Compressible Flow - TME085

Compressible Flow - TME085 Lecture 13 Niklas Andersson Chalmers University of Technology Department of Mechanics and Maritime Sciences Division of Fluid Mechanics Gothenburg, Sweden niklas.andersson@chalmers.se

Adressed Learning Outcomes 6 Define the special cases of calorically perfect gas, thermally perfect gas and real gas and explain the implication of each of these special cases Niklas Andersson - Chalmers 2 / 59

Chapter 16 Properties of High-Temperature Gases Niklas Andersson - Chalmers 3 / 59

Chapter 16.2 Microscopic Description of Gases Niklas Andersson - Chalmers 4 / 59

Molecular Energy Vy y Vx x Vz z Translational kinetic energy thermal degrees of freedom: 3 Rotational kinetic energy thermal degrees of freedom: 2 for diatomic gases 2 for linear polyatomic gases 3 for non-linear polyatomic gases Vibrational energy (kinetic energy + potential energy) thermal degrees of freedom: 2 Electronic energy of electrons in orbit (kinetic energy + potential energy) O C O CO2 linear polyatomic molecule H O H2O non-linear polyatomic molecule H Translational energy Rotational energy (only for molecules - not for mono-atomic gases) Vibrational energy Electronic energy Niklas Andersson - Chalmers 5 / 59

Molecular Energy cont. The energy for one molecule can be described by ε = ε trans + ε rot + ε vib + ε el Results of quantum mechanics have shown that each energy is quantized i.e. they can exist only at discrete values Niklas Andersson - Chalmers 6 / 59

Molecular Energy - Translation ( ) ε trans = h2 n 2 1 8m a 2 + n2 2 1 a 2 + n2 3 2 a 2 3 n 1 n 3 quantum numbers (1,2,3,...) a 1 a 3 linear dimensions that describes the size of the system h Planck s constant m mass of the individual molecule Niklas Andersson - Chalmers 7 / 59

Molecular Energy - Rotation ε rot = h2 8π 2 J(J + 1) I J rotational quantum number (0,1,2,...) I moment of inertia (tabulated for common molecules) h Planck s constant Niklas Andersson - Chalmers 8 / 59

Molecular Energy - Rotation ( ε vib = hν n + 1 ) 2 n vibrational quantum number (0,1,2,...) ν fundamental vibrational frequency (tabulated for common molecules) h Planck s constant Niklas Andersson - Chalmers 9 / 59

Molecular Energy - Comments No simple expression for electronic energy ε el The lowest quantum numbers defines the zero-point energy for each mode for rotational energy the zero-point energy is exactly zero ε transo is very small but finite Niklas Andersson - Chalmers 10 / 59

Energy States three cases with the same rotational energy different direction of angular momentum quantum mechanics - different distinguishable states Niklas Andersson - Chalmers 11 / 59

Energy Distribution The Boltzmann distribution: N j = N g je ε j/kt Q where Q = f(t, V) is the state sum defined as Q j g j e ε j/kt g j is the number of degenerate states, ε j = ε j ε o, and k is the Boltzmann constant Niklas Andersson - Chalmers 12 / 59

Energy Distribution The Boltzmann distribution: N j = N g je ε j/kt Q For molecules or atoms of a given species, quantum mechanics says that a set of well-defined energy levels ε j exists, over which the molecules or atoms can be distributed at any given instant, and that each energy level has a certain number of energy states, g j. For a system of N molecules or atoms at a given T and V, N j are the number of molecules or atoms in each energy level ε j when the system is in thermodynamic equilibrium. Niklas Andersson - Chalmers 13 / 59

Energy Distribution P E At temperatures above 5K, molecules are distributed over many energy levels, and therefore the states are generally sparsely populated (N j g j ) Higher energy levels become more populated as temperature increases Niklas Andersson - Chalmers 14 / 59

Internal Energy The internal energy is calculated as E = NkT 2 ( ln Q T The internal energy per unit mass is obtained as ) V e = E M = NkT 2 Nm ( ) ln Q T V = { } ( ) k ln Q m = R = RT 2 T V Niklas Andersson - Chalmers 15 / 59

Internal Energy - Translation ( ) 2πmkT 3/2 Q trans = V h 2 ln Q trans = 3 2 ln T + 3 2 ln 2πmk h 2 + ln V ( ) ln Qtrans T V = 3 1 2 T e trans = RT 2 3 2T = 3 2 RT Niklas Andersson - Chalmers 16 / 59

Internal Energy - Rotation Q rot = 8π2 IkT h 2 ln Q rot = ln T + ln 8π2 Ik h 2 ( ) ln Qrot T V = 1 T e rot = RT 2 1 T = RT Niklas Andersson - Chalmers 17 / 59

