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

Ru • Equation of State: p = ρRgT −  Rg = 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)

Ru • Equation of State: p = ρRgT −  Rg = 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)

Ru • Equation of State: p = ρRgT −  Rg = 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

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

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 ⎛ 2 ⎞ Ae (pexit − p∞ ) * = ⎜ ⎟ + * P0 A T0 Rg ⎝ γ + 1⎠ A P0

• For isentropic ﬂow 1/2 ⎡ Texit ⎤ Vexit = 2cp ⎡T0 − Texit ⎤ = 2cpT0 ⎢1− ⎥ ⎣ exit ⎦ exit T ⎣⎢ 0 exit ⎦⎥ • Also for isentropic ﬂow γ γ −1 p ⎡T ⎤γ −1 T ⎛ p ⎞ γ 2 = 2 exit = exit ⎢ T ⎥ T ⎜ P ⎟ p1 ⎣ 1 ⎦ 0 exit ⎝ 0 exit ⎠ 30 MAE 5540 - Propulsion Systems Rocket Thrust Equation, revisited (cont’d)

• Subbing into velocity equation 1/2 ⎡ γ −1 ⎤ ⎛ ⎞ γ ⎢ pexit ⎥ V = 2c ⎡T − T ⎤ = 2c T 1− exit p ⎣ 0 exit exit ⎦ p 0 exit ⎢ ⎜ P ⎟ ⎥ ⎝ 0 exit ⎠ ⎣⎢ ⎦⎥ • Subbing into the thrust equation

1/2 ⎡ γ −1 ⎤ ⎛ ⎞ γ ⎢ pexit ⎥ 2c T 1− p 0 exit ⎢ ⎜ ⎟ ⎥ γ +1 ⎝ P0 ⎠ ⎢ exit ⎥ (γ −1) Thrust ⎣ ⎦ γ ⎛ 2 ⎞ Aexit (pexit − p∞ ) * = ⎜ ⎟ + * = p0 A T0 Rg ⎝ γ + 1⎠ A P0 1/2 ⎡ γ −1 ⎤ γ +1 ⎛ ⎞ γ (γ −1) ⎢ pexit ⎥ 2cpγ ⎛ 2 ⎞ Aexit (pexit − p∞ ) 1− + ⎢ ⎜ P ⎟ ⎥ R ⎜ 1⎟ A* P ⎝ 0 exit ⎠ g ⎝ γ + ⎠ 0 ⎣⎢ ⎦⎥ 31 MAE 5540 - Propulsion Systems Thrust Coefﬁcient 2 2cpγ 2cpγ 2γ 2γ • Simplifying = = = R c c 1 1 g p − v 1− γ − γ P = P • Finally, for an isentropic nozzle 0 exit 0

CF = 1/2 γ +1 ⎡ γ −1 ⎤ Thrust 2 ⎛ 2 ⎞ (γ −1) ⎛ p ⎞ γ A (p − p ) = γ ⎢1− exit ⎥ + exit exit ∞ * ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ * P0 A γ − 1⎝ γ + 1⎠ ⎝ P0 ⎠ A P0 ⎣⎢ ⎦⎥

• Non-dimensionalized thrust is a function of Nozzle pressure ratio and back pressure only …. “Thrust coefﬁcient” 32 MAE 5540 - Propulsion Systems Thrust Coefﬁcient (cont’d)

1/2 γ +1 ⎡ γ −1 ⎤ Thrust 2 ⎛ 2 ⎞ (γ −1) ⎛ p ⎞ γ A (p − p ) C = = γ ⎢1− exit ⎥ + exit exit ∞ F * ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ * P0 A γ − 1⎝ γ + 1⎠ ⎝ P0 ⎠ A P0 ⎣⎢ ⎦⎥

γ +1 γ +1 (γ −1) γ ⎛ 2 ⎞ ⎛ P0 ⎞ A ⎝⎜ γ + 1⎠⎟ ⎝⎜ p ⎠⎟ exit = exit A* ⎛ 2 ⎞ ⎡ (γ −1) ⎤ ⎛ P ⎞ γ ⎜ γ − 1⎟ ⎢ 0 1⎥ ⎝ ⎠ ⎢⎜ ⎟ − ⎥ ⎝ pexit ⎠ ⎣⎢ ⎦⎥ 33 MAE 5540 - Propulsion Systems Thrust Coefﬁcient(cont’d)

