Transcript Slides
ME 475/675 Introduction to Combustion Lecture 6 Constant pressure and volume adiabatic flame temperature, Chemical Equilibrium, Mixture entropy evaluation Announcements • Monday, Labor Day Holiday • HW 2 due Wednesday, Ch 2 (25, 30, 31, 32, 35) • Due Now (before lecture starts) • Extra Credit example Adiabatic (𝑄 = 0) Flame Temperature Complete Combustion Products CCO2 HH2O PP = PR, T = TAd Stoichiometric Reactants TR PR 𝑄𝐼𝑁 = 0 𝑊𝑂𝑈𝑇 = 0 • 1st Law, Steady Flow Reactor • 𝑄𝐼𝑁 − 𝑊𝑂𝑈𝑇 = 0 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅 • All chemical energy goes into heating the products • To find adiabatic flame temperature use • PP = PR and ℎ𝑃 = ℎ𝑅 Constant pressure Evaluation method 𝑇𝐴𝑑,𝑃 • 𝐻𝑅 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑃 • 𝑜 𝑜 𝑁 ℎ + Δ ℎ 𝑠 𝑅 𝑖 𝑓 • • For 𝑇𝑅 = 𝑇𝑟𝑒𝑓 𝑜 𝑁 ℎ 𝑅 𝑖 𝑓 𝑖− 𝑖 = 298.15 𝐾 • 𝐻𝑐 = • 𝑜 𝑜 𝑁 ℎ + Δ ℎ 𝑠 𝑃 𝑖 𝑓 𝑜 𝑁 ℎ 𝑃 𝑖 𝑓 𝑜 𝑁 ℎ 𝑅 𝑖 𝑓 − 𝑖 = 𝑜 𝑁 ℎ 𝑃 𝑖 𝑓 𝑖 𝑜 𝑁 Δ ℎ 𝑠 𝑃 𝑖 𝑖 𝑖 • Molecular constant-pressure heat of combustion • 𝐻𝑐 = 𝑃 𝑁𝑖 Δℎ𝑠𝑜 𝑖 Could iterate Problem 2.32 𝑇 Δℎ𝑠𝑜 𝑖 = 𝑇 𝐴𝑑,𝑃 𝑐𝑝,𝑖 𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑃 =𝑇 𝑅 𝑟𝑒𝑓 • 𝐻𝑐 = • = 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑃 𝐻𝑐 𝑇𝐴𝑑,𝑃 − 𝑇𝑅 = 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 − 𝑇𝑅 𝑖 − 𝑇𝑅 Constant Volume Adiabatic Flame Temperature 𝑄=0 V, m 𝑊=0 • 𝑉𝑃 = 𝑉𝑅 = 𝑉 • 𝑄 − 𝑊 = 0 = 𝑈2 − 𝑈1 = 𝑈𝑃 − 𝑈𝑅 so 𝑈𝑅 = 𝑈𝑃 • Use definition: 𝑈 = 𝐻 − 𝑃𝑉 (since standard internal energy U is not tabulated) • 𝐻𝑅 − 𝑃𝑅 𝑉 = 𝐻𝑃 − 𝑃𝑃 𝑉 • Idea gas: 𝑃𝑅 𝑉 = 𝑁𝑅 𝑅𝑢 𝑇𝑅 ; 𝑃𝑃 𝑉 = 𝑁𝑃 𝑅𝑢 𝑇𝑃 = 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑣 , • 𝐻𝑅 𝑇𝑅 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑣 − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑.𝑣 • Only 𝑇𝐴𝑑,𝑣 and 𝐻𝑅 𝑇𝐴𝑑,𝑣 are unknown Constant volume evaluation methods • 𝐻𝑅 𝑇𝑅 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑉 • 𝑜 𝑜 𝑁 ℎ + Δ ℎ 𝑠 𝑅 𝑖 𝑓 • 𝑜 𝑁 ℎ 𝑅 𝑖 𝑓 • 𝐻𝑐 = 𝑖 − 𝑖 𝑜 𝑁 ℎ 𝑅 𝑖 𝑓 𝑖 • 𝐻𝑐 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝑜 𝑜 𝑁 ℎ + Δ ℎ 𝑠 𝑃 𝑖 𝑓 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝑜 𝑁 ℎ 𝑃 𝑖 𝑓 − 𝑖 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝑜 𝑁 ℎ 𝑃 𝑖 𝑓 𝑖; 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 • 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 = − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 = 𝑇𝐴𝑑,𝑉 𝐻𝑐 −𝑁𝑅 𝑅𝑢 𝑇𝑅 +𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 • 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 = − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 (𝑇𝑅 = 𝑇𝑟𝑒𝑓 ) 𝑖 𝑜 𝑁 Δ ℎ 𝑠 𝑖 𝑃 𝑖 𝑇𝐴𝑑,𝑉 𝑜 Δℎ𝑠 𝑖 = 𝑇 =𝑇 𝑐𝑝,𝑖 𝑅 𝑟𝑒𝑓 • 𝐻𝑐 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 + 𝑇𝑅 • 𝑇𝐴𝑑,𝑉 = − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢 𝐻𝑐 −𝑁𝑅 𝑅𝑢 𝑇𝑅 +𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢 𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢 = 𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 − 𝑃 𝑁𝑖 𝑅𝑢 = 𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑣,𝑖,𝑎𝑣𝑔 Chemical Equilibrium 𝑄=0 V, m 𝑊=0 𝑄𝐼𝑁 = 0 𝑊𝑂𝑈𝑇 = 0 • “Ideal” combustion assumes CCO2 HH2O • CxHy + a(O2+3.76N2) _b_CO2 + _c_ H2O + _d_ N2 + _e_ CXHY + _f_ O2 • Specifies product mole fractions in the products • To find Tad (1st law) • Constant Volume: 𝑈𝑅 = 𝑈𝑃 • Constant Pressure: 𝐻𝑃 = 𝐻𝑅 • In real high temperature combustion the product species may dissociate (break apart) yielding: H2, OH, CO, H, O, N, NO, …. (no CxHy?) • How to find actual final molar fractions of each component for different temperatures and pressures? • Chemical Equilibrium (2nd Law) Constant Volume Reaction Constraints • Conservation of atoms (C, H, O, N) • For Constant Volume and no interaction with surroundings • Q=W=0 • Time-dependent Ideal Gas Equation of State: P(t)V = N(t)RuT(t) • Initially: T1 P1 and 𝜒𝑖 1 (initial mole fraction of each specie) • Find: T2 P2 and 𝜒𝑖 2 • Before we specified final mole fractions 𝜒𝑖 • Now we find them. 2 • Chemical Equilibrium • 1st and 2nd laws of Thermodynamics • Reaction stops when entropy reaches a maximum V, m 𝑄=𝑊=0 Example: Carbon Dioxide Formation/Dissociation • 𝐶𝑂 1 + 𝑂2 2 → 𝐶𝑂2 • At high temperature some of the 𝐶𝑂2 will 1 dissociate and form 𝐶𝑂 + 𝑂2 𝛼 2 2 • −𝛼𝐶𝑂2 + 𝛼𝐶𝑂 + 𝑂2 • 𝐶𝑂 1 + 𝑂2 2 → 1 − 𝛼 𝐶𝑂2 + 𝛼𝐶𝑂 + 𝛼 𝑂 2 2 • Were 𝛼 =? is the fraction of 𝐶𝑂2 that dissociates • 0≤𝛼≤1 • How do we find 𝛼? • First step: Find 𝑇𝐴𝑑 = 𝑓𝑛(𝛼) [as a function of 𝛼] • Find 𝑁𝑖 = 𝑓𝑛(𝛼) and 𝜒𝑖 𝑃 = 𝑓𝑛(𝛼) • Use 1st Law (𝑈𝑃 = 𝑈𝑅 ) to find 𝑇𝐴𝑑 = 𝑓𝑛(𝛼) • Use ideal gas law to find 𝑃𝑃 = 𝑃𝑅 𝑇𝐴𝑑 𝑇𝑅 = 𝑓𝑛(𝛼) • How much dissociation (𝛼) will actually take place? 1−𝛼 Find Entropy of Product Mixture • Entropy is a function of composition, temperature, and pressure • Product Mixture • 𝐶𝑂 1 + 𝑂2 2 → 1 − 𝛼 𝐶𝑂2 + 𝛼𝐶𝑂 • Extensive Property 𝛼 + 𝑂2 2 • 𝑆𝑀𝑖𝑥 𝑇𝑓 , 𝑃 = 𝑛𝑖=1 𝑁𝑖 𝑠𝑖 𝑇𝑓 , 𝑃𝑖 𝛼 = 1 − 𝛼 𝑠𝐶𝑂2 𝑇𝑓 , 𝑃𝐶𝑂2 + 𝛼 𝑠𝐶𝑂 𝑇𝑓 , 𝑃𝐶𝑂 + 𝑠𝑂2 𝑇𝑓 , 𝑃𝑂2 2 • Where partial pressure of species 𝑖 is: 𝑃𝑖 = 𝜒𝑖 𝑃𝑓 • 𝑃𝐶𝑂2 = • 𝑃𝐶𝑂 = • 𝑃𝑂2 = 1−𝛼 𝛼 𝑃𝑓 = 1−𝛼 +𝛼+ 2 𝛼 1−𝛼 𝛼 1+ 2 𝑃𝑓 𝛼𝑃 1+ 2 𝑓 𝛼 2 𝛼 1+ 2 𝑃𝑓 • How does 𝑆𝑀𝑖𝑥 𝑇𝑓 , 𝑃 𝑇𝑓 vary with 𝑇𝑓 ? Need 𝑠𝑖 𝑇𝑓 , 𝑃𝑖 Gibb’s nd 2 equation • 𝑇𝑑𝑠 = 𝑑 ℎ − 𝑣𝑑𝑃 • 𝑑𝑠 = 𝑑ℎ 𝑇 𝑣 𝑇 − 𝑑𝑃 = 𝑐𝑃 (𝑇) 𝑑𝑇 𝑇 − 𝑑𝑃 𝑅𝑢 𝑃 • For Ideal Gases: 𝑑ℎ = 𝑐𝑃 (𝑇)𝑑𝑇; 𝑃𝑣 = 𝑅𝑢 𝑇 • 𝑇,𝑃 𝑑𝑠 𝑇𝑅𝑒𝑓, 𝑃𝑜 = 𝑇 𝑐𝑃 (𝑇) 𝑑𝑇 𝑇𝑅𝑒𝑓 𝑇 • 𝑠 𝑇, 𝑃 = 𝑠 𝑇𝑅𝑒𝑓, 𝑃 𝑜 + − 𝑃 𝑑𝑃 𝑅𝑢 𝑃𝑜 𝑃 𝑇 𝑐𝑃 (𝑇) 𝑑𝑇 𝑇𝑅𝑒𝑓 𝑇 • Standard entropy • 𝑠 𝑜 (𝑇) = 𝑠 𝑇𝑅𝑒𝑓, 𝑃𝑜 + • Appendix A, pp 687-699 𝑇 𝑐𝑃 (𝑇) 𝑑𝑇 𝑇𝑅𝑒𝑓 𝑇 − 𝑅𝑢 ln 𝑃 𝑃𝑜 𝑜 = 𝑠 (𝑇) − 𝑅𝑢 ln 𝑃 𝑃𝑜 Equilibrium • Plot 𝑆𝑀𝑖𝑥 𝑇𝑓 versus 1 − 𝛼 • Maximum at 1 − 𝛼 = 0.51 • For Q = W = 0 (constant volume) and constant mass, 2nd law requires 𝑑𝑠 > 0 • Once a system gets to 𝑆𝑀𝑖𝑥 𝑚𝑎𝑥 it can no longer change spontaneously (on