AERO4442 Lunar Ascent Module design assignment

Sultan Anick Islam (100822163) | AERO 4442A Transatmospheric Propulsion | April 7, 2016
AERO 4442A: Design
Assignment
APOLO’S LUNAR MODULE ASCENT STAGE DESIGN

Table of Contents
1.0 Introduction....................................................................................................................................... 1
2.0 Ascent Propulsion System ............................................................................................................. 1
3.0 Preliminary Design for the Bi-propellant System ........................................................................ 1
3.1 Engine Mass and Size................................................................................................................2
3.2 Exit Mach Number and Specific impulse .................................................................................3
3.3 Oxidizer and Fuel Tank Pressure .............................................................................................7
3.4 Propellant Mass and Volume.....................................................................................................7
3.5 Chamber l/D Ratio...................................................................................................................... 8
3.6 Pressurant Tank Sizing ............................................................................................................. 9
4.0 Conclusion ....................................................................................................................................... 9
5.0 Appendix A: Excel Data File.........................................................................................................10
Table of Figures
Figure 1 Engine Dimensions versus Maxiumum Vacuum Thrust..........................................................2
Figure 2 Engine Thrust-to-Weight Ratio versus Maximum Vacuum Thrust.........................................2
Figure 3 Flame Temperature versus Oxidizer-To-Fuel Ratio ............................................................... 4
Figure 4 Molecular Mass versus Oxidizer-To-Fuel Ratio........................................................................5
Figure 5 Isentropc Parameter vesus Oxidizer-To-Fuel Ratio.................................................................5
Figure 6 Vacuum Specific Impulse versus Oxidizer-To-Fuel-Ratio.......................................................5

1.0 Introduction
In the past couple of weeks we have covered the basic of rocket propulsion from the basics
of orbital mechanics to rocketry equation and the different type of spacecraft propulsion. As a
design exercise and as an assignment, we are expected to do a preliminary design for a
liquid propulsion system. Specifically we are to design the Apollo’s Lunar Module Ascent
Stage. This assignment requires the use of past mission data as well as rocketry equations
to complete the preliminary design. This report is broken up into two sections, where the first
section covers the Ascent Propulsion system requirements and the second goes into the
detail of designing it.
2.0 Ascent Propulsion System
The rocket engine of the Ascent stage which is known as the Ascent Propulsion System was
designed as hypergolic bipropellant system. More specifically it was a bipropellant system
which needed no igniters with Aerozine 50 as the fuel and Nitrogen Tetroxide (N2O4) as an
oxidizer. Aerozine 50 is a 50-50 mixture by weight of hydrazine and unsymmetrical dimethyl
hydrazine (UDHM), however for this design assignment it is assumed that the fuel is 100%
UDHM as its properties are closest to Aerozine 50. With that assumption in mind, the design
requirements and given values are listed below:
 ΔV = 2,057 m/s
 Payload Mass = 2,350 kg
 Initial Mass = 4,700 kg
 Minimum F/W = 2
 Fuel = UDHM
 Oxidizer = N2O4
 Pressurant Gas = H2 (stored at 30 Mpa)
 Chamber Pressure = 0.7Mpa
 Cooling: Ablative
 go Earth = 9.80665 m/s2
 go Moon = 1.6249 m/s2
3.0 Preliminary Design for the Bi-propellant System
There are 7 specific design questions that is needed to be answered for this assignment.
The questions are as followed:
1. Based on historical trends, determine the size and mass of the engine (excluding
tanks).
2. For an expansion ratio of 100, determine the exit Mach number and specific impulse.
3. Determine the fuel tank and oxidizer tank pressures.
4. Determine the propellant mass.
5. Determine the fuel tank and oxidizer tank volume.
6. Determine the chamber L/D ratio, assuming Mach 0.2 in the chamber and L∗ = 0.9.
7. Determine the pressurant tank volume (note: iterations are needed).

