ME 475/675 Introduction to Combustion Lecture 43 Review for Final
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ME 475/675 Introduction to Combustion Lecture 43 Review for Final Announcements • HW17 Ch. 10 (5, 6, 8) • Due now, return by Wednesday • Final: • Friday, December 11, 2015, 2:45-4:45 PM • Trying to see if we can postpone to 3-5 PM • Hasib will have a tutorial review on Thursday (he will email time) • Course Evaluations: • Log in with your NetID to www.unr.edu/evaluate • Please complete before 11:59 PM on Wed (dead day) • If a student completes ALL teaching evaluations, they can see grades in real-time as they have done in the past. • If they leave even one evaluation incomplete, they will be unable to view grades through MyNevada Grade Viewing until the Tuesday after grades are due (December 22 this year). • I am curious how you feel about these components of the course: • Lecture slides, Worked examples, Extra-credit example problems, Long tests, Project Final • Friday, December 11, 2015, 2:45-4:45 PM (2 hours) • Chapters 8, 9, 10 and 16 • Even though Chapter 8 was partially covered on Midterm II • Use skills from earlier chapters as needed • 3-4 problems with parts • Open book with bookmarks + 1 page of notes • Study all: • HW problems • Lectures and extra-credit problems • Text reading assignments Ch. 8 Laminar Premixed Flames πΌ π£π’ • • • • ππΏ πΌ Bunsen Burner Inner Cone angle, πΌ ππΏ is the laminar flame speed relative to the premixed reactants π£π’ is the unburned reactant speed If π£π’ > ππΏ , then a cone will form whose angel πΌ adjusts so that ππΏ = π£π’ sin πΌ • The angel πΌ and its sine sin πΌ = ππΏ π£π’ decrease as π£π’ increases (inner cone length increases) • If π£π’ < ππΏ , then flame will flash back to air holes (unless quenched in tube). Tube filled with stationary premixed Oxidizer/Fuel Products shoot out as flame burns in ππΏ Laminar Flame Speed Burned Products πΏ Unburned Fuel + Oxidizer • Flame reference frame: π£π , ππ π£π’ = ππΏ , ππ’ , πΏ~1 ππ • 1 ππ πΉπ’ + π ππ ππ₯ → 1 + π ππ ππ • Conservation of mass: π" = ππ’ π£π’ = ππ π£π ; so π£π π£π’ = ππ ππ’ • Diffusion of heat and species cause flame to propagate • Estimate the laminar flame speed ππΏ and thickness πΏ. ≈ 2100πΎ 300πΎ = 7, π£π’ = ππΏ =? • Depends on the pressure, fuel, equivalence ratio, heat and mass diffusion,… Heat Flux with diffusion • Heat: Energy transfer at a boundary due to temperature difference • When there is a large species gradient, diffusion contributes to heat flux • ππ₯′′ = ππ −π ππ₯ • For πΏπ = πΌ π = ′′ ππ,πππππ’π πππ βπ + π ππππ ≈ π(1), true for most combustion mixtures • Shvab-Zeldovich form: ππ₯′′ = −ππ πβ ππ₯ =− π πβ ππ ππ₯ • Approximate Solution (Midterm II) • ππΏ = 2πΌ 1+π ππ’ • πΏ= 2πΌππ’ ′′′ 1+π −ππΉ −ππΉ′′′ ; π is air/fuel mass ratio = 2πΌ ππΏ (Fast flames are thin) • ππΉ′′′ = ππΉ πππΉ • ππΉ at average π that is closer to ππ than ππ’ , and average values of πΉπ’ππ and ππ₯ Pressure and temperature dependence of SL and πΏ ππΏ ππΏ ππΏ 2πΌ 1 + π • For methane: Φ ππ’ π • ππΏ = ′′′ −ππΉ ππ’ ~π0 ππ’ π 0.375 ππ−1 ππ₯π • Actually decreases as P increases: ππΏ cm ππΏ s •πΏ= πΏ cm s = 43 π [ππ‘π] Φ πΈπ π π’ − 2ππ ≠ ππ π (for HC fuels, N=2) Not reliable Better • Increases with temperature: = 10 + 3.71 × 10−4 ππ’2 πΎ • Decreases for Φ above or below 1.05 (because that decreases temperature) 2πΌ πΈ π ~π−1 π 0.375 ππ1 ππ₯π π π’ ππΏ 2ππ • Fast (high temperature) flame are thin Dependence on Fuel ππΏ ππΏ,πΆ3 π»8 ππ • Table 8.2, P = 1 atm, Φ = 1, Tu = Room temperature • Figure 8.17 presents ratio of maximum flame speed (at Φπ ≈ 1.05) of some hydrocarbons to propane speed (C3H8) versus ππ at maximum ππΏ (at Φπ ≈ 1.05). • C3-C6 follow same trend • hydrogen H2 and acetylene C2H2 are much faster Flame Speed Correlations for Selected Fuels • Hint: Be aware of what fuels are in this and other tables so you’ll know the easiest way to find the results you need • ππΏ = ππΏ,πππ ππ’ ππ’,πππ πΎ π ππππ π½ 1 − 2.1ππππ • ππ’ > ~350πΎ, ππππ = 298 πΎ, ππππ = 1 ππ‘π • ππΏ,πππ = π΅π + π΅2 Φ − Φπ 2 • πΎ = 2.18 − 0.8 Φ − 1 • π½ = −0.16 + 0.22 Φ − 1 • RMFD-303 is a research fuel that simulates gasoline Flame Mixture Flammability, Ignition, Quenching • What does it take to ignite a mixture? • What does it take to extinguish a flame? • “Williams Criteria” (rule of thumb) • Ignition will occur if enough energy is added to a slab of thickness πΏ (laminar flame thickness) to raise it to the adiabatic flame temperature, Tad. • A flame will be sustained if its rate of chemical heat release insides a slab is roughly equal to heat loss by conduction out of the slab Flammability Limits • Flames only propagate within certain equivalence ratio ranges • Φπππ < Φ < Φπππ₯ , • Φπππ = Φπππ€ππ = Φππππ • Φπππ₯ = Φπ’ππππ = Φπππβ • See page 291, Table 8.4 for limits • Φ= π΄ πΉ π π‘πππ π΄ πΉ • ππΆπ»4 ,πΏπππ = = ππΉ ππ΄,ππ‘ ππΉ,ππ‘ ππ΄ ππΆπ»4 ππΆπ»4 +ππ΄ππ πΏπππ = 1 π΄ πΉ π π‘πππ 4 1+ ππ π΄ππ ΦπΏπππ πππΆπ» ; ππΆπ»4 ,π ππβ = 1 πππΆπ» 1+ ππ 4 π΄ππ π΄ πΉ π π‘πππ Φπ ππβ Ignition ππ ππ₯ π πΆπππ‘ • The minimum electrical spark energy capable of igniting a flammable mixture. • It is dependent on the temperature, pressure and equivalence ratio of the mixture • Critical (minimum) radius of a spark that will propagate • Order of Magnitude: π ππππ‘ ≥ 6 πΌ ππΏ = 6 πΏ 2 = 1.22 πΏ • Energy to bring critical volume to Tb • πΈπππ = 61.56π πΌ 3 ππ ππ −ππ’ ππΏ π π ππ ~π−2 • Need lots of energy at low pressure • Agrees with measurements at low pressure • πΈπππ decreases as Tu increases • Table 8.5 page 298 Different fuels Data • Very low for Hydrogen, Acetylene, and Ethylene • Depends on fuel, equivalence ratio, initial (unburned) temperature, and pressure (next slide) • Cold Wall Quenching