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HEAT PROCESSES
Thermodynamics
processes and cycles
Thermodynamics fundamentals. State variables, Gibbs phase rule, state equations,
internal energy, enthalpy, entropy. First law and the second law of thermodynamics.
Phase changes and phase diagrams. Ts and hs diagrams (example: Ts diagrams for
air). Thermodynamic cycles Carnot, Clausius Rankine, Ericson, Stirling,
thermoacoustics.
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
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FUNDAMENTALS of
THERMODYNAMICS
Estes
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BASIC NOTIONS
Subsystem flame zone =
opened
SYSTEM
• Insulated- without mass or energy transfer
Subsystem candlewick =
opened
•Closed (without mass transfer)
• Opened (mass and heat transport through boundary).
Thermal units operating in continuous mode (heat
exchangers, evaporators, driers, tubular reactors, burners)
are opened systems
Subsystem candle = opened with
moving boundary
Subsystem stand = closed
Thermal units operating in a batch mode (some chemical
reactors) are closed systems
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StaTE VARIABLES
state of system
is characterized
by
THERMODYNAMIC STATE VARIABLES related with directly
measurable mechanical properties:
 T [K], p [Pa=J/m3], v [m3/kg] (temperature, pressure, specific volume)
Thermodynamické state variables related to energy (could be
derived from T,p,v):
 u [J/kg] internal energy
 s [J/kg/K] specific entropy
 h [J/kg] enthalpy
 g [J/kg] gibbs energy
 e [J/kg] exergy
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Gibbs phase rule
Not all state variables are independent. Number of independent variables
(DOF, Degree Of Freedom) is given by Gibbs rule
NDOF = Ncomponents – Nphases + 2
 1 component, 1 phase (e.g.gaseous oxygen) NDOF=2 . In this case
only two state variables can be selected arbitrarily, e.g. p,v, or p,T or v,T.
 1 component, 2 phases (e.g. equilibrium mixture of water and steam at
the state of evaporation/condensation). In this case only one state variable
can be selected, e.g. pressure (boiling point temperature is determined by p)
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State EquATIONS p-v-T
Van der Waals equation isotherms
RT
a
p~
 ~2
v b v
RTc
p 2a
 ~3  ~
0
2
~
v v c (vc  b)
2 RTc
2 p
6a



0
2
4
3
~
~
~
v
v c (vc  b)
Above critical temperature Tc the
substance exists only as a gas
(liquefaction is not possible even at
infinitely great pressure)
Critical point, solution of these
two equations give a,b
parameters as a function of
critical temperature and critical
pressure
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Internal energy u [J/kg]
u-all forms of energy of matter inside the system (J/kg), invariant with respect to
coordinate system (potential energy of height /gh/ and kinetic energy of motion of
the whole system /½w2/ are not included in the internal energy). Internal energy is
determined by structure, composition and momentum of all components, i.e. all
atoms and molecules.
 Nuclear energy (nucleus)
~1017J/kg
 Chemical energy of ionic/covalent bonds in molecule
~107 J/kg
 Intermolecular VdW forces (phase changes)
~106 J/kg
Only the following item (thermal energy) is often included into the internal
energy concept (sometimes distinguished as the sensible internal energy)
Thermal energy (kinetic energy of molecules)
~104 J/kg
It follows from energy balances that the change of
internal energy of a closed system at a constant volume
equals amount of heat delivered to the system
du = dq (heat added at isochoric change)
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Enthalpy h [J/kg]
h=u+pv
enthalpy is always greater than the internal energy. The added term pv
(pressure multiplied by specific volume) simplifies energy balancing of
continuous systems. The pv term automatically takes into account
mechanical work (energy) necessary to push/pull the inlet/outlet material
streams to/from the balanced system.
It follows from energy balances that the change of
enthalpy of a closed system at a constant pressure
equals amount of heat delivered to the system
dh = dq (heat added at isobaric change)
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Entropy s [J/kg/K]
Thermodynamic definition of entropy s by Clausius
ds  (
dq
) rev
T
where ds is the specific entropy change of system corresponding to the heat
dq [J/kg] added in a reversible way at temperature T [K].
Boltzmann’s statistical approach: Entropy represents probability of a
macroscopic state (macrostate is temperature, concentration,…). This
probability is proportional to the number of microstates corresponding to a
macrostate (number of possible configurations, e.g. distribution of molecules
to different energy levels, for given temperature).
It follows from energy balances that the change of
entropy of a closed system at a constant temperature
equals amount of heat delivered to the system / T
Tds = dq (heat added at an isothermal and reversible change)
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Entropy s [J/kg/K] statistical (Boltzmann)
taken from Wikipedia
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Laws of thermodynamics
Modigliani
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Laws of thermodynamics
First law of thermodynamics (conservation of energy)
δq = heat added to
system
δw = work done by system
δq = du + δw
expansion work (p.dV) in case of compressible fluids,
surface work (surface tension x increase of surface),
shear stresses x displacement, but also electrical
work (intensity of electric field x current). Later on we
shall use only the p.dV mechanical expansion work.
Second law of thermodynamics (entropy of closed
insulated system increases)
Tds  δq
δq = heat added to system is Tds
only in the case of reversible process
Combined first and second law of thermodynamics
Tds = du+pdv
Entropy and Helmoltz energy
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Gough-Joule effect – stretching rubber band
TdS  dU  fdL
Derived Maxwell
equation using
Helmholtz energy A
stretch
 U 
 S   U 
 f 
 f 
f 
 T   
 T   T 
 L T
 L T  L T
 T T
 T T
A  u  Ts
dA  du  Tds  sdT  fdL  s dT
A
A

