homework 2 - webmae-nas.eng.usu.edu/.../assignments/assignment2_solution.pdf · 2019. 1. 25. ·...
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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