Chemical Reactions

Chemical Reactions
Objectives:
•Apply the conservation of mass to reacting systems to
determine balanced reaction equations.
•Defined the parameters used in combustion analysis, such
as air-fuel ratio, percent theoretical air, and dew point
temperature.
•Apply energy balance to reacting systems for both steady-
flow control volumes and fixed mass systems.
•Calculate enthalpy of reaction, enthalpy of combustion,
and the heating value of fuels.
•Determine the adiabatic flame temperature for reacting
mixture.
Combustion of fuel Heat Heat Engine Work
Combustion of fuel Chemical Reaction
How much fuel is needed? How about the air used?
How high the combustion flame temperature will be?
Steam Generator
Rocket Engine
Jet Engine
Piston Engine
1

Fuels
• Fossil Fuels: Coal, Petroleum, Natural Gases (LNG:
liquid natural gas, CNG: Compressed Natural Gas)
• Petroleum: Gasoline, Kerosene, Diesel, Fuel Oil ,
Liquid Petroleum Gas (LPG)
Fuel
Approx. single Hydrocarbon
Gasoline ~ Octane (C 8
H 18 )
Diesel
~ Dodecane (C 12 H 26 )
Methanol ~ Methyl Alcohol (CH 3
OH)
Natural gas~ Methane (CH 4
)
Combustion Air
• Combustion is a chemical reaction during which a fuel is
oxidized and a large quantity of energy is released.
• Oxidizer = Oxygen
• Commonly AIR is used as oxidizer (free)
• By volume: AIR = 21% O 2
+ 79% N 2
• Therefore, 1 mole of O 2
N 2
= 79/21 = 3.76 mole
1 kmol O 2 + 3.76 kmol N 2 = 4.76 kmol of Air
Assumption at normal combustion :
• N 2 absolutely inert (no reaction)
• Water is also inert
2

1 kmol C + 1 kmol O 2 1 kmol of CO 2
C + O 2 CO 2 (15.2)
Reactants
Combustion
Products
To have combustion reaction, it must:
• T > ignition temp
• Fuel : air ratio must proper
• To get good combustion 3T
– Temperature (high)
– Turbulent (good mixing between fuel and air)
– Time (enough to reach complete combustion)
•Balancing of chemical reaction: Conservation of mass
principle
• Air fuel ratio, AF = m air
air /m fuel
fuel (15.3)
3

Example 15.1 Balancing the combustion equation
One kmol of octane (C 8 H 18 ) is burned with air that
contains 20 kmol of O 2 , Assuming the products
contain only CO 2 , H 2 O, O 2 , and N 2 , determine the
mole number of each gas in the products and the airfuel
ratio for this combustion process.
Solution Chemical reaction equation:
C 8 H 18 +20 (O 2 + 3.76 N 2 ) xCO 2 + yH 2 O + zO 2 + wN 2
Balance of each element : @ ReactantS = @ ProductS
C: 8 = x x = 8
H: 18 = 2y y = 9
O: (20x2) = 2x +y + 2z z = 7.5
N 2 : 20*3.76 = w w = 75.2
Air Fuel Ratio, AF = m air /m fuel = (NM) air /(NM) fuel
= 24.2 kg air/kg fuel
Theoretical and Actual Combustion
Processes
Complete combustion: 1) all C burns to CO 2
, and 2) all
HC burns to H 2
O
Incomplete combustion: Products contains Unburned
fuel or components สาเหตุหลัก: 1)Insufficient O 2
2)
Insufficient mixing. 3) Dissociation
Stoichiometric Air or Theoretical Air:
• = minimum amount of air needed for complete
combustion.
• no unburned and no O 2
left in products
• = Chemically correct amount of air or 100%
theoretical air
4

