Chemical reaction .pdf

8/20/2019 Chemical reaction .pdf

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Reference : Thermodynamics: An Engineering Approach, 5th edition

by Yunus A. Çengel and Michael A. Boles

Chemical Reactions:

Combustion reaction


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The combustion process is a chemical reaction whereby fuel is oxidized and energy

is released.

Fuels are usually composed of some compound or mixture containing carbon, C, andhydrogen, H2.

Examples of hydrocarbon fuels are

CH4 Methane

C8H18 OctaneCoal Mixture of C, H2, S, O2, N2 and non-combustibles

Initially, we shall consider only those reactions that go to completion. The

components prior to the reaction are called reactants and the components after the

reaction are called products.


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Reactants Products

C O CO

H O H O

2 2

2 2 2

1

2

For complete or stoichiometric combustion, all carbon is burned to carbon dioxide

(CO2) and all hydrogen is converted into water (H2O). These two complete

combustion reactions are as follows:

Example 15-1

A complete combustion of octane in oxygen is represented by the balanced combustionequation. The balanced combustion equation is obtained by making sure we have the

same number of atoms of each element on both sides of the equation. That is, we

make sure the mass is conserved.

C H A O B CO D H O8 18 2 2 2

Note we often can balance the C and H for complete combustion by inspection.

C H A O CO H O8 18 2 2 28 9


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The amount of oxygen is found from the oxygen balance. It is better to conserve

species on a monatomic basis as shown for the oxygen balance.

O A

A

: ( ) ( ) ( )

.

2 8 2 9 1

12 5

C H O CO H O8 18 2 2 212 5 8 9 .

Note: Mole numbers are not conserved, but we have conserved the mass on a total

basis as well as a specie basis.

The complete combustion process is also called the stoichiometric combustion, andall coefficients are called the stoichiometric coefficients.

In most combustion processes, oxygen is supplied in the form of air rather than pure

oxygen.

Air is assumed to be 21 percent oxygen and 79 percent nitrogen on a volume basis.For ideal gas mixtures, percent by volume is equal to percent by moles. Thus, for

each mole of oxygen in air, there exists 79/21 = 3.76 moles of nitrogen. Therefore,

complete or theoretical combustion of octane with air can be written as

C H O N

CO H O N

8 18 2 2

2 2 2

12 5 3 76

8 9 47

. ( . )


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Air-Fuel Ratio

Since the total moles of a mixture are equal to the sum of moles of each component,

there are 12.5(1 + 3.76) = 59.5 moles of air required for each mole of fuel for the

complete combustion process.

Often complete combustion of the fuel will not occur unless there is an excess of air

present greater than just the theoretical air required for complete combustion.

To determine the amount of excess air supplied for a combustion process, let us define

the air-fuel ratio AF as

AF kmol air

kmol fuel

Thus, for the above example, the theoretical air-fuel ratio is

AF kmol air kmol fuel

th

12 5 1 3761

595. ( . ) .


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On a mass basis, the theoretical air-fuel ratio is

AF kmol air

kmol fuel

kg air

kmol air

kg fuel

kmol fuel

kg air

kg fuel

th

595

2897

8 12 18 1

1512

.

.

[ ( ) ( )]

.

Percent Theoretical and Percent Excess Air

In most cases, more than theoretical air is supplied to ensure complete combustion and

to reduce or eliminate carbon monoxide (CO) from the products of combustion. The

amount of excess air is usually expressed as percent theoretical air and percent excess

air.

Percent theoretical air AF

AF

actual

th

100%

Percent excess air AF AF

AF

actual th

th

100%


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Show that these results may be expressed in terms of the moles of oxygen only as

Percent theoretical air N

N

O actual

O th

2

2

100%

Percent excess air N N

N

O actual O th

O th

2 2

2

100%

Example 15-2

Write the combustion equation for complete combustion of octane with 120 percent

theoretical air (20 percent excess air).