d d • Quenching distance d • Smallest dimension d that allows flame to pass • Experimentally determined • By shutting off flow of a premixed stabilized flame • dtube = (1.2 to 1.5) dslot • Order of magnitude analysis: π < πΌ 2 ππΏ π=πΏ π • where πΏ = πππππππ πππππ π‘βππππππ π ; b ≥ 2 Quenching Distance Data, Table 8.4 page 291 • Chapter 16 Detonations • Flames (deflagrations) are subsonic • Detonations are supersonic • π£π₯ > π = πΎπ π = • Mach Number: ππ π π’ πΎ π, πππππ₯ π£π₯ = >1 π Increases with temperature T and πΎ = ππ ππ£ >1 • Sound speed in room temperature air • π= 1.4 ππ π3 ππππ πΎ ππ 28.85 ππππ 8315 298πΎ = π 347 π = ππππ 776 βπ • A detonation is a shockwave sustained by energy released by combustion • πΆππππππ π π πππ π βπππ → βππβ π‘πππππππ‘π’ππ → πππππ’π π‘πππ → πππππππ π πππ π βπππ • Mass flux rates ′′ • ππ·ππ ~ 1 • ′′ ππ·ππ‘ ππ π2 π ππ ~ 2000 2 π π Confined planar and spherical systems Hot Compression Shock Combustible Mixture π£π₯1 Combustion Zone Products of Combustion High Temperature T, Pressure P and density, r Combustible Mixture π£π₯2 • Closed tube or expanding spark (both have confined hot products) • Expansion of hot gases behind shock may be much faster than laminar or turbulent flame speed • Flame Frame 2 1 Shock Detonation • Unknown: π2 , π2 , π2 , π2 = πΎπ π2 , ππ2 (π£π₯,2 ), and π£π₯,1 = π£π· detonation velocity Typical Properties • Rankine-Hugoniot Curve • Conservation of momentum and mass • π2 = π1 − π′′ 2 π£2 − π£1 • Conservation of energy • πΎ πΎ−1 π2 π£2 − π1 π£1 − π2 −π1 2 π£2 + π£1 − π=0 • π2 versus π£2 is a hyperbola (for given π1 , π£1 , πΎ and π) • Need to be given π′′ to find state 2 One-dimensional detonation analysis π1 , π£1 , π£π₯,1 π2 , π£2 , π£π₯,2 = π2 = π2 >>π1 2 1 πΎπ π2 Detonation • Real 1-D solutions must satisfy • mass, momentum and energy conservation • If we also know π′′ = π1 π£π₯,1 = π2 π£π₯,2 • Then we can find state 2 • Need to find the Detonation velocity π£π· = π£1,π₯ • Book says π£π₯,2 = π2 = πΎπ π2 • π′′ = π1 π£π₯,1 = π1 π£π· = π2 π£π₯,2 = π2 πΎπ π2 , need π2 πππ π2 • We found: π£π· = 2 1 + πΎ2 πΎ2 π 2 ππ1 ππ2 π1 + π ππ2 • Mostly use state 2 • Need ππ1 , ππ2 , πΎ2 , π 2 , π • • ππ = πΎ2 = πΆπ π π’ ππ πππ ; π 2 = ππ ; ππ = ππ = ππ2 ;π ππ£2 π£2 2 = ππ2 − π 2 ππ πΆππ ππ [T1 is known, but T2 and TAvg = (T1 +T2)/2 must be guessed] Ch. 9 Laminar Diffusion Flames (not premixed) • Non-reacting, constant-density laminar fuel jet in quiescent air • Assume • Temperature and Pressure are constant • πππΉ = ππππ₯ = ππ • ππΉ = πππ₯ = π π • Schmidt number, ππ = ππ = ππ = π(1) Fuel ππ , π£π , π ππΉ = π£π ππ 2 ππΉ = ππ π£π ππ 2 Centerline: Dimensionless π£ Speed, π₯,0 π£π Fuel Mass Fraction ππΉ Constant in Core Then decrease due to spreading Axial Speed Profiles π£π₯,0 = ππΉ versus r π£π • before πΏπ = πΌ π =1 • Axial diffusion is small compared to advection outside core