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Cold Gas Thruster Project 1 Cold Gas Thruster Design, Build, Test AAE 439: Extra Credit Project Team Dorado Saphal Adhikari, Claire Alexander, Alicia Benhidjeb-Carayon, and Julian Wang

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Page 1: Team Dorado Cold Gas Thruster Report Final

Cold Gas Thruster Project

1

Cold Gas Thruster Design, Build, Test

AAE 439: Extra Credit Project

Team Dorado

Saphal Adhikari, Claire Alexander, Alicia Benhidjeb-Carayon, and Julian Wang

Page 2: Team Dorado Cold Gas Thruster Report Final

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Table of Contents Design and Thruster Analysis: ......................................................................................... 3

Design Requirements .................................................................................................. 3

Known Conditions and Assumptions .......................................................................... 3

Design Background ..................................................................................................... 3

Change in Altitude ...................................................................................................... 4

Nozzle Design ............................................................................................................ 7

Stress analysis .............................................................................................................. 8

Overall Thickness ....................................................................................................... 9

Thrust Chamber Design .............................................................................................. 9

Overall Thruster Design and Dimensions ................................................................. 11

Test Results: .................................................................................................................. 13

Test Analysis: ................................................................................................................ 13

Theoretical Test Analysis: .............................................................................................. 15

Conclusions and Considerations: ................................................................................... 16

Conclusions about Design from Testing ................................................................... 16

Comparison of Cold Gas Thruster and Hydrogen Peroxide Thruster ........................ 16

Consider injecting reacting gas into the thrust chamber ............................................. 17

Aluminum Thruster Heat Transfer Analysis .............................................................. 21

Steel Thruster Heat Transfer Analysis ....................................................................... 22

Heat Transfer Analysis Discussion ............................................................................ 23

Use as a main propulsion system ............................................................................... 23

References: .................................................................................................................... 24

Appendix: ..................................................................................................................... 25

Value Table(s) ........................................................................................................... 25

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Design and Thruster Analysis:

Design Requirements

Design Requirements

Propellant Air

Operating Temperature TBD

Operating Altitude 86km – 97km

Thrust 10 lbf

Chamber Pressure 100 psia

Thruster Diameter Less than 3 in

Thruster Length Less than 3 in

Nozzle geometry 80% bell nozzle

Known Conditions and Assumptions

Working Propellant: Air (k = 1.4, R = 1716 ft/s)

P1 = Chamber Pressure = 100 psi

P2 = Exit Pressure = 0.4*Pambient = 5.88 psi

T1 = Chamber Temperature = 536.4 R

Max Dimensions: 3x3x3 inches cubed

Altitude (km) Pressure (Pa) Pressure (psia)

80 0.945 1.37x10-4 86 0.669 9.70x10-5

87 0.623 9.04x10-5

88 0.577 8.37x10-5

89 0.531 7.70x10-5

90 0.485 7.03x10-5

91 0.439 6.37x10-5

92 0.393 5.70x10-5

93 0.347 5.03x10-5

94 0.301 4.37x10-5

95 0.255 3.70x10-5

96 0.209 3.03x10-5

97 0.163 2.36x10-5

100 0.025 3.63x10-6

Table 1: Atmospheric pressure as a function of altitude (Source: NASA GSFC)1

Design Background

The design of the thruster is to perform optimally at an altitude range of 86 km to 97 km.

However since the thruster is also expected to undergo testing at sea level conditions, we must consider if a thrust capable of performing optimally at an altitude range of 86 km to 97 km is even possible.

In order for the thruster to operate optimally at those conditions, the exit pressure of the thrust must equal to that of the ambient conditions. In this case, the thruster would have to accelerate its propellant to near vacuum conditions.

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Based on what Professor Anderson said in class, a thruster can have backflow occur if the thruster is too over expanded. Therefore if the Pexit < 0.4*Pambient, our thruster flow would face separation. Thus any test of our optimally expanded thruster would not provide any useful data at sea level conditions because it would be too over expanded.

Our design will be 100% efficient and from the cold gas operation test, we can then determine what is the C* and Cf efficiencies.

Also looking at literature of previous cold gas thruster test, the temperature of the supply air was at room temperature (298 K = 536.4 R).

Change in Altitude

Assuming that flow from the chamber to exit is isentropic; we can assume that the chamber pressure is stagnating while the exit pressure is static.

Exit Mach = 2.4970

Using Mach number, we can use the isentropic area expansion equation to find expansion ratio:

Eqn. 1

Ae/At = 2.6294

However since we are ranging from an altitude of 86 km to 97 km, pressure will change and therefore affect our performance.

Knowing how pressure changes with altitude, we can calculate C*, Cf, Isp, and Mass Flow.

