Section 5, Lecture 1:! Review of Idealized Nozzle Theory Summary of Fundamental Properties and Relationships Taylor Chapter 4, Material also taken from Sutton and Biblarz, Chapter 3 1 MAE 5540 - Propulsion Systems Thermodynamics Summary R • Equation of State: p = ρR T − R = u g g M w - Ru = 8314.4126 J/°K-(kg-mole) - Rg (air) = 287.056 J/°K-(kg-mole) • Relationship of Rg to specific heats cp = cv + Rg • Internal Energy and Enthalpy h = e + Pv ⎛ de ⎞ ⎛ dh ⎞ cv = ⎜ ⎟ cp = ⎜ ⎟ ⎝ dT ⎠ v ⎝ dT ⎠ p 2 MAE 5540 - Propulsion Systems Thermodynamics Summary (2) R • Equation of State: p = ρR T − R = u g g M w - Ru = 8314.4126 J/°K-(kg-mole) - Rg (air) = 287.056 J/°K-(kg-mole) • Relationship of Rg to specific heats cp = cv + Rg • Internal Energy and Enthalpy h = e + Pv ⎛ de ⎞ ⎛ dh ⎞ cv = ⎜ ⎟ cp = ⎜ ⎟ ⎝ dT ⎠ v ⎝ dT ⎠ p 3 MAE 5540 - Propulsion Systems Thermodynamics Summary (3) R • Equation of State: p = ρR T − R = u g g M w - Ru = 8314.4126 J/°K-(kg-mole) - Rg (air) = 287.056 J/°K-(kg-mole) • Relationship of Rg to specific heats cp = cv + Rg • Internal Energy and Enthalpy h = e + Pv ⎛ de ⎞ ⎛ dh ⎞ cv = ⎜ ⎟ cp = ⎜ ⎟ ⎝ dT ⎠ v ⎝ dT ⎠ p 4 MAE 5540 - Propulsion Systems Thermodynamics Summary (4) • Speed of Sound for calorically Perfect gas c = γ RgT • Mathematic definition of Mach Number V M = γ RgT 5 MAE 5540 - Propulsion Systems Thermodynamics Summary (5) • First Law of Thermodynamics, reversible process de = dq − pdv dh = dq + vdp • First Law of Thermodynamics, isentropic process (adiabatic, reversible) de = − pdv dh = vdp 6 MAE 5540 - Propulsion Systems Thermodynamics Summary (6) • Second Law of Thermodynamics, reversible process ⎡T ⎤ ⎡ p ⎤ s − s c ln 2 − R ln 2 2 1 = p 2 ⎢ ⎥ g ⎢ ⎥ Tds = dh − vdp ⎣ T1 ⎦ ⎣ p1 ⎦ • Second Law of Thermodynamics, isentropic process (adiabatic, reversible) ------> s2 - s1 = 0 γ γ γ −1 ⎡ p ⎤ ⎡ ρ ⎤ p2 ⎡T2 ⎤ 2 2 = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ p1 ⎣ T1 ⎦ ⎣ p1 ⎦ ⎣ ρ1 ⎦ 7 MAE 5540 - Propulsion Systems One Dimensional Compressible Flow Approximations • Many Useful and practical Flow Situations can be Approximated by one-dimensional flow analyses • Flow Characterized by motion only along longitudinal axis 8 MAE 5540 - Propulsion Systems Distinction Between True ! 