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Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
Turbojet Design Project
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Turbojet Design Project

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Thermodynamic design of a Turbojet engine for the given conditions of altitude and speed. Blade geometry was taken into consideration for this project. A Matlab program was written to calculate the …

Thermodynamic design of a Turbojet engine for the given conditions of altitude and speed. Blade geometry was taken into consideration for this project. A Matlab program was written to calculate the best compressor ratio, temperatures and geometry to obtain maximum thrust. Inlet and Nozzle were drafted using CREO Parametric.

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  • 1. Turbojet Design Carlos J Gutiérrez Román 54543 Ramón Natal Ramos 70286 ME4935 Aircraft Propulsion October 31, 2013
  • 2. Problem Definition Design a turbojet engine under the following conditions:  Altitude of 37000 ft.  The pressure ratio across the compressor πc.  The Mach number M0=2.0.  Maximum enthalpy ratio τλ=7.0  Fuel type is hydrocarbon with QR = 42800 kJ/kg For a range of compressor pressure ratios, namely 2<πc<40, calculate: a) The pressure and temperatures at each principal state in kPa and K. b) Air mass flow rate. c) The velocity at the nozzle exit. d) Power of the turbo-components. e) Fuel flow rate necessary to power the engine. f) Performance parameters (SFC, TSFC, thermal efficiency, propulsive efficiency). Defined problem gives us a start point for assumptions & correct analysis. ME4935 FA-13 ME 5930 – FA12 2
  • 3. Technology Assessment  http://en.wikipedia.org/wiki/Turbojet Simple explanations on how a turbojet works.  http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node85.html Very specific explanations of the thermodynamic cycle of a turbojet.  http://www.stanford.edu/~cantwell/AA283_Course_Material/AA283_Course_Notes/Ch _04_Turbojet_Cycle.pdf Complete description and formulas for turbojet design, from thermodynamics to compressor geometry. It’s always a good idea to understand the current trends to better guide the design. ME4935 FA-13 ME 5930 – FA12 3
  • 4. Applicable Engineering Principles  Thermodynamics Analysis. The gas generator is composed of a compressor, a turbine, a pair of nozzles and a burner. All are analyzed thermodynamically to find the total pressures and temperatures. With these data pressure and temperatures ratios, air fuel ratio and power of the turbine and compressor are calculated.  Fluid dynamics The compressor geometry depends on the angles the fluid must take when approaching the blades. A fluid dynamics analysis was made to obtain the mean radius of the compressor blades. With this data nozzles areas and mass flow are calculated. Previously learned topics are the key elements to analyze the given problem. ME4935 FA-13 ME 5930 – FA12 4
  • 5. Engineering Requirements  The engine must be able to operate at an altitude of 37000 ft.  The pressure ratio across the compressor πc must produce the maximum thrust for each pound of fuel.  The engine must be able to operate at a Mach number of M0=2.0.  Maximum enthalpy ratio possible for this engine is τλ= 7.0  Fuel type it will use is hydrocarbon with QR = 42800 kJ/kg.  It can not violate any fluid mechanics or thermodynamics laws.  Turbine inlet temperature must not exceed 2000K. Constraints creates the boundaries for the design requirements. ME4935 FA-13 ME 5930 – FA12 5
  • 6. Analysis and Synthesis ME4935 FA-13 ME 5930 – FA12 6