C* was calculated using the following equation:

Eqn. 2

The figure below details the change in C* with respect to altitude.

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Figure 1: C* vs Altitude

C* will be constant because it is only a function of Tchamber and Tchamber = 536.4 R.

Cf was calculated using the following equation:

Eqn 3.

The figure below details the change in Cf with respect to altitude.

Figure 2: Cf vs Altitude

Cf changes with altitude because P3 (ambient pressure) changes with altitude. The higher the altitude, the lower the pressure, thus the term P2-P3 becomes larger, making Cf larger.

Isp was calculated using the following equation:

86 88 90 92 94 96 981400

1400.5

1401

1401.5

1402

1402.5

Altitude [km]

C*

[ft/

s]

86 88 90 92 94 96 981.4084

1.4084

1.4084

1.4084

1.4084

1.4084

1.4084

1.4084

1.4084

1.4084

1.4084

Altitude [km]

Thru

st

Coeff

icie

nt

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Eqn. 4

The figure below details the change in Isp with respect to altitude.

Figure 3: Isp vs Altitude

Since Isp is directly related to Cf, as Cf increases Isp increases. Therefore as altitude increases, Isp also increases. From the figure above, the change in Isp isn’t much because the change in pressure of the atmosphere is negligible since its already so thin at those altitude.

Mass flow can be calculated using the following equation:

Eqn. 5

The figure below details the change in Thrust with respect to altitude.

Figure 4: Mass Flow vs Altitude

Since mass flow is directly related to Isp and since thrust is constant, as Isp goes up, mass flow goes down. Therefore you need less mass flow to achieve the same desired

86 88 90 92 94 96 9861.2844

61.2844

61.2844

61.2844

61.2844

61.2844

61.2844

61.2844

Altitude [km]

Isp [

secs]

86 88 90 92 94 96 980.1632

0.1632

0.1632

0.1632

0.1632

0.1632

0.1632

Altitude [km]

Mass F

low

[lb

m/s

]

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thrust. That means if someone wants to use a thruster to maintain a spacecraft’s altitude, they better operate the station keeping maneuvers higher rather than lower in the atmosphere, otherwise they will be wasting more propellant.

Nozzle Design

Since the thruster is designed to be operating at a range of 86 to 97, the design point of the thruster will be at 92 km, the average of the two boundary conditions.

With a design point of 92 km, the performance of the thruster can be determined:

Performance at Operating Point

Cf 1.4084

C* 1401.14 ft/s

Isp 61.2844 secs

Mass Flow 0.1532 lbm/s

Table 2: Performance of Thruster at Operating Point

The geometry of the thruster can then be determined using the following equations:

Eqn. 6

Eqn .7

Thruster Geometry

At 0.071 in^2

Rt 0.15033 in

Ae 0.18669 in^2

Re 0.2437 in

Lcone 0.3485 in

Table 3: Thruster Geometry

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Stress analysis

Assumptions:

o Thin wall pressure vessel o Radius and thickness remains constant o Stress distributions throughout wall thickness is constant

Figure 5: Visual of stresses in a thin walled pressure vessel

Variables:

o σa = axial stress (longitudinal stress) o σh = hoop stress (latitudinal stress) o Pc = chamber pressure o Rc = chamber radius o t = thickness of vessel o CR = contraction ratio o Ac = Area of the chamber o At = Area of the throat o Θ = angle of pressure force

Use the maximum chamber pressure: Pc=100psia

The force of the chamber pressure acts on the wall at an angle perpendicular to the wall in the longitudinal direction. (see figure below)

Figure 6: visual of finding axial stress

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Find the Axial stress using Trigonometry:

𝜎𝑎 = 𝑃𝑐 tan(𝜃) Eqn. 8

𝜎𝑎 = 100𝑝𝑠𝑖𝑎 tan(45°)

𝝈𝒂 = 𝟏𝟎𝟎𝒑𝒔𝒊𝒂

Find hoop stress (equation from Textbook: Mechanics of Materials):

𝜎ℎ = 2𝜎𝑎 Eqn. 9

𝜎ℎ = 2 ∗ 100𝑝𝑠𝑖𝑎

𝝈𝒉 = 𝟐𝟎𝟎𝒑𝒔𝒊𝒂

Overall Thickness

Thickness determined based on the stress analysis.

(from Mechanics of Materials):

𝜎ℎ = 𝑃𝑐∗𝑟𝑐

𝑡, 𝜎𝑎 =

𝑃𝑐∗𝑟𝑐

2∗𝑡 Eqn. 10 &11

Find CR (contraction ratio) for thickness calculation:

𝐶𝑅 = 𝐴𝑐

𝐴𝑡 Eqn. 12

We decided to go with a Contraction Ratio of 10. A larger Combustion Ratio is necessary based on the small overall size of the thruster and the fact that the fuel that will be used is cold gas.

