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)
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MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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*
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MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
Project Overview (3)
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• 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
MAE 5450 - Propulsion Systems
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)
MAE 5540 - Propulsion Systems
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
⎛⎝⎜
⎞⎠⎟
MAE 5450 - Propulsion Systems
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 …
*
**
*
*
*
MAE 5450 - Propulsion Systems
Mass Fraction Relationships (2)
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…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
MAE 5450 - Propulsion Systems
Mass Fraction Relationships (3)
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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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
AP Composite Propellant -- Key Physical Properties
• Powdered Elemental Al, micron-scale
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Al Powder acts both as a fuel and combustion catalyst
(molecular weight) kg / kg −molΔH f0
= 0KJ /mol
MAE 5450 - Propulsion Systems
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 -
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
Key Physical Properties (4)
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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
MAE 5450 - Propulsion Systems
Key Physical Properties (5)
• 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
MAE 5450 - Propulsion Systems
Key Physical Properties (6)
• Other Important Physical Properties
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ρHTPB = 0.930 gcm3
ρNH4ClO4= 1.950 g
cm3
ρAl Powder = 2700 kgm3
MAE 5450 - Propulsion Systems
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”
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
CEA Analysis Results
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P0=3000 kPa Results for Combustor Properties
85% Limit Line
Best C* à 25% AL/HTPB Ratio, O/F = 4
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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*
MAE 5450 - Propulsion Systems
Calculate Propellant Density
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25% AL/HTPB ratio, O/F = 4
= 1694.73 kg/m3=
2700
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
Original Project 2 Solution (2)
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= 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
MAE 5450 - Propulsion Systems
Revised Project 2 Solution (2)
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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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
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
MAE 5450 - Propulsion Systems
Project 3 Solution
3
• 11) Plot Minimum length contour over actual nozzle contour
Prandtl Meyer Function
nexit = 51.9515o
qmax_cone = nexit /2 = 25.98o
MAE 5450 - Propulsion Systems
Project 3 Solution
4
• 11) Plot Minimum length contour over actual nozzle contour
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