1

Chemical ReactionsChemical ReactionsObjectives:••Apply the conservation of mass to reacting systems to Apply the conservation of mass to reacting systems to determine balanced reaction equations.determine balanced reaction equations.••Defined the parameters used in combustion analysis, such Defined the parameters used in combustion analysis, such as airas air--fuel ratio, percent theoretical air, and dew point fuel ratio, percent theoretical air, and dew point temperature.temperature.••Apply energy balance to reacting systems for both steadyApply energy balance to reacting systems for both steady--flow control volumes and fixed mass systems.flow control volumes and fixed mass systems.••Calculate enthalpy of reaction, enthalpy of combustion, Calculate enthalpy of reaction, enthalpy of combustion, and the heating value of fuels.and the heating value of fuels.••Determine the adiabatic flame temperature for reacting Determine the adiabatic flame temperature for reacting mixture.mixture.

Combustion of fuel Heat Heat Engine WorkCombustion of fuel Chemical ReactionHow much fuel is needed? How about the air used?How high the combustion flame temperature will be?

Jet Engine

Piston Engine

Rocket EngineSteam Generator

2

FuelsFuels• Fossil Fuels: Coal, Petroleum, Natural Gases (LNG:

liquid natural gas, CNG: Compressed Natural Gas)• Petroleum: Gasoline, Kerosene, Diesel, Fuel Oil ,

Liquid Petroleum Gas (LPG)

Fuel Approx. single HydrocarbonFuel Approx. single HydrocarbonGasoline ~Gasoline ~ Octane (COctane (C88HH1818))Diesel ~Diesel ~ DodecaneDodecane (C(C1212HH2626))Methanol ~Methanol ~ Methyl Alcohol (CHMethyl Alcohol (CH3 3 OH)OH)Natural gas~Natural gas~ Methane (CHMethane (CH44))

•• Combustion is a chemical reaction during which a fuel is Combustion is a chemical reaction during which a fuel is oxidized and a large quantity of energy is released.oxidized and a large quantity of energy is released.

•• Oxidizer = OxygenOxidizer = Oxygen•• Commonly AIR is used as oxidizer (free)Commonly AIR is used as oxidizer (free)•• By volume: AIR = 21% OBy volume: AIR = 21% O22 + 79% N+ 79% N22

•• Therefore, 1 mole of OTherefore, 1 mole of O22 NN22 = 79/21 = 3.76 mole= 79/21 = 3.76 mole1 1 kmolkmol OO22 + 3.76 + 3.76 kmolkmol NN22 = 4.76 = 4.76 kmolkmol of Airof Air

Assumption at normal combustion :• N2 absolutely inert (no reaction)• Water is also inert

CombustionCombustion AirAir

3

1 1 kmolkmol C + C + 1 1 kmolkmol OO22 1 1 kmolkmol of COof CO22

C + C + OO22 COCO22 (15.2)(15.2)

To have combustion reaction, it must:To have combustion reaction, it must:•• T > ignition tempT > ignition temp•• Fuel : air ratio must properFuel : air ratio must proper

•• To get good combustion To get good combustion 3T3T–– Temperature (high)Temperature (high)–– Turbulent (good mixing between fuel and air)Turbulent (good mixing between fuel and air)–– Time (enough to reach complete combustion)Time (enough to reach complete combustion)

Reactants Products

CombustionCombustion

•Balancing of chemical reaction: Conservation of mass principle

•• Air fuel ratio, AF = Air fuel ratio, AF = mmairair/m/mfuelfuel (15.3)(15.3)

4

Example 15.1 Balancing the combustion equation

One kmol of octane (C8H18) is burned with air that contains 20 kmol of O2, Assuming the products contain only CO2, H2O, O2, and N2, determine the mole number of each gas in the products and the air-fuel ratio for this combustion process.Solution Chemical reaction equation:

C8H18 +20 (O2 + 3.76 N2) xCO2 + yH2O + zO2 + wN2Balance of each element : @ ReactantS = @ ProductS

C: 8 = x x = 8H: 18 = 2y y = 9O: (20x2) = 2x +y + 2z z = 7.5N2: 20*3.76 = w w = 75.2