Each question is answered in its own specific subsection outlining the method and
sample calculation. All calculations were done using Excel, if needed the excel
spreadsheet can be found in Appendix A.
3.1 ENGINE MASS AND SIZE
So for this part we need to rely of past data points to estimate an engine mass and
size. In order to do this lets reference these two graphs for bipropellant systems.
Figure 1 Engine Dimensions versus Maxiumum Vacuum Thrust
Figure 2 Engine Thrust-to-Weight Ratio versus Maximum Vacuum Thrust

The two figures above are graphs based on historical data of bipropellant liquid
propulsion engines. Figure 1 is used to size the engine while Figure 2 is used to find the
mass.
First we need to find the maximum vacuum thrust, in order to do this we use the F/W
given in the design problem with our initial mass.
𝐹 =
𝐹
𝑊
× 𝑚𝑖 𝑔0 EQ1
Where go is the gravitational acceleration in vacuum or for our case the moon. Thus the
maximum vacuum thrust is given to be:
𝐹 = 2 × (4700)(1.6249) = 15, 274.06 𝑁
Using this and the equation given in figure 1, the size of the engine is determined to be:
𝐿 𝐸 = 0.00504(15274.06) + 31.92 = 108.9 𝑐𝑚
𝐷 𝐸 = 0.0024(15274.06) + 0.6392 = 69.01 𝑐𝑚
Which when rounded gives us a Length of 1.1 m and a diameter of 0.7 m. For the engine
mass, we first need to find the engine F/W and then using the gravitational acceleration
of earth, find the mass (since Weight = mass x acceleration). Thus the engine mass is
determined to be:
𝐹
𝑊
= 0.0006098(15274.06) + 13.44 = 22.754
𝑚 𝐸 =
15274.06
22.754 × 9.80665
= 68.45 𝑘𝑔
3.2 EXIT MACH NUMBER AND SPECIFIC IMPULSE
For this part we need to determine the exit mach number and specific impulse at an
expansion ratio of 100. Before we do that, let’s determine some other parameters
(such as Gamma and R for the gases) as we will need them for this and subsequent
parts.
Referencing Appendix B from our notes, density of the propellants can be found
using table 1:

Table 1 Properties of Liquid Propellants
Where the density of UDHM is 789 kg/m3 and the density of N204 is 1440 kg/m3.
The isentropic parameter (gamma), flame temperature, vacuum Isp, oxidizer to fuel
ratio and molecular mass is found using the three figure bellow.
Figure 3 Flame Temperature versus Oxidizer-To-Fuel Ratio

Figure 4 Molecular Mass versus Oxidizer-To-Fuel Ratio
Figure 5 Isentropc Parameter vesus Oxidizer-To-Fuel Ratio
Figure 6 Vacuum Specific Impulse versus Oxidizer-To-Fuel-Ratio