𝑪𝑹 = 10

Use design parameters to find dimensions of the chamber:

𝐴𝑐 = 𝐴𝑡 ∗ 𝐶𝑅 Eqn. 13

𝑨𝒄 = 𝟎. 𝟕𝟏 𝒊𝒏𝟐

𝑟𝑐 = √𝐴𝑐

𝜋 Eqn. 14

𝒓𝒄 = 𝟎. 𝟒𝟕 𝒊𝒏

𝑫𝒄 = 𝟎. 𝟗𝟓 𝒊𝒏

Re-arrange stress formulas to find thickness:

𝑡 = 𝑃𝑐∗ 𝑟𝑐

𝜎ℎ Eqn. 15

𝑡 = 100𝑝𝑠𝑖𝑎 ∗ 0.47𝑖𝑛

200𝑝𝑠𝑖𝑎

𝒕 = 𝟎. 𝟐𝟑𝟓 𝒊𝒏

Thrust Chamber Design

Stated in the project parameters, we must design the thruster to be no more than 3 inches in length total. Therefore, with a nozzle length of 0.3485 inches, the chamber

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length can only be as much as 2.6515 inches. We will round this number down and have a chamber length of 2.5 in.

𝑳𝒄 = 𝟐. 𝟓 𝒊𝒏

We can now find the combustor aspect ratio:

𝐶𝐴𝑅 = 𝐿𝑐

𝐷𝑡 Eqn. 17

When Dt is the diameter of the throat.

𝐷𝑡 = 2 ∗ 𝑟𝑡 Eqn. 18

𝐷𝑡 = 0.15 𝑖𝑛 ∗ 2 = 0.3 𝑖𝑛

Therefore:

𝐶𝐴𝑅 = 2.5 𝑖𝑛

0.3 𝑖𝑛= 𝟖. 𝟑𝟑

Injector Design

The design of the Injector was based on the parameters given for the gas supply hose as well as the dimensions for the chamber. The inside of the injector chamber will be threaded using a ½’’ NPT threading. The gas supply hose will have a diameter of 0.5 inches. Therefore, the injector chamber diameter is as follows:

𝑫𝒊𝒄 = 𝟎. 𝟓 𝒊𝒏

The injector will then expand to the diameter of the chamber and fasten around the outside of the chamber. The two will be fastened using a NPT threading. Therefore, the overall injector diameter is as follows:

𝐷𝑖𝑜 = 2 ∗ 𝑟𝑐 + 2 ∗ 𝑡 Eqn. 19

𝑫𝒊𝒐 = 𝟏. 𝟒𝟏 𝒊𝒏

An illustration is provided below (injector = red, chamber = blue):

Figure 7: Injector Diagram

The horizontal length of the injector parts will not be counted towards the overall 3 inch parameter because the injector will mostly be surrounding the camber and the supply hose. Therefore, the length of the injector parts is determined based on need rather than calculations.

𝑳𝟏 = 𝟎. 𝟓 𝒊𝒏

𝑳𝟐 = 𝟏 𝒊𝒏

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Overall Thruster Design and Dimensions

Figure 6: Side View

Figure 7: Isometric View

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Nozzle:

θi – inflection angle

θe – exit angle N/A

Lbell 0.3485 in

Approach radius 1.5rt 0.225 in

Expansion radius 0.4rt 0.06 in

At 0.071 in2

rt 0.15 in

Ae 0.186 in2

re 0.244 in

Lnozzle 0.3485 in

t 0.235 in

Table 6: Nozzle Dimensions

Chamber:

Ac 0.5 in2

rc 0.43 in

Lc 2.5 in

t 0.239 in

Table 7: Chamber Dimensions

Injector:

L1 0.5 in

L2 1 in

Dio 1.41 in

Dic 0.5 in

t 0.235 in

Table 8: Injector Dimensions

Page 13: Team Dorado Cold Gas Thruster Report Final

Test Results:

The following is the test procedure for the cold gas thruster.

Figure 8: Testing Setup

i. Align and mount the cold gas thruster on the thrust stand as shown in the picture above. It should be checked carefully if the thruster is secured properly or not.

ii. Clear any tool and instruments on the thrust stand before running the test. iii. Check the pressure valve of the pressure supply. If the valve is open then close it. iv. Attach the pressure supply to the cold gas thruster via NPT fittings. v. Turn on the pressure supply valve and maintain a steady pressure of 100 psi. vi. Record the exact pressure value supplied. vii. Record the mass flow from the supply tank viii. Record the thrust data recorded by the thrust stand. ix. Turn off the pressure supply. x. Repeat the procedure 5 times.

