1. Fanno Flow -1 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Fanno Flow - Thermodynamics Steady, 1-d, constant area, adiabatic flow with no external work but with friction Conserved quantities –since adiabatic, no work: h o =constant –since A=const: mass flux=  v=constant L pTvMpTvM  x, f ? –combining: h o =h+ (  v) 2 /2  =constant On h-s diagram, can draw Fanno Line –line connecting points with same h o and  v

2. Fanno Flow -2 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Fanno Line Velocity change (due to friction) associated with entropy change h s hoho v=0 s max v max M<1 M>1 (v)(v) (v)(v) M=1 Friction can only increase entropy Two solutions given (  v,h o,s): subsonic & supersonic –change mass flux: new Fanno line –friction alone can not allow flow to transition between sub/supersonic –can only approach M=1

3. Fanno Flow -3 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Fanno Line - Choking Total friction experienced by flow increases with length of “flow”, e.g., duct length, L For long enough duct, M e =1 (L=L max ) What happens if L>L max –flow already “choked” –subsonic flow: must move to different Fanno line (– – –), i.e., lower mass flux –supersonic flow: get a shock ( – – – ) h s hoho M<1 M>1 (v)(v) M=1 M1M1 MeMe L

4. Fanno Flow -4 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Fanno Line – Mach Equations Simplify (X.4-5) for  q=dA=0 (X.7) (X.6) (X.8) (X.9) (X.10) can write each as only f(M) p o loss due to entropy rise

5. Fanno Flow -5 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Property Variations Look at signs of previous equations to see how properties changed by friction as we move along flow –(1-M 2 ) term makes M 1 M<1M>1 s p o M h,T p  v Friction increases s,  p o drop Friction drives M  1 h o,T o const: h,T opposite to M p,  same as T (like isen. flow)  v=const: v opposite of 

6. Fanno Flow -6 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 A Solution Method Need to integrate (X.6-10) to find how properties change along length of flow (fdx/D) –can integrate or use tables of integrated values function of Reynolds number (v) and surface roughness Mach number variation 1) use avg. f 2) to tabularize solution, use reference condition: M 2 =1, L 2 =L max M1M1 M2M2 x1x1 x2x2

7. Fanno Flow -7 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Use of Tables To get change in M, use change in f L max /D (like using A/A*) Find values in Appendix E in John so if you know f L/D and M 1, 1) look up f L max /D at M 1 2) calculate f L max /D at M 2 3) look up corresponding M 2 (X.11) M1M1 M2M2 L max,2 L max,1 M=1 L

8. Fanno Flow -8 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 TD Property Changes To get changes in T, p, p o,... can also use M=1 condition as reference condition (*) Integrate (X.7-10), e.g., M1M1 M2M2 p 2, T 2, p o2 L max,2 L max,1 M*=1 p 1, T 1, p o1

9. Fanno Flow -9 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Fanno Flow Property Changes Summarize results in terms of reference conditions In terms of initial and final properties (X.12) (X.15) (X.13) (X.14) (X.16) (T o =const) (X.19) (X.17) (X.18)

10. Fanno Flow -10 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example Given: Exit of supersonic nozzle connected to straight walled test section. Test section flows N 2 at M test =3.0, T o =290. K, p o =500. kPa, L=1m, D=10 cm, f=0.005 Find: - M, T, p at end of test section - p o,exit /p o,inlet - L max for test section Assume: N 2 is tpg/cpg,  =1.4, steady, adiabatic, no work M test L

11. Fanno Flow -11 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution Analysis: –Me–Me –T–T another solution is M=0.605, but since started M>1, can’t be subsonic (X.11) M 1 =M test L M 2 =M e (Appendix E) (T o const)

12. Fanno Flow -12 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution (con’t) –p o,e / p o,test 25% loss in stagnation pressure due to friction (X.18) (X.17) M 1 =M test L M 2 =M e –p–p

13. Fanno Flow -13 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution (con’t) 10 m long section would have M=1 at exit M 1 =M test L M 2 =M e –L max

14. Fanno Flow -14 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 L<L max, Back Pressure Last problem (supersonic duct), what would happen if calculated exit pressure (p e,f ) did not match actual back pressure (p b ) p e,f < p b <p e,sh : oblique shocks outside duct (overexpanded) shock inside p e,sh shock at exit O U p*/p o x 1 p e,f p/p o1 p e,sh <p b : shocks inside duct (until shock reaches ~throat) p b <p e,f : expansion outside duct (underexpanded) M 1 =M test L M 2 =M e pepe pbpb p o1

15. Fanno Flow -15 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 L>L max, Back Pressure Can’t have flow transition to subsonic with pure Fanno flow  shock in duct p*/p o x 1 pbpb p/p o1 Shock location determined by back pressure –raise p b –shock moves upstream until shock reaches M=1 location in nozzle M 1 =M test L M 2 =M e pepe pbpb p o1