Using Figures 3-6, we first find the Oxidizer-To-Fuel Ratio by finding the correct oxidizer
and fuel Curve (N204 and UDHM) and find the point where Isp is maximum. Thus for a
maximum Isp of 310 second, the Oxidizer-To-Fuel ratio for N2O4/UDHM is 1.75. Similarly,
using the Oxidizer-To-Fuel ratio and the correct curve, the Isentropic Parameter
(Gamma) is given to be 1.225, with a molecular mass of 21 kg/mol and a flame
temperature (T0) of 3200 K. The specific gas constant R is given as 395.9 (since R=Ideal
Gas Constant/Molar Mass).
Before solving for the exit Mach number and Isp, the values for a0 (speed of sound) and
c* were calculated bellow.
𝑎0 = √ 𝛾𝑅𝑇0 = √1.225 × 395.9 × 3200 = 1245.764 𝑚/𝑠 EQ4
Γ′(𝛾) = 𝛾 (
2
𝛾+1
)
𝛾+1
2(𝛾−1
= 1.225(0.8988764)4.944
= 0.723 EQ5
𝑐∗
=
𝑎0
Γ′(𝛾)
=
1245.764
0.723
= 1722.771 𝑚/𝑠 EQ5
Now we can determine exit Mach number and the Isp for an expansion ratio of 100. The
best way to do this is to evaluate ME, Expansion ratio and Isp for a range of PE (exit
pressure) based on ranges given by historical data. Referencing our course notes PE
varies between 200 Pa to 50 KPa, the chamber pressure P0 was given in the problem to
be 0.7 Mpa. First step is to calculate ME by the following equation,
𝑀 𝐸 = √
2
𝛾−1
[ √
𝑃 𝐸
𝑃0
𝛾
1−𝛾
− 1] EQ6 .
Next the expansion ratio is determined by equation 7 as,
𝜖 =
1
𝑀 𝐸
(
2
𝛾+1
(1 +
𝛾−1
2
𝑀 𝐸
2
))
𝛾+1
2(𝛾−1)
EQ7.
Finally the Isp is determined by using the following equation,
𝐼𝑠𝑝 = 𝜆 [
𝐶∗ 𝛾
𝑔0
{
2
𝛾−1
(
2
𝛾+1
)
𝛾+1
𝛾−1 ⌊1 −
𝑃 𝐸
𝑃0
𝛾−1
𝛾
⌋}
1/2
+
(𝑃 𝐸−𝑃 𝑎)𝑐∗ 𝜖
𝑃0 𝑔0
] EQ8.
So utilizing equations 6 to 8 for ranges of Pe, it was found the optimal solution was found
by using a PE of 0.406 Pa.
𝑀 𝐸 = √
2
1.225 − 1
[ √
0.406
0.7 × 106
1.225
1−1.225
− 1] = 5.104
𝜖 =
1
5.104
(
2
1.225 + 1
(1 +
1.225 − 1
2
5.1042
))
1.225+1
2(1.225−1)
= 100.014

𝐼𝑠𝑝 = 0.98 [
1722.771 × 1.225
9.80665
{
2
1.225 − 1
(
2
1.225 + 1
)
1.225+1
1.225−1 ⌊1 −
0.406
0.7 × 106
1.225−1
1.225
⌋}
1/2
+
(0.406 − 0)(1722.771 × 100.014)
(0.7 × 106)(9.80665)
] = 330.4759 𝑠𝑒𝑐
Thus, from the above calculation, it is seen that using an exit pressure of 0.406 Pa
an expansion ratio of 100 is attainable with an exit Mach number of 5.104 and an Isp
of 330.5 sec.
3.3 OXIDIZER AND FUEL TANK PRESSURE
In order to figure out the tank pressures, let’s first figure out the pressure losses of
each system. For the fuel tank:
𝐼𝑛𝑗𝑒𝑐𝑡𝑜𝑟 𝐿𝑜𝑠𝑠 = ∆𝑃𝐼𝑁𝐽 = 0.2𝑃0 = 0.2(0.7 × 106) = 140,000 𝑃𝑎
𝐹𝑒𝑒𝑑 𝑆𝑦𝑠𝑡𝑒𝑚 𝐿𝑜𝑠𝑠 = ∆𝑃𝐹𝑒𝑒𝑑 = 50,000 𝑃𝑎
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐿𝑜𝑠𝑠 = ∆𝐷 𝐷𝑌𝑁 =
1
2
𝜌𝑉2
= 0.5(789)(102) = 39, 450 𝑃𝑎
Similarly for the oxidizer tank:
𝐼𝑛𝑗𝑒𝑐𝑡𝑜𝑟 𝐿𝑜𝑠𝑠 = ∆𝑃𝐼𝑁𝐽 = 0.2𝑃0 = 0.2(0.7 × 106) = 140,000 𝑃𝑎
𝐹𝑒𝑒𝑑 𝑆𝑦𝑠𝑡𝑒𝑚 𝐿𝑜𝑠𝑠 = ∆𝑃𝐹𝑒𝑒𝑑 = 50,000 𝑃𝑎
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐿𝑜𝑠𝑠 = ∆𝐷 𝐷𝑌𝑁 =
1
2
𝜌𝑉2
= 0.5(1440)(102) = 72, 000 𝑃𝑎
The tank pressure for the fuel and oxidizer tanks were determined to be:
𝐹𝑢𝑒𝑙 𝑇𝑎𝑛𝑘 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃0 + ∆𝑃𝐼𝑁𝐽 + ∆𝑃𝐹𝑒𝑒𝑑 + ∆𝐷 𝐷𝑌𝑁 = 929, 450 𝑃𝑎
𝑂𝑥𝑖𝑑𝑧𝑖𝑒𝑟 𝑇𝑎𝑛𝑘 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃0 + ∆𝑃𝐼𝑁𝐽 + ∆𝑃𝐹𝑒𝑒𝑑 + ∆𝐷 𝐷𝑌𝑁 = 962, 000 𝑃𝑎
3.4 PROPELLANT MASS AND VOLUME
In order to calculate the delta m we need to determine the inert mass fraction which is in
term of the inert mass which itself is in terms of delta m. This is why this portion of the
calculation requires an iteration. It was calculated the delta m should be 2, 208 kg, the
calculation in question is shown below.
∆𝑚 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 = 2,208 𝑘𝑔
𝑓𝑖𝑛𝑒𝑟𝑡 =
𝑚𝑖𝑛𝑒𝑟𝑡
𝑚𝑖𝑛𝑒𝑟𝑡 + ∆𝑚
=
𝑚𝑖 − 𝑚 𝑝𝑎𝑦𝑙𝑜𝑎𝑑 − ∆𝑚
(𝑚𝑖 − 𝑚 𝑝𝑎𝑦𝑙𝑜𝑎𝑑 − ∆𝑚) + ∆𝑚
=
4700 − 2350 − 2208
(4700 − 2350 − 2208) + 2208
= 0.0604
∆𝑚 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =
𝑚 𝑝𝑎𝑦𝑙𝑜𝑎𝑑 {𝑒
∆𝑣
𝐼𝑠𝑝𝑔0 − 1} (1 − 𝑓𝑖𝑛𝑒𝑟𝑡)
1 − (𝑓𝑖𝑛𝑒𝑟𝑡 × 𝑒
∆𝑣
𝐼𝑠𝑝𝑔0)
=
2350 {𝑒
2057
(330.5)(9.80665) − 1} (1 − 0.0604)
1 − (0.0604 × 𝑒
2057
(330.5)(9.80665))
= 2208.96 𝑘𝑔

