Chapter 14 Chemical Reactions

14-16 Ethane is burned with an unknown amount of air during a combustion process. The AF ratio and the

percentage of theoretical air used are to be determined.

Assumptions 1 Combustion is complete. 2 The combustion products contain C02, H2O, O2, and N2 only.

Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively

(Table A-I).

Analysis (a) The combustion equation in this case can be written as

~H6"+c{O2+3.76'1J ~ 2CO2+3H2O+202+3.76cN2

O2 balance: a = 2 + 15 + 2 a =55~

Substituting, C2H6 +55[02 +3.76N2] ~ 2C02 +3H2O+202 +20.68N2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,

AF = ~ = (~:5 x 4. 76kmolX29kg/kmol ) = 25.3 kgair/kgfuel

mfuel (2kmoIX12kg/kmol)+ (3kmoIX2kg/kmol)

(b) To find the percent theoretical air used, we need to know the theoretical amount of air. which isdetermined from the theoretical combustion equation of ClH6,

~H6 + Oth[ O2 + 3.7Et'J 2] --+ 2CO2 + 3H2O + 3. ~N 2

O2 balance: Clth = 2+ 15 Cith = 35--+

Then,

Percent theoretical air = -~ = 157 %

-35

14-4

Chapter 14 Chemical Reactions

14-22 Butane is burned with air. The masses of butane and air are given. The percentage of theoretical airused and the dew-point temperature of the products are to be determined.

Assumptions 1 Combustion is complete. 2 The combustion products contain COz, HzO, and Nz only. 3Combustion gases are ideal gases.

Properties The molar masses of C, Hz, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively(Table A-l ).

Analysis (a) The theoretical combustion equation in this case can be written as

C4H1O + 0t!1[ O2 + 3.7EX'\12] r 4CO2 + 5H2O + 3. 7~N 2

where Qlh is the stoichiometric coefficient for air. It is determined from

O2 balance: Clth = 4+ 25 Oth=65--4

The air-fuel ratio for the theoretical reaction is determined by laking lhe ratio of the mass of lhe air lo themass of lhe fuel for .

mair,lh

mfuel

AFth == , (~;5 x 4.76k~oIX29kg/kmol) .

(4kmoIX12kg/kmol)+ (5kmoIX2kg/kmol) = 15.5kgalr/kgfuel

The actual air-fuel ratio used is

25 kg

I kgAF =~-act -

mfuel

= 25 kg air I kg fuel

(b) The combustion is complete, and thus products will contain only C02, H2O, O2 and N2° The air-fuelratio for this combustion process on a mole basis is

-AF Nair mair /M air (25 kgy(29 kg/kmol) 50 k 1 ° /k If 1= -= -= mo air mo ue

Thus,

T dp = Tsar @8571 kPa = 42.7°C

14-9,

j

NfueJ mfuel/Mfuel .(1 kg)/(58 kg/kmol)

Thus the combustion equation in this case can be written as

C4H1O +(50/4.76XO2 +3.76N2]~4CO2 +5H2O+4.002 +39.5N2

The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the

product gases corresponding to its partial pressure. That is,

Chapter 14 Chemical Reactions

14-42 The enthalpy of combustion of liquid octane at a 25°C and 1 atm is to be determined using the datafrom Table A-26 and to be compared to the value listed in Table A-27.

Assumptions The water in the products is in the liquid phase.

Analysis The stoichiometric equation for this reaction is

C8Hl8 +12.5[02 +3.76N2]~8C02 +9H2O(f.)+47N2

Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, N2 and O2

are stable ele.ments, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C8HI8

becomes

Using 11; values from Table A-26.

hc = (8kmoIX- 393.520kJ/kmol)+ (9kmoIX- 285.830kJ/kmol)

-(lkmolX- 249.950kJlkmol)

= -5,470,680 kJ

The listed value in Table A-27 is -5,512,200 kJ/kmol for gaseous octane. The hc value for liquid octane is

obtained by adding ~ = 41 ,460 kJ/kmol to it, which yields -5,470, 740 kJ .Thus the two values are

practically identical. Since the water in the products is assumed to be in the liquid phase, this h, valuecorresponds to the higher heating value of CsH!s.