dL 
dT
L
T
A
A
f 
s 
L
T
 f 
 s 




 
 T  L
 L T
 S 
  0
 L T
f
L
use your lips to feel the
temperature increase
during stretching
f+df
L+dL
entropy decreases with
stretching
df>0 ds<0
f
Lmax
s=0 (only 1 microstate -possible arrangement)
 f 

 0
 T  L
tension f increases with
temperature at fixed length (fibre
contracts with increasing
temperature!)
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Energies and Temperature
The temperature increase increases thermal energy (kinetic energy of molecules).
For constant volume (fixed volume of system) internal energy change is proportional
to the change of thermodynamic temperature (Kelvins)
du = cv dT
where cv is specific heat at constant volume
For constant pressure (e.g. atmospheric pressure) the enthalpy change is also
proportional to the thermodynamic temperature
dh = cp dT
where cp is specific heat at constant pressure.
Specific heat at a constant pressure is always greater than the specific heat at a
constant volume (it is always necessary to supply more heat to increase temperature
at constant pressure, because part of the delivered energy is converted to the
volume increase, therefore to the mechanical work). Only for incompressible
materials it holds cp=cv.
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u(T,v) internal energy change
How to evaluate internal energy change? Previous relationship du=cvdT holds only at
a constant volume. However, according to Gibbs rule the du should depend upon a
pair of state variables (for a one phase system). So how to calculate du as soon as
not only the temperature (dT) but also the specific volume (dv) are changing?
Solution is based upon the 1st law of thermodynamic (for reversible changes)
du  Tds  pdv  T ((
s
s
s
s
) v dT  ( )T dv)  pdv  T ( ) v dT  (T ( )T  p)dv
T
v
T c
v
ds
v
where du is expressed in terms dT and dv (this is what we need), coefficient at dT is
known (cv), however entropy appears at the dv term. It is not possible to measure
entropy directly (so how to evaluate ds/dv?), but it is possible to use Maxwell
relationships, stating for example that
s
p
(
v
)T  (
T
)v
Instead of exact derivation I can give you only an idea based upon dimensional analysis: dimension of s/v is
Pa/K and this is just the dimension of p/T!
This term is zero for
So there is the final result
ideal gas (pv=RT)
p
du  cv dT  (T (
T
) v  p )dv
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h(T,p) enthalpy changes
The same approach can be applied for the enthalpy change. So far we can calculate
only the enthalpy change at constant pressure (dh=cpdT). Using definition h=u+pv
and the first law of thermodynamics
c
p
s
s
dh  du  pdv  vdp  Tds  vdp  T ( ) p dT  (T ( ) T  v)dp
T
p
Tds
And the same problem how to express the entropy term ds/dp by something that is
directly measurable. Dimensional analysis: s/p has dimension J/(kg.K.Pa)=m3/(kg.K)
and this is dimension of v/T. Corresponding Maxwells relationship is
(
s
v
) T  ( ) p
p
T
After substuting we arrive to the final expression for enthalpy change
dh  c p dT  (T (
v
) p  v)dp
T
Negative sign in the Maxwell equation is probably confusing, and cannot be derived from dimensional
analysis. Correct derivation is presented in the following slide.
Maxwell relationships
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just for information
Volumetric work
For incompressible elongation (f –
tensile force, positive when stretched)
For first law
Gibbs
g  h  Ts
(
v
s
) p  ( )T
T
p
Helmholtz
  u  Ts
(
internal energy
p
s
) v  ( )T
T
v
T
p
) s  ( ) v
v
s
u
(
h
T
v
(
)s  ( ) p
p
s
enthalpy
f
S
(
) L  ( )T
T
L
Stretch
decreases
entropy
For magnetisation
Thermal motion or
volume expansion
increases entropy
Magnetisation
decreases
entropy
H-intensity of magnetic. field
-specific magnetisation
0-magnetic permeability of vacuum
(
s

