energy balance and thermo presentation

7/29/2019 Energy Balance and Thermo Presentation

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Energy Balances and

Thermodynamics

Florabel R. Toro-Rodrguez, Esq, PE


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Agenda

I. Terminology Associated with the EnergyBalances and ThermodynamicsII. Types of Energy A. Work (W) B. Heat (Q) C. Kinetic Energy (KE) D. Potential Energy (PE)

E. Internal Energy (U)

F. Enthalpy (H) G. Entropy (S)


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Agenda (cont.)

III. Energy Balances for Processes without Reaction A. Closed, Unsteady-State Systems B. Energy Balances for Closed, Steady-State Systems C. Energy Balances for Open, Unsteady-State Systems D. Energy Balances for Open, Steady-State Systems E. Summary

IV. Calculation of Enthalpy Changes A. Phase Transitions B. Heat Capacity Equations C. Tables and Charts to Retrieve Enthalpy Values


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Agenda (cont.)

V. Applications of Energy Balances A. Solving Energy Balances Problems in the Absenceof Chemical Reactions

B. Solving Energy Balances Problems with ChemicalReactions

C. Applications of Energy Balances in Processes thatInclude Reactions

VI. Volumetric Properties of Pure Fluids A. PVT Behavior of Pure Substances B. Single Phase Region


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Agenda (cont.)

VII. Ideal Processes, Efficiency, andCyclesVIII. Gibbs Energy, Equilibrium Constant,and Solution Thermodynamics A. Gibbs Energy (G) B. Equilibrium Constant (K) C. Solution Thermodynamics D. Heat Effects of Mixing Processes


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I. Terminology

Refer to your notes


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II. Types of Energy

Energy is the capacity to do work.Types of Energy A. Work (W)

Mechanical work Electrical work

Shaft work Flow work Path function

B. Heat (Q) Conduction Convection Radiation Path function


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II. Types of Energy (cont.)

C. Kinetic Energy (KE) KE = State function

D. Potential Energy (PE) PE = mgh State function

2mv21


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E. Internal Energy (U) Takes into account all of the molecular, atomic,and subatomic energies

State function

dVVUdT

TUdU

TV

+

=

=2

1

T

TVdTCU


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F. Enthalpy (H) H = U + PV

State function

For an ideal gas CP = C V + R

dPPH

dTTH

dHTP

+

=

=

2

1

T

TPdTCH

V

p

C

Ck =


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G. Entropy (S) Unusable energy or the measure of amaterials disorder

S = Q/T S = C P ln(T2 /T1) - Rln(P 2 /P 1), for ideal gases State function


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III. Energy Balances for Processes without Reaction

A. Closed, Unsteady-State System Closed does not interchange mass with thesurroundings. W and Q can be interchanged.

Unsteady-state the state of the material changesinside the system.

Refer to Figures III.A(1a) and (1b) of your notes. Examples batch processes used to manufacture

specialized pharmaceuticals and polymer products Low production quantities


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III. Energy Balances for Processeswithout Reaction (cont.)

B. Energy Balances for Closed, Steady-StateSystems Closed does not interchange mass with the

surroundings. W and Q can be interchanged.

Steady-state - the state of the material does notchange inside the system.

All W done on a closed, steady-state system must betransferred out as Q. However, the reverse is false.

Refer to Figures III.B(1a) and (1b) of your notes.


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C. Energy Balances for Open, Unsteady-StateSystems Open does interchange mass with the surroundings.

W and Q can also be interchanged. Unsteady-state the state of the material changes

inside the system. PV work work done by the surroundings to put amass of matter into or out of the system. Enthalpy (H) the form of energy related to mass

includes the PV work term.

Examples filling of a fixed volume tank, batchdistillation without replacement of feed


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D. Energy Balances for Open, Steady-State Systems Open does interchange mass with the

surroundings. W and Q can also beinterchanged.

Steady-state - the state of the material doesnot change inside the system.

Examples refining and chemical industries High-volume products


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Summary - refer to your notes


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IV. Calculation of Enthalpy Changes

A. Phase Transitions For a single phase U = fn(T)

H = fn(T) Sensible heat T 0 Latent heat T = 0


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IV. Calculation of Enthalpy Changes (cont.)

