Transcript Internal energy

CH5
CHEMISTRY E182019
Heat and
energies
This course is approximately at this level
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
CH5
Systems
SYSTEM
 Insulated – without mass, energy, and heat transfer through the
system boundary (no exchange with environment)
 Closed – without mass transport (impermeable boundary for mass
transfer), but heat and energy exchange with surrounding is possible
(e.g.batch chemical reactor, Papine’s pot)
 Opened – both mass and energy exchange between system and
surroundings is possible.
State of matter inside a system is characterised by
state variables
T,p,v
Internal energy u [J/kg]
Enthalpy h [J/kg]
Entropy s [J/kg.K]
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Energy balance of Closed Systems
q
du
-internal
energy increase
w
=p.dv mechanical work done by
system (volume increased by dv)
heat
transferred to
system
First law of thermodynamics
δq = du + δw
Only du-is a state variable, while q (heat) and w (work)
depend upon previous history and are NOT state variables.
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Energy balance of Closed Systems
at constant VOLUME
q
du
-internal
energy increase
heat
transferred to
system
Closed system and constant volume (no mechanical work).
Internal energy increase is related to temperature increase
δq = du
du = cv dT
cv – specific heat capacity at constant volume [J/kg.K]
CH5
Energy balance of Closed Systems
at constant PRESSURE
q
du
-internal
energy increase
w
=p.dV mechanical work done by
system (volume increased by dv, dp=0)
heat
transferred to
system
Closed system and constant pressure. Amount of heat is
expressed in terms of enthalpy h=u+pv
δq = du+pdv=dh-vdp=dh
dh = cp dT
cp – specific heat capacity at constant pressure [J/kg.K]
CH5
Energy balance of Closed Systems
at constant TEMPERATURE
q
heat
transferred to
system
ds
-entropy
increase dT=0
w
=p.dV mechanical work done by system
(at constant temperature dT=0).
However w can be also surface work (surface tension x
increase of surface), shear stresses acting at surface x
displacement, or even electrical work (intensity of electric field x
electrical current). Later on we shall consider only the p.dV
term representing the mechanical expansion work of
compressible substances (e.g. gases).
Closed system and constant temperature. Amount of heat is
expressed in terms of entropy
δq = du+pdv=Tds
ds = (du+pdv)/T
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Internal energy
du = cv dT
...heat dq transferred to system at constant volume
(more correctly, as soon as no other form of work is
done).
Internal energy changes with changing temperature (small, kinetic energy)
Internal energy changes during phase changes (medium, VdW forces)
Internal energy changes during chemical reactions (high, bonding forces)
CH5
Internal energy
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)
Chemical energy of ionic/covalent bonds in molecule
Intermolecular VdW forces (phase changes)
Pressure forces
Thermal energy (kinetic energy of molecules)
~1017J/kg
~107 J/kg
~106 J/kg
~105 J/kg
~104 J/kg
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Internal energy
Internal energy is a state variable and according to Gibb’s phase rule it
depends not only upon temperature T, but also upon specific volume or
pressure (even in the case of no phase or chemical changes). The previous
statement du=cv.dT holds only at the particular case, when volume is
constant (or dv=0). More general formula
p
du  cv dT  (T ( ) v  p )dv
T
describes variation of internal energy with temperature and specific volume.
The second term is zero not only if the volume is constant, but also in the
case that the substance can be characterized by state equation of ideal gas:
T(
p
R
)v  p  T  p  p  p  0
T
v
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Internal energy/enthalpy
Internal energy u is suitable for design (and energy balancing) of closed
batch systems, e.g. reactors operating periodically. Volume of system is
constant no mechanical work is done and amount of transferred heat is
equivalent to internal energy change.