  • 7. Proposed Concepts  For the given conditions we will design a turbojet engines that delivers the maximum thrust possible.  Instead of using three different compressor pressure ratios we write a Matlab script program that varies the compressor ratio for the following range of values: 2<πc<40  Our program utilizes the compressor temperature ratio to calculate the mean radius of the compressor and the corresponding inlet area.  Since the temperature ratio depends of the pressure ratio, the program returns the optimal mean radius for a given πc.  For all the different values of πc our program also optimizes the corresponding inlet and nozzle areas, inlet mass flow and fuel ratio.  Plotting the most relevant data, we can decide which compressor pressure ratio returns the best performance from the turbojet engine. Technology and logical assumptions ease the design process from the start ME4935 FA-13 ME 5930 – FA12 7
  • 8. Engineering Analysis • • • • • • • • • • • • • • • • • • • clear all clc %TurboJet Engine Analysis Pidmax = 1; T0 = 273.15-56.5; M0 = 2; R = 285; kc = 1.4; kh = 1.33; V0 = sqrt(kc*R*T0)*M0; TaoA = 7; QR = 42800000; Cpc = 1004; Cph = 1156; nm = 0.99; h = (37000*0.3048); P0 = 101325*(1-0.0000225577*(h))^5.25588; Pt0 = P0*(1+((kc-1)/2)*M0^2)^((kc/(kc-1))); Tt0 = T0*(1+((kc-1)/2)*M0^2); • • % Inlet Diffuser Pid = Pidmax*(1-0.075*(M0-1)^1.35); Specification 5008B Pt2 = Pt0*Pid; Tt2 = Tt0; • • ME4935 FA-13 ME 5930 – FA12 %Maximum Inlet Total Pressure Recovery (ideal) %Temperature at initial conditions %Mach number at initial conditions %Cold Section ONLY %Hot Section ONLY %Speed at initial conditions %Tao Lambda %Mechanical Efficiency %Altitude 37,000 ft. converted to meters %Air Pressure and Altitude above Sea Level formula %Total pressure at initial conditions %Total temperature at initial conditions % nr = (1-0.075*(M0-1)^1.35 ... Pid = Pidmax x nr ---> Military %Temperature doesnt change in ducts 8
  • 9. Engineering Analysis • • • • • • • • • • • • • • • • • • • • • • • • • • % Compressor Pic = 2:40; value ) ec = 0.9; Pt3 = Pt2.*Pic; Tt3 = Tt2.*(Pt3./Pt2).^((kc-1)/(kc.*ec)); TaoC = Tt3./Tt2; nc = (Pic.^((kc-1)./kc)-1)./(TaoC-1); %Compressor pressure ratio. Pic can vary from 2 to 40 ( 40 max %Compressor Polytropic efficiency %Pressure at compressor exit / combustor entrance %Temperature at compressor exit / combustor entrance %Compressor temperature ratio %Compressor efficiency % Burner (Combustors) Mb = 0.2; E = 1; nb = 0.98; Pib = 1-(E.*(0.5.*kc).*Mb.^2); Pt4 = Pt3.*Pib; Tt4 = (TaoA.*T0.*Cpc)./Cph; h0 = T0.*Cpc; TaoR = Tt2./T0; f = (TaoA-(TaoR.*TaoC))./(((QR.*nb)./h0)-TaoA); %Burner efficiency %Burner pressure ratio %Pressure at combustor exit / turbine entrance %Temperature at combustor exit / turbine entrance %Ambient enthalpy h0 = Cpc*T0 %Compressor temperature ratio %fuel-air ratio % Turbine et = 0.9; %Turbine polytropic efficiency TaoT = 1-((TaoR.*(TaoC-1))./((1+f).*TaoA)); %Turbine temperature ratio Tt5 = Tt4.*TaoT; %Temperature at tubine exit / nozzle entrance PiT = (TaoT).^((kh)./((kh-1).*et)); %Turbine Pressure ratio Pt5 = Pt4.*(Tt5./Tt4).^((kh)./((kh-1).*et));%Pt4*PiT; Pressure at turbine exit / nozzle entrance nt = (1-TaoT)./(1-TaoT.^(1./et)); %Turbine adiabatic efficiency ME4935 FA-13 ME 5930 – FA12 9
  • 10. Engineering Analysis • • • • • • • • • • • • % Outlet Nozzle P9 = P0; NPR = Pt5./P9; • • • • • • • • • • %Inlet Design & Dimensions D2 = rm*4; %Diameter at end of nozzle section = 4 times Compressor mean radius A2 = (pi./4).*D2.^2; %Area at End of inlet (2) M1 = M0; %For Supersonic flight Ati = A2./((1./M2).*((1+((kc-1)./2).*M2.^2)./((kc+1)./2)).^((kc+1)./(2.*(kc-1)))); %Inlet throat area A1 = Ati.*((1./M0).*((1+((kc-1)./2).*M0.^2)./((kc+1)./2)).^((kc+1)./(2.*(kc-1)))); %Inlet area (1) A0 = A1; %For supersonic flight intet area = A0 m0 = A0.*((P0./(R.*T0)).*V0); %Mass flow D1 = ((4./pi).*A1).^(1/2); %Inlet Diameter Dti = ((4./pi).*Ati).^(1/2); %Inlet throat diameter %Perfectly Expanded %Nozzle Pressure Ratio %Compressor Design & Dimensions M2 = 0.4; %Assuming subsonic speed before compressor w = 30000; %Angular Velocity Beta = -25; %Beta angle Alfa = 12; %Alfa angle T2 = Tt2./(1+((kh-1)./2).*M2.^2); %Temperature at inlet exit U = ((TaoC-1)./(1+.6.*(tan(Beta)-tan(Alfa))).*Cpc.*Tt2).^(1/2); rm = U./(w.*(2.*pi./60)); %Mean Radius ME4935 FA-13 ME 5930 – FA12 10