)T   0 (
)H
H
T
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s(T,v) s(T,p)
entropy changes
Changes of entropy follow from previous equations for internal energy and enthalpy
changes
p
) v dv
T
v
Tds  dh  vdp  c p dT  T ( ) p dp
T
Tds  du  pdv  cv dT  T (
Special case for IDEAL GAS (pv=RmT, where Rm is individual gas constant)
dT
dv
 Rm
T
v
dT
dp
ds  c p
 Rm
T
p
ds  cv
Please notice the difference between universal and individual gas constant. And
the difference between molar and specific volume.
pv~  RT
pv  RmT
~
v molar volu me, m3 / mol
v m3 / kg
R  8.314 J/(mol.K)
v  v/M M is molecular mass
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u,h,s finite changes (without phase changes or reactions)
Previous equations describe only differential changes. Finite changes must be
calculated by their integration. This integration can be carried out analytically for
constant values of heat capacities cp, cp and for state equation of ideal gas
u 2  u1  cv (T2  T1 )
h2  h1  c p (T2  T1 )
s 2  s1  cv ln
T2
v
 Rm ln 2
T1
v1
s 2  s1  c p ln
T2
p
 Rm ln 2
T1
p1
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u,h,s finite changes during phase changes
During phase changes (evaporation, condensation, melting,…) both temperature T
and pressure p remain constant. Only specific volume varies and the enthalpy/entropy
changes depend upon only one state variable (for example temperature). These
functions are tabulated (e.g. h-enthalpy of evaporation) as a function of temperature
(see table for evaporation of water), or approximated by correlation
 T T 
h  r  c

 Tc  T1 
n
Tc=647 K, T1=373 K,
r=2255 kJ/kg, n=0.38
for water
Pressure corresponding to the phase change
temperature is calculated from Antoine’s equation
B
ln p  A 
C T
T[0C]
0
593
2538
50
12335
2404
100
101384
2255
200
1559120
1898
300
8498611
1373
C=-46 K, B=3816.44,
A=23.1964 for water
Entropy change is calculated directly from the enthalpy change
h
s 
T
h[kJ/kg]
p[Pa]
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h,s during phase changes (phase diagram p-T)
Melting hSL>0, sSL>0,,
p
L-liquid
Evaporation hLG>0, sLG>0,
S-solid
G-gas
Sublimation hSG>0, sSG>0,
T
Phase transition lines in the p-T diagram are described by the Clausius
Clapeyron equation
dp
h

