Download - Section 4.3: The RamJet Propulsion Cycle
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MAE 6530 - Propulsion Systems II
Section 4.3: The RamJet Propulsion Cycle
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MAE 6530 - Propulsion Systems II
Background 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 flight
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MAE 6530 - Propulsion Systems II
Control Volume for a Ramjet
p∞T∞V∞ρ∞
peTeVeρe
p2V2T2
ρ1V1
Fthrust =!mair + !mfuel( )⋅Vexit− !mair( )⋅V∞+ Aexit ⋅ pexit− p∞( )
p∞ ⋅ A0=
Fthrust = !mair ⋅V∞f +1f
⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟⎟⋅VexitV∞−1
⎡
⎣⎢⎢
⎤
⎦⎥⎥+AexitA0⋅pexitp∞−1
⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟⎟
f =!mair!mfuel
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MAE 6530 - Propulsion Systems II
Ideal Ramjet Thermodynamic Cycle Analysis Step 1-2 Step3 Step4
Region Process Ideal Behavior RealBehavior
A to 1(inlet) Isentropic flow P0,T0 constant P0 drop
∞-1-2 (diffuser) Adiabatic Compression
P,T increaseP0 drop
P0 drop
2-3 (burner) Heat Addition P0 constant, T0s Increase
P0 drop
3-4 (nozzle) Isentropic expansion
T0,P0 constant s IncreaseT0 drop 4
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MAE 6530 - Propulsion Systems II
Ideal Ramjet Cycle Analysis (2)
T-s Diagram
5
Step Process1) Intake (suck) Isentropic Compression2) Compress the Air (squeeze) Adiabatic Compression3) Add heat (bang) Constant Pressure Combustion4) Extract work (blow) Isentropic Expansion in Nozzle5) Exhaust Heat extraction by surroundings
Step1-2
Step3
Step4
Step5
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet
• Propulsive Power Output--> work perform by system in step 4minus work required for step 1-2
• Net Heat Input --> heat input during step 3 (combustion)- heat lost in exhaust plume
6
6
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (2)
7
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (3)
8
Add and Subtract From Right Hand Side
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (4)
• Assume … calorically perfect gasses à h ~ Cp•T
9
• For conceptual simplicity … let … f >> 1 … and … Cpair ~ Cpproducts
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (5)
• From C-->D flow is isentropic …(CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
10
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (6)
• Approximate as Adiabatic compression across diffuser (CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (7)
(CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
• Sub
into efficiencyequation
12
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (8)
(CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
13
Factor Out
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (9)
(CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
1. Cycle Efficiency Proportional to Inlet Pressure ratio, PB/PA
2. Cycle Efficiency Proportional to combustor temperature difference TC-TB
3. As inlet total pressure ratio (P0B/P0A) goes down …Cycle Efficiency Drops
4. Characteristic of a Brayton process, the cycle efficiency is anchored by the inlet compression process.
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MAE 6530 - Propulsion Systems II
Cycle Efficiency of Ideal Ramjet (10)
(CreditNarayananKomerath,GeorgiaTech)
Step1-2 Step3 Step4
1. High Inlet Compression Ratio Desirable2. Low Stagnation Pressure Loss Desirable3. Large Temperature Change across Combustor Desirable4. Ramjets Cannot Start from Zero Velocity
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MAE 6530 - Propulsion Systems II
Ideal Ramjet Example: Inlet and Diffuser
• Take a Rocket motor and “lop the top off”m•fuel
m• air m• air+fuelnormal
shockwaveM > 1 M < 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 lip
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MAE 6530 - Propulsion Systems II
Ideal Ramjet Example: Inlet and Diffuser (2)
m•fuel
m• air m• air+fuelnormal
shockwaveM > 1 M < 1
• Mechanical Energy is Dissipated into Heat
• Huge Loss in Momentum
M¥ M2
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MAE 6530 - Propulsion Systems II
Ideal Ramjet Example: Inlet and Diffuser (3)
• So … we put a spike in front of the inletm•fuel
m• air m• air+fuel
Much weaker normalshockwave
M > 1 M < 1
Obliqueshockwave
M∞ M2 • How does this spike Help?
• By forming an ObliqueShock wave ahead of the inlet
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example
M1=4.0
b=40°
B
B
P0B
P0B
•CompareMB andP0 behindnormalshockwaves
Assumeg=1.4
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (3)
M1=4.0B
P0B
• From Normal Shock wave solver
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (4)
=40deg.
B
•AcrossObliqueShockwave
•M1n=M1sin 1 = =2.5714 π180
40⎝ ⎠⎛ ⎞sin M2n=0.5064
tan θ( ) = 2 M12 sin2 β( ) −1{ }
tan β( ) 2 + M12 γ + cos 2β( )⎡⎣ ⎤⎦⎡⎣ ⎤⎦
→180π
2 42π180
40⎝ ⎠⎛ ⎞sin2 1−⎝ ⎠
⎛ ⎞
π180
40⎝ ⎠⎛ ⎞tan⎝ ⎠
⎛ ⎞ 2 42 1.4π180
2 40⋅⎝ ⎠⎛ ⎞cos+⎝ ⎠
⎛ ⎞+⎝ ⎠⎛ ⎞
⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞
atan
=26.2 deg
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (5)
•AcrossObliqueShockwave
=2.123M 2n = 0.5064→M2=M2n
sin(β1-θ)=
0.5064π180
40 26.2−( )⎝ ⎠⎛ ⎞sin
P02/P0∞ =0.4711
=40deg.
B
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (6)
0.3126 1 1.4 1−2
42+⎝ ⎠⎛ ⎞
1.41.4 1−( )
⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞
1 1.4 1−2
0.5578532+⎝ ⎠⎛ ⎞
1.41.4 1−( )
⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞
=38.422
•AcrossObliqueShockwave
=40deg.
B
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (7)
M1=4.0
•Compare
M1=4.0
•SpikeaidsinincreasingTotalPressurerecoveryReducing“ramdrag”
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (8)
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
•Computeefficiency….Obliqueshockinlet
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
• Compute TC, TB
TC =
TB = 1325.1ºK
856.61 ºK =
= 0.4652
• Compute efficiency …. Oblique shock inlet
M1=4.0
1
38.4221.4 1−( )−1.4 1325.1 0.3126
1.4 1−( )1.4
⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞
856.61−⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞
1325.1 856.61−( )−
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MAE 6530 - Propulsion Systems II
2-D Ramjet Inlet Example (cont’d)
• Compute efficiency …. Oblique shock inlet
M1=4.0
= 0.4652
M1=4.0 = 0.2328
200% increase in efficiency!
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MAE 6530 - Propulsion Systems II
Multi-stage Compression always Works Best For Stagnation Pressure Recovery!
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MAE 6530 - Propulsion Systems II
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MAE 6530 - Propulsion Systems II 34
Questions??