Lect- 231 Lect-23 In this lecture... • Tutorial on axial flow turbines Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay 2 Prof. determine (a) the specific work done (b) the Mach number leaving the nozzle (c) the axial velocity (d) total to total efficiency (e) stage reaction. the total pressure and temperature are 311 kPa and 850oC respectively. At stage entry. A M Pradeep. Prof.Lect-23 Problem # 1 • A single stage gas turbine operates at its design condition with an axial absolute flow at entry and exit from the stage. The exhaust static pressure is 100 kPa. Department of Aerospace. Bhaskar Roy. IIT Bombay 3 .87 and mean blade speed is 500 m/s. The absolute flow angle at the nozzle exit is 70 deg. the total to static efficiency is 0. Assuming constant axial velocity through the stage. Bhaskar Roy. Department of Aerospace. Prof. IIT Bombay 4 . A M Pradeep.Lect-23 Problem # 1 α1 C1 1 Stator/Nozzle β2 α2 2 C2 V2 U Rotor α3 β3 V3 3 C3 Prof. the Mach number is M2 = C 2 / γRT2 From the velocity triangle. Department of Aerospace. IIT Bombay 5 . C w3 = 0.248 = 276 kJ / kg ] (b) At the nozzle exit.11)0. w t = η ts c p T01 1 − (P3 / P01 )( γ − 1) / γ [ ] = 0. Prof. A M Pradeep.Lect-23 Solution: Problem # 1 We know that total − to − static efficiency. Bhaskar Roy. w t = UC w 2 ∴ C w 2 = w t / U = 276 × 10 3 / 500 = 552 m / s Prof.87 × 1148 × 1123 × 1 − (1 / 3. wt η ts = c p T01 1 − (P3 / P01 )( γ − 1) / γ [ ] [ ∴ Specific work is. Bhaskar Roy. IIT Bombay 6 .93 Prof. A M Pradeep.static efficiency as : C 32 1 1 200 2 1 = − = − = 1.97 (c) The axial velocity. M2 = 588 / 1.0775 3 0. Department of Aerospace.87 2 × 276 × 10 η tt η ts 2w t ∴ η tt = 0.to . Prof.33 × 287 × 973 = 0.Lect-23 Solution: Problem # 1 C 2 = C w 2 / sin α 2 = 588 m / s We know that T 2 = T01 − 12 C 22 / c p = 973 K Hence. C a = C 2 cos α 2 = 200 m / s (d) The total .to .total efficiency is related to the total . Prof. IIT Bombay 7 . A M Pradeep. Department of Aerospace. Bhaskar Roy.Lect-23 Solution: Problem # 1 P01 P1 T 01 P2 1 P03 2 03 2s 03s 3 P3 C 32 2c p 3s s Expansion process in a turbine stage Prof. Lect-23 Solution: Problem # 1 (e) Degree of reaction.451 Prof. Bhaskar Roy. R x = 1 − 12 (C a / U)(tan β 3 − tan β 2 ) From the velocity triangle. A M Pradeep. IIT Bombay 8 . Department of Aerospace.2745 / 1000 = 0. Prof. tan β 3 = U / C a and tan β 2 = tan α 2 − U / C a ∴ R x = 1 − 12 (C a / U) tan α 2 = 1 − 200 × 0. Department of Aerospace.500 rpm. Prof. a flow coefficient of 0. Bhaskar Roy.12 m and the shaft speed is 10. (c) the static temperature and pressure at nozzle exit assuming a nozzle efficiency of 0.0 bar. Prof.Lect-23 Problem # 2 Combustion gases enter the first stage of a gas turbine at a stagnation temperature and pressure of 1200 K and 4. At the mean radius the stage operates with a reaction of 50%. the blade height is 0. Determine (a) the relative and absolute flow angles for the stage. A M Pradeep. (b) the velocity at nozzle exit. The rotor blade tip diameter is 0. IIT Bombay 9 .5.7 and a stage loading coefficient of 2.75m.96 and the mass flow. Bhaskar Roy. Department of Aerospace. IIT Bombay 10 . Prof. A M Pradeep.Lect-23 Solution: Problem # 2 1 C3 3 C2 U V3 2 α2 V2 C2 β2 V2 U Stator/Nozzle β3 V3 Rotor Prof. Prof. Department of Aerospace. α 2 = β 3 = 68. Bhaskar Roy.98 o Prof.2 o and α 3 = β 2 = 46. tan β 3 = (ψ / 2 + R) /(C a / U) and tan β 2 = (ψ / 2 − R) /(C a / U) Substituti ng values of ψ.2 o and β 2 = 46. (C a / U) = φ and R x β 3 = 68.Lect-23 Solution: Problem # 2 (a) The stage loading is given by ψ = Δh0 / U2 = (Vw3 + Vw 2 ) / U = (C a / U)(tan β 3 + tan β 2 ) Also.98 o For a 50% reaction stage. A M Pradeep. R x = (C a / U)(tan β 3 − tan β 2 ) / 2 Simplifyin g the above equations. IIT Bombay 11 . Department of Aerospace. rm = (0.315m the blade speed.86 m / s (c) The static temperature at the nozzle exit. Bhaskar Roy.86 2 /(2 × 1160) = 1016. T2 = T02 − C 22 / 2c p = 1200 − 652.3 K Prof.36 m / s The axial velocity. IIT Bombay 12 . Um = (10500 / 30) × π × 0. C a = φUm = 242.45 / cos 68. Prof.Lect-23 Solution: Problem # 2 (b) At the mean radius. A M Pradeep.2 = 652. C 2 = C a / cos α 2 = 242.315 = 346.12) / 2 = 0. velocity at the nozzle exit.75 − 0.45 m / s and Therefore. 