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Solution Manual for Combustion 4th Edition Irvin Glassman Department of Mechanical and Aerospace Engineering Princeton University Princeton, New Jersey Richard Yetter Department of Mechanical and Nuclear Engineering The Pennsylvania State University University Park, PA This solution manual is a revision to the solution manual for "Combustion, 3rd. Ed." that was written in its final form by Prof. Queiroz and Dr. Black of Brigham Young University.

1 © 2008 Glassman and Yetter.

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A Solution Manual for: COMBUSTION. 4th Ed. by I. Glassman, and R.A. Yetter CHAPTER 1 Problem 1 Calculate the heat of reaction when methane and air in stoichiometric proportions are brought into a calorimeter at 500 K. The product composition is brought to the ambient temperature (298 K) by the cooling water. The pressure in the calorimeter is assumed to remain at 1 atm, but the water formed has condensed. Solution: The heat of reaction can be calculated using Eqn. (10) in the text. However, as also suggested in the textbook, this equation can be simplified for the general case when H 0T0 = H 0298 . The modified equation becomes: −Q = ΔH =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

⎤ − ∑ n j ⎡ H 0T − H 0298 ⎦ j react ⎣ 1

298 i

)

298

+ ( ΔH 0f )

298

⎤ ⎦j

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. The general equation for the complete combustion of one mole of methane in air is: CH 4 + a th ( O 2 + 3.76N 2 ) → a CO 2 + b H 2 O + c N 2

where ath is a coefficient describing how much air is needed for stoichiometric combustion. Balancing the above reaction one has: CH 4 + 2 ( O 2 + 3.76N 2 ) → CO 2 + 2 H 2 O + 7.52 N 2

In order to apply Eqn. (1) to the above reaction, the following properties from Tables in the textbook are needed (note that for this problem T2 = 298 K and T1 = 500 K): gases

CH4 O2 N2 H2O(liq) CO2

ΔH 0f (kJ/mol)

H 0T1 − H 0298 (kJ/mol)

H 0T2 − H 0298 (kJ/mol)

−74.90 0.0 0.0 −286.04 −393.77

8.21 6.09 5.92 0.0 0.0

0.0 0.0 0.0 0.0 0.0

Equation (1) then becomes: 2 © 2008 Glassman and Yetter.

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−Q = 1( −393.77 + 0.0 )CO + 2 ( −286.04 + 0.0 )H O + 7.52 ( 0.0 + 0.0 ) N − 1( −74.90 + 8.21)CH 2

2

−2 ( 0.0 + 6.09 )O − 7.52 ( 0.0 + 5.92 ) N 2

2

4

2

Q=955.85 kJ

or

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Problem 2 Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ratio of 0.5. For this problem assume there is no dissociation of the stable products formed. All reactants are at 298 K and the system operates at 1-atm pressure. Compare your results with those given in the graphs in the text. Explain any differences. Solution: We will assume that adiabatic conditions exist and that the water in the products is in the vapor phase. In this case, Eqn. (10) in the text can be used in an iterative approach to solve for the adiabatic flame temperature. However, this equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified, the solution approach is to find T2 such that the above equality will be satisfied. The general equation for the complete combustion of one mole of octane with excess air is: C8 H18( liq ) + ( m )( a th )( O 2 + 3.76N 2 ) → a CO 2 + b H 2 O + c O 2 + d N 2 where ath is the coefficient describing how much air is needed for stoichiometric combustion and m is the coefficient used to represent the amount of excess air, respectively. For the case when c = 0 and m = 1 (no excess air), a simple mass balance yields ath = 12.5. Therefore, for an excess air of 100% (m = 2) or an equivalence ratio of 0.5, the chemical reaction becomes: C8 H18( liq ) + 25 ( O2 + 3.76N 2 ) → 8 CO 2 + 9 H 2 O + 94 N 2 + 12.5 O 2 Since the conditions of the reactants are specified (T1 = 298 K), Table 2 in the text and Appendix A provide the following thermodynamic properties: gases

C8H18(liq) O2 N2

ΔH 0f (kJ/mol)

H 0T1 − H 0298 (kJ/mol)

−250.12 0.0 0.0

0.0 0.0 0.0

The enthalpy of the reactants (Hr, given by the right-hand side of Eqn (1)) can be calculated as: H r = 1( −250.12 + 0.0 )C H + 11( 0.0 + 0.0 )O + 41.36 ( 0.0 + 0.0 ) N = −250.12kJ 8

18

2

2

For the products it follows from thermodynamic properties in the text that: 4 © 2008 Glassman and Yetter.

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ΔH 0f (kJ/mol) 0.0 0.0 −242.00 −393.77

gases

O2 N2 H2O(vap) CO2

Therefore, the expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1) in terms of the enthalpy of the individual gases in the products becomes let ΔH = H 0T2 − H 0298 :

(

)

(

) (

)

(

)

(

H p = 8 −393.77 + ΔH CO2 + 9 −242.00 + ΔH H2O + 12.5 0.0 + ΔH O2 + 94 0.0 + ΔH N2

)

Simplifying the above equation yields: 8 ΔH CO2 + 9 ΔH H2O + 12.5ΔH O2 + 94ΔH N2 = 5, 078.04 A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 123.5 = 8 + 9 + 12.5 + 94): 123.5 ΔH N2 = 5,078.04

so

ΔH N2 = 41.12 kJ/mol

therefore T2 = 1577

We will then assume that T2 is between 1500 and 1600 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ): T2 1500 1550 1506

ΔH CO2 (kJ/mol) 61.76 64.69 62.11

Interpolate

ΔH H2O 48.13 50.51 48.41

ΔH O2 40.64 42.48 40.86

1500 T2 1550

ΔH N2 38.43 40.18 38.64

Hp – Hr (kJ) -30.37 202.00 -2.55

-30.37 00 202.00

T2 − 1500 50

30.37 232.37

=

so

T2 = 1506 K

The enthalpy difference for this temperature was also calculated in the above table to show that 5 © 2008 Glassman and Yetter.

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once you have a good idea of the temperature interval where the adiabatic temperature should be in, linear interpolation does a good job of predicting the temperature. Comparing the results for the adiabatic flame temperature found above to that found using Fig. 3 on page 22 of the textbook is done as follows: Enthalpy per gram C8H18

=

−250.12 96 + 18

=

2.194 kJ/gm

H/C =

18 = 2.25 8

With these values Fig. 3 yields T2 ≅ 1505 K, which is very close to the calculated adiabatic temperature.

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Problem 3

Carbon monoxide is oxidized to carbon dioxide in an excess of air (1 atm) in an afterburner so that the final temperature is 1300 K. Under the assumption of no dissociation determine the airfuel ratio required. Report the results on both a molar and mass basis. For the purposes of this problem assume that air has the composition of 1 mole of oxygen to 4 moles of nitrogen. The carbon monoxide and air enter the system at 298 K. Solution: We will assume that adiabatic conditions exist. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1300 K. This equation can be simplified for the general case when H 0T0 = H 0298 For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

i prod

0 T2

)

− H 0298 + ( ΔH 0f )

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (T1 = 298 K and T2 = 1300 K), the problem simplifies to finding x in the reaction shown below, which will satisfy Eqn. (1) above. CO + x ( O 2 + 4 N 2 ) = CO 2 + ( x − 0.5 ) O 2 + 4 x N 2

The following table can be developed from information in the textbook. Gases CO2 CO O2 N2

ΔH 0f (kJ/mol)

H 0T1300 − H 0298 (kJ/mol)

−393.77 −110.62 0.0 0.0

50.19 Not needed 33.37 31.52

Substituting these values in Eqn. (1) one has: 2x −1 ( 0.0 + 33.37 )O2 + 4 x ( 0.0 + 31.52 ) N2 2 2 −1( −110.62 + 0.0 )CO − x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N = 0.0

1( −393.77 + 50.19 )CO +

2

2

−343.58 + 33.37 x − 16.69 + 126.08 x + 110.62 = 0.0 159.45 x = 249.65 x = 1.566 Therefore the total number of moles of air needed to react with one mole of CO is 1.566 * (1 + 7 © 2008 Glassman and Yetter.

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4) = 7.83 moles (which is the A/F ratio on a molar basis). The air/fuel ratio on a mass basis is: AF =

m air 7.83 moles of air × 29 kg / mol = m fuel 1 mole of CO × 28 kg / mol

AF = 8.11

so

8 © 2008 Glassman and Yetter.

kg of air kg of CO

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Problem 4 The exhaust of a carbureted engine, which is operated slightly fuel rich, has an efflux of unburned hydrocarbons entering the exhaust manifold. One can assume that all the hydrocarbons are equivalent to ethylene (C2H4) and all the remaining gases are equivalent to inert nitrogen (N2). On a molar basis there are 40 moles of nitrogen for every mole of ethylene. The hydrocarbons are to be burned over an oxidative catalyst and converted to carbon dioxide and water vapor only. In order to accomplish this objective, ambient (298 K) air must be injected into the manifold before the catalyst. If the catalyst is to be maintained at 1000 K, how many moles of air per mole of ethylene must be added? Take the temperature of the manifold gases before air injection as 400 K. Assume the composition of air to be 1 mole of oxygen to 4 moles of nitrogen. Solution: We will assume that adiabatic conditions exist and that no dissociation of the stable products will take place. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1000 K. This equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the

modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H T0 − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (Tl = 400 K and T2 = 1000 K), the problem simplifies to finding x in the reaction shown below, which will satisfy equation (1) above. C2H4 + 40N2 + x(O2 + 4N2) → 2 CO2 + 2 H2O + (x - 3) O2 + (40 + 4x) N2

(2)

The following table can be developed from information in the text. Gases

C2H4 N2 O2 CO2 H2O(vap)

ΔH 0f (kJ/mol

H 0T400 − H 0298 (kJ/mol)

H 0T1000 − H 0298 (kJ/mol)

52.34 0.0 0.0 −393.77 −242.00

4.89 2.97 3.03 Not needed Not needed

Not needed 21.47 22.72 33.43 26.00

Substituting these values in Eqn. (1) with the coefficients from Eqn. (2) one has:

9 © 2008 Glassman and Yetter.