Enthalpy h=u+pv is more suitable for energy balancing of continuous
(opened) systems operating at steady state. The term pv automatically
takes into account mechanical work necessary for delivering/removal of
matter to inlet/outlet streams. Therefore energy balance reduces to
 2 h2  m
 1h1
Q  m
H 1  m 1h1
 1, m
 2 Mass flowrates at
m
inlet/outlet streams [kg/s]
Q [W ]
H 2  m 2 h2
H 1 , H 2 Enthalpy flows at
inlet/outlet streams [W]
Q [W ] Heating power delivered
to system through heat
transfer surface [W]
Enthalpy is preferred for energy balancing of processes carried out at
constant pressure, e.g. chemical reactions, phase changes,…when volume
is changing but pressure remains constant (typical situation in technologies)
CH5
Enthalpy
dh = cp dT
h = u + pv
...heat dq transferred to system at constant pressure
Enthalpy changes with changing temperature (small, kinetic energy)
Enthalpy changes during phase changes (medium, VdW forces)
Enthalpy changes during chemical reactions (high, bonding forces)
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Enthalpy changes with T
Enthalpy is a state variable and according to Gibb’s phase rule it depends
not only upon temperature T, but also upon specific volume or pressure
(even in the case of no phase or chemical changes). The previous
statement dh=cp.dT holds only at the particular case, when pressure is
constant (or dp=0). More general formula
dh  c p dT  (T (
v
) p  v)dp
T
describes variation of enthalpy with temperature and pressure. The second
term is zero not only if pressure is constant, but also in the case that the
substance can be characterized by state equation of ideal gas:
T(
v
R
) p  v  T  v  v  v  0
T
p
CH5
Enthalpy of phase changes
Evaporation of liquids and melting of solids are examples of processes
when heat must be added. The heat is necessary to break the VdW forces
and make molecules „free“. Reverse processes are condensation of steam
and solidification of liquids – heat must be removed from the system to
complete the phase change. In both cases phase changes proceed at
constant temperature and constant pressure. Therefore heat
supplied/removed is the enthalpy change of phase changes.
hLG = hG – hL
hSL = hL – hS
enthalpy of evaporation (>0 heat must be supplied)
enthalpy of melting (>0 heat must be supplied)
hGL = hL – hG
hLS = hS – hL
enthalpy of condensation (<0 heat must be removed)
enthalpy of freezing (<0 heat must be removed)
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Enthalpy of phase changes
Phase changes proceed at equlibrium between two phases and according
to Gibb’s rule there is only one degree of freedom (2-phases, 1-component).
Therefore enthalpy change depends only upon one state variable, e.g.
upon temperature T (temperature of boiling, …) and corresponding pressure
is determined by some thermodynamic equations.
Antoine’s equation describes relationship
between T,p of saturated vapours
B
ln p  A 
C T
Parameters A,B,C can be found
in tables for most substances.
ln p  A 
Later on Clausius Clapeyron
equation will be derived, enabling
to evaluate enthalpy of phase
changes, for example as
B
C T
RT 2 d ln p
hLG 
M dT
Boiling temperature of water at
atmospheric pressure 100 kPa is 1000C
T  1004 p
0
100
200
300
T [ 0C]
CH5
Enthalpy of phase changes
Maybe that you are little bit confused by previous slide (Antoine’s
and Clausius Clapeyron equations). Details will be discussed later
when analysing equilibrium states using different techniques. So far it
is sufficient to know, that
 Temperature of phase changes increases with pressure
 Enthalpy of evaporation decreases with pressure. The higher
is pressure, the higher is boiling point temperature and the lower is
amount of heat necessary for evaporation of 1 kg of substance. As
soon as temperature is increased up to the critical temperature Tc,
the enthalpy of evaporation drops to zero.
 Example: Enthalpy of evaporation of water is
hLG = 2500 kJ/kg at 00C
hLG = 2400 kJ/kg at 500C
hLG = 2250 kJ/kg at 1000C
hLG = 1900 kJ/kg at 2000C
Enthalpy of chem.reactions
CH5
During chemical reactions some covalent bonds are released and some
are created. Knowing Lewis structure of reactants and products it is
possible to estimate energy released during chemical reaction.
-C
-N
-O
-H
=C
=N
=O
C
N
Example: CH4+2O22H2O+CO2
C
348
292
351
413
611
615
741
837
891
CH4
(4single bonds 413)=1652 kJ.mol-1
N
292
161
200
391
615
418
481
891
946
2O2
2(1double bond 498)=996 kJ.(2mol)-1
O
351
200
139
463
741
481
498
H
413
391
463
436
2H2O 2(2single bond 463)=1852 kJ.(2mol)-1
CO2 (2 double bonds 741)=1482 kJ.mol-1
This example demonstrates that breaking the bonds of the reactants (one mole of CH4 and two moles of O2)
requires 2648=1652+996 kJ, while when the bonds of the product are formed 3334=1852+1482 kJ of energy are
released. The net profit is 3334-2648=686 kJ, which is approximately 15% less than the molar enthalpy change
predicted before (802.8 kJ.mol-1).
Because most chemical reactions proceed at a constant pressure the heat
exchanged in the course of reactions is equal to the enthalpy change hR. If
enthalpy of products is less than the enthalpy of reactants heat is released
(hR<0) and reaction is called EXOTHERMIC. The reaction when the heat
must be supplied is ENDOTHERMIC (hR>0) .
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Enthalpy of chem.reactions
The enthalpy change depends on pressure, temperature and also on the
physical state of the reactants and products (s-solid, l-liquid, g-gas). Data
corresponding to the standard state (p=100 kPa at constant temperature
T=298 K), are published, e.g., by the National Institute of Standards and
Technology. The enthalpy change for different or varying temperature must be recomputed by an
integration of the specific (or molar) heat capacities of the reactants and products.
