The document analyzes the performance of the Pratt & Whitney JT8D turbofan engine under varying operating conditions using EES software. It includes four parts: (1) a baseline check calculates take-off thrust, (2) altitude effects on thrust, efficiency and cost are examined, (3) the impact of bypass ratio on specific thrust is analyzed, (4) the effect of pressure ratio on specific thrust is determined. The optimal conditions found are a cruise altitude of 14,000 meters, bypass ratio of 38:1, and compressor factor of 1.778.

1. UNIVERSITY OF MINNESOTA - DULUTH
JT8D Jet Engine Analysis
Justin Rees
12/15/2015
The analysisof the JT8D engine involvesusingEESsoftware tounderstandthe performanceof the
engine whenoperatingconditionsare changing.The analysis includesfourparts:BaselineCheck,Effect
of Altitude,Effectof BypassRatioandEffectof Pressure Ratio.

2. Introduction
The purpose of the analysisistoillustrate how the performance of the jetengine isaffected by
variable operatingconditions.The variable operatingconditions accountedforinthe analysisinclude jet
speed,cruise altitude,bypassratioandcompressorfactor.Eachof these parametersisanalyzed
individuallytorecognize the change inperformance. PrattandWhitney’sJT8DTurbofanengine willbe
usedinthe analysis.The JT8D is a popularandreliable commercialaviationengine usedonthe
McDonnell DouglasDC-9,Boeing727 and 737 aircrafts.
Methodology
In orderto successfullyanalyze the JT8Djetengine,some assumptionsneededtobe made.The
primaryand secondaryflownozzlesoperate “ondesign”,meaningthe engine producesthrustonly by
meansof momentum.Additionallyinthe stage Icompressor,the pressure change acrossthe fanis
included inthe pressure ratio.The engineisassumedtobe in a steadystate and steadyflow scenario.
The analysiswascompletedinthe programEES (EngineeringEquationSolver).The programwasused
for itsthermodynamicfunctions:density, entropyandenthalpy.The EEScode can be locatedin
Appendix A. Figure 1belowillustratesthe differentstatesof the engineusedinthe analysis withthe
same numberingsystem. Asseeninthe diagram, the high-pressure turbine spinsthe rearcompressor
and the low-pressure turbinespinsboth the frontcompressorandthe fan.
Figure 1: JT8D Cutaway Detailed States
Part I: Baseline Check
The Baseline Checkwasusedtocalculate the take-off thrustforone engine andthe enthalpyvsentropy
(h-s) diagramat the basicparameters,whichincludes:
 No Bypass,B=0
 CruisingAltitude@sealevel,z=0
 AircraftSpeed,Vaircraft=0
 CompressorFactor,x=1

3. The take-off thrustcalculatedatthese parametersis107,893 N (24,255 lbf).Thisisan acceptable value
since Pratt andWhitney’sJT8Dspecslista take-off thrustof 14,000 - 21,700 lbf of thrust whichincludes
a bypassratio of approximatelyone.The take-off thrustwasrecalculatedwithabypassof one and
foundto be 15,485 lbf thrust,whichconcurs with Prattand Whitney’s specs.
Figure 2: Actual h-s Diagram for Baseline Parameters
Table 1: Actual Values for Baseline h-s Diagram
Actual Values for Baseline h-s
State
h
(kJ/kg)
s
(kJ/kg*K)
P
(kPa)
2 288.4 5.661 101.3
13 377.7 5.692 233
2.5 456.9 5.71 425.6
3 699.6 5.757 1617
4 1819 6.776 1556
4.5 1583 6.796 853.6
5 1419 6.812 535.9
9 911.4 6.823 101.3
Part II: Effectof Altitude
The analysisinthissectioninvolvesthe cruisingaltitudetobe a variable between0and14,000 meters.
Most commercial aircraftoperate at a cruisingaltitude of 30,000 ft.(9144 meters).The analysisinvolved
a bypassratio of five,jetspeedof 295 m/s andthe compressorfactorto be one.

