:
28
Heat exchangers
Throttling Valves
29
Throttling valves are any kind of flow-
restricting devices that cause a
significant pressure drop in the fluid.
The pressure drop in the fluid is often
accompanied by a large drop in
temperature, and for that reason
throttling devices are commonly used in
refrigeration and air-conditioning
applications.
30
31
32
The temperature of an ideal gas does
not change during a throttling
(h = constant) process since h = h(T).
During a throttling process, the enthalpy
of a fluid remains constant. But internal
and flow energies may be converted to
each other.
Energy
balance
33
Heat exchangers
Heat exchangers are devices where two
moving fluid streams exchange heat without
mixing. Heat exchangers are widely used in
various industries, and they come in various
designs.A heat exchanger can be as
simple as two concentric pipes.
34
Mass and energy balances for the
adiabatic heat exchanger in the figure
is:
35
The heat transfer associated with a heat
exchanger may be zero or nonzero depending
on how the control volume is selected.
36
Mass and Energy balances for a steady-flow process
A water
heater in
steady
operation.
Mass balance
37
WEEK 6
THE SECOND LAW OF
THERMODYNAMICS
2
Objectives• Introduce the second law of thermodynamics.
• Identify valid processes as those that satisfy both the first and second
laws of thermodynamics.
• Discuss thermal energy reservoirs, reversible and irreversible
processes, heat engines, refrigerators, and heat pumps.
• Describe the Kelvin–Planck and Clausius statements of the second law
of thermodynamics.
• Discuss the concepts of perpetual-motion machines.
• Apply the second law of thermodynamics to cycles and cyclic devices.
• Apply the second law to develop the absolute thermodynamic
temperature scale.
• Describe the Carnot cycle.
• Examine the Carnot principles, idealized Carnot heat engines,
refrigerators, and heat pumps.
• Determine the expressions for the thermal efficiencies and coefficients
of performance for reversible heat engines, heat pumps, and
refrigerators.
3
INTRODUCTION TO THE SECOND LAW
A cup of hot coffee
does not get hotter in
a cooler room.
Transferring
heat to a wire
will not
generate
electricity.
Transferring
heat to a
paddle wheel
will not cause
it to rotate.
These processes
cannot occur
even though they
are not in violation
of the first law.
4
Processes occur in a
certain direction, and not
in the reverse direction.
A process must satisfy both
the first and second laws of
thermodynamics to proceed.
MAJOR USES OF THE SECOND LAW
1. The second law may be used to identify the direction of processes.
2. The second law also asserts that energy has quality as well as quantity.
The first law is concerned with the quantity of energy and the
transformations of energy from one form to another with no regard to its
quality. The second law provides the necessary means to determine the
quality as well as the degree of degradation of energy during a process.
3. The second law of thermodynamics is also used in determining the
theoretical limits for the performance of commonly used engineering
systems, such as heat engines and refrigerators, as well as predicting the
degree of completion of chemical reactions.
5
THERMAL ENERGY RESERVOIRS
Bodies with relatively large thermal
masses can be modeled as thermal
energy reservoirs.
A source
supplies
energy in the
form of heat,
and a sink
absorbs it.
• A hypothetical body with a relatively large thermal energy capacity (mass x
specific heat) that can supply or absorb finite amounts of heat without
undergoing any change in temperature is called a thermal energy reservoir,
or just a reservoir.
• In practice, large bodies of water such as oceans, lakes, and rivers as well as
the atmospheric air can be modeled accurately as thermal energy reservoirs
because of their large thermal energy storage capabilities or thermal masses.
6
HEAT ENGINES
Work can always
be converted to
heat directly and
completely, but the
reverse is not true.
Part of the heat
received by a heat
engine is
converted to work,
while the rest is
rejected to a sink.
The devices that convert heat to
work.
1. They receive heat from a high-
temperature source (solar energy,
oil furnace, nuclear reactor, etc.).
2. They convert part of this heat to
work (usually in the form of a
rotating shaft.)
3. They reject the remaining waste
heat to a low-temperature sink
(the atmosphere, rivers, etc.).
4. They operate on a cycle.
Heat engines and other cyclic
devices usually involve a fluid to
and from which heat is
transferred while undergoing a
cycle. This fluid is called the
working fluid.
7
A steam power plant
A portion of the work output
of a heat engine is consumed
internally to maintain
continuous operation.
8
Thermal efficiency
Some heat engines perform better
than others (convert more of the
heat they receive to work).
Schematic of
a heat engine.
Even the most
efficient heat
engines reject
almost one-half
of the energy
they receive as
waste heat.
9
Can we save Qout?
A heat-engine cycle cannot be completed without
rejecting some heat to a low-temperature sink.
In a steam power plant,
the condenser is the
device where large
quantities of waste
heat is rejected to
rivers, lakes, or the
atmosphere.
Can we not just take the
condenser out of the
plant and save all that
waste energy?
The answer is,
unfortunately, a firm
no for the simple
reason that without a
heat rejection process
in a condenser, the
cycle cannot be
completed.
Every heat engine must waste
some energy by transferring it to a
low-temperature reservoir in order
to complete the cycle, even under
idealized conditions.
10
The Second Law of
Thermodynamics:
Kelvin–Planck Statement
A heat engine that violates the
Kelvin–Planck statement of the
second law.
It is impossible for any device
that operates on a cycle to
receive heat from a single
reservoir and produce a net
amount of work.
No heat engine can have a thermal
efficiency of 100 percent, or as for a
power plant to operate, the working fluid
must exchange heat with the
environment as well as the furnace.
The impossibility of having a 100%
efficient heat engine is not due to
friction or other dissipative effects. It is a
limitation that applies to both the
idealized and the actual heat engines.
11
REFRIGERATORS AND HEAT PUMPS• The transfer of heat from a low-
temperature medium to a high-
temperature one requires special
devices called refrigerators.
• Refrigerators, like heat engines,
are cyclic devices.
• The working fluid used in the
refrigeration cycle is called a
refrigerant.
• The most frequently used
refrigeration cycle is the vapor-
compression refrigeration cycle.
Basic components of a
refrigeration system and
typical operating conditions.
In a household refrigerator, the freezer compartment
where heat is absorbed by the refrigerant serves as
the evaporator, and the coils usually behind the
refrigerator where heat is dissipated to the kitchen
air serve as the condenser.
12
Coefficient of Performance
The objective of a refrigerator is to
remove QL from the cooled space.
The efficiency of a refrigerator is expressed
in terms of the coefficient of performance
(COP).
The objective of a refrigerator is to remove
heat (QL) from the refrigerated space.
Can the value of COPR be
greater than unity?
13
Heat
Pumps
The objective
of a heat
pump is to
supply heat
QH into the
warmer
space. The work
supplied to a
heat pump is
used to extract
energy from the
cold outdoors
and carry it into
the warm
indoors.
for fixed values of QL and QH
Can the value of COPHP
be lower than unity?
What does COPHP=1
represent?
14
When installed backward,
an air conditioner
functions as a heat pump.
• Most heat pumps in operation today have a
seasonally averaged COP of 2 to 3.
• Most existing heat pumps use the cold outside air
as the heat source in winter (air-source HP).
• In cold climates their efficiency drops considerably
when temperatures are below the freezing point.
• In such cases, geothermal (ground-source) HP
that use the ground as the heat source can be
used.
