PH508: Propulsion Systems.

Spring 2011: [F&S, Chapter 6]

Derivation of escape velocity: I

Q; What velocity, v, do I need to just escape the gravitational pull of the planet? (the escape velocity).

A: Think about the energies involved!

Initial state:

Kinetic energy = 0 (planet) +

Gravitational potential energy =

Derivation of escape velocity: II

2

21 mv

rGMm

Final state:

Kinetic energy = 0 (planet) + 0 (spacecraft)

Gravitational potential energy =

Initial state energy must equal final state energy

Derivation of escape velocity: III

0

GMm

Therefore:

Derivation of escape velocity: IV00

21 2

rGMmmv

rGMv

rGMv

rGMv

rGMmmv

2

22121

2

2

2

LEARN THIS DERIVATIONAND THE FINAL EQUATION!

Conceptually The various phases of a space mission from

‘concept’ through to ‘end-of-life’ phase. An appreciation of some of the details of each of

these phases and how financial, engineering and science constraints etc. affect mission design.

How a spacecraft’s environment changes from ground level, near earth orbit and deep space.

How these environments (radiation, thermal, dust etc.) feedback into the final mission design.

What you should now know at this point!

Mathematically

Understand how to use the drag equation to work out the force on a body as it travels through the atmosphere

Calculate the solar constant for Earth and (other bodies) making justifiable assumptions.

Derive the escape velocity of a body.

What you should now know at this point!

PH508: Propulsion Systems.

Spring 2011: [F&S, Chapter 6]

Propulsion systems: I

4 major tasks:

1. Launch

2. Station/trajectory acquisition

3. Station/trajectory keeping (staying where it should be, or going in the correct direction).

4. Attitude control (pointing in the correct direction)

Propulsion systems: II

Launch Need lift-off acceleration, a, to be greater

than gravitational acceleration, g. (“a>g”) for an extended period.

This implies a very high thrust for a long duration. E.g., the shuttle main engine: 2 x 106 N for 8 minutes.

Typical Δv ≥ 9.5 km s-1 (including drag and gravity losses).

Propulsion systems: III

Launch phase (continued) Still difficult to achieve with current

technology Only achievable with chemical rockets Massive launch vehicles required for

relatively small payloads Major constraint for spacecraft and mission

design is the mass cost: £1000s - £10,000s per kilogram.

Propulsion systems: IV

Station/trajectory acquisition Apogee motors (apogee = ‘furthest point’)

◦ Orbit circularisation ◦ Inclination removal◦ Requires a force of ~75 kN for 60 seconds. Δv = 2 km

s-1. Perigee motors (perigee=‘nearest point’)

◦ Orbit raising◦ Payload Assist modules (‘PAM’)◦ Interial Upper Stages (‘IUS’)◦ Δv ~4.2 km s-1 (30° inclination parking orbit ->

equatorial geostationary).

Propulsion systems: V

Earth Escape Δv ~ 7.6 km s-1 (Mars flyby) Δv ~ 16 km s-1 (Solar system escape

velocity)◦ Without using gravity assist manoeuvres.

Station/trajectory keeping Low thrust levels required (mN – 10s N)

pulsed for short durations. Δv ~ 10s – 100s m s-1 over duration of

mission.

Propulsion systems: VI

Attitude control (‘pointing’)

Very low thrust levels for short duration Small chemical rockets Reaction wheels (diagram).

Principle of operation of all propulsion systems is Newton’s third law

“...for every action, there is an equal and opposite reaction...”

Propulsion systems: VII

Rocket equation: I

Derivation: Need to balance exhaust (subscript ‘e’) momentum with rocket momentum.

∑momenta = 0 (Conservation of linear momentum)(Recall: momentum = mass x velocity)

∴ m dV = -dm Ve dV = -Ve dm/m

So, now some maths...

Rocket equation: II

mmVVV

mmVVVmmVVV

mVV

mdmVdV

oeo

oeo

oeo

mme

VV

V

V

m

me

oo

o o

ln

lnln]ln[ln

ln

Tsiolkovsky’s Equation (the rocket equation).

•dm is the mass ejected

•dV is the increase in speed dueto the ejected mass (dm)

•Ve is the exhaust velocity (ie. thevelocity of the ejected mass relative to the rocket)

•m is the rocket mass (subscript‘o’ denotes initial values)

In practice, drag reduces Vmax by~0.3 – 0.5 km s-1.