• Thrust Coefﬁcient is a function only of Combustion process

{P0, γ}, the Nozzle expansion (pexit), and the back pressure, (p∞)

⎧ ⎫ ⎪ γ +1 ⎪ −γ 1/2 ⎛ p ⎞ γ +1 ⎪ γ −1 exit ⎪ ⎡ ⎤ (γ −1) γ ⎜ ⎟ 2 ⎛ 2 ⎞ ⎪ ⎛ p ⎞ γ − 1 ⎝ P0 ⎠ ⎡ p p ⎤⎪ C =γ ⎢1− exit ⎥ + exit − ∞ F ⎜ ⎟ ⎨⎢ ⎜ ⎟ ⎥ (γ −1) ⎢ ⎥⎬ γ − 1⎝ γ + 1⎠ ⎪ ⎝ P0 ⎠ 2γ ⎡ ⎤ ⎣ P0 P0 ⎦⎪ ⎣⎢ ⎦⎥ ⎛ p ⎞ −γ ⎪ ⎢ exit 1⎥ ⎪ ⎢⎜ ⎟ − ⎥ ⎪ ⎝ P0 ⎠ ⎪ ⎩⎪ ⎣⎢ ⎦⎥ ⎭⎪

34 MAE 5540 - Propulsion Systems Characteristic Velocity, C* • • Solving for mexit Vexit * P0 A

1/2 • γ +1 ⎡ γ −1 ⎤ γ −1 γ mexit V Thrust A (p − p ) 2 ⎛ 2 ⎞ ( ) ⎛ p ⎞ exit = − exit exit ∞ = γ ⎢1− exit ⎥ * * * ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ P0 A P0 A A P0 γ − 1⎝ γ + 1⎠ ⎝ P0 ⎠ ⎣⎢ ⎦⎥

• Letting the nozzle expand until γ −1 γ ⎛ pexit ⎞ 1 >> ⎜ ⎟ ⎝ P0 ⎠

35 MAE 5540 - Propulsion Systems Characteristic Velocity, C* (cont’d)

γ +1 * ⎛ γ − 1⎞ • V ⎛ * ⎞ (γ −1) ⎛ * ⎞ exit ⎜ ⎟ mexit Vexit 2 ⎛ 2 ⎞ * P0 A ⎝ 2 ⎠ ⎜ * ⎟ = γ → C = ⎜ • ⎟ = ⎜ P A ⎟ γ − 1⎝⎜ γ + 1⎠⎟ γ +1 0 ⎝ mexit ⎠ ⎝ ⎠ infinite expansion 2 ⎛ 2 ⎞ (γ −1) γ γ − 1⎝⎜ γ + 1⎠⎟ • Measure of Combustion performance Independent of Nozzle design … function of combustion Only See tables 5.5 - 5.6 In Sutton and Biblarz

• From earlier for a choked throat

• γ +1 R T * γ g 0 c m 1 γ ⎛ 2 ⎞ (γ −1) → C = = 0 γ +1 γ +1 * = ⎜ ⎟ 1 1 A P0 T0 Rg ⎝ γ + 1⎠ ⎛ 2 ⎞ (γ − ) ⎛ 2 ⎞ (γ − ) γ γ ⎝⎜ γ + 1⎠⎟ ⎝⎜ γ + 1⎠⎟ 36 MAE 5540 - Propulsion Systems Characteristic Velocity, C* (cont’d)

• The characteristic velocity is a figure of thermo-chemical merit for a particular propellant and may be considered to be Indicative of the combustion efficiency.

γ RgT0 c γ R T → C* = = 0 = u o γ +1 γ +1 γ +1 MW ⎛ 2 ⎞ (γ −1) ⎛ 2 ⎞ (γ −1) ⎛ 2 ⎞ (γ −1) γ γ γ ⎝⎜ γ + 1⎠⎟ ⎝⎜ γ + 1⎠⎟ ⎝⎜ γ + 1⎠⎟

• Lower Molecular Weight Propellants Produce Higher C*

37 MAE 5540 - Propulsion Systems Ideal Isp of a Combustion Process • from Earlier

1/2 γ +1 ⎡ γ −1 ⎤ Thrust 2 ⎛ 2 ⎞ (γ −1) ⎛ p ⎞ γ A (p − p ) = γ ⎢1− exit ⎥ + exit exit ∞ * ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ * P0 A γ − 1⎝ γ + 1⎠ ⎝ P0 ⎠ A P0 ⎣⎢ ⎦⎥