How is the overall H for a pure substance calculated?

Solid Liquid Vapor

Tm Tv Temperature, T

Enthalpy, H

AB

C D

EF

Figure IV.A(1) Enthalpy changes of a pure substance as a function of temperature.


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Refer to your notes for relationships betweenthe slope of the vapor pressure (P ) curve andthe molar heat of vaporization ( ). Clapeyron equation Clausius-Clapeyron equation Chens equation Watson equation

vapH


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Example IV.A(1)The vapor pressure of 2,2,4-trimethylpentane at several temperatures is

given.Calculate the heat of vaporization (Btu/lbmole) of this compound at 77F.


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IV. Calculation of Enthalpy Changes (cont.)B. Heat Capacity Equations

Heat capacity amount of energy required to increasethe temperature of a substance by one degree, (J/mol-K, Btu/lbm-R)

Generally, C P = C V for liquids and solids. H = C P T. Why is C P frequently used? CP = fn(T). Most equations are empirical. For mixtures, the heat capacity is the mole-weighted

average of the heat capacity of the components

Pin

1iiavgP CxC

=

=


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Example IV.B(1)What is the enthalpy change for 1 kg ofwater from -30C to 130C using the

following data: Tf = 0C, T vaporization = 100C, Hfusion@0C = 6.01 kJ/gmol, Hvaporization@100C =

40.65 kJ/gmol,

Pice = 23.7 J/gmol-C, Pliquid water = 75.4 J/gmol-C, Pvapor water = 33.9 J/gmol-C.

C

CC


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C. Tables and Charts to Retrieve EnthalpyValues Experimental data of pure compounds are organized

in tables , T and P being the independent variables.

Steam tables are the most common ones. Steam tables practice exercise. Find H at 150C and375 kPa.

Enthalpy differences can be calculated, H = H f - Hi. For compressed liquids, properties of saturated liquid

can be used as a good approximation.


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V. Applications of Energy Balances

A. Solving Energy Balance Problems inthe Absence of Chemical Reactions Assumptions for energy balance problems

saturated liquid vapor pressure of the liquid gas one atmosphere EB: in out = accumulation - generation


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V. Applications of Energy Balances (cont.)

Example V.A(1)An insulated heat exchanger is carryingsaturated steam at a pressure of 2.0 bar. Anoxygen stream at a pressure of 1.5 bar andflowing at the rate of 150 kmol/h is to beheated from 25C to 220C using the heatexchanger. The average heat capacity ofoxygen is constant over the range at 30.07kJ/kmol-C. Calculate the steam consumption.


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B. Solving Energy Balance Problemswith Chemical Reactions Exothermic reaction products have less

energy than reactants, Q out . Endothermic reaction products have moreenergy than reactants, Q in.


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- standard heat of formation. denotes standard state, 25C and 1 atm. It is defined as one mole of a compound formed

from its constituent elements in their standard state.

The standard heat of formation is negative forexothermic reactions. The reaction can be a fictitious process. Remember

that H is a state function.

The heat of formation is zero in the standard state foreach stable element.

o

f H


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- standard heat of reaction denotes standard state, 25C and 1 atm. It is defined as one mole of a compound that reacts

when stoichiometric quantities of reactants in the

standard state react completely to produce products inthe standard state.

i species i stoichiometric coefficient ratio

o

rxnH

oif,

reactants

i

io

if,

products

i

iorxn H H H =


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Thermodynamics routes

Reactants,T

Reactants,H =

H(25 C) - H(T)

Products,T

Reactants,25C

Hrxn, 25 C Products,25C

Products,H=

H(T) - H(25 C)

Hrxn, T

Figure V.B(1) Steps for the calculation of the enthalpies comprising Hrxn


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Example V.B(1)Public concern about the increase in the carbon dioxide inthe atmosphere has led to numerous proposals tosequester or eliminate the carbon dioxide. An inventorbelieves he has developed a new catalyst that can makethe gas phase reaction

CO 2(g) + 4H 2(g) 2H 2O(g) + CH 4(g)proceed with 100% conversion of CO 2. The source of thehydrogen would be from the electrolysis of water usingelectricity generated from solar cells. Assume

stoichiometric amounts of the reactants enter the reactor.Calculate the heat of reaction if the gases enter andleave at 1 atm and 500C.


