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Rocket Propulsion

by: Demond Hoppe

Rocket Propulsion AE 6450

Demond Hoppe

GPA 3.79


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This 0 page Class Notes was uploaded by Demond Hoppe on Monday November 2, 2015. The Class Notes belongs to AE 6450 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/234309/ae-6450-georgia-institute-of-technology-main-campus in Aerospace Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15
AE 6450 Lecture 2 Thrust Rocket Equation Speci c Impulse Mass Ratio Thrust Equation for a General Jet Propulsion System in 39 Qomrgl Eldridge Treact10n to thrust I H A I I 1 IKf quot Engine I FIn p1 L Eli T e e gt Ipe gt 39 Momentum Conservation ZForceS 2 I pudV I Um ndA gives cv 03 d St d 0 68 y dt T2m8u8maul1e1lAe Rocket No air mass flow rate T mee P6 POD16 Note on Thrust Generation Thrust comes from a Increase in momentum of the propellant fluid momentum thrust b Pressure at the exit plane being higherthan the outside pressure pressure thrust Where does the thrust act In the rocket engine the force is felt on the nozzle and the combustorwalls and is transmitted through the engine mountings to the rest ofthe vehicle Effective Exhaust Velocity A Ce Ue epe pa 7 Rocket Trajectory Basics First a quick review of undergraduate material Please see httpwwwadlgatecheduclassesdcispacedci10htmlSPACEFLGHT Propellant gets used up as the vehicle produces thrust to accelerate Consider a rocket with effective exhaust velocity ce Effective exhaust velocity is a way of expressing the thrust in a simple manner by adding the momentum thrust and the pressure thrust and dividing it by the mass flow rate Also as the propellant is blasted out the exhaust nozzle the mass ofthe vehicle decreases This is substantial in the case of the rocket as compared to airbreathing engines because all the propellant comes from inside the vehicle Development of the Rocket equation T Ref Hill amp Peterson Chapter 10 u i The rate of acceleration of the vehicle is du 2 Ce d 2dr gCostt M M Where Ce is the equivalent exhaust velocity of the vehicle Ce is the thrust divided by the exhaust mass flow rate Drag Term in the Rocket Equation D 2050U2AfCD where atmospheric density above Earth varies roughly as 002 aExpbhl3915 With density in kgm3 and h in meters a 12 and b 29 X 105 Roughly density at 30000 meters is about 1 of its sealevel value Dra Coefficientst ical Note drag coefficient peak is reached at around Mach 12 Inclination deg to fit direction CD lowspeed CD peak 12 CD Mach 2 0 006 015 013 4 0185 016 8 023 02 Gravity Term 2 R Reh At 100 miles above the surface the change from the surface is still only about 5 gzge Development of the Rocket equation Ref Hill amp Peterson Chapter 10 The rate of acceleration of the vehicle is du 2 Ce d 2dr gCostt M M u Neglecting the air drag and gravity terms we get the Ideal Rocket Equation Rocket Equation 2 du 2 c8 M Integrating M2 AUZ Ce I V M1 Or M AU 2 ca loge 1 M2 M1 is initial mass which includes the propellant M2 is the mass after the propellant has been used up to achieve the velocity increment AU Note It is more usual to use the term AV deltavee to denote the velocity increment which is used as a measure of the energy requirements ofdifferent orbital missions We will use U or V as context dictates Specific Impulse Define the Specific Impulse ofthe propulsion system 0 SP 6 go where 90 is the standard value of acceleration due to gravity at sealevel 98m52 Note that the unit of Specific Impulse is seconds Using this definition M AU 2 Spgo loge V1 2 This is also called the Ideal Rocket Equation It does not include the effects of drag or inclination ofthe trajectory Mass Ratio and Specific Impulse Thus the Mass Ratio of a rocket is AU M1 Ispgo M2 Or AV M1 Ispgo M2 Note that for missions such as a launch from Earth39s surface to a