1/2 ⎡ γ +1 ⎡ γ −1 ⎤ ⎤ Thrust P A* ⎢ 2 ⎛ 2 ⎞ (γ −1) ⎛ p ⎞ γ A (p − p )⎥ I = = 0 γ ⎢1− exit ⎥ + exit exit ∞ sp • • ⎢ γ − 1⎜ γ + 1⎟ ⎢ ⎜ P ⎟ ⎥ A* P ⎥ g m g m ⎢ ⎝ ⎠ ⎝ 0 ⎠ 0 ⎥ o o ⎣ ⎣⎢ ⎦⎥ ⎦ • Assuming an inﬁnitely expanded nozzle in a vacuum

38 MAE 5540 - Propulsion Systems Ideal Isp of a Combustion Process (cont’d) ⎡ γ +1 ⎤ C* ⎢ 2 ⎛ 2 ⎞ (γ −1) ⎥ I = γ ⇒ ( sp )ideal g ⎢ γ − 1⎝⎜ γ + 1⎠⎟ ⎥ o ⎢ ⎥ ⎣ ⎦ maximum possible specific impulse for given propellants • Propellants that burn Hot ⎡ γ +1 ⎤ C* ⎢ 2 ⎛ 2 ⎞ (γ −1) ⎥ and have a low I = γ ⇒ I = ( sp )ideal g ⎢ γ − 1⎝⎜ γ + 1⎠⎟ ⎥ ( sp )ideal product Molecular o ⎢ ⎥ ⎣ ⎦ weight … have better I γ Ru To sp γ +1 MW ⎛ 2 ⎞ (γ −1) 1 γ ⎜ ⎟ ⎡ γ + ⎤ ⎝ γ + 1⎠ ⎢ 2 ⎛ 2 ⎞ (γ −1) ⎥ 1 2γ R T γ = u o g ⎢ γ − 1⎝⎜ γ + 1⎠⎟ ⎥ g γ − 1 M o ⎢ ⎥ o W ⎣ ⎦ 39 MAE 5540 - Propulsion Systems SSME Computational Example • Space Shuttle Main Engine … • Unlike other propellants, the optimum mixture ratio for liquid oxygen and liquid hydrogen is not necessarily that which will produce the maximum specific impulse. Because of the extremely low density of liquid hydrogen, the propellant volume decreases significantly at higher mixture ratios.

• Maximum specific impulse typically occurs at a mixture ratio of around 3.5, however by increasing the mixture ratio to, say, 5.5 the storage volume is reduced by one-fourth. This results in smaller propellant tanks, lower vehicle mass, and less drag, which generally offsets the loss in performance that comes with using the higher mixture ratio. In practice, most liquid oxygen/liquid hydrogen engines typically operate at mixture ratios from about 5 to 6.

MAE 5540 - Propulsion Systems What is the Stoichiometric Mixture

Ratio of LOX/LH2?

M w LH 2 → 2.016kg/kg− mol

M w LO2 → 31.999kg/kg− mol

1 LO × M LO 31.999 MR = mol 2 w 2 = = 7.936 2mol LH 2 × M w LH 2 2 × 2.016

MR=6.0 (What the shuttle operates at) --> “Rich Mixture”

MAE 5540 - Propulsion Systems Compare Tank Volumes • Space Shuttle has the following mass fraction characteristics

• Shuttle has 721,000 kg of propellant in main tank on pad

MAE 5540 - Propulsion Systems Compare Tank Volumes (cont’d) ⎡ ⎤ 7.396 1 3 MR = 7.936 → 721,000kg ⎢ + ⎥ = 15315m 1140 3 67.78 3 ⎣⎢ kg/m kg/m ⎦⎥ “best compromise” ⎡ ⎤ 6.000 1 3 MR = 6.000 → 721,000kg ⎢ + ⎥ = 14432m 1140 3 67.78 3 ⎣⎢ kg/m kg/m ⎦⎥ ⎡ ⎤ 6.000 1 3 MR = 3.5 → 721,000kg ⎢ + ⎥ = 12850m 1140 3 67.78 3 ⎣⎢ kg/m kg/m ⎦⎥

