ME 440 Aerospace Engineering Fundamentals

Fall 00 ME 440 Aerospace Engineering Fundamentals Propulsion Examples Example: Compressor Turbine Determine the outlet temperature and pressure for a turbine whose purpose is to power the compressor described below: Compressor Inlet Condition: 80 K and 80 kpa Compressor Outlet Pressure: 400 kpa Compressor Efficiency: 0.78 Turbine Inlet Conditions: 500 K and 400 kpa Turbine Efficiency: 0.9 Solution: For a compressor turbine we have w t = w c From the first law we will have w t = c P (T 4 -T ) so we will need to calculate the work of the compressor. It will be determined from wc,ideal wc = ηc The ideal compressor work is given by w c,ideal = c p T P ( γ ) / γ For the properties we will use γ =.4 c P = 005 J/(kg K) and (.4 ) /.4 400 wc,ideal = (005)(80) = 4,87 J/kg 80 4,87 w c = = 0,4 J/kg 0.78

ME 440 Aerospace Engineering Fundamentals Fall 00 The actual exit temperature for the turbine is calculated w t 0,4 T4 = T - = 500 - = 90 K cp 005 The exit pressure will be determined from the isentropic relationship γ /( γ ) T4,ideal P4 = P T which requires us to determine the ideal outlet temperature. It will come from the st law, w t,ideal T4, ideal = T - cp where w t 0,4 w t.ideal = = =,478 J/kg ηt 0.9,478 T 4, ideal = 500 - = 75 K 005 so that P.4 /(.4 ) 4 = 75 = (400) 500 kpa Example: Simple Turbojet Engine Consider a jet aircraft flying at 00 m/s at an altitude of,000 m. The jet operates with a simple, ideal turbojet engine. The burner of the engine operates at 00 kpa and 800 K. Determine a. specific thrust produced b. propulsive efficiency Solution: We begin with a block diagram of the system. 4 5 Diffuser Compressor Burner Turbine Nozzle

ME 440 Aerospace Engineering Fundamentals Fall 00 Now we set up our table. Node T(K) P(kPa) V (m/s) 9 9. 00 4 8. 0 00 0 4 800 00 0 5 49 7.9 0 779 9. 97 Bold values are calculated We begin by entering our operating information. For state we go to the standard atmosphere table and find T = 9 K and P = 9. kpa This will also be our pressure at the exhaust. We now traverse the cycle. - Diffuser: Isentropic The st law gives c (T - T ) P = V V (00) T T = + = 9 + = 4 K cp (005) Using the isentropic relationship we have P γ /( γ ).4 /(.4 ) 4 = (9.) = 9 T = P T - Compressor: Isentropic Our isentropic condition is ( γ ) / γ 8. kpa T = T P However, at this point we do not know P, so we must just move on -4 Burner: Isobaric P = P 4 = 00 kpa

ME 440 Aerospace Engineering Fundamentals Fall 00 4-5 Turbine: Isentropic Since this the purpose of the compressor is to power the turbine, we have w t = w c But we have not done the compressor calculation, so we must move on. 5- Nozzle: Isentropic Our isentropic condition is ( γ ) / γ T = T5 P5 But we do not know T 5 or P 5 and we must move on. Second Pass - Diffuser: Isentropic Done - Compressor: Isentropic Our isentropic condition is T = T P We now know P, so T ( γ ) / γ (.4 ) /.4 = 00 = 4 8. -4 Burner: Isobaric Done K 4-5 Turbine: Isentropic Since this the purpose of the compressor is to power the turbine, we have w t = w c w t = c P (T -T ) = (005)(-4) = 09,4 J/kg From conservation of energy w t = c P (T 4 -T 5 ) so that w t 09,4 T5 = T4 - = 800 - = 49 K cp 005 Using the isentropic relationship we have 4

ME 440 Aerospace Engineering Fundamentals Fall 00 P γ /( γ ).4 /(.4 ) 5 49 5 = (00) = 4 800 T = P4 T 5- Nozzle: Isentropic Our isentropic condition is ( γ ) / γ (.4 ) /.4 = P 9. T = T5 = (49) P 5 7.9 The velocity is then calculated from the st law or h 5 = h V + V = cp 5 7.9 kpa 779 K ( T - T ) = (005)(49-779) 97 m/s = System Calculations Specific thrust ft = V - V =97-00 = 897 m/s = 897 N s/kg Propulsive efficiency p η p = W & Q& in Propulsive power W& = V - V V = (97-00)(00) = 9, p ( ) W aircraft Heat transfer rate in Q& in = mc & P T4 - T = (05)(800 - ) =,84,0 9, η p = = 0. or %,84,0 ( ) W 5

ME 440 Aerospace Engineering Fundamentals Fall 00 Example: Turbofan Engine Thrust Determine the specific thrust produced with a turbofan engine operating under the following conditions. Bypass Ratio:. Outlet Pressure: 45 kpa Aircraft Speed: 00 m/s Turbojet Air Pressure: 50 kpa Temperature: 00 K Area: 5% of exhaust area Bypass Air Pressure: 00 kpa Temperature: 00 K Area: 5% of exhaust area Solution: Let s begin with s block diagram of the system Bypass Turbojet Jet Pipe Nozzle Exhaust The specific thrust will be given by ft = Vexh - Vaircraft The exhaust velocity will come from Vexh = cp( T - T4 ) The temperature leaving the nozzle will come from ( γ ) / γ T = 4 4 T P The temperature and pressure coming into the nozzle must come form the jet pipe (mixing tank) analysis where & c T + c T = c T mbypass p bypass turbojet p turbojet out p Abypass Pbypass + AturbojetPturbojet = AoutP We will divide by the outlet mass flow rate and relate the ratios to the bypass ratio. bp bp β = = = + + / β β + out bp tj

ME 440 Aerospace Engineering Fundamentals Fall 00 tj out = bp tj + tj = β + β T = Tbp + T β + β + For pressure we have Abp Abp P = Pbp + Ptj A A. = (00) + (00). +. + tj = = (0.5)(00) + (0.5)(50) = 88kPa for the nozzle (.4 ) /.4 4 = 45 T = (99) K 88 So that V exh = (005) ( 99 - ) = 78 m/s and = 78-00 = 58 Nt s/kg f t 99K Example: Required Aircraft Speed for Ramjet Operation Determine the required aircraft speed and Mach number for a compression stage pressure ratio of in a ramjet engine. The aircraft is flying at km. Solution: Conservation of energy on the ramjet diffuser gives V cp (T - T ) = The inlet temperature may be obtained for the standard atmosphere table and is found as T =.78 K P =.7 kpa a = 95. m/s The exit temperature may be obtained by assuming that the diffuser is isentropic. T ( γ ) / γ = (.78) = P = T P (.4 ) /.4 ( ) K 7

ME 440 Aerospace Engineering Fundamentals Fall 00 the intake velocity is found as V = (005)( -.78) = 59 m/s M = V /a = 59/95. =.8 8