Preliminary Design of a Turbofan Engine MAE 112 Propulsion Justin Oyas ID#43026527 University of California, Irvine Henry Samueli School of Engineering Department of Mechanical and Aerospace Engineering
Abstract The objective of this project is to design a turbofan engine that is needed for a passenger airplane with one engine having a minimum of 25,000 N thrust and a thrust specific fuel consumption of less than or equal to 0.025 kg/(kn*s). This airplane will fly at steady state condition at an altitude of 35,000 feet at a flying speed of 0.85 Mach. Using the computer software Matlab, iterations of parameters such as compression ratio, fan pressure ratio, inlet turbine temperature, bypass ratio, and inlet diameter were studied to choose such parameters that meet the specified flight requirements. From the results, each parameter was chosen such that the engine produced a thrust value of 40065 N and a thrust specific fuel consumption of.0124 kg/kn*s. Introduction A turbofan jet consists of six main sections which are the fan, inlet, compressor, combustor, turbine, and the nozzle. There are two main parts of the inlet which are the bypass inlet and the core engine inlet; furthermore, the bypass inlet directs air around the engine for which it does not go through the core engine processes. The function of the bypass is to increase the overall mass flow rate; in addition, the bypass also helps reduces the noise of the common turbojet engine. After air passes through the fan, the air heads into the diffuser which brings the air down to a slower velocity for which it prepares the air to enter the compressor stage. Air is compressed to higher pressures which are directly determined by the pressure ratio from the compressor s design. The compressor is made up many rotor and stator blades and usually contains more blades than a turbine as it is more difficult to compress air. Once air is compressed to high pressures, it passes through the combustion chamber and it is mixed with fuel and is ignited to increase the air to a high temperature. The hot compressed air is then passed along through a turbine which extracts work from the system and is used to power the fan, compressor and other systems in the aircraft. Directly after the turbine is the nozzle for which the air is a bit more compressed up until it exits with high velocity where it meets the air from the bypass inlet.
The parameters and efficiencies of these various stages are stated below: Component Efficiency Average Specific Heat Ratio Inlet/Diffuser η d =0.95 γ d =1.40 Compressor Polytropic Efficiency η c =0.90 γ c =1.70 Fan Adiabatic Efficiency η f =0.92 γ f =1.40 Burner Effiency η b =0.97 γ b =1.35 Burner Pressure Recovery π b =0.95 γ b =1.35 Turbine η t = 0.91 γ t =1.33 Primary Nozzle η n = 0.98 γ n =1.36 Fan Nozzle η n = 0.99 γ n =1.40 Design Method From the given steady state flight conditions, the airspeed can be found with the following equation: Diffuser Section: U = M R T a γ air T a, γ d, and M are known and total temperature (T02) can be determined: T 02 = T a (1 + γ d 1 M 2 ) 2 With the Value of T02, Pa can be found with the following equation: Mass Flow Rate: P 02 = P a (1 + η d ( T 02 1)) T a γ d γ d 1 With the values of T 02, P 02, P a, and the given T a, the mass flow rate can be found using the given equations (P a and T a are at sea level): θ 0 = T 02 T a and δ 0 = P 02, P a