dT T v
hLG
Specific volume changes, e.g. vG-vL
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SUMMARY
State equation p,v,T. Ideal gas pV=nRT (n-number of moles, R=8.314 J/mol.K)
First law of thermodynamics (and entropy change)
Tds  du  pdv
Internal energy increment (du=cv.dT for constant volume dv=0)
du  cv dT  (T (
p
) v  p )dv
T
Enthalpy increment (dh=cp.dT for constant pressure dp=0)
v
dh  c p dT  (T ( ) p  v)dp
T
These terms are zero for ideal gas
(pv=RT)
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Check units
It is always useful to check units – all terms in equations must have the same
dimension. Examples
Tds  du  pdv
K
J
kg  K
N
J
Pa  2  3
m
m
J
kg
s
( )T
v
J

p
( )v
T
Pa
J
 3
K
m K
J
kg.K
 3
3
m
m K
kg
p~
v  RT
Pa 
J
m3
m3
mol
J
mol  K
K
m3
kg
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Important values
cv=cp ice
= 2 kJ/(kg.K)
cv=cp water = 4.2 kJ/(kg.K)
cp steam
= 2 kJ/(kg.K)
cp air
= 1 kJ/(kg.K)
Δhenthalpyof evaporation water = 2.2 MJ/kg
R = 8.314 kJ/(kmol.K)
Rm water = 8.314/18 = 0.462 kJ/(kg.K)
Example: Density of steam at 200 oC and pressure 1 bar.
p
10 5
kg


 0.457 [ 3 ]
RmT 462  (273  200)
m
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THERMODYNAMIC
DIAGRAMS
Delvaux
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DIAGRAM T-s
isobars
s  s0  c p ln
T
T0
Critical point
isochors
s  s0  cv ln
Left curveliquid
Right curvesaturated steam
Implementation of previous equations in the T-s diagram with isobars and
isochoric lines.
T
T0
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DIAGRAM h-s
Critical point
Left curve
liquid
Right curve
saturated steam
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Thermodynamic processes
Basic processes in thermal apparatuses are
 Isobaric dp=0 (heat exchangers, ducts, continuous reactors)
 Isoentropic ds=0 (adiabatic-thermally insulated apparatus, ideal
flow without friction, enthalpy changes are fully converted to
mechanical energy: compressors, turbines, nozzles)
 Isoenthalpic dh=0 (also adiabatic without heat exchange with
environment, but no mechanical work is done and pressure energy
is dissipated to heat: throttling in reduction valves)
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Thermodynamic processes
STEAM expansion in a
turbine the enthalpy decrease is
transformed to kinetic energy, entropy is
almost constant (slight increase
corresponds to friction)
Expansion of saturated
steam in a nozzle the same as
s
s
h
T
turbine (purpose: convert enthalpy to
kinetic energy of jet)
Steam compression power
h
T
s
s
h
T
consumption of compressor is given by
enthalpy increase
s
Throttling of steam in a valve
s
h
T
or in a porous plug. Enthalpy remains
constant while pressure decreases. See
next lecture Joule Thomson effect.
s
s
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Thermodynamic processes
Superheater of steam.
Pressure only slightly decreases
(friction), temperature and enthalpy
increases. Heat delivered to steam is
the enthalpy increase (isobaric process).
The heat is also hatched area in the Ts
diagram (integral of dq=Tds).
Boiler (evaporation at the
boiling point temperatrure)
s
mixing is to generate a saturated steam
from a superheated steam. Resulting
state is determined by masses of
condensate and steam (lever rule).
s
h
T
constant temperature, pressure. Density
decreases, enthalpy and entropy
increases. Hatched area is the enthalpy
of evaporation.
Mixing of condensate and
superheated steam purpose of
h
T
s
s
h
T
s
s
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Thermodynamic cycles
Periodically repeating processes with working fluid (water, hydrocarbons,
CO2,…) when heat is supplied to the fluid in the first phase of the process
followed by the second phase of heat removal (final state of the working
medium is the same as the initial one, therefore the cycle can be repeated
infinitely many times). Because more heat is supplied in the first phase
than in the second phase, the difference is the mechanical work done by
the working medium in a turbine (e.g.). It follows from the first law of
thermodynamics.
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Thermodynamic cycles
Carnot cycle
2
3
T
2
3
1
4
Mechanical work
W  (s4  s1 )(T2  T1 )
Q  (s4  s1 )T2

