ME 475/675 Introduction to

CombustionLecture 2

Announcement

• On Friday I will modify HW 1, which will be due on Monday

Ideal Stoichiometric Hydrocarbon Combustion• CxHy + a(O2+3.76N2) (x)CO2 + (y/2) H2O + 3.76a N2

• a = number of oxygen molecules per fuel molecule, • Number of air molecules per fuel molecule is a(1+3.76)

• If a = aST = x + y/4, then the reaction is Stoichiometric• No O2 or Fuel in products • This mixture produces nearly the hottest flame temperature

• If a < x + y/4, then reaction is fuel-rich (oxygen-lean, fuel left over)• If a > x + y/4, then reaction is fuel-lean (oxygen-rich, O2 left over)• Equivalence Ratio

• Stiochiometric

• Fuel Lean

air

Air to fuel mass ratio [kg air/kg fuel] of reactants

• Need to find molecular weights

Molecular Weight of a Pure Substance• Only one type of molecule: • AxByCz…

• Molecular Weight• MW = x(AWA) + y(AWB) + z(AWC) + …

• AWi = atomic weights• Inside front cover of book

• Examples

•

• See page 701 for fuels

xx

xx

x

x

x

xx

Mixtures containing n components

• number of moles of species

• Total number of moles in system•

• Mole Fraction of species i

• mass of species • Total Mass

• Mass Fraction of species i

• Useful facts:

• but

• Mixture Molar Weight: • (weighted average)

• Example

• • Remember and/or write inside front cover of your

book

• Relationship between and

xx

xx

x

xo o

o

o

x

xx

Stoichiometric Air/Fuel Mass Ratio

• For Hydrocarbon fuel CxHy

•

• aSt = x + y/4

• For , • Constraints on y/x later

Equivalence Ratio

• Stiochiometric

• Fuel Lean

• CxHy + a(O2+3.76N2)

• % Stoichiometric Air (%SA)• % Excess Oxygen (%EO) = (%SA)-100%

Example• For extra credit, this problem may be clearly reworked and turned in at the

beginning of the next class period.• Problem 2.11, Page 91: In a propane-fueled truck, 3 percent (by volume)

oxygen is measure in the exhaust stream of the running engine. Assuming “complete” combustion without dissociation, determine the air-fuel ratio (mass) supplied to the engine. • Also find • Equivalence Ratio: • % Stoichiometric air: %SA• % Excess Oxygen: %EA

• ID: Are reactants Fuel Rich, Fuel Lean, or Stoichiometric?• Work on the board

Thermodynamic Systems (reactors)• Closed systems• Rigid tanks, piston/cylinders

• 1 = Initial state; 2 = Final state• Mass: • Chemical composition changes• But atoms are conserved

• 1st Law•

• How to find internal energy for mixtures, and change when composition changes due to reactions (not covered in Thermodynamics I)

m, E𝑄21❑

𝑊 21❑

Open Systems (control volume)• Steady State, Steady Flow (SSSF)

• Fixed volume, no moving boundaries

• Properties are constant and uniform • Inside CV (Dm=DE=0), and

• At ports ()

• One inlet and one outlet: • Atoms are conserved

• Energy: • Composition and temperature of inlet and outlet are not the same due to

reaction • Need to find (not covered in Thermodynamics I)

Dm=DE=0

Inlet i Outlet o

�̇�𝐶𝑉

�̇�𝑖 (𝑒+𝑃𝑣 )𝑖 �̇�0 (𝑒+𝑃𝑣 )𝑜

�̇� 𝐶𝑉

Combustion Thermochemistry • We use thermodynamics to evaluate the internal energy, enthalpy and entropy of a

systems at different states• The difference in energy and enthalpy between the products and reactants is used

with the first law to predict the heat of combustion (energy released during combustion, due to changes in chemical bonds) if the products are assumed to be at the same temperature as the reactants• The adiabatic flame temperature (product temperature assuming all reaction heat

release stays in the system) can be determined by assuming the products have the same energy level as the reactants• The steady state product composition for a reaction may be determined from

entropy considerations• Need to find

• The internal energy, enthalpy and entropy of mixtures of gases, and • How to account for effects of chemical bonds• These are not covered in ME 311 Thermodynamics I, but we we’ll cover them now

Ideal Gas Equation of State

• Universal Gas Constant

• Inside book front cover• kJ = kN*m= kPa*m3

• Specific Gas Constant• R =• MW = Molecular Weight of that gas

• Number of molecules• N*NAV

• Avogadro's Number, •

• Number of molecules in 12 kg of C12

Partial Pressure and volume fraction of a specie in a mixture of pressure and volume V

• Each specie acts as if it was the only component at the given V and T• Specie : • Mixture: • Ratio:

• Volume Fraction• Each specie acts as if it was the only

component at the given P and T• Specie :• Mixture: • Ratio:

End 2015

Extensive and Intensive System Properties

• Extensive thermodynamic properties depend on System Size (extent)• Examples• Volume V [m3]• Internal Energy E [kJ]• Enthalpy H = E + PV [kJ]

• Test: cut system in half• Denoted with CAPITAL letters

• Intensive Properties• Independent of system size• Examples

• Per unit mass (lower case)• v = V/m [m3/kg]• u = U/m [kJ/kg]• h = H/m [kJ/kg]

• Denoted using lower-case letters• Exceptions

• Temperature T [°C, K]• Pressure P [Pa]

• Molar Basis (use bar )• V = vm = • U = um = • H = hm =

• N number of moles in the system• Useful because chemical equations deal with

the number of moles, not mass

For ideal gases• Differentials (small changes)

• For ideal gas• = 0;

• For ideal gas• = 0;

• Specific Heat [kJ/kg C]• Energy input to increase temperature of

one kg of a substance by 1°C at constant volume or pressure

• How are and measured?

• Calculate • = ; =

m, TQ

𝑐𝑣

V = constant

m, TQ

w

P = wg/A = constant

𝑐𝑝 T [K] Q [joules]

Calorific Equations of State for a pure substance

Molar Specific Heat Dependence on Temperature

• Monatomic molecules: Nearly independent of temperature • Only translational kinetic energy

• Multi-Atomic molecules: Increase with temperature and number of molecules • Also possess rotational and vibrational kinetic energy

Internal Energy and Enthalpy

• Once cp(T) and cv(T) are known, internal energy change can be calculated by integration

• Appendix A (pp. 687-699, bookmark)

• Note = • =

Mixture Properties• Enthalpy

• Internal Energy

• Use these relations to calculate mixture enthalpy and internal energies (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• Individual gas properties are on

pp. 687-699 as functions of gas and T

Standardized Enthalpy and Enthalpy of Formation

• Needed for chemically-reacting systems because energy is required to form and break chemical bonds• Not considered in Thermodynamics I• Needed to find and

• Standard Enthalpy at Temperature T = • Enthalpy of formation at standard reference state: Tref and P° + • Sensible enthalpy change in going from Tref to T =

• Appendices A and B pp 687-702

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

• = 0

• Use these as bases to tabulate the energy for form of more complex compounds• Example: • At 298K (1 mole) O2 + 498,390 kJ (2 mole) O

• To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond• • for other compounds are in Appendices A and B