  • 11. Engineering Analysis • • • • • • • • %Nozzle Design & Dimensions mg = m0.*(1+f); M9 = (((NPR).^((kh-1)./kh)-1)./((kh-1)./2)).^(1/2); %Mach number at nozzle outlet T9 = Tt5./(1+((kh-1)./2).*M9.^2); %Temperature at nozzle outlet V9 = M9.*(kh.*R.*T9).^(1/2); %Speed at nozzle outlet A9 = mg./((P9./(R.*T9)).*V9); %Exhaust mass flow Atn = A9./((1./M9).*((1+((kh-1)./2).*M9.^2)/((kh+1)/2)).^((kh+1)/(2.*(kh-1)))); %Nozzle throat area Dtn = ((4./pi).*Atn).^(1/2); %Nozzle throat Diameter • • • • • • • • • • • • • %Other Parameters mf = mg-m0; Pt = m0.*(1+f).*Cph.*(Tt4-Tt5); Pc = Pt.*nm; %Fuel Flow Rate %Turbine Power %Compressor Power %Performance Parameters RD = m0.*V0; GT = (m0+mf).*V9; F = GT-RD; TSFC = mf./F; nT = ((m0+mf).*V9.^2-m0.*V0.^2)./(2.*mf.*QR); nP = 2./(1+(V9./V0)); nO = (F.*V0)./(mf.*QR); %Ram Drag %Gross Thrust %Uninstalled Trust %Thrust Specific Fuel Consumption %Thermal Efficiency %Propulsive Efficiency %Overall Efficiency ME4935 FA-13 ME 5930 – FA12 11
  • 12. Engineering Analysis • • • • • • • • • • • • • • • • • • • • • • • • %Plots Thrust vs Compressor Pressure Ratio subplot(2,4,1) plot(Pic,F); title('Thrust vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Thrust (lb)'); axis([2 40 0 45000]) grid on %Plots Tt3 and Tt5 vs Compressor Pressure Ratio subplot(2,4,2) plot(Pic,Tt3,Pic,Tt5); title('Tt3 and Tt5 vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Tt3 and Tt5'); grid on %Plots Mass flow vs Compressor Pressure Ratio subplot(2,4,3) plot(Pic,m0); title('Mass flow vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Mass flow'); grid on ME4935 FA-13 ME 5930 – FA12 12
  • 13. Engineering Analysis • • • • • • • • • • • • • • • • • • • • • • • %Plots Fuel to Air ratio vs Compressor Pressure Ratio subplot(2,4,4) plot(Pic,f); title('Fuel to Air ratio vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Fuel to Air ratio'); grid on %Plots TSFC vs Compressor Pressure Ratio subplot(2,4,5) plot(Pic,TSFC); title('TSFC vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('TSFC'); grid on %Plots Thermal Efficiency vs Compressor Pressure Ratio subplot(2,4,6) plot(Pic,nT); title('Thermal Efficiency vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Thermal Efficiency'); grid on ME4935 FA-13 ME 5930 – FA12 13
  • 14. Engineering Analysis • • • • • • • • • • • • • • • %Plots Propulsive Efficiency vs Compressor Pressure Ratio subplot(2,4,7) plot(Pic,nP); title('Propulsive Efficiency vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Propulsive Efficiency'); grid on %Plots Fuel Mass Flow vs Compressor Pressure Ratio subplot(2,4,8) plot(Pic,mf); title('Fuel Mass Flow vs Compressor Pressure Ratio') xlabel('Compressor Pressure Ratio'); ylabel('Fuel mass flow'); grid on ME4935 FA-13 ME 5930 – FA12 14
  • 15. Engineering Analysis 4 Thrust vs Compressor Pressure Ratio x 10 Compressor Di 1 4 Compressor Diameter 0.9 Thrust (N) 3 2 1 0.8 0.7 0.6 0.5 0.4 0 5 10 15 20 25 Compressor Pressure Ratio 30 35 40 1 1.5 Co Compressor Pressure Ratio vs Mass flow Compress 40 Compressor Pressure Ratio Compressor Pressure Ratio 40 30 20 10 0 0 20 40 60 80 100 Mass flow 120 140 160 180 30 20 10 0 0.005 0.01 Graphical Analysis helps to identify the highest performing values in our design. ME4935 FA-13 ME 5930 – FA12 15