C H O N

CO H O O N

8 18 2 2

2 2 2 2

12 12 5 376

8 9 0 2 12 5 12 47

. ( . ) ( . )

( . )( . ) . ( )

Note that (1)(12.5)O2 is required for complete combustion to produce 8 kmol of

carbon dioxide and 9 kmol of water; therefore, (0.2)(12.5)O2 is found as excess

oxygen in the products.

C H O N

CO H O O N

8 18 2 2

2 2 2 2

12 12 5 376

8 9 2 5 12 47

. ( . ) ( . )

. . ( )


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Second method to balance the equation for excess air (see the explanation of this

technique in the text) is:

C H A O N

CO H O A O A N

O A A

A

th

th th

th th

th

8 18 2 2

2 2 2 2

12 376

8 9 0 2 12 376

12 2 8 2 9 1 0 2 2

12 5

. ( . )

. . ( . )

: . ( ) ( ) ( ) . ( )

.

Incomplete Combustion with Known Percent Theoretical Air

Example 15-3

Consider combustion of C8H18 with 120 % theoretical air where 80 % C in the fuel goes

into CO2.

C H O N CO CO H O X O N

8 18 2 2

2 2 2 2

12 12 5 37608 0 2 9 12 47

. ( . ) ( . ). (8) . (8) . ( )


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O balance gives

O X

X

: . ( . )( ) . (8)( ) . (8)( ) ( ) ( )

.

12 12 5 2 08 2 0 2 1 9 1 2

33

Why is X > 2.5?

Then the balanced equation is

C H O N

CO CO H O O N

8 18 2 2

2 2 2 2

12 12 5 376

6 4 16 9 33 12 47

. ( . ) ( . )

. . . . ( )

Combustion Equation When Product Gas Analysis Is Known

Example 15-4

Propane gas C3H8 is reacted with air such that the dry product gases are 11.5 percent

CO2, 2.7 percent O2, and 0.7 percent CO by volume. What percent theoretical air was

supplied? What is the dew point temperature of the products if the product pressure is100 kPa?

We assume 100 kmol of dry product gases; then the percent by volume can be

interpreted to be mole numbers. But we do not know how much fuel and air were

supplied or water formed to get the 100 kmol of dry product gases.


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X C H A O N

CO CO O B H O A N

3 8 2 2

2 2 2 2

376

115 0 7 2 7 376

( . )

. . . ( . )

The unknown coefficients A, B, and X are found by conservation of mass for each

species.

C X X

H X B B

O A

B A

N A

: ( ) . ( ) . ( ) .

: (8) ( ) .

: ( ) . ( ) . ( )

. ( ) ( ) .

: ( . ) .

3 115 1 0 7 1 4 07

2 16 28

2 115 2 0 7 1

2 7 2 1 22 69

376 85312

The balanced equation is

4 07 22 69 376

115 0 7 2 7 16 28 8531

3 8 2 2

2 2 2 2

. . ( . )

. . . . .

C H O N

CO CO O H O N

Second method to find the coefficient A:

Assume the remainder of the 100 kmol of dry product gases is N2.

kmol N 2 100 115 0 7 2 7 851 ( . . . ) .


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Then A is

A fairly good check 851

37622 65

.

.. ( )

These two methods don’t give the same results for A, but they are close.

What would be the units on the coefficients in the balanced combustion equation?

Later in the chapter we will determine the energy released by the combustion process

in the form of heat transfer to the surroundings. To simplify this calculation it is

generally better to write the combustion equation per kmol of fuel. To write the

combustion equation per unit kmol of fuel, divide by 4.07:C H O N

CO CO O H O N

3 8 2 2

2 2 2 2

557 376

2 83 017 0 66 4 0 20 96

. ( . )

. . . . .

The actual air-fuel ratio is

AF kmol air

kg air

kmol air

kmol fuel kg fuel

kmol fuel

kg air

kg fuel

actual

(5. )( . ) .