Theoretical and Actual Combustion
Processes
Stoichiometric Combustion or Theoretical Combustion:
• Complete combustion with 100% theoretical air ; ex
CH 4
+ 2(O 2
+ 3.76N 2
) CO 2
+ 2H 2
O + 7.52N 2
Actual combustion processes: need excess air to
complete combustion
% Excess air: = % more air needed than theoretical air
80% excess air = 180% theoretical air
Deficiency of air: = amount of air used < theoretical air
(10% deficiency of air = 90% theoretical air)
Theoretical and Actual Combustion
Processes
Equivalent Ratio = (AF actual
)/(AF stoich
)
การประเมิน/ตรวจสอบความสมบูรณของการการเผาไหม
Orsat Gas Analyzer = อุปกรณวิเคราะหองคประกอบของ combustion
gas (products) ปจจุบันไมคอยนิยมใชแลว – ใช gas analyzer
5

Example 15.2 dew point temperature of combustion products
Ethane (C 2 H 6 ) is burned with 20% excess air during
combustion process. Assuming complete combustion
and a total pressure of 100 kPa, determine (a) the
air fuel ratio and (b) the dew point temperature of
the products.
Solution
(a) write the chemical reaction
equation at 120% theoretical air
C 2 H 6 +1.2a (O 2 + 3.76 N 2 )2CO 2 + 3H 2 O + 0.2aO 2 +(1.2x3.76)aN 2
From O 2 : 1.2a = 2+ 3/2+0.2a then a = 3.5
C 2 H 6 +4.2(O 2 + 3.76 N 2 ) 2CO 2 + 3H 2 O + 0.7O 2 + 15.79N 2
Air Fuel Ratio, AF = m air /m fuel = (NM) air /(NM) fuel
= (4.2x4.76 kmol)(29 kg/kmol)/[(2kmolx12 kg/kmol)x(3 kmol x 2 kg/kmol)]
= 19.3 kg air/kg fuel
C 2 H 6 +4.2(O 2 + 3.76 N 2 ) 2CO 2 + 3H 2 O + 0.7O 2 + 15.79N 2
(b) T dp = T sat @ P v (H 2 O)
Ideal gas mixture P i /P m = N i /N m
P v = (3 kmol/21.49 kmol)(100 kPa)
P v = 13.96 kPa T dp = 52.3 o C
6

Example 15.3 combustion of gasous fuel with moist air
A certain natural gas has the following volumetric analysis:
72% CH 4 , 9% H 2 , 14%N 2 , 2%O 2 and 3%CO 2 . The gases is
now burn with stochiometic amount of air that enters
combustion chamber at 20 o C, 1 atm, and 80%RH. Assume
complete combustion and a total pressure of 1 atm,
determine the dew point temperature of the products.
Solution
0.72CH 4 +0.09H2+ 0.02O 2 + 0.14N 2 + 0.03CO 2
ath(O 2 +3.76N 2 )+ xCO 2 + yH 2 O+zN 2
Example 15.3
1. write the chemical reaction equation
at 100% theoretical air (use dry air)
2. moles of air per kmol of fuel can be determined
3. then extra moles water vapor (20%RH at inlet conditions)
can be calculated
4. Rewrite the chemical reaction by adding the water vapor
into both side.
5. N of each component in products are known
6. T dp = T sat @ P v (H 2 O) (Ideal gas mixture P i /P m = N i /N m )
P v = 20.88 kPa T dp = 60.9 o C
7

Chemical Energy from Process
This session deal with the chemical energy within the
molecules of a closed system that involve a chemical
reaction. During a chemical reaction, some chemical
bonds that bind the atoms into molecules are broken
and new ones are formed. The chemical energy
associated with this process is usually different for the
reactants and products.
393,520 kJ
1 kmol C
at 25 o C, 1 atm
Combustion
Chamber
CO 2
at 25 o C, 1 atm
1 kmol H
at 25 o C, 1 atm
241,820 kJ
Combustion
Chamber
H 2
O (g)
at 25 o C, 1 atm
1 kmol O 2
at 25 o C, 1 atm
1 kmol O 2
at 25 o C, 1 atm
First Law of Thermodynamics
The first law of thermodynamics states that in any
closed system, energy is conserved. Which means that
energy cannot be created nor destroyed, but it can only
change forms. Meaning:
∆E sys = 0 and ∆E products =-∆E reactants
The molecules of a closed system possess energy in
various forms such as sensible and latent energy,
chemical energy, and nuclear energy. All of these forms
must balance out in the reactants and products to give
the system a net energy of zero.
8