4. Firstthe thrust developed byasingle enginewasplottedversusaltitude.The thrustdevelopedbythe
engine decreasesasthe elevationincreases.Thisisdue tothe densityof airdecreasingathigher
altitudes.Densityisafunctionof temperature andpressure whichbothdecrease withahigheraltitude.
The thrust decreasesbecause the massflow rate of airis decreasedwiththe lowerdensityof air.
Figure 3: Altitude vs Thrust for a Single Engine Plot
The cost to flyone-wayfromNewYorkto San Francisco(3500 miles) atthe varyingaltitudeswasfound
usingtwoengineslike mostcommercial planes.The costwasfoundusinga fuel costof $0.48/kg of JP-4
jetfuel.The flightcostislowerathigherelevations.Thisistobe expectedsince the massflow rate of air
islowerwhichresultsinthe fuel flowrate toalsobe lower.
Figure 4: Altitude vs. Cost (2 Engines New York to San Francisco)

5. The thermal efficiencyisalsoinfluencedbythe varyingaltitude. Thermal efficiency will increaseby
eitherloweringthe atmosphericpressure/temperatureorincreasingthe turbine inlet
pressure/temperature.Inthiscase the thermal efficiencyisincreasedbyloweringthe atmospheric
conditions. The thermal efficiencyincreasesathigheraltitudessince the pressureandtemperature
allowfora higheramountof work to be producedbythe compressor, turbine andnozzle.
Figure 5: Altitude vs. Thermal Efficiency
Part III: Effectof Bypass Ratio
The nextstage of the analysisinvolvesadjustingthe bypassratiotochange the thrustproduced. The
analysisutilizes ajetspeedof 295 m/s, cruisingaltitude of 11,300 meters,compressorfactorof one and
varyingthe bypassratio from 0 to 15.
As showninFigure 6, the specificthrustincreaseswiththe bypassratio.Thisisdue to the highermass
flowrate of airthrough the bypasssectionof the engine eventhoughthe velocityleavingisnotashigh
as the core velocity.There isalimitforthe bypassratioof approximately38:1 forthisanalysis.The core
requiresacertainmass flowrate of air inorder to spinthe fanwhichdrawsthe air intothe bypass.
Therefore if the core doesn’treceiveenoughairthenthe low-pressure turbine (4.5-5) cannotproduce
enoughworkto spinthe frontcompressor(2-2.5) and the fan (2-13). Accordingto Pratt and Whitney,
the JT8D isoperatedwitha bypassratioof 1.0 – 1.7.

6. Figure 6: Bypass Ratio vs. Specific Thrust
Part IV: Effectof Pressure Ratio
The last variable analyzedwasthe effectof variable pressure ratio. The controllingvariableswere jet
speedof aircraftat 295 m/s,cruisingaltitude of 11,300 meters,bypassratioof six andthe compressor
factor varyingfromone to two.
The specificthrustincreaseswiththe pressure ratiountilthe compressorfactorreaches1.778. The
specificthrustthenfallsandeventuallythe plane will stopflyingata compressorfactorof 2.2. The
reasonthe curve behaveslike thisisthatthe compressorsrequire more worktohave themcompress
the air to a greaterpressure.The specificthrustincreasesuntil the turbinescannotprovide enoughwork
to spinthe compressorstotheirdesiredpressure ratio. Inorderforthe specificthrusttokeepincreasing
withthe pressure ratio,the inlettemperature of the high-pressure turbine (4-4.5) needstobe
increased. Table 2,showsthe increasesinpressure from(2-4) of the engine andthe high-pressure
turbine inlettemperature (Tmax) isconstantthroughout.

7. Figure 7: Effect of Pressure Ratio vs. Specific Thrust
Table 2: Parametric Table from EES for variable compressor factor, x
Conclusion
The analysisdemonstrateshowthe engine performance canbe enhancedwithchanging
operatingconditions.At14,000 metersthe engine hasthe greatestthermal efficiencyandhaslesscost
but alsodevelopslessthrustatthe higheraltitudes.The higherthe bypassratio,the more thrustthe
bypasscan produce until approximately38:1ratio whenthe core doesn’texperience enoughairflowto
spinthe fan.The pressure ratiowill produce more thrustthroughthe bypassbutlessthrustthroughthe
core andan optimal value of 1.778 was foundto produce the max specificthrust.Fromthisanalysisit
can be concludedforthe purpose of thisstudythatthe mostoptimal operatingconditionsare at14,000
metercruisingaltitude,a38:1 bypassratio anda compressorfactorof 1.778. Accordingto Pratt and
Whitney,the engine isoperatedata bypassof 1.0-1.7, fanpressure ratioof 1.92-2.21 and an overall
pressure ratioof 15.8-21.0.

