Project 3, CEA Exercise -...

MAE 5450 - Propulsion Systems

Project 3, CEA Exercise• This Programing Assignment is due on Beginning of Final Exam Period-- 11:30 AM MDT, Wednesday May 3.

• We are going to build a chemistry table for an AP-composite rocketPropellant, and investigate the effects due to increasing metallization of the grain.

• Look at important effects on flame temperature, molecular weights, andC* (infinitely expanded nozzle)

1


Project Background

2

Need atLeast 15% minimumtotal HTPB binder by mass for “cake” to stick together

0.150≤MHTPB

MAl + MAP+ MHTPB( )Adiabatic Flame Temperature of AP/HTPB/Al Composite Propellant as function of Mass


Project Background (2)

• Investigate Mixture effects

Plot T0,g ,C* for both Chamber & Throat vs O/F for increasing aluminization levels

3

• Using equilibrium properties at Throat, plot data for various mixture ratios and determine optimal operating mixture ratio (based on C*) .. Assume T0 is constant through out motoe• Based on flow properties from nozzle throat … Update AMW L-700 model for “Best case” Formulation Properties

Assume that St. Roberts burn /erosive burn/Bates Grain parameters are same as previously used

O/F


Project Overview• Down Load the CEAGUI from the NASA Glenn Research center Web site

(recommend you use the windows GIU as it seems to be bug free …Linix code has a bunch of compile errors that need to be fixed)

http://www.grc.nasa.gov/WWW/CEAWeb/ceaRequestForm.htm

• Set up input file to run as “Rocket” Problem with a combustion pressureof 3000 kPa (30 bars)

• Look AP Composite propellant with Mixture ratio of AP/HTPB/AL

• Run code in “equilibrium”, with “infinite” combustor contraction ratio

• Use results for molecular weight(Mw), ratio of specific heats (g), and combustion temperature to calculate and C*

1) Based on Chamber, g, Mw.2) Based on Throat (*), g, Mw.

…. Assume that T0chamber = T0*

4


Project Overview (2)

5

• Apply what we learn to AMW-L700 Motor Analysis

.. Look at both Cylindrical Port with both erosive burn and Bates grain .. Compare original and “improved” propellant” formulation (be sure to re-evaluate the propellant density based on formulation) Using new g, Rg, Mw, T0 ….......

Compare time history plots of chamber pressure thrustregression rate

Calculate and compare … total impulse .. effective Isp’s

… assume that Saint Robert’s burn parameters (a, n} and the erosionparameters (k, Mcrit) remain unchanged for new propellant


Project Overview (3)

6

• Based on Equilibrium flow “Frozen” at nozzle throat …

Compare to project 2 solution including erosive burn model for cylindrical grainand /Bates grain model

• Using equilibrium combustor properties from CEA


Equilibrium Properties at Chamber and Throat (example)

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CHAMBER THROAT

P0/Pstatic 1.0000 1.7428P, BAR 31.000 17.787T, K 2925.58 2744.19RHO, KG/CU M 3.65500 2.2555 0H, KJ/KG 0.00000 -454.46U, KJ/KG -848.15 -1243.08G, KJ/KG -26683. 9 -25484.0S, KJ/(KG)(K) 9.1209 9.1209

MW, (1/n) 28.680 28.932(dLV/dLP)t -1.01044 -1.00776(dLV/dLT)p 1.2400 1.1912Cp, KJ/(KG)(K) 3.1727 2.9098GAMMAs 1.1495 1.1526SON VEL,M/SEC 987.4 953.4MACH NUMBER 0.000 1.000

Assume P0, T0 constant throughout motorCalculate C* based on local g (throat), Mw (throat) , T0 (chamber)


Modified Ballistic Equation for Shifting Equilibrium

When “shifting equilibrium” from chamber to throat and then frozen flow at throat … is used … the chamber pressure equation for solid motor must be modified as …

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∂P0

∂t=AburnaPo

n

Vcρ pRgT0 − P0⎡⎣ ⎤⎦ − P0

A*

Vcγ *Rg

*T0* 2γ * +1

⎛⎝⎜

⎞⎠⎟

γ *+1γ *−1( )

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

( )*→ condition at nozzle throat ... γ *,Rg

*,T0*,Mw

*{ }Rg

* =RuMw

* γ * =Cp

*

Cv* T0

* = T * ⋅ γ* +12

⎛⎝⎜

⎞⎠⎟


Characteristic Velocity, C*• The characteristic velocity is a figure of thermo-chemical merit for a particular propellant and may be considered to beIndicative of the combustion efficiency.