Internal Energy - Vibration Q vib = 1 1 e hν/kt ln Q vib = ln(1 e hν/kt ) ( ) ln Qvib T V = hν/kt 2 e hν/kt 1 e vib = hν/kt e hν/kt 1 RT lim T hν/kt e hν/kt 1 = 1 e vib RT Niklas Andersson - Chalmers 18 / 59

Specific Heat e = e trans + e rot + e vib + e el e = 3 hν/kt RT + RT + 2 e hν/kt 1 RT + e el C v ( ) e T V Niklas Andersson - Chalmers 19 / 59

Specific Heat cont. Molecules with only translational and rotational energy e = 3 2 RT + RT = 5 2 RT C v = 5 2 R C p = C v + R = 7 2 R γ = C p C v = 7 5 = 1.4 Niklas Andersson - Chalmers 20 / 59

Specific Heat cont. Mono-atomic gases with only translational and rotational energy e = 3 2 RT C v = 3 2 R C p = C v + R = 5 2 R γ = C p C v = 5 3 = 12 3 1.67 Niklas Andersson - Chalmers 21 / 59

Calorically Perfect Gas In general, only translational and rotational modes of molecular excitation Translational and rotational energy levels are sparsely populated, according to Boltzmann distribution (the Boltzmann limit) Vibrational energy levels are practically unpopulated (except for the zero level) Characteristic values of γ for each type of molecule, e.g. mono-atomic gas, di-atomic gas, tri-atomic gas, etc He, Ar, Ne,... - mono-atomic gases (γ = 5/3) H 2, O 2, N 2,... - di-atomic gases (γ = 7/5) H2 O (gaseous), CO 2,... - tri-atomic gases (γ < 7/5) Niklas Andersson - Chalmers 22 / 59

Calorically Perfect Gas cont. p = ρrt e = C v T h = C p T h = e + p/ρ C p C v = R γ = C p /C v C v = C p = R γ 1 γr γ 1 γ, R, C v, and C p are constants γp a = ρ = γrt Niklas Andersson - Chalmers 23 / 59

Thermally Perfect Gas In general, only translational, rotational and vibrational modes of molecular excitation Translational and rotational energy levels are sparsely populated, according to Boltzmann distribution (the Boltzmann limit) The population of the vibrational energy levels approaches the Boltzmann limit as temperature increases Temperature dependent values of γ for all types of molecules except mono-atomic (no vibrational modes possible) Niklas Andersson - Chalmers 24 / 59

Thermally Perfect Gas cont. p = ρrt e = e(t) C v = de/dt h = h(t) C p = dh/dt h = e + p/ρ C p C v = R γ = C p /C v C v = C p = R γ 1 γr γ 1 R is constant γ, C v, and C p are variable (functions of T) γp a = ρ = γrt Niklas Andersson - Chalmers 25 / 59

High-Temperature Effects Example: properties of air 2000 K T Thermally perfect gas: e and h are non-linear functions of T the temperatur range represents standard atmospheric pressure (lower pressure gives lower temperatures) region of variable γ thermally perfect gas 600 K region of constant γ (γ=1.4) calorically perfect gas 50 K Niklas Andersson - Chalmers 26 / 59

High-Temperature Effects cont. For cases where the vibrational energy is not negligible (high temperatures) lim e vib = RT C v = 7 T 2 R However, chemical reactions and ionization will take place long before that Translational and rotational energy are fully excited above 3 K Vibrational energy is non-negligible above 600 K Chemical reactions begin to occur above 2000 K Niklas Andersson - Chalmers 27 / 59

High-Temperature Effects cont. Example: properties of air (continued) T 9000 K O O + + e (start of ionization) 4000 K 2500 K N 2 2N (start of dissociation) O 2 2O (start of dissociation) no reactions With increasing temperature, the gas becomes more and more mono-atomic which means that vibrational modes becomes less important Niklas Andersson - Chalmers 28 / 59

High-Temperature Effects cont. For temperatures T > 2500K Air may be described as being in thermodynamic and chemical equilibrium (Equilibrium Gas) reaction rates (time scales) low compared to flow time scales reactions in both directions (example: O 2 2O) Tables must be used (Equilibrium Air Data) or special functions which have been made to fit the tabular data Niklas Andersson - Chalmers 29 / 59

Thermodynamic Equilibrium Molecules are distributed among their possible energy states according to the Boltzmann distribution (which is a statistical equilibrium) for the given temperature of the gas extremely fast much faster than flow time scales in general (not true inside shocks) Global thermodynamic equilibrium there are no gradients of p, T, ρ, v, species concentrations, etc true thermodynamic equilibrium Local thermodynamic equilibrium gradients can be neglected locally this requirement is fulfilled in most cases (since the time and length scales of the molecular processes are extremely small) Niklas Andersson - Chalmers 30 / 59

