ME 354 Tutorial #10
Winter 2001
Reacting Mixtures
Problem 1:
Determine the mole fractions of the products of combustion when octane, C 8H18,
is burned with 200% theoretical air. Also, determine the dew-point temperature of
the products if the pressure is 0.1 MPa.
Step 1: Draw a diagram to represent the system
C8 H18
Air
Combustion
Chamber
CO2
H 2O
O2
(0.1 MPa)
N2
( 200% Theoretical)
Step 2: State your assumptions
Assumptions:
1) Combustion is complete
2) Combustion gases are ideal gases
3) Steady operating conditions exist
Step 3: Calculations
Before we start the solution to this problem let us examine the general cases of
complete combustion to develop a procedure for balancing the reaction
equations.
1) General Case for 100% Theoretical Air and Complete Combustion
C x H y  vO2 (O2  3.76 N 2 )  vCO2 CO2  v H 2O H 2 O  v N 2 N 2
where the vi’s are the stoichiometric coefficients.
Performing a mass balance on each of the elements i.e. requiring that the total
mass or mole number of each ELEMENT in the products should be equal to that
in the reactants.
Mass Balance
C’s  vCO2  x
H2’s  v H 2O 
y
2
O2’s  vO2  vCO2 
v H 2O
N2’s  v N 2  3.76vO2
2
 x
y
4
1
2) General Case for Excess Air and Complete Combustion
C x H y  vO2 (1  a)(O2  3.76 N 2 )  vCO2 CO2  v H 2O H 2 O  avO2 O2  v N 2 N 2
Let (1+a) = the percent of theoretical air, where “a” is the percentage of excess
air. For example, for 300% theoretical air (200% excess air), (1+a)=3 so a=2.
Mass Balance
C’s  vCO2  x
H2’s  v H 2O 
y
2
O2’s  (1  a)vO2  vCO2 
v H 2O
N2’s  v N 2  3.76(1  a)vO2
2
 avO2  vO2  x 
y
4
In the problem we are told octane, C8H18, is combusted with 200% theoretical air
(or 100% excess air) so a =1. Using the formulas developed for the general case
with excess air:
C’s  vCO2  x  8
y 18

9
2 2
y
18
 x 8
 12.5
4
4
 3.76(1  a)vO2  3.76(2)(12.5)  94
H2’s  v H 2O 
O2’s  vO2
N2’s  v N 2
Therefore, the balanced reaction equation is
C8 H18  25(O2  3.76 N 2 )  8CO2  9H 2 O  12.5O2  94 N 2
The total moles of the products, Nproducts, is
Nproducts=(8+9+12.5+94)=123.5
The mole fractions, yi, of each component are:
yCO2 = NCO2/Nproducts = (8/123.5) = 6.47%
yH2O = NH2O/Nproducts = (9/123.5) = 7.29%
Answer Problem#1 a)
yO2 = NO2/Nproducts = (12.5/123.5) = 10.12%
yH2 = NN2/Nproducts = (94/123.5) = 76.12%
Recall the dew-point temperature of the products is the temperature at which the
water vapor in the products starts to condense as the products are cooled. This
is the saturated temperature corresponding to the pressure of H2O (i.e. the partial
pressure of H2O in the mixture). Since we have assumed that the combustion
2
gases behave as ideal gases, we can determine the partial pressure of H2O from
the total pressure multiplied by the mole fraction of H2O, yH2O.
PH2O = yH2OP=(0.0729)(0.1 [MPa]) = 7.29 kPa
From Table A-4, interpolating in between 35C and 40C
Tdp  35 7.287  5.628

40  35 7.384  5.628
Tdp = 39.72C
Answer Problem#1 b)
Problem #2:
A small gas turbine uses C8H18 (l) for fuel, with 400% theoretical air. The air and
fuel enter at 25C and the products of combustion leave at 900K. The output of
the engine and the fuel consumption are measured, and it is found that the
specific fuel consumption is 0.25 kg/s of fuel per megawatt output. Determine the
heat transfer from the engine per kmol of fuel. Assume complete combustion.
Step 1: Draw a diagram to represent the system
Q
C8 H18
25 C
Air
Combustion
Chamber
(400% Theoretical)
CO2
H 2O
900K
O2
N2
Step 2: State your assumptions
Assumptions:
1) Combustion is complete
2) Combustion gases are ideal gases
3) ke, pe  0
4) steady operating conditions exist
Step 3: Calculations
Using the procedures developed in Problem #1, we can balance the reaction
equation for 400% theoretical air (a=3).
C’s  vO2  x  8
y 18