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Example V.B(2)One mol of CO 2 is entering at 800 K andreacts with 4 mol of H 2 entering at 298 K.

Only 70% conversion of the CO 2occurred. The products exited from thereactor at 1000 K. Calculate the heattransfer to or from the reactor.


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- standard heat of combustion denotes standard state, 25C and 1 atm. The compound is oxidized with oxygen or some other

substance to the products CO 2(g), H 2O(l), HCl(aq),

and so on. Zero values of are assigned to certain of the oxidationproducts as, for example, CO 2(g), H 2O(l), HCl(aq), andto O 2(g) itself.

Stoichiometric quantities are assumed to reactcompletely.

o

cH


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The relation between and is

=

where n i is the species i number of moles.

o

rxnH o

cH

products

i

reactants

i

oic,i

oic,i HnHn

orxnH


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Heating value the negative of thestandard heat of combustion for a fuel,such as coal or oil.

Depending upon the state of the water inthe combustion products LHV [lower (net) heating value] vapor HHV [high (gross) heating value] - liquid


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Example V.B(3)For CO(g) + H 2O(g) H2(g) + CO 2(g)calculate the from the heat of

combustion data.

orxnH


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Example V.B(4)In a waste-to-energy facility, aconcentrated solution of urea is sprayed

into the furnace in the selective non-catalytic reduction of nitrogen oxide,CO(NH 2)2(g) +2NO(g) +1/2O 2(g)

2N2(g) + CO 2(g) + 2H 2O(g).Calculate the ideal-gas heat of reactionat 77F (Btu/lb mole urea).


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Example V.B(5)Orimulsion is a tar-like bitumen that uses anemulsifying agent to produce a liquid. This blend(15wt% emulsifying agent) can be substituted for oil

in utility burners. Data on orimulsion (exclusive ofemulsifying agent): heating value 16,800 Btu/lb (dryand ash-free) content (%wt): ash 12.1%, moisture10.9%, organic 77.0%

The emulsifying agent has no heating value.Calculate the value (Btu/lb) of the as-deliveredemulsifying fuel.


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Example V.B(6)A combustor currently operates on high-sulfur wastecoal (gob coal). To reduce sulfur emissions, somepelletized waste is used to displace 10% by weight

of the gob coal. The heating values of pelletizedwaste and gob coal are 6,550 Btu/lb and 5,500Btu/lb, respectively.Calculate the ratio of pounds of pellets fired topounds of gob coal fired to maintain the same heatoutput.


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Example V.B(7)A boiler is fired with a gas (MW = 16.36) that has thefollowing composition by volume: methane 97%,nitrogen 3%. The boiler is designed for a volumetric

fuel rate of 60,000 standard cubic feet (scf) per hour(60F and 1 atmosphere). Methane has a grossheat of combustion (HHV) of 1,013 Btu/scf and a netheat of combustion (LHV) of 913 Btu/scf.

Calculate the firing rate (Btu/hr) to the boiler (withwater as a vapor in the flue gas and 0% excess air).


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Example V.B(8)An incinerator is designed to process600 tons/day of municipal solid waste

(MSW). The heating value of the MSWis 1,100 Btu/lb. The incinerator iscurrently being operated at 75% of

capacity with an efficiency of 50%.Calculate the heat liberated (10 9Btu/day) by the incinerator.


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C. Applications of Energy Balances inProcesses that Include Reactions The adiabatic reaction (theoretical flame,

combustion) temperature is defined as thetemperature obtained when the process is adiabatic and the limiting reactants react completely.


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Example V.C(1)Calculate the theoretical flametemperature for CO gas burned at

constant pressure with 100% excess air,when the reactants enter at 100C and 1atm.


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Example V.C(2)Propylene glycol is produced by the hydrolysis ofpropylene oxide (PO) in the liquid phase: CH 3CHOCH 2 +H2O CH 3CHOHCH 2OHThis exothermic reaction occurs readily at roomtemperature when catalyzed by sulfuric acid. The feed isa stream of 24,000 kg/hr with a composition of 10 wt%propylene oxide, 1 wt % sulfuric acid, 89 wt% water.