Enthalpy Change
Enthalpy is the system we use to measure that change
in energy of a closed system due to chemical bonds
being broken. Reaction enthalpies are real physical
quantities for which numeric values can be calculated or
measured. In order to put the calculation into algebraic
form, chemists use the defined equation:
∆H = Σ∆H f (products) -Σ∆H f (reactants)
The reaction enthalpy, which is the enthalpy change that
occurs in the reaction, is always calculated as the sum
of the enthalpies of the products minus the sum of the
enthalpies of the reactants.
Standard Reference State
Since composition of a system at the end of a process is
no longer the same as that at the beginning of the
process, there’s a need to use a standard condition in
which to make the measurements from. This standard
condition is called the standard reference point, which
are 25 o C and 1 atm.
The superscript “ ° ” is used to indicate property values at
the standard state. The defined equation above, under
standard conditions, becomes:
∆H ° = Σ∆H ° f(products) -Σ∆H ° f(reactants)
9

Enthalpy of Combustion
The enthalpy of reaction in a combustion process is called
the enthalpy of combustion (symbolized by h c ). The
calculation for an enthalpy of combustion is done for 1
kmol (1 kg) of fuel is burned completely at a specified
temperature and pressure and can be expressed:
h c =H prod –H react
Example 15.5 Evaluation of the Enthalpy of Combustion
Determine the enthalpy of combustion of liquid octane
(C 8 H 18 ) at 25 o C and 1 atm, using enthalpy-of-formation data
from table A-26. Assume the water in the products is in the
liquid form.
Solution
1. Write chemical reaction equation
based on 1 kmol of octane
C 8
H 18 +a(O 2
+ 3.76 N 2
) 8 CO 2
+ 9H 2
O(l) + 3.76aN 2
Balance of oxygen : get a = 12.5
2. Energy analysis h C = H P –H
_ _ R
o
o
H R = ∑N h f ,R = (Nh
_ _ f ) C8H18
_
o
o
o
H P = ∑N h f = Nh f ) + Nh f )
,P
(Nh CO2
(Nh
,P
H2O
10

Table A-26 A
Enthalpy of Formation
_
o
h f ,CO 2
= -393,520 kJ/kmol
kmol
_
= -285,830 kJ/kmol
kmol
h f
o
_
h f
o
,H2O
,C8H18
= - 249,950 kJ/kmol
kmol
Enthalpy of combustion = - 5,471,100 kJ/kmol C 8 H 18
= -47,891 kJ/kg C 8 H 18
Disscussion: This is the HHV of liquid C 8 H 18 in Table A-27
Enthalpy of formation has 2 values : 1. for vapor vapor phase 2. for liquid vapor
phase -the different = laten heat of vaporization
11

Enthalpy of Formation
The enthalpy of formation is defined as the
enthalpy of a substance at a specified state due
to its chemical composition. This property makes
analyzing easier because it represents chemical
energy of an element or a compound at the
standard reference state.
The property values are obtained by first
assigning all of the elements in its chemically
stable form at the standard reference state a
value of zero (such N2, O2, N2, C).
Enthalpy of Formation
So we can use this concept to find the enthalpy of
formation of individual compounds by adding up the
enthalpy for each reaction it takes to react some of
the chemically stable elements to get the compound.
•Consider the formation of CO 2 (a compound) from
elements C and O 2 at 25 o C, 1 atm. during SSSF
process
1 st law: Q cv + ΣH i = ΣH e
Q cv = H P –H R
Q cv = - 393,520 kJ
H R = 0 ; elements @ ref. state
Enthalpy of Formation of CO 2
= - 393,520 kJ/kmol
12

