1. ME 475/675 Introduction to Combustion
Lecture 3

2. Thermodynamic Systems (reactors)
m, E
1 𝑄 2
1 𝑊 2
Dm=DE=0
Inlet i
Outlet o
𝑄 𝐶𝑉
𝑚 𝑖 𝑒+𝑃𝑣 𝑖
𝑚 0 𝑒+𝑃𝑣 𝑜
𝑊 𝐶𝑉
Closed systems
1 𝑄 2 − 1 𝑊 2 =𝑚 𝑢 2 − 𝑢 𝑣 − 𝑣 𝑔 𝑧 2 − 𝑧 1
Open Steady State, Steady Flow (SSSF) Systems
𝑄 𝐶𝑉 − 𝑊 𝐶𝑉 = 𝑚 ℎ 𝑜 − ℎ 𝑖 + 𝑣 𝑜 2 2 − 𝑣 𝑖 𝑔 𝑧 𝑜 − 𝑧 𝑖
How to find changes, 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖 , for mixtures when temperatures and composition change due to reactions (not covered in Thermodynamics I)

3. Calorific Equations of State for a pure substance
𝑢=𝑢 𝑇,𝑣 =𝑢(𝑇)≠𝑓𝑛(𝑣)
ℎ=ℎ 𝑇,𝑃 =ℎ(𝑇)≠𝑓𝑛(𝑃)
For ideal gases
Differentials (small changes)
𝑑𝑢= 𝜕𝑢 𝜕𝑇 𝑣 𝑑𝑇+ 𝜕𝑢 𝜕𝑣 𝑇 𝑑𝑣
𝑑ℎ= 𝜕ℎ 𝜕𝑇 𝑃 𝑑𝑇+ 𝜕ℎ 𝜕𝑃 𝑇 𝑑𝑃
For ideal gas
𝜕𝑢 𝜕𝑣 𝑇 = 0; 𝜕𝑢 𝜕𝑇 𝑣 = 𝑐 𝑣 𝑇
𝒅𝒖= 𝒄 𝒗 𝑻 𝒅𝑻
𝜕ℎ 𝜕𝑃 𝑇 = 0; 𝜕ℎ 𝜕𝑇 𝑃 = 𝑐 𝑃 𝑇
𝒅𝒉= 𝒄 𝑷 𝑻 𝒅𝑻
Specific Heats, 𝑐 𝑣 and 𝑐 𝑃 [kJ/kg K]
Energy input to increase temperature of one kg of a substance by 1°C at constant volume or pressure
How are 𝑐 𝑣 𝑇 and 𝑐 𝑃 𝑇 measured?
Calculate 𝑐 𝑝 𝑜𝑟 𝑣 = 𝑄 𝑚Δ𝑇 𝑝 𝑜𝑟 𝑣
Molar based
𝑐 𝑝 = 𝑐 𝑝 ∗𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 ∗𝑀𝑊
m, T Q w
P = wg/A = constant
𝑐 𝑝
m, T Q
𝑐 𝑣
V = constant

4. Molar Specific Heat Dependence on Temperature
𝑐 𝑝 𝑇
𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾
𝑇 [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

5. Specific Internal Energy and Enthalpy
Once 𝑐 𝑣 𝑇 and 𝑐 𝑝 𝑇 are known, specific enthalpy h(T) and internal energy u(T) can be calculated by integration
𝑢 𝑇 = 𝑢 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑣 𝑇 𝑑𝑇
ℎ 𝑇 = ℎ 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑝 𝑇 𝑑𝑇
Primarily interested in changes, i.e. ℎ 𝑇 2 − ℎ 𝑇 1 = 𝑇 1 𝑇 2 𝑐 𝑝 𝑇 𝑑𝑇 ,
When composition does not change 𝑇 𝑟𝑒𝑓 and ℎ 𝑟𝑒𝑓 are not important
Tabulated: Appendix A, pp , for combustion gases
bookmark (show tables)
Curve fits, Page 702, for Fuels
Use in spreadsheets
𝑐 𝑣 = 𝑐 𝑝 − 𝑅 𝑢 ;
𝑐 𝑝 = 𝑐 𝑝 /𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 /𝑀𝑊

6. Mixture Properties Extensive Enthalpy 𝐻 𝑚𝑖𝑥 = 𝑚 𝑖 ℎ 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥
𝐻 𝑚𝑖𝑥 = 𝑚 𝑖 ℎ 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥
𝒉 𝒎𝒊𝒙 (𝑻)= 𝑚 𝑖 ℎ 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝒀 𝒊 𝒉 𝒊 (𝑻)
𝐻 𝑚𝑖𝑥 = 𝑁 𝑖 ℎ 𝑖 = 𝑁 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥
𝒉 𝒎𝒊𝒙 (𝑻)= 𝑁 𝑖 ℎ 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝝌 𝒊 𝒉 𝒊 (𝑻)
Specific Internal Energy
𝒖 𝒎𝒊𝒙 (𝑻)= 𝒀 𝒊 𝒖 𝒊 (𝑻)
𝒖 𝒎𝒊𝒙 𝑻 = 𝝌 𝒊 𝒖 𝒊 𝑻
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

7. Standardized Enthalpy and Enthalpy of Formation
Needed to find 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖 for chemically-reacting systems because energy is required to form and break chemical bonds
Not considered in Thermodynamics I
ℎ 𝑖 𝑇 = ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠,𝑖 (𝑇)
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 = 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑝 𝑇 𝑑𝑇
Normally-Occurring Elemental Compounds
Examples: O2, N2, C, He, H2
Their enthalpy of formation at 𝑇 𝑟𝑒𝑓 =298 K are defined to be ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 = 0
Use these compounds as bases to tabulate the energy to form other compounds

8. 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) O ,390 kJ  (2 mole) O
ℎ 𝑓,𝑂 𝑜 𝑇 𝑟𝑒𝑓 = 498,390 kJ 2 𝑘𝑚𝑜𝑙 𝑂 =+ 249,195 𝑘𝐽 𝑘𝑚𝑜𝑙 𝑂
ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 for other compounds are in Appendices A and B, pp
To find enthalpy of O at other temperatures use
ℎ 𝑂 2 𝑇 = ℎ 𝑓, 𝑂 2 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠, 𝑂 2 (𝑇)

9. 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 ( 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 K of different bases
Problem 2.15: Repeat for T = 500 K

10. Standard Enthalpy of Isooctane
Coefficients 𝑎 1 to 𝑎 8 from Page 702
𝜃= 𝑇 [𝐾] 1000 𝐾 ;
ℎ 𝑜 𝑘𝐽 𝑘𝑚𝑜𝑙𝑒 =4184( 𝑎 1 𝜃+ 𝑎 2 𝜃 𝑎 3 𝜃 𝑎 4 𝜃 4 4 − 𝑎 5 𝜃 + 𝑎 6 )
Spreadsheet really helps this calculation

11. Enthalpy of Combustion (or reaction)
Reactants
K, P = 1 atm
Stoichiometric
Products
Complete Combustion
CCO2 HH2O
K, 1 atm
𝑄 𝐼𝑁 <0
𝑊 𝑂𝑈𝑇 =0
How much energy can be released if product temperature and pressure are the same as those of the reactant?
Steady Flow Reactor
𝑄 𝐼𝑁 − 𝑊 𝑂𝑈𝑇 = 𝐻 𝑃 − 𝐻 𝑅 = 𝑚 ℎ 𝑃 − ℎ 𝑅
𝑄 𝑂𝑈𝑇 = 𝐻 𝑅 − 𝐻 𝑃 = 𝑚 ℎ 𝑅 − ℎ 𝑃