Air Fuel Ratio, AF = mair/mfuel = (NM)air/(NM)fuel= 24.2 kg air/kg fuel

Theoretical and Actual Combustion Theoretical and Actual Combustion ProcessesProcesses

Complete combustionComplete combustion: : 1) all C burns to CO2, and 2) all HC burns to H2O

Incomplete combustionIncomplete combustion: Products contains Unburned fuel or components สาเหตุหลัก: 1)Insufficient O2 2) Insufficient mixing. 3) Dissociation

StoichiometricStoichiometric Air Air or Theoretical AirTheoretical Air:• = minimum amount of air air needed for complete

combustion. • no unburnedunburned and no OO22 left in products• = Chemically correct amount of air or 100% 100%

theoretical airtheoretical air

5

Theoretical and Actual Combustion Theoretical and Actual Combustion ProcessesProcesses

StoichiometricStoichiometric CombustionCombustion or Theoretical CombustionTheoretical Combustion:• Complete combustion with 100% theoretical air ; ex

CH4 + 2(O2 + 3.76N2) CO2 + 2H2O + 7.52N2

Actual combustion processesActual combustion processes: : need excess air to complete combustion

% Excess air% Excess air: = % more air needed than theoretical air80% excess air = 180% theoretical air

Deficiency of air:Deficiency of air: = amount of air used < theoretical air air used < theoretical air ((10% deficiency of air = 90% theoretical air)

Equivalent RatioEquivalent Ratio = (AFactual)/(AFstoich)การประเมินการประเมิน//ตรวจสอบความสมบรูณของการการเผาไหมตรวจสอบความสมบรูณของการการเผาไหมOrsatOrsat Gas Analyzer Gas Analyzer = อุปกรณวิเคราะหองคประกอบของ combustion

gas (products) ปจจุบันไมคอยนิยมใชแลว – ใช gas analyzergas analyzer

Theoretical and Actual Combustion Theoretical and Actual Combustion ProcessesProcesses

6

Example 15.2 dew point temperature of combustion productsEthane (C2H6) is burned with 20% excess air during combustion process. Assuming complete combustion and a total pressure of 100 kPa, determine (a) the air fuel ratio and (b) the dew point temperature of the products. Solution(a) write the chemical reaction equation at 120% theoretical airC2H6 +1.2a (O2 + 3.76 N2) 2CO2 + 3H2O + 0.2aO2 +(1.2x3.76)aN2

From O2: 1.2a = 2+ 3/2+0.2a then a = 3.5C2H6 +4.2(O2 + 3.76 N2) 2CO2 + 3H2O + 0.7O2 + 15.79N2

Air Fuel Ratio, AF = mair/mfuel = (NM)air/(NM)fuel= (4.2x4.76 kmol)(29 kg/kmol)/[(2kmolx12 kg/kmol)x(3 kmol x 2 kg/kmol)]

= 19.3 kg air/kg fuel

(b) Tdp = Tsat @ Pv (H2O)Ideal gas mixture Pi/Pm = Ni/Nm

Pv = (3 kmol/21.49 kmol)(100 kPa)Pv = 13.96 kPa Tdp = 52.3 oC

C2H6 +4.2(O2 + 3.76 N2) 2CO2 + 3H2O + 0.7O2 + 15.79N2

7

Example 15.3 combustion of gasous fuel with moist airA certain natural gas has the following volumetric analysis: 72% CH4 , 9% H2 , 14%N2 , 2%O2 and 3%CO2. The gases is now burn with stochiometic amount of air that enters combustion chamber at 20oC, 1 atm, and 80%RH. Assume complete combustion and a total pressure of 1 atm, determine the dew point temperature of the products. Solution

0.72CH4 +0.09H2+ 0.02O2 + 0.14N2 +0.03CO2ath(O2+3.76N2)+ xCO2 + yH2O+zN2

1. write the chemical reaction equation at 100% theoretical air100% theoretical air (use dry air)

2. moles of air2. moles of air per kmol of fuel can be determined3. then extra moles water vapormoles water vapor (20%RH at inlet conditions) can be calculated4. Rewrite4. Rewrite the chemical reaction by adding the water vapor into both side.5. N of each component in products are known6. Tdp = Tsat @ Pv (H2O) (Ideal gas mixture Pi/Pm = Ni/Nm)

Pv = 20.88 kPa Tdp = 60.9 oC

Example 15.3

8

Chemical EnergyChemical Energy from Processfrom Process

This session deal with the chemical energy within the molecules of a closed system that involve a chemical reaction. During a chemical reaction, some chemical bonds that bind the atoms into molecules are broken and new ones are formed. The chemical energy associated with this process is usually different for the reactants and products.