If you change the delta m assumed to a higher number, the new calculated delta m would
not converge, hence why 2,208 kg was chosen for the delta m. Now that we have the
delta m, propellant mass for the fuel and oxidizer can be calculated as followed:
𝑚 𝑓𝑢𝑒𝑙 =
∆𝑚
𝑂𝑥𝑖𝑑𝑖𝑧𝑒𝑟
𝐹𝑢𝑒𝑙
+ 1
=
2208
1.75 + 1
= 802.9 𝑘𝑔
𝑚 𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟 =
∆𝑚(
𝐹𝑢𝑒𝑙
)
𝐹𝑢𝑒𝑙
+ 1
=
2208(1.75)
1.75 + 1
= 1405.09 𝑘𝑔
Since both the density and mass of the oxidizer and fuel is known, the volume for each
can now be calculated as:
𝑉𝑓𝑢𝑒𝑙 = (
𝑚 𝑓𝑢𝑒𝑙
𝜌 𝑓𝑢𝑒𝑙
) × 1.1 = (
802.9
789
) × 1.1 = 1.119 𝑚3
𝑉𝑜𝑥𝑖𝑑𝑧𝑖𝑒𝑟 = (
𝑚 𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟
𝜌 𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟
) × 1.1 = (
1405.09
1440
) × 1.1 = 1.073 𝑚3
3.5 CHAMBER L/D RATIO
In order to find the chamber L/D, we need to find both L0 and D0. First we need to
find A0/At, which can be found using a modified version of equation 7 where ME is
replaced with M0 (Since the expansion ratio is just a ratio of AE/At). Thus, the ratio of
chamber and throat area is calculated as:
𝐴0
𝐴 𝑡
=
1
0.2
(
2
1.225 + 1
(1 +
1.225 − 1
2
0.22
))
1.225+1
2(1.225−1)
= 3.0179
Now, we need to find the actual chamber area, let’s first find the throat area:
𝐴 𝑡 =
𝐹𝑐∗
𝑃0 𝑔0 𝐼𝑠𝑝
=
(15274.06)(1722.771)
(0.7 × 106)(9.80665)(330.5)
= 0.0116 𝑚3
Next the chamber Area can be calculated to be:
𝐴0 =
𝐴0
𝐴 𝑡
𝐴 𝑡 = (3.0179)(0.0116) = 0.035 𝑚3
Now chamber length and diameter can be determined. Using L*, the chamber
length is calculated to be:
𝐿0 = 𝐿∗
(
𝐴 𝑡
𝐴0
) = 𝐿∗
((
𝐴0
𝐴 𝑡
)−1
) = (0.9)(3.0179−1) = 0.298 𝑚
Next using the chamber area, the chamber diameter is calculated to be:
𝐷0 = 2√
𝐴0
𝜋
= 2√
0.035
𝜋
= 0.211 𝑚
Thus the chamber L/D is:

𝐿
𝐷
=
𝐿0
𝐷0
=
0.298
0.211
= 1.41
3.6 PRESSURANT TANK SIZING
The final question of the design assignment asks us to size the pressurant tank. From the
design problem an initial pressure is given to be 30 Mpa, the gas is Helium with an
isentropic ratio 𝛾 = 1.66 and an initial tank temperature of 273 K. The final tank pressure
and temperature were calculated to be:
𝑃𝑓 =
𝑃𝐹𝑢𝑒𝑙 𝑇𝑎𝑛𝑘 + 𝑃𝑂𝑥𝑖𝑑𝑖𝑧𝑒𝑟 𝑇𝑎𝑛𝑘
2
=
929450 + 962000
2
= 945, 725 𝑃𝑎
𝑇𝑓 = 𝑇𝑖 (
𝑃𝑓
𝑃𝑖
)
𝛾−1
𝛾
= 273 [(
945725
30 × 106
)
1.66−1
1.66
] = 69.06 𝐾
Now that the final pressure and temperature is known, the volume of the pressurant tank
can be obtained via iteration. The iterative step is outlined below:
 Assume a Vpt = 0 m3
 𝑉𝑝𝑔 = 𝑉𝑝𝑟𝑜𝑝𝑒𝑙𝑙𝑎𝑛𝑡 + 𝑉𝑝𝑡 Where Vpropellant is the volume of fuel and oxidizer added
together.
 𝑚 𝑝𝑔 = {
𝑉𝑝𝑔 𝑃 𝑓 𝑀
𝑇 𝑓 𝑅 𝑢
} 1.05
 𝑉𝑝𝑡 =
𝑚 𝑝𝑔 𝑇 𝑖 𝑅 𝑢
𝑀𝑃 𝑖
 Repeat above step until solution converges to the same Vpt.
It took about 10 iteration for the solution to fully converge, the 10th iteration calculation is
shown below:
 Assume a Vpt = 0.33 m3
 𝑉𝑝𝑔 = 2.193 + 0.33 = 2.52 𝑚3
 𝑚 𝑝𝑔 = {
(2.52)(945725)(21)
(69.06)(8314)
} 1.05 = 91.62 𝑘𝑔
 𝑉𝑝𝑡 =
(91.62)(273)(8314)
(21)(30×106)
= 0.33 𝑚3
 Solution converged to a pressurant tank volume of 0.33 m3
4.0 Conclusion
To conclude, the 7 design questions were answered and the calculations were illustrated
above. Though not required, a quick google search showed the performance of the ascent
module, the relevant numbers seem to be close to what was obtained from this exercise.
Sources of errors are due to rounding and the assumption that the fuel is 100% UDHM whilst
in the real ascent propulsion system it was a 50-50 by weight ratio of UDHM and Hydrazine.