1-D Flow and Quasi 1-D Flow • In “true” 1-D flow Cross sectional area is strictly constant • In quasi-1-D flow, cross section varies as a Function of the longitudinal coordinate, x • Flow Properties are assumed constant across any cross-section • Analytical simplification very useful for evaluating Flow properties in Nozzles, tubes, ducts, and diffusers Where the cross sectional area is large when compared to length 9 MAE 5540 - Propulsion Systems Rocket Thrust Equation • Thrust = me Ve + ( peAe − p∞ Ae ) • • Thrust + Oxidizer enters combustion mi = 0 Chamber at ~0 velocity, combustion Adds energy … High Chamber pressure Accelerates flow through Nozzle Resultant pressure forces produce thrust 10 MAE 5540 - Propulsion Systems Review: Continuity Equation for ! Quasi 1-D Control Volume dS −> −> Control Surface ⎛ ρV• ds⎞ = 0 ∫∫ ⎝ ⎠ dS C.S. V1 dS p 1 V2 Control Volume T1 p2 A • Upper and Lower Surfaces … 1 T2 no flow across boundary A −> −> 2 V• ds = 0 dS • Inlet (properties constant across Cross section) --> −> −> −> −> ⎛ ρV• ds⎞ = ρ V • ds = ρ V cos(180o ) ds = −ρ V A ∫∫ ⎝ ⎠ 1 1 ∫∫ 1 1 ∫∫ 1 1 1 1 1 1 • Inlet (properties constant across Cross section) --> −> −> −> −> ⎛ ρV• ds⎞ = ρ V • ds = ρ V cos(0o ) ds = ρ V A ∫∫ ⎝ ⎠ 2 2 ∫∫ 2 2 ∫∫ 2 2 2 ρ V A = ρ V A 2 2 2 1 1 1 2 2 2 11 MAE 5540 - Propulsion Systems Review: Momentum Equation for ! Quasi 1-D Control Volume • Similar Analysis for Momentum Equation Yields dS Control Surface dS V1 dS • Newton’s Second law-- p 1 V2 Time rate of change of momentum Control Volume T1 p2 Equals integral of external forces A 1 T2 −> −> −> −> A ⎛ ρV• ds⎞ V = − p dS → 2 ∫∫ ⎝ ⎠ ∫∫ ( ) C.S. C.S. dS 2 −> −> p A V + ρ V 2 A + pds• i =p A V + ρ V 2 A 1 1 1 1 1 1 ∫ x 2 2 2 2 2 2 1 • Because of duct symmetry the “Z-axis” Component of pressure integrated to zero −> “Unit vector” x-direction ix 12 MAE 5540 - Propulsion Systems Review: Momentum Equation for ! Quasi 1-D Control Volume • Similar Analysis for Momentum Equation Yields dS Control Surface dS V1 dS • Newton’s Second law-- p 1 V2 Time rate of change of momentum Control Volume T1 p2 Equals integral of external forces A 1 T2 −> −> −> −> A ⎛ ρV• ds⎞ V = − p dS → 2 ∫∫ ⎝ ⎠ ∫∫ ( ) C.S. C.S. dS 2 −> −> p A V + ρ V 2 A + pds• i =p A V + ρ V 2 A 1 1 1 1 1 1 ∫ x 2 2 2 2 2 2 1 • Because of duct symmetry the “Z-axis” Component of pressure integrated to zero 13 MAE 5540 - Propulsion Systems Review: Energy Equation for ! Quasi 1-D Control Volume dS Control Surface dS V1 dS p 1 V2 Control Volume T1 p2 A 1 T2 A 2 2 • −> −> ⎛ V ⎞ −> −> Q− ∫∫ (pd S ) • V = ∫∫ ρ⎜ e + ⎟ V • d S → C.S. C.S. ⎝ 2 ⎠ dS V 2 V 2 Δq+ h + 1 = h + 2 1 2 2 2 14 MAE 5540 - Propulsion Systems Fundamental Results of 1-D Compressible Flow • Energy Equation for Adiabatic Flow: Stagnation temperature is a measure of the Kinetic Energy of the flow Field. • Largely responsible for the high Level of heating that occurs on high speed aircraft or reentering space Vehicles … T0 (γ − 1) 2 = 1+ M T 2 15 MAE 5540 - Propulsion Systems Stagnation Pressure for the! Isentropic Flow of a Calorically Perfect Gas • Energy Equation for Isentropic Flow: Temperature T(x), pressure p(x), and a velocity V(x) T(x) x p(x) V(x) γ γ “stagnation” γ −1 γ −1 p0 ⎡T0 ⎤ ⎡ (γ − 1) 2 ⎤ (total) pressure: = ⎢ ⎥ = ⎢1+ M ⎥ Constant throughout p ⎣ T ⎦ ⎣ 2 ⎦ Isentropic flow field 16 MAE 5540 - Propulsion Systems Thrust Equation • Steady, Inviscid, One-Dimensional Flow Through Ramjet −> −> −> −> − p dS = ⎛ ρV• ds⎞ V ∫∫ ( ) ∫∫ ⎝ ⎠ C.S. C.S. • • Thrust = me Ve − mi Vi + ( peAe − p∞ Ae ) 17 MAE 5540 - Propulsion Systems Fundamental Properties of Supersonic! and Supersonic Flow 18 MAE 5540 - Propulsion Systems … Hence the Shape of the De-Laval Rocket Nozzle 19 MAE 5540 - Propulsion Systems What is a NOZZLE • FUNCTION of rocket nozzle is to convert thermal energy in propellants into kinetic energy as efficiently as possible • Nozzle is substantial part of the total engine mass. • Many of the historical data suggest that 50% of solid rocket failures stemmed from nozzle problems. The design of the nozzle must trade off: 1. Nozzle size (needed to get better performance) against nozzle weight penalty. 2. Complexity of the shape for shock-free performance vs. cost of fabrication 20 MAE 5540 - Propulsion Systems Temperature/Entropy! Diagram for a Typical Nozzle • 2 2 V1 V2 ⎛ dh ⎞ q+ h1 + = h2 + cp = ⎜ ⎟ 2 2 ⎝ dT ⎠ p Tds = δq + dsirrev • q cp Isentropic Nozzle 21 MAE 5540 - Propulsion Systems Nozzle Mass Flow per Unit Area • maximum Massflow/area Occurs when When M=1 • m T γ M 0 R = A p g γ +1 c 0 2(γ −1) ⎡ (γ − 1) 2 ⎤ ⎢1+ M ⎥ ⎣ 2 ⎦ 22 MAE 5540 - Propulsion Systems Nozzle Mass Flow per Unit Area (concluded) • maximum • • Massflow/area ⎛ ⎞ ⎛ ⎞ m T0 m T0 Occurs when ⎜ Rg ⎟ = ⎜ * Rg ⎟ = When M=1 ⎜ Ac p0 ⎟ ⎜ A p0 ⎟ ⎝ ⎠ max ⎝ ⎠ • Effect known γ +1 γ ⎛ 2 ⎞ (γ −1) as Choking in a = γ → Duct or Nozzle γ +1 ⎝⎜ γ + 1⎠⎟ ⎡ (γ − 1)⎤ 2(γ −1) 1+ • i.e. nozzle will ⎢ 2 ⎥ Have a mach 1 ⎣ ⎦ throat • γ +1 • Choking m γ ⎛ 2 ⎞ (γ −1) p 0 Massflow * = ⎜ ⎟ A Rg ⎝ γ + 1⎠ T0 Equation 23 MAE 5540 - Propulsion Systems Isentropic Nozzle