m = 231.8 ( δ 0 θ 0 ) A A = π ( d 2 )2 [Area of the engine in m 2 ] Fan/Bypass Section: The bypass exit pressure can be found using the relation: P 08 = π f P 02 π f = fan pressure ratio which is a chosen design parameter With the π f design parameter chosen and the given η f =0.92 and γ f =1.40, the bypass exit temperature can be found with the following: T 08 = T 02 (1 + 1 η f (π f γ f 1 γ f 1)) With R=287 kj kg K, we can determine the fan specific heat: C pf = We can then find the exit velocity of the fan: Compressor Section: γ f γ f 1 R U ef = 2 η f C pf T 08 (1 P γ f 1 a γ ) f P 08 With the following relation, the compressor exit pressure along with exit temperature can be determined by: P 03 = π c P 02 T 03 = T 02 (1 + 1 η c (π c γ c 1 γ c 1)) π c = compressor pressure ratio which is a chosen design parameter Combustor Section: To find out the amount of fuel required for the combustor, we first find C pb = γ f γ f 1 R
We can then find the amount of fuel with the equation: Turbine Section: f = T 04 T 03 1 Q r T 03 C pb T 04 T 03 Total Temperature and Pressure at the turbine exit can be found using these equations: T 05 = T 04 T 03 T 02 β(t 08 T 02 ) P 05 = P 04 (1 1 (1 T 05 )) η t T 04 γ t γ t 1 Note: This Turbofan design does not have an afterburner which would be after the Turbine Section Nozzle Section: After the air passes through the turbine, we assume that the exit pressure expands to the ambient pressure for which exit temperature can be found: T e = T 06 (1 η n (1 ( P e )) P 06 γ n γ n 1 The last equations are needed to complete data for the air exiting the nozzle C pn = U e = 2 η n C pn T 06 (1 P γ e n 1 ) γ n P 08 γ n γ n 1 R Thrust: T = m (1 + f) Ue + (β Uef) ((1 + β) U) Specific Thrust: ST =. 001(1 + f)u e + (β U ef ) (1 + β)u 1 + β Thrust Specific Fuel Consumption: TSFC = f ST(1 + β)
Calculations and Analysis of Results Constant Diameter 1.8m, Constant Fan Pressure Ratio 1.6, - Changing Temperature
Constant Temp, Constant Fan Pressure Ratio, Changing Diameter
Constant Temp, Constant Diameter, Changing Fan Pressure Ratio Using the equations stated above, the each section of the turbine is analyzed from diffuser to the nozzle. A code was written that cycles through the equations and parameters with the given conditions and efficiencies to meet the design parameter. The compressor ratio was iterated from the values 16 to 40 and the bypass ratio was iterated from 0 to 10. The diameter, fan pressure ratio, and inlet turbine (T04) temperature were varied with the given design parameters. From the graphs, it can be seen that an increase in the inlet turbine temperature affects the thrust levels of the engine, the higher the temperature gives more thrust. The diameter affects the mass flow rate of the engine and the larger diameter, the larger the thrust the
engine produces. The fan pressure ratio affects the TSFC and the higher the fan pressure ratio, the lower the TSFC and from the design restriction, the engines TSFC has to be less than.025 kg/kn*s. From the graphs and restrictions, the chosen engine specifications are calculated using the equations presented above: Airspeed: U = 0.85 287 218.94 1.4 =252.10 m/s Diffuser Section: T a, γ d, and M are known and total temperature (T02) can be determined: T 02 = 218.94(1 +.95 1 0.85 2 ) = 250.57 K 2 With the Value of T02, Pa can be found with the following equation: Mass Flow Rate: P 02 = 23908.5(1 + 0.95 ( 250.57 1.4 1)) 1.4 1 = 37,504 Pa 218.94 With the values of T 02, P 02, P a, and the given T a, the mass flow rate can be found using the given equations (P a and T a are at sea level): θ 0 = 250.57 218.94 =.8696 and δ 0 = 37504 23908.5 =.3701 A = π ( 1.7 2 )2 = 2.2698 m 2 m = 231.8 (.3701 ) 2.2698 = 208.835 kg/s. 8696 Fan/Bypass Section: The bypass exit pressure can be found: m a = 208.835 = 18.98 kg/s (1 + 10) π f = 1.5