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1( 52.34 + 4.89 )C H + 40 ( 0.0 + 2.97 ) N + x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N − 2

4

2

2

2

2 ( −393.77 + 33.43)CO − 2 ( −242.00 + 26.00 )H O − ( x − 3)( 0.0 + 22.72 )O − 2

( 40 + 4 x )( 0.0 + 21.47 ) N

2

2

= 0.0 2

57.22 + 118.91 + 720.68 + 432.00 − 22.72 x + 68.17 − 858.96 − 85.90 x = 0.0 108.62 x = 538.01 and x = 4.953 Therefore, the total number of moles of air needed to react with one mole of C2H4 is 4.953 x (1 + 4) = 24.75 moles.

10 © 2008 Glassman and Yetter.

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Problem 5 A combustion test was performed at 20-atm pressure in a hydrogen-oxygen system. Analysis of the combustion products, which were considered to be in equilibrium, was as follows: Compound H2O H2 O2 O H OH

Mole fraction 0.4855 0.4858 0.0 0.0 0.0216 0.0069

What must the combustion temperature have been in the test? Solution: Since all the products are in equilibrium, any equilibrium reaction between the products can be used to solve for the temperature. The simplest reaction to choose (and the only one that is an equilibrium reaction of formation) is:

1 H2 ⇔ H 2 The expression for the equilibrium constant of formation is then given by 1/ 2

K P,f ( H )

n ⎛ P ⎞ = H1/ 2 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

Since values are given in mole fraction

K P,f ( H ) =

∑ n = 1 . Therefore,

0.0216 1/ 2 20 ) = 0.138593 1/ 2 ( 0.4858

log10 K P,f ( H ) = −0.85826 Looking up the temperature in Table 2 in the appendix for H which corresponds to this value yields T ~ 2959 K If one desires to use the other possible equilibrium reaction given by 1 H 2 + OH ⇔ H 2 O 2 11 © 2008 Glassman and Yetter.

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The equilibrium constant is given by KP =

n H 2O 1/ 2

n H2 n OH

1/ 2

⎛ P ⎞ ⎜⎜ ⎟⎟ ⎝ ∑n ⎠

=

0.4855

( 0.0069 ) ( 0.4858)

1/ 2

1 = 22.573349 20

or log10 KP = 1.353596 Then, in order to solve for the final temperature of the mixture one would have to calculate the value for the equilibrium constant for the above reaction and find the corresponding temperature as shown below. KP =

K P,f ( H2O ) K P,f ( OH )

Temp.

K P,f ( OH )

K P,f ( H2O )

KP

Log10 KP

2900 K 3000 K

1.1066 1.1614

31.2608 22.0293

28.2494 18.9679

1.4510 1.2780

Interpolating for the temperature between 2900 and 3000 K that would give log10 KP = 1.353596 yields: T ~ 2956 K

12 © 2008 Glassman and Yetter.

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Problem 6 Whenever carbon monoxide is present in a reacting system it is possible for it to disproportionate into carbon dioxide according to the equilibrium 2CO ⇔ Cs + CO2. Assume that such an equilibrium can exist in some crevice in an automotive cylinder or manifold. Determine whether raising the temperature decreases or increases the amount of carbon present. Determine the KP for the reaction equilibrium and the effect of raising the pressure on the amount of carbon formed. Solution: One simply considers how the equilibrium is shifted by the temperature and pressure to solve the problem. (i) Temperature effect

2 CO ⇔ Cs + CO2 KP =

(1)

pCO2

(2)

2 pCO

in terms of equilibrium constants of formation for C + O2 ⇔ CO2 one has K P,f ( CO2 ) =

p CO2

(3)

p O2

1 and for C + O 2 ⇔ CO 2 K P,f ( CO ) =

pCO p1O22

(4)

For Eqn. (1) it follows that

KP =

K P,f ( CO2 ) K 2P,f ( CO )

or log10 K P = log10 K P,f ( CO2 ) − 2 log10 K P,f ( CO ) Evaluating this log equation in two different temperatures one has (a) At 1000 K

13 © 2008 Glassman and Yetter.

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log10 K P,f ( CO2 ) = 20.68 and 2 log10 K P,f ( CO ) = 2 (10.46 ) = 20.92 so

log Kp = −0.24 or KP = 0.58 (b) At 2000 K log10 K P,f ( CO2 ) = 10.35 and 2 log10 K P,f ( CO ) = 2 ( 7.47 ) = 14.94 so

log KP = −4.59 or KP = 2.57 × 10−5 Therefore, KP drops as temperature increases. Equation (1) and (2) show that as KP decreases, the reaction given by Eqn. (1) is shifting to the left, thus less carbon forms as the temperature increases. (ii) Pressure effect KP =

p CO2 2 p CO

n CO2 ⎛ P ⎞ = ⎜ ⎟ n CO 2 ⎜⎝ ∑ n ⎟⎠

−1

Thus, as the pressure increases more CO2 forms, shifting the reaction from left to right and thus more carbon forms as the pressure is increased.

14 © 2008 Glassman and Yetter.

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Problem 7 Determine the equilibrium constant KP at 1000 K for the following reaction

2 CH4 ⇔ 2 H2 + C2H4 Solution: As explained on page 14 of the text, the equilibrium constant (given in terms of the equilibrium constant of formation) is:

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 ) K 2P,f ( CH4 )

At 1000 K it follows from Table 2 in the appendix that (note Table 2 provides the logl0 of K): K P,f ( H2 ) = 100 = 1 K P,f ( C2 H4 ) = 10−6.213 = 6.1235 ×10−7 K P,f ( CH4 ) = 10−1.011 = 0.0975 so evaluating Eqn. (1) for KP one has KP =

1× ( 6.1235 ×10−7 )

( 0.0975)

2

= 6.442 ×10−5 K P = 6.442 × 10-5

so

15 © 2008 Glassman and Yetter.

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Problem 8 The atmosphere of Venus is said to contain 5% carbon dioxide and 95% nitrogen by volume. It is possible to simulate this atmosphere for Venus reentry studies by burning gaseous cyanogen (C2N2) and oxygen and diluting with nitrogen in the stagnation chamber of a continuously operating wind tunnel. If the stagnation pressure is 20 atm, what is the maximum stagnation temperature that could be reached and still have Venus atmosphere conditions? If the stagnation pressure were 1 atm, what would be the maximum temperature? Assume all gases enter the chamber at 298 K. Take the heat of formation of cyanogen as ( ΔH 0f ) = 381.84 k/mol. 298

Solution: In order to maintain Venus atmosphere, the composition of 100 moles of products would have to be 5 CO2 + 95 N2. Eqn. (10) in the textbook can be used to calculate the stagnation temperature of this system. This equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified the solution approach is to find T2 such that the above equality will be satisfied. (a) for Venus atmosphere the general combustion reaction would be a C2N2 + b O2 + c N2 → 5 CO2 + 95 N2 which can be balanced for one mole of C2N2 as follows: C2N2 + 2 O2 + 36 N2 → 2 CO2 + 38 N2, In preparation to apply Eqn. (1) the following tables can be generated. gases

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

C2N2 O2 N2

381.84 0.0 0.0

0.0 0.0 0.0

An expression for the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the reactants becomes: H r = 1( 381.84 + 0.0 )C N + 2 ( 0.0 + 0.0 )O + 36 ( 0.0 + 0.0 ) N = −381.84 kJ 2

2

2

2

16 © 2008 Glassman and Yetter.

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For the products it follows that:

gases

ΔH 0f ( kJ mol )

N2 CO2

0.0 −393.77

Thus, an expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the products becomes let ΔH = H 0T2 − H 0298 :

(

(

)

(

H p = 2 −393.77 + ΔH CO2 + 38 0.0 + ΔH N2

)

)

Equating Hr and Hp yields: 2 ΔH CO2 + 38 ΔH N 2 = 1,169.37 A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 40 = 2 + 38): 40 ΔH N2 = 1,169.37 so ΔH N 2 = 29.23kJ mol , therefore, T2 ≈ 1232 K We will then assume that T2 is between 1200 and 1300 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ): T2

ΔH CO2 ( kJ mol )

ΔH N2

Hp − Hr (kJ)

44.51 50.19

28.13 31.52

−11.51 128.87

1200 1300 Interpolate

1200 T2 1300

−11.51 00 128.31

T2 − 1200 11.51 = 100 139.88 T 2 = 1208 K

so

Since K P,f ( CO2 ) = 1.75 × 1017 at 1200 K, no significant dissociation of the products occurs. Therefore, pressure has no effect and T2 would remain the same at all pressures. 17 © 2008 Glassman and Yetter.

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Problem 9 A mixture of 1 mole of N2 and 0.5 mole O2 is heated to 4000 K at 0.5 atm pressure and results in an equilibrium mixture of N2, O2, and NO only. If the O2 and N2 were initially at 298 K and the process is one of steady heating, how much heat was required to bring the final mixture to 4000 K on the basis of one initial mole of N2? Solution: The first step is to find the equilibrium concentration of products. Once this is done, one can use Eqn. (1.10) to calculate the amount of heat necessary to bring the final mixture to 4000 K. Consider the following reaction at equilibrium.

1 1 N 2 + O 2 → a N 2 + b O 2 + c NO 2

(1)

balancing this reaction yields the following relationship between a, b and c. N: 2 = 2 a + c 0: 1 = 2 b + c Solving these two equations for b and c in terms of a it follows that b=

2a − 1 2

and c = 2 − 2a

Therefore, Eqn. ( 1) becomes: 1N 2 + 0.5O 2 → a N 2 +

2a − 1 O 2 + ( 2 − 2a ) NO 2 1 1 N 2 + O 2 ⇔ NO , the expression for the equilibrium 2 2

For the equilibrium reaction given by constant of formation is given by: 1−1 2 −1 2

K P,f ( NO )

⎛ P ⎞ n NO = ⎟ 1/ 2 1/ 2 ⎜ n N2 n O2 ⎜⎝ ∑ n ⎟⎠

2a − 1 + 2 − 2a = 3 . At 4000 K, 1og10 KP,f(NO) = 2 - 0.524 or KP,f(NO) = 0.29923. Squaring Eqn. (3) would give the following equation for the variable a: The total number of moles is given by

∑n = a +

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( 2 − 2a )

2

2a − 1 a 2

− K 2P,f ( NO ) = 0,

Solving for a in the above equation and then using the mass balance equations to solve for b and c yields: a = 0.909, b = 0.409 and c = 0.182. Equation (2) now becomes 1 N2 + 0.5 O2 → 0.909 N2 + 0.409 O2 + 0.182NO The amount of heat required to bring the above mixture to 4000 K is calculated using Eqn. (1.10) in the text by taking T2 = 298 K and T1 = 4000 K: −Q =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(4)

from tables in the text one has: gases

O2 N2 NO

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

H 0T2 − H 0298 ( kJ mol )

0.0 0.0 90.43

139.01 130.16 132.76

0.0 0.0 0.0

so the equation for Q becomes −Q = 0.182 ( 90.43 + 132.76 ) NO + 0.409 ( 0.0 + 139.01)O + 0.909 ( 0.0 + 130.16 ) N − 2

1( 0.0 + 0.0 ) N − 1/ 2 ( 0.0 + 0.0 )O 2

2

2

Q = - 215.79 kJ

so if no dissociation occurred the value of Q would be −Q = 1/ 2 ( 0.0 + 139.01)O + 1( 0.0 + 130.16 ) N − 2

2

1( 0.0 + 0.0 ) N − 1/ 2 ( 0.0 + 0.0 )O 2

2

so

Q = - 199.67 kJ

Therefore, more energy is required to raise the temperature of the mixture to 4000 K due to dissociation. 20 © 2008 Glassman and Yetter.