84052 ηn ∴ P2 = 4 × 0. IIT Bombay 13 .1 kg / s ∴m Prof.986 bar = ρ 2 A 2 C a = (P2 / RT2 )A 2 C a The mass flow rate is m = (1. Prof.986 × 105 / 287 × 1016.0303 = 1.2375 × 242.45 = 39. A M Pradeep. Bhaskar Roy.84052 4.Lect-23 Solution: Problem # 2 1 − T2 / T01 h01 − h 2 = The nozzle efficiency. ηn = h01 − h 2s 1 − (P2 / P01 )( γ − 1) / γ ∴ (P2 / P01 )( γ − 1) / γ = 1 − 1 − T2 / T01 = 0.3) × 0. Department of Aerospace. 9. The mean blade speed is 298 m/s and the mass flow rate is 18. IIT Bombay 14 . Prof.Lect-23 Problem # 3 A single stage axial flow turbine operates with an inlet temperature of 1100 K and total pressure of 3. Bhaskar Roy. A M Pradeep. Prof. If the convergent nozzle is operating under choked condition determine (a) blade-loading coefficient (b) pressure ratio of the stage and (c) flow angles.4 bar.75 kg/s. Department of Aerospace. The turbine operates with a rotational speed of 12000 rpm. The total temperature drop across the stage is 144 K and the isentropic efficiency of the turbine is 0. Department of Aerospace. Bhaskar Roy. A M Pradeep. Prof.Lect-23 Problem # 3 α1 C1 1 Stator/Nozzle β2 α2 2 C2 V2 U Rotor α3 β3 V3 3 C3 U Prof. IIT Bombay 15 . 8615 ψ= 298 2 U2 (b) T02 = T01 = 1100 K T03 = T01 − ΔT0 = 1100 − 144 = 956 K The isentropic efficiency of a turbine. Department of Aerospace.Lect-23 Problem # 3 (a) The blade loading is defined as c p ΔT0 1148 × 144 = = 1.875 and P03 = 1. A M Pradeep.533 or η P01 T t 01 The pressure ratio of the turbine is P01 = 1. Bhaskar Roy. IIT Bombay 16 .813 bar P03 Prof. η t = = T01 − T03 T01 − T03s ΔT0 T01 1 − (P03 / P01 )( γ − 1) / γ [ ] γ /( γ − 1) ΔT0 P03 = 1 − = 0. Prof. A M Pradeep. Prof. C 2 = and γRT2 T02 γ +1 = = 1. C 2 = 600. The axial velocity C a = Uφ = 298 × 0.95 = 283 m / s.165 T2 2 The static temperature at the nozzle exit is T2 = 944.2 K. IIT Bombay 17 .Lect-23 Problem # 3 (c) Since the nozzle is choked. cos α 2 = C a / C 2 = 283 / 600 = 0. the exit Mach number is unity.4716 and α 2 = 62 o Prof. The absolute velocity of the gases leaving the choked nozzle is therefore. Bhaskar Roy. Department of Aerospace. From the velocity triangles.3 m / s. Therefore. 132 φ and therefore. A M Pradeep. Prof. Bhaskar Roy.54 o U = tan β 3 − tan α 3 Ca or tan β 3 = 1 + tan α 3 = 1. w t = c p ΔTo = UC a (tan α 2 + tan α 3 ) or tan α 3 = c p ΔTo UC a − tan α 2 = 1148 × 144 − 1.0793 298 × 283 or α 3 = 4.6 o φ The turbine specific work. IIT Bombay 18 .54 o Prof.8807 = 0.Lect-23 Problem # 3 U 1 = tan α 2 − tan β 2 = Ca φ 1 tan β 2 = tan α 2 − = 0. β 3 = 48. Department of Aerospace.828 or β 2 = 39. Mean blade speed is 250 m/s and the axial velocity is 150 m/s and is a constant across the turbine.Lect-23 Problem # 4 A multi-stage axial turbine is to be designed with impulse stages and is to operate with an inlet pressure and temperature of 6 bar and 900 K and outlet pressure of 1 bar. Estimate the number for stages required for this turbine. The isentropic efficiency of the turbine is 85 %. Prof. IIT Bombay 19 . All the stages are to have a nozzle outlet angle of 75o and equal inlet and outlet rotor blade angles. Department of Aerospace. A M Pradeep. Bhaskar Roy. Prof. IIT Bombay 20 .Lect-23 Problem # 4 1 Cw3 Cw2 C3 2 C2 α3 C β3 2 Vw3 U α2 V3 Vw2 3 V2 V2 β3 β2 V3 α2 β2 U Ca Stator/Nozzle Rotor Prof. Department of Aerospace. Prof. A M Pradeep. Bhaskar