;

TEKNOFESTHAVACILIK, UZAY VE TEKNOLOJI FESTIVALI

JET MOTOR TASARIM YARISMASI

KRITIK TASARIM RAPORU

TAKIM ADI: BLADE BENDERS

TAKIM ID: T3-12248-154

TAKIM LIDERI ADI SOYADI : BERKCAN GULSEN

TAKIM UYELERI : DERYA SARMISAKNESRIN NUR TOZLUSEMRA SULTAN UZUNYUSUFCAN TASMAZ

AKADEMIK DANISMAN: PROF. DR. OGUZ UZOL

Table of Contents

1 Introduction 2

2 CFD Analysis 22.1 CFD Analysis Preparation Process . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1.1 Geometry and Computational Domain . . . . . . . . . . . . . . . . . . . . 22.1.2 Analyzed Blade Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1.3 Computational Grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.4 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.5 Solution Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 CFD Analysis Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 NGV Cooling Design Details 17

4 Hand Calculation for External Flow / 1-D Flow Analysis Results 214.1 Thermal Boundary Conditions of External Flow . . . . . . . . . . . . . . . . . . 224.2 Method for External Flow HTC Calculation and HTC Results . . . . . . . . . . . . 29

5 Hand Calculation for Internal Flow / 1-D Flow Network Analysis Results 315.1 Details of the Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5.1.1 Mass Flow Rate Calculation Method and Initial Mass Flow Rates . . . . . 325.1.2 Coolant Flow Properties Calculation Method . . . . . . . . . . . . . . . . 33

5.2 Results for Mass Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.3 Results for Pressure, Temperature, Velocity . . . . . . . . . . . . . . . . . . . . . 385.4 Back Flow Margin (BFM) Calculations . . . . . . . . . . . . . . . . . . . . . . . 40

6 Hand Calculation of Heat Transfer / 1-D Analysis Results 416.1 Details of the Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.2 Metal Temperature Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.2.1 Pressure Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2.2 Suction Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.2.3 Temperature Distribution at different sections along the span of NGV . . . 49

6.3 Comments on Wall Temperature Results . . . . . . . . . . . . . . . . . . . . . . . 536.3.1 Cooling Effectiveness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7 Hand Calculation for Cold Flow Test/1-D Heat Transfer Analysis Results 54

8 Lifetime Prediction 558.1 Stress Calculation Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558.2 Lifetime Calculation Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

9 Appendix 58

References 76

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1 IntroductionThe objective of this project is to develop a proper cooling design system for a turbine nozzle guidevane. This report contains the details of our design and analysis effort to satisfy the requirementsfor the Critical Design Report (KTR) of the Jet Engine Design Competition in TEKNOFEST 2020.In preliminary design report, we have already presented the methods that we developed in orderto calculate the wall temperature of nozzle guide vane by hand.However, we have made a fewimprovements on both external and internal flow calculations thanks to further literature survey onthe topic of heat transfer. The calculation methodology is explained in detail within the content ofthis report.

Since the delivery of Preliminary Design Report, we have performed conjugate heat transferanalysis of NGVs containing different cooling systems using ANSYS CFX tool. The rationalebehind performing these analysis is to observe the temperature distributions on NGV surface and todecide on the final choice of the cooling design. Moreover, there is another reason driving us forperforming CFD analysis. The hand calculations made for external heat transfer coefficients makeuse of the results from these analysis. In order to clarify how we perform them, the CFD process isexplained in detail. Both the results of hand calculations and CFD analysis for final cooling designare included in this report.

In addition to these efforts, we have also made calculations regarding back flow margin valuesof coolant ejection locations like film cooling holes and trailing edge slots. Furthermore, a lifetimeprediction method is applied to our cooling design and we have estimated the time that the NGVwill operate in hours. The calculation details and the result can be seen in the final part of this report.

2 CFD AnalysisIn this section, results of 3D cascade CFD Analysis performed for various design cases will bepresented. These designs are made considering the trade off studies mentioned in PreliminaryDesign Report (OTR). When the cascade analysis is constructed, it is a standard assumption that;

• Radial variations in velocity can be eliminated or ignored.