MAE 5540 - Propulsion Systems SSME Computational Example (cont’d) • Space Shuttle Main Engine …

• LOX/LH2 Propellants, 6.03: 1 Mixture ratio

P0=186.92 atm =18940 Kpa

MAE 5540 - Propulsion Systems Ref: http://www.braeunig.us/space/ SSME Computational Example (cont’d) • Space Shuttle Main Engine …

T0 ~ 3615°K

MAE 5540 - Propulsion Systems SSME Computational Example (cont’d) • Space Shuttle Main Engine …

Mw ~ 13.6 kg/kg-mol

MAE 5540 - Propulsion Systems SSME Computational Example (cont’d) • Space Shuttle Main Engine …

γ ~ 1.196

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine

• The Space Shuttle Main Engines Burn LOX/LH2 for Propellants with A ratio of LOX:LH2 =6:1

• The Combustor Pressure, p0 Is 18.94 Mpa, combustor

temperature, T0 is 3615°K, throat diameter is 26.0 cm

• What propellant mass ﬂow rate is required for choked ﬂow in the Nozzle?

• Assume no heat transfer Thru Nozzle no frictional losses, γ=1.196

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

-- By product ~ Burns rich, byproduct

is water vapor + GH2

MW ~ 13.6 kg/kg-mole

-- Rg = 8314.4126 /13.6 = 611.35 J/°K-kg

• γ +1 (γ −1) m γ ⎛ 2 ⎞ p0 * = ⎜ ⎟ = A Rg ⎝ γ + 1⎠ T0

(1.196 + 1) 0.5 ⎛ ⎞ 6 ⎜ 1.196 ⎛ 2 ⎞ 1.196 − 1 ⎟ 18.94⋅10 ⎜ ⎝ ⎠ ⎟ ⎝611.35 1.196 + 1 ⎠ (3615) 0.5

= 8252.59 kg/sec-m2 MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Massﬂow rate • Compute Throat Area 2 ⎛ 26 ⎞ π =0.05297 m2 ⎝100⎠ 4

• Mass ﬂow = ⎛ • ⎞ m * ⎜ * ⎟ × A =8252.598252.59 ×0.052970.04714 = 389.03kg/sec A ⋅ =437.1 kg/sec ⎝⎜ ⎠⎟

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (concluded)

• The nozzle expansion ratio is 77.5 -- what is the exit mach number

γ +1 A 1 ⎡⎛ 2 ⎞ ⎛ γ − 1 ⎞ ⎤ 2(γ −1) 77.5 1 ( ) M 2 * = = ⎢⎜ ⎟ ⎜ + ⎟ ⎥ A M ⎣⎝ γ + 1⎠ ⎝ 2 ⎠ ⎦

• Non -linear function of mach number

• Solution methods i) Plot A/A* versus mach ii) Numerical Solution

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Exit Mach Number Expansion ratio = 77.5

γ +1 A 1 ⎡⎛ 2 ⎞ ⎛ γ − 1 ⎞ ⎤ 2(γ −1) 1 ( ) M 2 = * = ⎢⎜ ⎟ ⎜ + ⎟ ⎥ A M ⎣⎝ γ + 1⎠ ⎝ 2 ⎠ ⎦ 1.196 + 1 2 1.196 − 1 2 (1.196 − 1) ⎛ ⎛ ⎞ ⎛1 + (4.6770842) ⎞ ⎞ ⎝ ⎝1.196 + 1⎠ ⎝ 2 ⎠ ⎠ 4.677084

= 77.49998 ----> Mexit = 4.677084

Newton Solver comes in handy here!

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Exit Temperature

T0 (γ − 1) 2 Mexit = 4.677084 = 1+ M T 2

T0 Texit = = (γ − 1) 2 1+ M 2 exit 1.196 − 1 − 1 3615 ⎛1 + (4.6770842) ⎞ ⎝ 2 ⎠ =1149.90 °K

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Exit Velocity

Mexit = 4.677084 Vexit = M exit γ RgTexit =

4.677084 (1.196⋅611.35⋅1149.9) 0.5

= 4288.61 m/sec

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Exit Pressure M = 4.677084 exit P0 Pexit = γ = γ −1 ⎛ (γ − 1) 2 ⎞ 1+ M ⎝⎜ 2 exit ⎠⎟