Roy. 9 K = 6 0.85(900 − 576.33 / 1. ΔT0overall = η t (T01 − T0es ) = 0. A M Pradeep. C 2 = C a / cos α 2 = 150 / cos 75 = 579.5 2 / 2 × 1148 = 753.33 Hence.5 m / s T02 = T2 + C 22 / 2c p → T2 = T02 − C 22 / 2c p Since.Lect-23 Solution: Problem # 4 Since the overall pressure ratio is known. there is no change in stagnation temperature in the nozzle.9) = 274. the degree of reaction. IIT Bombay 21 .6 K From the velocity triangles. T2 = T01 − C 22 / 2c p = 900 − 579. Prof.7 K Since this is an impulse turbine. ( γ − 1) / γ P T01 = 01 T0es P0e ∴ T0es = 576. Department of Aerospace. R x = 0 Rx = h 2 − h3 h01 − h03 or h 2 = h3 → T2 = T3 = 753. Bhaskar Roy.7 K Prof. 14 2 / 2 × 1148 = 805. tanβ 2 = (C 2 sin α 2 − U) / C a = (579.065 ∴ β 2 = 64.15 m / s We can see that V2 = V3 = C 3 for cons tan t axial velocity.28 K The temperature drop per stage is T01 − T03 = 900 − 805.28 = 94. T03 = T2 + C 32 / 2c p = 753.Lect-23 Solution: Problem # 4 From the velocity triangles at rotor entry. Prof. A M Pradeep.6 / 94. Bhaskar Roy.89 ≅ 3 stages.7 K The number of stages required for the turbine is ΔT0overall /(T01 − T03 ) = 274. Therefore.16 o V2 = C a / cos β 2 = 344. Department of Aerospace.5 sin 75 − 250) / 150 = 2.7 = 2.7 + 344. Prof. IIT Bombay 22 . Lect-23 Exercise Problem # 1 An axial flow turbine operating with an overall stagnation pressure of 8 to 1 has a polytropic efficiency of 0. If. Prof. determine the inlet total temperature. in addition. Bhaskar Roy.85. 86. If the exhaust Mach number of the turbine is 0. Determine the total-to-total efficiency of the turbine. IIT Bombay 23 . the exhaust velocity of the turbine is 160 m/s.3.17%. Department of Aerospace.6 K Prof. A M Pradeep. Ans: 88%. 1170. determine the total-to-static efficiency. β2=7.Lect-23 Exercise Problem # 2 The mean blade radii of the rotor of a mixed flow turbine are 0.000 rev/min and the turbine is required to produce 430kW. The rotor rotates at 20.02 deg. β3=50.3 m at inlet and 0. Determine the absolute and relative flow angles and the absolute exit velocity if the gas flow is 1 kg/s and the velocity of the through-flow is constant through the rotor. Ans: α2=70 deg. Prof. The flow velocity at nozzle exit is 700 m/s and the flow direction is at 70° to the meridional plane. A M Pradeep.37 deg Prof.1 m at outlet. IIT Bombay 24 .4 deg. Bhaskar Roy. α3=18. Department of Aerospace. Prof. Bhaskar Roy. Department of Aerospace. (c) the total-to-static efficiency. 0. determine (a) the nozzle exit velocity. 0. At the stage entry the stagnation pressure and temperature are 772 kPa and 727°C. A M Pradeep.128 Prof.2 kg/s. Ans: 488m/s.36MW at a mass flow rate of 27.1 m/s. (b) the blade speed. Assuming the axial velocity is constant across the stage and the gas enters and leaves the stage without any absolute swirl velocity. (d) the stage reaction. respectively.83.Lect-23 Exercise Problem # 3 An axial flow gas turbine stage develops 3. IIT Bombay 25 . 266. The static pressure at exit from the nozzle is 482 kPa and the corresponding absolute flow direction is 72° to the axial direction. Ans: 13550 rpm. A M Pradeep.6. (b) stage pressure ratio (c) flow coefficient (d) degree of reaction and (e) the power delivered by the turbine. If the stage efficiency is 0.Lect-23 Exercise Problem # 4 A single stage axial turbine has a mean radius of 30 cm and a blade height at the stator inlet of 6 cm. The gases enter the turbine stage at 1900 kPa and 1200 K and the absolute velocity leaving the stator is 600 m/s and inclined at an angle of 65 deg to the axial direction. The relative angles at the inlet and outlet of the rotor are 25 deg and 60 deg respectively.41. 2.88.6 MW Prof. 34. 0. IIT Bombay 26 . 0. calculate (a) the rotor rotational speed. Department of Aerospace.346. Prof. Bhaskar Roy.