Recall (in zero g):

Now add gravity: (diagram)

Rocket equation: III (with gravity)

dtdmV

dtdVm e

Bfo

o

s tmm

me

V

e

e

e

gdtdmm

VdV

gdtdmm

VdV

gdtdm

mV

dtdV

mgdtdmV

dtdVm

00

1

1

1

Bfo

oes

Bfooes

Bmm

mes

tmm

me

V

gtmm

mVV

gtmmmVV

gtmVV

gdtdmm

VdV

fo

o

Bfo

o

s

ln

lnln

ln

1

00

Rocket equation: IV (with gravity)Integrating previous equation:

Define R as:

Rocket equation: V (with gravity)

fo

o

mmmR

e

Bs

es

e

Bses

e

Bsees

Bses

VtgRR

RVV

VtgRVV

VtgVRVV

tgRVV

exp

ln

expln

ln

ln

R’ is the “effective mass ratio

•Vs =spacecraft velocity•Ve= exhaust velocity•gs = accl. of gravity acting on spacecraft•tB = rocket burn time•mf= mass of fuel

Therefore, want a short burn time as possible to minimise gravitational losses.

Gravitational losses reduce V by ~1 km s-1

This conflicts with the requirements to reduce drag effects at low atmosphere (low speed at low altitude)

Resolve conflict by using non-vertical ascent.

Rocket equation: VI (with gravity)

Rocket equation: VII (with gravity)

Retarding gravitational force = ge cos θ

ge = net downward accl. = gravity - centrifugal

Typical launch sequence:◦ Lift-off (straight up!)

◦ Clear tower

◦ Roll to correct heading

◦ Pitch to desired trajectory

◦ Recall space shuttle launch sequence. Rolls and pitches almost immediately after clearing tower. Reason to minimise loss dues to drag and gravity!

Rocket equation: VIII (with gravity)

Assume a single stage, liquid propellant chemical rocket.◦ Fuel – kerosene◦ Oxidiser – liquid oxygen◦ Typical of fuel used for Atlas, Thor, Titan and

Saturn rockets.

Ve ~2.5 km s-1, assume mass: 20% structure, 80% fuel

Recall,

Rocket equation: an example

5

fo

o

mmmR (In this case)

Rocket equation: an example (cont.)

4lnmax RVV e km s-1

•Compare with Earth’s escape velocity ~ 11 km s-1!

•Velocity required for 300 km altitude Earth orbit ~7.8 km s-1.

•Taking into account drag and gravity losses implies arequired Vmax ≥ 9.3 km s-1

•Best performance from a fully cryogenic fuel system get Vmax ~9.5 km s-1.

SOLUTION: Multi-Staging!

Parallel staging◦ Partially simultaneous operation (e.g. Space

Shuttle) Series staging

◦ Sequential operation (e.g. Ariane, Saturn V etc.)

Principle: jettison inert mass to reduce load for subsequent rocket stages.

Stage velocity, Vs = Vmax – Vo = Ve ln R = -Ve ln (1-R)

Multi-stage rockets: I

Stage velocity, Vs = Vmax – Vo = Ve ln R = -Ve ln (1-R)

Jettisoning structure from stage ‘n’ increases R (the mass ratio) and thus Vs for the subsequent stages, n+1, n+2 etc.

However, since Vs ∝ ln R, improvement is slow with R

Multi-stage rockets: II

Assume a simple rocket where: mf = mass of propellant ms = mass of structure mp = mass of payload mo = mf + ms + mp Define mass ratio, R: Payload ratio, P: Structure ratio, S:

Multi-stage rockets: III – an example

RSSRP

mm

mmm

S

mmP

mmm

mmmR

s

f

s

sf

p

o

fo

o

sp

o

1

1

Assume a liquid propellant chemical rocket.◦ Fuel – kerosene◦ Oxidiser – liquid oxygen◦ Typical of fuel used for Atlas, Thor, Titan and

Saturn rockets.

Ve ~2.5 km s-1, assume: 1t structure, 8t fuel, 1t payload

Recall definition of payload ratio, R

Mutli-staging: an example

5210

8)811(811

fo

o

mmmR

(In this case)

RSSRP

mm

mmm

S

mmP

mmmR

mmmm

s

f

s

sf

p

o

sp

o

spf

1

1

0

Mutli-staging: an exampleNow calculate the payload ratio, P and the structure ratio, S.Recall: mo = 1 + 8 +1= 10, mf = 8, ms = 1 tons.