Calculate the reactor exit temperature if an adiabaticreactor is used to achieve a fractional conversion of 0.8for PO with a feed temperature of 20C.


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VI. Volumetric Properties of Pure Fluids

The quantities of Q and W necessary tocarry out industrial processes arecalculated from thermodynamics

properties.These properties, such as U, H, and S,are often evaluated from molar volumes,

V = fn(T, P), yielding equations of state.


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VI. Volumetric Properties of Pure Fluids (cont.)

A. PVT Behavior of Pure Substances

solid

vapor

liquid

1

2

3Fluid region

Gas region

Pressure

Temperature

Triple point

C

Figure VI.A(1) PT diagram for a pure substance


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Pressure

Volume

S L

Liquid/vapor

C

Gas

Vapor

Fluid

Tc

Solid/vapor

Figure VI.A(2) PV diagram for a pure substance


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Figure VI.A(3) TS diagram for a pure substance

T

S

S L

Liquid/vapor

C

Gas

Vapor

Fluid

Solid/vapor

P = const


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Quality (x)

D = D f + xD fg

mixtureof amounttotalvaporof amountx =

f g

f

DDDD

x

=


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Refer to Table VI.A(1) Equipmentsenergy balances


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Example VI.A(1)

A stream consisting of saturated steam at 356F with abulk velocity of 130 ft/sec and a flow rate of 250 lbm/hrgoes through a boiler that adds heat to the stream at therate of 300 Btu/lbm. The superheated steam

subsequently expands through a back-pressure turbinewhere it develops 60 hp of shaft work and exits through adiffuser with a velocity of 1 ft/sec. The elevation changebetween the inlet and the outlet of the system is 220 ft.Assume negligible pressure drop due to friction.Calculate the quality of the steam at the outlet of thediffuser.


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Example VI.A(2)Spent steam from a turbine is cooled in aheat exchanger.

When 5,000 lbm/min of steam enters theheat exchanger, 2,000,000 Btu/min ofheat is removed. Calculate the quality(% vapor) of the steam exiting the heatexchanger.


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B. Single Phase Region For regions of the diagram where a single

phase exists a correlation connecting P, V,and T, expressed analytically as fn(P, V, T) isknown as a PVT equation of state.

The ideal gas equation PV = RT

For real gases, Z is the compressibility factor ZRTPV =


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Example VI.B(1)Dry air and vinyl chloride (C 2H3Cl) are mixed toproduce a saturated mixture at 3739 mm Hgand 20C. At this temperature, the vapor

pressure of vinyl chloride is 2580 mm Hg. Theaverage molecular weights of vinyl chlorideand dry air are 62.50 g/mol and 28.96 g/mol,respectively. Assume the ideal gas lawapplies. Calculate the mass of vinyl chloride in100 cm 3 of saturated mixture.


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VII. Ideal Process, Efficiency,and Cycles

Energy is conserved but how is thisenergy converted from one form toanother and how efficiently is this done?Ideal vs actual processesReversible vs irreversible processesCycle number of processes applied to

a substance that leaves it in its originalstate.


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VII. Ideal Process, Efficiency,and Cycles (cont.)

Efficiencies definitions - refer to yournotes, Table VII(2) Efficienciesdefinitions


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Carnot cycle

T

S

1

4

2

3

Figure VI(1) Carnot cycle

Q in

Qout


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Rankine cycle

T

S

1

4

2

3

Figure VI(2) Rankine cycle

P const


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Efficiency relationships - refer to yournotes, Table VI(2) Important work andefficiency relationships.


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Equipment processes refer to yournotes, Figure VI(3) Adiabatic expansionprocess in a turbine or expander and

Figure VI(4) Adiabatic compressionprocess.


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Example VI(1)

A steam power plant operates on a reheat Rankine cycleat a mass flow rate of 1 kg/s. The steam enters the high-pressure turbine at 12 MPa and 612C and is condensedat a pressure of 12 kPa. The saturated liquid is thenpumped to the boiler. The temperature of the fluid atstate 5 is the same as the temperature at state 3. Themoisture content in the low-pressure turbine is not toexceed 8%. Assume all devices operate with 100%efficiency. Calculate the thermal efficiency of the

Rankine cycle operating at these conditions.