14-20

Chapter 14 Chemical Reactions

25°C2~=3+2+1.~ ~ ~=5

Air--.12°C

C3Hg(f.)+12.5(O2 +3.76N2)~3CO2 +4H2O+7.502 +47N2

(a) The air-fuel ratio for this combustion process is

mair {12.5 x 4. 76kmol X 29kg/kmol )AF=

-= 39.2kgair/kgfuel

~

mfuel

-QOU! = IN p(h; + h-ho)p -IN R(h; + h -ho)R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

II; h285 K h298 K h1200 KSubstance

-r"' TT / n, ~~~~! k.J/kmol k.J/kmol k.J/kmol-C3Hs (e) -I 18,91D

O2 0 8296.5 8682 38,447N2 0 8286.5 8669 36, 777

H2O (g) -241,820 9904 44,380

C02 -393,520 9364 53,848The hi of liquid propane is obtained by adding ~ of propane at 25°C to hi of gas propane.

Substituting,

-QOUI= (3X- 393,520+ 53,848- 9364)+ (4X- 241,820 + 44,380- 9904)+ (7.5Xo + 38,447 -8682)

+ (47Xo + 36,777- 8669)- (lX-118,910 + h298 -h298 ) -(12.5Xo + 8296.5- 8682)

-(47Xo + 8286.5- 8669 )= -190,464kJ/kmoIC3H8

or

9o1i = 100.464 kJ /krrol C3HB

Then the rate of heat transfer for a mass flow rate of 0.1 kg/min for the propane becomes

QOUI =NQou, =(~)QOUI =(~}190,464kJ/kmol)=S194 kJ/min

Products

1200K

wnere ath IS the stoichiometric coefficient and is Q

determined from the O2 balance. CJH8

Chapter 14 Chemical Reactions

Products

Tp

I

C2H2

O2

N2

H2O (g)

C02

226, 730

0

0

-241,820

-393,520

8682

8669

9904

9364

8736

8723

Thus;

~

-75,000 = (2)(- 393,520 + hco2 -9364)+ (1)(- 241,820 + hH2O -9904)

+ (0.75)(0 + hO2 -8682)+ (12.22)(0 + hN2 -8669 )- (1)(226,730)

-(3.25)(0 + 8736- 8682) + (12.22)(0 + 8723- 8669)

It yields 2~2+ ~20 + Q 75~2 + 12Z2.~2 = 1.':r2;1184 kJ

The temperature of the product gases is obtained from a trial and error solution. A first guess is obtained

by dividing the right-hand side of the equation by the total number of moles, which yields 1,321 , 184/(2 + I+ 0.75 + 12.22) = 82,729 kJlkmol. This enthalpy value corresponds to about 2500 K for N2. Noting that

the majority of the moles are N2, T p will be close to 2500 K, but somewhat under it because of the higher

specific heats of CO2 and H2O.

At 2350 K:

2hco2 +hH2O +0.75ho2 +12.22hN2 =(2)(122,091)+(1)(100,846)+(0.75)(81,243)+(12.22)(77,496)

= 1,352,961kJ ( Higher than 1,321 ,184 kJ )At 2300 K:

14-40

= 1,320,517kJ (Lowerthanl,321,184kJ)

By interpolation, T p = 2301 K

C2H2 +1.3athl02 +3.76N2)~2C02 +H2O+0.3ath02 +(1.3)(3.76)athN2 75.000kJ/kmol

where ath is the stoichiometric coefficient and is determinedfrom the O2 balance, ~

25°C1..'X:ltt1 = 2+ 0.5+ 0..'X:ltt1 ~ Clth = 2.5 .

AIrThus, -+-

30% excess airC2H2 +3.25(02 +3.76N2)~2C02 +H2O+0.7502 +12.22N2 27°C

Under steady-flow conditions the energy balance Ein -Eout = Msystem applied on the combustion chamber

with W = O reduces to

-QOUI =LNp(h; +h-ho)p -LNR(h; +h-ho)R

Assuming the air and the combustion products to be ideal gases, we have h = h(7). From the tables.