The analysed results which are obtained from ANSYS CFX tool are compared. Then the mostappropriate design for required temperature distribution is chosen. Then these results are improvedby decreasing the required mass flow rate for cooling flow. The analysis results of the final case arecompared with the results with of hand calculations performed by the MATLAB tool. Also variousresults from ANSYS have been referred in our hand calculations while solving for external flowheat transfer rates.These calculations will be presented in detail in the related sections.

2.1 CFD Analysis Preparation Process

2.1.1 Geometry and Computational Domain

The cascade is consisting two blade passages as shown in Figure 1. As we investigated in OTR thefluid inlet domain is modeled far enough away from vane to provide a uniform flow upstream of

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NGV. Similarly, the domain outlet is modeled far enough away from the NGV to capture symmetryin the flow and not disrupt the flow in the downstream.

Figure 1: 2D representation of designed cascade geometry with dimensions.

The 3D form of the cascade is shown in Figure 2 in isometric perspective. The given NGVgeometry is placed within the cascade with our cooling design. The origin placed at the leadingedge of the NGV geometry.

Figure 2: Isometric view of the 3D cascade domain.

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2.1.2 Analyzed Blade Designs

In this section, the details of the cooling design decided as a result of the analyzes made in thecritical design phase is introduced. The various views of the blade shown in Figure 3 .

Figure 3: Various views of NGV with cooling configuration.

The cooling configuration consists of 2 cooling channels. The first channel has a jet impingementand film holes cooling configurations on the pressure side of the blade. The angle of the film holes(10 pieces spaced equally) was determined as 45°as a result of the literature research. In the secondchannel, there is a row of film holes on the pressure side of the blade (9 pieces spaced equally). Inaddition, a slot was placed to cool the blade trailing edge. The close view of the channels is shownin the Figure 4. The geometrical parameters of cooling channels are shown in Table ?? with 1mmNGV thickness. Details of NGV cooling design drawing is given in appendix.

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Figure 4: Cooling channels close view.

First Cooling ChannelInlet Surface Area 7.082 mm2

Film HolesHole Number 10Hole Angle 45°

Hole Diameter 0.8 mmImpingement Holes

Hole Number 8Hole Angle 45°

Hole Diameter 0.8 mm

Second Cooling ChannelInlet Surface Area 36.96 mm2

Film HolesHole Number 9Hole Angle 45°

Hole Diameter 0.8 mmTE Slot

Width 18.45 mmLength 6.81 mmHeight 0.3 mm

2.1.3 Computational Grid

In order to increase accuracy, we performed analysis with very fine mesh. The information aboutthe total grid and node numbers will be given in this section in Table 1. The computational grids foreach domain is shown in Figures 5, 6, 7, 8.

5

Figure 5: Computational grid for Hot Domain.

Figure 6: Computational grid for Smooth Channel Domain.

6

Figure 7: Computational grid for Jet Channel Domain.

Figure 8: Computational grid for NGV.

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Table 1: Grid and node numbers for different mesh sizes.

Domain Elements NodesBlade 545198 112401Hot Gas 12742695 2261314Jet Channel 260172 51724Smooth Channel 177631 36386All Domains 13725696 2461825

2.1.4 Boundary Conditions

In order to perform a reliable CFD analysis, the boundary conditions should be given accurately.Theboundary conditions are introduced for each domain and resultant boundary condition tables areplaced at the end of each section. Cooling flow mass flow rate is calculated according to theTEKNOFEST specifications and divided to two cooling channels. In order to increase the efficiencyof turboshaft engine, the given 7% mass flow rate is further decreased and total 5% mass flow rateis used in CFX analysis.

2.1.4.1 Hot Domain

The surfaces shown Figure 9 are described as no slip wall, meaning that relative velocity of themoving fluid on these surfaces is equal to zero.

Figure 9: Wall boundary surfaces of NGVs.

In Figures 10 and 11, the boundary conditions for upstream, downstream and domain sides areshown. These boundary conditions are described as translational periodicity. The NGVs are placed

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in a repeated row physically, and this boundary condition provides realistic and more accuratemodelling. Translational periodicity prevents the losses due to the surfaces around the domain.

Figure 10: Hot gas upstream and downstream boundaries.

