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Lecture 1 Conservation Equations

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Lecture 1 Conservation Equations Summer 2013 Lecture 1 Conservation Equations Moshe Matalon Conservation equations for a pure substance We focus on a material volume Vm(t) (made of the same set of material n n points). d ⇢dV =0 dt Vm(t1) Vm(t2) VZ(t) Mass Conservation The time rate of a change of a quantity φ(x,t) over a given volume V,moving with velocity VI is d @ φdV = dV + φ(V n) dS dt @t I · VZ(t) VZ(t) SZ(t) This is the multidimensional analogue of Leibnitz’s theorem from calculus; it is a kinematic relation, not a law of fluid mechanics. Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 1 Summer 2013 When VI = v,i.e.,theCVmoveswiththefluidvelocity,theresultisthe Reynolds’ transport theorem D @ φdV = dV + φ v n dS Dt @t · VZ(t) VZ(t) S(Zt) D @ φdV = + (φv) dV Dt @t r· ZV(t) ZV(t) ⇢ D Dφ φdV = + φ v dV Dt Dt r· ZV(t) ZV(t) ⇢ and the common notation for the “time rate of change” is D/Dt,theconvective or material derivative. For example: time rate of change of the D⇢ @⇢ = + v ⇢ density of a fluid particle Dt @t ·r acceleration of a fluid Dv @v = + v v particle Dt @t ·r Moshe Matalon Conservation of mass D @⇢ ⇢dV =0 + ⇢v dV =0 Dt @t r· ZV(t) ) ZV ⇢ @⇢ + ⇢v =0 @t r· @⇢ + v ⇢ +⇢ v =0 D⇢ @t ·r r· + ⇢ v =0 ) Dt r· D⇢/Dt | {z } 1 D⇢ v = r· −⇢ Dt rate of compression or rate of dilation D⇢ =0 v =0 Dt )r· Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 2 Summer 2013 Conservation of momentum Newton’s second law can be applied directly to a material C.V. The time rate of change of momentum in V is the sum of the forces acting on V. D ⇢v dV = ⇢f dV + t dS Dt ZV(t) ZV(t) ZS(t) body force surface force per unit volume per unit area | {z } | {z } t(x,t)=n σ · σ is the stress tensor; σ = σ { ij} σ is a symmetric tensor, i.e σij = σji. D ⇢v dV = ⇢f dV + n σdS Dt · ZV(t) ZV(t) ZS(t) @(⇢v) + (⇢vv) ⇢f σ dV =0 @t r· − r· ZV(t) ⇢ ⇢vv is the momentum flux tensor, Moshe Matalon contains nine components ⇢vivj @(⇢v) + (⇢vv)=⇢f + σ @t r· r· Dv ⇢ = ⇢f + σ Dt r· Constitutive relation σ = pI + ⌃ stress tensor Newtonian Fluid − pressure viscous stress tensor 2 ⌃ = µ[( v)+( v)T]+( µ)( v)I r r − 3 r· bulk shear viscosity viscosity The first term is a measure of the deformation of the fluid element and the second term is the rate of change of its volume Dv ⇢ = p + ⌃ + ⇢f Dt r r· Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 3 Summer 2013 Conservation of energy Applied to a material control volume Vm Total energy (internal + kinetic) e = e + 1 v v T 2 · D rate of work done rate of internal ⇢e dV = + Dt T on V by external forces energy flux across S ZV(t) heat flux vector positive for outward transport D ⇢e dV = ⇢f v dV + t v dS q n dS Dt T · · − · ZV(t) ZV(t) ZS(t) ZS(t) + n (σ v) dS + · · q dV ZS(t) r· ZV(t) + (σ v) dV r· · ZV(t) @(⇢e ) T + (⇢e v) ⇢f v (σ v)+ q dV =0 @t r· T − · r· · r· ZV(t) ⇢ Moshe Matalon @(⇢e ) T + (⇢e v)=⇢f v + (σ v) q @t r· T · r· · r· De ⇢ T = ⇢f v + (σ v) q Dt · r· · r· rate of work