[ ( ) ( )]

.

57 1 376 28 97

1 3 12 8 1

17 45


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The theoretical combustion equation is

C H O N

CO H O N

3 8 2 2

2 2 2

5 376

3 4 0 1880

( . )

. .

The theoretical air-fuel ratio is

AF

kmol air kg air

kmol air

kmol fuel kg fuel

kmol fuel

kg air

kg fuel

th

(5)( . ) .

[ ( ) ( )]

.

1 376 28 97

1 3 12 8 1

1566

The percent theoretical air is

Percent theoretical air

AF

AF

actual

th

100%

17 45

1566100 111%

.

.


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or

Percent theoretical air

N

N

O actual

O th

2

2

100%

5575

100 111%.

The percent excess air is

Percent excess air

AF AF

AF

actual th

th

100%

17 45 1566

1566100 11%

. .

.

Dew Point Temperature

The dew point temperature for the product gases is the temperature at which the

water in the product gases would begin to condense when the products are cooled at

constant pressure. The dew point temperature is equal to the saturation temperature

of the water at its partial pressure in the products.


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T T P y P

y N

N

v

v

e

dp sat v products

water

products

at

Example 15-5

Determine dew point temperature of the products for Example 15-4.

products

dp sat

4 0.13982.83 0.17 0.66 4 20.96

0.1398(100 )

13.98

at13.98kPa

=52.31

v

v v

o

y

P y P kPa

kPa

T T

C

What would happen if the product gases are cooled to 100oC or to 30oC?


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Example 15-6

An unknown hydrocarbon fuel, C X HY is reacted with air such that the dry product

gases are 12.1 percent CO2, 3.8 percent O2, and 0.9 percent CO by volume. What is

the average makeup of the fuel?

We assume 100 kmol (do you have to always assume 100 kmol?) of dry product

gases; then the percent by volume can be interpreted to be mole numbers. We do

not know how much air was supplied or water formed to get the 100 kmol of dry

product gases, but we assume 1 kmol of unknown fuel.

C H A O N CO CO O B H O D N

X Y

( . ). . .

2 2

2 2 2 2

376121 0 9 38

The five unknown coefficients A, B, D, X , and Y are found by conservation of mass

for each species, C , H , O, and N plus one other equation. Here we use the

subtraction method for the nitrogen to generate the fifth independent equation for the

unknowns.C H A O N

CO CO O B H O D N

X Y

( . )

. . .

2 2

2 2 2 2

376

121 0 9 38


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The unknown coefficients A, B, D, X , and Y are found by conservation of mass for

each species. Here we assume the remainder of the dry product gases is nitrogen.

N D2 100 121 0 9 38 832: ( . . . ) .

O A D

O A B

B

C X

X

H Y B

Y

2

376

832

376

2213

2 121 2 0 9 1 38 2 1

1154

1 121 1 0 9 1

130

1 2

2308

:

.

.

.

.

: ( ) ( . )( ) ( . )( ) ( . )( ) ( )

.

: ( ) . ( ) ( . )( )

.

: ( ) ( )

.

The balanced equation is

C H O N

CO CO O H O N

13 23 08 2 2

2 2 2 2

2213 376

121 0 9 38 1154 832

. . ( . )

. . . . .


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Enthalpy of Formation

When a compound is formed from its elements (e.g., methane, CH4, from C and H2),

heat transfer occurs. When heat is given off, the reaction is called exothermic.

When heat is required, the reaction is called endothermic. Consider the following.

The reaction equation is

C H CH 2 2 4The conservation of energy for a steady-flow combustion process is

E E

Q H H

Q H H

in out

net

net

Reactants Products

Products Reactants


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Q N h N h

Q h h h

net e e i i

net CH C H

Products Reactants

1 1 24 2

( )

A common reference state for the enthalpies of all reacting components is

established as

The enthalpy of the elements or their stable compounds is defined to

be ZERO at 25oC (298 K) and 1 atm (or 0.1 MPa).

