homework 1 - webmae-nas.eng.usu.edu/.../homework1_solution_rev2011.pdf · mae 5540 - propulsion...

MAE 5540 - Propulsion Systems!

Homework 1 • Space Shuttle has the following mass fraction characteristics

• 1) Calculate the actual propellant mass fraction as the shuttle sits on the pad

1

Pmf =Mpropellant

M"dry" + Mpayload


Homework 1 (cont’d) • Assume that Shuttle is being launched on a Mission to the International Space Station (ISS)

• ISS orbit altitude is approximately 375 km above Mean sea level (MSL), assume that Shuttle Pad 41A altitude is exactly Sea level, Latitude is 28.5 deg. , ISS Orbit Inclination is 51.6 deg.

• Assume that the Earth is a perfect sphere with a radius of 6371 km

• Calculate 2) The required Orbital Velocity 3) The “Boost Velocity” of the Earth

at the Pad 41A launch site 4) Equivalent “Delta V” required to lift

the shuttle to altitude 5) Total “Delta V” required to reach the ISS orbit 2

µ = 3.986044 !105km3

sec2


Homework 1 (cont’d) • The 2 SRB’s each burn for approximately 123 seconds and produce 2,650,000 lbf thrust at sea level

• The 3 SSME engines each burn for ~509.5 seconds and each produces 454,000 lbf thrust at sea level

• Each of the SSME’s consume 1040 lbm/sec of propellant

6) Calculate the average specific impulses of the SRB’s, the SSME’s, and the Effective specific impulse of the Shuttle Launch System as a whole during the First 123 seconds of flight (ignore altitude effects)

3

Isp =Fthrust

0

Tburn

! "dt

!mpropellant0

Tburn

! "dt=Impulse( )totalM propellant

Hint:

g0

g0

g0

g0


Homework 1 (cont’d) 7) Based on the calculated “Delta V” requirements for the mission, what would be the required propellant mass fraction For the space shuttle to reach orbit in a single stage assuming the mean launch specific impulse?

-- base this calculation on the mean Isp for the system during the first 123 seconds after launch

8) How does the shuttle manage to reach orbit? …. ?

4


Homework 1 (cont’d) …. Next evaluate estimate launch conditions by breaking calculation into two “stages” .. That is

i) Stage 1 … first 123 seconds … SRB’s and SSME’s burning ii) Stage 2 … after SRB’s jettisoned .. Only SSME’s burning

5

“stage 1” “stage 2”:

!Vtotal = !Vstage1 + !Vstage2 + !Vstage3... = !Vstage _ ii=1

# ofstages

"


Homework 1 (cont’d) i) Stage 1 … first 123 seconds … SRB’s and SSME’s burning

6

“stage 1” Flight is vertical

-- Assume shuttle flys ~ “vertically” during Stage 1 flight. …

9) Calculate “Available Delta V” for “stage 1” Based On Mean Isp, and Pmf (ignore altitude effects)

-- Include “gravity losses” and assume an 8% drag loss in the available propulsive “Delta V” … assume g(t) ~ g0 = 9.8067m/sec2

!V( )available = g0 " Isp " lnMinitial

M final

#

$%&

'() !V( )gravity

loss) !V( )drag

!V( )drag * 0.10 + g0 " Isp " lnMinitial

M final

#

$%&

'(

Stephen Whitmore

0.08

0.08


Homework 1 (cont’d) … Break Calculation into two “stages” .. That is

ii) Stage 2 … flight time from 123 seconds to SSME burnout

7

“stage 2” flight is horizontal

-- Assume shuttle flys ~ “horizontally” during Stage 2 flight. …

-- 10) Calculate “Available Delta V” Based On SSME Isp, and remaining Pmf after the SRB’s Have been Jettisoned

-- Assume no drag losses for stage 2 burn

-- 11) Compute total available delta V .. Compare to mission requirements


How About the Shuttle?