18.94⋅106 ⎛ 1.196 ⎞ =17.45511 kPa 1.196 − 1 ⎝1.196 − 1⎠ ⎛1 + (4.6770842) ⎞ ⎝ 2 ⎠

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

* Compute C γ RgT0 → C* = = γ +1 ⎛ 2 ⎞ (γ −1) γ ⎝⎜ γ + 1⎠⎟

(1.196⋅611.35⋅3616) 0.5 =2295.35 m/sec ⎛1.196 + 1⎞ 0.5 ⎛ ⎝ ⎠⎞ ⎜ ⎛ 2 ⎞ 1.196 − 1 ⎟ 1.196 ⎜ ⎝ ⎠ ⎟ ⎝ 1.196 + 1 ⎠

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

Compute Idealized Isp ⎡ γ +1 ⎤ C* ⎢ 2 ⎛ 2 ⎞ (γ −1) ⎥ I = γ = ( sp )ideal g ⎢ γ − 1⎝⎜ γ + 1⎠⎟ ⎥ o ⎢ ⎥ ⎣ ⎦

0.5 ⎛ ⎛ 1.196 + 1⎞ ⎞ ⎝ ⎠ ⎜ 2 ⎛ 2 ⎞ 1.196 − 1 ⎟ 2295.35⋅1.196 ⎜ ⎝ ⎠ ⎟ ⎝ 1.196 − 1 1.196 + 1 ⎠ 9.806

= 529.69 sec

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Idealized Thrust

• Thrust = m I g = ideal ( sp )ideal 0

529.69⋅437.14⋅9.806 = 2.271 mNt 106

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Effective Exhaust Velocity (Vacuum)

Thrust Aexit * (pexit − p∞ ) Ce = • = Vexit + * A • = m A m

77.5⋅0.0529708 (17.455⋅1000) 4288.61 + 437.14

= 4452.53 m/sec

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Thrust (Vacuum)

• Thrust = mCe =

437.14⋅4452.53 = 1.9464 mNt 106

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

Compute True Isp (Vacuum) (ignore nozzle Losses)

Ce Isp = = g0

4803.891 = 454.06 sec 9.806

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Effective Exhaust Velocity (Sea level)

Thrust Aexit * (pexit − p∞ ) Ce = • = Vexit + * A • = m A m

77.5⋅0.0529708 (17.455⋅1000 − 101325) 4288.61 + 437.14

= 3500.98 m/sec

P sea Level =101.325 kPa

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) Compute Thrust (Seal level) (ignore nozzle Losses)

• Thrust = mCe =

437.14⋅3500.976 = 1.540 mNt 106

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

Compute True Isp (Seal level) (ignore nozzle Losses)

Ce Isp = = g0

3500.976 = 357.024 sec 9.806

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d)

Summary:

Ideal Calc. Calc. Actual Actual Vac. S.L. Vac. S.L ______

Isp 529.69 454.06 357.03 452.5 363 (sec):

Thrust: 2.271 1946.37 1530.42 2.10 1.67 (mNt)

• Obviously Out estimate of throat area is a bit small …. … but you get the point …

MAE 5540 - Propulsion Systems Example: SSME Rocket Engine (cont’d) • When we automate the process … … It appears that A* ~ 0.05785 … or a Throat diameter Of ~ 27.2 cm!

MAE 5540 - Propulsion Systems Plot Flow Properties Along SSME Nozzle Length • A/A*

MAE 5540 - Propulsion Systems Plot Flow Properties Along SSME Nozzle Length (cont’d) ^ • Mach Number ^ ^ F(M ( j) ) M ( j +1) = M ( j) − ⎛ ∂F ⎞ ⎜ ⎟ ⎝ ∂M ⎠ |( j)

MAE 5540 - Propulsion Systems Plot Flow Properties Along SSME Nozzle Length (cont’d) • Temperature T T (x) = 0 γ − 1 1+ M (x)2 2

T0 = 3615°K Tthroat = 3292.3 °K MAE 5540 - Propulsion Systems Plot Flow Properties Along SSME Nozzle Length (concluded)

P0 • Pressure P(x) = γ γ −1 ⎛ γ − 1 2 ⎞ 1+ M (x) ⎝⎜ 2 ⎠⎟

P0 = 18.94 Mpa Pthroat = 11.71 MPa MAE 5540 - Propulsion Systems