5.0 Appendix A: Excel Data File
Given Data
Chemical: UDHM+N2O4
delta v [m/s] 2057
Payload Mass [kg] 2350
Initial Mass [kg] 4700
Minimum Force/Weight 2
Pressurant Tank of H [MPa] 30
Chamber Pressure [MPa] 0.7
G_o of earth [m/s^2] 9.80665
G_o of moon [m/s^2] 1.6249
Data read from Appendix:
Vacuum Specific Impulse [s] 310
M_ox/M_fuel 1.75
Isentropic Parameter 1.225
Molecular Mass of Combustion [kg/mol] 21
Flame Temperature T0 [K] 3200
Specific Gas Constant R 395.9
Sig(gamma) 0.723116404
a0 [m/s] 1245.764023
c* 1722.771072
Part A Initial Sizing
F=(F/W)*(mi*go) [N] 15274.06
Engine Length [cm] 108.9012624
Engine Diameter [cm] 69.0083942
Engine F/W 22.75412179
Engine Mass [kg] 68.45004507

Part B: Exit Mach and ISP
Pe [Kpa] Pe/Po Me Epsilon ISP
0.2 0.00028
6
5.55908
1
174.571
8
335.676
0.4 0.00057
1
5.11336
9
101.188
5
330.593
2
0.401 0.00057
3
5.11179
8
100.990
6
330.573
5
0.402 0.00057
4
5.11023
2
100.793
5
330.553
9
0.403 0.00057
6
5.10866
9
100.597
3
330.534
4
0.404 0.00057
7
5.10711 100.402 330.514
8
0.405 0.00057
9
5.10555
6
100.207
5
330.495
4
0.406 0.00058 5.10400
5
100.013
9
330.475
9
Part C: Oxidizer and Fuel Tank Pressure
Find Losses to system and then find pressure
Fuel Tank:
Injector Loss 140000
Feed System Loss 50000
Dynamic Pressure Loss 39450
Oxidizer Tank:
Injector Loss 140000
Feed System Loss 50000
Dynamic Pressure Loss 72000
Tank Fuel Pfuel [Pa] 929450
Tank Oxidizer Pox [Pa] 962000

Part C: Oxidizer and Fuel Tank
Pressure
Fuel Tank:
Injector Loss [Pa] 140000
Feed System Loss [Pa] 50000
Dynamic Pressure Loss [Pa] 39450
Oxidizer Tank:
Injector Loss [Pa] 140000
Feed System Loss [Pa] 50000
Dynamic Pressure Loss [Pa] 72000
Tank Fuel Pfuel [Pa] 929450
Tank Oxidizer Pox [Pa] 962000
Part D & E: Propellant Mass and Volume
Finding Delta M M_assumed Finert M_calculated
2208 0.06042553 2208.964548
Finding Mass M_fuel [Kg] M_ox [kg]
802.9090909 1405.09091
Finding Volume V_fuel [m^3] V_ox [m^3] V_prop
[m^3]
1.119391635 1.07333333 2.192724968
Part F: Chamber L/D Ratio
Mo L* Ao/At
0.2 0.9 3.017888671
Lo Ao Do
0.298221737 0.03500473 0.211168215
Lo/Do
1.412247277

Part G:
Pressuran
t Tank
Volume
Finding Pf
and Tf
Pf [Pa] Tf [K]
945725 69.061874
5
Iterate
untill Vpt
matches
Iteration Vpt Vpg Mpg Vpt_new
1 0 2.1927249
7
79.6359642
9
0.28690714
3
2 0.28690714
3
2.4796321
1
90.0559336
6
0.32444751
4
3 0.32444751
4
2.5171724
8
91.4193347
6
0.32935948
5
4 0.32935948
5
2.5220844
5
91.5977290
1
0.33000219
2
5 0.33000219
2
2.5227271
6
91.6210710
1
0.33008628
7
6 0.33008628
7
2.5228112
5
91.6241251
9
0.33009729
7 0.33009729 2.5228222
6
91.6245248
2
0.33009873
8 0.33009873 2.5228237 91.6245771
1
0.33009891
8
9 0.33009891
8
2.5228238
9
91.6245839
5
0.33009894
3
10 0.33009894
3
2.5228239
1
91.6245848
4
0.33009894
6
11 0.33009894
6
2.5228239
1
91.6245849
6
0.33009894
6

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