• Lower Molecular Weight Propellants Produce Higher C*

• For this calculation based value on g, Mw

at the nozzle throat …

*

**

*

*

*


Mass Fraction Relationships

10

O/F

O/F

O/F

O/F

O/F

( O/F O/FO/F


Mass Fraction Relationships (2)

11

…i.e. Show that based on 85% max solid constraint

O / FMax =

MAl + MAP

MHTPB+ MAl + MAP

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟− fAl

1−MAl + MAP

MHTPB+ MAl + MAP

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=0.85( )− fAl1− 0.85( )

fAl =MAl

MHTPB+ MAl

MAl

MHTPB

=fAl

1− fAl

Required .... MAl + MAP

MHTPB+ MAl + MAP

≤0.85


Mass Fraction Relationships (3)

12

fAl =MAl

MHTPB+ MAl

= 0.175 →MAl

MHTPB

=fAl

1− fAl= 0.2121 ... O / F =

MAP

MHTPB+ MAl

= 4.5

Required .... 100% ⋅MAl + MAP

MHTPB+ MAl + MAP

≤85%

MAl + MAP

MHTPB+ MAl + MAP

=

MAl

MAP

+1

MHTPB+ MAl

MAP

+1=

MAl

O / F ⋅ MHTPB+ MAl( )+1

1O / F

+1=

MAl / MHTPB( )1+ MAl / MHTPB( )( )

+O / F

O / F+1=

O / F+

fAl1− fAl

1+fAl

1− fAl

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

O / F+1=

O / F+ fAl1− fAl + fAl( )

O / F+1=O / F+ fAlO / F+1

MAl + MAP

MHTPB+ MAl + MAP

=O / F+ fAlO / F+1 =

…e.g. look at à 25% AL/HTPB ratio @ O/F = 5

Too High of Al Fraction! .. Must Lower O/F to get p

> 0.85


AP Composite Propellant Key Physical Properties

• AP chemical formula, NH4 Cl O4

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Fine white powder usedas propellant oxidizer

Ammonium perchlorate crystals decompose before melting

When AP is mixed with a fuel (like a Al powder) or polymeric binder (like HTPB) it can generate considerable het release and allows self-sustained combustion once lit

ρNH4ClO4= 1.950 g

cm3

MwNH4ClO4= 117.49kg/kg−mol


AP Composite Propellant -- Key Physical Properties

• Powdered Elemental Al, micron-scale

14

Al Powder acts both as a fuel and combustion catalyst

(molecular weight) kg / kg −molΔH f0

= 0KJ /mol


Key Physical Properties, cont’d (2)

• HTPB -- (ARCO R-45TM, polymerization ~ 50)• Full Chemical Formula

• Butadiene C4H6, with n-50 degree of polymerization with hydroxyl termination (hydroxyl makes polymerized rubber more hydrophobic)

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(C4H6)50 OH2 à Molecular Weight ~ 2766 kg/kg-mol

- OH

OH -


Key Physical Properties (3)

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ρHTPB = 0.930 gcm3

C4H6( )⋅ OH( )0.40.04

• “Reduced” Chemical Formula w Broken Polymer Bonds

Molecular Weight ~ 54.68 kg/kg-mol

O2 H2 TARs

C4H6 (butadiene gas)

• Main “fuel” for combustion reaction

• Enthalpy of Formation Accounts for Energy Required to Break Polymer Chains

ΔH f0~ 23.99KJ /g−mol = 23.99×10

3KJ /kg−mol



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• Insert HTPB PropertiesHere (Not in CEA tables)

• CEA calculates mass properties based on entered chemical formula

• HTPB Properties not listed in CEA tables



• Other Important Physical Properties

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Need to calculate your propellant density for“best mixture”

ρHTPB = 0.930 gcm3

ρNH4ClO4= 1.950 g

cm3

ρAl Powder = 2700 kgm3



• Other Important Physical Properties

19

ρHTPB = 0.930 gcm3

ρNH4ClO4= 1.950 g

cm3

ρAl Powder = 2700 kgm3


CEA Input File ….. Example < file >.inp

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problem o/f=1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7.,7.5,8,9, rocket equilibrium frozen nfz=1 tcest,k=3000 p,bar=30,

react oxid=NH4CLO4(I) wt=100 t,k=298 fuel=AL(cr) wt=25 t,k=298 fuel=HTPB wt=75 t,k=298 h,kj/mol=23.99 C 4 H 6 O 0.04 H 0.04

output plot p t rho m cp gam end

Setup file by default written to < file > input whenCode is saved using > “save as”


Solution

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******************************************************************************* NASA-GLENN CHEMICAL EQUILIBRIUM PROGRAM CEA2, MAY 21, 2004

BY BONNIE MCBRIDE AND SANFORD GORDONREFS: NASA RP-1311, PART I, 1994 AND NASA RP-1311, PART II, 1996

******************************************************************************* problem o/f=1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7.,7.5,8,9,

rocket equilibrium frozen nfz=1 tcest,k=3000 p,bar=30,

react oxid=NH4CLO4(I) wt=100 t,k=298 fuel=AL(cr) wt=25 t,k=298 fuel=HTPB wt=75 t,k=298

h,kj/mol=23.99 C 4 H 6 O 0.04 H 0.04 output

plot p t rho m cp gam end


CEA Analysis Results

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… Look at 25% AL/HTPB ratio, O/F = 5

Too High!MAl + MAP

MHTPB+ MAl + MAP

=O / F+ fAlO / F+1

=

… Look at 20% AL/HTPB ratio, O/F = 5MAl + MAP

MHTPB+ MAl + MAP

=O / F+ fAlO / F+1

= Still Too High!