Heating Value of Fuel
= the amount of heat released when a fuel is burned completely in
SSSF process and the products returned to the state of the
reactants…..
= absolute value of the enthalpy of combustion of the fuel:
Heating value = |h C | kJ/kg fuel
Higher heating value, HHV,:
H 2 O in products is in LIQUID form
Lower heating value, LHV,:
H 2 O in products is in Vapor form
HHV = LHV + (mh fg ) H2O (15.7)
m = mass fo water in the products per unit mass of fuel
h fg = the enthalpy of vaporization of water at the specified temp.
13

15.4 First Law Analysis of Reacting System
SSSF Process: (see Chapter 4 the first law) ∆KE~0,
∆PE~0
Q
Q
o
cv
cv
(i = Reactants, and
where
(scrip
Q
cv
o
o
+ Σ n
i
+ ΣN
_
_
i
_
h
_
i
h
i
_
o
f
h = h
_
o
f
h = h
= W
o
cv
= W
+ ( h−
h
e = Products)
)
stand for reference state which is 25 C,1atm)
_
Pr od
cv
_
_
+ ∆ h
o
+ Σ n
_
o
e
+ ΣN
e
Ract
_
T →To
−Wcv
= ΣNe
he
- ΣNi
hi
= H − H
h ........(15-8) per rate of mole eqn.
_
e
_
h ......(15-9) per unit mole eqn.
e
o
.........(15 -11)
Closed System ∆KE~0,
∆PE~0
from
Then
or
Q
Q
12
12
Q there is no u
= W
= W
U = N{ h
U = N{ h
_
o
f
12
12
+ ( U
+ ( U
U = H − PV
_
o
f
_
o
f
Prod
+ ( h−
h
_
2
_
−U
_
o
+ ( h−
h
_
o
)} − PV
Provided in tables
1
)
−U
React
_
)........(15.....)
) − P v}...........(15....)
14

Example 15.6 First Law Analysis of Steady-Flow Combustion
Liquid propane (C 3 H 8 ) enters a combustion chamber at 25 o C
at a rate of 0.05 kg/min where it is mixed and burned with
50% excess air that enters the combustion chamber at 7 o C,
as shown in the figure. An analysis of the combustion gases
reveals that all the hydrogen in the fuel burns to H 2 O but
only 90% of carbon burns to CO 2 , with the remaining 10%
forming CO. If the exit temperature of the combustion
gases is 1,500 K, determine (a) the mass flow rate of the air
and (b) the rate of heat transfer from the combustion
chamber.
Solution: Concepts
1. Write chemical reaction equation based on 1 kmol of propane
1.1 Theoretical Air-Fuel ratio (Stochiometic)
1.2 with 150% theoretical air
+ incomplete burned CO AF mass flow rate of air
2. Energy balance: SSSF Q = H P –H R
- Stochiometic combustion equation based on 1 kmol of propane
C 3
H (l)+a(O 8 2
+ 3.76 N 2
) 3CO 2
+ 4H 2
O + 3.76aN 2
Balance of oxygen : get a = 5
- 150% theoretical air combustion equation based on 1 kmol of propane
with 90%CCO 2 + 10%CCO
C 3
H (l)+(1.2*5)(O
8 2 + 3.76 N 2 ) (0.9*3)CO
2
+ (0.1*3)CO + 2.65O 2
+ 4H 2
O + 28.2N 2
(a) AF = m air /m fuel = (NM) air /(NM) fuel = ………………. = 25.53 kg air/kg fuel
mass flow rate of air, m dot,air = m dot,fuel AF
= (0.05 kg fuel/min)(25.53 kg air/kg fuel) = 1.18 kg air/min
Answer
15