Q h

h

net CH

CH

1 1 0 2 04

4

( ( ) ( ))

h f o

h f o

This heat transfer is called the enthalpy of formation for methane, . The

superscript (o) implies the 1 atm pressure value and the subscript (f ) implies 25oC

data, is given in Table A-26.

During the formation of methane from the elements at 298 K, 0.1 MPa, heat is given

off (an exothermic reaction) such that

Q h kJ

kmol net f

o

CH CH

4

4

74 850,


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The enthalpy of formation is tabulated for typical compounds. The enthalpy of

formation of the elements in their stable form is taken as zero. The enthalpy of

formation of the elements found naturally as diatomic elements, such as nitrogen,

oxygen, and hydrogen, is defined to be zero. The enthalpies of formation for several

combustion components are given in the following table.

h f o

h f oSubstance Formula M kJ/kmol

Air 28.97 0

Oxygen O2 32 0

Nitrogen N2 28 0

Carbon dioxide CO2 44 -393,520

Carbon monoxide CO 28 -110,530

Water (vapor) H2Ovap 18 -241,820

Water (liquid) H2Oliq 18 -285,830

Methane CH4 16 -74,850

Acetylene C2H2 26 +226,730

Ethane C2H6 30 -84,680

Propane C3H8 44 -103,850

Butane C4H10 58 -126,150

Octane (vapor) C8H18 114 -208,450

Dodecane C12H26 170 -291,010


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The enthalpies are calculated relative to a common base or reference called the

enthalpy of formation. The enthalpy of formation is the heat transfer required to form

the compound from its elements at 25oC (77 F) or 298 K (537 R), 1 atm. The

enthalpy at any other temperature is given as

h h h h f o

T

o

( )Here the term is the enthalpy of any component at 298 K. The enthalpies at the

temperatures T and 298 K can be found in Tables A-18 through A-25. If tables are

not available, the enthalpy difference due to the temperature difference can be

calculated from

h o

Based on the classical sign convention, the net heat transfer to the reacting system is

Q H H

N h h h N h h h

net P R

e f

o

T

o

e i f

o

T

o

i

[ ( )] [ ( )]Products Reactants

In an actual combustion process, is the value of Qnet positive or negative?


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Example 15-7

Butane gas C4H10 is burned in theoretical air as shown below. Find the net heat

transfer per kmol of fuel.

Balanced combustion equation:

C H O N

CO H O N

4 10 2 2

2 2 2

65 376

4 5 24 44

. ( . )

.

The steady-flow heat transfer is

Q H H

N h h h N h h h

net P R

e f

o

T

o

e i f

o

T

o

i



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Reactants: T R = 298 K

h f o

hT ho N h h hi f

o

T

o

i[ ( )] Comp N ikmol/kmol

fuel

kJ/kmol kJ/kmol kJ/kmol kJ/kmol fuel

C4H10 1 -126,150 -- -- -126,150O

2 6.5 0 8,682 8,682 0

N2

24.44 0 8,669 8,669 0

H N h h h

kJ kmolC H

R i f

o

T

o

i

[ ( )]

,

Reactants

126 1504 10Products: T P = 1000 K

h f o

hT ho N h h he f

o

T

o

e[ ( )] Comp N ekmol/kmol

fuel

kJ/kmol kJ/kmol kJ/kmol kJ/kmol fuel

CO2

4 -393,520 42,769 9,364 -1,440,460

H2O 5 -241,820 35,882 9,904 -1,079,210

N2

24.44 0 30,129 8,669 +524,482


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H N h h h

kJ

kmol C H

P e f

o

T

o

e

[ ( )]

, ,

Products

19951884 10

Q H H kJ

kmol C H

net P R

1869 0384 10

, ,

Adiabatic Flame Temperature

The temperature the products have when a combustion process takes placeadiabatically is called the adiabatic flame temperature.