Examples of chemical reactions at standard state:
1
H2 ( g)  O2 ( g)  H2 O( g)
2
1
H2 ( g)  O2 ( g)  H2 O(l )
2
~
( h 0 ) 298  24183
. kJ . mole 1
~
( h 0 ) 298  28584
. kJ . mole 1
Standard temperature T=298K
Condensed water
Standard pressure p=100kPa
The difference between the standard enthalpy changes is the standard molar enthalpy of vaporisation.
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Hess law
The enthalpy change of the reaction is a function of
state and is independent of the path, that is, of any
intermediate reaction that may have occurred.
Example: Combustion of carbon (symbols s,g,l means solid, gas, liquid)
1
C( s)  O2 ( g)  CO( g)
21
CO( g)  O2 ( g)  CO2 ( g)
2
~
( h 0 ) 298  111 kJ . mole 1
~
( h 0 ) 298  283 kJ . mole 1
Two consecutive exothermic reactions yield the same result as the summary reaction
C ( s)  O2 ( g )  CO2 ( g )
~
( h 0 ) 298  394 kJ . mole 1
Consequence: While specific reaction mechanism is important for evaluation
of chemical reaction rate, it can be substituted by any other sequence of
intermediate reactions (may be fictive) when calculating enthalpy cheanges.
CH5
Enthalpy of formation
It can be assumed that the analysed chemical reaction proceeds in two
steps, the first is the decomposition of all reactants to free elements, and the
second step is the addition of the free elements into the products. Both the
steps can be considered as chemical reactions accompanied by the enthalpy
changes, corresponding to the formation of compounds from elements in
their standard state. This is the standard
enthalpy of formation
~0
( h f ,compound ) 298
The standard molar enthalpy of formation of free elements, e.g. C, or
molecules O2, N2, H2, etc., is equal to zero (molecule O2 and not the radical O is
assumed to be in the most stable state).
The standard enthalpy change of the calculated reaction is
~
~
( Hr0 )298   P ( h f0, P )298  R ( h f0, R )298
P
R
where  R  0, P  0 are stoichiometric coefficients of reactants (R) and
products (P).
Combustion of Methane
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Example: Gas burner operating at atmospheric pressure consumes mass
flowrate of CH4 0.582 kg/s. Calculate mass flowrate of flue gas. Assuming
temperature of fuel and air 298K and temperature of flue gases 500 K
calculate heating power.
Air
Q
fg-flue gas
CH4
(h )
0
R 298
CH  2O  CO  2H O
4
2
2
2
CH 4  1,
O 2  2,
CO 2  1,  H 2O  2
(h 0f ,CH 4 ) 298  75, (h f0,CO2 )298  393, (h f0, H 2O )298  242
 i (h )
0
f ,i 298
(hR0 )298
802
MJ
hR 

 50
M CH4
0.016
kg
 75  1 393  2  242  802 kJ / mol CH 4
i
Flue gas production 1kg CH
m fg  mCH4  mO2  mN2  mCH4  mO2 (1 
needs 4 kg O2.
Mass fraction of oxygen in air is 0.233.
N
N
0.767
)  mCH (5  4
)  0.582(5  4
)  10.57kg / s
O
O
0.233
4
2
2
4
2
2
Heat removed in heat exchanger
Q  hR mCH4  m fg c p (Tfg  T0 )  50  0.582  0.01057  (500  298)  27MW
cp0.001 MJ/kgK
CH5
Combustion Baloon
Example: Calculate fuel consumption (propane burner) of a baloon
filled by hot air. D=20m, T=600C, Te=200C, heat transfer coefficient
estimated to =1 W/m2.K.
Reaction
C H 5O  3CO  4H O
3 8
2
2
2
D
C 3 H 8  1, O 2  5, CO 2  3,  H 2O  4
~
~
~
(h f0,C3 H 8 ) 298  103, (h f0,CO2 ) 298  393, (h f0, H 2O ) 298  242
~
~
(hR0 ) 298   i (h f0,i ) 298  103  3  393  4  242  2044 kJ / mol C3 H 8
i
~
(hR0 ) 298
2044
MJ
hR 

 46.5
M C3 H 8
0.044
kg
m
Heat losses
Q  D 2 (T  Te )    400  40  50kW
Fuel consumption
mf 
Q
50

 0.001kg / s
hR 46500
CH5
Combustion Baloon
The burner unit gasifies liquid propane, mixes it with air,
ignites the mixture, and directs the flame and exhaust into
the mouth of the envelope. Burners vary in power output;
each will generally produce 2 to 3 MW of heat