Figure 11: Hot gas domain side boundaries.

For the inlet boundary of the hot domain shown in Figure 12, the overall temperature distributionfactor should be taken into consideration.Since turbine NGV’s are generally designed to withstandthe maximum temperature, which is a consequence of OTDF, the whole inlet temperature is given

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as 1780 K.We calculated this value by below equation where Tt3 = 700K, Tt4 = 1600K andOTDF = 0.2.

Tt4,OTDF = (Tt4 − Tt3)×OTDF (1)

The boundary condition for outlet surface shown in same figure is described as a relative pressurevalue. This value is chosen to satisfy the chocking condition at the smallest area between two NGVsurfaces. Therefore, relative pressure value is set to 350 kPa at the outlet to provide chocked flow.

Figure 12: Hot gas domain inlet and outlet surfaces.

Table 2: Boundary conditions for hot gas domain.

Section Name Described Boundary ConditionBlades No-slip wallUpstream Region Translational PeriodicityDownside Region Translational PeriodicityDomain Sides Translational Periodicity

InletTotal Temperature=1780 K,Total Pressure=1079kPa(gage with respect to seal level atmospheric)

Outlet Gage Static Pressure 350 kPa

2.1.4.2 Smooth Channel Domain

In Figure 13, the surface where the flow enters the smooth channel is shown. Boundary conditionfor this inlet is set to 1099 kPa and 700 K at first. After several tries, no converging solution can beobtained at all. Therefore, we decided to convert the boundary condition to mass flow rate. Totalcoolant mass flow rate is divided into cooling channels to satisfy the metal temperature requirements

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while also keeping the coolant pressure at 1200 kPa and coolant temperature at 700 K at the inlet ofboth channels. After several trial and errors MFR is set to msmooth = 0.0066 kg/s to provide tosmooth channel. The boundary conditions for smooth channel can be seen in Table 3.

Figure 13: Smooth channel domain inlet surface.

Outlet surface of the smooth channel is shown in Figure 14. Boundary condition is chosen as noslip wall for this surface. The trailing edge part and film exits of the smooth channel is described asan interface between smooth channel cold gas domain and external hot gas domain.

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Figure 14: Smooth channel domain no-slip wall surfaces.

Table 3: Boundary conditions for smooth channel domain.

Section Name Described Boundary ConditionInlet Total Temperature=700 K, MFR=0.0066 kg/sOutlet No-slip Wall

2.1.4.3 Jet Channel Domain

The inlet surface for jet channel domain is shown in Figure 15. The boundary condition for jetchannel inlet is described as mjet = 0.0014 kg/s.

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Figure 15: Jet channel domain inlet surface.

In Figure 16, the outlet surfaces of the jet channel is shown. Boundary conditions are chosen asno slip wall for the outlet surface. Film hole exits of the jet channel are set as interface between jetchannel cold gas domain and external hot gas domain. The boundary conditions for jet channel areshown in Table 4.

Figure 16: Jet channel domain no-slip wall surfaces.

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Table 4: Boundary conditions for jet channel domain.

Section Name Described Boundary ConditionInlet Total Temperature=700 K, MFR=0.0014 kg/sOutlet No-slip Wall

2.1.4.4 Solid NGV Domain

Our main focus is to investigate the heat transfer process in ANSYS CFX. Therefore, heat transferoption is activated for all blade interfaces. The material property of NGV is given as CM247 LC DSsuperalloy. Since no detailed information can be found in literature, similar superalloy MarM247 ischosen as the NGV material for solid NGV domain. The material properties of MarM247 is shownin Table 5 [1]. MarM247 superalloy is also used as a material for turbomachinery applications.

Table 5: Material properties of MarM247 superalloy at 650 °C reference temperature.

Density, ρ 8500 kg/m3

Melting Temperature 1260 °CSpecific Heat, Cp 465 J/kgKThermal Conductivity, k 17.7 W/mKYoung’s Modulus, E 1.68× 1011

Thermal Expansion, α 17.7× 10−6 K−1

2.1.5 Solution Details

2.1.5.1 Solver Details

ANSYS CFX solver is used for 3D Cascade analysis for our NGV design. Solver details are shownin Table 6.

