Rockets and Launch Vehicles

7/23/2019 Rockets and Launch Vehicles

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AE 3310Introduction to Aerospace Vehicle Performance

18. Rocket Engines and Launch Vehicles18.1 Momentum and Thrust Equation18.2 Specific Impulse

18.3 Rocket Equation18.4 Rocket Mass Relationships



Rocket Performance – Thrust Equation

We will derive the thrust equation by starting with conservationof momentum using an example of a static thrust test for arocket on a test stand.

Assumptions:

1.Static (test stand is not moving)

2.Steady exit velocity (ue) and steady propellant flow rate ( )

3.Quasi-1D flow

m&




Sum of forcesapplied to theCV (Solid bodyon the fluid)

=Change in total momentuminstantaneously contained

within CV+

Net flow-rateof momentumleaving the CV

(CV itself is not accelerating for astatic test stand)

0

(steady exit velocity andpropellant flow rate)

(only care about x-direction for 1D-flow)

x x x

CV CS

d F u dV u dm

dt ρ = +∑ ∫ ∫ &

emu&




1D-Force Summation (for left-hand side of momentumequation):

Substitute into Momentum Equation:

Optimum expansion when pe = p0 and 2nd term goes to zero

• Over-expansion: pe < p0 (degrades thrust)

• Under-expansion: pe > p0 (improves thrust? No, because ue impacted)

( )0 x e e F T p p A= + −∑

( )0

0

( )e e e

e e e

T p p A mu

T mu p p A+ − == + −

&

&



Rocket Performance – Specific Impulse

Sometimes convenient to define equivalent exhaust velocity:

Where thrust has units of Newtons:

Thrust is an instantaneous measure of the ability of a rocket toaccelerate a mass.Impulsemeasures the total momentumgain over the course of an entire burn:

Where impulse has units of Newton-seconds:

If we normalize theimpulse by the propellant mass, we get ameasure of efficiency (impulse per unit mass of propellant):

With units:

dt

0eeq e e

eq

p pu u A

m

T mu

− = + ÷

=&

& 2

kg m[N]

s

× =

[ ]kg m

N s s

× × =

eq p

I T u

m m= =

&

N s m

kg s

×

=

eq p eq I Tdt mu m u= = =

∫ ∫ &



Rocket Performance – Specific Impulse

By convention, we typically define aspecific impulse which is

theimpulse per weight of fuelon Earth:

Where specific impulse has units of seconds:

The gravity term just provides a common reference point tocompare rocket fuel efficiencies. An efficient fuel minimizes themass of fuel required to achieve a certain impulse, and thushigherIsp is desired.

and

eq

sp

p E E

u I I

m g g

= =[ ]

2

N ss

mkgs

×

=

× 2

m9.8

s E g =



Rocket Performance – Rocket Equation

Static thrust test is a simplified model because

the engine mass doesn’t change.

In reality, propellant mass (especially for launchvehicles) accounts for a majority of the initial

mass, and the vehicle mass varies a lot over thecourse of a burn.

At initial time,t0, the mass of the vehicle isM and

the velocity isu.

At timet0 +dt, the mass of the vehicle isM –dm,

the velocity isu +du, and the incrementalpropellant massdm has an exhaust velocityue.

Mg E

u

ue

dm

θ




Recall the momentum equation, now aligned intheu-direction:

Mg E

u

ue

dm

θ

Since the vehicle is no longer fixed to the stand,we define the net forces on the system in theu-direction:

Force due to

pressure

differential at

nozzle exit

Component of

weight opposing

thrust in the u-

direction

Aerodynamic

drag force

The instantaneous change in momentum inside

theCV is no longer 0:

negligible

u

CV CS

d F udV udmdt ρ = +∑ ∫ ∫ &

( )0 cosu e e E F p p A D Mg θ = − − −

∑

( ) ( )( )

( )

CV

CV

CV

d d udV M dm u du Mu

dt dt

d d udV Mu Mdu udm dmdu Mu

dt dt

d duudV M mudt dt

ρ

ρ

ρ

= − + −

= + − − −

= −

∫

∫

∫ &




Recall the momentum equation, now aligned intheu-direction:

Mg E

u

ue

dm

θ

Since the vehicle is moving, the net flow-rate ofmomentum leaving theCV is as follows:

Substituting terms into the momentum equation:

Solving for the acceleration term:

u

CV CS

d F udV udmdt ρ = +∑ ∫ ∫ &

eCS

udm mu mu= −

∫ & & &

( )0cose e E e

du p p A D Mg M mu mu mu

dt θ − − − = − + −& & &

( )0

1cos

e e e E

du Dmu p p A g

dt M M θ = + − − − &

eqmu= &

coseq E

du m Du g

dt M M

θ = − −&




Multiply through bydt:

Mg E

u

ue

dm

θ

To find required ∆u (or “delta-V”), integrate.For simplified formulation make the followingassumptions:• Negligible drag (D = 0)

• Negligible gravity (Remove gE term)

• Equivalent exhaust velocity (ueq) approximately constant

Substitute definition ofspecific impulse:

The RocketEquation

Here, M without a subscript denotes

the initial mass of the rocket.

coseq E

du m Du g

dt M M θ = − −

&

cos cos

Sign flip because:

eq E eq E

dm D dM Ddu u g dt u g dt

M M M M

dm dM

θ θ = − + = − − +

= −

ln ln f f u M f

eq eq equ M

f

M dM M du u u u u M M M

= ∆ = − = − = +∫ ∫

ln sp E

f

M u I g

M

∆ =



Rocket Mass Relationships

Often when comparing performance of rocket engines, itbecomes useful to decompose the mass terms:

Additionally, it can be useful to define other mass ratios,including the payload ratio (π) and the structural coefficient (λ):

Substituting these terms back into the rocket equation:

As an exercise, try to make substitutions to convince yourselfthe above expression is true.