• Such heat pumps are more expensive to install,
but they are also more efficient.
• Air conditioners are basically refrigerators whose
refrigerated space is a room or a building instead
of the food compartment.
• The COP of a refrigerator decreases with
decreasing refrigeration temperature.
• Therefore, it is not economical to refrigerate to a
lower temperature than needed.
Energy efficiency rating (EER): The amount of heat removed from the
cooled space in Btu’s for 1 Wh (watthour) of electricity consumed.
15
The Second Law of Thermodynamics:
Clasius Statement
It is impossible to construct a device that
operates in a cycle and produces no effect
other than the transfer of heat from a lower-
temperature body to a higher-temperature
body.
It states that a refrigerator cannot operate unless
its compressor is driven by an external power
source, such as an electric motor.
This way, the net effect on the surroundings
involves the consumption of some energy in the
form of work, in addition to the transfer of heat
from a colder body to a warmer one.
To date, no experiment has been conducted that
contradicts the second law, and this should be
taken as sufficient proof of its validity.
A refrigerator that
violates the Clausius
statement of the second
law.
16
Equivalence of the Two Statements
Proof that the
violation of the
Kelvin–Planck
statement leads
to the violation
of the Clausius
statement.
The Kelvin–Planck and the Clausius statements are equivalent in their
consequences, and either statement can be used as the expression of the
second law of thermodynamics.
Any device that violates the Kelvin–Planck statement also violates the
Clausius statement, and vice versa.
17
REVERSIBLE AND IRREVERSIBLE PROCESSES
Two familiar
reversible processes.
Reversible processes deliver the most and consume
the least work.
Reversible process: A process that can be reversed without leaving any trace
on the surroundings.
Irreversible process: A process that is not reversible.
• All the processes occurring in nature are irreversible.
• Why are we interested in reversible processes?
• (1) they are easy to analyze and (2) they serve as
idealized models (theoretical limits) to which actual
processes can be compared.
• Some processes are more irreversible than others.
• We try to approximate reversible processes. Why?
18
Irreversibilities
Friction
renders a
process
irreversible.
Irreversible
compression
and
expansion
processes.
(a) Heat
transfer
through a
temperature
difference is
irreversible,
and (b) the
reverse
process is
impossible.
• The factors that cause a process to be
irreversible are called irreversibilities.
• They include friction, unrestrained expansion,
mixing of two fluids, heat transfer across a finite
temperature difference, electric resistance,
inelastic deformation of solids, and chemical
reactions.
• The presence of any of these effects renders a
process irreversible.
19
Internally and Externally Reversible Processes
A reversible process
involves no internal and
external irreversibilities.
• Internally reversible process: If no irreversibilities occur within the boundaries of
the system during the process.
• Externally reversible: If no irreversibilities occur outside the system boundaries.
• Totally reversible process: It involves no irreversibilities within the system or its
surroundings.
• A totally reversible process involves no heat transfer through a finite temperature
difference, no nonquasi-equilibrium changes, and no friction or other dissipative
effects.
Totally and internally reversible heat
transfer processes.
20
21
PERPETUAL-MOTION MACHINES
A perpetual-motion machine that
violates the first law (PMM1).
A perpetual-motion machine that
violates the second law of
thermodynamics (PMM2).
Perpetual-motion machine: Any device that violates the first or the second
law.
A device that violates the first law (by creating energy) is called a PMM1.
A device that violates the second law is called a PMM2.
Despite numerous attempts, no perpetual-motion machine is known to have
worked. If something sounds too good to be true, it probably is.
22
Proof of the first Carnot principle.
Week 7
THE CARNOT CYCLE
2
REVERSIBLE AND IRREVERSIBLE PROCESSES
Two familiar
reversible processes.
Reversible processes deliver the most and consume
the least work.
Reversible process: A process that can be reversed without leaving any trace
on the surroundings.
Irreversible process: A process that is not reversible.
• All the processes occurring in nature are irreversible.
• Why are we interested in reversible processes?
• (1) they are easy to analyze and (2) they serve as
idealized models (theoretical limits) to which actual
processes can be compared.
• Some processes are more irreversible than others.
• We try to approximate reversible processes. Why?
3
Irreversibilities
Friction
renders a
process
irreversible.
Irreversible
compression
and
expansion
processes.
(a) Heat
transfer
through a
temperature
difference is
irreversible,
and (b) the
reverse
process is
impossible.
• The factors that cause a process to be
irreversible are called irreversibilities.
• They include friction, unrestrained expansion,
mixing of two fluids, heat transfer across a finite
temperature difference, electric resistance,
inelastic deformation of solids, and chemical
reactions.
• The presence of any of these effects renders a
process irreversible.
4
Internally and Externally Reversible Processes
A reversible process
involves no internal and
external irreversibilities.
• Internally reversible process: If no irreversibilities occur within the boundaries of
the system during the process.
• Externally reversible: If no irreversibilities occur outside the system boundaries.
• Totally reversible process: It involves no irreversibilities within the system or its
surroundings.
• A totally reversible process involves no heat transfer through a finite temperature
difference, no nonquasi-equilibrium changes, and no friction or other dissipative
effects.
Totally and internally reversible heat
transfer processes.
5
THE
CARNOT
CYCLE
Reversible Isothermal Expansion (process 1-2, TH = constant)
Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)
Reversible Isothermal Compression (process 3-4, TL = constant)
Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)
Execution of
the Carnot
cycle in a
closed
system.
6
P-V diagram of the Carnot cycle. P-V diagram of the reversed
Carnot cycle.
The Reversed Carnot Cycle
The Carnot heat-engine cycle is a totally reversible cycle.
Therefore, all the processes that comprise it can be reversed,
in which case it becomes the Carnot refrigeration cycle.
7
THE CARNOT
PRINCIPLES
1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.
2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.
8
All reversible heat
engines operating
between the same two
reservoirs have the
same efficiency.
THE THERMODYNAMIC TEMPERATURE SCALE
The arrangement of heat
engines used to develop
the thermodynamic
temperature scale.
A temperature scale that is
independent of the
properties of the
substances that are used
to measure temperature is
called a thermodynamic
temperature scale.
Such a temperature scale
offers great conveniences
in thermodynamic
calculations.
9
For reversible cycles, the
heat transfer ratio QH /QL
can be replaced by the
absolute temperature ratio
TH /TL.
A conceptual experimental setup
to determine thermodynamic
temperatures on the Kelvin
scale by measuring heat
transfers QH and QL.
This temperature scale is
called the Kelvin scale,
and the temperatures on
this scale are called
absolute temperatures.
10
THE CARNOT HEAT ENGINE
The Carnot
heat engine
is the most
efficient of
all heat
engines
operating
between the
same high-
and low-
temperature
reservoirs.
No heat engine can have a higher
efficiency than a reversible heat engine
operating between the same high- and
low-temperature reservoirs.Any heat
engineCarnot heat
engine
11
The Quality of Energy
The fraction of heat that
can be converted to work
as a function of source
temperature.
The higher the temperature
of the thermal energy, the
higher its quality.
How do you increase the
thermal efficiency of a Carnot
heat engine? How about for
actual heat engines?
Can we use
C unit for
temperature
here?
12
THE CARNOT REFRIGERATOR
AND HEAT PUMP
No refrigerator can have a higher COP
than a reversible refrigerator operating
between the same temperature limits.