Combustion Chamber

1 kmol Hat 25oC, 1 atm

1 kmol O2at 25oC, 1 atm

H2O (g)at 25oC, 1 atm

241,820 kJ

Combustion Chamber

1 kmol Cat 25oC, 1 atm

1 kmol O2at 25oC, 1 atm

CO2at 25oC, 1 atm

393,520 kJ

FirstFirst LawLaw ofof ThermodynamicsThermodynamics

The first law of thermodynamics states that in anyclosed system, energy is conserved. Which means thatenergy cannot be created nor destroyed, but it can onlychange forms. Meaning:

∆Esys = 0 and ∆Eproducts = -∆Ereactants

The molecules of a closed system possess energy invarious forms such as sensible and latent energy,chemical energy, and nuclear energy. All of these formsmust balance out in the reactants and products to givethe system a net energy of zero.

9

Enthalpy ChangeEnthalpy Change

Enthalpy is the system we use to measure that change in energy of a closed system due to chemical bonds being broken. Reaction enthalpies are real physical quantities for which numeric values can be calculated or measured. In order to put the calculation into algebraic form, chemists use the defined equation:

∆H = Σ∆Hf (products) - Σ∆Hf (reactants)

The reaction enthalpy, which is the enthalpy change thatoccurs in the reaction, is always calculated as the sumof the enthalpies of the products minus the sum of theenthalpies of the reactants.

Standard Reference StateStandard Reference State

Since composition of a system at the end of a process isno longer the same as that at the beginning of theprocess, there’s a need to use a standard condition inwhich to make the measurements from. This standardcondition is called the standard reference point, whichare 2525ooC and 1 C and 1 atmatm..

The superscript “°” is used to indicate property values atthe standard state. The defined equation above, understandard conditions, becomes:

∆H° = Σ∆H°f(products) - Σ∆H°f(reactants)

10

EnthalpyEnthalpy ofof CombustionCombustion

The enthalpy of reaction in a combustion process is calledthe enthalpy of combustion (symbolized by hc). Thecalculation for an enthalpy of combustion is done for 1kmol (1 kg) of fuel is burned completely at a specifiedtemperature and pressure and can be expressed:

hc = Hprod – Hreact

Example 15.5 Evaluation of the Enthalpy of CombustionDetermine the enthalpy of combustion of liquid octane (C8H18) at 25oC and 1 atm, using enthalpy-of-formation data from table A-26. Assume the water in the products is in the liquid form.Solution

1. Write chemical reaction equation based on 1 1 kmolkmol of octane

CC88HH1818 +aa(O2 + 3.76 N2) 8 CO2 + 9H9H22O(l) O(l) + 3.76aN2 Balance of oxygen : get a = 12.52. Energy analysis hC = HP – HR

HP = ∑N = + hhffoo

__,P,P (NhNhff)

oo__

H2OH2O(NhNhff)oo

__CO2CO2

HR = ∑N = hhffoo

__,R,R (NhNhff)

oo__

C8H18C8H18

11

hhffoo

__,H2O,H2O

hhffoo

__,C8H18,C8H18

hhffoo

__,CO,CO22

Table ATable A--26 Enthalpy of Formation26 Enthalpy of Formation

= = --393,520 kJ/393,520 kJ/kmolkmol

= = --285,830 kJ/285,830 kJ/kmolkmol

= = -- 249,950 kJ/249,950 kJ/kmolkmol

Enthalpy of combustion = - 5,471,100 kJ/kmol C8H18

= -47,891 kJ/kg C8H18

Disscussion: This is the HHV of liquid C8H18 in Table A-27

Enthalpy of formation has 2 values : 1. for vapor vapor phase 2. for liquid vapor phase -the different = laten heat of vaporization

12

EnthalpyEnthalpy ofof FormationFormation

The enthalpy of formation is defined as theenthalpy of a substance at a specified state dueto its chemical composition. This property makesanalyzing easier because it represents chemicalenergy of an element or a compound at thestandard reference state.The property values are obtained by firstassigning all of the elements in its chemicallystable form at the standard reference state a value of zero (such N2, O2, N2, C).