Example 15-8

Liquid octane C8

H18

(liq) is burned with 400 percent theoretical air. Find the adiabatic

flame temperature when the reactants enter at 298 K, 0.1 MPa, and the products

leave at 0.1MPa.


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The combustion equation is

C H O N

CO O H O N

8 18 2 2

2 2 2 2

4 12 5 376

8 37 5 9 188

( . ) ( . )

.The steady-flow heat transfer is

Q H H

N h h h N h h h

Adiabatic Combustion

net P R

e f

o

T

o

e i f

o

T

o

i

[ ( )] [ ( )]

( )

Products Reactants

0

Thus, H P = H R for adiabatic combustion. We need to solve this equation for T P .


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Since the temperature of the reactants is 298 K, ( )i = 0,h hT o

H N h

kJ kmol C H

R i f

o

i

Reactants

1 249 950 4 125 0 4 125 376 0

249 9504 10

( , ) ( . )( ) ( . )( . )( )

,

Since the products are at the adiabatic flame temperature, T P > 298 K

2

2

2

2

2 2

2 2

Products

, ,

, ,

4 10

[ ( )]

8( 393, 520 9364)

9( 241, 820 9904)

37.5(0 8682)

188(0 8669)

( 7, 443,845 8 9

37.5 188 )

P

P

P

P

P

P P

P P

o o

P e f T e

T CO

T H O

T O

T N

T CO T H O

T O T N

H N h h h

h

h

h

h

h h

kJ h h

kmolC H


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Thus, setting H P = H R yields

N h h h h he T eoducts

T CO T H O T O T N P P P P P ,

Pr

, , , ,.

, ,

8 9 37 5 188

7 193 895

2 2 2 2

To estimate T P , assume all products behave like N2 and estimate the adiabatic flametemperature from the nitrogen data, Table A-18.

242 5 7 193 895

29 6655

985

2

2

2

. , ,

, .

,

,

h

h kJ

kmol N

T K

T N

T N

p

P

P

Because CO2 and H2O are triatomic gases and have specific heats greater than

diatomic gases, the actual temperature will be somewhat less than 985 K. Try T P =

960 K and 970K.h K 960 h K 970

N he T e P ,Produts

N e

CO2 8 40,607 41,145

H2O 9 34,274 34,653

O2

37.5 29,991 30,345

N2

188 28,826 29,151

7,177,572 7,259,362


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Interpolation gives: T P = 962 K.

Example 15-9

Liquid octane C8H18(liq) is burned with excess air. The adiabatic flame temperature is

960 K when the reactants enter at 298 K, 0.1 MPa, and the products leave at0.1MPa. What percent excess air is supplied?

Let A be the excess air; then combustion equation is

C H A O N

CO A O H O A N 8 18 2 2

2 2 2 2

1 12 5 376

8 12 5 9 1 12 5 376

( )( . ) ( . )

. ( )( . )( . )


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28


Q H H

N h h h N h h h

Adiabatic combustion

net P R

e f

o

T

o

e i f

o

T

o

i

[ ( )] [ ( )]

( )

Products Reactants

0Here, since the temperatures are known, the values of are known. The product

gas mole numbers are unknown but are functions of the amount of excess air, A.

The energy balance can be solved for A.

hT P

A 3

Thus, 300 percent excess, or 400 percent theoretical, air is supplied.

Example 15-10

Tabulate the adiabatic flame temperature as a function of excess air for the complete

combustion of C3H8 when the fuel enters the steady-flow reaction chamber at 298 K

and the air enters at 400 K.

The combustion equation is

C H A O N

CO A O H O A N

3 8 2 2

2 2 2 2

1 376

3 5 4 1 376

( )(5) ( . )

( )(5)( . )


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where A is the value of excess air in decimal form.