Any refrigerator or heat pump
Carnot refrigerator or heat pump
13
Summary• The Carnot cycle
The reversed Carnot cycle
• The Carnot principles
• The thermodynamic temperature scale
• The Carnot heat engine
The quality of energy
• The Carnot refrigerator and heat pump
14
WEEK 9
ENTROPY
2
Objectives
• Apply the second law of thermodynamics to processes.
• Define a new property called entropy to quantify the second-
law effects.
• Establish the increase of entropy principle.
• Calculate the entropy changes that take place during
processes for pure substances, incompressible substances,
and ideal gases.
• Examine a special class of idealized processes, called
isentropic processes, and develop the property relations for
these processes.
• Introduce and apply the entropy balance to various systems.
3
ENTROPY AND THE SECOND LAW
The second law of thermodynamics can be stated in 3
equivalent ways:
(1)Heat flows spontaneously from a hotter to a colder object,
but not vice versa.
(2)No heat engine that cycles continuously can change all its
heat-in to useful work-out.
(3) If a system undergoes spontaneous change, it will change in
such a way that its entropy will increase or, at best, remain
constant.
The second law tells us: the manner in which the spontaneous change
will occur.
The first law tells us: the conservation of energy.
The second law deals with the dispersal of energy.
4
Entropy (S) is a state variable and a
characteristic for a system in
equilibrium . By this is meant that S
is always the same for the system
when it is in equilibrium state.
5
Entropy is a measure of disorder.
A state that can occur in only one way is a state of
high order. For example : one arrangement of its
molecules.
But a state that can occur in many ways is a more
disordered state. One way to associate a number
with disorder is to take the disorder of a state as
being proportional to W, the number of ways a state
can occur.
6
Spontaneous processes in systems that
contain many molecules always occur in a
direction from a
State that
can exist
in only a
few ways
State that
can exist
in many
ways
7
Hence when left to themselves, systems
retain their original state of order or else
increase their disorder.
The most probable state of a system is
the state with the largest entropy.
It is also the state with the most disorder
and the state that can occur in the
largest number of ways.
8
ENTROPY
Evolution of the Different Perspectives of
the Second Law
· some events are naturally impossible
· Kelvin-Planck statement (no perfect engine)
· Clausius Statement(no perfect refrigerator)
· Carnot Principles (Carnot efficiency is the limit)
· Clausius Inequality (concept of entropy)
9
The 1st Law deals with the property energy and its
conservation.
The 2nd Law leads to the definition of a new property
called Entropy.
Entropy is somewhat abstract concept and is difficult to
give a physical description of it.
It is best understood by investigating its uses in
commonly encountered processes and problems.
10
Clausius Inequality
n Another corollary of the 2nd Law.
n Now we will deal with increments of heat
and work, Q and W, rather than Q and W.
n We will employ the symbol
which means to integrate over all the
parts of the cycle .
Q The cyclic integral of the heat transfer
11
Corollary – hal, akibat atau kesimpulan
yang pasti berlaku mengikut urutan
logik.
12
S. kJ/K
T, K
13
Any heat transfer to or from a system can be
considered to consist of differential amounts
of heat transfer.
Then the cyclic integral of Q/T can be
viewed as
‘The Differential Amounts of Heat Transfer
DIVIDED by the TEMPERATURE T at the
boundary.
14
Clausius Inequality is expressed as;
valid for all cycles, reversible or irreversible:
always less than or equal to zero.
German Physicist
R.J.E Clausius –
(1822-1888)
The cyclic integral of Q/T
15
System Considered in the
Development of Clausius
Inequality
At constant
thermodynamic
temperature TR
(absolute).
Heat received by the cyclic device from
the reservoir and supplies heat
to the system.
Work produced by the
reversible cyclic device.
Dashed lines.
QR
Work produced by the system as a
result of this heat transfer.
Energy balance of combined system:
[or total work of the combined system]
WC = Wrev + Wsys
Hence from first law for combined system
17
Change in total energy of combined system
Q – W = dU
Q – dU = W
but
Considering that the cyclic device is a reversible one
But for a cyclic process, dE = 0
WC net work for the combined cycle.
WC is produced by the combined cycle by
exchange heat with a single reservoir
18
this is against the Kelvin-Planck Statement
therefore WC cannot be work output (must be
negative)
-since TR is always positive
This equation is the Clausius Inequality
19
Let say the system & the cyclic device are reversible
the cycle of the combined system is internally reversible
(all quantities will have the same magnitude but the
opposite sign)
Wc , which cannot be positive earlier, now cannot be
negative.
Therefore it follows that WC int rev = 0
Hence
20
Let’s Look at a Simple Reversible Cycle
V is a property must be a property
We’ll define a new property, entropy, as:
21
s = S/m : intensive property
22
A Special Case: Internally Reversible
Isothermal Heat Transfer Processes
This equation is particularly useful for determining
the entropy changes of thermal energy reservoirs.
23
Reversible Heat Engine
For a reversible heat engine operating between a high
temperature reservoir at TH and a low-temperature reservoir
at TL,HIGH TEMPERATURE RESERVOIR
LOW TEMPERATURE RESERVOIR
REVERSIBLE
HE
QH
QL
Wnet
L
L
H
H
L
L
H
HL
L
H
H
rev
T
Q
T
Q
QT
QTT
Q
T
Q
T
Q
11
24
for REVERSIBLE HEAT ENGINE
L
L
H
H
T
Q
T
Q
0
revT
Q
25
IRREVERSIBLE HEAT ENGINE
For an irreversible heat engine operating between
the same reservoirs (as in reversible HE):
HIGH TEMPERATURE RESERVOIR
LOW TEMPERATURE RESERVOIR
IRREVERSIBLE
HE
QH
QL
Wnet
26
00
)()(
L
diff
L
diff
L
revL
H
H
L
irrL
H
H
irrev
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
From The Carnot principle
(Wnet)irr < (Wnet)rev
and (QL)irr > (QL)rev
can write : (QL)irr = (QL)rev + Q diff
where Qdiff = (QL)irr - (QL)rev and Qdiff>0
27
The Increase of
Entropy Principle
28
The Increase of Entropy Principle
From Clausius inequality
or
for reversible processfor irreversible process
29
S = S2 – S : entropy change of the system
=for reversible process
entropy transfer with heat
For irreversible process
or S = S2 – S1 = + Sgen
Sgen is always positive or zero
Depends on process (not a property)
If entropy transfer is zero, S = Sgen
30
For isolated system (adiabatic closed system), where heat transfer
is zero,
S 0
the entropy of an isolated system during a process always
increase or in the limited case of a reversible process, remains
constant
---INCREASE OF ENTROPY PRINCIPLE----
Isolated system – no interaction with surrounding
-may consist of subsystems
The entropy change of an isolated
system is the sum of the entropy
changes of its components, and is
never less than zero.
31
A system and its surroundings can be viewed as two subsystems of an
isolated system
since isolated system involves no entropy transfer
Sgen = Stotal = Ssys + Ssur 0
Note : equality for the
reversible process
Note : equality is for the reversible process
32
Since no process is reversible, some entropy is
generated during a process –entropy of the universe
is continuously increasing.
Sgen > 0 : irreversible process
Sgen = 0 : reversible process
Sgen < 0 : impossible process
The entropy of an isolated system will increase until
the entropy of the system reaches a maximum value
– state of equilibrium.