Chemical Equilibrium Composition of gas (species concentrations) is fixed in time forward and backward rates of all chemical reactions are equal zero net reaction rates Global chemical equilibrium, together with global thermodynamic equilibrium all gradients are zero Local chemical equilibrium gradients can be neglected locally not always true, since chemical reactions may be either slow or fast, depending on the case studied Niklas Andersson - Chalmers 31 / 59

Thermodynamic and Chemical Equilibrium Most common cases: Thermodynamic Equilibrium Chemical Equilibrium Gas Model 1 local thermodynamic equilibrium local chemical equilibrium equilibrium gas 2 local thermodynamic equilibrium chemical non-equilibrium finite rate chemistry 3 local thermodynamic equilibrium frozen composition frozen flow 4 thermodynamic non-equilibrium frozen composition vibrationally frozen flow length and time scales of flow decreases from 1 to 4 vibrationally frozen flow gives the same gas relations as calorically perfect gas! no chemical reactions and unchanged vibrational energy Niklas Andersson - Chalmers 32 / 59

Equilibrium Gas How do we obtain a thermodynamic description? p = p(r, T) e = e(ν, T) h = h(p, T) h = e + p ρ C v = C p = ( ) e T ν ( ) h T p 1 + 1 ( ) e a 2 p ν e = γrt ( ) h 1 ρ p T T γ = C p C v = ( ) h T ( ) e T p ν RT = p ρ Note: R is not a constant here i.e. this is not the ideal gas law Niklas Andersson - Chalmers 33 / 59

Chapter 17 High-Temperature Flows: Basic Examples Niklas Andersson - Chalmers 34 / 59

Chapter 17.2 Equilibrium Normal Shock Wave Flows Niklas Andersson - Chalmers 35 / 59

Equilibrium Normal Shock Wave Flows Question: Is the equilibrium gas assumption OK? Answer: for hypersonic flows with very little ionization in the shock region, it is a fair approximation not perfect, since the assumption of local thermodynamic and chemical equilibrium is not really true around the shock however, it gives a significant improvement compared to the calorically perfect gas assumption Niklas Andersson - Chalmers 36 / 59

Equilibrium Normal Shock Wave Flows cont. Basic relations (for all gases), stationary normal shock: ρ 1 u 1 = ρ 2 u 2 ρ 1 u 2 1 + p 1 = ρ 2 u 2 2 + p 2 h 1 1 2 u2 1 = h 2 + 1 2 u2 2 For equilibrium gas we have: ρ = ρ(p, h) T = T(ρ, h) (we are free to choose any two states as independent variables) Niklas Andersson - Chalmers 37 / 59

Equilibrium Normal Shock Wave Flows cont. Assume that ρ 1, u 1, p 1, T 1, and h 1 are known Also u 2 = ρ ( ) 2 1u 1 ρ 1 u 2 ρ1 1 + p 1 = ρ 2 u 1 + p 2 ρ 2 ρ 2 p 2 = p 1 + ρ 1 u 2 1 ( 1 ρ ) 1 ρ 2 h 1 + 1 2 u2 1 = h 2 + 1 2 ( h 2 = h 1 + 1 2 u2 1 ( ) 2 ρ1 u 1 ρ 2 1 ( ρ1 ρ 2 ) 2 ) Niklas Andersson - Chalmers 38 / 59

Equilibrium Normal Shock Wave Flows cont. initial guess ρ 1 ρ 2 when converged: calculate p 2 and h 2 ρ 2 = ρ(p 2, h 2 ) T 2 = T(ρ 2, h 2 ) update ρ 1 ρ 2 ρ 2 = ρ(p 2, h 2 ) ρ 2, u 2, p 2, T 2, h 2 known no ρ 2 ρ 2old < ε yes stop Niklas Andersson - Chalmers 39 / 59

Equilibrium Air - Normal Shock Tables of thermodynamic properties for different conditions are available For a very strong shock case (M 1 = 32), the table below (Table 17.1) shows some typical results for equilibrium air calorically perfect gas (γ = 1.4) equilibrium air p 2 /p 1 1233 1387 ρ 2 /ρ 1 5.97 15.19 h 2 /h 1 206.35 212.80 T 2 /T 1 206.35 41.64 Niklas Andersson - Chalmers 40 / 59

Equilibrium Air - Normal Shock cont. Analysis: Pressure ratio is comparable Density ratio differs by factor of 2.5 Temperature ratio differs by factor of 5 Explanation: The chemical reactions taking place in the shock region lead to an absorption of energy into chemical energy drastically reducing the temperature downstream of the shock this also explains the difference in density after the shock Niklas Andersson - Chalmers 41 / 59