Q
(i = Reactants, and
Q
cv
cv
where
2
_
+ ΣN
h = W
i
_
_
= ΣN
h − ΣN
h
_
o
f
Remark : h
e
i
e
_
o
f
cv
+ ΣN
i
_
e = Products, W
_
_
o
h = h + ( h−
h ),
_
o
f
_
o
f
( g)
= h
_
i
298K
_
( l)
+ h
e
_
h
e
_
o
( h
cv
= 0)
_
= h
Assume air and combustion gases are ideal gases, get data from the property tables
h
h
Substitute, Q cv
= 363,880 kJ/kmol fuel 8,270 kJ/kg fuel Q dot
= m dot
Q = 6.89 kW
_
280K
_
298K
1500K
Substance (kJ/kmol) (kJ/kmol) (kJ/kmol) (kJ/kmol)
C3H8(
l)
-118,910 0 NA NA
O2
0 8,682 8,150 49,292
N2
0 8,669 8,141 47,073
H 2O(
g)
- 241,820 9,904 NA 57,999
CO -393,520 9,364 NA 71,078
CO -110,530 8,669 NA 47,517
fg
h
h
)
Example 15.7 First Law Analysis of Combustion in a Bomb
Constant volume tank contains 1 kmol of methane (CH 4 ) gas
and 3 kmol of O 2 at 25 o C and 1 atm. The contents of the
tank are ignited, and the methane gas burns completely. If
the final temperature is 1,000K, determine (a) the final
pressure in the tank and (b) the heat transfer during this
process.
Solution: Concepts
1. Write chemical reaction equation
Assume ideal gas for both reactants and
products: PV = NR u T P 2
2. Energy balance: SSSF
Q = U P –U R
= (H p -P P V) - (H R -P R V)
16

(a) Combustion equation:
CH 4 (g)+ 3O 2 CO 2 + 2H 2 O + O 2
N react = 1 + 3 = 4 kmol, N react = 1 + 2 + 1 = 4 kmol, N 1 = N 2
Assume ideal gas for all gases:
State 1 (Reactants) P 1 V = N 1 R u T 1 (1)
State 2 (Products) P 2 V = N 2 R u T 2 (2)
eqn(2)/eqn(1) P 2 = (T 2 /T 1 )*P 1 = (1,000K/298K)(1atm) = 3.36 atm
answer
First law :
W
12
from
or
Ideal gas P v = R T
then
= 0 :
U = N{ h
U = N{ h
from property tables get valus of h f , h
substitute in equation above - - >
Amount of heat transfer out = - Q
or = 717,590/16 =
= ( U
U = H − PV = N{ h
_
u
Q
Q
_
o
f
12
12
+ ( h−
h
_
o
f
= W
_
_
12
_
o
+ ( h−
h
+ ( U
_
) − P v}
_
o
44,850 kJ/kg CH
+ ( h−
h
)
Pr React
odo
f
) − R T}
_
o
12
2
−U
−U
_
_
_
o
)
)} − PV
of each gas and hat each state
= 717,590 kJ/kmol CH
4
u
1
_
o
_
Answer
4
17

15.5 Adiabatic Flame Temperature
Adiabatic Flame Temperature = Maximum limit of combustion
gas temperature of each Air – Fuel mixture
(Adiabatic Flame Temperature = Combustion Temperature)
Q cv =0,W cv =0 ,∆KE=,
KE=∆PE=0 :
1 st law
_
_
Qcv
+ ΣNi
hi
= Wcv
+ ΣNe
he
ΣN
R
{
_
o
h f
_
ΣN
i
_
o
_
h
+ ( h−
h )}
i
= ΣN
R
e
= ΣN
_
h
P
e
{
_
o
h f
To Calculate the adiabatic flame temperature, T P
1. Write the combustion equation
2. Apply energy balance (1 st law)
3. Solving by trial-and
and-error technique by assume a value of T P
get values…and
and
substitute in (2) ….LHS = RHS ..if not try new T P
….. (in good procedure we can
interporate the former value to get the right value of T P
_
_
o
+ ( h−
h )}
P
• What is your first guess of T
• What should be the 2 nd trial.
• How about the 3 rd , 4 th ......
• When/how to interporate
a
b
Trail and error procedure
LHS - RHS = Error
T
T 2
T c
Interporation
T b
T a
c
T 2
E 2
T a -E a
T b -E b
T 2
0.0
T c +E c
T 2 = 342 o C
m i =1.263 kg
-E a -E b E = 0
+E c
Error
18