The increase of
entropy principle
33
2
1T
δQ
Entropy Increase Principle
S = S2 – S1 = where Sgen 0
Isolated system
+ Sgen
Notes:
•A process can take place in the direction that complies with
the increase of entropy princ. , Sgen≥0
•The entropy of the universe is continuously increasing,
more disorganized and is approaching chaotic.
•The Sgen is due to the existence of irreversibilities. For a
device, higher irreversibilities lower the efficiency.
34
Example.
Source 800 K
Sink 200K
Q = 2000 kJ
Show that the heat can not transfer from the low-temperature
sink to the high-temperature source based on the increase
entropy principle.
35
Ans:
∆S(source) = Q/T = 2000 kJ/ 800 K = 2.5 kJ/K
∆S(sink) = Q/T = - 2000 kJ/ 500 K = - 4 kJ/K
Sgen = ∆S(source) + ∆S(sink) = (2.5 – 4) kJ/K = -1.5 kJ/K < 0 ,
…. Impossible
•Based on the entropy increase principle
•heat cannot be transferred from low-temp to high-temp (without
external work)
Source 800 K
Sink 500K
Q = 2000 kJSgen = ∆S(source) + ∆S(sink)
36
Entropy generation, Sgen (for closed system)
System
System
Sgen0
Heat
inHeat
out
entropy balance for any system undergoing any process
Sin – Sout + Sgen = Ssystem
For closed system
12systemgen
k
k SSΔSST
Q
adiabatic closed system: Sgen = Ssystem
system + surroundings: Sgen = S = Ssystem +Ssurr
or Sin – Sout + Sgen= Ssystem
---- entropy balance37
Entropy Balance
System
Esystem
Ssystem
Sgen0
Ein
Sin
systemtheof
entropytotal
theinChange
generated
entropy
Total
leaving
entropy
Total
entering
entropy
Total
38
Entropy Balance:
the entropy change of a system during a
process is equal to the net entropy transfer
through the system boundary and the entropy
generated within the system.
Mechanisms of entropy transfer
•heat transfer (for closed and open system)
•mass flow (for open system only)
39
Tb = 400KQ = 500 kJ
Sheat = Q/T
= 1.25 kJ/kg
system
T
QS
thea
If T is constant
Entropy transfer by heat transfer (for closed system)
If T is not constant
k
k
2
1
heatT
Q
T
δQS
Entropy transfer by work : Swork = 0
Entropy : a measure of molecular disorder or molecular randomness
Sgas > Sliquid> Ssolid 40
Random motion
Oderly motion
Molecular disorders or disorganized form of energy(heat)
Organized form of energy (work)
Note:
41
Energy : quantity is preserved during process ( 1st Law)
quality is bound to decrease (2nd Law)
decrease in quality is accompanied by an increase in entropy
When energy is transferred as heat (Q) it is accompanied by
entropy transfer (Q/T)
When energy is transferred as work (W) there is no entropy
transfer
1st Law: no distinction between heat transfer and work
2nd Law: gives the distinction between heat transfer and work
an energy interaction that is accompanied by entropy transfer is
heat transfer;
an energy interaction that is not accompanied by entropy
transfer is work.
42
Q/TsurrSgen
Q/Tsys
Q Q
Tsys
Tsurr
System surrounding
Heat transfer
Entropy transfer
43
Some Remarks about Entropy
The entropy change of a
system can be negative,
but the entropy generation
cannot.
1. Processes can occur in a certain direction
only, not in any direction. A process must
proceed in the direction that complies with
the increase of entropy principle, that is,
Sgen ≥ 0. A process that violates this
principle is impossible.
2. Entropy is a nonconserved property, and
there is no such thing as the conservation of
entropy principle. Entropy is conserved
during the idealized reversible processes
only and increases during all actual
processes.
3. The performance of engineering systems is
degraded by the presence of irreversibilities,
and entropy generation is a measure of the
magnitudes of the irreversibilities during that
process. It is also used to establish criteria
for the performance of engineering devices.
44
45
RECAP
46
THE INCREASE OF ENTROPY PRINCIPLE
A cycle composed of a
reversible and an
irreversible process.
The equality holds for an internally
reversible process and the inequality
for an irreversible process.
Some entropy is generated or created during an irreversible process,
and this generation is due entirely to the presence of irreversibilities.
The entropy generation Sgen is always a positive quantity or zero.
Can the entropy of a system during a process decrease?
47
ENTROPY
The system considered in
the development of the
Clausius inequality.
Clausius
inequality
The equality in the Clausius inequality holds
for totally or just internally reversible cycles
and the inequality for the irreversible ones.
Formal
definition
of entropy
48
Entropy is an extensive
property of a system.
The net change
in volume (a
property) during
a cycle is
always zero.
The entropy change between two
specified states is the same whether
the process is reversible or irreversible.
A quantity whose cyclic
integral is zero (i.e., a
property like volume)
A Special Case: Internally Reversible
Isothermal Heat Transfer Processes
This equation is particularly useful for determining
the entropy changes of thermal energy reservoirs.
49
The entropy change of an isolated
system is the sum of the entropy
changes of its components, and is
never less than zero.
A system and its surroundings
form an isolated system.
The increase
of entropy
principle
50
Summary
• Entropy
• Clausius Inequality
• The Increase of entropy principle
WEEK 10
ENTROPY (continuation)
2
ENTROPY CHANGE OF PURE SUBSTANCES
The entropy of a pure substance
is determined from the tables
(like other properties).
Schematic of the T-s diagram for water.
Entropy is a property, and thus the
value of entropy of a system is fixed
once the state of the system is fixed.
Entropy change
3
5:
4
from the definition of entropy
revT
Qds
TdsQ or 2
1
TdsQhence
On a T-S diagram, the area under the curve
represents the heat transfer
5
The Tds equation
1st law for a closed system : Q - W = dU
and W = PdV
substituting: Q = dU + PdV
for a reversible process Q = TdS
TdS = dU + PdV …….1st TdS eq.
Enthalpy : H = U + PV
dH = d( U + PV )
= dU + PdV + VdP
= TdS + VdP
or TdS = dH – VdP………2nd TdS eq.
6
Entropy change for an incompressible substance(incompressible: dV 0)
from Q = dU + PdV
divide by T: Q/T = dU/T + PdV/T
dS = dU/T + PdV/T
change in internal energy :
dU = CV dT
dS = Cv dT / T + 0
or
1
2
2
1
12 lnT
TC
T
dTCSSS VV
7
Example:
Aluminum at 100oC is placed in a large,
insulated tank having 10 kg of water at a
temperature of 30oC. If the mass of the
aluminum is 0.5 kg, find,
•the final temperature of the aluminum and
water
•the entropy change of the aluminum
•the entropy change of the water
•the total entropy change of the universe
(Given specific heat capacity of water and aluminium are
4.177 kJ/kgK and 0.941 kJ/kgK respectively)
8
Answer:
water
Aluminum
System: Constant volume, adiabatic and no work done
From the conservation of energy (1st law)
Q – W = Usys = UW + UAL
But Q = W = 0
at equilibrium (T2)W=(T2)AL= T2
9
Entropy change for water and aluminum
Sgen>0 : irreversible process
Total Entropy change of the universe
10
Reversible and irreversible processes: Calculation of entropy
Irreversible process (large temperature differences).