Table 6: Details in the ANSYS CFX solver.

Turbulence Model SST (Standard Wall Function)Transition Model Gamma Theta ModelInlet Boundary Condition Inlet (Specified total pressure and temperature)

Outlet boundary condition(Outlet Static pressure (to prevent backflow validationanalyzes were run))

Advection Scheme Higher resolutionTransient Scheme Second order (Fully implicit backward Euler)Transient inner loop coefficients Up to 1 iteration per time stepConvergence Criteria r.m.s residual level e-04Relaxation Parameters Solver relaxation fluids (0.4)Working fluid Air Ideal Gas

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2.1.5.2 Turbulence Model

The Shear Stress Transport (SST) turbulence model combines the k-omega turbulence model withthe k-epsilon turbulence model, making highly accurate predictions for the amount of the flowseparation under adverse pressure gradients. Therefore, SST turbulence was chosen as the turbulenceoption. In addition, percentage of 10 intensity factor was selected as the model initialization for hotgas, smooth channel and jet channel domains. TI=10% is chosen to provide more realistic solutionfor cold flow, since it is taken from compressor through pipes. The transition model is set to GammaTheta model in solver with defaults of CFX.

2.2 CFD Analysis Results3D Cascade CFD analysis using CFX solver results in terms of static temperature distribution overthe NGV body is shown in Figures 17, 18, 19 and 20. Temperature range over the NGV is alsoshown in the color bars. Temperature distribution over NGV with maximum around 1190 K, isappropriate in terms of operational safety described in TEKNOFEST specifications.

Figure 17: Temperature distribution over the NGV, side view.

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Figure 18: Temperature distribution over the NGV, pressure side view.

Figure 19: Temperature distribution over the NGV, other side view.

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Figure 20: Temperature distribution over the NGV, suction side view.

3 NGV Cooling Design DetailsIn this section the NGV designs that we have considered so far including our final design are shown.As a result of the analyzes made on the cooling configuration decided in the preliminary designphase, it was seen that some changes should be made in our cooling design. A variety of designs forcooling configurations that we have considered so far will be shown in this section.

The Figure 21 shows the cooling design that was decided during the preliminary design process.As stated in the preliminary design report, a trade-off study was conducted for ribs on the secondchannel. In this study, it was observed that the temperatures did not change much when there wereno ribs. As a result, it was decided not to use cooling method using ribs in the second channel.

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Figure 21: Cooling Design decided in Preliminary Design phase.

The second cooling design configuration is shown in the Figure 22. In this configuration, asstated in TEI online training, the diameter of the film holes has been changed to 0.8. The secondcooling design configuration shows an updated version of the first configuration. In the secondcooling configuration, in addition to the previous configuration, the blade trailing edge is designedas a slot.

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Figure 22: Cooling Design result from trade-off study for ribs.

After making the necessary changes in the blade trailing edge section and the second channeldesigns, many analyzes were made to use the film holes more effectively. In these analyzes,examinations were made on the angle, position, and number of film holes. Trade-off study wascarried out considering the position of the temperatures we calculated on the blade and the pressurevalues in the cooling channels. Different configurations have been tried for the film cooling such asthe shower head (shown in Figure 23), the pressure side of the first channel (shown in Figure 24)and the suction side of the first channel (shown in Figure 25).

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Figure 23: Cooling Design with using Shower Head Film Cooling configuration.

Figure 24: Cooling Design with film holes placed at suction side at jet channel.

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Figure 25: Cooling Design with film holes placed at pressure side at jet channel.

As a result, the cooling configuration, which had a row of film holes on the pressure side ofthe first channel and a row of film holes on the second channel, was decided (Figure 25) as finalNGV cooling design. It was deemed suitable for the temperatures taken on the blade and coolingefficiency.

4 Hand Calculation for External Flow / 1-D Flow Analysis Re-sults

All the hand calculations related to external flow were done by simply treating our blade sections asa combination of one circle and two flat plates.The configuration created can be seen below.Theadvantages of this approach will clearly be understood in the following parts of the current section.

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Figure 26: The new airfoil configuration representing the actual airfoil shape

Below figure is included in order to clarify how we convert the blade surface to flat plate.In ourcalculations, we used 50 data points on both suction and pressure side.The properties of externalflow at any data point on the blade surface are assumed to belong to the corresponding point on theflat plate.