π₯ > π₯π • Potential core not affected by viscosity Spreads out as x increases • Jet volume flow rate: ππΉ = π£π ππ 2 Max magnitude 2 ππ 2 • Jet initial momentum: π½ = π π£ π π π Decreases “Similarity” Solution • π£π₯ = 3 π½π 8π ππ₯ 1+ π3 1/2 3π½π 1 π− 4 2 16πππ π₯ π2 1+ −2 π2 ; π£π = 4 4 • π= • π£π₯ π£π = 3ππ π½π 1 2 1 π 16π ππ₯ π 0.375π ππ π₯ Similarity variable (Greek letter Xi) 1 π2 + 4 −2 ππ π£π π π • Jet Reynolds number π ππ = • For Schmidt number ππ = • ππΉ = π 0.375π ππ π₯ 1 π2 + 4 π π = −2 = π£π₯ π£π π ππ =π 1 Jet “half” radius Fast π1 • π1 2 = radius where π£π₯ π£π₯,0 = • Jet spreading half-angle πΌ; • π1 2 π₯ • πΌ= = 2.97 π ππ 1 2 π1 = tan πΌ −1 2.97 tan , π ππ angle decreases as π£π and π ππ increase 2 2 Slow Burning Fuel Jet (Diffusion Flame) • Laminar Diffusion flame structure • T and Y versus r at different x • Flame shape • Assume flame surface is located where Φ ≈ 1, stoichiometric mixture • No reaction inside or outside this • Products form in the flame sheet and then diffuse inward and outward • No oxidizer inside the flame envelop • Assume over-ventilated: enough oxidizer to burn all fuel Fuel ππ , π£π , π ππΉ = π£π ππ 2 ππΉ = ππ π£π ππ 2 Flame length (a measurable quantity) • Flame length πΏπ : • Φ π = 0, π₯ = πΏπ = 1; ππΉ = ππΉ,π π‘ • For un-reacting fuel jet (no buoyancy), with Schmidt number ππ = 3 8 • πΏπΉ = π ππ π ππΉ,π π‘ = 3 ππ π£π π π π 8 π ππΉ,π π‘ π = 3 ππ ππΉ 8π πππΉ,π π‘ π π = 1, • Increases with flow rate ππΉ = π£π ππ 2 (not dependent on π£π ππ π separately) • Buoyancy accelerates the flame but increases radial diffusion • These effects on the flame length “tend” to cancel (within order of magnitude) Experimentally-Confirmed Numerical Solutions • Roper Correlations pp. 336-9; Table 9.3, Equations 9.59 to 9.70 • Subscripts: • thy = Theoretical • expt = Experimental (use these) • Experimental results • round nozzles, • πΏπ,ππ₯π = 1330 ππΉ π∞ ππΉ ln(1+1 π) • square nozzles, • Inverse Gaussian error function “inverf” from Table 9.4 • πΏπ,ππ₯π = 1045 ππΉ π∞ ππΉ πππ£πππ 1+π −0.5 2 • Metric units (m, m3/s) • S = Molar Stoichiometric ratio • For pure CxHy/air: S = 4.76*(x+y/4) • Temperatures: • π∞ oxidizer, ππΉ Fuel, ππ mean-flame Slot Burners • Slot burners are dependent on Froude number • πΉππ = π£π πΌππΉ,π π‘πππ 0.6π 1500πΎ π∞ 2 −1 πΏπ = ππππ‘πππ πππ‘ ππππππ‘π’π ππ’ππ¦ππππ¦ • πΉππ β« 1: Momentum-Controlled, πΉππ ~ 1: Mixed (transitional), πΉππ βͺ 1: Buoyancy-Controlled • πΌ = 1 for plug nozzle velocity profile; πΌ = 1.5 for parabolic • Need to iterate since we are trying to find πΏπ (Hint: Initially assume πΏπ = 1m) 1 • Slot Nozzle Experimental results (stagnant oxidizer); π½ = • • 1 4 πππ£πππ 1+π 2 π∞ 2 4 ππ½ ππΉ Momentum Controlled: πΏπ,ππ₯π = 8.6 β 10 ~ππΉ βπΌππ,ππ‘πππ ππΉ 4 1 3 4 π½ 4 ππΉ4 π∞ 3 3 Buoyancy Controlled: πΏπ,ππ₯π = 2 β 10 ~π 4 πΉ πβ4 ππΉ • Transitional: πΏπ = πΏπ,π΅ 4 πΏπ,π 