Q H H

N h h h N h h h

Adiabatic combustion

net P R

e f

o

T

o

e i f

o

T

o

i

[ ( )] [ ( )]( )

Products Reactants

0

Percent Excess

Air

Adiabatic Flame

Temp. K

0 2459.3

20 2191.9

50 1902.5

100 1587.1

217 1200


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Enthalpy of Reaction and Enthalpy of Combustion

When the products and reactants are at the same temperature, the enthalpy of

reaction h R , is the difference in their enthalpies. When the combustion is assumed

to be complete with theoretical air supplied the enthalpy of reaction is called the

enthalpy of combustion h C . The enthalpy of combustion can be calculated at anyvalue of the temperature, but it is usually determined at 25oC or 298 K.

h H H when T T C K

N h N h

C P R P R

o

e f

o

e i f

o

i

25 298

Products Reactants

Heating Value

The heating value, HV, of a fuel is the absolute value of the enthalpy of combustion

or just the negative of the enthalpy of combustion.

HV hC

The lower heating value, LHV, is the heating value when water appears as a gas in

the products.

LHV h h with H O in productsC C gas 2


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The lower heating value is often used as the amount of energy per kmol of fuel

supplied to the gas turbine engine.

The higher heating value, HHV, is the heating value when water appears as a liquid

in the products.

HHV h h with H O in productsC C liquid 2

The higher heating value is often used as the amount of energy per kmol of fuel

supplied to the steam power cycle.

See Table A-27 for the heating values of fuels at 25

o

C. Note that the heating valuesare listed with units of kJ/kg of fuel. We multiply the listed heating value by the molar

mass of the fuel to determine the heating value in units of kJ/kmol of fuel.

The higher and lower heating values are related by the amount of water formed

during the combustion process and the enthalpy of vaporization of water at the

temperature. HHV LHV N h H O fg H O 2 2


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Example 15-11

The enthalpy of combustion of gaseous octane C8H18 at 25oC with liquid water in the

products is -5,500,842 kJ/kmol. Find the lower heating value of liquid octane.

8 18 8 18 2 2

2

8 18 8 18 2

8 18

5,500,842 9 (44,010)

5,104,752

gas gasC H C H H O fg H O LHV HHV N hkJ kmol H O kJ

kmol C H kmol C H kmol H O

kJ

kmol C H

8 18 8 18 8 18

8 18

8 18

(5,104752 41,382)

5,063,370

liq gasC H C H fg C H

liq

LHV LHV h

kJ

kmol C H

kJ

kmol C H

Can you explain why LHVliq< LHVgas?

Cl d S A l i


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Closed System Analysis

Example 15-12

A mixture of 1 kmol C8H18 gas and 200 percent excess air at 25oC, 1 atm, is burned

completely in a closed system (a bomb) and is cooled to 1200 K. Find the heattransfer from the system and the system final pressure.

Apply the first law closed system:

A th t th t t d d t id l th


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Assume that the reactants and products are ideal gases; then

PV NR T u

The balanced combustion equation for 200 percent excess (300 percent theoretical)

air isC H O N

CO O H O N

8 18 2 2

2 2 2 2

3 12 5 376

8 25 9 141

( )( . ) ( . )

Q 8 393 520 53 848 9364 8 314 1200( ( ))


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35

Q

h h

kJ

kmol C H

net CO

H O

O

N

K

o

C H

O

N

8 393 520 53 848 9364 8 314 1200

9 241820 44 380 9904 8 314 1200

25 0 38 447 8682 8 314 1200

141 0 36 777 8669 8 314 1200

1 208 450 8 314 298

37 5 0 8682 8682 8 314 298

141 0 8669 8669 8 314 298

112 10

2

2

2

2

8 18

2

2

298

6

8 18

( , , . ( ))

( , , . ( ))

( , . ( ))

( , . ( ))

( , . ( ))

. ( . ( ))

( . ( ))

.

To find the final pressure, we assume that the reactants and the products are ideal-

gas mixtures.