Weight (lb) Gross lift-off . . . . . . . 4,500,000 External Tank (full) . . . . 1,655,600 External Tank (Inert) . . . . 66,000 SRBs (2) each at launch . . . 1,292,000 SRB inert weight, each . . . . 192,000

Pmf burn

= M fuel

+ oxidizerM

dry + M payload

M fuel+ oxidizer

= M fuel+ oxidizer

external tank

+ M fuel+ oxidizer

SRB

=

1,655,600 -66,000 + 2 1,292,000 - 192,000 ! 3,789,600 lbs

Pmf launch

= M fuel

+ oxidizerM

dry + M payload

= 3,789,6004,500,000 - 3,789,600 ! 5.33

Mass Fraction

8


Mass Fraction, Definition 2

Lf =Mp

Mgross

=Mp

M p + Mdry

=1

1+Mdry

M p

=1

1+ 1Pmf

=Pmf

Pmf +1=

5.331 5.33+

=0.842

9

“Load mass fraction”


Homework 1 (cont’d) • Orbital Velocity …

10

3.9860044 5!10

6371 375+" #$ %& ' 0.5

= 7.6868 km/sec Vorbit =µ

Rearth + horbit( ) =

=0.28857 km/sec

• “Boost” Velocity … along the direction of flight

6371 0+( ) 2!23 3600" 56 60" 4.1+ +

!180

51.6# $% &cos

V"boost " = R! + hlaunch( ) "#! " cos iorbit =


Homework 1 (cont’d)

11

• “Lift” Delta V… equivalent change in potential energy

!Vgravity = 2 µ "horbitRearth " Rearth + horbit( ) = 2

3.9860044 5!10( ) 375( )

6371 6371 375+( )" #$ %& ' 0.5

= 2.6374 km/sec

!Vtotal( )required = Vorbit "V"boost "( )2 + !Vgravity( )2 =

= 7.8542 km/sec

7.6868 0.28857!( ) 2 2.63722+( ) 0.5


SRB Mean Isp

Isp =Impulse

g0Mpropellant

=Fthrust

0

t

! dt

g0 m•propellant dt

0

t

!=

2,650,000 lbf "123sec

32.1742 ftsec2

"1,292,000 #192,000( )lbm

32.1742 lbm# ftlbf #sec2

$

%

&&&

'

(

)))

= 296.3sec

12


SSME Isp • Calculate Instantaneous of “operational” performance based on data given

Isp =1g0

Fthrust•

mpropellant

=

454,000 lbf

32.1742 ftsec2

! 1040 lbm/sec 32.1742 lbm" ft

lbf "sec2

#

$

%%%

&

'

(((

= 436.5sec

13


Effective Isp • Shuttle has multiple !engines at launch!

Fundamental definition!

14

g0

g0

g0

g0


Effective Isp

So … the solids cost you .. Why use them?

2 2.65 106123! ! 3 0.454 106123! !+2 1100000! 3 1040 123! !+

=317.14 sec

15

g0

g0

g0


Need more thrust than weight to get off of the ground

• Shuttle Gross Lift-off weight: 4,500,000 lbm

• SSME thrust: = 1,362,000 lbf

• SRB thrust: = 5,300,000 lbf

• Total thrust: = 6,662,000 lbf

3 454000!2 2650000!

• That’s why the shuttle “leaps” off of the pad when the SRB’s are fired

G’s = = 1.48 3 454000! 2 2650000!+

450000016


Compute Propellant Mass Fraction Needed for Shuttle to Reach Orbit

Pmf = e!Vg0 Isp "1 = e

7608m/sec9.806

m/sec2355.97sec

#

$

%%

&

'

(("1 = 7.8422

> 5.33 (actual mass fraction of propellant)

• How does the shuttle manage to reach orbit?

76089.806 317.14!" #

$ %exp 1&=10.548

17

Stephen Whitmore

Stephen Whitmore

7854.2

=11.496

7854.2

=11.496


Compute Propellant Mass Fraction Needed for Shuttle to Reach Orbit

> 0.842 (actual Load mass fraction of propellant)

• How does the shuttle manage to reach orbit?