9 πΏπ,π 3 1 + 3.38 πΏπ,π πΏπ,π΅ 3 2 3 −1 • Note: Surprisingly, these are not dependent on diffusion coefficient π∞ = π Geometry and Flow Rate Dependence • Page 341, Fig. 9.9, • Methane • Same areas • πΏπ increases with ππΉ • Circular: πΏπ ~ππΉ 4 • Slot, πΉππ βͺ 1: Buoyancy-Controlled: πΏπ ~ππΉ 3 • πΏπ decreases for large aspect ratios Oxidizer (Surroundings) π2 Content (ππ2 < or > 21%) • Circular tube • • • • ππΉ π∞ ππΉ πΏπ,ππ₯π = 1330 ln(1+1 π) 1 π¦ π= π₯+ ππ2 4 2 ππΆπ»4 = ππ2 πΏπ πΏπ,21% = ln(1+1 π21% ) ln 1+1 π = ln(1+0.21 2) ln 1+ππ2 2 • Increasing ππ2 in oxidizer decreases flame length • Circular tube Fuel Dependence • πΏπ,ππ₯π = 1330 πΆ3 π»8 • π= πππ₯ ππΉπ’ ππ‘ π¦ 4 π2 + 3.76π2 → β― π π2 1 =π ππΆπ₯ π»π¦ π2 1 ππ‘ =π π2 π¦ π₯+4 • Alkane Fuels: πΆπ»4 , πΆ2 π»6 , πΆ3 π»8 ,… πΆπ₯ π»2(π₯+1) 2(π₯+1) 4 = 4.76 1.5π₯ + 0.5 • ππΆπ»4 = 9.52 • For a given flow rate ππΉ and air oxidizer πΆπ»4 • π»2 ππ πΆπ π¦ • If air is the oxidizer, then ππ2 = 0.21 and π = 4.76 π₯ + 4 π¦ • If the oxidizer is pure π2 , then ππ2 = 1 and π = π₯ + 4 • π = 4.76 π₯ + πΆ2 π»6 (increases with S) • For stoichiometric reaction of a generic HC fuel • πΆπ₯ π»π¦ + π₯ + πΆ4 π»10 ππΉ π∞ ππΉ ln(1+1 π) πΏπ πΏπ,πΆπ»4 = ln(1+1 ππΆπ»4 ) ln(1+1 π) , • Heavier fuels require more air, and so more time and distance (πΏπ ) to reach the stoichiometric condition • Light fuels, π = 4.76 2 = 2.33 (short flame) 1 • π»2 + 2 π2 + 3.76π2 → π»2 π + 1.88π2 1 • πΆπ + 2 π2 + 3.76π2 → πΆπ2 + 1.88π2 Stoichiometric Factors • When nozzle gas is pure πΆπ₯ π»π¦ fuel and ambient gas is air with π2 mole fraction ππ2 • π= 1 ππ2 ππ2 ππΆπ₯ π»π¦ = ππ‘ 1 ππ2 π₯+ π¦ 4 = πππ’ππ (pure fuel) • Now, generalize the Roper Correlations to include: • Air or inter gas added to fuel (fuel is “aerated” or “diluted”) • Revise the definition of π = ππ΄ππππππ‘ ππππ§π§ππ ππ‘ • Number of moles of the ambient gas per mole of nozzle gas when fuel to O2 ratio is stoichiometric. • Adding air to fuel shortens the fuel length and makes it more premixed • bluer and less sooty • Keep A/F ratio low enough so that equivalence ration Φ = A/F flashback π π‘ A/F is above rich limit to avoid • Adding inert gas such π2 or products of combustion reduces flame temperature and oxides of nitrogen and increases flame length Primary Aeration (of nozzle) • ππ΄ππ Ambient = 1 − ππππ πππ’ππ • • ππ΄ππππππ‘ 1−ππππ πππ’ππ 1 πππ’ππ π= = ππππ§π§ππ ππ‘ 1+ππππ πππ’ππ 1 πππ’ππ 1−ππππ π= 1 πππ’ππ +ππππ πΏπ ln(1+1 πππ’ππ ) ln(1+1 πππ’ππ ) = = 1 π +π πΏπ,ππ’ππ ln 1+1 π ln 1+ ππ’ππ πππ 1−ππππ • For methane πππ’ππ = 9.52 Fuel+Oxidizer π¦ πΆπ₯ π»π¦ + ππππ π₯ + π2 + 3.76π2 4 ππππ π¦ ππππ§ = 1 + π₯+ = 1 + ππππ πππ’ππ ππ2 4 π¦ π₯+ 4 Fuel Dilution •π= • Ambient ππ΄ππ = πππ’ππ π¦ π₯+4 = ππ2 • ππ΄ππππππ‘ ππππ§π§ππ ππ‘ π¦ 2 1+ππ· ππ· 1 1 ππ· = = ; 1+ππ· 1 ππ· +1 ππ2 π ππ· = π· 1−ππ· π¦ π₯+ 1−ππ· 4 •π= • = ππ πΏπ πΏπ,ππ’ππ = = ππ2 1+ππ· 1 ππ· + 1; ππ· = 