Example 15.8 Adiabatic Flame Temperature in Steady Combustion
Liquid octane (C 8 H 18 ) enters the combustion chamber of a
gas turbine steadily at 1 atm and 25 o C, and it is burns with
air that enters the combustion chamber at the same state,
as shown in the figure. Determine the adiabatic flame
temperature for (a) complete combustion at 100%
theoretical air, (b) complete combustion at 400%
theoretical air and (c) incomplete combustion (some CO in
the products) with 90% theoretical air.
Asumptions:
1. SSSF process
2. Adiabatic
3. No work
4. ∆KE=∆PE=0
5. Air and combustion gases are ideal gas
1. Combustion equation equation based on 1 kmol of octane
C 8
H 18
+12.5(O 2
+ 3.76 N 2
) 8 CO 2
+ 9H 2
O + 47N 2
at T R = 298K at T P = ?
P
_
_
Qcv
+ ΣNi
hi
= Wcv
+ ΣNe
he
_ _
ΣNi
hi
= ΣNe
he
_ _ _
_ _ _
o o
o o
ΣN
R { h f + ( h−
h )} R = ΣN
P { h f + ( h−
h )} P
_ _
o
Qreactants
are at referencestate ( h−
h ) R = 0
_
_ _ _
o o o
ΣN
R { h f } R = ΣN
P { h f + ( h−
h )}
2. Energy balance: H R = H P
19

(b) 400% theoretical air : combustion equation:-.
Combustion equation equation based on 1 kmol of octane
C 8
H 18
+4.0x12.5(O 2
+ 3.76 N 2
) 8CO 2
+ 9H 2
O + (3.0x12.5)O 2
+ 4.0x47N 2
at T R = 298K at T P = ?
by trial and error of T P
Adiabatic flame temperature = 962 K ……………………..answer
(C) 90% theoretical air : combustion equation:-.
Combustion equation equation based on 1 kmol of octane
C 8
H 18
+0.9x12.5(O 2
+ 3.76 N 2
) aCO 2
+bCO + 9H 2
O + 0.9x47N 2
at T R = 298K at T P = ?
C and O balance a = 5.5 and b = 2.5
by trial and error of T P
Adiabatic flame temperature = 2,236K ……………………..answer
• What is your first guess of T
• What should be the 2 nd trial.
• How about the 3 rd , 4 th ......
• When/how to interporate
a
b
Trail and error procedure
LHS - RHS = Error
T
T 2
T c
Interporation
T b
T a
c
T 2
E 2
T a -E a
T b -E b
T 2
0.0
T c +E c
T 2 = 342 o C
m i =1.263 kg
-E a -E b E = 0
+E c
Error
20

Assume air and combustion gases are ideal gases, get data from the property tables
Substance (kJ/kmol) (kJ/kmol) (kJ/kmol) (kJ/kmol)
C8H18(
l)
- 249,950 0 NA NA
O2
0 8,682 ............ ............
N2
0 8,669 ............ ............
H 2O(
g)
- 241,820 9,904 ............ ............
CO -393,520 9,364 ............ ............
2
CO -110,530 8,669 ............. ............
Remark : h
ΣN
_
o
f
( Nh
)
_
o
f
_
o
R { h f } R
C8H18
_
o
f
h
_
o
f
( g)
= h
P
_
o
f
= { N(
h
( l)
+ h
_
o
f
= ΣN
{ h
+ ( h−
h )}
+ h−
h )}
CO2
1x(
−249,950)
= {8( −393,520)
+ h−
9,364)
_
_
h
_
298K
_
o
_
o
+ {47(0 + h−
8,669)}
_
_
fg
_
_
P
_
h
+ { N(
h
N2
xxxxK
_
o
f
CO2
_
_
h
yyyyK
_
o
+ h−
h )}
H 2O
+ { N(
h
+ h−
h )}
+ {9( −241,820)
+ h−
9,904)}
_
_
o
f
_
_
o
N2
H 2O
(a) Adiabatic flame temperature = 2,395 K ………………..answer
21