1
4
s
W
T
 1 1
Q
T2
3
Clausius Rankine cycle
2
3
T
Cycle makes use phase changes.
Example POWERPLANTS.
1-2 feed pump
2-3 boiler and heat exchangers
3-4 turbine and generator
4-5 condenser
1
2
1
4
4
s
3
Ericsson cycle
John Ericsson designed (200 years
ago) several interesting cycles working
1
with only gaseous phase. Reversed
cycle (counterclock orientation) is
applied in air conditioning – see
Brayton cycle shown in diagrams.
2
3
T
4
2
4
1
s
Stirling machine
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Stirling ENGINE
Stirling cycle
Gas cycle having thermodynamic
efficiency of Carnot cycle. Casn be
used as engine or heat pump (Stirling
machnines fy.Philips are used in
cryogenics).
Efficiency can be increased by heat
regenerator (usually a porous insert in
the displacement channel capable to
absorb heat from the flowing gas).
β-Stirling
1.
Compression and transport of
cool gas to heater
2.
Expansion of hot gas
3.
Displacement of gas from hot to
cool section
4.
Compression (phase 1)
Stirling HEAT PUMP
1-2 isothermal compression
2
T
1
2-3 cooling in
regenerator
3
4-1 displacement v=const.
and heating in regenerator
4
3-4 isothermal expansion
s
-Stirling with
regenerator
Thermoacoustical engine
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Thermoacoustic analogy of Stirling engine
Standing waves – mutually shifted
pressure and velocity waves (90o)
u,p
Velocity
amplitude
Wave equation for pressure,
velocity. C is speed of sound
u
u
 c2 2
2
t
x
2
2
c2 
x
pressure
cp p
cv 
Cold
HE
Stack
(regeneratir)
Hot HE
Very simple design can be seen on Internet video engines. Cylinder can be a
glass test tube with inserted porous layer (stack). Besides toys there exist
applications with rather great power driven by solar energy or there exist
equipments for cryogenics – liquefaction of natural gas.
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Thermoacoustics
Phenomena and principles of thermoacoustics are more than hundred years old.
Singing Rijke tube
Sondhauss tube
Taconis oscillations
Rijke P.L. Annalen der Physik 107 (1859), 339
Sondhauss C. Annalen der Physik 79 (1850), 1
Taconis K.W. Physica 15 (1949) 738
Generated sound
waves
Thin tube
Heated wire
screen
Heated bulb
Thin tube
inserted into a
cryogenic
liquid
Liquid helium
Air flowing through
an empty tube
Lord Rayleigh
| author=Lord Rayleigh | title=The explanation of certain acoustical phenomena | journal=Nature (London) |
year=1878 | volume=18 | pages=319–321]
formulated principles as follows:
thermoacoustic oscillations are generated as soon as
Heat is supplied to the gas at a place of greatest condensation (maximum density)
Heat is removed at a place of maximum rarefaction (minimum pressure)
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Thermoacoustics
Scott Backhaus and Greg Swift: New varieties of thermoacoustic engines,
2002
Thermoacoustic machines can operate either as heat pumps (usually refrigerators) when forced oscillations
are driven by an oscillating membrane (usually loadspeaker), or as an engine (prime mover) that turns heat
into mechanical energy (sound).
Spontaneous oscillations (engine mode) exist only if the axial temperature gradient in stack is high enough so
that the Rayleigh criterion will be satisfied (see fig. showing axial gradient Tstack in a stack and Tcrit in a gas
parcel oscillating in the x-direction at a large distance from the wall of stack).
Babaei H.,Siddiqui K.: Design and optimisation of thermoacoustic
devices. Energy Conversion and Management, 49 (2008), 3585-3598
Refrigerator (heat pump if  Tstack   Tcrit )
T
Hot
HE
loadspeaker
Axial temperature
of stack Tstack
stack
Tcrit 
Tp
c p u
-frequency, -therm.expansion [1/K],
p-mean pressure amplitude, u-mean
velocity amplitude, -mean density
Engine (oscillations generated if  Tstack   Tcrit
Heat supplied to gas
parcel at max.density
Hot
HE
Cold
HE
Critical gradient Tcrit
(insulated parcel)
Piston or
piezocrystal
Heat removed
from gas to stack
x
Gas parcel oscillating back and forth in accordance with pressure (motion
left – increasing pressure – compression – gas temperature increases heat is removed from gas to stack and work is consumed by gas parcel)
)
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Thermoacoustics
How to understand the expression for the critical axial temperature gradient?
Tcrit 
 Tpa
 c pua
-frequency, -therm.expansion [1/K],
p-mean pressure amplitude, u-mean
velocity amplitude, -mean density
T describes temperature difference during pressure oscillation of a thermally
insulated gas parcel (adiabatic compression/expansion)
Tds  dh  vdp
Tcrit
T
T

