Preliminary Design of a Turbofan Engine · 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

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 𝜂𝑑=0.95 𝛾𝑑=1.40

Compressor Polytropic Efficiency 𝜂𝑐=0.90 𝛾𝑐=1.70

Fan Adiabatic Efficiency 𝜂𝑓=0.92 𝛾𝑓=1.40

Burner Effiency 𝜂𝑏=0.97 𝛾𝑏=1.35

Burner Pressure Recovery 𝜋𝑏=0.95 𝛾𝑏=1.35

Turbine 𝜂𝑡= 0.91 𝛾𝑡=1.33

Primary Nozzle 𝜂𝑛= 0.98 𝛾𝑛=1.36

Fan Nozzle 𝜂𝑛′= 0.99 𝛾𝑛′=1.40

Design Method

From the given steady state flight conditions, the airspeed can be found with the following

equation:

𝑈 = 𝑀√𝑅 ∗ 𝑇𝑎 ∗ 𝛾𝑎𝑖𝑟

Diffuser Section:

𝑇𝑎, 𝛾𝑑, and 𝑀 are known and total temperature (T02) can be determined:

𝑇02 = 𝑇𝑎(1 +𝛾𝑑 − 1

2𝑀2)

With the Value of T02, Pa can be found with the following equation:

𝑃02 = 𝑃𝑎(1 + 𝜂𝑑 (𝑇02

𝑇𝑎− 1))

𝛾𝑑𝛾𝑑−1

Mass Flow Rate:

With the values of 𝑇02, 𝑃02, 𝑃𝑎, and the given 𝑇𝑎 , the mass flow rate can be found using the

given equations (𝑃𝑎 𝑎𝑛𝑑 𝑇𝑎 𝑎𝑟𝑒 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙):

𝜃0 =𝑇02

𝑇𝑎 and 𝛿0 =

𝑃02,

𝑃𝑎

�̇� = 231.8 ( 𝛿0

𝜃0

) ∗ 𝐴

𝐴 = 𝜋 ∗ (𝑑

2)2 [Area of the engine in m2]

Fan/Bypass Section:

The bypass exit pressure can be found using the relation:

𝑃08 = 𝜋𝑓 ∗ 𝑃02

𝜋𝑓 = fan pressure ratio which is a chosen design parameter

With the 𝜋𝑓 design parameter chosen and the given 𝜂𝑓=0.92 and 𝛾𝑓=1.40, the bypass exit

temperature can be found with the following:

𝑇08 = 𝑇02(1 +1

𝜂𝑓(𝜋𝑓

𝛾𝑓−1

𝛾𝑓 − 1))

With R=287 𝑘𝐽

𝑘𝑔∗𝐾, we can determine the fan specific heat:

𝐶𝑝𝑓 =𝛾𝑓

𝛾𝑓 − 1∗ 𝑅

We can then find the exit velocity of the fan:

𝑈𝑒𝑓 = √2 ∗ 𝜂𝑓 ∗ 𝐶𝑝𝑓 ∗ 𝑇08(1 −𝑃𝑎

𝑃08)

𝛾𝑓−1

𝛾𝑓

Compressor Section:

With the following relation, the compressor exit pressure along with exit temperature can be

determined by:

𝑃03 = 𝜋𝑐 ∗ 𝑃02

𝑇03 = 𝑇02(1 +1

𝜂𝑐(𝜋𝑐

𝛾𝑐−1𝛾𝑐 − 1))

𝜋𝑐 = 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 𝐶𝑝𝑏 = 𝛾𝑓

𝛾𝑓−1∗ 𝑅

We can then find the amount of fuel with the equation:

𝑓 =

𝑇04

𝑇03− 1

𝑄𝑟

𝑇03 ∗ 𝐶𝑝𝑏−

𝑇04

𝑇03

Turbine Section:

Total Temperature and Pressure at the turbine exit can be found using these equations:

𝑇05 = 𝑇04 −𝑇03

𝑇02− 𝛽(𝑇08 − 𝑇02)

𝑃05 = 𝑃04(1 −1

𝜂𝑡(1 −

𝑇05

𝑇04))

𝛾𝑡𝛾𝑡−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:

𝑇𝑒 = 𝑇06(1 − 𝜂𝑛(1 − (𝑃𝑒

𝑃06))

𝛾𝑛𝛾𝑛−1

The last equations are needed to complete data for the air exiting the nozzle 𝐶𝑝𝑛 = 𝛾𝑛

𝛾𝑛−1∗ 𝑅

𝑈𝑒 = √2 ∗ 𝜂𝑛 ∗ 𝐶𝑝𝑛 ∗ 𝑇06(1 −𝑃𝑒

𝑃08)

𝛾𝑛−1𝛾𝑛

Thrust: 𝑇 = �̇� ∗ (1 + 𝑓) ∗ 𝑈𝑒 + (𝛽 ∗ 𝑈𝑒𝑓) − ((1 + 𝛽) ∗ 𝑈)

Specific Thrust:

𝑆𝑇 =. 001(1 + 𝑓)𝑈𝑒 + (𝛽 ∗ 𝑈𝑒𝑓) − (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:

𝑇𝑎, 𝛾𝑑, and 𝑀 are known and total temperature (T02) can be determined:

𝑇02 = 218.94(1 +.95−1

20.852) = 250.57 K

With the Value of T02, Pa can be found with the following equation:

𝑃02 = 23908.5(1 + 0.95 (250.57

218.94− 1))

1.4

1.4−1 = 37,504 Pa

Mass Flow Rate:

With the values of 𝑇02, 𝑃02, 𝑃𝑎, and the given 𝑇𝑎 , the mass flow rate can be found using the

given equations (𝑃𝑎 𝑎𝑛𝑑 𝑇𝑎 𝑎𝑟𝑒 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙):

𝜃0 =250.57

218.94= .8696 and 𝛿0 =

37504

23908.5= .3701

𝐴 = 𝜋 ∗ (1.7

2)

2= 2.2698 𝑚2

�̇� = 231.8 ( .3701

. 8696) ∗ 2.2698 = 208.835 𝑘𝑔/𝑠

𝑚𝑎̇ =208.835

(1 + 10)= 18.98 𝑘𝑔/𝑠

Fan/Bypass Section:

The bypass exit pressure can be found:

𝜋𝑓 = 1.5

𝑃08 = 1.5 ∗ 37,504 Pa = 56,257 Pa

With the 𝜋𝑓 design parameter chosen and the given 𝜂𝑓=0.92 and 𝛾𝑓=1.40, the bypass exit

temperature can be found with the following:

𝑇08 = 250.57(1 +1

0.92(1.5

1.4−1

1.4 − 1)) = 281.66 K

With R=287 𝑘𝐽

𝑘𝑔∗𝐾, we can determine the fan specific heat:

𝐶𝑝𝑓 =1.4

1.4−1∗ 287 = 1004.5 kJ/kg*k

We can then find the exit velocity of the fan:

𝑈𝑒𝑓 = √2 ∗ .92 ∗ 1004.5 ∗ 288.66(1 −23908.5

𝑃08)

1.4−1

1.4 = 348.57 m/s

Compressor Section:

With the following relation, the compressor exit pressure along with exit temperature can be

determined by:

𝜋𝑐 = 28

𝑃03 = 28 ∗ 37,504 = 1050100 Pa

𝑇03 = 250.57 (1 +1

. 90(28

1.70−11.70 − 1)) = 683.92 𝐾

𝜋𝑐 = compressor pressure ratio which is a chosen design parameter

Combustor Section:

𝐶𝑝𝑏 = 1081.4𝑘𝐽

𝑘𝑔∗𝐾

We can then find the amount of fuel with the equation:

𝑓 =

1700683.92

− 1

45000𝐸3683.92 ∗ 1081.4

−1700

683.92

= .0261

Turbine Section:

Total Temperature and Pressure at the turbine exit can be found using these equations:

𝑇05 = 1700 −683.92

250.57− (10)(281.66 − 250.57) = 966.79 𝐾

𝑃05 = 997600(1 −1

. 99(1 −

966.79

1700))

1.331.33−1 = 74922 𝑃𝑎

No Afterburner:

𝑇06 = 𝑇05 = 966.79 𝐾

𝑃06 = 𝑃05 = 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:

𝑃𝑒 = 𝑃𝑎 = 23908.5 𝑃𝑎

𝑇𝑒 = 966.79 (1 − .98(1 − (23908.5

74922))

1.361.36−1

The last two equations are needed to complete data for the air exiting the nozzle

𝐶𝑝𝑛 = 𝛾𝑛

𝛾𝑛−1∗ 𝑅 = 1084.2

𝑘𝐽

𝑘𝑔∗𝐾

𝑈𝑒 = √2 ∗ 0.98 ∗ 1084.2 ∗ 966.79 ∗ (1 −23908.5

56,257)

1.36−11.36

= 1362.3𝑚

𝑠

Thrust:

𝑇 = �̇� ∗ (1 + 𝑓) ∗ 𝑈𝑒 + (𝛽 ∗ 𝑈𝑒𝑓) − ((1 + 𝛽) ∗ 𝑈) = 40065 𝑁

Specific Thrust:

𝑆𝑇 =.001(1+𝑓)1362.3+(10∗348.57 )−(1+10)252.10

1+10= 0.1918 kN*s/kg

Thrust Specific Fuel Consumption:

𝑇𝑆𝐹𝐶 =0.0261

0.1918(1+10)= .0124 kg/kN*s

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.

Performance Data

Parameters Values

Inlet Diameter 1.7 m

Compression Ratio 28

Inlet Turbine Temperature (T04) 1700 K

Fan Pressure Ratio 1.5

Bypass Ratio 10

Mass Flow Rate 208.835 𝑘𝑔/𝑠

Core Engine Exit Velocity 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 end; for j=1:1:21 plot(ST(:,j,k),TSFC(:,j,k),'Color','b') hold on end; else if (k == 3) for i=1:1:13 plot(ST(i,:,k),TSFC(i,:,k),'Color','b') hold on end; for j=1:1:21 plot(ST(:,j,k),TSFC(:,j,k),'Color','b') hold on end; else for i=1:1:13 plot(ST(i,:,k),TSFC(i,:,k),'Color','b') hold on end; for j=1:1:21 plot(ST(:,j,k),TSFC(:,j,k),'Color','b') hold on end; end; end; 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)') end;

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