511

10

91811

10110

sp

o

s

f

p

o

mmmR

mm

S

mmP

For our kerosene rocket:Ve ~ 2.5 km s-1 , R =5

Recall, Vs = Ve ln R = 2.5 x ln 5 ~ 4 km s-1

∴ Vs = 4 km s-1 – suborbital!

Now consider this 10 ton rocket to be a payload (i.e. a stage) of a larger rocket…

Multi-stage rockets: IV – an example

Therefore, assume that the mass/fuel ratio is the same for the second stage, and thus we can use the same ratios (ie, this stage is just a scaled up version of the original stage):

∴ mp= 10t and thus,

mo= P x mp = 10 x 10t = 100t

Now our 1 ton original payload can reach: 4 + 4 = 8 km s-1 – orbital, just…

using a 100t rocket!

Multi-staging: example continued

Now consider this 100 ton rocket to be a payload (i.e. a stage) of an even larger rocket…

Therefore, assume that the mass/fuel ratio is the same for the previous stage, and thus we can use the same ratios:

∴ mp= 100t and thus,

mo= P x mp = 10 x 100t = 1000t

Multi-staging: example continued

Now our 1 ton original payload can reach: 4 + 4 + 4= 12 km s-1 – escape velocity

using a 1000t rocket!

Therefore a 3-stage kerosene rocket can put a payload into orbit, and reach Earth escape velocity, whereas a single stage could not!

Multi-stage rockets: V – an example

In general:

◦Vmax = ∑Vs

Maximum rocket velocity is the total of the stage velocities.

Using conventional definitions (i.e. 1st stage is the first to burn etc.), the payload ratio of the ith stage is, Pi:

Multi-stage rockets: VI

1

oi

oii m

mP

(ie. The payload ratio of stage 1 = mass of stage 1/ mass of stage 2)

Thus the total payload ratio, P is:

The structural payload, S is:

And the mass ratio, R is:

Multi-stage rockets: VII

np

o PPPPmmP 321

1

si

sifii m

mmS

fioi

oii mm

mR

Therefore,

And if all stages have the same Ve

(Generally, however, this is not the case)

Multi-stage rockets: VIII ieis RVVV lnmax

n

e

ne

ie

RRRRRRVV

RRRVV

RVV

321

max

21max

max

lnlnlnln

ln

Stage

Propellant Ve (km s-

1)

mf (tons)

ms (tons)

thrust(tons wgt)

burn time(secs)

1st Kerosene + O2 2.32 2160 140 3400 1502nd H2(l) + O2(l) 4.10 420 35 450 3903rd H2(l) + O2(l) 4.25 100 10 90 480

1st Stage

2nd Stage

3rd Stage

Fuel consumption (tons/sec) 14.40 1.08 0.21For a payload, mp, of 100 tonsmoi 2965 665 210Ri 3.69 2.72 1.91Pi 4.46 3.16 2.10Vsi (km s-1) 3.03 4.10 2.75

Multi-stage rockets: IX – the Saturn V

Note high thrust of 1st stage High efficiency of stages 2 and 3 P for a 3 stage kerosene rocket ~90 P for the Saturn V ~ 30 (more efficient).

Such a rocket could lift:◦ 100 tons into a low earth orbit◦ 40 tons into earth escape (i.e. to the moon)◦ 1 ton payload to Mars!◦ For Apollo, the Saturn V lifted the Command module to

LEO.◦ The command module went to the moon and back.

Multi-stage rockets: X – the Saturn V

Optimisation of number of stages:Q: What is the optimum number of stages?Theoretically...recall:

Multi-stage rockets: XI

22

2

11

121

321

321

11

1

1

RSS

RSSRRP

RSSRP

RRRRRPPPPP

RSSRP

ii

iii

n

n

P=payload ratio

R=mass ratio

S=structure ratio

Now, R=R1R2...Rn and if R1=R2=...=Rn

Then:

And if S1=S2=...=Sn, then:

Multi-stage rockets: XII

ni RR

1

n

nRS

SRP

1

1

(Effectively all this says is that the mass ratios of each stage are thesame, just scaled versions of each other).

Optimisation of number of stages involves minimising P

∴ want to minimise:

∴want to maximise 1/n, i.e., n→∞.

Q: Why don’t we see systems with very large numbers of small stages?

Multi-stage rockets: XIII

nRS

S1

1

Each stage requires:◦ Engine and nozzles◦ Ignition mechanism◦ Separation mechanism◦ Fuel pumps (for liquid propellants)◦ Small stages have worse P, R and S.

Therefore greater cost and complexity Thus a ‘trade-off’. ‘n=3’ is usually the maximum number of

stages (some ‘n=4’, but rare).