If the experiment was performed, the experimental data would have consisted of thrust values for corresponding chamber pressure and mass flow

Test Analysis:

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The chamber will attain the same pressure as the pressure supply. It was assumed that there will be no pressure drop between the camber and the supply.

The experimental thrust values will be compared to the design thrust value of 10 lbf. The difference between experimental thrust and theoretical thrust will determine the accuracy of our design. Furthermore, performance parameters will be calculated at test conditions and will be scaled to the flight conditions.

By using the chamber pressure (p1), throat area (At) and the total mass flow (�̇�) the value of effective exhaust velocity (c*) will be found.

Eqn. 20

Then, the value of thrust coefficient (CF) will be calculated by using the experimental thrust value (F), chamber pressure (p1) and throat area (At).

Eqn. 21

The Isp of thruster is then calculated by using effective exhaust velocity (c*), thrust coefficient (CF) and standard gravity (g0).

Eqn. 22

The performance of the thruster at different altitude (design altitude) where the flow is perfectly expanded (p2 = p3) is calculated by scaling the thrust coefficient value.

Eqn. 23 Where, p2 = p3 = pambient

Effective exhaust velocity (c*) is a function of geometry and will not change with altitude. Then, the scaled Isp is calculated by using new value of thrust coefficient.

There will be inefficiencies in the thruster due to pressure loss, friction loss, etc. The effective exhaust velocity efficiency, thrust coefficient efficiency and overall efficiency will be calculated by using experimental values and theoretical values. The theoretical values are calculated by using theoretical design parameters.

𝜂𝑐∗ =𝑐∗,𝑎𝑐𝑡𝑢𝑎𝑙

𝑐∗,𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 Eqn. 24

𝜂𝐶𝐹 =𝐶𝐹,𝑎𝑐𝑡𝑢𝑎𝑙

𝐶𝐹,𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 Eqn. 25

𝜂𝑜𝑣𝑒𝑟𝑎𝑙𝑙 =𝐼𝑠𝑝,𝑎𝑐𝑡𝑢𝑎𝑙

𝐼𝑠𝑝,𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 Eqn. 26

We were not able to test our design due to the semester time constraint as well as some difficulty we ran into when trying to get our design manufactured. Therefore, we do not have any data or conclusions about the design to discuss.

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Theoretical Test Analysis:

Due to time constraints, the thruster was not able to undergo experimental testing. Therefore a theoretical test was conducted to examine what kind of performance could have been expected if the experimental test was conducted.

Therefore using the same geometry and mass flow rate but setting the ambient pressure to 14.7 psi, the following values are the expected outcome of the experimental test:

Theoretical Performance at SL

C* 1400.00 ft/s

Cf 1.12

Isp 48.70 secs

Thrust 7.46 lbf

Table 9: Theoretical SL Performance

However the theoretical performance values obtained would be the ideal performance at sea level. So the theoretical and experimental values will be off by a factor of .90~.98% due to efficiency factors.

Comparing the performance at the operating point and at sea level, it can be seen that performance across every spectrum is worse at sea level than at the operating point. The main contributor to this decline in performance is the ambient pressure. As ambient pressure decreases, the thruster performance will increase. Therefore is it unsurprising that at the operating point, the thruster will perform better.

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Conclusions and Considerations:

Conclusions about Design from Testing

We were not able to test our design due to the semester time constraint as well as some difficulty we ran into when trying to get our design manufactured. Therefore, we do not have any data or conclusions about the design to discuss.

Comparison of Cold Gas Thruster and Hydrogen Peroxide Thruster (HPT) The cold gas thruster and the HPT were compared to examine which system is better. In order to begin the comparison, the cold gas thruster must be resized to match that of the HPT. The HPT case study thruster has a different operating temperature, pressure, and exit pressure, all of which will contribute to a different performance. Therefore the new cold gas thruster must match the case study. The new cold gas thruster will have the flowing performance parameters:

Resized Cold Gas Thruster Parameters

F 10 lbm

P1 150 psi

P2 5 psi

T1 850 K

Table 10: Resized Cold Gas Thruster Parameters

The new operating temperature was centered on the HPT’s max operating temperature. Based on the information above, the cold gas thruster new performance was examined below:

Resized Cold Gas Thruster Performance

C* 2370 ft/s

Cf 1.54 psi

Isp 113 psi

mair 0.0884 lbm/s

Table 11: Resized Cold Gas Thruster Performance

Comparing the new cold gas thruster’s performance with the HPT case study’s performance, it is clear that the HPT case study outperforms the cold gas thruster based on Isp. Given the same operating conditions, the cold gas thruster’s Isp was still ~75% of the HPT’s Isp. The reason why the HPT outperformances the cold gas thruster is due its specific heat ratio. Examining the equations for Cf and C*, there are two main gas properties that affect Isp, the molecular weight of a gas and its specific heat constant. A greater molecular weight will cause a decrease in Isp performance while a decrease in specific heat constant will cause an increase in Is Since H2O2 has a greater molecular weight than air, the specific heat ratio of H2O2 must be lower

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than that of air to a significant point that H2O2 is a better option than air. However no data can be found on the specific ratio of H2O2, so this theory cannot be verified.

