Homework 2 - Webmae-nas.eng.usu.edu/.../Assignments/Assignment2_solution.pdf · 2019. 1. 25. · Homework 2 2 • Initial Launch Angle 60 degrees (consider constant while rocket is

MAE 5540 - Propulsion Systems

Homework 2

1

• Consider Sub-Orbital Rocket Launch on Moon’s Surface

MAE 5540 - Propulsion Systems

Homework 2

2

• Initial Launch Angle 60 degrees (consider constant while rocket is burning)

• Total Launch Mass, 20 kg

• Initial Propellant Mass, 5 kg

• Thrust 1000 N

• Isp =250 sec

• Acceleration of Lunar Gravity (assume Constant)

• Standard Earth GravityAcceleration

gmoon =1.622 msec2

g0⊕ = 9.8067 msec2

• Calculate:1. Burnout Altitude and Velocity2. Apogee Altitude (Note Vapogee != 0)3. Impact Downrange (ignore surface curvature)4. Time to Impact 5. Plot Flight Path Altitude vs Time & Downrange6. Velocity vs Time & Downrange

Assume Point Mass Calculations

MAE 5540 - Propulsion Systems

Applicable Equations

3

tburn =Mprop ⋅ g0⊕ ⋅ Isp

F

Assume θ launch = constant, V0 = 0→ at time t :

V t( ) = g0⊕⋅ Isp ln

Minitial

Minitial − !m ⋅ t⎛⎝⎜

⎞⎠⎟− g0moon

⋅sinθ launch ⋅ t⎛

⎝⎜⎞

⎠⎟

In the absence of dissipative (drag, etc) forces … total mechanical energy of rocket remains constant following motor burnout

9

“MotorBurnout”

“ApogeePoint”

AtMotorBurnout…

AtApogeePoint…

• After Burnout EmechM final ⋅ g0

=Vburnout( )22 ⋅ g0

+ hburnout = Const

• During Burn

∂!V∂t=

!F∑

Mburnout

∂h∂t=V t( )⋅sinθ ∂X

∂t=V t( )⋅cosθ

∂X∂t=Const

Vhoriz ≠ 0 ... @ apogee!

MAE 5540 - Propulsion Systems 4

Questions??

MAE 5540 - Propulsion Systems

Solution

5

• First Calculate Burnout Parameters of Rocket Motor

sec

kg/sec

kg

MAE 5540 - Propulsion Systems

Solution (2)

6

• … Calculate Burnout Parameters of Rocket Motor

burn m/sec

m

m

MAE 5540 - Propulsion Systems

Solution (3)

7

• … Calculate Burnout Parameters of Rocket Motor

J/kg

J/kg

J/kg

MAE 5540 - Propulsion Systems

Solution (4)

8

• … Burnout Parameters Summary

MAE 5540 - Propulsion Systems

Solution (5)

9

• … Next Calculate Apogee Conditions

m

sec

…. Uphill against gravitySee following slide for derivation

MAE 5540 - Propulsion Systems

Solution (6)

10

• … from previous slide After burnout

dVv (t)dt=−gmoon

→dVv (t)dtburnout

apogee

∫ ⋅dt=− gmoon ⋅dtburnout

apogee

∫→Vv (apogee) =Vv (burnout ) −gmoon ⋅ tapogee− tburn( )

10

sec

@apogee→Vv (apogee) = 0=Vv (burnout ) −gmoon ⋅ tapogee− tburnout( )

→Vv (burnout )gmoon

= tapogee− tburnout( )→ tapogee =Vv (burnout )gmoon

+ tburnout

Vv (burnout ) = 595.898m/sec

gmoon =1.622m/sec2

tburnout =12.2584sec

→ tapogee =

MAE 5540 - Propulsion Systems

Solution (7)

11

• … Apogee Conditions

m

J/kg

J/kg

MAE 5540 - Propulsion Systems

Solution (8)

12

• … Impact time … After apogee at time t ...

0Vv ( t ) =Vv (apogee) −gmoon ⋅ t− tapogee( )

h(impact )

= hapogee+ −gmoon ⋅ t− tapogee( )⋅dttapogee

timpact

∫ = hapogee−gmoon12⋅ timpact− tapogee( )2

from previous ... tapogee =Vv (burnout )

gmoon+ tburnout

→ h(impact )

= hapogee−gmoon12⋅ timpact−

Vv (burnout )

gmoon− tburnout

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

2

MAE 5540 - Propulsion Systems

Solution (9)

13

h(impact )

= 0

→ hapogee = gmoon12⋅ timpact−

Vv (burnout )gmoon

− tburnout⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

2

Solve for timpact → timpact = 2 ⋅hapogeegmoon

+Vv (burnout )gmoon

+ tburnout

2 112935·1.622⎝ ⎠

⎛ ⎞0.5 595.898

1.62212.2584+ +

= 752.811 sec

Vv (burnout ) = 595.898m/sec

gmoon =1.622m/sec2

tburnout =12.2584sec

→ timpact =

MAE 5540 - Propulsion Systems

Solution (10)

14

… simplify

m/sec

344.042 m/sec

m/sec

m/sec

→ timpact = 2 ⋅hapogeegmoon

+Vv (burnout )gmoon

+ tburnout

→ timpact =2 112935·1.622⎝ ⎠

⎛ ⎞0.5 595.898

1.62212.2584+ +

= 752.811 sec

MAE 5540 - Propulsion Systems

Solution (11)

15

… Apogee and Impact Condition Summary

MAE 5540 - Propulsion Systems

Solution (12)

16

… Trajectory Plots

MAE 5540 - Propulsion Systems

What About Acceleration?

17

• Must Consider Whole Moon

MAE 5540 - Propulsion Systems

What About Acceleration (2)

18