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Example VI(2)

The compressor of a simple-cycle gas turbine operateswith an efficiency of 87% while the turbine operates withan efficiency of 90%. The compressor pressure ratio is 5and the pressure drop in the combustion chamber is 6%.The turbine inlet temperature is 1500F. Air enters thecompressor at 70F and 1 atm. The air has an isentropicexponent of 1.40 and a constant heat capacity of 0.24Btu/lbm-R. The combustion gas has a heat capacity of0.2744 Btu/lbm-R and an isentropic exponent of 1.333.

Calculate the thermal efficiency of this simple-cycle gasturbine.


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Example VI(3)

An ideal gas flowing at a rate of 15 mol/s, a temperatureof 298 K, and at 120 kPa is to be compressed to 1600kPa using a three-stage isentropic compressor. Thecompressor system has an efficiency of 90% and isdriven by an electric motor with 70% efficiency. Thecompression ratio is equal for each stage. Assume aconstant outlet temperature of 298 K for the intercoolers.Calculate the compression power required by the three-stage compressor.

VIII. Gibbs Energy, Equilibrium


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gy q

Constant, and SolutionThermodynamics

A. Gibbs Energy (G) G is the maximum amount of non-expansion

(without increasing the total volume) workwhich can be extracted from a closed

(isothermal, isobaric) system, and thismaximum can be attained only in acompletely reversible process.

A reaction is the result of a system trying tominimize its Gibbs energy.



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Constant, and SolutionThermodynamics

Gibbs energy is defined asG = H - TS

orxn

orxn

orxn STHG =

o if,

reactants

iio if,

products

iiorxn G G G =



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Constant, and SolutionThermodynamics (cont.)

Example VIII.A(1)Liquid methylhydrazine (CH 6N2) reactswith liquid dinitrogen tetroxide (N 2O4) to

give gaseous carbon dioxide (CO 2),gaseous water (H 2O), and gaseousnitrogen (N 2) at 25C and 1 atm.

Calculate the Gibbs energy for thereaction at 25C.



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B. Equilibrium Constant The equilibrium constant expresses the

equilibrium of a reaction with the finalconcentrations of the products and reactants.

For a typical reaction aA + bB pP + qQ thegeneral form of the equilibrium constant forliquids is

[ ][ ][ ] [ ]bB

aA

qQ

pp

eqCC

CCK =



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For gases assuming ideal gases and DaltonsLaw of partial pressures

The tendency for a reaction to reach anequilibrium is driven by the Gibbs energy.

The K can be calculated from = -RTlnK.

[ ][ ][ ] [ ]

baqpYb

B

a

A

qQ

pp

P PKPyPy

PyPyK

+==

G



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The change of K with T can be determined bythe vant Hoff equation

where His the change in enthalpy of reaction(heat of reaction) at equilibrium and it is constant.

=

21T2

T1

T1

T1

RH

KK

lno



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Example VIII.B(1)One (1) mol of carbon monoxide (CO) reactswith 3 mol of hydrogen (H 2) at 100C and 1atm to give methane (CH 4) and water (H 2O).For the reaction

CO(g) + 3H 2(g) CH 4(g) + H 2O(g),Calculate the numerical value of theequilibrium constant for the reaction at 100C.



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C. Solution Thermodynamics The thermodynamics of pure substances is

different to that of mixtures. At common P and T, the vapor phase can

usually be considered ideal and the vapor-liquid equilibrium can usually be consideredideal and the vapor-liquid expression can besimplified to a modified version of the Lewis-

Randall relationship Pypx ioiii =



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In non-ideal liquid problems the pure liquidphase vapor pressure, p o, is usually given inthe form of Antoine equation

It should be noted that the p o = fn(T) while =fn(x). The usual problem dealing with non-ideal

liquids is to find the composition (x) at anotherstate.

B/TAlnp o +=



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Two types of equations for binary mixturesare Van Laar equations Three Suffix Margules equations Refer to eqns. 40a, 40b, 41a, 41b of your notes.

Two conditions are usually studied azeotrope and infinite dilutions.

