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Solve Problems – Ideal Cycle Analysis of Air Breathing Engines 34 1 Lect-34 In this lecture ... • Solve problems – Ideal cycle analysis of air breathing engines 2 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Problem # 1 • The following data apply to a turbojet flying at an altitude where the ambient conditions are 0.458 bar and 248 K. • Speed of the aircraft: 805 km/h • Compressor pressure ratio: 4:1 • Turbine inlet temperature: 1100 K • Nozzle outlet area 0.0935 m2 • Heat of reaction of the fuel: 43 MJ/kg Find the thrust and TSFC assuming cp as 1.005 kJ/kgK and γ as 1.4 3 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Ideal cycle for jet engines Combustion chamber/burner Diffuser Compressor Turbine Nozzle a 1 2 3 4 5 6 7 Afterburner Schematic of a turbojet engine and station numbering scheme 4 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Ideal cycle for jet engines 4 T 5 3 7 2 a s Ideal turbojet cycle (without afterburning) on a T-s diagram 5 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 • Speed of the aircraft = 805x1000/3600=223.6 m/s • Mach number = 223.6/√(γRT) = 223.6/ √(1.4x287x248) = 0.708 • Intake: γ −1 1.4 −1 T = T 1+ M 2 = 2481+ 0.7082 = 272.86 K 02 a 2 2 γ /(γ −1) T = 02 = 1.4/(1.4−1) = P02 Pa 0.458(272.86 / 248) 0.639 bar Ta 6 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 • Compressor: P03 = π c P02 = 4×0.639 = 2.556 bar (γ −1)/γ (1.4−1)/1.4 T03 = T02 (π c ) = 272.86(4) = 405.63K • Combustion chamber: From energy balance, h04 = h03 + fQR T /T −1 or, f = 04 03 QR / c pT03 −T04 /T03 1100 / 405.63−1 = = 0.017 (43×106 /1005× 405.63) −1100 / 405.63 7 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 • Turbine: Since the turbine produces work to drive the compressor, Wturbine = Wcompressor m tc p (T04 −T05 ) = m ac p (T03 −T02 ) T05 = T04 − (T03 −T02 ) /(1+ f ) =1100 − (405.63− 272.86) /(1+ 0.017) = 969.45K γ /(γ −1) T = 05 = 1.4/(1.4−1) Hence, P05 P04 2.556(969.45/1100) T04 =1.642 bar 8 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 • Nozzle: we first check for choking of the nozzle. • The nozzle pressure ratio is P05/Pa=1.642/0.458=3.58 • The critical pressure ratio is γ /(γ −1) 1.4/(1.4−1) P γ +1 1.4 +1 05 = = = * 1.893 P 2 2 • Therefore the nozzle is choking. • The nozzle exit conditions will be determined by the critical properties. 9 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 * 2 2 T7 = T = T05 = 969.5 = 807.92 K γ +1 1.4 +1 1 1.642 = * = = = P7 P P05 * 0.867 P04 / P 1.893 5 3 ρ7 = P7 / RT7 = 0.867×10 /(287×807.92) = 0.374kg/m Therefore, ue = γRT7 = 1.4× 287×807.92 = 569.75 m/s The mass flow rate is, m = ρ7 A7ue =19.92 kg/s 10 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 1 * The thrust developed is ℑ = m [(1+ f)ue − u]+ A7 (P − Pa ) =19.92[(1+ 0.017)569.75 − 223.6] + 0.0935(0.867 − 0.458)×105 =10.912 kN Fuel flow rate, m f = f × m a = 0.017×19.92 = 0.3387 kg/s −5 Therefore, TSFC = m f / ℑ = 3.1×10 kg/Ns = 0.111 kg/N h 11 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Problem # 2 • The following data apply to a twin spool turbofan engine, with the fan driven by the LP turbine and the compressor by the HP turbine. Separate hot and cold nozzles are used. • Overall pressure ratio: 19.0 • Fan pressure ratio: 1.65 • Bypass ratio: 3.0 • Turbine inlet temperature: 1300 K • Air mass flow: 115 kg/s • Find the sea level static thrust and TSFC if the ambient pressure and temperature are 1 bar and 288 K. Heat of reaction of the fuel: 43 MJ/kg 12 