EnthalpyEnthalpy ofof FormationFormation

So we can use this concept to find the enthalpy offormation of individual compounds by adding up theenthalpy for each reaction it takes to react some ofthe chemically stable elements to get the compound.

•Consider the formation of CO2 (a compound) from elements C and O2 at 25 oC, 1 atm. during SSSF process1st law: Qcv + ΣHi = ΣHe

Qcv = HP – HRQcv = - 393,520 kJHR = 0 ; elements @ ref. stateEnthalpy of Formation of COEnthalpy of Formation of CO22

= - 393,520 kJ/kmol

13

= the amount of heat released when a fuel is burned completely in SSSF process and the products returned to the state of the reactants…..

= absolute value of the enthalpy of combustion of the fuel:Heating value = |hC| kJ/kg fuel

Higher heating value, HHV,: H2O in products is in LIQUID form

Lower heating value, LHV,: H2O in products is in Vapor form

HHV = LHV + (mhfg)H2O (15.7)m = mass fo water in the products per unit mass of fuelhfg = the enthalpy of vaporization of water at the specified temp.

Heating Value of FuelHeating Value of Fuel

14

15.4 First Law Analysis of Reacting System15.4 First Law Analysis of Reacting System

SSSF Process: (see Chapter 4 the first law) SSSF Process: (see Chapter 4 the first law) ∆∆KE~0, KE~0, ∆∆PE~0PE~0

11)-15.........( -

) 1 ,25 is which state referencefor stand (scrip

)(

where

Products) e and Reactants, (ieqn. moleunit per 9)-(15 ......

eqn. mole of rateper 8)-5........(1

Pr

__

o

____

___

__

__

Ractod

iieecvcv

o

oof

ToTof

eecviicv

eeoo

cviioo

cv

HHhNhNWQ

atmC

hhhh

hhh

hNWhNQ

hnWhnQ

−=ΣΣ=−

−+=

∆+=

==Σ+=Σ+

Σ+=Σ+

→

Closed System Closed System ∆∆KE~0, KE~0, ∆∆PE~0PE~0

in tables Provided no is there

..(15....)}.........)(N{ or

})(N{ Then

from5.....)........(1)(

)(

_

____

___

RePr1212

121212

of

oof

oof

actod

u

vPhhhU

PVhhhU

PVHUUUWQ

UUWQ

Q

−−+=

−−+=

−=−+=

−+=

15

Example 15.6 First Law Analysis of Steady-Flow CombustionLiquid propane (C3H8) enters a combustion chamber at 25oC at a rate of 0.05 kg/min where it is mixed and burned with 50% excess air that enters the combustion chamber at 7oC, as shown in the figure. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 90% of carbon burns to CO2, with the remaining 10% forming CO. If the exit temperature of the combustion gases is 1,500 K, determine (a) the mass flow rate of the air and (b) the rate of heat transfer from the combustion chamber.

- StochiometicStochiometic combustioncombustion equation based on 1 1 kmolkmol of propaneCC33HH88 (l)+aa(O2 + 3.76 N2) 3CO2 + 4H4H22O O + 3.76aN2

Balance of oxygen : get a = 5

- 150% theoretical150% theoretical airair combustion equation based on 1 1 kmolkmol of propane with 90%C CO2 + 10%C CO

CC33HH88 (l)+(1.2*5)(1.2*5)(O2 + 3.76 N2) (0.9*3)(0.9*3)COCO22 + (0.1*3)CO+ (0.1*3)CO + 2.65O2 + 4H4H22O O + 28.2N2

(a) AF = mair/mfuel = (NM)air/(NM)fuel = ………………. = 25.53 kg air/kg fuelmass flow rate of air, mdot,air = mdot,fuelAF