Refer to Table VIII.C(1) Summary of activitycoefficients of your notes.



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Raoults assumptions are the vapor phase is an ideal gas, PV = RT, and the liquid phase is an ideal solution, = 1.

When the liquid phase is considered pure, x = 1, then

When it is necessary to calculate the mole fraction of a

species dissolved in liquid the Henrys law can beapplied

Pp

y

oi

i =

Pyx ii =i H



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Example VIII.C(1)Dry air and vinyl chloride (C 2H3Cl) arecombined to produce a saturated mixture at14.3 lbf/in 2 and 77F. At this temperature, the

vapor pressure of vinyl chloride is 5.77 lbf/in2

.The average molecular weight of vinyl chlorideand dry air are 62.50 lbm/lbmol and 28.96lbm/lbmol, respectively. Assume the ideal gas

law applies. Calculate the mass ratio of vinylchloride to dry air in the saturated mixture.



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Example VIII.C(2)

A vent stream consisting of acetone, air, and a negligibleamount of moisture flows at 200 ft 3 /min at standardconditions. The acetone is to be separated from theremaining vapors by condensation. This process requiresa refrigerated condenser system. The inlet volume

fraction of acetone in the vent stream is 0.385. Therequired removal efficiency is 90%. The relationshipbetween the temperature in degrees Celsius and thevapor pressure of the acetone in mm Hg is

Calculate the temperature required to achieve the desiredremoval efficiency.

C229.664T

C1210.5957.117logp

o

oo

mmHgac,+

=



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Example VIII.C(3)

A combustion chamber receives a feed thatconsists of a gaseous waste containingbenzene 1000 ppm (by volume), chloroform1000 ppm (by volume), and air at 77F. Themolecular weight of the waste gas is 28.97lbm/lbmol. At 77F, the volumetric heat ofcombustion is 3616 Btu/ft 3 for benzene and

705 Btu/ft3

for chloroform. Calculate the heatof combustion of the waste gas stream.



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Example VIII.C(4)The Henrys law constant for carbondioxide in seawater at 25C is 35.4

atm/(mol/kg). This constant is based onconcentration (c) rather than activity inthe liquid phase.

Calculate the equilibrium concentration(mol/kg) of CO 2 (aq) with CO 2 partialpressure at 0.000354 atm.

VIII. Gibbs Energy, EquilibriumConstant and Solution


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Example VIII.C(5)

Water (60 lb moles), benzene (20 lb moles), andtoluene (20 lb moles) equilibrate as a liquid mixtureat a pressure of 0.9 atm. The water (W) and theorganics are immiscible, and the benzene (B) andthe toluene (T) form an ideal mixture.Calculate the saturation pressure (atm) of water, ,at the bubble-point temperature and the specified

system pressure.

oWP

VIII. Gibbs Energy, EquilibriumConstant and Solution


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D. Heat Effects of Mixing Processes

For ideal solutions

Heat of solution ( ) heat effect due tosolids or gases dissolved in liquids

Heat of mixing heat effect due to the mixingof liquid-liquid.

Pi

n

1iiavgP CxC

=

=

solutionH

VIII. Gibbs Energy, EquilibriumConstant, and Solution


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For any solution

where is the integral heat of formation ofsolution at standard conditions per mole of solute,. is the heat of formation of the solutionitself per mole of solute, and is the heat

of formation of the solute. is not included.

osolution

osolutef,

osolutionf, HHH +=

osolutiof,H

osolutionH o

solutef,H

oOHf, 2

H



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Example VIII.D(1)Calculate the heat of formation of LiCl in12 mol of H 2O at 25C.



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,Thermodynamics (cont.)

Example VIII.D(3)

Two pure process streams are fed to a perfectly mixedtank kept at 77F. One stream consists of purecompound M. The other stream is water (H 2O). Thesolution in the tank is produced at a flow rate of 28,800lbm/hr of a 15% (by weight) solution. The molecularweight component M is 42.39 lbm/lbmol and that of H 2Ois 18.015 lbm/lbmol. In the evaporator, 18,000 lbm/hr ofH2O are removed. Calculate the rate of the heat producedin the tank.


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The End

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energy balance and thermo presentation

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