Equilibrium Air - Normal Shock cont. Additional notes: For a normal shock in an equilibrium gas, the pressure ratio, density ratio, enthalpy ratio, temperature ratio, etc all depend on three upstream variables, e.g. u 1, p 1, T 1 For a normal shock in a thermally perfect gas, the pressure ratio, density ratio, enthalpy ratio, temperature ratio, etc all depend on two upstream variables, e.g. M 1, T 1 For a normal shock in a calorically perfect gas, the pressure ratio, density ratio, enthalpy ratio, temperature ratio, etc all depend on one upstream variable, e.g. M 1 Niklas Andersson - Chalmers 42 / 59

Equilibrium Gas - Detached Shock calorically perfect gas equilibrium gas M = 20 M = 20 shock moves closer to body Niklas Andersson - Chalmers 43 / 59

Chapter 17.3 Equilibrium Quasi-One-Dimensional Nozzle Flows Niklas Andersson - Chalmers 44 / 59

Equilibrium Quasi-1D Nozzle Flows First question: Is chemically reacting gas also isentropic (for inviscid and adiabatic case)? entropy equation: Tds = dh νdp Quasi-1D equations in differential form (all gases): momentum equation: dp = ρudu energy equation: dh + udu = 0 Niklas Andersson - Chalmers 45 / 59

Equilibrium Quasi-1D Nozzle Flows cont. udu = dp ρ = νdp Tds = udu νdp = udu + udu = 0 Isentropic flow! ds = 0 Niklas Andersson - Chalmers 46 / 59

Equilibrium Quasi-1D Nozzle Flows cont. Second question: Does the area-velocity relation also hold for a chemically reacting gas? Isentropic process gives M = 1 at nozzle throat stil holds da A = (M2 1) du u Niklas Andersson - Chalmers 47 / 59

Equilibrium Quasi-1D Nozzle Flows cont. For general gas mixture in thermodynamic and chemical equilibrium, we may find tables or graphs describing relations between state variables. Example: Mollier diagram h p = constant T = constant for any point (h, s), we may find p, T, ρ, a, s Niklas Andersson - Chalmers 48 / 59

Equilibrium Quasi-1D Nozzle Flows cont. h p o For steady-state inviscid adiabatic nozzle flow we have: T o h 1 + 1 2 u2 1 = h 2 + 1 2 u2 2 = h o h o assume h o is known p 1 T 1 where h o is the reservoir enthalpy 1 p 2 T 2 2 isentropic process s Niklas Andersson - Chalmers 49 / 59

Equilibrium Quasi-1D Nozzle Flows cont. At point 1 in Mollier diagram we have: 1 2 u2 1 = h o h 1 u 1 = 2(h o h 1 ) Assume that u 1 = a 1 (sonic conditions) gives ρ 1 u 1 A 1 = ρ a A At any point along isentropic line, we have u = 2(h o h) and ρ, p, T, a etc are all given which means that ρu is given A A = ρ a ρu may be computed for any point along isentropic line Niklas Andersson - Chalmers 50 / 59

Equilibrium Quasi-1D Nozzle Flows cont. Equilibrium gas gives higher T and more thrust During the expansion chemical energy is released due to shifts in the equilibrium composition T equilibrium gas calorically perfect gas 1 A/A Niklas Andersson - Chalmers 51 / 59

Equilibrium Quasi-1D Nozzle Flows - Reacting Mixture Real nozzle flow with reacting gas mixture: T equilibrium gas real case calorically perfect gas 1 A/A Niklas Andersson - Chalmers 52 / 59

Large Nozzles High T o, high p o, high reactivity Real case is close to equilibrium gas results Example: Ariane 5 launcher, main engine (Vulcain 2) H 2 + O 2 H 2 O in principle, but many different radicals and reactions involved (at least 10 species, 20 reactions) T o 3600 K, p o 120 bar Length scale a few meters Gas mixture is quite close to equilibrium conditions all the way through the expansion Niklas Andersson - Chalmers 53 / 59

Ariane 5 Ariane 5 space launcher extreme high temperature and high speed flow regime Niklas Andersson - Chalmers 54 / 59

Vulcain Engine Vulcain engine: first stage of the Ariane 5 launcher Niklas Andersson - Chalmers 55 / 59

Space Shuttle Launcher SSME Niklas Andersson - Chalmers 56 / 59

Space Shuttle Launcher SSME Niklas Andersson - Chalmers 57 / 59

Equilibrium Quasi-1D Nozzle Flows - Reacting Mixture Very fast chemical reactions equilibrium gas flow thermodynamic and chemical equilibrium Very slow chemical reactions frozen gas flow no chemical reactions take place if also the vibrational energy of molecules have no time to change, we obtain vibrationally frozen gas flow calorically perfect gas! Niklas Andersson - Chalmers 58 / 59

Small Nozzles Low T o, low p o, lower reactivity Real case is close to frozen flow results Example: Small rockets on satellites (for manoeuvring, orbital adjustments, etc) Niklas Andersson - Chalmers 59 / 59