8. Bibliography
JT8D ENGINE.(2012, October1). RetrievedDecember15, 2015, from
http://www.pw.utc.com/JT8D_Engine
(n.d.).RetrievedDecember15, 2015, from http://media-cache-
ak0.pinimg.com/736x/ab/b0/a0/abb0a0f955bbbecd26864cf7bc4406ef.jpg

9. Appendix A: EES Baseline Code
The baseline code was adjusted for each part of the analysis and used the same
equations throughout the analysis.

10. {Part I: Baseline Check}
{Operating Conditions}
Tatm = 288 - (0.0065*z) {Temperature as a function of altitude}
Patm_bar = 1.01325 - 0.000112*z + ((3.8*10^(-9))*z^2) {Pressure as a function of altitude}
Patm = Patm_bar * 100 {Converting bar to kPa}
T_max = 1650 {Max temperature at Turbine Inlet in Kelvin}
LHV = 44900 {kJ/kg}
Fuel_cost = 0.48 {$/kg}
Qdot_total = 85 {m^3/sec} {Inlet air volumetric flow rate, entering at location (1) of engine}
n_c = 0.87 {Compressor adiabatic efficiency}
n_t = 0.89 {Turbine adiabatic efficiency}
n_n = 0.98 {Nozzle adiabatic efficiency}
n_comb = 0.93 {Combustor adiabatic efficiency}
rp1 = 4.2*x {Pressure ratio in stage I Compressor}
rp2 = 3.8*x {Pressure ratio in stage II Compressor}
rp0 = 2.3*x {Pressure ratio in Fan Stage}
P_loss = 0.038 {Pressure loss in Combustor}
{Baseline Check Parameters}
V_aircraft = 0
z = 0
B = 0
x = 1
rho=Density(Air,T=Tatm,P=Patm)
mdot_total = Qdot_total * rho
mdot_secondary = B * mdot_core
mdot_core = mdot_total - mdot_secondary
P2 = Patm
T2 = Tatm
h2 = Enthalpy(Air, T=T2)
s2a = Entropy(Air, T=T2, P=P2)
ho2 = h2 + ((V_aircraft)^2)/2000
{Fan Stage}
P13 = rp0 * P2
s13s =s2a
s13s = Entropy(Air, T=T13, P=P13) {Use entropy function to find T13 isentropic}
h13s = Enthalpy(Air, T=T13)
n_c = (h13s - ho2) / (h13a - ho2) {Use efficiency to find h2 actual}
T13a = Temperature (Air, h=h13a)
s13a = Entropy(Air, T=T13a, P=P13)
w_fan = mdot_secondary * (h13a - ho2)
{Stage 1 Compressor}
P2.5 = rp1 * P2
s2.5s = s2a
s2.5s = Entropy(Air, T=T2.5s, P=P2.5) {Use entropy function to find T2.5 isentropic}
h2.5s = Enthalpy(Air, T=T2.5s)
n_c = (h2.5s - ho2) / (h2.5a - ho2) {Use efficiency to find h2.5 actual}
T2.5a = Temperature(Air, h=h2.5a)
s2.5a = Entropy(Air, T=T2.5a, P=P2.5)
w_comp1 = mdot_core * (h2.5a - ho2)