9
2 2
y
18
 x 8
 12.5
4
4
H2’s  v H 2O 
O2’s  vO2
3
N2’s  v N 2  3.76(1  a)vO2  3.76(4)(12.5)  188
 C8 H18  50(O2  3.76 N 2 )  8CO2  9H 2 O  37.5O2  188N 2
In order to determine the heat transfer from the system we need perform a first
law analysis on the combustion chamber. From Cengel and Boles (14-9) on
p777, the first law for combustion analysis (on a “per mole of fuel basis”) is as
shown in Eq1.
Qin  Win   N r (h f  h  h  ) r  Qout  Wout   N p (h f  h  h  ) p
(Eq1)
We will assume the heat transfer and work are out of the system so the Q in & W in
terms are zero. We will first determine the appropriate enthalpies for the
reactants. Note: the circle superscript is used to denote that the property value is
referenced to a 25C, 1atm reference state.
Reactants (C8H18, O2, and N2)
Enthalpy of formation, h f
For stable elements O2 and N2, the enthalpy of formation is zero
 h f O 2 = 0
 h f N 2 = 0
We can find the enthalpy of formation for C8H18 (l), from Table A-26
 h f = –249950 kJ/kmol
Ideal Gas Enthalpy relative to reference state, h  h 
In the h  h  term, the h term is the ideal gas enthalpy at the temperature of
interest and h  is the ideal gas enthalpy at the reference temperature. By
subtracting h  from h we are, in effect, referencing the ideal gas enthalpy at the
temperature of interest to an ideal gas enthalpy of zero at the reference state. By
doing this for all the reactants and products we ensure that all the enthalpy
values are referenced to the same reference state (25C, 1atm) and are thus
comparable.
Since the reactants are already at the reference state, h  h  = 0 for all the
reactants.
 h  h  O 2 =0
 h  h  N 2 =0
 h  h  C 8H 18 =0
4
Products (CO2, H2O, O2, and N2)
Enthalpy of formation, h f
For stable elements O2 and N2, the enthalpy of formation is zero
 h f O 2 = 0
 h f N 2 = 0
We can find the enthalpy of formations for CO2 and H2O (g), from Table A-26
 h f CO 2 = –393520 kJ/kmol
 h f H 2O = –241820 kJ/kmol
Ideal Gas Enthalpy relative to reference state, h  h 
Since the products are at 900K, we must determine h at 900K and subtract the
h at the reference state of 25C/298K (this is h  ) to ensure our enthalpies are all
calculated with respect to the same reference state.
For CO2,
 h@ 900K = 37405 kJ/kmol (Table A-20)
 h   h@ 298K = 9364 kJ/kmol (Table A-20)

 h h

CO 2
= 28041 kJ/kmol
For H2O (g),
 h@ 900K = 31828 kJ/kmol (Table A-23)
 h   h@ 298K = 9904 kJ/kmol (Table A-23)

 h h

H 2O
= 21924 kJ/kmol
For O2,
 h@ 900K = 27928 kJ/kmol (Table A-19)
 h   h@ 298K = 8682 kJ/kmol (Table A-19)

 h h

O2
= 19246 kJ/kmol
For N2,
 h@ 900K = 26890 kJ/kmol (Table A-18)
 h   h@ 298K = 8669 kJ/kmol (Table A-18)
5

 h h

N2
= 18221 kJ/kmol
We are also given the specific fuel consumption is 0.25 kg/s of fuel per megawatt
output of the engine. We can use this information to find the Wout as shown
below.
Wout
 MJ 
1

 s   114.24 kg fuel  456.96 MJ   456960 kJ 





1[kmol fuel ]
 kg fuel 
 kmol fuel 
 kmol fuel 


0.25

 s 


Note: The molar mass of C8H18 was determined from the molar mass of C
(12.012kg/kmol) and H (1.008kg/kmol)
 Mfuel=8(12.012)+18(1.008) = 114.24kgfuel/kmol.
Eq1 can be rearranged to isolate for Qout as shown in Eq2.
Qout   N r (h f  h ) r  Wout   N p (h f  h ) p
(Eq2)
Substituting in the values determined above into Eq2 we determine the heat
transfer from the engine. Note: we can determine the number of moles of each
product using our balanced reaction equation: NCO2=8, NH2O=9, NO2 =37.5, and
NN2 = 188.
 kJ 
 kJ   8(393520  28041)  9(241820  21924) 

Qout  (249950) 
  (456960) 
  
 kmol fuel 
 kmol fuel    37.5(19246)  188(18221)

 Qout =(-249950-456960+755623) kJ/kmolfuel
 Qout = 48713 kJ/kmolfuel
Answer Problem #2)
6