Turbojet Design Project
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Transcript of Turbojet Design Project
Turbojet Design
Carlos J Gutiérrez Román Ramón Natal Ramos
54543 70286
ME4935 Aircraft Propulsion
October 31, 2013
2ME 5930 – FA12
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).
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Defined problem gives us a start point for assumptions & correct analysis.
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Technology Assessment
http://en.wikipedia.org/wiki/TurbojetSimple 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
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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 dynamicsThe 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
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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
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Analysis and Synthesis
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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 startME4935 FA-13
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Engineering Analysis
•clear all•clc•%TurboJet Engine Analysis•Pidmax = 1; %Maximum Inlet Total Pressure Recovery (ideal)•T0 = 273.15-56.5; %Temperature at initial conditions•M0 = 2; %Mach number at initial conditions•R = 285;•kc = 1.4; %Cold Section ONLY•kh = 1.33; %Hot Section ONLY•V0 = sqrt(kc*R*T0)*M0; %Speed at initial conditions•TaoA = 7; %Tao Lambda•QR = 42800000;•Cpc = 1004;•Cph = 1156;•nm = 0.99; %Mechanical Efficiency•h = (37000*0.3048); %Altitude 37,000 ft. converted to meters•P0 = 101325*(1-0.0000225577*(h))^5.25588; %Air Pressure and Altitude above Sea Level formula•Pt0 = P0*(1+((kc-1)/2)*M0^2)^((kc/(kc-1))); %Total pressure at initial conditions•Tt0 = T0*(1+((kc-1)/2)*M0^2); %Total temperature at initial conditions
• % Inlet Diffuser • Pid = Pidmax*(1-0.075*(M0-1)^1.35); % nr = (1-0.075*(M0-1)^1.35 ... Pid = Pidmax x nr ---> Military
Specification 5008B• Pt2 = Pt0*Pid;• Tt2 = Tt0; %Temperature doesnt change in ducts
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Engineering Analysis
• % Compressor• Pic = 2:40; %Compressor pressure ratio. Pic can vary from 2 to 40 ( 40 max
value )• ec = 0.9; %Compressor Polytropic efficiency• Pt3 = Pt2.*Pic; %Pressure at compressor exit / combustor entrance• Tt3 = Tt2.*(Pt3./Pt2).^((kc-1)/(kc.*ec)); %Temperature at compressor exit / combustor entrance• TaoC = Tt3./Tt2; %Compressor temperature ratio• nc = (Pic.^((kc-1)./kc)-1)./(TaoC-1); %Compressor efficiency• • % Burner (Combustors)• Mb = 0.2;• E = 1;• nb = 0.98; %Burner efficiency• Pib = 1-(E.*(0.5.*kc).*Mb.^2); %Burner pressure ratio• Pt4 = Pt3.*Pib; %Pressure at combustor exit / turbine entrance• Tt4 = (TaoA.*T0.*Cpc)./Cph; %Temperature at combustor exit / turbine entrance• h0 = T0.*Cpc; %Ambient enthalpy h0 = Cpc*T0• TaoR = Tt2./T0; %Compressor temperature ratio• f = (TaoA-(TaoR.*TaoC))./(((QR.*nb)./h0)-TaoA); %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
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Engineering Analysis
• % Outlet Nozzle• P9 = P0; %Perfectly Expanded• NPR = Pt5./P9; %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
• %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
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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; %Fuel Flow Rate• Pt = m0.*(1+f).*Cph.*(Tt4-Tt5); %Turbine Power• Pc = Pt.*nm; %Compressor Power• • %Performance Parameters• RD = m0.*V0; %Ram Drag• GT = (m0+mf).*V9; %Gross Thrust• F = GT-RD; %Uninstalled Trust• TSFC = mf./F; %Thrust Specific Fuel Consumption• nT = ((m0+mf).*V9.^2-m0.*V0.^2)./(2.*mf.*QR); %Thermal Efficiency• nP = 2./(1+(V9./V0)); %Propulsive Efficiency• nO = (F.*V0)./(mf.*QR); %Overall Efficiency
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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
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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
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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