Flow: Area Mach Relationship • Consider a “choked-flow” Nozzle … (I.e. M=1 at Throat) • Then comparing the massflow /unit area at throat to some Downstream station γ +1 • m T ⎛ 2 ⎞ (γ −1) 0 R γ +1 * g ⎜ ⎟ 2(γ −1) A p0 A ⎝ γ + 1⎠ 1 ⎡⎛ 2 ⎞ ⎛ (γ − 1) 2 ⎞ ⎤ = = = 1+ M • * 1 ⎢⎜ ⎟ ⎜ ⎟ ⎥ T A M ⎣⎝ γ + 1⎠ ⎝ 2 ⎠ ⎦ m 0 γ +1 Rg 2(γ −1) A p0 ⎡ (γ − 1) 2 ⎤ ⎢1+ M ⎥ ⎣ 2 ⎦ 24 MAE 5540 - Propulsion Systems Isentropic Nozzle Flow: Area Mach Relationship (cont’d) • A/A* Directly related to Mach number γ +1 A 1 ⎡⎛ 2 ⎞ ⎛ γ − 1 ⎞ ⎤ 2(γ −1) 1 ( ) M 2 * = ⎢⎜ ⎟ ⎜ + ⎟ ⎥ A M ⎣⎝ γ + 1⎠ ⎝ 2 ⎠ ⎦ • “Two-Branch solution: Subsonic, Supersonic • Nonlinear Equation requires Numerical Solution • “Newton’s Method” ^ ^ ^ F(M ( j) ) M ( j +1) = M ( j) − ⎛ ∂F ⎞ ⎜ ⎟ ⎝ ∂M ⎠ |( j) 25 MAE 5540 - Propulsion Systems Numerical Solution for Mach • Abstracting to a “jth” iteration ^ ^ ^ F(M ( j) ) Iterate until convergence M ( j +1) = M ( j) − ⎛ ∂F ⎞ j={0,1,….} ⎜ ⎟ ⎝ ∂M ⎠ |( j) • Drop from loop when γ +1 1 ⎡⎛ 2 ⎞ ⎛ (γ − 1) ^ 2 ⎞ ⎤ 2(γ −1) A ^ ⎢ 1+ M ( j +1) ⎥ − * ⎝⎜ γ + 1⎠⎟ ⎝⎜ 2 ⎠⎟ A M ( j +1) ⎣ ⎦ < ε A A* MAE 5540 - Propulsion Systems 26 Numerical Solution for Mach (cont’d) γ +1 ^ 1 ⎡⎛ 2 ⎞ ⎛ (γ − 1) ^ 2 ⎞ ⎤ 2(γ −1) A F(M ( j) ) = ^ ⎢ 1+ M ( j) ⎥ − * ⎝⎜ γ + 1⎠⎟ ⎝⎜ 2 ⎠⎟ A M ( j) ⎣ ⎦ ⎛ γ +1 ⎞ ^ 2 2(γ −1) ⎛ ∂F ⎞ ∂ ⎜ 1 ⎡⎛ 2 ⎞ ⎛ (γ − 1) ⎞ ⎤ ⎟ ⎜ ⎟ = ^ ^ ⎢ 1+ M ( j) ⎥ = ⎝ ∂M ⎠ ⎜ ⎝⎜ γ + 1⎠⎟ ⎝⎜ 2 ⎠⎟ ⎟ |( j) ∂ M ( j) ⎝ M ( j) ⎣ ⎦ ⎠ ⎛ γ +1 ⎞ ^ 2 ⎜ ⎟ ⎛ ⎞ ⎛ γ − 1 ^ 2 ⎞ ⎝ 2(γ −1)⎠ ⎛ 1− 3γ ⎞ M ( j) − 1 ( ) ⎛ ⎞ ⎜ ⎟ 1+ M ( j) ⎝⎜ 2−2γ ⎠⎟ ⎝ ⎠ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ ⎟ ^ 2 ^ 2 ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ ⎜ γ + 1 ⎟ M ( j) ⎢2 + M ( j) (γ − 1)⎥ ⎣ ⎦ ⎝ ⎠ MAE 5540 - Propulsion Systems 27 Rocket Thrust Equation, revisited • Thrust = me Ve + ( peAe − p∞ Ae ) • • Thrust + Oxidizer enters combustion mi = 0 Chamber at ~0 velocity, combustion Adds energy … High Chamber pressure Accelerates flow through Nozzle Resultant pressure forces produce thrust 28 MAE 5540 - Propulsion Systems Rocket Thrust Equation, revisited • Thrust = mVexit + Aexit (pexit − p∞ ) • Non dimensionalize as • Thrust mV A (p − p ) = exit + exit exit ∞ P0 Athroat P0 Athroat Athroat P0 • For a choked throat γ +1 • γ +1 m 1 γ ⎛ 2 ⎞ (γ −1) = Thrust V γ ⎛ 2 ⎞ (γ −1) A (p − p ) * ⎜ ⎟ = exit + e exit ∞ A P0 T0 Rg ⎝ γ + 1⎠ * ⎜ ⎟ * P0 A T0 Rg ⎝ γ + 1⎠ A P0 29 MAE 5540 - Propulsion Systems Rocket Thrust Equation, revisited (cont’d) γ +1 (γ −1) Thrust Vexit γ ⎛