Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

This course is approximately at this level

CHEMISTRY E182019 CH5

Heat and energies

Systems CH5

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 variablesT,p,vInternal energy u [J/kg]Enthalpy h [J/kg]Entropy s [J/kg.K]

Energy balance of Closed Systems CH5

δq = du + δw

qheat transferred to system

w=p.dv mechanical work done by system (volume increased by dv)

du-internal energy increase

First law of thermodynamics

Only du-is a state variable, while q (heat) and w (work) depend upon previous history and are NOT state variables.

Energy balance of Closed Systemsat constant VOLUME

CH5

δq = du

qheat transferred to system

du-internal energy increase

Closed system and constant volume (no mechanical work). Internal energy increase is related to temperature increase

du = cv dTcv – specific heat capacity at constant volume [J/kg.K]

Energy balance of Closed Systemsat constant PRESSURE

CH5

δq = du+pdv=dh-vdp=dh

qheat transferred to system

du-internal energy increase

Closed system and constant pressure. Amount of heat is expressed in terms of enthalpy h=u+pv

dh = cp dTcp – specific heat capacity at constant pressure [J/kg.K]

w=p.dV mechanical work done by system (volume increased by dv, dp=0)

Energy balance of Closed Systemsat constant TEMPERATURE

CH5

δq = du+pdv=Tds

qheat transferred to system

ds-entropy increase dT=0

Closed system and constant temperature. Amount of heat is expressed in terms of entropy

ds = (du+pdv)/T

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).

Internal energyCH5

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)

Internal energyCH5

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/kgChemical energy of ionic/covalent bonds in molecule ~107 J/kgIntermolecular VdW forces (phase changes) ~106 J/kgPressure forces ~105 J/kgThermal energy (kinetic energy of molecules) ~104 J/kg

Internal energyCH5

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

dvpTpTdTcdu vv ))((

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:

0)( ppp

vRTp

TpT v

Internal energy/enthalpyCH5

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

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)

1122 hmhmQ

][WQ111 hmH 222 hmH

][WQ

21,mm

21,HH

Mass flowrates at inlet/outlet streams [kg/s]

Enthalpy flows at inlet/outlet streams [W]

Heating power delivered to system through heat transfer surface [W]

EnthalpyCH5

...heat dq transferred to system at constant pressure

dh = cp dTh = u + pv

Enthalpy changes with changing temperature (small, kinetic energy)Enthalpy changes during phase changes (medium, VdW forces)Enthalpy changes during chemical reactions (high, bonding forces)

Enthalpy changes with TCH5

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

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:

dpvTvTdTcdh pp ))((

0)(

vvvpRTv

TvT p

Enthalpy of phase changesCH5

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 enthalpy of evaporation (>0 heat must be supplied)hSL = hL – hS enthalpy of melting (>0 heat must be supplied)

hGL = hL – hG enthalpy of condensation (<0 heat must be removed)hLS = hS – hL enthalpy of freezing (<0 heat must be removed)

Enthalpy of phase changesCH5

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 TC

BAp

ln

T p1004

TCBAp

ln

0 100 200 300 T [ 0C]

Boiling temperature of water at atmospheric pressure 100 kPa is 1000C

dTpd

MThLG

ln2R

Parameters A,B,C can be found in tables for most substances.

Later on Clausius Clapeyron equation will be derived, enabling to evaluate enthalpy of phase changes, for example as

Enthalpy of phase changesCH5

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 1000ChLG = 1900 kJ/kg at 2000C

Enthalpy of chem.reactionsCH5

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+2O22H2O+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 2H2O 2(2single bond 463)=1852 kJ.(2mol)-1

H 413 391 463 436 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) .

Enthalpy of chem.reactionsCH5

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:

H g O g H O g h kJ mole2 2 20

29811

224183( ) ( ) ( ) ( ~ ) . .

H g O g H O l h kJ mole2 2 20

29811

228584( ) ( ) ( ) ( ~ ) . .

Standard temperature T=298K

Standard pressure p=100kPaCondensed water

The difference between the standard enthalpy changes is the standard molar enthalpy of vaporisation.

Hess lawCH5

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.

C s O g CO g h kJ mole( ) ( ) ( ) ( ~ ) . 12

11120

2981

CO g O g CO g h kJ mole( ) ( ) ( ) ( ~ ) . 12

2832 20

2981

C s O g CO g h kJ mole( ) ( ) ( ) ( ~ ) . 2 2

0298

1394

Example: Combustion of carbon (symbols s,g,l means solid, gas, liquid)

Two consecutive exothermic reactions yield the same result as the summary reaction

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.

Enthalpy of formationCH5

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

( ~ ),h f compound0

298

The standard enthalpy change of the calculated reaction is

( ) ( ~ ) ( ~ ), , H h hr P f PP

R f RR

0298

0298

0298

where are stoichiometric coefficients of reactants (R) and products (P).

R P 0 0,

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).

CH5 Combustion of MethaneExample: 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.

4 2 2

4 2 2 2

0 0 0, 298 , 298 , 298

1, 2, 1, 2

( ) 75, ( ) 393, ( ) 242

2 24 2 2 2

CH O CO H O

f CH f CO f H Oh h h

CH O CO H O

4

00 0 298

298 , 298 4( ) 802( ) ( ) 75 1 393 2 242 802 / 50

0.016R

R i f i Ri CH

h MJh h kJ mol CH hM kg

Flue gas production 1kg CH4 needs 4 kg O2. Mass fraction of oxygen in air is 0.233.

2 2

4 2 2 4 2 4

2 2

0.767(1 ) (5 4 ) 0.582(5 4 ) 10.57 /0.233

N Nfg CH O N CH O CH

O O

m m m m m m m kg s

Heat removed in heat exchanger

4 0( ) 50 0.582 0.01057 (500 298) 27R CH fg p fgQ h m m c T T MW

cp0.001 MJ/kgK

Air

CH4

fg-flue gasQ

CH5 Combustion BaloonExample: 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.

D

m

Reaction

242)~( ,393)~( ,103)~(

4 ,3 ,5 ,1

2980,298

0,298

0,

22283

2283

24232583

OHfCOfHCf

OHCOOHC

hhh

OHCOOHC

kgMJ

Mhh

HCmolkJhh

HC

RR

iifiR

5.46044.0

2044)~(

/ 204424243933103)~

()~

(

83

2980

832980,298

0

Heat losseskWTTDQ e 5040400)(2

Fuel consumption50 0.001 /

46500fR

Qm kg sh

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