Multi-stage rockets: XIV

Because the Earth revolves on its axis from West to East once every 24 hours (86400 secs) a point on the Earth’s equator has a velocity of 463.83 ms-1.

Reason: radius of the Earth, RE = 6.3782 x 106 metres.

Earth’s circumference = 2πRE = 4.007 x 107 mEquatorial velocity = 4.007 x 107 / 86400 =

463.83 ms-1

Geographical velocity boost: I

Therefore, a spacecraft launched eastwards from the Earth’s equator would gain a free increment of velocity of 463.83 ms-1.

Away from the equator the Earth has a smaller circumference which is determined by multiplying the equatorial circumference by the cosine of the latitude in degrees.

For example, the Russian Baikonur Cosmodrome is at 45° 55’ north. The Earth’s rotational velocity at that point is: 322.69 m s-1.

Geographical velocity boost: II

System classification:◦ Various possible schemes (see F&S, Fig. 6.1)◦ Other ‘exotic’ systems possible

Function:◦ “Primary propulsion” – launch◦ “Secondary propulsion”

Station/trajectory acquisition and keeping Attitude control

Recall: vastly different requirements for different purposes:◦ ΔV of m s-1 – km s-1

◦ Thrust of mN – MN◦ Accelerations of μg - >10g

Different technologies applicable to different functions/regimes.

Propulsion systems: overview

Principle:◦ Combustion of propellants at high pressure in a small

confined volume produces high temperature gas.◦ Expansion through nozzles convert random thermal

energy to directed kinetic energy: “thrust”. Propellants:

◦ And fuel and oxidiser undergoing exothermic reaction producing gaseous products.

Considerations: ◦ Specific energy content, rate of heat release,

storage, handling, etc.

Chemical rockets: I

Chemical rocket types:◦ Solid propellant◦ Liquid propellant◦ Hybrid (usually solid fuel and a liquid oxidiser)

Solid propellant rockets:◦ Oldest rocket technology – Chinese 12th Century.◦ Very simple – no moving parts (nozzles?)◦ Only needs an igniter and a douser◦ Fuel stored in combustion chamber◦ Relatively cheap

Chemical rockets: II

Solid propellant rockets (continued) Advantages:

◦ Simple and cheap◦ Reliable◦ High thrust◦ High energy density propellant thus small volume

Disadvantages:◦ Only limited throttling◦ Generally only single burn (a firework effectively).

Chemical rockets: III

Solid propellant rockets (continued) Propellant is a fuel and oxidiser matrix with

aluminium powder regulator. Cast directly into casing of rocket Thrust is proportional to burn rate “Cigarette mode” – long duration, low thrust

because of small combustion area. Axial ignition used to increase burn area

and increase thrust.

Chemical rockets: IV

Solid propellant rockets (continued) Burn area and thrust defined by ‘grain’, produced

by the mandrel during casting of the fuel. This gives a limited amount of “pre-programmed”

throttling of the propellant. Ignition is via a pyrotechnic device which ignites

the propellant in the igniter. Axial burn. Applications:

◦ Early launch vehicles (missiles)◦ Launch vehicle strap-on boosters (e.g. Titan, Ariane,

Shuttle)◦ Secondary propulsion

Chemical rockets: V

Liquid propellant rockets First flight 16th March 1926.

◦ Robert Goddard using liquid oxygen and gasoline Max altitude = 12.5 metres Flight time = 2.5 seconds Engine thrust ~ 40 N Vmax ~ 96 km/hour Landed 56 metres from launch site

Advantages◦ Long burn time◦ Controllability

Throttling On-off-on operation Emergency shutdown Redundant systems

Chemical rockets: VI

Liquid propellants (continued) Disadvantages

◦ Complexity and reliability◦ Cost◦ Mass

Requirements◦ Separate storage of fuel and oxidiser remote from

combustion chamber ◦ Thus need propellant pump and feed system◦ Chamber injector and mixer◦ Igniter, combustion chamber and exit nozzle

Chemical rockets: VII

Liquid propellants◦ (Kerosene/ethanol) + liquid oxygen + N2O4◦ Monomethyl hydrazine (‘MMH’)◦ Unsymmetrical dimethyl hydrazine (‘UDHM’)◦ Aerozine50 (50/50 mix of hydrazine and UDMH)◦ Liquid hydrogen and liquid oxygen (cryogenic propellants)

Some combinations require ignition, others (known as ‘hypergols’) are self-igniting as soon as the fuel + oxidiser mix (e.g. Aerozine50 and N2O4).