Figure 27: Point projection method

4.1 Thermal Boundary Conditions of External FlowIn this section, the method that we followed in order to obtain the thermal boundary conditions ofexternal flow will be explained.

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Figure 28: Thermal Boundary Layer [2]

We know that the static pressure at an arbitrary point on the blade surface remains constant alongy direction inside the boundary layer (i.e dP

dy= 0).Hence, from CFD analysis, we first obtained the

static pressure distribution on the blade surface namely, on the boundary layer edge.At this point itis assumed that;

• The total pressure of the flow outside the boundary layer is constant and equal to the turbineinlet total pressure (Pt4).

Therefore, Mach number distribution at the edge of boundary layer could easily be obtained bythe following isentropic relation.

M2∞ =

[(Pt∞P∞

)(γ/(γ−1))

− 1

]2

γ − 1(2)

Since the flat plate Nusselt number correlations include temperature dependent parameters likeReynolds and Prandtl number, it is reasonable to calculate mean temperatures inside the boundarylayer in order to obtain more accurate results for those numbers.Equation 3 was used for thispurpose.This equation includes the terms like adiabatic wall temperature (Taw), wall temperature(Tw) and the static temperature at the edge of boundary layer (T∞).Although the main objective ofcalculating heat transfer coefficients is to obtain wall temperature of NGV, we should make an initialestimation for wall temperature (Tw) in order to be able to calculate the mean temperature.Thisresults in an iterative process.When hand calculations are done, the final wall temperatures shouldbe compared with the initial estimation. If difference between them is in acceptable range, theprocess can be finished.Otherwise, the hand calculations should be restarted by taking the last walltemperature as the new estimation.

T ∗ = T∞ + 0.58× (Tw − T∞) + 0.19× (Taw − T∞) (3)

Then, since we have Mach number distribution at the edge of thermal boundary layer, thecorresponding static temperature values could be obtained.At this point, we assumed that;

• The total temperature of the flow outside the boundary layer is constant and equal to the totaltemperature at the turbine inlet (Tt4,OTDF ).

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T∞ = Tt∞/

(1 +

γ − 1

2M2∞

)(4)

Another term that we needed to plug in equation 3 was the adiabatic wall temperature.Adiabaticwall temperature is the temperature that actually governs the heat transfer between the external flowand blade surface.In the absence of film cooling holes, adiabatic wall temperature calculation canbe done by equation 5, where r is the recovery factor and is equal to

√Pr and Pr1/3 for laminar

and turbulent flows, respectively.Therefore, at each data point on the flat plate, we first calculatedReynolds number and determined the type of the flow.External flow was considered as laminar atpoints where Reynolds number was below 5× 105.

Taw = T∞(1 + rγ − 1

2M2∞) (5)

Calculation of Taw with the effect of film cooling

Figure 29: A film cooling hole

At the points which are downstream of the film cooling holes, the calculation of adiabatic walltemperature is totally different.The calculation method of adiabatic wall temperature is based on thepaper of Goldstein written in 1971 [3].In the following calculations, it is assumed that;

• The secondary flows generated by film cooling holes do not mix with the flows generated byother holes located downstream of them.

We first needed to calculate a reference temperature and all the properties should be calculatedat that temperature within the boundary layer.This temperature can be calculated by equation6.However, since this equation includes adiabatic wall temperature, we needed to make an initialestimation for Taw.

T ∗ = T∞ + 0.72(Taw − T∞) (6)

Then, the adiabatic film cooling effectiveness can be calculated by equation 7

ηr = 1 + (cp∞/cp2)[0.33(4 + ξ∗)0.8 − 1]−1 (7)

where

ξ∗ = (x/Ms)(Re2 µ2/µ∗)−0.25(ρ∗/ρ∞) (8)

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wherex: distance between the point of injection and any data point downstream of the holeM: blowing ratios: slot length along flow direction

The dimensionless blowing ratio M is the ratio of the velocity of the injected fluid to themainstream mass velocity [3]. This parameter is calculated by the below equation.

