### Slides

```ME 475/675 Introduction to
Combustion
Lecture 3
Thermodynamic Systems (reactors)
12
12
Inlet i
+
Outlet o

Dm=DE=0
0  +

m, E

• Closed systems
•
12
− 12 =  2 − 1 +
22
2
−
12
2
+  2 − 1
•  −  =  ℎ − ℎ +
2
2
−
2
2
+   −
• How to find changes, 2 − 1 and ℎ − ℎ , for mixtures when temperatures and
composition change due to reactions (not covered in Thermodynamics I)
Calorific Equations of State for a pure substance
•  =  ,  = () ≠ ()
• ℎ = ℎ ,  = ℎ() ≠ ()
For ideal gases
• Differentials (small changes)
•  =
• ℎ =

ℎ

• For ideal gas
•

=

0;
+
+

ℎ

=
ℎ

=
•  =
•
ℎ
=

0;
•  =
• Specific Heats,  and  [kJ/kg K]
• Energy input to increase temperature of one
kg of a substance by 1°C at constant volume
or pressure
• How are   and   measured?

w
Q
m, T
V = constant
Q
m, T
P = wg/A = constant
• Calculate    =

Δ
• Molar based
•  =  ∗ ;  =  ∗
T [K]
Q [joules]
Molar Specific Heat Dependence on Temperature

[K]
• Monatomic molecules: Nearly independent of temperature
• Only possess translational kinetic energy
• Multi-Atomic molecules: Increase with temperature and number of molecules
• Also possess rotational and vibrational kinetic energy
Specific Internal Energy and Enthalpy
• Once   and   are known, specific enthalpy h(T) and internal energy u(T)
can be calculated by integration
•   =  +
• ℎ  = ℎ +

• Primarily interested in changes, i.e. ℎ 2 − ℎ 1 =
2

1
,
• When composition does not change  and ℎ are not important
• Tabulated: Appendix A, pp. 687-699, for combustion gases
• bookmark (show tables)
• Curve fits, Page 702, for Fuels
•  =  −  ;
•  = /;  = /
Mixture Properties
• Extensive Enthalpy
•  =  ℎ =   ℎ
•  () =
•  =
ℎ

=
()
ℎ =  ℎ
•  () =
ℎ

=
()
• Specific Internal Energy
•  () =   ()
•   =
• Use these relations to calculate
mixture specific enthalpy and internal
energy (per mass or mole) as functions
of the properties of the individual
components and their mass or molar
fractions.
• u and h depend on temperature, but
not pressure
Standardized Enthalpy and Enthalpy of Formation
• Needed to find 2 − 1 and ℎ − ℎ for chemically-reacting systems because
energy is required to form and break chemical bonds
• Not considered in Thermodynamics I

• ℎ  = ℎ,
+ Δℎ, ()
• Standard Enthalpy at Temperature T =
• Enthalpy of formation from “normally occurring elemental compounds,” at standard
reference state: Tref = 298 K and P° = 1 atm
• Sensible enthalpy change in going from Tref to T =

• Normally-Occurring Elemental Compounds
• Examples: O2, N2, C, He, H2

• Their enthalpy of formation at  = 298 K are defined to be ℎ,
= 0
• Use these compounds as bases to tabulate the energy to form other compounds
Standard Enthalpy of O atoms
• To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy
input to break O-O bond (initial and final T and P are same)
• At 298K (1 mole) O2 + 498,390 kJ  (2 mole) O
498,390 kJ

• ℎ,  =
= + 249,195
2

• ℎ,
for other compounds are in Appendices A and B, pp 687-702
• To find enthalpy of O at other temperatures use

• ℎ 2  = ℎ,
2  + Δℎ, 2 ()
Example:
• Problem 2.14, p 71: Consider a stoichiometric mixture of isooctane and
air. Calculate the enthalpy of the mixture at the standard-state
temperature (298.15 K) on a per-kmol-of-fuel basis (kJ/kmolfuel), on a
per-kmol-of-mixture basis (kJ/kmolmix), and on a per-mass-of-mixture
basis (kJ/kgmix).
• Find enthalpy at 298.15 K of different bases
• Problem 2.15: Repeat for T = 500 K
Standard Enthalpy of Isooctane
T [K]
298.15
theta
0.29815
h [kJ/Kmol]
-224108.82
a1
a2
-0.55313 181.62
-0.16492 8.072412
a3
a4
a5
-97.787 20.402 -0.03095
-0.8639 0.040304 0.103807
a6
-60.751
-60.751
• Coefficients 1 to 8 from Page 702
[]
;
1000

ℎ
=

• =
•
4184(1
2
+ 2
2
3
+ 3
3
• Spreadsheet really helps this calculation
+
4
4
4
−
5

+ 6 )
a8
20.232
Enthalpy of Combustion (or reaction)
Reactants
298.15 K, P = 1 atm
Stoichiometric
< 0

Products
Complete Combustion
CCO2 HH2O
= 0 298.15 K, 1 atm
• How much energy can be released if product temperature and pressure are the
same as those of the reactant?