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Problem 10 Calculate the adiabatic decomposition temperature of benzene under the constant pressure condition of 20 atm. Assume that benzene enters the decomposition chamber in the liquid state at 298 K and decomposes into the following products: carbon (graphite), hydrogen, and methane. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are obtained. Consider the following reaction at equilibrium: C6H6 (1) → a C (s) + b H2 + c CH4

(1)

balancing this reaction yields the following relationship between a, b and c. C: 6 = a + c H: 6 = 2 b + 4 c solving these two equations for a and b in terms of c it follows that b=

6 − 4c = 3 − 2c 2

and a = 6 − c

Therefore, Eqn. (1) becomes: C6 H 6(1) → ( 6 − c ) C(s ) +

6 − 4c H 2 + c CH 4 2

(2)

Now consider the equilibrium reaction given by C(s) + 2 H2 ⇔ CH4 For this equilibrium reaction, the expression for the equilibrium constant of formation of CH4 is 1− 2

K P,f ( CH4 )

n CH ⎛ P ⎞ = KP = 2 4 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

(3)

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6 − 4c + c = 3 − c (note that the number of moles of 2 graphite are not included in the calculation for the total number of moles at equilibrium). Equation (3) then becomes:

The total number of moles is given by

∑n =

c b+c − KP = 0 b 2 20 expressing b in terms of c gives:

c

( 3 − 2c )

2

3+ c − KP = 0 20

(4)

In order to calculate the adiabatic flame temperature one has to apply Eqn. (1.11) in the text taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: gases

C6H6 (1)

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

49.03

0.0

And the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (5)) can be calculated as: Hr = 1 (49.03)C6H6 (1) = 49.03 kJ Let us first guess that T2 = 1000 K; therefore, logl0 KP = - 1.011 or KP = 0.097499. Solving Eqn. (4) for c using this value of KP and then solving the mass balance equations for a and b results in: c = 0.9943, b = 1.0114, and a = 5.0057 For a product temperature of 1000 K, application of Eqn. (5) using the values of a, b, and c found above yields: T2 1000

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

8.47

20.70

38.20

26.87

−22.16

22 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

If T2 = 1100 K, logl0 Kp. = - 1.44 or KP = 0.03631. Solving Eqn. (4) for c using this value of KP and then solving the mass balance equations for a and b gives: c = 0.7409, b = 1.5182, and a = 5.2591 Following the same procedure as previously outlined yields: T2 1100

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

14.01

23.74

45.58

88.01

38.98

Interpolation for the value of temperature where Hp − Hr = 0 yields T2 ~ 1036 K. At T2 = 1036K, logl0 Kp = -1.1654 or Kp = 0.06832, c = 0.9101, b = 1.1798, a = 5.0899 and T2 1036

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

10.47

21.79

40.86

48.02

−1.01

Since Hp − Hr ~0

T 2 ~ 1036 K

23 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 11 Calculate the flame temperature and the product composition of liquid ethylene oxide decomposing at 20-atm pressure by the irreversible reaction:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4 The four products are as specified. The equilibrium known to exist is 2 CH4 ⇔ 2 H2 + C2H4 The heat of formation ΔH 0f of liquid ethylene oxide is = - 76.62 kJ/mol. It enters the decomposition chamber at 298 K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4 balancing this reaction yields the following equations for a, b, c and d. C: 2 = a + b +2 d O: 1 = a H: 4 = 4 b + 2 c + 4 d simplifying it follows that: b+2d=1

(1)

4b+2c+4d=4

(2)

The other equation for the solution of b, c, and d will come from the Kp expression for the given equilibrium reaction. 2 CH4 ⇔ 2 H2 + C2H4 For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is, 24 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 ) K 2P,f ( CH4 )

n 2H2 n C2 H4 ⎛ P ⎞ = ⎜⎜ ⎟⎟ 2 n CH ⎝ ∑n ⎠ 4

2 +1− 2

(3)

or in terms of b, c, and d ( with P = 20 atm.) one has KP =

c2d 20 2 b 1+ b + c + d

(4)

By combining the equation above with the mass balance equations the following expression can be written for KP: KP

(1 − b ) = b

2

2

20 5−b

(5)

The approach now is to guess T2, look up KP and solve for the three unknowns in Eqns. (1), (2), and (5). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(6)

With Eqns. (5) and (6) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: gases

C2H4O(1) CO CH4 C2H4

ΔH 0f ( kJ mol )

−76.62 −110.62 −74.90 52.34

(

)

Applying Eqn. (6) yields let ΔH = H 0T2 − H 0298 :

(

) (

) (

−76.62 = a ( −110.62 + ΔH CO ) + b −74.90 + ΔH CH4 + c 0.0 + ΔH H2 + d 52.34 + ΔH C2 H4 First guess that T2 = 1400 K, from Tables in the text one has:

25 © 2008 Glassman and Yetter.

)

(7)

1/25/2011 9:27 AM

log10 K P,f ( H2 ) = 1 log10 K P,f ( C2 H4 ) = −5.664 log10 K P,f ( CH4 ) = −2.372 so from Eqn. (3), log10 KP = 0.12023. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives: a = 1.0, b = 0.7557, c = 0.2443, and d = 0.1221 Application of Eqn. (7) using T2 = 1400 K and the values of a, b, c, and d just calculated yields: T2 1400

ΔHCO (kJ/mol) 35.36

ΔH CH4 69.66

ΔH H2 33.08

ΔH C2 H4 91.26

Hp (kJ) -53.60

If we now take T2 = 1300 K, from Tables in the text one has: log10 K P,f ( H2 ) = 1 log10 K P,f ( C2 H4 ) = −5.766 log10 K P,f ( CH4 ) = −2.107 so from Eqn. (3), log10 KP = - 1.552 or KP = 0.028054. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives: a = 1.0, b = 0.8398, c = 0.1603, and d = 0.0801 Application of Eqn. (7) using T2 = 1300 K and the values of a, b, c, and d just calculated yields: T2 1300

ΔHCO (kJ/mole)

ΔH CH4

ΔH H2

ΔH C2 H4

Hp (kJ)

31.89

61.34

29.93

80.63

-74.65

Extrapolation to the value of T2 where Hp ~ Hr (−76.62) yields: 2 ~ 1291 K T

26 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 12 Liquid hydrazine (N2H4) decomposes exothermically in a monopropellant rocket operating at 100-atm chamber pressure. The products formed in the chamber are N2, H2, and ammonia (NH3) according to the irreversible reaction

N2H4(1) → a N2 + b H2 + c NH3 Determine the adiabatic decomposition temperature and the product composition a, b, and c. Take the standard heat of formation of liquid hydrazine as 50.03 kJ/mole. The hydrazine enters the system at 298 K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

N2H4(1) → a N2 + b H2 + c NH3 balancing this reaction yields the following equations for a, b, and c. N: 2 = 2 a + c

(1)

H: 4 = 2 b + 3 c

(2)

The other equation necessary for the solution of a, b, and c will come from the Kp for the following equilibrium reaction. 1 3 N 2 + H 2 ⇔ NH 3 2 2 For this equilibrium reaction, the equilibrium constant of formation of NH3 can be expressed in terms of the mixture composition at equilibrium as shown: 1−1 2 −3 2

K P,f ( NH3 )

n NH ⎛ P ⎞ = K p = 1 2 33 2 ⎜ ⎟ n H2 n N2 ⎜⎝ ∑ n ⎟⎠

(3)

or in terms of a, b, and c (with P = 100 atm) one has:

27 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Kp =

c a b

1/ 2 3/ 2

a+b+c 100

(4)

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

)

− H 0298 + ( ΔH f0 )

0 T2

i prod

(

)

⎤ − n j ⎡ H 0T1 − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j∑ ⎣ react

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic properties: ΔH 0f ( kJ mol )

gases

N2H4(1) N2 H2 NH3

50.03 0.0 0.0 −46.22

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

) (

) (

50.03 = a 0.0 + ΔH N2 + b 0.0 + ΔH H2 + c −46.22 + ΔH NH3

)

(6)

Let us first guess that T2 = 900 K, from Tables in the text it follows that: log10 K P,f ( NH3 ) = −2.915

Solving Eqns. (l), (2) and (4) results in: a = 0.9487, b = 1.8462, and c = 0.1026 For a product temperature of 900 K, application of Eqn. (6) using the values of a, b, and c, found above, yields (all ΔH's are in kJ/mol and Hp is in kJ): T2 900 K

ΔH N2 ( kJ mol )

ΔH H2

ΔH NH3

Hp (kJ)

18.23

17.69

27.09

47.98

First guess that T2 = 1000 K, from Tables in the text it follows that: 28 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

log10 K P,f ( NH3 ) = −3.233

Solving Eqns. (1), (2) and (4) one has: a = 0.9739, b = 1.9218, and c = 0.05216 For a product temperature of 1000 K, application of Eqn. (6) using values of a, b, and c, found above yields: T2 1000 K

ΔH N2 ( kJ mole )

ΔH H2

ΔH NH3

Hp (kJ)

21.47

20.70

32.60

60.00

Interpolation to the value of T2 where Hp ~ Hr (50.03) yields: T 2 ~ 917 K

29 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 13 Gaseous hydrogen and oxygen are burned at l-atm pressure under the rich conditions designated by the following combustion reaction:

O2 + 5 H2 → a H2O + b H2 + c H The gases enter at 298 K. Calculate the adiabatic flame temperature and the product composition a, b, and c. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known.

Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium: O2 + 5 H2 → a H2O + b H2 + c H balancing this reaction yields the following equations for a, b, and c. O: 2 = a

(1)

H: 10 = 2 a + 2 b + c

(2)

The other equation necessary for the solution of a, b, and c will come from the KP expression for the following equilibrium reaction. 1 H2 ⇔ H 2 For this equilibrium reaction, the equilibrium constant of formation of H can be expressed in terms of the mixture composition at equilibrium as shown: 1−1/ 2

K P,f ( H )

n ⎛ P ⎞ = K P = 1/H2 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

(3)

or in terms of a, b, and c (with P = 1 atm) one has KP =

c 1 b a+b+c

(4)

1/ 2

30 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that also satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic properties: gases

ΔH 0f ( kJ mol )

O2 H H2 H2O

0.0 218.09 0.0 −242.00

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

) (

)

0.0 = a −242.00 + ΔH H2O + b 0.0 + ΔH H2 + c ( 218.09 + ΔH H ) Let us first guess that T2 = 2500 K, from Table 2 it follows that, log10 Kp,f(H) = −1.601 Solving Eqns. (1), (2) and (4) one has: a = 2.0, b = 2.9516, and c = 0.0967 For a product temperature of 2500 K application of Eqn. (6) using the values of a, b, and c, yields (all ΔH's are in kJ/mole and Hp is in kJ): T2 2500 K

ΔH H2O ( kJ mol )

ΔH H2

ΔHH

Hp (kJ)

99.03

70.54

45.80

−52.21

Let us first guess that T2 = 2700 K, from Table A2 it follows that 31 © 2008 Glassman and Yetter.

(6)

1/25/2011 9:27 AM

log10 Kp,f(H) = −1.247 Solving Eqns. (1), (2) and (4) one has: a = 2.0, b = 2.8912, and c = 0.2176 For a product temperature of 2700 K application of Eqn. (6) using the values of a, b, and c, found above yields (all ΔH's are in kJ/mol and Hp is in kJ): T2 2700 K

ΔH H2O ( kJ mol )

ΔH H2

ΔHH

Hp (kJ)

109.89

77.77

49.96

18.96

Interpolation to the value of T2 where Hp ~ Hr (Hr = 0.0) yields: T2 ~ 2647 K

32 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 14 The liquid propellant rocket combination nitrogen tetroxide (N2O4) and UDMH (unsymmetrical dimethyl hydrazine) has optimum performance at an oxidizer to fuel weight ratio of 2 at a chamber pressure of 67 atm. Assume that the products of combustion of this mixture are N2, CO2, H2O, CO, H2, O H, OH, and NO. Set down the equations necessary to calculate the adiabatic combustion temperature and the actual product composition under these conditions. These equations should contain all the numerical data in the description of the problem and in the tables in the appendices. The heats of formation of the reactants are: N 2 O 4(1)

ΔH f0 298 = −2.90 kJ mol

UDMH (1)

ΔΗ 0f 298 = +53.17 kJ mol

The propellants enter the combustion chamber at 298 K. Solution: The solution for this problem consists of finding the correct number of equations necessary to solve for the unknown quantities in this problem. There are 9 unknown species coefficients and the temperature is unknown. Therefore, 10 equations must be found to solve the problem. These equations consist of the energy, 4 mass balance, and 5 equilibrium equations.

To specify the general reaction equation, the mole proportion of reactants must be found first. This can be done as follows. The molecular weights of the reactants are: for N2O4,

MW = 92

for C2N2H8,

MW = 60

From the problem statement one has: mass UDMH =2 mass N 2 O 4 By assuming 1 mole N2O4 this equation becomes mass UDMH =2 60 or

nUDMH (92) = 120 33 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

So nUDMH = 1.304 The general equation for the reaction then becomes: C2N2H8 + 1.304 N2O4 → a N2 + b CO2 + c H2O + d CO + e H2 + f O + g H + h OH + i NO The mass balance equations can be written as: C: 2 = b + d N: 4.609 = 2a + i H: 8 = 2c + 2e + g + h O: 5.216 = 2b + c + d + f + h + i

(1) (2) (3) (4)

The energy equation would be the equation for adiabatic flame temperature (Eq. (11)) found in the text.

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

The following information can be used to simplify the equation ΔH fN2O4 = −2903 kJ mol ΔH f UDMH = 53.172 kJ mol The right side of the equation is: 1(53.172) + 1.304 (-2.093) = 50.44 kJ Therefore, the energy equation becomes: 50.44 =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

⎤ ⎦

(5)

298 i

Some of the possible equilbrium reactions include: H2 ⇔ 2 H CO + O ⇔ CO2 N2 + 2 O ⇔ 2 NO H2 + 2 OH ⇔ 2 H2O 34 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

CO + 2 OH ⇔ H2O + CO2 These reactions give the following equilibrium equations: K p1 =

n 2H n H2

⎛ P ⎞ ⎜⎜ ⎟⎟ ⎝ ∑n ⎠

n 2NO ⎛ P ⎞ K p2 = ⎜ ⎟ n N2 n O2 ⎜⎝ ∑ n ⎟⎠

−1

n CO2 ⎛ P ⎞ K p3 = ⎜ ⎟ n CO n O ⎜⎝ ∑ n ⎟⎠

−1

n 2H2O ⎛ P ⎞ K p4 = 2 ⎜ ⎜ ∑ n ⎟⎟ n H2 n OH ⎝ ⎠

∑n

i

g 2 ⎛ 67 ⎜ e ⎜⎝ ∑ n i

b = df

⎛ ∑ ni ⎜⎜ ⎝ 67

(6)

(7)

⎞ ⎟⎟ ⎠

(8)

=

c2 ⎛ ∑ n i ⎜ e h 2 ⎜⎝ 67

⎞ ⎟⎟ ⎠

(9)

=

c b ⎛ ∑ ni ⎜ d h 2 ⎜⎝ 67

⎞ ⎟⎟ ⎠

(10)

−1

n H O n CO ⎛ P ⎞ K p5 = 2 2 2 ⎜ ⎟ n CO n OH ⎜⎝ ∑ n ⎟⎠

where

⎞ ⎟⎟ ⎠ i2 ⎛ ∑ ni ⎞ = 2 ⎜⎜ ⎟ a f ⎝ 67 ⎟⎠ =

−1

= a + b + c + d + e + f + g + h + i, and the Kp's are evaluated at the adiabatic flame

temperature.

35 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 15 Consider a fuel burning in inert airs and oxygen where the combustion requirement is only 0.21 moles of oxygen. Order the following mixtures as to their adiabatic flame temperatures with the given fuel.

a) b) c) d)

pure O2 0.21 O2 + 0.79 N2 0.21 O2+0.79 Ar 0.21 O2 + 0.79 CO2

(air)

Solution: The temperatures order with respect to the Cp of the inert and excess oxygen. The lower the Cp of the inert gas the higher the temperature. Thus, Ar being monoatomic has the lowest Cp, O2 and N2 are about the same, but higher than Ar since they are diatomic and have vibrational and rotational degrees of freedom, and CO2 being triatomic has the highest Cp. Thus the order from highest to lowest is

c, a and/or b, d

36 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 16 Propellant chemists have proposed a new high energy liquid oxidizer, penta-oxygen O5, which is also a monopropellant. Calculate the monopropellant decomposition temperature at a chamber pressure of 10 atm if it assumed the only products are O atoms and O2 molecules. The heat of formation of the new oxidizer is estimated to be very high, + 1025 kJ / mol. Obviously the amounts of O2 and O must be calculated for one mole of O5 decomposing. The O5 enters the system at 298 K. Hint: The answer will lie somewhere between 4000 and 5000K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

O5→ a O2 + b O balancing this reaction yields the following equation for a. O: 5 = 2a + b or b = 5 - 2a

(1)

The other equation necessary for the solution of a and b will come from the Kp expression for the following equilibrium reaction. 1 O2 ⇔ O 2 For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is, KP =

K P,f ( O ) K1/P,f2( O2 )

n ⎛ P ⎞ = 1/O2 ⎜ ⎟ n O2 ⎜⎝ ∑ n ⎟⎠

2 −1

(2)

or in terms of a and b ( with P = 10 atm.) one has 1/ 2

b ⎡ 10 ⎤ K p = 1/ 2 ⎢ a ⎣ a + b ⎥⎦

(3)

By combining the equation above with the mass balance equation the following expression can be written for Kp: 37 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

5 − 2a ⎡ 10 ⎤ a1/ 2 ⎢⎣ 5 − 1 ⎥⎦

1/ 2

Kp =

(4)

The approach now is to guess T2 look up Kp and solve for the two unknowns in Eqns. (1) and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K.

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: ΔH 0f ( kJ mol )

gases

O O2 O5

249.36 0 1025

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

)

1025 = a 0.0 + ΔH O2 + b ( 249.36 + ΔH O )

(6)

First guess that T2 = 4500 K, from Tables in the text one has: log10 K P,f ( O2 ) = 1 log10 K P,f ( O ) = 0.543 so from Eqn. (2), log10 Kp = 0.543 or Kp = 3.491. Solving Eqn. (4) for a and then Eqn. (1) for b gives: a = 1.29, b = 2.42 Application of Eqn. (6) using T2 = 4500 K and the values of a and b just calculated yields: T2

ΔH O2 ( kJ mol )

ΔHO

Hp (kJ/mol)

4500

159.9

88.45

1023.77

38 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Since the guessed value of T2 give Hp ~ H (1025 kJ/mol) the decomposition temperature is T 2 ~ 4500 K

39 © 2008 Glassman and Yetter.

View more...
Solution Manual for Combustion 4th Edition Irvin Glassman Department of Mechanical and Aerospace Engineering Princeton University Princeton, New Jersey Richard Yetter Department of Mechanical and Nuclear Engineering The Pennsylvania State University University Park, PA This solution manual is a revision to the solution manual for "Combustion, 3rd. Ed." that was written in its final form by Prof. Queiroz and Dr. Black of Brigham Young University.