done on rate of heat the fluid by body transferred to the fluid; and viscous forces | {z } D(e + 1 v2) | {z } positive for ⇢ 2 outward transfer Dt D( 1 v2) ⇢ 2 = ⇢f v + v ( σ) Dt · · r· De ⇢ =[ (σ v) v ( σ)] q Dt r· · − · r· r · σ : v = p v +Φ r − r· | {z } De ⇢ = p v +Φ q Dt − r· r· Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 4 Summer 2013 Thermodynamics State of the gas is uniquely determined by two variables: p = p(⇢,T ),e= e(p, T ), etc. Other thermodynamic variables are the enthalpy (per unit mass) h = e + p/⇢ and the entropy (per unit mass) s defined from the relations 1 Tds = dh ⇢− dp − 1 Tds = de + pd(⇢− ) Ideal gas: ρ T =8.3145 J/kmol K universal gas constant p = R R · W = /W specific gas constant R R Also for an ideal gas: e = e(T ) only. h = e(T )+ T h = h(T ) only. R ) Moshe Matalon Introducing the specific heats at constant pressure/volume c ,c (in units kJ /kg K) and the ratio γ = c /c p v · p v T dh o = cp h = h + cpdT dT ) T o Z T de o = cv e = e + cvdT dT ) o ZT reference temperature dh de d p = + c = c + /W dT dT dT ⇢ p v R ✓ ◆ ) γ = cp/cv > 1 γ γ c = R = R p γ 1 γ 1 W − − ✓ ◆ 1 c = R = R v γ 1 γ 1 W − − ✓ ◆ Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 5 Summer 2013 Other forms of the energy equation De ⇢ = p v +Φ q Dt − r· r· Dh De 1 Dp p D⇢ h = e + p/⇢ = + ) Dt Dt ⇢ Dt − ⇢2 Dt Dh Dp ⇢ = +Φ q Dt Dt r· 1 Ds Dh 1 Dp Tds = dh ⇢− dp T = − ) Dt Dt − ⇢ Dt Ds 1 ⇢ = (Φ q) Dt T r· Moshe Matalon Ds 1 ⇢ = (Φ q) Dt T r· For a reversible process, Φ= 0 and ∆s heat acquired/T ⇠ In a reversible transition from one equilibrium state to a neighboring one, the increase in entropy is proportional to the heat given to the fluid. For an irreversible process, Φ > 0 and ∆S is larger than when Φ= 0 Isentropic flow ds = 0, i.e., adiabatic and reversible. dh = dp/⇢ dp d⇢ γ = γ p⇢− = const. dh = cpdT ) p ⇢ Speed of sound a, defined as the ratio of pressure to density change under isentropic conditions, is @p a = = γp/⇢ = γ T @⇢ R s✓ ◆s p q Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 6 Summer 2013 Conservation equations seven variables: v,p,⇢, h, T @⇢ + ⇢v =0 @t r· Dv ⇢ = p + ⌃ + ⇢f Dt r r· Dh Dp ⇢ = +Φ q Dt Dt r· ⇢ T p = R ,dh= c dT W p Constitutive Relations 2 ⌃ = µ[( v)+( v)T]+( µ)( v)I for a Newtonian fluid r r − 3 r· 1 @v @v 2 2 Φ= µ i + j +( µ)( v)2 2 @x @x − 3 r· ✓ j i ◆ q = λ T Fourier law (neglecting radiation, etc...) − r Moshe Matalon Conservation equations for multicomponent, chemically reacting, gas mixtures The mixture Gas mixture consists of N species (i =1, 2,...,N) - Chemical symbol of species i (e.g. H ,O ,CH , ) Mi 2 2 4 ··· The quantitative description of the chemical transformation between species requires a precise definition of concentration. The quantitative description of the chemical transformation between species requires a precise definition of concentration, Ci, defined as the number of moles of species i per unit volume, ni Ci = i =1, 2,...,N V ni - number of moles of species i in the mixture. From the definition of the Avogadro number, there are 6.0236 1023 molecules in a 1 