ME 475/675 Introduction to

CombustionLecture 14

Midterm I Review

Announcements• HW 5 Due Now

• Solutions will be posted today

• Midterm 1• Monday, September 28, 2015• 8-10 AM, PE 104• In-class review Today• Hasib Tutorial: Saturday, Sept. 26, 5 PM, in PE 101 (who will come?)

Midterm I• See me after class today for any special accommodations

• Confirm by email (greiner@unr.edu)

• Open book with bookmarks, 1 page of notes• 3-4 HW-like problems• Coverage

• Chapter 1-3, HW 1-5• All examples in class and book

• Handout • Last year’s Midterm 1• Intended to show problem style (like HW)• Please do not ask Hasib or me to work with you on them or show you how

they are done.

Likely Problem Types• Mixtures and their properties, air/fuel mass ratio• Heat of Combustion

• for a specified temperature, TProd = TReact

• Adiabatic Flame Temperature • no dissociation• constant Pressure or Volume

• Chemical Equilibrium • for specified T and P

• Simple Combustion Equilibrium Products • water/shift reaction

• Diffusion • Stefan (planar) or radial (droplet)• mass flow rates and mass fraction profiles • Boundary mass fraction• Dependence of diffusion coefficient on T and P• Average density, dependence on temperature, pressure and composition

Ideal Stoichiometric Hydrocarbon Combustion

• CxHy + a(O2+3.76N2) (x)CO2 + (y/2) H2O + 3.76a N2 (no dissociation) • 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)• If a > x + y/4, then reaction is fuel-lean (oxygen-rich)• Air to fuel mass ratio [kg air/kg fuel] of reactants

• (generally ~20)• Need to find molecular weights

air

Equivalence Ratio

• (compared to stoichiometric fuel)• Stiochiometric

• Fuel Lean

• CxHy + a(O2+3.76N2)

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

Molecular Weight of a Pure Substance

• Only one type of molecule: • AxByCz…

• MW = x(AWA) + y(AWB) + z(AWC) + …• AWi = atomic weights

• Inside front cover of book

• Hint: put in front cover of book

•

• Fuels • Bookmark 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:

• Hint: Put inside front cover of book

• Relationship between and

xx

xx

x

xo o

o

o

x

xx

Ideal Gas Equation of State

• Universal Gas Constant

• Inside book front cover• kJ = kPa*m3

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

• Number of molecules• N*NAV

• Avogadro's Number, •

Thermodynamic Systems (reactors)

• Closed systems

• Open Steady State, Steady Flow (SSSF) Systems

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

m, E𝑄21❑

𝑊 21❑

Dm=DE=0

Inlet i Outlet o

�̇�𝐶𝑉

�̇�𝑖 (𝑒+𝑃𝑣 )𝑖 �̇�0 (𝑒+𝑃𝑣 )𝑜

�̇� 𝐶𝑉

For a pure substances• Mass and Molar () Bases

• N number of moles in the system

• and • Appendix A, pp. 687-699, for combustion gases

• bookmark

• =

• =; =• For mixtures

• , • ,

Standardized Enthalpy and Enthalpy of Formation• Needed to find and for chemically-reacting matter because energy is required to

form and break chemical bonds

• Standard Enthalpy of substance 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 are defined to be = 0• Use these compounds as bases to tabulate the energy to form other compounds

Mixture Example: Stoichiometric Acetylene Combustion

• C2H2 + 2.5 (O2 + 3.76 N2) 2 CO2 + 1 H2O + 9.4N2

• Reactant standard enthalpy

• At , • (page 701, >0, Add energy to Standard Elemental Compounds) • Energy that must be added to 2C + H2 + 2.5 O2 + 9.4 N2 at to form Reactants

• Product standard enthalpy

• At ,

• Energy added to 2C + H2 + 2.5 O2 + 9.4 N2 at to form Products

• Which one has more energy? Reactants or Products? • Constant pressure heat of combustion at

• Energy is released (Exothermic)

Enthalpy of Combustion (or reaction)

• How much energy is released from a reaction if the product and reactant temperatures and pressures are the same?

• 1st Law, Steady Flow Reactor

• and Enthalpy of Reaction (< 0 for combustion)• Dependent on T and P of reaction

• Heat of Combustion • For exothermic reactions

Reactants298.15 K, P = 1 atm

Stoichiometric

ProductsComplete Combustion

CCO2 HH2O298.15 K, 1 atm

�̇�𝐼𝑁<0 �̇� 𝑂𝑈𝑇=0

Example: Stoichiometric Methane Combustion

• CH4 + 2 (O2 + 3.76 N2) 1 CO2 + 2 H2O + 7.52 N2

• Heat of reaction into system for TR = TP = 25°C and 1 kmol CH4

p 688 p 692 p 701 (< 0, exothermic)

Water Vapor

Other Bases• Per kg fuel

• Heat of Combustion• (Heat out for TR = TP)• See page 701, LHV = Lower Heating Value • Corresponds to water vapor in the products

• p 692 p 692

• • p. 701: Higher Heating Value = HHV = 55,528 (slightly larger than book, not due to dissociation since temp is low)

Per kg of reactant mixture

• LHV =

Adiabatic () Flame Temperature,

• 1st Law, Steady Flow Reactor

• All chemical energy goes into heating the products

• To find adiabatic flame temperature use• PP = PR and

• To evaluate using specific heats

• will be lower if we include dissociation

StoichiometricReactants

TR PR

Complete Combustion ProductsCCO2 HH2OPP = PR, T = TAd

�̇�𝐼𝑁=0 �̇� 𝑂𝑈𝑇=0

Constant Volume Adiabatic Flame Temperature

• • Use definition: (since standard internal energy U is not tabulated)