1 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

A Solution Manual for: COMBUSTION. 4th Ed. by I. Glassman, and R.A. Yetter CHAPTER 1 Problem 1 Calculate the heat of reaction when methane and air in stoichiometric proportions are brought into a calorimeter at 500 K. The product composition is brought to the ambient temperature (298 K) by the cooling water. The pressure in the calorimeter is assumed to remain at 1 atm, but the water formed has condensed. Solution: The heat of reaction can be calculated using Eqn. (10) in the text. However, as also suggested in the textbook, this equation can be simplified for the general case when H 0T0 = H 0298 . The modified equation becomes: −Q = ΔH =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

⎤ − ∑ n j ⎡ H 0T − H 0298 ⎦ j react ⎣ 1

298 i

)

298

+ ( ΔH 0f )

298

⎤ ⎦j

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. The general equation for the complete combustion of one mole of methane in air is: CH 4 + a th ( O 2 + 3.76N 2 ) → a CO 2 + b H 2 O + c N 2

where ath is a coefficient describing how much air is needed for stoichiometric combustion. Balancing the above reaction one has: CH 4 + 2 ( O 2 + 3.76N 2 ) → CO 2 + 2 H 2 O + 7.52 N 2

In order to apply Eqn. (1) to the above reaction, the following properties from Tables in the textbook are needed (note that for this problem T2 = 298 K and T1 = 500 K): gases

CH4 O2 N2 H2O(liq) CO2

ΔH 0f (kJ/mol)

H 0T1 − H 0298 (kJ/mol)

H 0T2 − H 0298 (kJ/mol)

−74.90 0.0 0.0 −286.04 −393.77

8.21 6.09 5.92 0.0 0.0

0.0 0.0 0.0 0.0 0.0

Equation (1) then becomes: 2 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

−Q = 1( −393.77 + 0.0 )CO + 2 ( −286.04 + 0.0 )H O + 7.52 ( 0.0 + 0.0 ) N − 1( −74.90 + 8.21)CH 2

2

−2 ( 0.0 + 6.09 )O − 7.52 ( 0.0 + 5.92 ) N 2

2

4

2

Q=955.85 kJ

or

3 © 2008 Glassman and Yetter.

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Problem 2 Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ratio of 0.5. For this problem assume there is no dissociation of the stable products formed. All reactants are at 298 K and the system operates at 1-atm pressure. Compare your results with those given in the graphs in the text. Explain any differences. Solution: We will assume that adiabatic conditions exist and that the water in the products is in the vapor phase. In this case, Eqn. (10) in the text can be used in an iterative approach to solve for the adiabatic flame temperature. However, this equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified, the solution approach is to find T2 such that the above equality will be satisfied. The general equation for the complete combustion of one mole of octane with excess air is: C8 H18( liq ) + ( m )( a th )( O 2 + 3.76N 2 ) → a CO 2 + b H 2 O + c O 2 + d N 2 where ath is the coefficient describing how much air is needed for stoichiometric combustion and m is the coefficient used to represent the amount of excess air, respectively. For the case when c = 0 and m = 1 (no excess air), a simple mass balance yields ath = 12.5. Therefore, for an excess air of 100% (m = 2) or an equivalence ratio of 0.5, the chemical reaction becomes: C8 H18( liq ) + 25 ( O2 + 3.76N 2 ) → 8 CO 2 + 9 H 2 O + 94 N 2 + 12.5 O 2 Since the conditions of the reactants are specified (T1 = 298 K), Table 2 in the text and Appendix A provide the following thermodynamic properties: gases

C8H18(liq) O2 N2

ΔH 0f (kJ/mol)

H 0T1 − H 0298 (kJ/mol)

−250.12 0.0 0.0

0.0 0.0 0.0

The enthalpy of the reactants (Hr, given by the right-hand side of Eqn (1)) can be calculated as: H r = 1( −250.12 + 0.0 )C H + 11( 0.0 + 0.0 )O + 41.36 ( 0.0 + 0.0 ) N = −250.12kJ 8

18

2

2

For the products it follows from thermodynamic properties in the text that: 4 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

ΔH 0f (kJ/mol) 0.0 0.0 −242.00 −393.77

gases

O2 N2 H2O(vap) CO2

Therefore, the expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1) in terms of the enthalpy of the individual gases in the products becomes let ΔH = H 0T2 − H 0298 :

(

)

(

) (

)

(

)

(

H p = 8 −393.77 + ΔH CO2 + 9 −242.00 + ΔH H2O + 12.5 0.0 + ΔH O2 + 94 0.0 + ΔH N2

)

Simplifying the above equation yields: 8 ΔH CO2 + 9 ΔH H2O + 12.5ΔH O2 + 94ΔH N2 = 5, 078.04 A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 123.5 = 8 + 9 + 12.5 + 94): 123.5 ΔH N2 = 5,078.04

so

ΔH N2 = 41.12 kJ/mol

therefore T2 = 1577

We will then assume that T2 is between 1500 and 1600 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ): T2 1500 1550 1506

ΔH CO2 (kJ/mol) 61.76 64.69 62.11

Interpolate

ΔH H2O 48.13 50.51 48.41

ΔH O2 40.64 42.48 40.86

1500 T2 1550

ΔH N2 38.43 40.18 38.64

Hp – Hr (kJ) -30.37 202.00 -2.55

-30.37 00 202.00

T2 − 1500 50

30.37 232.37

=

so

T2 = 1506 K

The enthalpy difference for this temperature was also calculated in the above table to show that 5 © 2008 Glassman and Yetter.

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once you have a good idea of the temperature interval where the adiabatic temperature should be in, linear interpolation does a good job of predicting the temperature. Comparing the results for the adiabatic flame temperature found above to that found using Fig. 3 on page 22 of the textbook is done as follows: Enthalpy per gram C8H18

=

−250.12 96 + 18

=

2.194 kJ/gm

H/C =

18 = 2.25 8

With these values Fig. 3 yields T2 ≅ 1505 K, which is very close to the calculated adiabatic temperature.

6 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 3

Carbon monoxide is oxidized to carbon dioxide in an excess of air (1 atm) in an afterburner so that the final temperature is 1300 K. Under the assumption of no dissociation determine the airfuel ratio required. Report the results on both a molar and mass basis. For the purposes of this problem assume that air has the composition of 1 mole of oxygen to 4 moles of nitrogen. The carbon monoxide and air enter the system at 298 K. Solution: We will assume that adiabatic conditions exist. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1300 K. This equation can be simplified for the general case when H 0T0 = H 0298 For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

i prod

0 T2

)

− H 0298 + ( ΔH 0f )

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (T1 = 298 K and T2 = 1300 K), the problem simplifies to finding x in the reaction shown below, which will satisfy Eqn. (1) above. CO + x ( O 2 + 4 N 2 ) = CO 2 + ( x − 0.5 ) O 2 + 4 x N 2

The following table can be developed from information in the textbook. Gases CO2 CO O2 N2

ΔH 0f (kJ/mol)

H 0T1300 − H 0298 (kJ/mol)

−393.77 −110.62 0.0 0.0

50.19 Not needed 33.37 31.52

Substituting these values in Eqn. (1) one has: 2x −1 ( 0.0 + 33.37 )O2 + 4 x ( 0.0 + 31.52 ) N2 2 2 −1( −110.62 + 0.0 )CO − x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N = 0.0

1( −393.77 + 50.19 )CO +

2

2

−343.58 + 33.37 x − 16.69 + 126.08 x + 110.62 = 0.0 159.45 x = 249.65 x = 1.566 Therefore the total number of moles of air needed to react with one mole of CO is 1.566 * (1 + 7 © 2008 Glassman and Yetter.

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4) = 7.83 moles (which is the A/F ratio on a molar basis). The air/fuel ratio on a mass basis is: AF =

m air 7.83 moles of air × 29 kg / mol = m fuel 1 mole of CO × 28 kg / mol

AF = 8.11

so

8 © 2008 Glassman and Yetter.

kg of air kg of CO

1/25/2011 9:27 AM

Problem 4 The exhaust of a carbureted engine, which is operated slightly fuel rich, has an efflux of unburned hydrocarbons entering the exhaust manifold. One can assume that all the hydrocarbons are equivalent to ethylene (C2H4) and all the remaining gases are equivalent to inert nitrogen (N2). On a molar basis there are 40 moles of nitrogen for every mole of ethylene. The hydrocarbons are to be burned over an oxidative catalyst and converted to carbon dioxide and water vapor only. In order to accomplish this objective, ambient (298 K) air must be injected into the manifold before the catalyst. If the catalyst is to be maintained at 1000 K, how many moles of air per mole of ethylene must be added? Take the temperature of the manifold gases before air injection as 400 K. Assume the composition of air to be 1 mole of oxygen to 4 moles of nitrogen. Solution: We will assume that adiabatic conditions exist and that no dissociation of the stable products will take place. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1000 K. This equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the

modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H T0 − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (Tl = 400 K and T2 = 1000 K), the problem simplifies to finding x in the reaction shown below, which will satisfy equation (1) above. C2H4 + 40N2 + x(O2 + 4N2) → 2 CO2 + 2 H2O + (x - 3) O2 + (40 + 4x) N2

(2)

The following table can be developed from information in the text. Gases

C2H4 N2 O2 CO2 H2O(vap)

ΔH 0f (kJ/mol

H 0T400 − H 0298 (kJ/mol)

H 0T1000 − H 0298 (kJ/mol)

52.34 0.0 0.0 −393.77 −242.00

4.89 2.97 3.03 Not needed Not needed

Not needed 21.47 22.72 33.43 26.00

Substituting these values in Eqn. (1) with the coefficients from Eqn. (2) one has:

9 © 2008 Glassman and Yetter.

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1( 52.34 + 4.89 )C H + 40 ( 0.0 + 2.97 ) N + x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N − 2

4

2

2

2

2 ( −393.77 + 33.43)CO − 2 ( −242.00 + 26.00 )H O − ( x − 3)( 0.0 + 22.72 )O − 2

( 40 + 4 x )( 0.0 + 21.47 ) N

2

2

= 0.0 2

57.22 + 118.91 + 720.68 + 432.00 − 22.72 x + 68.17 − 858.96 − 85.90 x = 0.0 108.62 x = 538.01 and x = 4.953 Therefore, the total number of moles of air needed to react with one mole of C2H4 is 4.953 x (1 + 4) = 24.75 moles.