Methods to Increase Performance

The performance of the cold gas thruster can be easily increased by pre-heating the chamber gas by transferring electrothermal heat from electrical resistors. By adding electrical resistors, the cold gas thruster transforms into an electric propulsion device known as a resistojet.

The increase in performance can be modeled increasing the chamber temperature and then calculating the new Isp of the thruster.

The increase in chamber temperature will lead to an increase in 𝐶∗. Since Isp is directly related

with 𝐶∗, an increase in chamber temperature will lead to an increase in Isp.

Assuming constant geometry, an operating altitude of 92 km where ambient pressure = 5.7e-05 psi, and an original chamber temperature of 536.4 Rankine, the following graph was created:

Figure 9: Isp vs Chamber Temperature

Analysis was done if the thruster was created out of either steel or aluminum. The design temperature of each material was taken into consideration in order to evaluate the max possible gain in Isp due to electrothermal heating.

From a starting Isp of 61.28 seconds, the aluminum thruster can have its Isp increase upwards to an Isp of 72.3 seconds, while the steel thruster can have its Isp increase upwards to an Isp of 93.09 seconds.

That means the aluminum thruster can experience a 17.98% boost in performance while the steel thruster can experience a 51.91% boost in performance!

Taking the overall systems into account, discuss whether the designed thruster would be a good alternative to HTP for the Nanolaunch 1200. Take into consideration the performance vs weight and performance vs cost for the required duty cycle.

Consider injecting reacting gas into the thrust chamber

Setup Conditions

500 600 700 800 900 1000 1100 1200 130060

65

70

75

80

85

90

95

Chamber Temperature [Rankine]

Isp [

secs]

Chamber Temp

Al Design Temp

Steel Design Temp

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Running CEA with LH2/LOX with the following conditions:

Chamber Pressure: 100 psi

Propellant: H2(L)

Oxidizer: O2(L)

O/F: Variable

Supersonic Area Ratio: 2.6294

O/F of the LH2/LOX system was based on the design temperature of the construction material.

Viewing the graph below:

Figure 10: Material Tensile Yield Stress vs Temperature

It is observed that Aluminum has a design temperature of ~ 350 F (809.7 R) and that Stainless Steel has a design temperature of ~800 F (1260 R).

Calculations for thruster performances were made at the design point altitude of 92 km where ambient pressure = 5.7e-05 psi and a desired thrust of 10 lbf is still maintained.

Running CEA

Since the thruster could be made of either aluminum or stainless steel, it is necessary to obtain the O/F ratios that correspond to these two design temperatures to see the maximum performance of each design. Therefore using CEA, a series of chamber temperature was created based on a variety of O/F ratios.

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Figure 11: Chamber Temperature vs OF Ratios

Examining the graph above, the aluminum thruster would have a max O/F = 0.454 and the steel thruster would have a max O/F = 0.7075

Performance Analysis

Aluminum Thruster Analysis

Examining the Aluminum Thruster first, with an O/F = 0.454, the thruster would have an:

Aluminum Thruster Properties

P2 7.836 psi

C* 5421 ft/s

Cf optimal 1.3301

Table 12: Aluminum Thruster Properties

Due to the exit pressure, the Cf correction term:

𝐶𝑓𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛=

𝑃2 − 𝑃3

𝑃1= 0.07836

Therefore,

𝐶𝑓𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙= 𝐶𝑓𝑜𝑝𝑡𝑖𝑚𝑎𝑙

+ 𝐶𝑓𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛= 1.4085

𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = (𝐶𝑓𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙

) (𝐶∗)

𝑔0= 237.12 𝑠𝑒𝑐𝑠

With our desired thrust = 10 lbf,

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4600

800

1000

1200

1400

1600

1800

2000

2200

2400

O/F Ratios

Cham

ber

Tem

pera

ture

[R

]

LH2/LOX System

Chamber Temp

Al Design Temp

Steel Design Temp

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𝐹 = (𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙)(�̇�)(𝑔0)

�̇� = 0.04217 𝑙𝑏𝑚

𝑠𝑒𝑐𝑠

Steel Thruster Analysis

Examining the Aluminum Thruster first, with an O/F = 0.7075, the thruster would have an