  • 16. Engineering Analysis essure Ratio Tt3 and Tt5 vs Compressor Pressure Ratio 1400 Tt3 and Tt5 1200 1000 800 600 25 re Ratio 30 35 400 40 0 5 Pressure Ratio 10 15 20 25 Compressor Pressure Ratio 30 35 40 35 40 Fuel to Air ratio vs Compressor Pressure Ratio 0.03 Fuel to Air ratio 0.025 0.02 0.015 0.01 25 re Ratio 30 35 40 0.005 0 5 10 15 20 25 Compressor Pressure Ratio 30 Optimum values for pressure ratio ME4935 FA-13 ME 5930 – FA12 16
  • 17. Engineering Analysis -5 6 TSFC vs Compressor Pressure Ratio x 10 Thermal Efficiency 0.6 0.55 Thermal Efficiency TSFC 5.5 5 4.5 0.5 0.45 0.4 0.35 4 0.3 3.5 0 5 10 15 20 25 Compressor Pressure Ratio 30 35 0.25 40 0 5 Propulsive Efficiency vs Compressor Pressure Ratio 1.6 0.9 1.4 Fuel mass flow 1.8 0.95 Propulsive Efficiency 15 Compre Fuel Mass Flow v 1 0.85 0.8 0.75 0.7 0.65 ME4935 FA-13 ME 5930 – FA12 10 1.2 1 0.8 0.6 0 5 10 15 20 25 Compressor Pressure Ratio 30 35 40 0.4 0 5 10 15 Compre 17
  • 18. Engineering Analysis essure Ratio Thermal Efficiency vs Compressor Pressure Ratio 0.6 Thermal Efficiency 0.55 0.5 0.45 0.4 0.35 0.3 25 re Ratio 30 35 0.25 40 0 5 ssor Pressure Ratio 10 15 20 25 Compressor Pressure Ratio 30 35 40 35 40 Fuel Mass Flow vs Compressor Pressure Ratio 1.8 Fuel mass flow 1.6 1.4 1.2 1 0.8 0.6 25 re Ratio 30 ME4935 FA-13 ME 5930 – FA12 35 40 0.4 0 5 10 15 20 25 Compressor Pressure Ratio 30 18
  • 19. πc = 40 Total πc = 17 10 10 40 20 10 10 100 πc = 2 Weighting Air mass flow rate Fuel flow rate Thrust TSFC Thermal efficiency Propulsive efficiency Baseline Criteria Problem Statement Concept Selection 1 -1 1 1 0 0 60 1 0 0 0 0 1 20 Analyzing 3 different cases of compressor pressure ratio makes us take a better judgment of which is the best for the steps to come and final design. ME4935 FA-13 ME 5930 – FA12 19
  • 20. Engineering Analysis  The compressor pressure ratio that develops the most thrust is πc = 17.  It’s also the best compressor pressure ratio to obtain the best TSFC.  The temperatures Tt5 and Tt3 are the same at πc = 14 which means that our method to approximate the mean radius is very good.  For our design we chose a πc = 17 and obtained the following results:            Funin = 42,705 N TSFC = 3.53 x10-5 ηT = 51% ηP = 75% ṁ0 = 108.87 kg/s ṁf = 1.51 kg/s ϝ = 0.0139 Pc = 6.15 x107 W Pt = 6.21 x107 W V9 = 966.89 m/s NPR = 15.12 Using the best compressor pressure ratio yields optimum results. ME4935 FA-13 ME 5930 – FA12 20
  • 21. Engineering Analysis  Temperature and pressures per stage: Stage Tt (K) Pt (KPa) Total Temperatures and Pressures per stage 0 2 3 4 387.97 387.97 958.73 1371 169.47 156.76 2664.9 2590.3 5 830 327.53 With all needed parameters, parts and/or components of the turbine can be designed. ME4935 FA-13 ME 5930 – FA12 21
  • 22. Final Design  INLET ME4935 FA-13 ME 5930 – FA12 22
  • 23. Final Design  NOZZLE ME4935 FA-13 ME 5930 – FA12 23
  • 24. Conclusions  Increasing the compressor pressure ratio not necessary increase the thrust output. Air fuel ratio started deceasing after the optimum πc.  When the output temperatures of the compressor and turbine are similar the engines returns more thrust per pound of fuel.  Increasing the air mass flow and fuel mass flow not necessary increase the thrust output (it also depends in other factors like pressure ratio).  We have to convert all the pressure that its not consumed by the turbine into speed to get the most thrust (perfectly expanded).  The diameter of the compressor wheel and blades depends on the thermodynamic calculations. ME4935 FA-13 ME 5930 – FA12 24
  • 25. Recommendations  Being able to acquire the knowledge in class to implement to our project the following:  Cooling of the turbine section.  Detailed design of the compressor and turbine blades.  Compare more than one type of propulsion (turbojet vs turbofan vs turboprop). ME4935 FA-13 ME 5930 – FA12 25
  • 26. Back Up Slides ME4935 FA-13 ME 5930 – FA12 26
  • 27. Backup INLET ME4935 FA-13 ME 5930 – FA12 NOZZLE 27

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