10 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 5 A combustion test was performed at 20-atm pressure in a hydrogen-oxygen system. Analysis of the combustion products, which were considered to be in equilibrium, was as follows: Compound H2O H2 O2 O H OH

Mole fraction 0.4855 0.4858 0.0 0.0 0.0216 0.0069

What must the combustion temperature have been in the test? Solution: Since all the products are in equilibrium, any equilibrium reaction between the products can be used to solve for the temperature. The simplest reaction to choose (and the only one that is an equilibrium reaction of formation) is:

1 H2 ⇔ H 2 The expression for the equilibrium constant of formation is then given by 1/ 2

K P,f ( H )

n ⎛ P ⎞ = H1/ 2 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

Since values are given in mole fraction

K P,f ( H ) =

∑ n = 1 . Therefore,

0.0216 1/ 2 20 ) = 0.138593 1/ 2 ( 0.4858

log10 K P,f ( H ) = −0.85826 Looking up the temperature in Table 2 in the appendix for H which corresponds to this value yields T ~ 2959 K If one desires to use the other possible equilibrium reaction given by 1 H 2 + OH ⇔ H 2 O 2 11 © 2008 Glassman and Yetter.

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The equilibrium constant is given by KP =

n H 2O 1/ 2

n H2 n OH

1/ 2

⎛ P ⎞ ⎜⎜ ⎟⎟ ⎝ ∑n ⎠

=

0.4855

( 0.0069 ) ( 0.4858)

1/ 2

1 = 22.573349 20

or log10 KP = 1.353596 Then, in order to solve for the final temperature of the mixture one would have to calculate the value for the equilibrium constant for the above reaction and find the corresponding temperature as shown below. KP =

K P,f ( H2O ) K P,f ( OH )

Temp.

K P,f ( OH )

K P,f ( H2O )

KP

Log10 KP

2900 K 3000 K

1.1066 1.1614

31.2608 22.0293

28.2494 18.9679

1.4510 1.2780

Interpolating for the temperature between 2900 and 3000 K that would give log10 KP = 1.353596 yields: T ~ 2956 K

12 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 6 Whenever carbon monoxide is present in a reacting system it is possible for it to disproportionate into carbon dioxide according to the equilibrium 2CO ⇔ Cs + CO2. Assume that such an equilibrium can exist in some crevice in an automotive cylinder or manifold. Determine whether raising the temperature decreases or increases the amount of carbon present. Determine the KP for the reaction equilibrium and the effect of raising the pressure on the amount of carbon formed. Solution: One simply considers how the equilibrium is shifted by the temperature and pressure to solve the problem. (i) Temperature effect

2 CO ⇔ Cs + CO2 KP =

(1)

pCO2

(2)

2 pCO

in terms of equilibrium constants of formation for C + O2 ⇔ CO2 one has K P,f ( CO2 ) =

p CO2

(3)

p O2

1 and for C + O 2 ⇔ CO 2 K P,f ( CO ) =

pCO p1O22

(4)

For Eqn. (1) it follows that

KP =

K P,f ( CO2 ) K 2P,f ( CO )

or log10 K P = log10 K P,f ( CO2 ) − 2 log10 K P,f ( CO ) Evaluating this log equation in two different temperatures one has (a) At 1000 K

13 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

log10 K P,f ( CO2 ) = 20.68 and 2 log10 K P,f ( CO ) = 2 (10.46 ) = 20.92 so

log Kp = −0.24 or KP = 0.58 (b) At 2000 K log10 K P,f ( CO2 ) = 10.35 and 2 log10 K P,f ( CO ) = 2 ( 7.47 ) = 14.94 so

log KP = −4.59 or KP = 2.57 × 10−5 Therefore, KP drops as temperature increases. Equation (1) and (2) show that as KP decreases, the reaction given by Eqn. (1) is shifting to the left, thus less carbon forms as the temperature increases. (ii) Pressure effect KP =

p CO2 2 p CO

n CO2 ⎛ P ⎞ = ⎜ ⎟ n CO 2 ⎜⎝ ∑ n ⎟⎠

−1

Thus, as the pressure increases more CO2 forms, shifting the reaction from left to right and thus more carbon forms as the pressure is increased.

14 © 2008 Glassman and Yetter.

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Problem 7 Determine the equilibrium constant KP at 1000 K for the following reaction

2 CH4 ⇔ 2 H2 + C2H4 Solution: As explained on page 14 of the text, the equilibrium constant (given in terms of the equilibrium constant of formation) is:

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 ) K 2P,f ( CH4 )

At 1000 K it follows from Table 2 in the appendix that (note Table 2 provides the logl0 of K): K P,f ( H2 ) = 100 = 1 K P,f ( C2 H4 ) = 10−6.213 = 6.1235 ×10−7 K P,f ( CH4 ) = 10−1.011 = 0.0975 so evaluating Eqn. (1) for KP one has KP =

1× ( 6.1235 ×10−7 )

( 0.0975)

2

= 6.442 ×10−5 K P = 6.442 × 10-5

so

15 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 8 The atmosphere of Venus is said to contain 5% carbon dioxide and 95% nitrogen by volume. It is possible to simulate this atmosphere for Venus reentry studies by burning gaseous cyanogen (C2N2) and oxygen and diluting with nitrogen in the stagnation chamber of a continuously operating wind tunnel. If the stagnation pressure is 20 atm, what is the maximum stagnation temperature that could be reached and still have Venus atmosphere conditions? If the stagnation pressure were 1 atm, what would be the maximum temperature? Assume all gases enter the chamber at 298 K. Take the heat of formation of cyanogen as ( ΔH 0f ) = 381.84 k/mol. 298

Solution: In order to maintain Venus atmosphere, the composition of 100 moles of products would have to be 5 CO2 + 95 N2. Eqn. (10) in the textbook can be used to calculate the stagnation temperature of this system. This equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified the solution approach is to find T2 such that the above equality will be satisfied. (a) for Venus atmosphere the general combustion reaction would be a C2N2 + b O2 + c N2 → 5 CO2 + 95 N2 which can be balanced for one mole of C2N2 as follows: C2N2 + 2 O2 + 36 N2 → 2 CO2 + 38 N2, In preparation to apply Eqn. (1) the following tables can be generated. gases

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

C2N2 O2 N2

381.84 0.0 0.0

0.0 0.0 0.0

An expression for the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the reactants becomes: H r = 1( 381.84 + 0.0 )C N + 2 ( 0.0 + 0.0 )O + 36 ( 0.0 + 0.0 ) N = −381.84 kJ 2

2

2

2

16 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

For the products it follows that:

gases

ΔH 0f ( kJ mol )

N2 CO2

0.0 −393.77

Thus, an expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the products becomes let ΔH = H 0T2 − H 0298 :

(

(

)

(

H p = 2 −393.77 + ΔH CO2 + 38 0.0 + ΔH N2

)

)

Equating Hr and Hp yields: 2 ΔH CO2 + 38 ΔH N 2 = 1,169.37 A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 40 = 2 + 38): 40 ΔH N2 = 1,169.37 so ΔH N 2 = 29.23kJ mol , therefore, T2 ≈ 1232 K We will then assume that T2 is between 1200 and 1300 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ): T2

ΔH CO2 ( kJ mol )

ΔH N2

Hp − Hr (kJ)

44.51 50.19

28.13 31.52

−11.51 128.87

1200 1300 Interpolate

1200 T2 1300

−11.51 00 128.31

T2 − 1200 11.51 = 100 139.88 T 2 = 1208 K

so

Since K P,f ( CO2 ) = 1.75 × 1017 at 1200 K, no significant dissociation of the products occurs. Therefore, pressure has no effect and T2 would remain the same at all pressures. 17 © 2008 Glassman and Yetter.

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18 © 2008 Glassman and Yetter.

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Problem 9 A mixture of 1 mole of N2 and 0.5 mole O2 is heated to 4000 K at 0.5 atm pressure and results in an equilibrium mixture of N2, O2, and NO only. If the O2 and N2 were initially at 298 K and the process is one of steady heating, how much heat was required to bring the final mixture to 4000 K on the basis of one initial mole of N2? Solution: The first step is to find the equilibrium concentration of products. Once this is done, one can use Eqn. (1.10) to calculate the amount of heat necessary to bring the final mixture to 4000 K. Consider the following reaction at equilibrium.

1 1 N 2 + O 2 → a N 2 + b O 2 + c NO 2

(1)

balancing this reaction yields the following relationship between a, b and c. N: 2 = 2 a + c 0: 1 = 2 b + c Solving these two equations for b and c in terms of a it follows that b=

2a − 1 2

and c = 2 − 2a

Therefore, Eqn. ( 1) becomes: 1N 2 + 0.5O 2 → a N 2 +

2a − 1 O 2 + ( 2 − 2a ) NO 2 1 1 N 2 + O 2 ⇔ NO , the expression for the equilibrium 2 2

For the equilibrium reaction given by constant of formation is given by: 1−1 2 −1 2

K P,f ( NO )

⎛ P ⎞ n NO = ⎟ 1/ 2 1/ 2 ⎜ n N2 n O2 ⎜⎝ ∑ n ⎟⎠

2a − 1 + 2 − 2a = 3 . At 4000 K, 1og10 KP,f(NO) = 2 - 0.524 or KP,f(NO) = 0.29923. Squaring Eqn. (3) would give the following equation for the variable a: The total number of moles is given by

∑n = a +

19 © 2008 Glassman and Yetter.