11. {Stage 2 Compressor}
P3 = rp2 * P2.5
s3s = s2.5a
s3s = Entropy(Air, T=T3, P=P3) {Use entropy function to find T3 isentropic}
h3s = Enthalpy(Air, T=T3)
n_c = (h3s - h2.5a) / (h3a - h2.5a) {Use efficiency to find h3 actual}
T3a = Temperature(Air, h=h3a)
s3a = Entropy(Air, T=T3a, P=P3)
w_comp2 = mdot_core*(h3a - h2.5a)
{Combustor}
T4 =T_max {Temperature leaving combustor as Max Turbine Inlet Temperature}
Qdot_in = mdot_fuel * LHV * n_comb {Qdot_in is the heat transfer rate going into the engine from the
combustor}
Qdot_in = (mdot_core + mdot_fuel) * (h4a) - (mdot_core*(h3a)) {Using the two Qdot_in equations to
solve for mdot_fuel}
h4a = Enthalpy(Air, T=T4)
P4 = P3 - (P3 * P_loss) {Combustor experiences 3.8% loss in pressure}
s4a = Entropy(Air, T=T4, P=P4)
{High Pressure Turbine}
{The High Pressure Turbine spins the Stage 2 Compressor, so work of High Pressure Turbine = work of
Stage 2 Compressor}
w_HPturb = w_comp2
w_HPturb = (mdot_fuel + mdot_core) * (h4a - h4.5a) {Use work to solve for h4.5a}
T4.5a = Temperature(Air, h=h4.5a)
n_t = (h4a - h4.5a) / (h4a - h4.5s)
T4.5s = Temperature(Air, h=h4.5s)
s4.5s = s4a
s4.5s = Entropy(Air, T=T4.5s, P=P4.5) {Use entropy function to solve for P4.5}
s4.5a = Entropy(Air, T=T4.5a, P=P4.5)
{Low Pressure Turbine}
{The Low Pressure Turbine spins the Stage 1 Compressor and Fan}
{So therefore the work of Low Pressure Turbine = work of Stage 1 Compressor + work of Fan}
w_LPturb = w_comp1 + w_fan
w_LPturb = (mdot_fuel + mdot_core) * (h4.5a - h5a) {Use work to solve for h5a}
T5a = Temperature(Air, h=h5a)
n_t = (h4.5a - h5a) / (h4.5a - h5s)
T5s = Temperature(Air, h=h5s)
s5s = s4.5a
s5s = Entropy(Air, T=T5s, P=P5) {Use entropy function to solve for P5}
s5a = Entropy(Air, T=T5a, P=P5)
{Nozzle}
ho9 = h5a {Stagnation enthalpy is constant through nozzle, no work in}
s9s = s5a
s9s = Entropy(Air, T=T9s, P=Patm) {Use to find T9s}
h9s = Enthalpy(Air, T=T9s) {Use to find h9s}
n_n = ((h5a - h9a) / (h5a - h9s)) {Use to find h9a}
T9a = Temperature(Air, h=h9a) {Use to find T9a}
s9a = Entropy(Air, T=T9a, P=Patm)
V9 = sqrt((ho9-h9a)*2000) {Need to use static enthalpy and stagnation enthalpy}

12. {Bypass Nozzle}
ho17 = h13a {Stagnation enthalpy is constant through nozzle, no work in}
s17s = s13a
s17s = Entropy(Air, T=T17s, P=Patm) {Use to find T17s}
h17s = Enthalpy(Air, T=T17s)
n_n = ((h13a - h17a) / (h13a - h17s)) {Use to find h17a}
T17a = Temperature(Air, h=h17a)
s17a = Entropy(Air, T=T17a, P=Patm)
V17 = sqrt((ho17-h17a)*2000)
{Thrust} {Divide by 4.48 to convert from N to lbf}
Thrust_core = (mdot_fuel+mdot_core) * (V9)
Thrust_bypass = ((mdot_secondary) * (V17))
Thrust_total = (Thrust_core + Thrust_bypass - ((mdot_core + mdot_secondary)*V_aircraft))
{Thermal Efficiency}
f = mdot_fuel / mdot_core
nth = (mdot_core*((1+f)*(((V9^2)/2000) - ((V_aircraft)^2/2000)))) / (mdot_core * f * LHV)
s_array[1..8] = [s2a, s13a, s2.5a, s3a, s4a, s4.5a, s5a, s9a]
h_array[1..8] = [h2, h13a, h2.5a, h3a, h4a, h4.5a, h5a, h9a]