Section 4.3: The RamJet Propulsion Cycle

1 MAE 6530 - Propulsion Systems IISection : The RamJetPropulsion CycleMAE 6530 - Propulsion Systems IIBackground on RamJets Ramjets are a very simple jet engine configuration that are capable of high speeds Ramjets cannot produce thrust at zero airspeed; they cannot move an aircraft from a standstill. A RamJet powered vehicle, therefore, requires an assisted take-off like a rocket assist to accelerate it to a speed where it begins to produce thrust. Inherently constrained to combined Cycle applications for flightMAE 6530 - Propulsion Systems IIControl Volume for a Ramjetp T V peTeVe ep2V2T2 1V1 Fthrust=!mair+!mfuel() Vexit !mair() V +Aexit pexit p ()p A0=Fthrust=!mair V f+1f VexitV 1 +AexitA0 pexitp 1 f=!mair!mfuelMAE 6530 - Propulsion Systems IIIdeal RamJet Thermodynamic Cycle AnalysisStep1-2 Step 3 Step 4 RegionProcessIdeal BehaviorRealBehaviorA to 1(inlet)Isentropic flowP0,T0constantP0drop -1-2 (diffuser)Adiabatic CompressionP,TincreaseP0dropP0drop2-3 (burner)Heat AdditionP0constant, T0sIncreaseP0drop3-4 (nozzle)Isentropic expansionT0,P0constantsIncreaseT0drop4 MAE 6530 - Propulsion Systems IIIdeal RamjetCycle Analysis (2)T-sDiagram5 StepProcess1) Intake (suck)Isentropic Compression2) Compress the Air (squeeze)Adiabatic Compression3) Add heat (bang)Constant Pressure Combustion4) Extract work (blow)Isentropic Expansion in Nozzle5)

2 ExhaustHeat extraction by surroundingsStep 1-2 Step 3 Step 4 Step 5 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet Propulsive Power Output--> work perform by system in step 4minus work required for step 1-2 NetHeat Input --> heat input during step 3 (combustion)-heat lost in exhaust plume66 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (2)7 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (3)8 Add and Subtract From Right Hand SideMAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (4) Assume ..calorically perfect gasses h~ Cp T9 Forconceptual simplicity .. let ..f>> 1 .. and .. Cpair~ CpproductsMAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (5) From C-->D flow is isentropic ..(Credit Narayanan Komerath, Georgia Tech)Step 1-2 Step 3 Step 410 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (6) Approximate as Adiabatic compressionacross diffuser(Credit Narayanan Komerath, Georgia Tech)Step 1-2 Step 3 Step 411 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (7)(Credit Narayanan Komerath, Georgia Tech)Step 1-2 Step 3 Step 4 Subinto efficiencyequation12 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (8)(Credit Narayanan Komerath, Georgia Tech)Step 1-2 Step 3 Step 413 Factor OutMAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (9)(Credit Narayanan Komerath, Georgia Tech)

3 Step 1-2 Step 3 Step Efficiency Proportional to Inlet Pressure ratio, Efficiency Proportional to combustor temperature difference inlet total pressureratio (P0B/P0A) goes down .. Cycle Efficiency of a Braytonprocess, the Cycle efficiency is anchored by the inlet compression process. 14 MAE 6530 - Propulsion Systems IICycle Efficiency of Ideal RamJet (10)(Credit Narayanan Komerath, Georgia Tech)Step 1-2 Step 3 Step Inlet Compression Ratio Stagnation Pressure Loss Temperature Change across Combustor Cannot Start from Zero Velocity15 MAE 6530 - Propulsion Systems IIIdeal RamJet Example: Inlet and Diffuser Take a Rocket motor and lop the top off m fuelm airm air+fuelnormalshockwaveM > 1M < 1 Works Ok for subsonic, but for supersonic flow .. can t cram enough air down the tube Result is a normal shock waveat the inlet lipMAE 6530 - Propulsion Systems IIIdeal RamJet Example: Inlet and Diffuser(2)m fuelm airm air+fuelnormalshockwaveM > 1M < 1 Mechanical Energy is Dissipated into Heat Huge Loss in MomentumM M2 MAE 6530 - Propulsion Systems IIIdeal RamJet Example: Inlet and Diffuser (3) So.

4 We put a spike in front of the inletm fuelm airm air+fuelMuch weaker normalshockwaveM > 1M < 1 ObliqueshockwaveM M2 How does this spike Help? By forming an ObliqueShock wave ahead of the inletMAE 6530 - Propulsion Systems II2-D RamJet Inlet ExampleM1= 40 BBP0BP0B Compare MBand P0behind normal shockwavesAssume g= 6530 - Propulsion Systems II2-D RamJet Inlet Example (3)M1= From Normal Shock wave solverMAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (4)= 40 Across Oblique Shock wave M1n= M1 sin 1== 18040 sinM2n= ()=2M12sin2 () 1{}tan ()2+M12 +cos2 () 180 242 18040 sin21 18040 tan 180240 cos+ + atan= 6530 - Propulsion Systems II2-D RamJet Inlet Example (5) Across Oblique Shock wave= M2=M2nsin( 1- )= () sinP02/P0 = 40 6530 - Propulsion Systems II2-D RamJet Inlet Example (6) 242+ () + ()

5 = Across Oblique Shock wave= 40 6530 - Propulsion Systems II2-D RamJet Inlet Example (7)M1= CompareM1= Spike aids in increasing Total Pressure recoveryReducing ram drag MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (8)MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d)MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d)MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d)MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d) Compute efficiency .. Oblique shock inletMAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d) Compute TC, TBTC= TB= K== Compute efficiency .. Oblique shock inletM1= () () () MAE 6530 - Propulsion Systems II2-D RamJet Inlet Example (cont d) Compute efficiency.

6 Oblique shock inletM1= increase in efficiency!MAE 6530 - Propulsion Systems IIMulti-stage Compression always Works Best For Stagnation Pressure Recovery!MAE 6530 - Propulsion Systems IIMAE 6530 - Propulsion Systems II34 Questions??