of Compressors and Turbines (AE 651) Autumn Semester 2009 - PowerPoint PPT Presentation

Aerodynamics of Compressors and Turbines (AE 651) Autumn Semester 2009 Instructor : Bhaskar Roy Professor, Aerospace Engineering Department I.I.T., Bombay e-mail : aeroyia@aero.iitb.ac.in 1

h-s of Axial Turbine Stage Axial Turbines 2 2 AE 651 - Prof Bhaskar Roy, IITB Lect - 12

Axial Turbines Isentropic Efficiencies ; ΔT Total-to-total η = ΔT / 0T efficiency , 0T 0T ΔT ΔT η = = T T Static-to-static T ΔT / ΔT / + ΔT / efficiency , T Stator Rotor AE 651 - Prof Bhaskar Roy, IITB 3 Lect - 12

Axial Turbines Isentropic Efficiencies ; Total-to-static ΔT ΔT efficiency, η = = 0T 0T TS ΔT ΔT + ΔT / / / T Stator Rotor Total-to-total ΔT T -T isentropic η = 0-Rotor 02 03 efficiency of 0-Rotor ΔT T -T / / the rotor only is, 0-Rotor 02 03 H = U C +C Work Done w w Th 2 3 AE 651 - Prof Bhaskar Roy, IITB 4 Lect - 12

Axial Turbines Degree of Reaction Theoretical DoR , Static isentropic enthalpy drop in rotor D.R = isen Static isentropic enthalpy drop in rotor and stator / / h h rotor rotor = = / / / h h + h rotor stator T Actual DoR , 2 2 V -V h h rotor rotor 3 2 D.R = = = actual h + h h 2h rotor stator T T AE 651 - Prof Bhaskar Roy, IITB 5 Lect - 12

Axial Turbines Degree of Reaction These are two definitions are related by , / // 2 1- D.R T -T actual 2 3 = 1.03, where, = 0 .97 - 0 .98 η T -T / 1- D.R isen T 2 3 where is the thermodynamic loss coefficient The change in DR due to real flow effects is shown below 0 0.1 0.25 0.35 0.45 0.5 DR 0.03 0.073 0.226 0.33 0.433 0.485 DR act AE 651 - Prof Bhaskar Roy, IITB 6 Lect - 12

Axial Turbines Total heat drop, assuming C 1 to be zero, i.e. flow starting from static condition 2 C 3 H = H + 2 0T T H is converted to turbine work T 2 C is the exhaust flow kinetic energy 3 2 C 2 3 H = H .w + 2 0T Th rotor H is the theoretical enthalpy drop Th Rotor loss coefficient, - ΔT / = ΔT w rotor 023-relative 023-relative AE 651 - Prof Bhaskar Roy, IITB 7 Lect - 12

Axial Turbines This Implies that , 2 2 V - V 3 2 D.R= 2 C3 2H .w 1+ rotor Th 2H .wrotor Th 2 2 C -C 2 3 H - Th 2 = 2 C3 2H .w 1+ rotor Th 2H .wrotor Th AE 651 - Prof Bhaskar Roy, IITB 8 Lect - 12

Axial Turbines This Implies that , 2 2 V - V 3 2 D.R= 2 C3 2H .w 1+ rotor Th 2H .wrotor Th 2 2 C -C 2 3 H - Th 2 = 2 C3 2H .w 1+ rotor Th 2H .wrotor Th AE 651 - Prof Bhaskar Roy, IITB 9 Lect - 12

Axial Turbines C C Assuming that entry velocity is not negligible , i.e. 1 3 2 2 C + C C - C w1 w2 a1 a2 1 - - 2.U 2H Th DR = 2 2 C - C 3 1 w 1+ 2H rotor Th A simplified DoR as Ψ C 1 = C 3 C + C / used in Compr is : t1 t2 DR = 1 - = 1 - C a1 =C a2 2.U 2 Combining the two above equations a simplified DR is written: 2 2 C C / - cos α 1 - 3 2 a1 DR = DR .w + DR actual actual rotor actual 3 2 2H C Th a2 AE 651 - Prof Bhaskar Roy, IITB 10 Lect - 12

Axial Turbines / Further simplification may be made DR / ω = DR DR DR Second term is zero, and If rotor Ca / DR DR 1 = 1- = 1- 2 2 C cos α ω .cos α a rotor 3 3 2 If, from experience, a certain value of DR is assumed, we 2 is selected and w rotor is get a value of C a1 /C a2 . If available from cascade data. The axial velocity ratio found here is the starting point of detailed analysis – and fixing of annular flow track AE 651 - Prof Bhaskar Roy, IITB 11 Lect - 12

Axial Turbines Net reversible polytropic expansion 1 work = 1-4-3-2-1 = v.dp 2 Total real expansion work 1 2 2 C -C 1 2 = v.dp+ L + R 2 2 AE 651 - Prof Bhaskar Roy, IITB Lect - 12

Axial Turbines Expansion process is accompanied by internal heat exchange (conversion of irreversible losses L R to Q R , Q 0 internally) and external heat exchange R Q 0 And external heat exchange 1 q v.dp Irreversible polytropic expansion work 2 k = 1 1 k -1 2 p.dv k k p 2 1 2 H = p .v 1- 2 T 1 1 poly k - 1 p p 1 1 ln 1 p 2 k = 2 v ln 1 v 2 AE 651 - Prof Bhaskar Roy, IITB Lect - 12

Axial Turbines Again, for all practical considerations, k 1 = k 2 =k . In aircraft turbine generally and or k = 1.28 or 1.29, which tend to go up in the rear stages, as temperature drops. The relation between and k is given by : ∂Q ± ∂Q -1 -1 R R q q 1+ . R.dT k = . ∂Q ± ∂Q R R q q 1+ -1 . R.dT AE 651 - Prof Bhaskar Roy, IITB Lect - 12

Axial Turbines ( 1- 2)= ∂Q = 0, ∂Q > 0, L = Q =1- 2 - 3 - 4 - 1 q R R R ( 1- 2 ) ∂Q > 0, ∂Q > 1- 2", ∂Q < 0, ∂Q > / 0 0 q q R R q q R R AE 651 - Prof Bhaskar Roy, IITB Lect - 12