its own) • At equilibrium for given U, V and m: 𝑑𝑠 𝑈,𝑉,𝑚 = 0 End 2015 Find equilibrium composition (reactants and products) for a given Temperature, Pressure & Mass 𝑊= 𝑃𝑑𝑉 G Q T,P • aA + bB + … eE + fF + … • Use Q and boundary work 𝑊 = 𝑃𝑑𝑉 to achieve P and T • Gibbs Function: 𝐺 𝑇, 𝑃 = 𝐻 𝑇 − 𝑇𝑆 𝑇, 𝑃 • 𝐺𝑀𝑖𝑥 = 𝑁𝑖 𝑔𝑖 • 𝑔𝑖 = ℎ𝑖 𝑇 + 𝑇𝑠𝑖 𝑇, 𝑃 = ℎ𝑖 𝑇 + 𝑇 𝑠𝑖𝑜 (𝑇) − 𝑅𝑢 ln 𝑃 𝑃𝑜 a = 𝑔𝑖𝑜 𝑇 + 𝑇𝑅𝑢 ln 𝑃 𝑃0 0 • 𝑔𝑖𝑜 𝑇 is Gibbs function at P = Po = 1 atm (what is difference between 𝑔𝑖𝑜 𝑇 and 𝑔𝑖,𝑇 ?) • For a constant temperature, pressure, and mass, and only boundary work (PdV) • It can be shown that second Law requires 𝑑𝐺 𝑇,𝑃,𝑚 ≤ 0 for spontaneous changes (no outside influence) (don’t spent too much time explaining this) • 𝑑𝐺 • 𝑑𝐺𝑀𝑖𝑥 = 𝑇,𝑃,𝑚 =0 0 𝑑𝑁𝑖 𝑔𝑖,𝑇 It can be shown… + 𝑃𝑖 𝑅𝑢 𝑇 ln 𝑜 𝑃 + 𝑁𝑖 𝑑 0 𝑔𝑖,𝑇 + 𝑃𝑖 𝑅𝑢 𝑇 ln 𝑜 𝑃 =0 For General Reactions • aA + bB + … eE + fF + … • dNA = -a(dk), dNB = -b(dk) • dNE = +e(dk), dNF = +f(dk) • 𝑃𝑖 + 𝑅𝑢 𝑇 ln 𝑜 = 0 𝑃 𝑃𝐴 𝑃𝐵 0 0 𝑔𝐴,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 −b(dk) 𝑔𝐵,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 − ⋯ 𝑃 𝑃 𝑃𝐸 𝑃𝐹 𝑜 𝑜 +e(dk) 𝑔𝐸,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 +f(dk) 𝑔𝐹,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 0 𝑑𝑁𝑖 𝑔𝑖,𝑇 • −a(dk) • 𝑅𝑢 𝑇 ln 𝑃𝐸 𝑒 𝑃𝐹 𝑓 … 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 … 𝑃𝑜 𝑃𝑜 𝑃 𝑃 =0 𝑜 𝑜 𝑜 𝑜 = − 𝑒𝑔𝐸,𝑇 + 𝑓𝑔𝐹,𝑇 + ⋯ − 𝑎𝑔𝐴,𝑇 − 𝑏𝑔𝐵,𝑇 −⋯ Solution Method • 𝑅𝑢 𝑇 ln 𝐾𝑃 = − Δ𝐺𝑇𝑜 • 𝐾𝑃 = exp −Δ𝐺𝑇𝑜 𝑅𝑢 𝑇 • Equilibrium Constant • 𝐾𝑃 = 𝑃𝐸 𝑒 𝑃𝐹 𝑓 … 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 … 𝑃𝑜 𝑃𝑜 = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 • Standard State Gibbs Function Change 𝑜 𝑜 𝑜 𝑜 • Δ𝐺𝑇𝑜 = 𝑒𝑔𝐸,𝑇 + 𝑓 𝑔𝐹,𝑇 + ⋯ − 𝑎𝑔𝐴,𝑇 − 𝑏𝑔𝐵,𝑇 − ⋯ = 𝑓𝑛(𝑇) 𝑜 𝑜 𝑜 𝑜 • Δ𝐺𝑇𝑜 = 𝑒𝑔𝑓,𝐸 + 𝑓𝑔𝑓,𝐹 + ⋯ − 𝑎𝑔𝑓,𝐴 − 𝑏𝑔𝑓,𝐵 − ⋯ = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑜 • In terms of Gibbs functions of formation 𝑔𝑓,𝑖 = 𝑓𝑛(𝑇) (tabulated in App. A and B) • Products are “favored” as 𝐾𝑃 increases, which happens when Δ𝐺𝑇𝑜 decreases Example 2.7 (p 43) Turn in next time for EC • Consider the dissociation of CO2 as a function of temperature and pressure: 1 𝐶𝑂2 ↔ 𝐶𝑂 + 𝑂2 . 