= (0.05 kg fuel/min)(25.53 kg air/kg fuel) = 1.18 kg air/min Answer

Solution: Concepts1. Write chemical reaction equation based on 1 1 kmolkmol of propane

1.1 Theoretical Air-Fuel ratio (Stochiometic)1.2 with 150% theoretical air + incomplete burned CO AF mass flow rate of air

2. Energy balance: SSSF Q = HP – HR

16

) ( ),( where

0) WProducts, e and Reactants, (i

298K______

__cv

__

hhhhhh

hNhNQ

hNWhNQ

ooof

iieecv

eecviicv

=−+=

−=

===+=+

ΣΣ

ΣΣ

Substitute, Qcv = 363,880 kJ/kmol fuel 8,270 kJ/kg fuel Qdot = mdotQ = 6.89 kW

___

2

2

2

2

8

1500K_

280K_

298K__

)()( :Remark

47,517NA 8,669 110,530- 71,078 NA 9,364 393,520- 57,999NA 9,904 241,820- )(

47,073 8,141 8,669 0 49,292 8,150 8,682 0

NA NA 0 118,910- )( Substance

fgof

of

3

of

hlhgh

COCO

gOHNO

lHC(kJ/kmol) (kJ/kmol) (kJ/kmol) (kJ/kmol)

hhhh

+=

Assume air and combustion gases are ideal gases, get data from the property tables

Example 15.7 First Law Analysis of Combustion in a BombConstant volume tank contains 1 kmol of methane (CH4) gas and 3 kmol of O2 at 25oC and 1 atm. The contents of the tank are ignited, and the methane gas burns completely. If the final temperature is 1,000K, determine (a) the final pressure in the tank and (b) the heat transfer during this process.

Solution: Concepts1. Write chemical reaction equation Assume ideal gas for both reactants and products: PV = NRuT P2

2. Energy balance: SSSF Q = UP – UR

= (Hp-PPV) - (HR-PRV)

17

(a) Combustion equation:CH4 (g)+ 3O2 CO2 + 2H2O + O2

Nreact = 1 + 3 = 4 kmol, Nreact = 1 + 2 + 1 = 4 kmol, N1 = N2

Assume ideal gas for all gases:State 1 (Reactants) P1V = N1RuT1 (1)State 2 (Products) P2V = N2RuT2 (2)eqn(2)/eqn(1) P2 = (T2 /T1)*P1 = (1,000K/298K)(1atm) = 3.36 atm

answer

Answer CH kJ/kg 44,850 717,590/16 orCH kJ/kmol 717,590 Q- out fer heat trans ofAmount

-- aboveequation in substitute

stateeach at and gaseach of , of get valus ablesproperty t from

})(N{ then

gas Ideal

})(N{ or

})(N{ from

)( :0 W)( :lawFirst

4

412

___

___

_

____

___RePr1212

121212

====

>

−−+=

=

−−+=

−−+=−=

−==−+=

hhh

TRhhhU

TRvP

vPhhhU

PVhhhPVHU

UUQUUWQ

oof

uoo

f

u

oof

oof

actod

18

Adiabatic Flame Temperature = Maximum limit of combustion = Maximum limit of combustion gas temperature of each Air gas temperature of each Air –– Fuel mixture Fuel mixture

(Adiabatic Flame Temperature = Combustion Temperature)(Adiabatic Flame Temperature = Combustion Temperature)QQcvcv=0,W=0,Wcvcv=0 ,=0 ,∆∆KE=KE=∆∆PE=0 : 1PE=0 : 1stst lawlaw

To Calculate the adiabatic flame temperature, TTo Calculate the adiabatic flame temperature, TPP

1. Write the combustion equation1. Write the combustion equation2. Apply energy balance (12. Apply energy balance (1stst law)law)3. Solving by trial3. Solving by trial--andand--error technique by assume a value of Terror technique by assume a value of TP P get valuesget values……and and substitute in (2) substitute in (2) …….LHS = RHS ..if not try new T.LHS = RHS ..if not try new TP P ……. (in good procedure we can . (in good procedure we can interporateinterporate the former value to get the right value of Tthe former value to get the right value of TPP