trajectory which will escape from earth39s gravitational field this Mass Ratio is large number Example Specific impulse of 390 s g0 98 ms2 and AV 11186 ms 36700 fps Mass ratio is 1867 This means that the rocket at launch time must be at least 1867 times as big as the spacecraft which is left after all the fuel is burned To get a high specific impulse like 390 s we have to use a costly system like liquid hydrogen liquid oxygen Velocity increment AV for Low Earth Orbit 25000 fps Escape from Earth39s gravitational field 36700fps Single Stage Rocket Altitude at burnout tb j Udt 0 Negleaing drag M U 02 InAI 0 gel If me is constant I Mt M0 7 M0 iMbt b Suundmgmcket nmqunesmg er age NASA Guddavd Spa HimWNW 351quot 2 thm 2an asa guv SingleStage Rocket cont d 1 r U c 1n 1 1 r e RM g6 HR 1 2 hb Cetb E Cetb getb Expression for Maximum Altitude Reached Equating kinetic energy at burnout with change in potential energy of the final mass Mb U 2 Mb 7b Mbge hmax hb 2 hm hb U b 2ge 2 2 1 R de R l Chemical Rockets Burn time of existing rockets is 30 to 200 seconds MO ML Mp Ms times Mb 2 ML Ms iimsimrmgs final em M0 M0 f i mpg lm Mazes Mb ML Ms Definitions M L M L Payload ratio A M0 ML Mp Ms Ms Mb M L Structure coefficient 8 z Mp Ms M0 ML 11 Thus g A Multistaging 1 Total initial mass of i th stage prior to firing include its effective M0 payload M Total mass of i th stage after burnout include its effective payload bi ML Payload of last stage M Structural mass of i th stage include engine controls instruments Multistaging 2 th Payload of 1 stage is mass of all subsequent stages th Structural coefflolent of 1 stage 5139 8139 139 M0i1 M0 th If 1 stage contains no propellant at burnout 8 sz39 M0il l Moi M0i1 Multistaging 3 Mass ratio of I m stage M01 R1 Mbl ie R 1 739 81 4 I l Un ZCel lan i1 Similarstages same 8 and 1 11 U C1 n e n8 Multistaging 4 214 11 wzH Q M02 21 M03 12 Lou n Dwsz H ML ML i1k 17 If are equal 1 wk M 2 ML A 01 quotn1n Multistaging 5 AU 21W1m 1eCe M0 M0 P Structural coefficient Mtank M engine 8 MpMtankM engine AU Mtank 1e Ce Mengine Mp MO AU 1e Ce 1Mtank Mengl39ne Mp MO 21 Apollo engines Source Hill Peterson page 479 Engine J2 H1 Thrust kN 1023 1023 Fuel Hydrogen Hydrocarbon Engine Mass Millions ofgrams 1622 0921 Engine Mass Fraction 0024 0014 Tank Mass Propellant Mass 0046 0016 Eq Exhaust Vel ms 4175 2891 Specific Impulse seconds 426 295 22 Saturn V Apollo 11 Flight Configuration Stage 1 2 3 Engine F l J2 J2 Fuel RP l Hydrocarbon LH2 LH2 Number of engines 5 5 1 Total thrust Newton 33 Million 445 Million 089 Million Total Initial Mass 2780 Million 0677 Million 0215 Million Kg Propellant mass kg 1997 Million 0429 Million 0109 Million Structure amp engine 0106 Million 00326 Million 00257 Million kg Structure mass 005 0071 0191 fraction Payload fraction 0321 0466 0603 23 Thrust Coefficient 1 F 1y P0 where At is nozzle throat area and po is chamber pressure Nm2 Thus ut 1gRTt 1 y l 2 For sonic conditions at the throat or 00 7 1 CF and 12 1W l 7 1W 2TT 2 F Atpo7R j 1 amp pepoAe 7 1 71 P0 24 Thrust Coefficient 2 Using isentropic flow relations 71 TtTO p0ijylj 2 ptR 71 i 2 w 1 02 71 002 y H H 12 2 2 7 1 p y FA 1 e A tpOy p6 pa 6 and Thrust Coefficient y H 74 12 y l CF 1amp y pepaAe 7 1 7 1 p0 pOAt Depends entirely on nozzle characteristics The thrust coefficient is used to evaluate nozzle performance 25 Characteristic Exhaust Velocity c Used to characterize the performance of propellants and combustion chambers independent of the nozzle characteristics 71 mp0Ar L m4 p0Arr 7RT0 71 7RT0 where Fis the quantity in brackets Note a0 V SO 190141 1quot a0 6 pOAt Characteristic exhaust velocity m Assuming steady quasi1dimensional perfect gas The condition for maximum thrust is ideal expansion nozzle exit static pressure being equal to the outside pressure In other words Pe 1 26


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