= 0.913

Mass Fraction, Definition 2

Lf =Pmf

Pmf +1=

10.54810.548 1+

18

11.496

11.496+1

= 0.92

11.496

11.496+1

= 0.92


Rocket System Staging

• By losing the solids part way up !

!Vtotal = !Vstage1 + !Vstage2 + !Vstage3... = !Vstage _ ii=1

# ofstages

"

!Vtotal = g0Isp ln Minitial

M final

"

#$$

%

&''

(

)*

+

,-stage _ i

i=1

# ofstages

.

19


Rocket System Staging I (cont’d) • Let’s do this calculation for the shuttle

• “two-stage” system 1) SRB’s+SSME’s 2) SSME’s alone

• Stage 1: Minitial = 4,500,000lbm ! M final = Minitial " Mpropellant _ consumed !Mpropellant _ consumed = 2 # SRBpropellant + 3# SSMEpropellant =

2 # 1,292,000 "192,000( )lbm + 3#1040 lbm/sec #123sec =

2,583,760lbm ! M final( )SRB"cutoff = 4,500,000 " 2,583,760 = 1,916,240lbm

At SRB burnout Mfinal includes dry weight of SRB’s

20


Rocket System Staging I (cont’d) • Stage 1:

21

= 1.2365 km/sec

!V( )available"# $%stage1 = g0 & Isp & lnMinitial

M final

'

()*

+,- g(t) & sin.dt

0

Tburn

/ - !V( )drag =

g0 0.92 0 &Isp & lnMinitial

M final

'

()*

+,- Tburn

"

#11

$

%223

8% drag loss

9.80673!

"10( ) 0.92 317.14#( )45000001916240$ %

& 'ln 123!$ %& '



Minitial = 1,916,240lbm ! 2Mdry SRB= 1,916,240 ! 2 "192,000 =

1,532,240 lbm

! M final = Minitial " Mpropellant _ consumed !two possible answers !

1)Mpropellant _ consumed = Mprop tank

-#MSSME @srb burnout=

1,655,600lbm " 66,000lbm[ ]" 3$1040lbm /sec $123sec[ ] = 1,205,840lbm

2) Mpropellant _ consumed = m•SSME$ tburn = 3$1040lbm /sec $ (509.5 "123)sec = 1,205,880lbm

Based on problem statement

22

there is a slight inconsistency due to roundoff error

there is a slight inconsistency due to roundoff error



Minitial = 1,916,240lbm ! 2Mdry SRB= 1,916,240 ! 2 "192,000 =

1,532,240 lbm

1)!Vstage2 = g0Isp ln Minitial

M final

"

#$$

%

&''

(

)*

+

,-stage2

=

9.806m /sec . 436.5 . ln 1, 532, 240

1, 532, 240 / 1, 205, 840"#$

%&'= 6618.9m /sec

Based on problem statement

23

9.8067 436.5 15322401532240 1205840!" #

$ %ln& =6619.40 m/sec


Rocket System Staging I (cont’d) • Total Available Delta V:

• Compare to mission Delta V requirement of approximately 7854.2 m/sec ~ dang close shave .. Actual shuttle has slightly Greater margins

• … shuttle engines throttle back at max Qbar (for dynamic loads) and the actual burn time is a bit longer … actual burn time is Around 522-525 seconds …and Isp increases with altitude

… but this example explains the big picture concepts

24

!V( )available"# $%total = !V( )available"# $%stage1 + !V( )available"# $%stage2 =

1236.50m /sec + 6619.40m /sec = 7855.9m /sec

MAE 5540 - Propulsion Systems 21

STS-114 Trajectory Example (concluded)

MAE 5540 - Propulsion Systems 19

STS-114 Trajectory Example

homework 1 - webmae-nas.eng.usu.edu/.../homework1_solution_rev2011.pdf · mae 5540 - propulsion...

Documents