Heating water from 293K to 373K
373K
Heat source
373K
Heat source
293K 293K373K 373K
break the irreversible process down into a series of reversible steps
11
293.1K
Heat source
293K 293.1K
293.2K
Heat source
293.2K
T+T K
Heat source
T K
373K
Heat source
373K373 K
12
T
dQdS
T
dTmcdS
T
dTcds avav ;
When the water is at a temperature T and it’s heated to T + T,
the heat entering (reversibly) is
dQ = mcavT.
and
the entropy change of the water at each reversible step is:
Therefore the total change in entropy
;;373
293
373
293
T
dTmcdSS
T
dTcdss avav
13
Examples:
Calculate the change in the total entropy of the Universe for the
following processes:
283
373T
dQSBlock
(1) A block of mass 1 kg, temperature 100°C and heat capacity 100 JK-1
is placed in a lake whose temperature is 10°C.
The change in entropy of the block, SBlock, is given by:
SBlok= - 27.61 JK-1
(2) The same block at 10°C is dropped into the lake from a height of 10 m.
Slake = mgh/Tlake = 1 x 9.81 x 10/283 = +0.35 JK-1
14
Entropy Change of an IDEAL GAS
Ts Diagrams
Integrate for the
area under the
process curve:
Two Special Cases:
15
For a Carnot Cycle, the Ts diagram
S1=S4 S2=S3 S (kJ/kgK)
1 2
34
QH
T (oC)
T1
T4
S1=S4 S2=S3 S (kJ/kgK)
1 2
34
QL
T (oC)
T1
T4
16
Remember each process:
1-2: Isothermal expansion
2-3: Adiabatic expansion (Isentropic)
3-4: Isothermal compression4-1: Adiabatic compression (Isentropic)
Also, since Q = TOS for an isothermal, internally
reversible process:
1-2: QH = area under 1-2 line and S = QH/TH
3-4: QL = area under 3-4 line and S = QL/TL
In Process 2-3, 4-1: s = 0 (Isentropic)
17
TdS = dU + PdV
TdS = dH – VdP
dS = dU/T + P/T dV
dS = 1/TdH – V/T dP
18
From TdS equation
ideal gas : u = (T)
h = (T)
du = cv(T) dT
dh = cP(T) dT
Pv = RT
by integration,
19
need to know the function of cv(T) and cp(T) in order
to complete the integration.
Cases with constant specific heats
The integration can be simplified
20
Ideal Gases in Isentropic Processes (s = 0)Assuming, Constant Specific Heats:
1
2
V1
2
v
vln
c
R
T
Tln
or 1
2
P1
2
P
Pln
c
R
T
Tln
Using ideal gas relationships: cP = cV + R and k = cP/cV
constantTVorV
V
T
T 1k
1k
2
1
0ΔS1
2
constantPVorV
V
P
P k
k
2
1
0ΔS1
2
constantTPorP
P
T
T k)/k(1
1)/k(k
1
2
0ΔS1
2
1k
2
1
1)/k(k
1
2
const.S1
2
v
v
P
P
T
T
Or
note: valid for ideal gas,
isentropic process,
constant specific heats.
21
Example:Air is compressed from an initial state of 100 kPa
and 300 K to 500 kPa and 360 K. Determine the
entropy change.
(cp = 1.003 kJ/kgK; Rair = 0.287 )
Answer:
Assume cp constant
kgKkJ
P
PR
T
Tcss p
/279.0100
500ln)287.0(
300
360ln003.1
)ln()ln(1
2
1
212
P1 = 100 kPa; P2 = 500 kPa
T1 = 300 K; T2 = 360 K
22
Summary of Entropy Change
Pure Substances:
Any process
s = s2 – s1
[kJ/(kgK)] (from table)
Isentropic process
s2 = s1
Incompressible substances:
Any process:
Isentropic process: T2 = T1
23
Ideal gasesconstant specific heat
1
2
1
2,12
1
2
1
2,12
lnln
lnln
P
PR
T
Tcss
v
vR
T
Tcss
avp
avv
24
THE ENTROPY CHANGE OF IDEAL GASES
From the first T ds relation From the second T ds relation
25
THE END…AT LAST!!
Week 11
GAS POWER CYCLES
2
Thermodynamic cycle :
1. power cycle – for power output (heat engine)
2. refrigeration cycle – for refrigeration effect
(refrigerator, air conditioner, heat pump)
Thermodynamic cycle :
1. gas cycle - working fluid remains in gaseous phase
2. vapour cycle – vapour phase in one part & fluid
phase in another
Thermodynamic cycle :
1. closed cycle – working fluid is returned to initial state
& recirculated
2. open cycle – working fluid is replaced by fresh one
3
Heat engine:
•internal combustion – burning the fuel inside the system
•external combustion – heat from the external sources
Gas
Power
Cycle
Compressor
Combustor
Turbine
Exhaust
Fuel
Air
v
sP
T
4
BASIC CONSIDERATIONS IN THE ANALYSIS
OF POWER CYCLES
Modeling is a
powerful
engineering tool
that provides
great insight and
simplicity at the
expense of some
loss in accuracy.
The analysis of many
complex processes can be
reduced to a manageable
level by utilizing some
idealizations.
Most power-producing devices operate on cycles.
Ideal cycle: A cycle that resembles the actual cycle
closely but is made up totally of internally reversible
processes.
Reversible cycles such as Carnot cycle have the
highest thermal efficiency of all heat engines
operating between the same temperature levels.
Unlike ideal cycles, they are totally reversible, and
unsuitable as a realistic model.
Thermal efficiency of heat engines
1
2
5
The idealizations and simplifications in the
analysis of power cycles:
1. The cycle does not involve any friction.
Therefore, the working fluid does not
experience any pressure drop as it flows in
pipes or devices such as heat exchangers.
2. All expansion and compression processes
take place in a quasi-equilibrium manner.
3. The pipes connecting the various
components of a system are well
insulated, and heat transfer /lost through
them is negligible.
Care should be exercised
in the interpretation of the
results from ideal cycles.
On both P-v and T-s diagrams, the area enclosed
by the process curve represents the net work of the
cycle.
On a T-s diagram, the ratio of the
area enclosed by the cyclic curve to
the area under the heat-addition
process curve represents the thermal
efficiency of the cycle. Any
modification that increases the ratio
of these two areas will also increase
the thermal efficiency of the cycle.
6
= 1 -
H
L
T
T
increases when TH
increases
increases when TL
decreases
8
For both ideal and actual cycles:
Thermal efficiency increases with an increase
in the average temperature at which heat is
supplied to the system
or
with a decrease in the average temperature at
which heat is rejected from the system.
9
A steady-flow Carnot engine.
10
P-v and T-s
diagrams of
a Carnot
cycle.
11
AIR-STANDARD ASSUMPTIONS
The combustion process is replaced by a heat-addition process in ideal cycles.
Air-standard assumptions:
1. The working fluid is air, which
continuously circulates in a closed loop
and always behaves as an ideal gas.
2. All the processes that make up the
cycle are internally reversible.
3. The combustion process is replaced by
a heat-addition process from an
external source.
4. The exhaust process is replaced by a
heat-rejection process that restores the
working fluid to its initial state.
12
Cold-air-standard assumptions:
When the working fluid is considered to
be air with constant specific heats at room
temperature (25°C).