mole. · Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 7 Summer 2013 The mass mi of all molecules of species i is related to the number of moles by mi = Wini i =1, 2,...,N where Wi is the molecular weight of species i. The density ⇢i is defined as the mass of species i per unit volume ⇢i = mi/V i =1, 2,...,N N 1 N m ⇢i = mi = = ⇢ density of the mixture i=1 V i=1 V X X N 1 N n Ci = ni = = C concentration of the mixture i=1 V i=1 V X X Moshe Matalon C n The mole fraction X = i = i i C n N 0 X 1 X =1 i i i=1 X mi mi/V ⇢i The mass fraction Yi = = = m m/V ⇢ N 0 Y 1 Y =1 i i i=1 X N N N N m = mi = niWi = n XiWi W = X W · i i i=1 i=1 i=1 i=1 X X X X W mixture molecular weight | {z } Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 8 Summer 2013 XiWi Yi = mi niWi/V XiWi W Yi = = = m nW/V W YiW ) Xi = Wi this shows that there is another way to calculate W N Y N X 1 N 1 i = i = X = W W W i W i=1 i i=1 i=1 X X X 1 N N Y − mixture molecular weight W = X W = i i i W i=1 "i=1 i # X X Moshe Matalon Mass balance of the chemical species vi the velocity of species i ⇢i = ⇢Yi the density of species i !i net rate of production of species i (per unit volume, per unit time) @(⇢Y ) i + ⇢Y v = ! for i =1,N @t r· i i i N v = Yivi mass average (mean) velocity of the mixture i=1 X N V = v v di↵usion velocity Y V =0 i i ) i i − i=1 X @ (⇢Y )+ [⇢Y (v + V )] = ! i =1, 2,...,N @t i r· i i i @⇢ Summing over all species + ⇢v =0 ) @t r· N because mass in neither created nor destroyed by chemical reaction !i =0 i=1 Moshe Matalon X Moshe Matalon University of Illinois at Urbana-Champaign 9 Summer 2013 @ (⇢Y )+ [⇢Y (v + V )] = ! @t i r· i i i @Y @⇢ ⇢ i + ⇢v Y + Y + ⇢v + (⇢Y V )=! @t ·r i i @t r· r· i i i ✓ ◆ DY ⇢ i + (⇢Y V )=! i =1, 2,...,N Dt r· i i i @⇢ + ⇢v =0 @t r· Note that there among the species equations there are only N 1independent equations because − N Yi =1 i=1 X Moshe Matalon Conservation of momentum the separate momentum balances for the separate species can be added to obtain the overall momentum equation for the mixture Dv N ⇢ = p + ⌃ + ⇢ Y f Dt r r· i i i=1 X 2 ⌃ = µ[( v)+( v)T]+( µ)( v)I r r − 3 r· the sum of the separate viscous stresses1 µ and are the mass averaged viscosity coefficients for the mixture gravity is the only external force on the mixture acting equally on all the species N the overall pressure is the sum of the partial pressures p = pi Dalton’s law i=1 X 1 Stresses due to di↵usion are nonlinear and therefore negligible Moshe Matalon Moshe Matalon University of Illinois at Urbana-Champaign 10 Summer 2013 Conservation of energy The energy balances for the separate species can be added to obtain the overall 1 2 energy equation for the total energy eT = e + 2 v of the mixture D⇢e N T = (σ v) q ++⇢ Y f (v+V ) Dt r· · r· i i · i i=1 X and when subtracting the time rate of change of the kinetic energy De N ⇢ = p v +Φ q + ⇢ Y f V Dt − r· r· i i · i i=1 X or, in terms of the enthalpy Dh Dp N ⇢ = +Φ q + ⇢ Y f V Dt Dt r· i i · i i=1 X Φis the viscous dissipation function for the mixture.
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