PH508: Propulsion Systems.

1. PH508: Propulsion Systems. Spring 2011: [F&S, Chapter 6]

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

3. Derivation of escape velocity: II A: Think about the energies involved! Initial state: Kinetic energy = 0 (planet) + Gravitational potential energy =

4. Derivation of escape velocity: III Final state: Kinetic energy = 0 (planet) + 0 (spacecraft) Gravitational potential energy = Initial state energy must equal final state energy

5. Derivation of escape velocity: IV • Therefore: LEARN THIS DERIVATION AND THE FINAL EQUATION!

6. What you should now know at this point! 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.

7. 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.

8. PH508: Propulsion Systems. Spring 2011: [F&S, Chapter 6]

9. Propulsion systems: I

10. Propulsion systems: II • 4 major tasks: • Launch • Station/trajectory acquisition • Station/trajectory keeping (staying where it should be, or going in the correct direction). • Attitude control (pointing in the correct direction)

11. Propulsion systems: III 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).

12. Propulsion systems: IV 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.

13. Propulsion systems: V 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).

14. Propulsion systems: VI 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.

15. Propulsion systems: VII 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...”

16. 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

17. Rocket equation: II So, now some maths... • dm is the mass ejected • dV is the increase in speed due • to the ejected mass (dm) • Ve is the exhaust velocity (ie. the • velocity 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. Tsiolkovsky’s Equation (the rocket equation).

18. Rocket equation: III (with gravity) • Recall (in zero g): • Now add gravity: (diagram)

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

20. Rocket equation: V (with gravity) • Define R as: • Vs =spacecraft velocity • Ve= exhaust velocity • gs = accl. of gravity acting • on spacecraft • tB = rocket burn time • mf= mass of fuel R’ is the “effective mass ratio

21. Rocket equation: VI (with gravity) • 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.

22. Rocket equation: VII (with gravity) ge = net downward accl. = gravity - centrifugal Retarding gravitational force = ge cos θ

23. Rocket equation: VIII (with gravity) • 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!

24. Rocket equation: an example • 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, (In this case)

25. Rocket equation: an example (cont.) 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 a • required Vmax ≥ 9.3 km s-1 • Best performance from a fully cryogenic fuel system • get Vmax ~9.5 km s-1. • SOLUTION: Multi-Staging!

26. Multi-stage rockets: I • 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)

27. Multi-stage rockets: II 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

28. Multi-stage rockets: III – an example 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:

29. Mutli-staging: an example • 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: 1tstructure, 8t fuel, 1t payload Recalldefinition of payload ratio, R (In this case)

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

31. Multi-stage rockets: IV – an example For our kerosene rocket: Ve ~ 2.5 km s-1 , R =5 Recall, Vs = VelnR = 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…

32. Multi-staging: example continued • 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!

33. 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

34. Multi-stage rockets: V – an example 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!

35. Multi-stage rockets: VI 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: (ie. The payload ratio of stage 1 = mass of stage 1/ mass of stage 2)

36. Multi-stage rockets: VII Thus the total payload ratio, P is: The structural payload, S is: And the mass ratio, R is:

37. Multi-stage rockets: VIII • Therefore, • And if all stages have the same Ve • (Generally, however, this is not the case)

38. Multi-stage rockets: IX – the Saturn V

39. Multi-stage rockets: X – 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.

40. Multi-stage rockets: XI Optimisation of number of stages: Q: What is the optimum number of stages? Theoretically...recall: P=payload ratio R=mass ratio S=structure ratio

41. Multi-stage rockets: XII • Now, R=R1R2...Rn and if R1=R2=...=Rn • Then: • And if S1=S2=...=Sn, then: (Effectively all this says is that the mass ratios of each stage are the same, just scaled versions of each other).

42. Multi-stage rockets: XIII • 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?

43. Multi-stage rockets: XIV • 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).

44. Geographical velocity boost: I 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 m Equatorial velocity = 4.007 x 107 / 86400 = 463.83 ms-1

45. Geographical velocity boost: II • 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.

46. Propulsion systems: overview • 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.

47. Chemical rockets: I • 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.

48. Chemical rockets: II • 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

49. Chemical rockets: III 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).

50. Chemical rockets: IV 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.