Air-standard cycle:
A cycle for which the air-standard
assumptions are applicable.
13
AN OVERVIEW OF RECIPROCATING ENGINES
Nomenclature for reciprocating engines.
• Spark-ignition (SI) engines
• Compression-ignition (CI) engines
14
Compression ratio
15
Mean effective
pressure
16
OTTO CYCLE
The Otto cycle is the ideal cycle for the
spark-ignition (internal combustion)
reciprocating engines, and it consists of
four internal reversible process;
•isentropic compression
•constant volume heat addition
•isentropic expansion
•constant volume heat rejection
17
18
Schematic of a two-stroke
reciprocating engine.
The two-stroke engines are
generally less efficient than
their four-stroke counterparts
but they are relatively simple
and inexpensive, and they
have high power-to-weight
and power-to-volume ratios.
T-s
diagram
of the
ideal Otto
cycle.
Four-stroke cycle
1 cycle = 4 stroke = 2 revolution
Two-stroke cycle
1 cycle = 2 stroke = 1 revolution
19
Mechanical
strokes
INTAKE STROKE
Air-fuel is drawn into the
cylinder
COMPRESSION STROKE:
air-fuel is compressed
COMBUSTION:
by spark-pluq, QinPOWER STROKE:
gas expands, delivering work
EXHAUST STROKE:
residual gases are
exhausted
12
34
5
20
21
OTTO CYCLE: THE IDEAL CYCLE FOR
SPARK-IGNITION ENGINES
Actual and ideal cycles in spark-ignition engines and their P-v diagrams.
22
ideal cycle
1 – 2 : isentropic compression
2 – 3 : constant volume heat transfer (Qin)
3 – 4 : isentropic expansion
4 – 1 : constant volume heat rejection (Qout)
Thermal efficiency ( ideal gas assumption)
23
14
23v
14v
in
out
in TT
TT1
)T(TmC
)T(TmC1
Q
Q1
Q
Wη
23
1
1
2
2
1
k
V
V
T
T1
4
3
3
4
k
V
V
T
T
process 1-2 & 3-4 : isentropic
&
but V2 = V3 & V1 = V4
3
4
2
1
T
T
T
T
2
1
2
32
1
41
1
1T
TT
1T
TT
1ηT
T
1k1k
max
min
1k
1
2
2
1
r
1
V
V
V
V
T
T
Thermal efficiency of Otto Cycle
1
11
kr
where : r : compression ratio
k : cp/cv
24
25
The thermal efficiency of the Otto
cycle increases with the specific
heat ratio k of the working fluid.
Thermal efficiency of the ideal
Otto cycle as a function of
compression ratio (k = 1.4).
In SI engines, the compression ratio is limited by auto-ignition or engine knock.
26
THE END
Week 12 (Part A)
THE DIESEL CYCLE
THE IDEAL CYCLE
FOR COMPRESSION-IGNITION ENGINES
Automobiles, Power generation,
diesel-electric locomotives and submarines
2
The Diesel Cycle
-The Diesel cycle is the ideal cycle for compression - ignition
reciprocating engines.
-very similar to the Otto cycle, except that the constant volume
heat-addition process is replaced by a constant pressure heat-
addition process.
-The fuel and the air are compressed separately, and bringing
them together at the time of combustion.
-the fuel is injected into the cylinder which contains
compressed air at a higher temperature than the self-ignition
temperature of the fuel.
-once injected, the fuel ignites on its own.
3
In diesel engines, only air is
compressed during the compression
stroke, eliminating the possibility of
autoignition (engine knock).
Therefore, diesel engines can be
designed to operate at much higher
compression ratios than SI engines,
typically between 12 and 24.
4
In diesel engines, the spark plug is replaced
by a fuel injector, and only air is compressed
during the compression process.
5
• 1-2 isentropic
compression
• 2-3 constant-
pressure heat
addition
• 3-4 isentropic
expansion
• 4-1 constant-
volume heat
rejection.
6
•Process 1 to 2 is isentropic compression
of the fluid (blue)
•Process 2 to 3 is reversible constant
pressure heating (red)
•Process 3 to 4 is isentropic expansion
(yellow)
•Process 4 to 1 is reversible constant
volume cooling (green)[
7
Consider: (i) Heat IN
(ii) Heat OUT
(iii) Efficiency
8
Cutoff ratio
for the same compression ratio
(ratio between the end and start volume
for the combustion phase)
9
Thermal
efficiency of the
ideal Diesel
cycle as a
function of
compression
and cutoff ratios
(k=1.4).
10
Otto and Diesel engines have the following
undesirable features:
•The up-and-down movement of the piston
stresses the bearings and limits the speed.
•A large fraction of the fuel energy is lost to
cooling water circulating through the cylinder
housing.
11
•Fuel combustion is not complete,
particularly near the cool cylinder walls.
•Their complexity requires a large
number of parts.
12
•The piston seals cause friction and
wear, thus limiting engine life.
•Upon exhaust, there is still significant
pressure in the cylinder which reduces
efficiency and creates significant
throttling noise.
13
Compression, combustion, and
expansion all occur in a single cylinder
which requires compromises.
For maximum efficiency, the
compression should be cool and the
combustion should be hot. A single
cylinder cannot be both hot and cold at
the same time.
14
THE END
Week 12 (Part B)
THE REFRIGERATION CYCLE
2
REFRIGERATION CYCLE
Refrigeration is the withdrawal of heat from a substance or space so that
temperature lower than that of the natural surroundings is achieved.
Refrigeration may be produced by
-thermoelectric means
-vapour compression systems
-expansion of compressed gases
-throttling or unrestrained expansion of gases.
The objective of a refrigerator is to remove heat (QL) from the cold medium
The objective of a heat pump is to supply heat (QH) to a warm medium.
3
4
High Temperature Reservoir
CONDENSER
EVAPORATOR
COMPRESSOR
EXPANSION
VALVE
Low Temperature Reservoir
(Refrigerated Area)
Win
QL
QH
process 1 - 2 :
compression of the fluid
(vapour) by the
compressor where the
temperature and pressure
are elevated.
Process 2 – 3:
condensing if the
vapour at higher
pressure, and the
resultant heat is
dissipated to the
surrounding.
Process 3 – 4 :
expansion of the fluid
by an expansion
valve through which
the fluid pressure is
lowered.
Process 4 – 1 :The
low-pressure fluid
enters the
evaporator and
evaporates by
absorbing heat from
the refrigerated
space, and reenters
the compressor.
4
32
1
The whole cycle is repeated.
5
process 1 - 2 : compression of the fluid (vapour) by the
compressors where the temperature and pressure are
elevated.
Process 2 – 3: condensing if the vapour at higher
pressure, and the resultant heat is dissipated to the
surrounding.
Process 3 – 4 :expansion of the fluid by an expansion
valve through which the fluid pressure is lowered.
Process 4 – 1 :The low-pressure fluid enters the
evaporator and evaporates by absorbing heat from the
refrigerated space, and reenters the compressor.
The whole cycle is repeated.
6
Components of a Refrigeration system
•Compressor
•Condenser
•Expansion valve
•Evaporator
7
An ordinary household refrigerator.
8
note:
Refrigeration: the transfer of heat from lower
temperature regions to the higher temperature
ones.
Refrigerators: devices that produce
refrigeration.