Mass Raio ! f propellant payload structure propellant

f f payload structure

M M M M M M

M M M M

+ + += =

+

payload M

M π =

structure

structure propellant

M

M M λ =

+

1ln ln ln(1 )

payload structure propellant

sp E sp E sp E

f payload structure

M M M M u I g I g I g M M M π π λ

+ + ∆ = = = + + −




Note that these results are without gravitational losses in the rocket equation!The situation is much worse for a launch vehicle!



Rocket Mass Relationships –Electric Propulsion

Solid Rocket

Propellant

Liquid

Hydrocarbon

Propellant

Liquid

Hydrogen

Propellant

Electric

PropulsionPropellant

Earth to LEO

TransferLEO to

Moon/Mars

Transfer ( )

1ln

1 sp E u I g

π π λ

∆ = + −




From the previous two plots, it would seem that the achievable

payloads are very limited for solid and liquid propellant types.So why is electric propulsion only used for in-space propulsionarchitectures and not launch vehicles?

We are missing a comparison of the thrust capability of thesetypes of propulsion architectures:

Electric propulsion devices feature high specific impulse, but atvery low thrust levels.




Another useful mass relationship is the propellant ratio (ξ):

To relate these three mass ratios, start with a trivial definition:

Rearranging terms:

Now recalling the definition of the structural coefficient:

1 propellant propellant f

payload structure propellant

M M M

M M M M M ξ = = = −

+ +

1 payload propellant structure structure M M

M M M M M M M M

π ξ = = + + = + +

1 structure M

M π ξ = − −

( )( )1

structure structure

structure propellant payload

payload structure

M M

M M M M

M M M

M M

λ

λ λ π

= =+ −

−⇒ = = −




Solving forπ:

Thus we have a single equation relating these three mass

ratios.

( )1 1 1 structure M

M

π ξ ξ λ π = − − = − − −

( )

1

1 1

1

1

π ξ λ λπ

π λ λ ξ

λ ξ π λ

= − − +

− = − −

− −=−




As can be seen, payload fraction varies linearly with propellantratio, with the slope set by the structural coefficient.



Example Problem 1

A launch vehicle places a massM = 125,000-kg with a payloadfractionπ = 0.3 into LEO. The transfer stage must supply a ∆V of 4.3-km/s to put the spacecraft on a Mars transfer orbit.

Some useful

equations:

>19,000 years!

1ln

(1 ) sp E u I g

π π λ

∆ = + −

1

1

λ ξ π λ

− −

=

−

p eq p sp E T m u m I g = =& &

propellant

burn

p

M t

m=

&



Lessons from Example Problem 1

1.This solid rocket motor is not a feasible transfer stage, because thespecific impulse is too low to achieve the necessary ∆v for a payloadfraction ofπ = 0.3.

2.The liquid rocket engine is the most feasible architecture, with areasonable structural coefficient ofλ = 0.11 and enough thrust toachieve a reasonably low burn time.

3.While the electric propulsion at first seems like a reasonablearchitecture choice due to its high specific impulse and low propellantratio, the thrust is too low and the burn time is much too long.

• Low-thrust electric propulsion architectures actually work by applyingcontinuous thrust as opposed to our instantaneous assumption (requiresdifferent performance equations)

• Can achieve high-speeds over a long period of time (deep-space missions)

• Also can be used for applications that require small ∆v, such as orbitalcorrection maneuvers or attitude control

• Chemical propulsion engines are typicallyenergy-limited, whereas electricpropulsion devices are typically power-limited



Example Problem 2

A satellite with a 500-kg gross mass is launched into an orbitwith an inclination angle of 60°. The satellite needs to transitionto an equatorial orbit at the apoapsis of the current orbit, whenthe velocity equals 10 km/s. What is the maximum mass thatcan be delivered to the equatorial orbit using a chemicalpropellant withIsp = 300-seconds? an electric propulsion

architecture withIsp = 1,000-seconds? When can each of these

be used?

If I sp = 300-s:

If I sp = 1000-s:

( )"0 2 km

2 sin 2 sin 2 sin #0 102 2 2 s

i vv v v v v

∆ ° ∆ = = = ° = = = ÷ ÷

( ) ( )2

10000 m$s

#00 s 9.8 m$s

(%00 kg)1"."" kg

sp E

f v

I g

M M

e e

∆= = =

( )

( )

2

10000 m$s

1000 s 9.8 m$s

(%00 kg)180.22 kg

sp E

f v

I g

M M

e e

∆= = =



Lessons from Example 2

1.If the only goal was to deliver the maximum mass to the

equatorial orbit, the electric propulsion architecture would bechosen every time.

2.Considering the lessons learned about burn-time fromExample Problem 1, we can guess the chemical propulsion

burn would take on the order of seconds, whereas theelectric propulsion burn would take on the order of months.

3.There is a tradeoff of payload fraction and burn timebetween these propulsion architectures:

–

If we need to deliver small payloads to orbit quickly, we wouldchoose chemical propulsion

– If we can wait longer, we can deliver larger payloads to orbit withelectric propulsion

Rockets and Launch Vehicles

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