P 08 = 1.5 37,504 Pa = 56,257 Pa With the π f design parameter chosen and the given η f =0.92 and γ f =1.40, the bypass exit temperature can be found with the following: With R=287 kj kg K T 08 = 250.57(1 + 1 0.92 (1.51.4 1 1.4 1)) = 281.66 K, we can determine the fan specific heat: We can then find the exit velocity of the fan: Compressor Section: C pf = 1.4 287 = 1004.5 kj/kg*k 1.4 1 U ef = 2.92 1004.5 288.66(1 23908.5 P 08 ) 1.4 1 1.4 = 348.57 m/s With the following relation, the compressor exit pressure along with exit temperature can be determined by: Combustor Section: π c = 28 P 03 = 28 37,504 = 1050100 Pa T 03 = 250.57 (1 + 1. 90 (281.70 1 1.70 1)) = 683.92 K π c = compressor pressure ratio which is a chosen design parameter C pb = 1081.4 kj kg K We can then find the amount of fuel with the equation: f = 1700 683.92 1 45000E3 683.92 1081.4 1700 =.0261 683.92 Turbine Section: Total Temperature and Pressure at the turbine exit can be found using these equations: T 05 = 1700 683.92 (10)(281.66 250.57) = 966.79 K 250.57
P 05 = 997600(1 1. 99 966.79 1.33 (1 )) 1.33 1 = 74922 Pa 1700 No Afterburner: T 06 = T 05 = 966.79 K P 06 = P 05 = 74922 Pa Nozzle Section: After the air passes through the turbine, we assume that the exit pressure expands to the ambient pressure for which exit temperature can be found: P e = P a = 23908.5 Pa T e = 966.79 (1.98(1 ( 23908.5 1.36 )) 1.36 1 74922 The last two equations are needed to complete data for the air exiting the nozzle C pn = γ n kj R = 1084.2 γ n 1 kg K U e = 2 0.98 1084.2 966.79 (1 23908.5 1.36 1 56,257 ) 1.36 m = 1362.3 s Thrust: T = m (1 + f) Ue + (β Uef) ((1 + β) U) = 40065 N Specific Thrust: ST = Thrust Specific Fuel Consumption:.001(1+f)1362.3+(10 348.57 ) (1+10)252.10 1+10 = 0.1918 kn*s/kg TSFC = 0.0261 =.0124 kg/kn*s 0.1918(1+10)
Summary From the results, it can be concluded that temperature, compressor ratio and inlet diameter enhances the engines thrust performance, however, it is at the cost of higher TSFC levels not meeting the design requirement. To effectively lower TSFC levels while still meeting the Thrust Criteria, a compromise must be met with a a fan pressure ratio and bypass ratio to help the engine lower its TSFC; however this comes at the expense of lower Thrust and Specific Thrust. The final results are tabulated below with the parameters and design meeting the mission requirements. Parameters Inlet Diameter Performance Data Values 1.7 m Compression Ratio 28 Inlet Turbine Temperature (T04) 1700 K Fan Pressure Ratio 1.5 Bypass Ratio 10 Mass Flow Rate Core Engine Exit Velocity 208.835 kg/s 1362.3 m/s Fan Exit Velocity 348.57 m/s Fuel Air Ratio.0261 TSFC. 0124 kg/kn*s ST. 1918 kn*s/kg Thrust 40,065 N
Appendix Design Iterations Matlab Code clear %Fixed Parameters h=35000; %altitude Nd=.95; %diffuser eff Yd=1.4; %diffuser gamma Nc=.90; %compressor eff Yc=1.70; %compressor gamma Nf=0.92; %fan eff Yf=1.40; %fan gamma Nb=0.97; %burner eff Yb=1.35; %burner gamma PIb=0.95; %burner pressure ratio Nt=0.91; %turbine eff Yt=1.33; %turbine gamma Nn=0.98; %nozzle eff Yn=1.36; %nozzle gamma Nfan=0.99; %Fan Nozzle eff Yfan=1.40; %Fan Nozzle gamma %Conditions M=0.85; %Mach Number R=287; %Gas Constant Qr=45000000; %Fuel Specific Heat T04=1600; %Kelvin - Chosen Parameter PIf=1.6; %Fan Pressure Ratio Psea=101325; %Sea Level Pressure Tsea=288.15; %Sea Level