1. MAE 5360: Hypersonic Airbreathing Engines
Rayleigh Flow and Fanno Flow Overview
Mechanical and Aerospace Engineering Department
Florida Institute of Technology
D. R. Kirk

2. Rayleigh Flow and Fanno Flow

3. Rayleigh Flow Property Summary
Heating
Cooling
Subsonic
Supersonic
Total Temperature, Tt ↑ ↓
Mach Number, M
Static Pressure, p
Density, r
Velocity, V
Total Pressure, Pt
Entropy, s
Static Temperature, T
Note 1
Note 2
Note 1: Increases up to M=1/√g and then decreases
Note 2: Decreases up to M=1/√g and then increases

4. Rayleigh Flow Summary Subsonic Inlet Heating:
No limit to heat addition
Flow chokes more and more as we add more heat with inlet velocity → 0.
Cooling:
No theoretical limit on amount of cooling allowed
Exit flow becomes slower and slower
Exit T → 0
Supersonic Inlet
Even if M1 → ∞, Tt1/Tt* =
If heat is added without limit to supersonic flow, normal shock wave adjustment required to accommodate property changes
Only finite amount of cooling can be allowed before exit Mach number → ∞
Tt2/Tt* =

5. Example 1 Fuel-Air mixture enters a combustor duct Inlet conditions:
V1=75 m/s
P1=150 kPa
T1=300 K
There is heat transfer to the fluid at a rate of 900 kJ/kg
Assume that the flow is inviscid
Calculate
All properties at duct exit
Plot Rayleigh line on a T-S diagram
Increase heating to 1400 kJ/kg

6. Example 1

7. Example 1

8. Example 2
Air enters a constant area duct of circular cross-section with diameter D = 20 cm
Inlet conditions:
M1=0.2
Pt1=100 kPa
Tt1=288 K
There is heat transfer to the fluid at a rate of 100 kW
Assume that the flow is inviscid
Calculate
Mass flow rate through the duct
The critical heat flux that would choke the duct for given M1
The exit Mach number, M2
The percentage total pressure loss
Entropy rise, DS
Static pressure drop
Plot Rayleigh line on a T-S diagram

9. Effect of Heat Transfer on Mach Number
Heating
Cooling
Heating
Cooling
D=20 cm, mdot= kg/s, M1=0.2, Max Q=3713 kW
Plot shows T and Tt vs. Mach number in the duct
Heating increases Tt and cooling decreases Tt . The maximum Tt occurs at M=1.0
Whether the inlet is subsonic or supersonic, heating drives the flow toward Mach 1
T increases from M=0 to M=1/√g and then decreases

10. Example 2

11. Example 2

12. Example 2

13. Example 3: SCRAMJet Combustor
Air enters a constant area combustion chamber
Inlet conditions:
M1=3.0
Pt1=100 kPa
Tt1=1,800 K
Fuels (assume f << 1)
n-Decane (C10H22): QR=48,000 kJ/kg
Methane (C10H22): QR=55,500 kJ/kg
Hydrogen (H2): QR=141,800 kJ/kg
Assume that the flow is inviscid
Calculate
Exit total temperature if the exit is choked
Maximum heat release per unit mass of air
Required fuel-to-air ratio to thermally choke the combustor exit
Total pressure loss in the supersonic combustor

14. Example 3: SCRAMJet Combustor

15. Example 3: SCRAMJet Combustor

16. Rayleigh Flow and Fanno Flow

17. Rayleigh Flow and Fanno Flow: T-s

18. Rayleigh Flow and Fanno Flow: T-s

19. Fanno Flow Example 1
Air enters a constant area duct of circular cross-section
D=10 cm
Length of the duct: L=20 m
Duct is insulated and flow inside of duct can be considered adiabatic
Average wall friction coefficient, cf=0.005
Inlet Mach number: M1=0.24
Calculate the following:
The choking length of the duct, L1*
The exit Mach number, M2
The percentage total pressure drop

20. Fanno Flow Example 1: Local Mach vs. Duct Length

21. Fanno Flow Example 1: T-s

22. Fanno Flow Example 2
Air enters a duct with rectangular cross section of 1 cm by 2 cm
Average wall friction coefficient, cf=0.005
Inlet Mach number: M1=0.5
Calculate the following:
The choking length of the duct
The new inlet conditions (M1 and mass flow) if Lnew=2.16 L

23. Fanno Flow Example 2: T-s

24. Example 3: Supersonic Duct Flow
Mach 3 flow enters isolator of hypersonic vehicle
Skin friction coefficient, cf=0.05
Consider several cases of isolator length, L
L/D=1
L/D=4
L/D=6
L/D=8 (case does not work – why?)

25. Fanno Flow Example 3: L/D = 1, (L/D)*=2.61

26. Fanno Flow Example 3: L/D = 4

27. Fanno Flow Example 3: L/D = 6

28. Fanno Flow Example 3: L/D = 8

29. Fanno Flow Example 3: L/D = 8
This case does not make physical sense (entropy decrease! negative Lx)
What is physically happening here?
Important to understand this phenomena for hypersonic vehicles