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× 𝑚𝑖𝑔0 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:

𝐿𝐸 = 0.00504(15274.06) + 31.92 = 108.9 𝑐𝑚

𝐷𝐸 = 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

𝑚𝐸 =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,

𝑀𝐸 = √2

𝛾−1[ √

𝑃𝐸

𝑃0

𝛾1−𝛾

− 1] EQ6 .

Next the expansion ratio is determined by equation 7 as,

𝜖 =1

𝑀𝐸(

2

𝛾+1(1 +

𝛾−1

2𝑀𝐸

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 −𝑃𝐸

𝑃0

𝛾−1

𝛾 ⌋}

1/2

+(𝑃𝐸−𝑃𝑎)𝑐∗𝜖

𝑃0𝑔0] 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.

𝑀𝐸 = √2

1.225 − 1[ √

0.406

0.7 × 106

1.2251−1.225

− 1] = 5.104

𝜖 =1

5.104(

2

1.225 + 1(1 +

1.225 − 1

25.1042))

1.225+12(1.225−1)

= 100.014

𝐼𝑠𝑝 = 0.98 [1722.771 × 1.225

9.80665{

2

1.225 − 1(

2

1.225 + 1)

1.225+11.225−1 ⌊1 −

0.406

0.7 × 106

1.225−11.225

⌋}

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 = ∆𝐷𝐷𝑌𝑁 =1

2𝜌𝑉2 = 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 = ∆𝐷𝐷𝑌𝑁 =1

2𝜌𝑉2 = 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 + ∆𝐷𝐷𝑌𝑁 = 929, 450 𝑃𝑎

𝑂𝑥𝑖𝑑𝑧𝑖𝑒𝑟 𝑇𝑎𝑛𝑘 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃0 + ∆𝑃𝐼𝑁𝐽 + ∆𝑃𝐹𝑒𝑒𝑑 + ∆𝐷𝐷𝑌𝑁 = 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.

∆𝑚𝑎𝑠𝑠𝑢𝑚𝑒𝑑 = 2,208 𝑘𝑔

𝑓𝑖𝑛𝑒𝑟𝑡 =𝑚𝑖𝑛𝑒𝑟𝑡

𝑚𝑖𝑛𝑒𝑟𝑡 + ∆𝑚=

𝑚𝑖 − 𝑚𝑝𝑎𝑦𝑙𝑜𝑎𝑑 − ∆𝑚

(𝑚𝑖 − 𝑚𝑝𝑎𝑦𝑙𝑜𝑎𝑑 − ∆𝑚) + ∆𝑚=

4700 − 2350 − 2208

(4700 − 2350 − 2208) + 2208

= 0.0604

∆𝑚𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =

𝑚𝑝𝑎𝑦𝑙𝑜𝑎𝑑 {𝑒∆𝑣

𝐼𝑠𝑝𝑔0 − 1} (1 − 𝑓𝑖𝑛𝑒𝑟𝑡)

1 − (𝑓𝑖𝑛𝑒𝑟𝑡 × 𝑒∆𝑣

𝐼𝑠𝑝𝑔0)

=

2350 {𝑒2057

(330.5)(9.80665) − 1} (1 − 0.0604)

1 − (0.0604 × 𝑒2057

(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:

𝑚𝑓𝑢𝑒𝑙 =∆𝑚

𝑂𝑥𝑖𝑑𝑖𝑧𝑒𝑟𝐹𝑢𝑒𝑙

+ 1=

2208

1.75 + 1= 802.9 𝑘𝑔

𝑚𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟 =∆𝑚(


)


+ 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 = (𝑚𝑓𝑢𝑒𝑙

𝜌𝑓𝑢𝑒𝑙

) × 1.1 = (802.9

789) × 1.1 = 1.119 𝑚3

𝑉𝑜𝑥𝑖𝑑𝑧𝑖𝑒𝑟 = (𝑚𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟

𝜌𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟

) × 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

𝐴𝑡

=1

0.2(

2

1.225 + 1(1 +

1.225 − 1

20.22))

1.225+12(1.225−1)

= 3.0179

Now, we need to find the actual chamber area, let’s first find the throat area:

𝐴𝑡 =𝐹𝑐∗

𝑃0𝑔0𝐼𝑠𝑝=

(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

𝐴𝑡

𝐴𝑡 = (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∗ (𝐴𝑡

𝐴0

) = 𝐿∗ ((𝐴0

𝐴𝑡

)−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−1𝛾

= 273 [(945725

30 × 106)

1.66−11.66

] = 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.

𝑚𝑝𝑔 = {𝑉𝑝𝑔𝑃𝑓𝑀

𝑇𝑓𝑅𝑢} 1.05

𝑉𝑝𝑡 =𝑚𝑝𝑔𝑇𝑖𝑅𝑢

𝑀𝑃𝑖

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

𝑚𝑝𝑔 = {(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 0.000286

5.559081

174.5718

335.676

0.4 0.000571

5.113369

101.1885

330.5932

0.401 0.000573

5.111798

100.9906

330.5735

0.402 0.000574

5.110232

100.7935

330.5539

0.403 0.000576

5.108669

100.5973

330.5344

0.404 0.000577

5.10711 100.402 330.5148

0.405 0.000579

5.105556

100.2075

330.4954

0.406 0.00058 5.104005

100.0139

330.4759

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: Pressurant Tank Volume

Finding Pf and Tf

Pf [Pa] Tf [K]

945725 69.0618745

Iterate untill Vpt matches

Iteration Vpt Vpg Mpg Vpt_new

1 0 2.19272497

79.63596429

0.286907143

2 0.286907143

2.47963211

90.05593366

0.324447514

3 0.324447514

2.51717248

91.41933476

0.329359485

4 0.329359485

2.52208445

91.59772901

0.330002192

5 0.330002192

2.52272716

91.62107101

0.330086287

6 0.330086287

2.52281125

91.62412519

0.33009729

7 0.33009729 2.52282226

91.62452482

0.33009873

8 0.33009873 2.5228237 91.62457711

0.330098918

9 0.330098918

2.52282389

91.62458395

0.330098943

10 0.330098943

2.52282391

91.62458484

0.330098946

11 0.330098946

2.52282391

91.62458496

0.330098946

aero4442 lunar ascent module design assignment

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