Steel Thruster Properties

P2 5.969 psi

C* 6257 ft/s

Cf optimal 1.348

Table 13: Steel Thruster Properties

Due to the exit pressure, the Cf correction term:

𝐶𝑓𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛=

𝑃2 − 𝑃3

𝑃1= 0.05969

Therefore,

𝐶𝑓𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙= 𝐶𝑓𝑜𝑝𝑡𝑖𝑚𝑎𝑙

+ 𝐶𝑓𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛= 1.4075

𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = (𝐶𝑓𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙

) (𝐶∗)

𝑔0= 273.5 𝑠𝑒𝑐𝑠

With our desired thrust = 10 lbf,

𝐹 = (𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙)(�̇�)(𝑔0)

�̇� = 0.03656 𝑙𝑏𝑚

𝑠𝑒𝑐𝑠

Geometry Discussion

Assuming the LH2/LOX system should also be able to undergo sea level testing where

P2 = 0.4*Pambient, the geometry can be the LH2/LOX system can be improved.

Since P2,Al and P2,Steel are both larger than the desired P2 = 5.88 psi, the expansion ratio would need to be increased to allow the exit pressure of both the Aluminum and Steel desgin equal to the desired exit pressure.

Other than the adjustment in expansion ratio, which would increase Isp performance, the reacting gas seems like a viable option for the current cold gas geometry. Isp increased significantly and the O/F were designed to allow the thrusters to thermally manage the heat loads.

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Aluminum Thruster Heat Transfer Analysis

Since the design temperature of the aluminum thruster was 800 Rankine, an O/F = 0.454 was used to satisfy the constraint. The transport properties were determined by the CEA output and assumed to be constant throughout the combustor length. In addition, the designed adiabatic wall temperature is assumed to be 80% of chamber temperature.

With those conditions, the following values were determined:

Figure 12: Diagram of Temperature & Velocity Profiles Across Chamber Length

ℎ𝑔𝐴(𝑇𝑎𝑤 − 𝑇𝑤,ℎ) =𝑘𝐴

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐) Eqn. 20

𝑇𝑔 = 809.2 𝑅

𝑇𝑎𝑤 = 𝑇𝑔𝑅𝐹; 𝑅𝐹 = 0.92 Eqn. 21

𝑇𝑎𝑤 = 744.46 𝑅

Using Bartz Equation to evaluate the heat transfer across the chamber wall,

ℎ𝑔 = [0.026

𝐷𝑡0.2 (

𝜇0.2𝐶𝑝

𝑃𝑟0.6 )𝑛𝑠

((𝑝𝑐)𝑛𝑠𝑔

𝑐∗ )0.8

(𝐷𝑡

𝑅)

0.1

] (𝐴𝑡

𝐴)

0.9

𝜎 Eqn. 22

𝜎 = 1

[1

2

𝑇𝑤𝑔

(𝑇𝑐)𝑛𝑠(1+

𝛾−1

2𝑀2)+

1

2]0.68

[1+ 𝛾−1

2𝑀2]

0.12 Eqn. 23

Aluminum Chamber Thruster Properties

Dt 0.3007 in

μ

1.44E-05

lbm/(in-s)

Cp 0.5973 Btu/(lbm-F)

(Pc)ns 100 psi

C* 5716 ft/s

R 0.2255 in

At 0.071 in^2

A 0.1867 in^2

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Twg 647.4 R

(Tc)ns 809.2 R

M 0

γ 1.392

Pr 0.6613

Table 14: Aluminum Chamber Properties

Calculations of the Bartz Equation can be examined in the Appendix section.

𝜎 = 1.004

ℎ𝑔 = 0.0007448𝐵𝑡𝑢

(𝑖𝑛2)(𝑠𝑒𝑐𝑠)(𝐹)

Examining aluminum, the thermal conductivity of Aluminum is 0.00317 Btu/(in-s-°F).

Since L=t=0.235 inches and Tw,h = Taw, Tw,c can be calculated:

This gives:

Tw, c= 642 °R

Steel Thruster Heat Transfer Analysis The same process repeated in order to perform a heat transfer analysis in the case of a thruster made of stainless steel. Since the design temperature of the aluminum thruster was 1260 Rankine, an O/F = 0.7075 was used to satisfy the constraint. The transport properties were determined by the CEA output and assumed to be constant throughout the combustor length. In addition, the designed adiabatic wall temperature is assumed to be 80% of chamber temperature.