15.5 Adiabatic Flame Temperature15.5 Adiabatic Flame Temperature

Poo

fPRoo

fR

eeii

eecviicv

hhhNhhhN

hNhN

hNWhNQ

)}({ )}({

______

__

__

−+=−+

=

+=+

ΣΣ

ΣΣ

ΣΣ

Trail and error procedureLHS - RHS = Error

T

Error

Ta

Tb

T2

Tc

-Ea -Eb +EcE = 0

Interporation -EaTa

+EcTc

0.0T2

-EbTb

E2T2

a

c

b

T2 = 342 oCmi =1.263 kg

• What is your first guess of T• What should be the 2nd trial.• How about the 3rd, 4th ......• When/how to interporate

19

Example 15.8 Adiabatic Flame Temperature in Steady CombustionLiquid octane (C8H18) enters the combustion chamber of a gas turbine steadily at 1 atm and 25oC, and it is burns with air that enters the combustion chamber at the same state, as shown in the figure. Determine the adiabatic flame temperature for (a) complete combustion at 100% theoretical air, (b) complete combustion at 400% theoretical air and (c) incomplete combustion (some CO in the products) with 90% theoretical air.

Asumptions: 1. SSSF process2. Adiabatic3. No work4. ∆KE=∆PE=05. Air and combustion gases are ideal gas

1. Combustion equation equation based on 1 1 kmolkmol of octaneC8H18 +12.5(O2 + 3.76 N2) 8 CO2 + 9H2O + 47N2

at TR = 298K at TP = ?

2. Energy balance: HR = HP

Poo

fPRofR

Ro

Poo

fPRoo

fR

eeii

eecviicv

hhhNhN

hh

hhhNhhhN

hNhN

hNWhNQ

)}({ }{

0)( state referenceat are reactants

)}({ )}({

____

__

______

__

__

−+=

=−

−+=−+

=

+=+

ΣΣ

ΣΣ

ΣΣ

ΣΣ

Q

20

(b) 400% theoretical air : combustion equation:-. Combustion equation equation based on 1 1 kmolkmol of octane

C8H18 +4.0x12.5(O2 + 3.76 N2) 8CO2 + 9H2O + (3.0x12.5)O2 + 4.0x47N2 at TR = 298K at TP = ?

by trial and error of TPAdiabatic flame temperature = 962 K ……………………..answer

(C) 90% theoretical air : combustion equation:-. Combustion equation equation based on 1 1 kmolkmol of octane

C8H18 +0.9x12.5(O2 + 3.76 N2) aCO2 +bCO + 9H2O + 0.9x47N2 at TR = 298K at TP = ?

C and O balance a = 5.5 and b = 2.5 by trial and error of TPAdiabatic flame temperature = 2,236K ……………………..answer

Trail and error procedureLHS - RHS = Error

T

Error

Ta

Tb

T2

Tc

-Ea -Eb +EcE = 0

Interporation -EaTa

+EcTc

0.0T2

-EbTb

E2T2

a

c

b

T2 = 342 oCmi =1.263 kg

• What is your first guess of T• What should be the 2nd trial.• How about the 3rd, 4th ......• When/how to interporate

21

2

22

222188

)}669,80(47{

)}904,9)820,241(9{)364,9)520,393(8{)950,249(1

)}({)}({)}({ )(

)}({ }{

)()( :Remark

............ ............. 8,669 110,530- ............ ............ 9,364 393,520- ............ ............ 9,904 241,820- )(

............ ............ 8,669 0 ............ ............ 8,682 0

NA NA 0 249,950- )( Substance

__

__

__________

____

___

2

2

2

2

18

yyyyK_

xxxxK_

298K__

N

OHCO

Noo

fOHoo

fCOoo

fHCof

Poo

fPRofR

fgof

of

8

of

h

hhx

hhhNhhhNhhhNNh

hhhNhN

hlhgh

COCO

gOHNO

lHC(kJ/kmol) (kJ/kmol) (kJ/kmol) (kJ/kmol)

hhhh

−++

−+−+−+−=−

−++−++−+=

−+=

+=

ΣΣ

Assume air and combustion gases are ideal gases, get data from the property tables

(a) Adiabatic flame temperature = 2,395 K ………………..answer