Applications◦ Most modern launch vehicles◦ Secondary propulsion systems

Chemical rockets: VIII

Manned spacecraft, mainly reusable (unlike Ariane)

Expensive “launch vehicle” Designed to be multi-purpose

◦ Laboratory (‘Spacelab’)◦ Recovery repair and return of satellites◦ Space station servicing◦ Launch of satellites◦ Just about to be retired!

Space shuttle transport system (STS) : I

Primary propulsion system, two elements:◦ External fuel tanks feeding SSME (‘Space Shuttle

Main Engines’, x3)◦ Two solid rocket strap-on boosters

Burn for 120 seconds, separate, parachute into ocean 300 km downrange for recovery and reuse.

◦ SSMEs use closed cycle combustion with a chamber pressure of 207 Bar (20.7 GPa) and a burn time of 480 seconds

◦ 100% thrust ~2.1 x 106 N.◦ External fuel tank jettisoned (and burns up) pre

orbital insertion.

Space shuttle transport system (STS) : II

The space shuttle combines liquid and solid propellants.

Solid propellants give a lower ‘specfic impulse’ (thrust per mass of propellant) but are compact, simple and stable.

Once ignited it burns continuously. Thrust can only be controlled by varying the burn area.

The liquid propellant in the STS combines hydrogen and oxygen and can be throttled to vary the thrust.

The STS uses 2 SRBM (‘Solid Rocket Booster Motors’) and the liquid propellant in the main external tank during launch.

Space shuttle transport system (STS) : III

Thrusters (secondary propulsion units). Once in space there is still a need for thrust. There are two main types:

◦ Sustained high thrust for orbital manoeuvring etc.◦ Low thrust for atitude control (rotate the spacecraft, or to

controls its spin rate etc.) Cold gas thrusters

◦ Take an inert gas (nitrogen or argon) stored at high pressure and connected to a series of valves.

◦ The thrusters are arranged off-axis to control spin/rotation. ◦ Specific impulse is low with low volumes of gas◦ Typical thrust ~10 mN in short bursts.

Space shuttle transport system (STS) : IV

Monopropellant The decomposition of hydrazine (N2H4)

generates heat. Expansion of the hot gas through nozzles

produces a specific impulse. Hydrazine is a liquid between 275 – 387 K

and is held under pressure in tanks. Can provide ~10 N for orbital control and

station keeping.

Space shuttle transport system (STS) : V

Bi-propellant E.g. MMH/nitrogen tetroxide Propellants burn on contact They are mixed in the thruster/apogee

motor and can provide sustained thrust. Can be used for orbital rising as well as

atitude control. Provides precise amounts of thrust on

demand.

Space shuttle transport system (STS) : VI

Solid propellant apogee motors To launch into GEO etc. usually launch to a LEO

and then boost with a final stage burn. To achieve a high, circular orbit at apogee, need

a high thrust, short duration burn Usually provided by a solid propellant apogee

motor. For a GEO satellite of 1000 kg need 900 kg of

propellant and a ΔV of ~2 km s-1. Burn for 40-60 seconds with an average thrust

of 50 – 75 kN

Space shuttle transport system (STS) : VII

Space shuttle transport system (STS) : VIII

Shuttle external tankand SRBs

The characteristic feature of the SRB’s thrust curve’s profile is the period of reduced thrust (to about 70% of max. 50 secs. into the flight) giving a ‘sway-backed’ appearance (below).

SRBs ‘sway-back’ thrust profile: I

Its purpose is to reduce the thrust while the shuttle is passing through the region of maximum dynamic pressure. This is when the product of velocity and air pressure is a maximum and when the possibility of damage by aerodynamic forces is greatest.

Therefore minimise risk by reducing thrust for a short period.

SRBs ‘sway-back’ thrust profile: II

Another term sometimes seen is the ‘thrust impulse’, Is.

It is simply defined as:

And has units of “seconds”. The larger IS the greater the effective thrust (recall Ve is the exhaust velocity).

Definition of thrust impulse

gVI e

S

Space shuttle transport system (STS) : IX

Amount of thrust generated is proportional to exposed (burning) surface area of propellant.

Space shuttle transport system (STS) : X

Space shuttle transport system (STS) : XI

Space shuttle transport system (STS) : XII

Space shuttle transport system (STS) : XIII

Cross-sections of various solid propellant castingsand associated thrust profiles.

Space shuttle transport system (STS) : XIV

Space shuttle transport system (STS) : XV