Using Bartz Equation to evaluate the heat transfer across the chamber wall,

Steel Chamber Thruster Properties

Dt 0.3007 in

μ

1.68E-05

lbm/(in-s)

Cp 2.093 Btu/(lbm-F)

(Pc)ns 100 psi

C* 6257 ft/s

R 0.2255 in

At 0.071 in^2

A 0.1867 in^2

Twg 1008 R

(Tc)ns 809.2 R

M 0

γ 1.38

Pr 0.6062

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Table 15: Steel Chamber Thruster Properties

𝜎 = 1.074

ℎ𝑔 = 0.002822 𝐵𝑡𝑢

(𝑖𝑛2)(𝑠𝑒𝑐𝑠)(𝐹)

The thermal conductivity of Stainless Steel 309 is 0.000187 Btu/(in-s-°F)

Since L=t=0.235 inches and Tw,h = Taw, Tw,c can be calculated:

This gives:

Tw,c= 469 °R

Heat Transfer Analysis Discussion Comparing the aluminum and steel heat transfer analysis, it would be optimal to choose the steel

thruster design over the aluminum thruster design.

Even though the steel thruster is running at a chamber temperature higher than the aluminum

thruster, the outer wall temperature of the steel thruster was still lower than the outer wall

temperature of the aluminum thruster. This meant that the steel thruster could operate at a higher

chamber temperature and require less cooling than the aluminum thruster. Higher chamber

temperature would mean higher thruster performance while less cooling would mean a less

complex thruster design. Therefore the steel thruster design is in superior to the aluminum

thruster design.

As a Main Propulsion System

We considered whether or not the design that we came up with could be used as a main propulsion system of a craft rather than just an attitude adjusting thruster.

Assuming the cold gas thruster has the same performance parameters previously calculated:

𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = 61.284 𝑠𝑒𝑐𝑠

Assuming the test system has a mass ratio, 𝑀𝑅 = 0.05

The ideal rocket equation can be applied to calculate the amount of ∆𝑉can be expected from a 10,000 lbm cold gas thruster.

∆𝑉𝑐𝑜𝑙𝑑 𝑔𝑎𝑠 = −(𝐼𝑠𝑝𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙)(𝑔0) ln(𝑀𝑅) = 5912 𝑓𝑡/𝑠

Assuming this launch system wants to go to LEO, which requires a ∆𝑉 = 23000 𝑓𝑡/𝑠, this cold gas launcher concept is not viable. The Isp of the cold gas launcher is simply too low.

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References: 1Hedin, A. E., “MSISE Model 1990”, NASA Goddard Space Flight Center, 1990

2Minzner, R. A., “The 1976 Standard Atmosphere Above 86km”, NASA SP-398, 1976

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Appendix:

Value Table(s)

Table 1: Atmospheric pressure as a function of altitude (Source: NASA GSFC)1

Altitude (km) Pressure (Pa) Pressure (psia)

80 0.945 1.37x10-4 86 0.669 9.70x10-5

87 0.623 9.04x10-5

88 0.577 8.37x10-5

89 0.531 7.70x10-5

90 0.485 7.03x10-5

91 0.439 6.37x10-5

92 0.393 5.70x10-5

93 0.347 5.03x10-5

94 0.301 4.37x10-5

95 0.255 3.70x10-5

96 0.209 3.03x10-5

97 0.163 2.36x10-5

100 0.025 3.63x10-6

Aluminum Heat Transfer Calculations

ℎ𝑔𝐴(𝑇𝑎𝑤 − 𝑇𝑤,ℎ) =𝑘𝐴

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐)

𝑇𝑔 = 809.2 𝑅

𝑇𝑎𝑤 = 𝑇𝑔𝑅𝐹; 𝑅𝐹 = 0.92

𝑇𝑎𝑤 = 744.46 𝑅

Using Bartz Equation to evaluate the heat transfer across the chamber wall,

ℎ𝑔 = [0.026

𝐷𝑡0.2 (

𝜇0.2𝐶𝑝

𝑃𝑟0.6)

𝑛𝑠

((𝑝𝑐)𝑛𝑠𝑔

𝑐∗)

0.8

(𝐷𝑡

𝑅)

0.1

] (𝐴𝑡

𝐴)

0.9

𝜎

𝜎 = 1

[12

𝑇𝑤𝑔

(𝑇𝑐)𝑛𝑠(1 +

𝛾 − 12 𝑀2) +

12]

0.68

[1 + 𝛾 − 1

2 𝑀2]0.12

𝐷𝑡 = 0.3007 𝑖𝑛, 𝜇 = 1.442𝐸 − 5 𝑙𝑏𝑚/(𝑖𝑛 − 𝑠), 𝐶𝑝 = 0.5973 Btu/(lbm − F)