Refrigeration cycle: the cycle on which the
refrigeration operates.
Refrigerants: the working fluids in refrigerators.
9
Coefficient of performance
Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.
Expansion
valve
10
Carnot Refrigerator and Heat Pump
Operate on reverse Carnot cycle
reasons for using reversed Carnot Refrigeration Cycle:
-easier to compress vapour only and not liquid-vapour mixture.
-cheaper to have irreversible expansion through an expansion
valve.
11
P-h diagram for ideal refrigeration cycle
Four processes of the ideal
vapour compression
refrigeration cycle
12
13
14
The actual vapour compression cycle.
15
Schematic and T-s diagram for
the ideal vapour-compression refrigeration cycle.
16
THE END
Week 14
THERMODYNAMIC
PROPERTY RELATIONS
2
Objectives• Develop fundamental relations between commonly
encountered thermodynamic properties and express the
properties that cannot be measured directly in terms of
easily measurable properties.
• Develop the Maxwell relations, which form the basis for
many thermodynamic relations.
• Develop the Clapeyron equation and determine the
enthalpy of vaporization from P, v, and T measurements
alone.
• Develop general relations for cv, cp, du, dh, and ds that
are valid for all pure substances.
• Discuss the Joule-Thomson coefficient.
• Develop a method of evaluating the ∆h, ∆u, and ∆s of
real gases through the use of generalized enthalpy and
entropy departure charts.
3
A LITTLE MATH—PARTIAL
DERIVATIVES AND
ASSOCIATED RELATIONS
The derivative of a function at a
specified point represents the slope of
the function at that point.
The state postulate: The state of
a simple, compressible substance
is completely specified by any two
independent, intensive properties.
All other properties at that state
can be expressed in terms of
those two properties.
The derivative of a function f(x)
with respect to x represents the
rate of change of f with x.
4
Partial Differentials
Geometric representation of
partial derivative (z/x)y.
The variation of z(x, y) with x when
y is held constant is called the
partial derivative of z with respect
to x, and it is expressed as
The symbol represents differential changes, just like the symbol d. They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables, whereas represents the partial differential change due to the variation of a single variable.
The changes indicated by d and are identical for independent variables, but not for dependent variables.
5
Geometric
representation of
total derivative dz for
a function z(x, y).
This is the fundamental relation for the
total differential of a dependent variable
in terms of its partial derivatives with
respect to the independent variables.
6
Partial Differential Relations
The order of differentiation is immaterial for
properties since they are continuous point
functions and have exact differentials. Thus,
Reciprocity
relation
Cyclic
relation
Demonstration of the
reciprocity relation for the
function z + 2xy 3y2z = 0.
7
THE MAXWELL RELATIONSThe equations that relate the partial derivatives of properties P, v, T, and s
of a simple compressible system to each other are called the Maxwell
relations. They are obtained from the four Gibbs equations by exploiting the
exactness of the differentials of thermodynamic properties.
Helmholtz function
Gibbs function
Maxwell relations
Maxwell relations are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T.
These Maxwell relations are limited to simple compressible systems.
8
THE
CLAPEYRON
EQUATION
9
The slope of the saturation curve
on a P-T diagram is constant at a
constant T or P.
The Clapeyron equation enables
us to determine the enthalpy of
vaporization hfg at a given
temperature by simply
measuring the slope of the
saturation curve on a P-T
diagram and the specific volume
of saturated liquid and saturated
vapor at the given temperature.
General form of the Clapeyron
equation when the subscripts 1
and 2 indicate the two phases.
Clapeyron
equation
10
At low pressures
Treating vapor
as an ideal gas
The Clapeyron equation can be simplified for liquid–vapor and solid–
vapor phase changes by utilizing some approximations.
Substituting these equations into the
Clapeyron equation
Integrating between two saturation states
The Clapeyron–Clausius
equation can be used to
determine the variation of
saturation pressure with
temperature.
It can also be used in the
solid–vapor region by
replacing hfg by hig (the
enthalpy of sublimation) of
the substance.
Clapeyron–
Clausius equation
11
GENERAL RELATIONS FOR du, dh, ds, cv, AND cp
• The state postulate established that the state of a simple
compressible system is completely specified by two independent,
intensive properties.
• Therefore, we should be able to calculate all the properties of a
system such as internal energy, enthalpy, and entropy at any state
once two independent, intensive properties are available.
• The calculation of these properties from measurable ones depends
on the availability of simple and accurate relations between the two
groups.
• In this section we develop general relations for changes in internal
energy, enthalpy, and entropy in terms of pressure, specific volume,
temperature, and specific heats alone.
• We also develop some general relations involving specific heats.
• The relations developed will enable us to determine the changes in
these properties.
• The property values at specified states can be determined only after
the selection of a reference state, the choice of which is quite
arbitrary.
12
Internal Energy Changes
13
14
Enthalpy Changes
15
16
Entropy Changes
17
18
Mayer
relation
19
Mayer
relation
The volume expansivity (also called the coefficient of
volumetric expansion) is a measure of the change in
volume with temperature at constant pressure.
Conclusions from Mayer relation:
1. The right hand side of the equation is
always greater than or equal to zero.
Therefore, we conclude that
2. The difference between cp and cv
approaches zero as the absolute
temperature approaches zero.
3. The two specific heats are identical for
truly incompressible substances since v
constant. The difference between the two
specific heats is very small and is usually
disregarded for substances that are nearly
incompressible, such as liquids and solids.
20
The internal energies and specific heats of ideal
gases and incompressible substances depend
on temperature only.
21
22
THE JOULE-THOMSON COEFFICIENT
The temperature of a fluid may increase,
decrease, or remain constant during a
throttling process.The development of an h = constant
line on a P-T diagram.
The temperature behavior of a fluid during a throttling (h = constant) process is
described by the Joule-Thomson coefficient
The Joule-Thomson coefficient
represents the slope of h = constant
lines on a T-P diagram.
23
Constant-enthalpy lines of a substance
on a T-P diagram.
A throttling process proceeds along
a constant-enthalpy line in the
direction of decreasing pressure,
that is, from right to left.
Therefore, the temperature of a
fluid increases during a throttling
process that takes place on the
right-hand side of the inversion line.
However, the fluid temperature
decreases during a throttling
process that takes place on the left-
hand side of the inversion line.
It is clear from this diagram that a
cooling effect cannot be achieved
by throttling unless the fluid is
below its maximum inversion
temperature.
This presents a problem for
substances whose maximum
inversion temperature is well below
room temperature.
24
25
The temperature of
an ideal gas remains
constant during a
throttling process
since h = constant
and T = constant lines
on a T-P diagram
coincide.
26
THE ∆h, ∆u, AND ∆s OF REAL GASES
• Gases at low pressures behave as ideal gases and obey
the relation Pv = RT. The properties of ideal gases are
relatively easy to evaluate since the properties u, h, cv,
and cp depend on temperature only.
• At high pressures, however, gases deviate considerably
from ideal-gas behavior, and it becomes necessary to
account for this deviation.
• In Chap. 3 we accounted for the deviation in properties P,
v, and T by either using more complex equations of state
or evaluating the compressibility factor Z from the
compressibility charts.
• Now we extend the analysis to evaluate the changes in
the enthalpy, internal energy, and entropy of nonideal
(real) gases, using the general relations for du, dh, and
ds developed earlier.