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( 2 − 2a )

2

2a − 1 a 2

− K 2P,f ( NO ) = 0,

Solving for a in the above equation and then using the mass balance equations to solve for b and c yields: a = 0.909, b = 0.409 and c = 0.182. Equation (2) now becomes 1 N2 + 0.5 O2 → 0.909 N2 + 0.409 O2 + 0.182NO The amount of heat required to bring the above mixture to 4000 K is calculated using Eqn. (1.10) in the text by taking T2 = 298 K and T1 = 4000 K: −Q =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(4)

from tables in the text one has: gases

O2 N2 NO

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

H 0T2 − H 0298 ( kJ mol )

0.0 0.0 90.43

139.01 130.16 132.76

0.0 0.0 0.0

so the equation for Q becomes −Q = 0.182 ( 90.43 + 132.76 ) NO + 0.409 ( 0.0 + 139.01)O + 0.909 ( 0.0 + 130.16 ) N − 2

1( 0.0 + 0.0 ) N − 1/ 2 ( 0.0 + 0.0 )O 2

2

2

Q = - 215.79 kJ

so if no dissociation occurred the value of Q would be −Q = 1/ 2 ( 0.0 + 139.01)O + 1( 0.0 + 130.16 ) N − 2

2

1( 0.0 + 0.0 ) N − 1/ 2 ( 0.0 + 0.0 )O 2

2

so

Q = - 199.67 kJ

Therefore, more energy is required to raise the temperature of the mixture to 4000 K due to dissociation. 20 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 10 Calculate the adiabatic decomposition temperature of benzene under the constant pressure condition of 20 atm. Assume that benzene enters the decomposition chamber in the liquid state at 298 K and decomposes into the following products: carbon (graphite), hydrogen, and methane. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are obtained. Consider the following reaction at equilibrium: C6H6 (1) → a C (s) + b H2 + c CH4

(1)

balancing this reaction yields the following relationship between a, b and c. C: 6 = a + c H: 6 = 2 b + 4 c solving these two equations for a and b in terms of c it follows that b=

6 − 4c = 3 − 2c 2

and a = 6 − c

Therefore, Eqn. (1) becomes: C6 H 6(1) → ( 6 − c ) C(s ) +

6 − 4c H 2 + c CH 4 2

(2)

Now consider the equilibrium reaction given by C(s) + 2 H2 ⇔ CH4 For this equilibrium reaction, the expression for the equilibrium constant of formation of CH4 is 1− 2

K P,f ( CH4 )

n CH ⎛ P ⎞ = KP = 2 4 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

(3)

21 © 2008 Glassman and Yetter.

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6 − 4c + c = 3 − c (note that the number of moles of 2 graphite are not included in the calculation for the total number of moles at equilibrium). Equation (3) then becomes:

The total number of moles is given by

∑n =

c b+c − KP = 0 b 2 20 expressing b in terms of c gives:

c

( 3 − 2c )

2

3+ c − KP = 0 20

(4)

In order to calculate the adiabatic flame temperature one has to apply Eqn. (1.11) in the text taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: gases

C6H6 (1)

ΔH 0f ( kJ mol )

H 0T1 − H 0298 ( kJ mol )

49.03

0.0

And the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (5)) can be calculated as: Hr = 1 (49.03)C6H6 (1) = 49.03 kJ Let us first guess that T2 = 1000 K; therefore, logl0 KP = - 1.011 or KP = 0.097499. Solving Eqn. (4) for c using this value of KP and then solving the mass balance equations for a and b results in: c = 0.9943, b = 1.0114, and a = 5.0057 For a product temperature of 1000 K, application of Eqn. (5) using the values of a, b, and c found above yields: T2 1000

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

8.47

20.70

38.20

26.87

−22.16

22 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

If T2 = 1100 K, logl0 Kp. = - 1.44 or KP = 0.03631. Solving Eqn. (4) for c using this value of KP and then solving the mass balance equations for a and b gives: c = 0.7409, b = 1.5182, and a = 5.2591 Following the same procedure as previously outlined yields: T2 1100

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

14.01

23.74

45.58

88.01

38.98

Interpolation for the value of temperature where Hp − Hr = 0 yields T2 ~ 1036 K. At T2 = 1036K, logl0 Kp = -1.1654 or Kp = 0.06832, c = 0.9101, b = 1.1798, a = 5.0899 and T2 1036

ΔH C(s) ( kJ mol )

ΔH H2

ΔH CH4

Hp

Hp − Hr (kJ)

10.47

21.79

40.86

48.02

−1.01

Since Hp − Hr ~0

T 2 ~ 1036 K

23 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 11 Calculate the flame temperature and the product composition of liquid ethylene oxide decomposing at 20-atm pressure by the irreversible reaction:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4 The four products are as specified. The equilibrium known to exist is 2 CH4 ⇔ 2 H2 + C2H4 The heat of formation ΔH 0f of liquid ethylene oxide is = - 76.62 kJ/mol. It enters the decomposition chamber at 298 K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4 balancing this reaction yields the following equations for a, b, c and d. C: 2 = a + b +2 d O: 1 = a H: 4 = 4 b + 2 c + 4 d simplifying it follows that: b+2d=1

(1)

4b+2c+4d=4

(2)

The other equation for the solution of b, c, and d will come from the Kp expression for the given equilibrium reaction. 2 CH4 ⇔ 2 H2 + C2H4 For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is, 24 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 ) K 2P,f ( CH4 )

n 2H2 n C2 H4 ⎛ P ⎞ = ⎜⎜ ⎟⎟ 2 n CH ⎝ ∑n ⎠ 4

2 +1− 2

(3)

or in terms of b, c, and d ( with P = 20 atm.) one has KP =

c2d 20 2 b 1+ b + c + d

(4)

By combining the equation above with the mass balance equations the following expression can be written for KP: KP

(1 − b ) = b

2

2

20 5−b

(5)

The approach now is to guess T2, look up KP and solve for the three unknowns in Eqns. (1), (2), and (5). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(6)

With Eqns. (5) and (6) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: gases

C2H4O(1) CO CH4 C2H4

ΔH 0f ( kJ mol )

−76.62 −110.62 −74.90 52.34

(

)

Applying Eqn. (6) yields let ΔH = H 0T2 − H 0298 :

(

) (

) (

−76.62 = a ( −110.62 + ΔH CO ) + b −74.90 + ΔH CH4 + c 0.0 + ΔH H2 + d 52.34 + ΔH C2 H4 First guess that T2 = 1400 K, from Tables in the text one has:

25 © 2008 Glassman and Yetter.

)

(7)

1/25/2011 9:27 AM

log10 K P,f ( H2 ) = 1 log10 K P,f ( C2 H4 ) = −5.664 log10 K P,f ( CH4 ) = −2.372 so from Eqn. (3), log10 KP = 0.12023. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives: a = 1.0, b = 0.7557, c = 0.2443, and d = 0.1221 Application of Eqn. (7) using T2 = 1400 K and the values of a, b, c, and d just calculated yields: T2 1400

ΔHCO (kJ/mol) 35.36

ΔH CH4 69.66

ΔH H2 33.08

ΔH C2 H4 91.26

Hp (kJ) -53.60

If we now take T2 = 1300 K, from Tables in the text one has: log10 K P,f ( H2 ) = 1 log10 K P,f ( C2 H4 ) = −5.766 log10 K P,f ( CH4 ) = −2.107 so from Eqn. (3), log10 KP = - 1.552 or KP = 0.028054. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives: a = 1.0, b = 0.8398, c = 0.1603, and d = 0.0801 Application of Eqn. (7) using T2 = 1300 K and the values of a, b, c, and d just calculated yields: T2 1300

ΔHCO (kJ/mole)

ΔH CH4

ΔH H2

ΔH C2 H4

Hp (kJ)

31.89

61.34

29.93

80.63

-74.65

Extrapolation to the value of T2 where Hp ~ Hr (−76.62) yields: 2 ~ 1291 K T

26 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 12 Liquid hydrazine (N2H4) decomposes exothermically in a monopropellant rocket operating at 100-atm chamber pressure. The products formed in the chamber are N2, H2, and ammonia (NH3) according to the irreversible reaction

N2H4(1) → a N2 + b H2 + c NH3 Determine the adiabatic decomposition temperature and the product composition a, b, and c. Take the standard heat of formation of liquid hydrazine as 50.03 kJ/mole. The hydrazine enters the system at 298 K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

N2H4(1) → a N2 + b H2 + c NH3 balancing this reaction yields the following equations for a, b, and c. N: 2 = 2 a + c

(1)

H: 4 = 2 b + 3 c

(2)

The other equation necessary for the solution of a, b, and c will come from the Kp for the following equilibrium reaction. 1 3 N 2 + H 2 ⇔ NH 3 2 2 For this equilibrium reaction, the equilibrium constant of formation of NH3 can be expressed in terms of the mixture composition at equilibrium as shown: 1−1 2 −3 2

K P,f ( NH3 )

n NH ⎛ P ⎞ = K p = 1 2 33 2 ⎜ ⎟ n H2 n N2 ⎜⎝ ∑ n ⎟⎠

(3)

or in terms of a, b, and c (with P = 100 atm) one has:

27 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Kp =

c a b

1/ 2 3/ 2

a+b+c 100

(4)

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎣⎡( H i

)

− H 0298 + ( ΔH f0 )

0 T2

i prod

(

)

⎤ − n j ⎡ H 0T1 − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j∑ ⎣ react

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic properties: ΔH 0f ( kJ mol )

gases

N2H4(1) N2 H2 NH3

50.03 0.0 0.0 −46.22

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

) (

) (

50.03 = a 0.0 + ΔH N2 + b 0.0 + ΔH H2 + c −46.22 + ΔH NH3

)

(6)

Let us first guess that T2 = 900 K, from Tables in the text it follows that: log10 K P,f ( NH3 ) = −2.915

Solving Eqns. (l), (2) and (4) results in: a = 0.9487, b = 1.8462, and c = 0.1026 For a product temperature of 900 K, application of Eqn. (6) using the values of a, b, and c, found above, yields (all ΔH's are in kJ/mol and Hp is in kJ): T2 900 K

ΔH N2 ( kJ mol )

ΔH H2

ΔH NH3

Hp (kJ)

18.23

17.69

27.09

47.98

First guess that T2 = 1000 K, from Tables in the text it follows that: 28 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

log10 K P,f ( NH3 ) = −3.233

Solving Eqns. (1), (2) and (4) one has: a = 0.9739, b = 1.9218, and c = 0.05216 For a product temperature of 1000 K, application of Eqn. (6) using values of a, b, and c, found above yields: T2 1000 K

ΔH N2 ( kJ mole )

ΔH H2

ΔH NH3

Hp (kJ)

21.47

20.70

32.60

60.00

Interpolation to the value of T2 where Hp ~ Hr (50.03) yields: T 2 ~ 917 K

29 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 13 Gaseous hydrogen and oxygen are burned at l-atm pressure under the rich conditions designated by the following combustion reaction:

O2 + 5 H2 → a H2O + b H2 + c H The gases enter at 298 K. Calculate the adiabatic flame temperature and the product composition a, b, and c. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known.

Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium: O2 + 5 H2 → a H2O + b H2 + c H balancing this reaction yields the following equations for a, b, and c. O: 2 = a

(1)

H: 10 = 2 a + 2 b + c

(2)

The other equation necessary for the solution of a, b, and c will come from the KP expression for the following equilibrium reaction. 1 H2 ⇔ H 2 For this equilibrium reaction, the equilibrium constant of formation of H can be expressed in terms of the mixture composition at equilibrium as shown: 1−1/ 2

K P,f ( H )

n ⎛ P ⎞ = K P = 1/H2 ⎜ ⎟ n H2 ⎜⎝ ∑ n ⎟⎠

(3)

or in terms of a, b, and c (with P = 1 atm) one has KP =

c 1 b a+b+c

(4)

1/ 2

30 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that also satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K. −Q = 0 =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic properties: gases

ΔH 0f ( kJ mol )

O2 H H2 H2O

0.0 218.09 0.0 −242.00

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

) (

)

0.0 = a −242.00 + ΔH H2O + b 0.0 + ΔH H2 + c ( 218.09 + ΔH H ) Let us first guess that T2 = 2500 K, from Table 2 it follows that, log10 Kp,f(H) = −1.601 Solving Eqns. (1), (2) and (4) one has: a = 2.0, b = 2.9516, and c = 0.0967 For a product temperature of 2500 K application of Eqn. (6) using the values of a, b, and c, yields (all ΔH's are in kJ/mole and Hp is in kJ): T2 2500 K

ΔH H2O ( kJ mol )

ΔH H2

ΔHH

Hp (kJ)

99.03

70.54

45.80

−52.21

Let us first guess that T2 = 2700 K, from Table A2 it follows that 31 © 2008 Glassman and Yetter.

(6)

1/25/2011 9:27 AM

log10 Kp,f(H) = −1.247 Solving Eqns. (1), (2) and (4) one has: a = 2.0, b = 2.8912, and c = 0.2176 For a product temperature of 2700 K application of Eqn. (6) using the values of a, b, and c, found above yields (all ΔH's are in kJ/mol and Hp is in kJ): T2 2700 K

ΔH H2O ( kJ mol )

ΔH H2

ΔHH

Hp (kJ)

109.89

77.77

49.96

18.96

Interpolation to the value of T2 where Hp ~ Hr (Hr = 0.0) yields: T2 ~ 2647 K

32 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 14 The liquid propellant rocket combination nitrogen tetroxide (N2O4) and UDMH (unsymmetrical dimethyl hydrazine) has optimum performance at an oxidizer to fuel weight ratio of 2 at a chamber pressure of 67 atm. Assume that the products of combustion of this mixture are N2, CO2, H2O, CO, H2, O H, OH, and NO. Set down the equations necessary to calculate the adiabatic combustion temperature and the actual product composition under these conditions. These equations should contain all the numerical data in the description of the problem and in the tables in the appendices. The heats of formation of the reactants are: N 2 O 4(1)

ΔH f0 298 = −2.90 kJ mol

UDMH (1)

ΔΗ 0f 298 = +53.17 kJ mol

The propellants enter the combustion chamber at 298 K. Solution: The solution for this problem consists of finding the correct number of equations necessary to solve for the unknown quantities in this problem. There are 9 unknown species coefficients and the temperature is unknown. Therefore, 10 equations must be found to solve the problem. These equations consist of the energy, 4 mass balance, and 5 equilibrium equations.

To specify the general reaction equation, the mole proportion of reactants must be found first. This can be done as follows. The molecular weights of the reactants are: for N2O4,

MW = 92

for C2N2H8,

MW = 60

From the problem statement one has: mass UDMH =2 mass N 2 O 4 By assuming 1 mole N2O4 this equation becomes mass UDMH =2 60 or

nUDMH (92) = 120 33 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

So nUDMH = 1.304 The general equation for the reaction then becomes: C2N2H8 + 1.304 N2O4 → a N2 + b CO2 + c H2O + d CO + e H2 + f O + g H + h OH + i NO The mass balance equations can be written as: C: 2 = b + d N: 4.609 = 2a + i H: 8 = 2c + 2e + g + h O: 5.216 = 2b + c + d + f + h + i

(1) (2) (3) (4)

The energy equation would be the equation for adiabatic flame temperature (Eq. (11)) found in the text.

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ = ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

The following information can be used to simplify the equation ΔH fN2O4 = −2903 kJ mol ΔH f UDMH = 53.172 kJ mol The right side of the equation is: 1(53.172) + 1.304 (-2.093) = 50.44 kJ Therefore, the energy equation becomes: 50.44 =

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH f0 )

i prod

⎤ ⎦

(5)

298 i

Some of the possible equilbrium reactions include: H2 ⇔ 2 H CO + O ⇔ CO2 N2 + 2 O ⇔ 2 NO H2 + 2 OH ⇔ 2 H2O 34 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

CO + 2 OH ⇔ H2O + CO2 These reactions give the following equilibrium equations: K p1 =

n 2H n H2

⎛ P ⎞ ⎜⎜ ⎟⎟ ⎝ ∑n ⎠

n 2NO ⎛ P ⎞ K p2 = ⎜ ⎟ n N2 n O2 ⎜⎝ ∑ n ⎟⎠

−1

n CO2 ⎛ P ⎞ K p3 = ⎜ ⎟ n CO n O ⎜⎝ ∑ n ⎟⎠

−1

n 2H2O ⎛ P ⎞ K p4 = 2 ⎜ ⎜ ∑ n ⎟⎟ n H2 n OH ⎝ ⎠

∑n

i

g 2 ⎛ 67 ⎜ e ⎜⎝ ∑ n i

b = df

⎛ ∑ ni ⎜⎜ ⎝ 67

(6)

(7)

⎞ ⎟⎟ ⎠

(8)

=

c2 ⎛ ∑ n i ⎜ e h 2 ⎜⎝ 67

⎞ ⎟⎟ ⎠

(9)

=

c b ⎛ ∑ ni ⎜ d h 2 ⎜⎝ 67

⎞ ⎟⎟ ⎠

(10)

−1

n H O n CO ⎛ P ⎞ K p5 = 2 2 2 ⎜ ⎟ n CO n OH ⎜⎝ ∑ n ⎟⎠

where

⎞ ⎟⎟ ⎠ i2 ⎛ ∑ ni ⎞ = 2 ⎜⎜ ⎟ a f ⎝ 67 ⎟⎠ =

−1

= a + b + c + d + e + f + g + h + i, and the Kp's are evaluated at the adiabatic flame

temperature.

35 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 15 Consider a fuel burning in inert airs and oxygen where the combustion requirement is only 0.21 moles of oxygen. Order the following mixtures as to their adiabatic flame temperatures with the given fuel.

a) b) c) d)

pure O2 0.21 O2 + 0.79 N2 0.21 O2+0.79 Ar 0.21 O2 + 0.79 CO2

(air)

Solution: The temperatures order with respect to the Cp of the inert and excess oxygen. The lower the Cp of the inert gas the higher the temperature. Thus, Ar being monoatomic has the lowest Cp, O2 and N2 are about the same, but higher than Ar since they are diatomic and have vibrational and rotational degrees of freedom, and CO2 being triatomic has the highest Cp. Thus the order from highest to lowest is

c, a and/or b, d

36 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Problem 16 Propellant chemists have proposed a new high energy liquid oxidizer, penta-oxygen O5, which is also a monopropellant. Calculate the monopropellant decomposition temperature at a chamber pressure of 10 atm if it assumed the only products are O atoms and O2 molecules. The heat of formation of the new oxidizer is estimated to be very high, + 1025 kJ / mol. Obviously the amounts of O2 and O must be calculated for one mole of O5 decomposing. The O5 enters the system at 298 K. Hint: The answer will lie somewhere between 4000 and 5000K. Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

O5→ a O2 + b O balancing this reaction yields the following equation for a. O: 5 = 2a + b or b = 5 - 2a

(1)

The other equation necessary for the solution of a and b will come from the Kp expression for the following equilibrium reaction. 1 O2 ⇔ O 2 For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is, KP =

K P,f ( O ) K1/P,f2( O2 )

n ⎛ P ⎞ = 1/O2 ⎜ ⎟ n O2 ⎜⎝ ∑ n ⎟⎠

2 −1

(2)

or in terms of a and b ( with P = 10 atm.) one has 1/ 2

b ⎡ 10 ⎤ K p = 1/ 2 ⎢ a ⎣ a + b ⎥⎦

(3)

By combining the equation above with the mass balance equation the following expression can be written for Kp: 37 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

5 − 2a ⎡ 10 ⎤ a1/ 2 ⎢⎣ 5 − 1 ⎥⎦

1/ 2

Kp =

(4)

The approach now is to guess T2 look up Kp and solve for the two unknowns in Eqns. (1) and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation. In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K.

∑ n ⎡⎣( H i

0 T2

)

− H 0298 + ( ΔH 0f )

i prod

(

)

⎤ − ∑ n j ⎡ H 0T − H 0298 + ( ΔH 0f ) ⎤ 298 ⎦ j ⎦ j react ⎣ 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties: ΔH 0f ( kJ mol )

gases

O O2 O5

249.36 0 1025

(

)

Applying Eqn. (5) yields let ΔH = H 0T2 − H 0298 :

(

)

1025 = a 0.0 + ΔH O2 + b ( 249.36 + ΔH O )

(6)

First guess that T2 = 4500 K, from Tables in the text one has: log10 K P,f ( O2 ) = 1 log10 K P,f ( O ) = 0.543 so from Eqn. (2), log10 Kp = 0.543 or Kp = 3.491. Solving Eqn. (4) for a and then Eqn. (1) for b gives: a = 1.29, b = 2.42 Application of Eqn. (6) using T2 = 4500 K and the values of a and b just calculated yields: T2

ΔH O2 ( kJ mol )

ΔHO

Hp (kJ/mol)

4500

159.9

88.45

1023.77

38 © 2008 Glassman and Yetter.

1/25/2011 9:27 AM

Since the guessed value of T2 give Hp ~ H (1025 kJ/mol) the decomposition temperature is T 2 ~ 4500 K

39 © 2008 Glassman and Yetter.

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