(𝑝𝑐)𝑛𝑠 = 100 𝑝𝑠𝑖, 𝑐∗ = 5716 𝑓𝑡/𝑠, 𝑅 = 0.2255 𝑖𝑛, 𝐴𝑡 = 0.071 in2 𝐴 = 0.1867 in2

𝑇𝑤𝑔 = 800 𝑅, (𝑇𝑐)𝑛𝑠 = 809.2 𝑅, 𝑀 = 0, 𝛾 = 1.392, Pr = 0.6613

𝜎 =1

[12

640809.2 (1 +

1.392 − 12 02) +

12]

0.68

[1 + 1.392 − 1

2 02]0.12 = 1.004

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ℎ𝑔 = [0.026

0.3007 0.2(

(1.442𝐸 − 5)0.2(0.5973)

0.66130.6)

𝑛𝑠

((100)(32.2)

5716)

0.8

(0.3007

0.2255)

0.1

] (0.071

0.1867)

0.9

1.004

ℎ𝑔 = 0.0007448𝐵𝑡𝑢

(𝑖𝑛2)(𝑠𝑒𝑐𝑠)(𝐹)

ℎ𝑔(𝑇𝑎𝑤 − 𝑇𝑤,ℎ) =𝑘

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐)

Examining aluminum, the thermal conductivity of Aluminum is 0.00317 Btu/(in-s-°F).

(0.0007448𝐵𝑡𝑢

𝑖𝑛2 − 𝑠𝑒𝑐𝑠 − 𝐹) (744.46 − 647.4) =

𝑘

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐)

Since L=t=0.235 inches and Tw,h = Taw, Tw,c can be calculated:

(0.0007448𝐵𝑡𝑢

𝑖𝑛2 − 𝑠𝑒𝑐𝑠 − 𝐹) (744.46 − 640) =

0.00317

0.235(647.4 − 𝑇𝑤,𝑐)

This gives:

Tw,c= 642 °R

Steel Heat Transfer Analysis Using Bartz Equation to evaluate the heat transfer across the chamber wall,

ℎ𝑔 = [0.026

𝐷𝑡0.2 (

𝜇0.2𝐶𝑝

𝑃𝑟0.6)

𝑛𝑠

((𝑝𝑐)𝑛𝑠𝑔

𝑐∗)

0.8

(𝐷𝑡

𝑅)

0.1

] (𝐴𝑡

𝐴)

0.9

𝜎

𝜎 = 1

[12

𝑇𝑤𝑔

(𝑇𝑐)𝑛𝑠(1 +

𝛾 − 12 𝑀2) +

12]

0.68

[1 + 𝛾 − 1

2 𝑀2]0.12

𝐷𝑡 = 0.3007 𝑖𝑛, 𝜇 = 1.682𝐸 − 5 𝑙𝑏𝑚/(𝑖𝑛 − 𝑠), 𝐶𝑝 = 2.093 Btu/(lbm − F)

(𝑝𝑐)𝑛𝑠 = 100 𝑝𝑠𝑖, 𝑐∗ = 6257 𝑓𝑡/𝑠, 𝑅 = 0.2255 𝑖𝑛, 𝐴𝑡 = 0.071 in2 𝐴 = 0.1867 in2

𝑇𝑤𝑔 = 1260 𝑅, (𝑇𝑐)𝑛𝑠 = 1008 𝑅, 𝑀 = 0, 𝛾 = 1.38, Pr = 0.6062

𝜎 =1

[12

10081260 (1 +

1.38 − 12 02) +

12]

0.68

[1 + 1.38 − 1

2 02]0.12 = 1.074

ℎ𝑔 = [0.026

0.3007 0.2(

(1.682𝐸 − 5)0.2(2.093)

0.60620.6)

𝑛𝑠

((100)(32.2)

6257)

0.8

(0.3007

0.2255)

0.1

] (0.071

0.1867)

0.9

1.074

ℎ𝑔 = 0.002822 𝐵𝑡𝑢

(𝑖𝑛2)(𝑠𝑒𝑐𝑠)(𝐹)

The thermal conductivity of Stainless Steel 309 is 0.000187 Btu/(in-s-°F)

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ℎ𝑔(𝑇𝑎𝑤 − 𝑇𝑤,ℎ) =𝑘

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐)

(0.002822𝐵𝑡𝑢

𝑖𝑛2 − 𝑠𝑒𝑐𝑠 − 𝐹) (1160 − 1008) =

𝑘

𝐿(𝑇𝑤,ℎ − 𝑇𝑤,𝑐)

Since L=t=0.235 inches and Tw,h = Taw, Tw,c can be calculated:

(0.002822𝐵𝑡𝑢

𝑖𝑛2 − 𝑠𝑒𝑐𝑠 − 𝐹) (1160 − 1008) =

0.000187

0.235(1008 − 𝑇𝑤,𝑐)

This gives:

Tw,c= 469 °R