--h; h298 K h300 K

Substance kJ/kmol kJ/kmol kJ/kmol

Chapter 14 Chemical Reactions

CsH1s(f.)+1.5ath(02 +3.76N2~8CO2 +9H2O+O.5atnO2 + (1.5X3.76~thN2

where arb is the stoichiometric coefficient and is determined from the T o = 298 K

O2 balance, 0CBH1B (e)

~

Air I

50% excess air

25°C

l.~ = 8+ 4.5+ O.~ Dth = 12.5~Products

25°C

CgH]g(t')+18.75(O2 +3.76N2) --+ 8CO2 +9H2O+6.2502 +70.5N2

Under steady-flow conditions the energy balance E;, -E,", = ~E,y",m applied on the combustion chamber

with W = O reduces to

-QO"' = l:::Np(h; +h -ho )p -LNR(h; +h -ho )R QO"' = l:::Nph;p -LNRh;J1

since all of the reactants are at 25°C Assuming the air and the combustion products to be ideal gases, wehave h = h(7) From the tables,

hoSubstance kJ /k r

-mol

CHRIs (I.) -249,950

O2 0

N2 0

R2O (I) -285,830

C02 -393,520

i,,

j

Substituting,

-Qout = (8X-393.520)+ (9X-285,830)+0+0-(IX-249,950)-0-0 = -5,470,680 kJ /krnol of C8HI8

or Qott = 5.470.~kJ/knDI of CSH1S

The CsHJs is burned at a rate of 0.4 kg/min or

.I.U~,

The heat transfer for this process is also equivalent to the enthalpy of combustion of liquid C8H18, which

could easily be de determined from Table A-27 to be ft, = 5,470,740 kJ/kmol C8H18.

14-47

Chapter 14 Chemical Reactions

(b) The entropy generation during this process is determined from

~

The CgH1g is at 25°C and I atm, and thus its absolute entropy is ~ H = 360.79 kJ/kmol.K (Table A-26)."\.;8 18

The entropy values listed in the ideal gas tables are for I atm pressure. Both the air and the product gases

are at a total pressure of I atm, but the entropies are to be calculated at the partial pressure of thecomponents which is equal to P i = yj P total, where yj is the mole fraction of component i. Also,

Si =Nisi(T,p;)=Ni~i.(T,Po)-Ru In(y,Pm))

The entropy calculations can be presented in tabular form as

Sp = 17,538 kJ/KThus,

and

Sgen =Nsgen =~.51xlO-3 krnol/min}17,799 kJlkmol.K)=62.47 kj/min.K

(c) The exergy destruction rate associated with this process is determined from

X destroyed =ToSgen =(298 KX62.47 kJ/min.K)=18,617 kJ/min=310.3 kW

14-48

Chapter 14 Chemical Reactions

,lcii,;

14-91 The highest possible temperatures that can be obtained when liquid gasoline is burned steadily withair and with pure oxygen are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic andpotential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic.

Analysis The highest possible temperature that can be achieved during a combustion process is the

temperature which occurs when a fuel is burned completely with stoichiometric amount of air in anadiabatic combustion chamber. It is determined from

L N p (h i + h -h o )p = L N R (h i + h -h o )R ~ L N p (hi + hr -h o )p = (Nh i )C8H18

since all the reactants are at the standard reference temperature of 25°C, and for O2 and N2. The theoreticalcombustion equation of CsHIs air is

CgHIg +12.5(02 +3.76N2)~8C02 +9H2O+47N2

From the tables.

ii; h29S KSubstance

kJ/k Imo kJ/kmol

CsHIs (l) -249,950

O2 0 8682

N2 0 8669

H2O (g) -241,820 9904

C02 -393,520 9364

Thus,

It yiel~s

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by

dividing the right-hand side of the equation by the total number of moles, which yields 5,646,081/(8 + 9 +47) = 88,220 kJ/kmol. This enthalpy value corresponds to about 2650 K for N2. Noting that the majority

of the moles are N2, T p will be close to 2650 K, but somewhat under it because of the higher specific heat

ofH2O.

At 2400 K: 8hco2 + 9hH20 + 47hN2 = (8Xl25,l52)+ (9 XlO3,508)+ (47X79,320)

= 5,660,828k1 (Higherthan5,646,O8lkJ )

8hco.. +9hH..o +47hN.. =(8Xl22,O9l)+(9XlOO,846)+(47X77,496)..-= 5,526,654k1 (Lowerthan5,646,O8lkJ)

At 2350 K:

By interpolation, T p = 2395 K

If the fuel is burned with stoichiometric amount of pure O2, the combustion equation would be

CsHIS + 12502 8CO2 + 9H2O

Thus,

14-66

It yields

Byextrapolation, Tp = 3597 K

14-67

Chapter 14 Chemical Reactions

14-93 Liquid octane is burned with 200 percent excess air during a steady-flow combustion process. The

heat transfer rate from the combustion chamber, the power output of the turbine, and the reversible work

and exergy destruction are to be determined.

Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion

gases are ideal gases. 4 Changes in kinetic and potential energies are negligible.