x
x 
ua
T  1

Mean velocity of gas parcel
times time of period
For ideal gas thermal
expansion coefficient =1/T
p
Ts  c p T 

pa
T  
cp
For adiabatic compression Ts=0
T temperature change during adiabatic compression of gas parcel
x displacement of a gas parcel during one oscillation
Magnetic refrigeration
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TZ2
Application of magnetic field upon ferromagnetic material causes orientation of magnetic spin of
molecules therefore decreases the magnetic entropy. Total entropy (sum of the magnetisation entropy
and the lattice entropy, thermal vibration of molecules in a crystal lattice) remains constant assuming a
thermally insulated (adiabatic) systém. Therefore the magnetic entropy decrease must be compensated
by the thermal entropy (and temperature) increase.
The first law of thermodynamic can be formulated in terms of internal energy as
du  Tds  pdv  0 Hd
(o magnetic permeability of vacuum, H intensity of magnetic field [T], specific magnetisation)
For constant volume the relationship between entropy, temperature and H is
ds 
c pH
T
dT  0

dH
T
s / H Maxwell
This equation enables to
construct the T-s diagrams and
demonstrate thermodynamic
cycles of refrigeration.
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TZ2
Laser cooling
Lasers illuminating crystals achieve extremely low cryogenic temperatures of 10-9 K.
Ruan X.L. et al. Entropy and efficiency in laser cooling solids. Physical Review B, 75 (2007), 214304
A phonon is a quantum of collective excitation in a periodic,
elastic arrangement of atoms or molecules in condensed matter.
Chu, Steven, Science 253, 861-866, 1991
Lasers to achieve extremely low temperatures has
advanced to the point that temperatures of 10-9 K
have been reached. Idea of Doppler effect
atom Na moving with
velocity about 570 m/s at
300K
collision with
photon from
behind has too low
energy
laser should be tuned below the resonance
frequency (difference f should be the
Doppler frequency given by velocity of atom)
v

f

f
0

c
Doppler
shift
HP2
EXAM
Thermodynamics
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What is important (at least for exam)
Gibbs phase rule
NDOF = Ncomponents – Nphases + 2
State equations Van der Waals and critical parameters
RT
a
p~
 ~2
v b v
First law of thermodynamics
Tds = du+pdv
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What is important (at least for exam)
Carnot
T
2
3
1
4
s
Clausius Rankine
3
T
2
1
4
s
3
Ericsson
T
2
4
1
s
HP2
What is important (at least for exam)
regenerator
STIRLING
displacing piston
T
41
3
Thermoacoustic
standing wave
compression/expansion wave
stack
2
Thermoacoustic
travelling wave
1
s
regenerator
N
AMR Active Magnetic
Refrigerator
S