1 1 ππ· −1 ππ2 = ln(1+1 πππ’ππ ) ln 1+1 π = ln(1+1 πππ’ππ ) ln 1+ • For methane πππ’ππ = 9.52 Fuel+Oxidizer πΆπ₯ π»π¦ + ππ· π· π₯+ 4 1 πππ’ππ +ππππ 1−ππππ Droplet Burning Fuel • 3 species: • 1ππ πΉ + π ππ₯ → 1 + π ππ ππ • Found temperature and mass fraction profiles ππΉ πππ π = ππ(π) general solutions • But there were 5 unknowns Products Pr Droplet rs rf ππΉ Oxidizer ππ • ππΉ , ππ , ππ (πππ¦ ππ πππππ€ ππ΅πππ ), ππΉ,π , ππ • Five constraints • At the flame assume the fuel and oxidize mass fractions are ππΉ,π = πππ₯,π = 0, and incoming fuel/oxidizer flow rates are stoichiometric. • Heat generated in flame is conducted to ππ ππΉ,π • droplet and surroundings • Fuel vapor mass fraction is governed by equilibrium at droplet/vapor interface ππ r Find ππ , ππΉ , ππ and ππΉ,π assuming ππ is known (guessed) • Method • Guess ππ (ππ΅πππ or below, or sometimes given) • Calculate π΅π,π , ππΉ,π , B and A, new value of • If necessary, find new value of ππ , and repeat • ππΉ = 4πππ ππ ππ,π ππ 1 + π΅π,π • Transfer number π΅π,π = • ππ = ππ + • ππ ππ = βππ +ππ→π βππ +ππ→π ππ΅π,π −1 ππ,π π+1 ππ 1+π΅π,π ππ • ππΉ,π = • ππ = Δβπ π−ππ,π ππ −π∞ π+1 π π΅π,π −1 π 1+π΅π,π π΅ ππ 1 ππΉ,π −1 πππΉπ’ π +1 ππππ π΄ (then iterate if necessary) Droplet Diameter • π·2 π‘ = π·02 − πΎπ‘ • Confirmed by experiments after initial transient • πΎ= 8ππ ππ ππ,π ππ 1 + π΅π,π • Droplet Lifetime: π‘π = π·02 πΎ • Property Evaluations (for best ππΉ estimate): • ∞ = ππππ π π‘ππππ; πΉ = πΉπ’ππ • π= ππ +ππ 2 ; • πππ = πππΉ π [use ππππ₯ π for best • ππ = 0.4ππΉ π + 0.6πππ₯ π • ππ = ππ ππ ππ ππ estimate?] Thank you for your attention in this class! • I learn more about combustion each time I teach it. • I also learn more about how to teach combustion. • I would appreciate your candid comments and suggestions on the course evaluations. Dropped in 2015 Ch 10 Simple Droplet Evaporation (no combustion) ππ = ππ΅πππ π ππΉ = π π βππ • In Chapter 3 we assumed liquid temperature is same as π∞ and known • Now assume πππ’πππππ < π∞ , so evaporation is controlled by Heat Transfer • Assumption • Find: • ππΉ π‘ → ππ π‘ → π‘π (droplet life) • π π = ππ ππ΅πππ , π∞ , ππΉ , ππ , ππ,π • π= ππ +ππ π΄π ππ = ππΉ βππ 1. 2. 3. 4. 5. Quiescent infinite medium Quasi-steady behavior Single compound liquid fuel ππ ≈ ππ΅πππ (constant and uniform), π∞ > ππ΅πππ Binary diffusion with π ≈ πΌ (πΏπ ≈ 1 ) • Shvab-Zeldovich energy equation 6. Constant average properties • Must be chosen carefully Evaporation Solution • Temperature Profile: π = • Z= ππ,π 4πππ ZππΉ ZππΉ − − π∞ −ππ΅πππ π π −π∞ π ππ +ππ΅πππ Zπ − π πΉ π 1−π = πΆπππ π‘πππ‘ • Evaporation Rate: ππΉ = 4πππ ππ ππ,π ππ π΅π + 1 • Spalding or Transfer Number: π΅π = • Droplet Lifetime: π‘π = • πΎ= 8π ππ ππ ππ,π π∞ −ππ΅πππ ππ,π βππ π·02 πΎ π΅π + 1 • Property Evaluations: ∞ = ππππ π π‘ππππ; πΉ = πΉπ’ππ • π= πππππ +π∞ ; 2 πππ = πππΉ π ; ππ = 0.4ππΉ π + 0.6π∞ π