• Idea gas: ; ,

• Only and are unknown

• To evaluate using specific heats:

V, m𝑄=0 𝑊=0

Find equilibrium composition (reactants and products) for a given Temperature, Pressure & Mass• aA + bB + … eE + fF + …• Use Q and boundary work to achieve P and T

• Equilibrium Constant• ,

• since

• Standard State Gibbs Function Change• )

• In terms of Gibbs functions of formation (tabulated in App. A and B)

• Products are “favored” • As increases, which happens when decreases• If NProd > NRact , as P decreases

T,PQ

𝑊=∫𝑃𝑑𝑉

Number of unknowns is the number of species in equilibrium

• May need additional Equations • Simple reactions of few species have fewer unknowns

• Use atomic balances if necessary• May need to solve system of equations

• May be quadratic, or learn to use calculator to solve non-linear

Equilibrium Products of Combustion• Combine Chemical Equilibrium (2nd law) & Adiabatic Flame

Temperature (1st law) • For Example: Propane and air combustion • Ideal Stoichiomectric

• Four products for a range of air/fuel ratios: • Now consider seven more possible dissociation products:

• What happens as air/fuel (equivalence) ratio changes

minor

Simple Product Calculation method

• No minor species

•

• Assume and are known

• What is a good assumption for lean or stoichiometric mixtures ?• • c = e = 0 (no CO or H2), but now include

• Mole Fractions• ; ; ;

1000 1500 2000 2500 3000 35000.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

T [K]

Kp

For Rich combustion

• ; no (or fuel)

• 4 unknowns: b, c, d and e• 3 Atom balances: C, H, O

• Need one more constraint

• Consider “Water-Gas Shift Reaction” equilibrium

• Not dependent on P since number of moles of products and reactants are the same• ;

• See plot from data on page 51• KP = 0.22 to 0.1635 for T = 2000 to 3500 K (on a test you may need to evaluate at temperature T)

Solution

• Since , use “-” root

• Products:

• Mole Fractions• ; ; ; ;

Stefan Problem (no reaction)

• One dimensional tube (Cartesian)• Gas B is stationary: • Gas A moves upward

• Want to find this

• but treat as constantY

x

YB

YA

L-𝑌 𝐴 , ∞

𝑌 𝐴 , 𝑖

A

B+A

Liquid-Vapor Interface Boundary Condition

• At interface need

• Saturation pressure at temperature T• For water, tables in thermodynamics textbook• Or use Clausius-Slapeyron Equation (page 18 eqn. 2.19)

• If given , we can use this to find • Page 701, Table B: , at

A+BVapor

𝑌 𝐴 , 𝑖

LiquidA

Mass Flux and fraction of evaporating liquid A

�̇�𝐴} = { { } rsub { }} over { } ln {{1− { } rsub { ,∞}} over {1− { } rsub { , }}𝜌 𝒟 𝐴𝐵 𝐿 𝑌 𝐴 𝑌 𝐴 𝑖 ¿

0 0.2 0.4 0.6 0.80

2

4

6

86.908

0

m Y( )

10 Y𝑌 𝐴 , 𝑖

�̇�𝐴}} over {{ { } rsub { }} over { }𝜌 𝒟 𝐴𝐵 𝐿 ¿¿

For

0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

10.99

0

YA x .05( )

YA x .1( )

YA x .5( )

YA x .9( )

YA x .99( )

10 x𝑥𝐿

=0.99=0.9

=0.5

=0.1=0.05

𝑌 𝐴 (𝑥 )=1− (1−𝑌 𝐴 , 𝑖 )( 11−𝑌 𝐴 , 𝑖

)𝑥𝐿

Possible Questions

• How long will it take for column to recede by given (measurable) amount?

• For a given , temperature, pressure and dimension, find • Find variation with temperature

Spherical Droplet Evaporation• A is evaporating, find and • B is stagnant

• If and do not change with time• Then decreases as decreases

• For

𝑌 𝐴 , ∞

𝑌 𝐴 , ∞

𝑌 𝐴 , 𝑠

0 10 20 30 40 500

1

2

3

43.932

0

m1 By( )

500 By

YA1 r YAs( ) 11 YAs( )

1 YAs( )1

1

r

�̇�𝐴

4𝜋 𝑟𝑠 𝜌𝒟𝐴𝐵

𝐵𝑌

2 4 6 8 100

0.2

0.4

0.6

0.8

1

4.652 10 3

YA1 r .05( )

YA1 r .1( )

YA1 r .5( )

YA1 r .9( )

YA1 r .95( )

111 r

Droplet Diameter versus time • Mass Conservation

• , • Evaporation Const.

• Constant slope for versus • Confirmed by experiment • Droplet life

Dependence of on Temperature and Pressure

• Fairly independent of T and P

Extra Slides

Flame temperature and major mole-fractions vs • Equivalence Ratio • At , O2, CO, H2 all present due to

dissociation. Not present in “ideal” combustion

• at .15• at .05

• and decrease for • For decreases faster• For decreases faster

Fuel RichGet CO, H2

Fuel LeanO2

%

Tad[K]

“Old”

“New”

Minor-Specie Mole Fractions

• NO, OH, H, O• < 4000 ppm = 0.4%• Peak near

ppm

1 %