Temperature Pa=23908.5; %Ambient Pressure -Table Ta=218.94; %Ambient Temperature -Table for x=1:13 PIc= 2*x+14 for y=1:21 B=0.5*y-0.5; for z=1:13 D=1.7; %Diameter, Max 2 PIf=1.4 T04=1700; %Kelvin - Max 1700 %Diffuser Section T02=Ta*(1+(Yd-1)/2*M^2); P02=Pa*(1+Nd*(T02/Ta-1))^(Yd/(Yd-1)); %Mass Flow Rate A=pi*(D/2)^2; d0=p02/psea; t0=(t02/tsea); mflow=a*231.8*((d0)/(sqrt(t0))); MFLOW=mflow/(1+B); % Bypass Section P08 = PIf*P02; T08 = T02*(1+(1/Nfan)*((PIf^((Yfan-1)/Yfan))-1)); Cpfan = Yfan/(Yfan-1)*R; %fan specific heat Cpc=R*((Yc)/(Yc-1)); %compressor specific heat P03=PIc*P02; T03=T02*(1+(1/Nf)*((PIc.^((Yf-1)/Yf)-1))); %Combuster Section Cpb=R*((Yb)/(Yb-1)); %combuster/burner specific heat f=(t04/t03-1)/(qr/(cpb*t03)-t04/t03); %Turbine P04=P03;
T05=T04-(T03-T02)/(1+f)-(B*(T08-T02)); P05=P04*(1-((1/Nt)*(1-(T05/T04))))^(Yt/(Yt-1)); %Neglect After Burner T06=T05; P06=P05; %Nozzle Section Cpn= Yn/(Yn-1)*R; Pe=Pa; %Velocities U=M*sqrt(Yd*R*Ta); %Air Velocity Ue=sqrt(2*Nn*Cpn*T06*(1-(Pa/P06)^((Yn-1)/Yn))); Uef=sqrt(2*Nfan*Cpfan*T08*(1-(Pa/P08)^((Yfan-1)/(Yfan)))); %Thrust Eqns Thrust(x,y,z)=MFLOW*((1+f)*Ue+B*Uef-(1+B)*U); ST(x,y,z)=.001*Thrust(x,y,z)/(MFLOW*(1+B)); TSFC(x,y,z)=f/(ST(x,y,z)*(1+B)); end end end for (k = 1:13) k = 4; if (k == 2) for i=1:1:13 plot(st(i,:,k),tsfc(i,:,k),'color','b') hold on for j=1:1:21 plot(st(:,j,k),tsfc(:,j,k),'color','b') hold on else if (k == 3) for i=1:1:13 plot(st(i,:,k),tsfc(i,:,k),'color','b') hold on for j=1:1:21 plot(st(:,j,k),tsfc(:,j,k),'color','b') hold on else for i=1:1:13 plot(st(i,:,k),tsfc(i,:,k),'color','b') hold on for j=1:1:21 plot(st(:,j,k),tsfc(:,j,k),'color','b') hold on STmin=25/mflow; xvaltsfc=[0,1]'; yvaltsfc=[.025,.025]'; plot(xvaltsfc,yvaltsfc,'--k') xvaltsfc=[stmin,stmin]'; yvaltsfc=[0,0.25]'; plot(xvaltsfc,yvaltsfc,'--k') axis([0,1,0.015,0.032]); title('tsfc versus ST Fan Diameter= 1.7m') xlabel('st (kn*s/kg)') ylabel('tsfc (kg/kn*s)')
Engine Performance Program clear clc %Fixed Parameters h=35000; %altitude Nd=.95; %diffuser eff Yd=1.4; %diffuser gamma Nc=.90; %compressor eff Yc=1.70; %compressor gamma Nf=0.92; %fan eff Yf=1.40; %fan gamma Nb=0.97; %burner eff Yb=1.35; %burner gamma PIb=0.95; %burner pressure ratio Nt=0.91; %turbine eff Yt=1.33; %turbine gamma Nn=0.98; %nozzle eff Yn=1.36; %nozzle gamma Nfan=0.99; %Fan Nozzle eff Yfan=1.40; %Fan Nozzle gamma %Conditions M=0.85; %Mach Number R=287; %Gas Constant Qr=45000000; %Fuel Specific Heat Psea=101325; %Sea Level Pressure Tsea=288.15; %Sea Level Temperature Pa=23908.5; %Ambient Pressure -Table Ta=218.94; %Ambient Temperature -Table D=1.7; T04=1700; PIf=1.5; PIc=28; B=10; U=M*sqrt(Yd*R*Ta) T02=Ta*(1+(Yd-1)/2*M^2) P02=Pa*(1+Nd*(T02/Ta-1))^(Yd/(Yd-1)) A=pi*(D/2)^2 d0=p02/psea t0=(t02/tsea) mflow=a*231.8*((d0)/(sqrt(t0))) mdot=mflow/(1+b) P08 = PIf*P02 T08 = T02*(1+(1/Nfan)*((PIf^((Yfan-1)/Yfan))-1)) Cpfan = Yf/(Yf-1)*R %fan specific heat Cpc=R*((Yc)/(Yc-1)) %compressor specific heat Uef=sqrt(2*Nfan*Cpfan*T08*(1-(Pa/P08)^((Yfan-1)/(Yfan)))) P03=PIc*P02 T03=T02*(1+(1/Nf)*(PIc.^((Yf-1)/Yf)-1)) Cpb=R*((Yb)/(Yb-1)); %combuster/burner specific heat f=(t04/t03-1)/(qr/(cpb*t03)-t04/t03) T05=T04-(T03-T02)/(1+f)-(B*(T08-T02)) P04=P03*PIb P05=P04*(1-((1/Nt)*(1-(T05/T04)))).^(Yt/(Yt-1)) T06=T05; P06=P05; Cpn= (Yn/(Yn-1))*R Pe=Pa; Ue=sqrt(2*Nn*Cpn*T06*(1-(Pe/P06))^((Yn-1)/Yn)) minst=25000/mflow
% Thrust=.001*mdot*((1+f)*Ue+B*Uef-(1+B)*U) % ST=Thrust/mdot % TSFC=1000*(f/((1+f)*Ue+B*Ue-(1+B))*U) Thrust=(mdot*(((1+f)*Ue)+(B*Uef)-((1+B)*U))) ST=.001*Thrust/(mdot*(1+B)) TSFC=(f*mdot)/Thrust*1000