2 • Find the composition of the mixture, i.e. the mole fractions of 𝐶𝑂2 , 𝐶𝑂 and 𝑂2 , that results from subjecting the originally pure 𝐶𝑂2 to various temperatures, and pressures • T = 1500, 2000, 2500 and 3000K • P = 0.1, 1, 20 and 100 atm • Find final values of: 𝜒𝐶𝑂2 , 𝜒𝐶𝑂 , 𝜒𝑂2 • Initially 𝜒𝐶𝑂2 =___? • MathCAD solution • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm Pressure and Temperature Dependence • For this reaction 1 • 𝐶𝑂2 ↔ 𝐶𝑂 + 𝑂2 2 𝑜 • Δ𝐺𝑇𝑜 = Δ𝐻 − 𝑇Δ𝑆 𝑜 > 0 (Δ𝐻 𝑜 > 0 exothermic) and Nproducts > Nreactants • 𝜒𝐶𝑂 (product) increases as T increases and P decreases Pressure and Temperature Dependence • 𝐾𝑃 = 𝑃𝐸 𝑒 𝑃𝐹 𝑓 … 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 … 𝑃𝑜 𝑃𝑜 = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 = exp −Δ𝐺𝑇𝑜 𝑅𝑢 𝑇 ; • 𝐾𝑃 Products are “favored” as KP increases and Δ𝐺𝑇𝑜 = Δ𝐻 𝑜 − 𝑇Δ𝑆 𝑜 decreases • Standard State Gibbs Function Change 𝑜 𝑜 𝑜 𝑜 • Δ𝐺𝑇𝑜 = 𝑒𝑔𝑓,𝐸 + 𝑓𝑔𝑓,𝐹 + ⋯ − 𝑎𝑔𝑓,𝐴 − 𝑏𝑔𝑓,𝐵 − ⋯ = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 Gibbs Function • Helps find equilibrium composition for given initial 𝑇1 , 𝑃1 and 𝜒𝑖 1 • 𝐺 = 𝐻 − 𝑇𝑆 = 𝑁𝑔 = 𝑁 ℎ − 𝑇𝑠 • For a fixed temperature, pressure and mass, all spontaneous changes in the absence of all work except boundary work, must cause changes to G such that 𝑑𝐺 𝑇,𝑃,𝑚 ≤ 0 (G must decrease) • At equilibrium 𝑑𝐺 𝑇,𝑃,𝑚 =0 • For mixtures • 𝐺𝑀𝑖𝑥 = 𝑁𝑖 𝑔𝑖 (𝑇, 𝑃𝑖 ) • 𝑔𝑖 𝑇, 𝑃𝑖 = ℎ𝑖 𝑇 − 𝑇𝑠𝑖 𝑇, 𝑃 = ℎ𝑖 𝑇 − 𝑇 𝑠𝑖 𝑇𝑅𝑒𝑓, 𝑃𝑜 + 𝑜 = 𝑔𝑖,𝑇 − 𝑇𝑅𝑢 ln 𝑃𝑖 𝑃𝑜 𝑐𝑃,𝑖 𝑇 𝑇 𝑇𝑅𝑒𝑓 𝑇 𝑑𝑇 − 𝑇𝑅𝑢 ln 𝑃𝑖 𝑃𝑜 Gibbs Function of pure species i as standard pressure • 𝑜 𝑔𝑖,𝑇 = ℎ𝑖 𝑇 − 𝑇 𝑠𝑖 𝑇𝑅𝑒𝑓, 𝑃 𝑜 + 𝑐𝑃,𝑖 (𝑇) 𝑇 𝑑𝑇 𝑇𝑅𝑒𝑓 𝑇