27
Enthalpy Changes of Real Gases
An alternative process path to evaluate
the enthalpy changes of real gases.
The enthalpy of a real gas, in
general, depends on the pressure
as well as on the temperature.
Thus the enthalpy change of a real
gas during a process can be
evaluated from the general relation
for dh
For an isothermal process dT = 0,
and the first term vanishes. For a
constant-pressure process, dP = 0,
and the second term vanishes.
28
Using a superscript asterisk (*) to denote an ideal-gas state, we can express
the enthalpy change of a real gas during process 1-2 as
The difference between h and h* is called the
enthalpy departure, and it represents the
variation of the enthalpy of a gas with pressure at
a fixed temperature. The calculation of enthalpy
departure requires a knowledge of the P-v-T
behavior of the gas. In the absence of such data,
we can use the relation Pv = ZRT, where Z is the
compressibility factor. Substituting,
29
Enthalpy
departure
factor
The values of Zh are presented in graphical form as a function of PR
(reduced pressure) and TR (reduced temperature) in the generalized
enthalpy departure chart.
Zh is used to determine the deviation of the enthalpy of a gas at a
given P and T from the enthalpy of an ideal gas at the same T.
from ideal gas tables
Internal Energy Changes of Real Gases
For a real gas
during a
process 1-2
Using the definition
30
Entropy Changes of Real Gases
An alternative process path to
evaluate the entropy changes of real
gases during process 1-2.
General relation for ds
Using the approach in the figure
During isothermal process
31
Entropy departure
Entropy
departure
factor
The values of Zs are presented in graphical form as a function of PR
(reduced pressure) and TR (reduced temperature) in the generalized
entropy departure chart.
Zs is used to determine the deviation of the entropy of a gas at a
given P and T from the entropy of an ideal gas at the same P and T.
For a real gas
during a
process 1-2
from the ideal gas relations
32
Summary• A little math—Partial derivatives and associated relations
Partial differentials
Partial differential relations
• The Maxwell relations
• The Clapeyron equation
• General relations for du, dh, ds, cv,and cp
Internal energy changes
Enthalpy changes
Entropy changes
Specific heats cv and cp
• The Joule-Thomson coefficient
• The ∆h, ∆ u, and ∆ s of real gases
Enthalpy changes of real gases
Internal energy changes of real gases
Entropy changes of real gases
WEEK 15
THERMODYNAMIC
PROPERTY RELATIONS
(cont.)
2
Internal Energy Changes
3
4
Enthalpy Changes
5
6
Entropy Changes
7
Specific Heats cv and cp
8
9
Mayer
relation
10
Mayer
relation
The volume expansivity (also called the coefficient of
volumetric expansion) is a measure of the change in
volume with temperature at constant pressure.
Conclusions from Mayer relation:
1. The right hand side of the equation is
always greater than or equal to zero.
Therefore, we conclude that
2. The difference between cp and cv
approaches zero as the absolute
temperature approaches zero.
3. The two specific heats are identical for
truly incompressible substances since v
constant. The difference between the two
specific heats is very small and is usually
disregarded for substances that are nearly
incompressible, such as liquids and solids.
11
The internal energies and specific heats of ideal
gases and incompressible substances depend
on temperature only.
12
13
THE JOULE-THOMSON COEFFICIENT
The temperature of a fluid may increase,
decrease, or remain constant during a
throttling process.The development of an h = constant
line on a P-T diagram.
The temperature behavior of a fluid during a throttling (h = constant) process is
described by the Joule-Thomson coefficient
The Joule-Thomson coefficient
represents the slope of h = constant
lines on a T-P diagram.
14
Constant-enthalpy lines of a substance
on a T-P diagram.
A throttling process proceeds along
a constant-enthalpy line in the
direction of decreasing pressure,
that is, from right to left.
Therefore, the temperature of a
fluid increases during a throttling
process that takes place on the
right-hand side of the inversion line.
However, the fluid temperature
decreases during a throttling
process that takes place on the left-
hand side of the inversion line.
It is clear from this diagram that a
cooling effect cannot be achieved
by throttling unless the fluid is
below its maximum inversion
temperature.
This presents a problem for
substances whose maximum
inversion temperature is well below
room temperature.
15
16
The temperature of
an ideal gas remains
constant during a
throttling process
since h = constant
and T = constant lines
on a T-P diagram
coincide.
17
THE ∆h, ∆u, AND ∆s OF REAL GASES
• Gases at low pressures behave as ideal gases and obey
the relation Pv = RT. The properties of ideal gases are
relatively easy to evaluate since the properties u, h, cv,
and cp depend on temperature only.
• At high pressures, however, gases deviate considerably
from ideal-gas behavior, and it becomes necessary to
account for this deviation.
• In Chap. 3 we accounted for the deviation in properties P,
v, and T by either using more complex equations of state
or evaluating the compressibility factor Z from the
compressibility charts.
• Now we extend the analysis to evaluate the changes in
the enthalpy, internal energy, and entropy of nonideal
(real) gases, using the general relations for du, dh, and
ds developed earlier.
18
Enthalpy Changes of Real Gases
An alternative process path to evaluate
the enthalpy changes of real gases.
The enthalpy of a real gas, in
general, depends on the pressure
as well as on the temperature.
Thus the enthalpy change of a real
gas during a process can be
evaluated from the general relation
for dh
For an isothermal process dT = 0,
and the first term vanishes. For a
constant-pressure process, dP = 0,
and the second term vanishes.
19
Using a superscript asterisk (*) to denote an ideal-gas state, we can express
the enthalpy change of a real gas during process 1-2 as
The difference between h and h* is called the
enthalpy departure, and it represents the
variation of the enthalpy of a gas with pressure at
a fixed temperature. The calculation of enthalpy
departure requires a knowledge of the P-v-T
behavior of the gas. In the absence of such data,
we can use the relation Pv = ZRT, where Z is the
compressibility factor. Substituting,
20
Enthalpy
departure
factor
The values of Zh are presented in graphical form as a function of PR
(reduced pressure) and TR (reduced temperature) in the generalized
enthalpy departure chart.
Zh is used to determine the deviation of the enthalpy of a gas at a
given P and T from the enthalpy of an ideal gas at the same T.
from ideal gas tables
Internal Energy Changes of Real Gases
For a real gas
during a
process 1-2
Using the definition
21
Entropy Changes of Real Gases
An alternative process path to
evaluate the entropy changes of real
gases during process 1-2.
General relation for ds
Using the approach in the figure
During isothermal process
22
Entropy departure
Entropy
departure
factor
The values of Zs are presented in graphical form as a function of PR
(reduced pressure) and TR (reduced temperature) in the generalized
entropy departure chart.
Zs is used to determine the deviation of the entropy of a gas at a
given P and T from the entropy of an ideal gas at the same P and T.
For a real gas
during a
process 1-2
from the ideal gas relations
23
Summary• A little math—Partial derivatives and associated relations
Partial differentials
Partial differential relations
• The Maxwell relations
• The Clapeyron equation
• General relations for du, dh, ds, cv,and cp
Internal energy changes
Enthalpy changes
Entropy changes
Specific heats cv and cp
• The Joule-Thomson coefficient
• The ∆h, ∆ u, and ∆ s of real gases
Enthalpy changes of real gases
Internal energy changes of real gases
Entropy changes of real gases