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Engineering Analysis
Graphical Analysis helps to identify the highest performing values in our design.ME4935 FA-13
5 10 15 20 25 30 35 400
1
2
3
4
x 104 Thrust vs Compressor Pressure Ratio
Compressor Pressure Ratio
Thr
ust
(N)
1 1.5 2 2.5 3 3.5
0.4
0.5
0.6
0.7
0.8
0.9
1Compressor Diameter vs Compressor Temperature Ratio
Compressor Temperature Ratio
Com
pres
sor
Dia
met
er
0 20 40 60 80 100 120 140 160 1800
10
20
30
40Compressor Pressure Ratio vs Mass flow
Mass flow
Com
pres
sor
Pre
ssur
e R
atio
0.005 0.01 0.015 0.02 0.025 0.030
10
20
30
40Compressor Pressure Ratio vs Fuel to Air ratio
Fuel to Air ratio
Com
pres
sor
Pre
ssur
e R
atio
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Engineering Analysis
Optimum values for pressure ratio
5 10 15 20 25 30 35 400
1
2
3
4
x 104 Thrust vs Compressor Pressure Ratio
Compressor Pressure Ratio
Thr
ust
(lb)
0 5 10 15 20 25 30 35 40400
600
800
1000
1200
1400Tt3 and Tt5 vs Compressor Pressure Ratio
Compressor Pressure Ratio
Tt3
and
Tt5
0 5 10 15 20 25 30 35 400
50
100
150
200Mass flow vs Compressor Pressure Ratio
Compressor Pressure Ratio
Mas
s flo
w
0 5 10 15 20 25 30 35 400.005
0.01
0.015
0.02
0.025
0.03Fuel to Air ratio vs Compressor Pressure Ratio
Compressor Pressure Ratio
Fue
l to
Air
ratio
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Engineering Analysis
0 5 10 15 20 25 30 35 403.5
4
4.5
5
5.5
6x 10
-5 TSFC vs Compressor Pressure Ratio
Compressor Pressure Ratio
TS
FC
0 5 10 15 20 25 30 35 400.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6Thermal Efficiency vs Compressor Pressure Ratio
Compressor Pressure Ratio
The
rmal
Eff
icie
ncy
0 5 10 15 20 25 30 35 400.65
0.7
0.75
0.8
0.85
0.9
0.95
1Propulsive Efficiency vs Compressor Pressure Ratio
Compressor Pressure Ratio
Pro
puls
ive
Eff
icie
ncy
0 5 10 15 20 25 30 35 400.4
0.6
0.8
1
1.2
1.4
1.6
1.8Fuel Mass Flow vs Compressor Pressure Ratio
Compressor Pressure Ratio
Fue
l mas
s flo
w
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Engineering Analysis
0 5 10 15 20 25 30 35 403.5
4
4.5
5
5.5
6x 10
-5 TSFC vs Compressor Pressure Ratio
Compressor Pressure Ratio
TS
FC
0 5 10 15 20 25 30 35 400.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6Thermal Efficiency vs Compressor Pressure Ratio
Compressor Pressure Ratio
The
rmal
Eff
icie
ncy
0 5 10 15 20 25 30 35 400.65
0.7
0.75
0.8
0.85
0.9
0.95
1Propulsive Efficiency vs Compressor Pressure Ratio
Compressor Pressure Ratio
Pro
puls
ive
Eff
icie
ncy
0 5 10 15 20 25 30 35 400.4
0.6
0.8
1
1.2
1.4
1.6
1.8Fuel Mass Flow vs Compressor Pressure Ratio
Compressor Pressure Ratio
Fue
l mas
s flo
w
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Concept Selection
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.
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Prob
lem
Sta
tem
ent
Wei
ghtin
g
πc =
2
πc =
17
πc =
40
Air mass flow rate 10 1 1Fuel flow rate 10 -1 0Thrust 40 1 0TSFC 20 1 0Thermal efficiency 10 0 0Propulsive efficiency 10 0 1
100 Total 60 20
Crite
ria
Bas
elin
e
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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
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Engineering Analysis
Temperature and pressures per stage:
With all needed parameters, parts and/or components of the turbine can be designed.
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Stage 0 2 3 4 5Tt (K) 387.97 387.97 958.73 1371 830
Pt (KPa) 169.47 156.76 2664.9 2590.3 327.53
Total Temperatures and Pressures per stage
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Final Design
22
INLET
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Final Design
23
NOZZLE
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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.
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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).
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Back Up Slides
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Backup
27
INLET NOZZLE
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