… Look at 25% AL/HTPB ratio, O/F = 4MAl + MAP

MHTPB+ MAl + MAP

=O / F+ fAlO / F+1

=OK!

… Look at 17.5% AL/HTPB ratio, O/F = 4.5MAl + MAP

MHTPB+ MAl + MAP

=O / F+ fAlO / F+1

= OK!

… Look at 10% AL/HTPB ratio, O/F = 5 OK!

85% LimitLine Calculations


CEA Analysis Results

23

P0=3000 kPa Results for Combustor Properties

85% Limit Line

Best C* à 25% AL/HTPB Ratio, O/F = 4


CEA Analysis Results (2)

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P0=3000 kPa Results for Throat Properties

85% Limit Line

Best C* à 25% AL/HTPB Ratio, O/F = 4


CEA Analysis Results (2)

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Results for Throat PropertiesResults for Combustor Properties

• Slight Enthalpy Loss (<1.19%) Due to Non-Adiabatic ExpansionBest C* à 25% AL/HTPB Ratio, O/F = 4

h0 combustor= cp ⋅T0( )

combustor=

1901.86J /kg−K ⋅2967.3K106 = 5.6433MJ /kg

h0 throat= cp ⋅T0( )

throat =

1877.93J /kg−K ⋅2969.15K106 = 5.5765MJ /kg

MAE 5450 - Propulsion Systems 26

Compare Chamber and Throat Temperatures

Chamber T0

Throat T0

Throat T*


Optimized CEA Result for

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25% AL/HTPB ratio, O/F = 4


Calculate Propellant Density

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25% AL/HTPB ratio, O/F = 4

= 1694.73 kg/m3=

2700


Original Project 2 Solution

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L0 = 35 cmD0= 7.6 cmD0=3 cm

= 1260 kg/m3

Fuel Grain Geometry

Nozzle GeometryA* = 1.887 cm 2Aexit/A* = 4.0qexit = 20 deg.

Animal WorksTM, L700 Motor Geometry

Part 1cylindrical port

Single propellantsegment

Assume ends are burn inhibited


Original Project 2 Solution (2)

33

= 1.18MW = 23 kg/kg-molT0= 2900 K

Combustion Gas Properties

Burn Parametersa= 0.12 cm/(sec-kPa n)n=0.16M crit = 0.3k = 0.2(cylindrical port only) 1260 kg/m3


Original Project 2 Solution (3)

34

C*=1588.36m/sec


Revised Project 2 Solution (2)

35

Chamber Throat= 1.2129 1.2162

MW = 24.906 kg/kg-mol 24.906T0= 2967.3 K T0

* =2969.15

Combustion Gas Properties

Burn Parametersa= 0.12 cm/(sec-kPa n)n=0.16M crit = 0.3k = 0.2(cylindrical port only)

25% AL/HTPB ratio, MR = 4

Part 1cylindrical port 1694.73 kg/m3

C*=1527.3 m/sec


3636

Cylindrical Port with Erosion Comparisons

Significant increase in Predicted Thrust, Total Impulse due to higher density, slight

Isp decrease

MAE 5450 - Propulsion Systems 3737

Bates Grain Comparisons

Significant increase in Predicted Thrust, Total Impulse due to higher

density, Also Isp increase


Project 3• 11) Plot Minimum length contour over actual nozzle contour

Original L700 Nozzle • Use nexit/2 rule

• What is the L-700 factor of safety ….

• Overlay Nozzle Contour Using

F.O.S =θwallmax −θexitactualθexitactual

θnozzle =23θwallmax

Rule


Project 3 Solution

2

• 11) Plot Minimum length contour over actual nozzle contour

Original L700 Nozzle • Use Conical Nozzle RuleExit Gas Properties @ Throat g = 1.2162 Mw = 24.906 kg/kg-molT0

* = 2969.15 KAexit/A* = 4.0

Mexit = 2.64385


Project 3 Solution

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Prandtl Meyer Function

nexit = 51.9515o

qmax_cone = nexit /2 = 25.98o


Project 3 Solution

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nexit = 51.9515o

qmax_cone = nexit /2 = 25.98o

F.O.S =θwallmax −θexitactualθexitactual

=25.9756−20

20= 29.88%

• Using 2/3rd’s safety margin

θnozzle =23iθwallmax =17.32

o


Project 3 Solution

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• 11) Compare Minimum length contours over actual nozzle contour

Project 3, CEA Exercise -...

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