PV N R T

PV N R T

u

u

1 1 1 1

2 2 2 2

h t t 1 i th t t f th i t f th t t b f th b ti


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where state 1 is the state of the mixture of the reactants before the combustion

process and state 2 is the state of the mixture of the products after the combustion

process takes place. Note that the total moles of reactants are not equal to the total

moles of products.

PV PV

N R T N R T

u

u

2 2

1 1

2 2

1 1

but V 2 = V 1.

S d L A l i f R ti S t


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Second Law Analysis of Reacting Systems

Second law for the open system

The entropy balance relations developed in Chapter 7 are equally applicable to both

reacting and nonreacting systems provided that the entropies of individual

constituents are evaluated properly using a common basis.

Taking the positive direction of heat transfer to be to the system, the entropy balance

relation can be expressed for a steady-flow combustion chamber asQ

T S S S S kJ k k

k

gen CV React Prod ( / )

For an adiabatic steady flow process the entropy balance relation reduces to


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For an adiabatic, steady-flow process, the entropy balance relation reduces to

S S S gen adiabatic, Prod React 0

The third law of thermodynamics states that the entropy of a pure crystalline

substance at absolute zero temperature is zero. The third law provides a common

base for the entropy of all substances, and the entropy values relative to this baseare called the absolute entropy.

The ideal-gas tables list the absolute entropy values over a wide range of

temperatures but at a fixed pressure of P o = 1 atm. Absolute entropy values at other

pressures P for any temperature T are determined from

s T P s T P R P

P kJ kmol K o o u

o

( , ) ( , ) ln [ / ( )]

For component i of an ideal-gas mixture, the absolute entropy can be written as

s T P s T P R

y P

P kJ kmol K i i io

o u

i m

o( , ) ( , ) ln [ / ( )]

where P i is the partial pressure, y i is the mole fraction of the component, and P m is

the total pressure of the mixture in atmospheres.

Example 15 13


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Example 15-13

A mixture of ethane gas C2H6 and oxygen enters a combustion chamber at 1 atm,

25oC. The products leave at 1 atm, 900 K. Assuming complete combustion, does the

process violate the second law?

The balanced combustion equation isC H O CO H O2 6 2 2 235 2 3 .

The mole fractions for the reactants and the products are


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The mole fractions for the reactants and the products are

y

y

y

y

C H

O

CO

H O

2 6

2

2

2

1

1 3 5

1

4 5

3 5

1 3 5

35

4 52

2 3

2

5

3

2 3

3

5

. .

.

.

.

.

Now calculate the individual component entropies.

For the reactant gases:


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S N s

kJ

kmol K

i i

C H

React

Reactants

1 242 0 35 2071

9669

2 6

( . ) . ( . )

.

For the product gases:


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For the product gases:

S N s


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S N s

kJ

kmol K

e e

C H

Prod

Products

2 2712 3 232 6

12402

2 6

( . ) ( . )

.

The entropy change for the combustion process is

S S kJ

kmol K

kJ kmol K

C H

C H

Prod React

( . . )

.

1240 2 966 9

2733

2 6

2 6

Now to find the entropy change due to heat transfer with the surroundings. The

steady-flow conservation of energy for the control volume is

Q H H N h h h N h h h

net sys P R

e f

o

T

o

e i f

o

T

o

i


Q CO 2 393 520 37 405 9364( )


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Q

h h

kJ

kmol C H

net sys CO

H O

K

o

C H

O

2 393 520 37 405 9364

3 241820 31828 9904

1 214 820

35 0 8682 8682

1306 10

2

2

2 6

2

298

6

2 6

( , , )

( , , )

( , )

. ( )

.

Q

T

Q

T

kJ kmolC H

K

kJ

kmol C H K

k

k

netsys

o

1306 10

25 273

4 383

6

2 6

2 6

.

( )

,

The entropy generated by this combustion process is


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The entropy generated by this combustion process is

Since Sgen, or Snet , is ≥ 0, the second law is not violated.

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