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Ideal turbofan engine 2’ 3’ Secondary 7’ Diffuser nozzle Fan Combustion chamber/burner Compressor Turbine Primary nozzle a 1 2 3 4 5 6 7 Schematic of an unmixed turbofan engine and station numbering scheme 13 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Since we are required to find the static thrust, the Mach number is zero. • Intake: γ −1 T = T 1+ M 2 = 288 K 02' a 2 γ /(γ −1) T = 02' = P02' Pa 1 bar Ta • Fan: Fan pressure ratio is known:π f = P03' / P02' P03' = π f P02' =1.65 bar (γ −1)/γ (1.4−1)/1.4 T03' = T02' (π f ) = 288(1.65) = 332.35 K 14 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Compressor: π c = Overall pressure ratio/1.65 =19 /1.65 =11.515 P03 = π c P02 =11.5151×1.65 =19.0 bar (γ −1)/γ (1.4−1)/1.4 T03 = T02 (π c ) = 332.35×(11.515) = 668.53 K • Combustion chamber: From energy balance, T /T −1 f = 04 03 QR / c pT03 −T04 /T03 1300 / 668.53−1 = = 0.01522 (43×106 /1005×668.53) −1300 / 668.53 15 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • High pressure turbine: m tc p (T04 −T05' ) = m aH c p (T03 −T02 ) Here, T05' is the temperature at the HPT exit. ∴T05' = T04 − (T03 −T02 ) /(1+ f ) =1300 − (668.53− 332.53) /(1+ 0.01522) = 969.04 K γ /(γ −1) 1.4/(1.4−1) T 969.04 = 05' = = Hence, P05' P04 19 6.79 bar T04 1300 16 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Low pressure turbine: m tc p (T05' −T05 ) = m aCc p (T03' −T02' ) Here, T05' is the temperature at the HPT exit/LPT inlet. m aC ∴ T05 = T05' − Β(T03' −T02' ) /(1+ f ), where, Β = m aH = 969.04 − 3×(332.35 − 288) /(1+ 0.01522) = 837.98 K γ /(γ −1) 1.4/(1.4−1) T 837.98 = 05 = = And, P05 P05' 6.79 4.08 bar T05' 969.04 17 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Primary nozzle: we first check for choking of the nozzle. • The nozzle pressure ratio is P05/Pa=4.08/1=4.08 bar • The critical pressure ratio is γ /(γ −1) 1.4/(1.4−1) P γ +1 1.4 +1 05 = = = * 1.893 P 2 2 • Therefore the nozzle is choking. • The nozzle exit conditions will be determined by the critical properties. 18 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 * 2 2 T7 = T = T05 = 837.98 = 698.32K γ +1 1.4 +1 1 4.08 = * = = = P7 P P05 * 2.155 bar P05 / P 1.893 Therefore, ue = γRT7 = 1.4× 287×698.32 = 529.7 m/s 19 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Secondary nozzle: • The nozzle pressure ratio is P03’/Pa=1.65/1=1.65 bar • The critical pressure ratio is γ /(γ −1) 1.4/(1.4−1) P γ +1 1.4 +1 05 = = = * 1.893 P 2 2 • Therefore the nozzle is not choking. (γ −1)/γ ∴uef = 2c pT03' [1− (Pa / P03' ) ] = 2×1005×332.35[1− (1/1.65)(1.4−1)/1.4 ] = 298.52 20 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Thrust, ℑ = m aH [(1+ f )ue − u]+ Βm aH (uef − u) assuming (Pe − Pa )Ae to be negligible. m aC / m aH = 3.0, m aH + m aC =115 kg/s ∴m aH =115/ 4 = 28.75 kg/s ℑ = 28.75[(1+ 0.01522)×529.7 − 0] + 3× 28.75(298.52 − 0) = 40.74 kN 21 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Solution: Problem # 2 • Exercise: calculate the thrust by factoring the pressure thrust term as well. Hint: you can calculate the exit area from mass flow, density and exhaust velocity. • TSFC, Fuel flow rate, m f = f × m a = 0.01522× 28.75 = 0.4376 kg/s −5 Therefore, TSFC = m f / ℑ =1.075×10 kg/Ns = 0.0388 kg/N h 22 Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay Lect-34 Problem # 3 • A helicopter using a turboshaft engine is flying at 300 km/h at an altitude where the ambient temperature is 5oC. Determine the specific power output and thermal efficiency.
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