Properties The molar mass ofCgH1g is 114 kg/kmol (Table A-1).

Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only C02,

H2O, N2, and some free O2. Considering 1 kmol of CgH1g, the combustion equation can be written as

CgH1g +3a1h(02 +3.76N2~8CO2 +9H2O+2athO2 +(3X3.76ath)N2

,

QC8H18 :

-.

25°C. 8 aIm

IAir

.?i:fttI = 8+ 4.5+ ~ Clth = 12.5~

Substituting,

1300 K

8 aIm

I:::J ~ wCombustion

gases

~\ -

950 K

2atm

Assuming the air and the combustion products to be ideal gases, wehave h = h(7). From the tables,

ii; ii5oo K ii298K iiI300 K ii950 KSubstance

~~~~ k.J/kmol k.J/kmol k.J/kmol k.J/kmol

C8HI8 (t) -249,950

O2 0 14,770 8682 42,033 26,652

N2 0 14,581 8669 40,170 28,501

H2O (g) -241,820 9904 48,807 33,841

C02 -393,520 9364 59,552 40,070

Substituting,

-QOUI = (8X- 393.520+ 59,522 -9364 )+ (9 X- 241,820+ 48,807 -9904 )

+(25Xo+42,033-8682)+ (141Xo+40,170- 8669 )

-(lX- 249,950+ hz98 -h298)- (37.5Xo+ 14.770-8682)- (141Xo+ 14,581- 8669 )

= -109,675kJ/kmo1C8HI8

The C8HI8 is burned at a rate of 0.8 kg/min or

.m 0.8 kg/min -3 ., -= (()( ) ( )()) =7.018xlO kmoVmm8 12 + 18 1 kg/kmol

N=-M

Thus,

QOUI = NQou, = (7.018 x 10-3 kmollminXl09,675 kJ/kmol) =770 kJ/min

14-70

where ath is the stoichiometric coefficient and is determined from the O2

balance,

200% excessCsHIS +37.5(02 +3.76N2~8C02 +9H20+2502 +14lN2 air

The heat transfer for this combustion process is determined from the

energy balance Ein -Eoul :::: l1Esystem applied on the combustion

chamber with W:::: 0,

Chapter 14 Chemical Reactions

Substituting,

Wout =: (8X59,522 -40,070)+ (9 X48,807 -33,841)+ (25X42,033- 29,652)+ (141X40,1 70- 28,501)

=: 2,320,164 kJlkmol CSH18

Thus the power output of the turbine is

WOU! = Nwou! = ~.018x 10-3 kmollmin}2,320,164 kJ/kmo1)= 16,283 kJ/min = 271.4 kW

(c) The entropy generation during this process is determined from

Sgen

where the entropy of the products are to be evaluated at the turbine exit state. The CsHIs is at 25°C and 1

atm, and thus its absolute entropy is 5.. H =360.79 kJ/krnol. K (Table A-26). The entropy values listed in

~ 18

the ideal gas tables are for 1 atrn pressure. The entropies are to be calculated at the partial pressure of the

components which is equal to Pi = YiP IOlaJ, where Yi is the mole fraction of component i. Also.

Si = Nisi (T.P; )= Ni~.(T.Po)- Ruln(y;p m))

The entropy calculations can be presented in tabular form as

Nj YjRuln(y;P m )s;(T,1 atm) N .s.

I I

C8HI8 1 1.00 360.79 17.288 343.50

O2 37.5 0.21 220.589 4.313 8,110.34

-B2- 141 0.79 206.630 15~329 26,~

-CO2 8 ()()L1~7 '"\££ AAA --SB= 35,427.28 kJ~~

CO2 8 0.0437 266.444 -20.260 2,293.63

H2O 9 0.0490 230.499 -19.281 2,248.02

O2 25 0.1366 241.689 -10.787 6,311.90

-B2- 141 0.7~226.389 3.~95 31,~~

Sp = 42,267.48 kJ/KThus,

Sgen =42,267.48-35,427.28+!Q9,67~ = 7208.2 kJ/K (per kIno])298 K

Then the rate of entropy generation becomes

Sgen == Nsgen == ~ .018 X 10-3kmol/min }7208.2 kJ/kmol. K) == 50.59 kJ/min .K

and

x destruction

Wrev

=ToSgen =(298 K)(50.59 kJ/min .K)=15,075 kJ/min =251.2kW

= W + X destruction = 271.4 + 251.2 = 522.6 k W

14- 71