M = ρ2V2/ρ∞V∞ (9)

The parameters with supercripts * must be calculated at reference temperature and with subcripts∞ must be calculated at T∞.The parameters with subscript 2 are calculated near the entry of filmcooling holes.In order to determine Reynolds number and dynamic viscosity of coolant flow, weneeded to obtain the temperature and velocity of the coolant there.In this regard, we made use ofthe CFD analysis results.Below, Reynolds number calculation can be seen.

Re2 = ρ2V2x/µ2 (10)

Below, in equation 11, adiabatic film cooling effectiveness is represented in a different way thanequation 7.This relation allows us to calculate adiabatic wall temperature of the external flow in thepresence of film cooling injection.Thanks to the effectiveness calculated by equation 7, adiabaticwall temperature can easily be calculated by simply plugging the values of Tr and Tr2 .Tr can bedefined as the adiabatic wall temperature of external flow if the film cooling effect was not an issueat that point.In other words, Tr is calculated by equation 5 at the film cooling hole location as if itwas not there.On the other hand, Tr2 is the recovery temperature of the coolant at the entry of thefilm cooling hole and can be calculated by equation 13 .At this point, we assumed that;

• The flow ejecting through film cooling holes is laminar and hence, recovery factor (r) is equalto√Pr.

ηr =Taw − TrTr2 − Tr

(11)

Taw = ηr(Tr2 − Tr) + Tr (12)

Tr2 = T2(1 + rγ − 1

2M2

2 ) (13)

Now, we have the adiabatic wall temperature of the external flow at each data point and thisvalue should be compared with the initial estimation.If difference between the values is greater than0.01, the process should be repeated by simply replacing the initial estimation with the latest valueof adiabatic wall temperature.

Boundary Conditions at the midspan of NGV

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Figure 30: Static pressure (P∞) at the edge of pressure side BL

Figure 31: Static pressure (P∞) at the edge of suction side BL

Figure 32: Mach number (M∞) at the edge of pressure side BL

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Figure 33: Mach number (M∞) at the edge of suction side BL

Figure 34: Static temperature (T∞) at the edge of pressure side BL

Figure 35: Static temperature (T∞) at the edge of suction side BL

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Figure 36: Adiabatic Wall Temperature (Taw) along the pressure side

Figure 37: Adiabatic Wall Temperature (Taw) along the suction side

Figure 38: Pressure side BL mean temperature (T ∗)

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Figure 39: Suction side BL mean temperature (T ∗)

4.2 Method for External Flow HTC Calculation and HTC ResultsEquation 14 is the correlation given by Churchill and Bernstein [4] and is used in order to calculatethe Nusselt number at the stagnation point.

Nud = 0.3 +0.62Re

1/2D Pr1/3[

1 + (0.4/Pr)2/3]1/4

[1 +

(ReD

282000

)5/8]4/5

(14)

This correlation is valid for the range PrReD ≥ 0.2. In addition, the following relationrepresents the equation used for calculating HTC for leading edge region. Here, kf stands forconduction coefficient of fluid at operating temperature and d stands for diameter of the leadingedge cylinder.

hd =(Nud × kf )

d(15)

In order to calculate the heat transfer coefficients over the flat plates representing suction andpressure sides, local Nusselt number correlations for both laminar and turbulent flow are used.Thiscorrelation are applicable for certain conditions.Hence, we assumed that;

• The flow is parallel to the flat plates.

• The heat flux is constant along flat plates.

Equation 16 and 17 are used for laminar and turbulent flows, respectively.In other words, atpoints where Reynolds number is greater than 5 × 105 equation 17 is used for Nusselt numberestimation.These correlations are taken from the book [2].

Nux = 0.453Re1/2x Pr1/3

Pr ≥ 0.6(16)

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Nux = 0.0308Re4/5x Pr1/3

0.6 ≤ Pr ≤ 60(17)

hx =Nuxkfx

(18)

All the temperature dependent parameters appearing in these correlations are calculated at themean temperature defined in the previous section.Since, Mach number distribution along the edgeof boundary layer is already obtained, the velocity distribution can easily be obtained in order toplug in Reynolds number calculation.

V∞ = M∞a∞ (19)

where

a∞ =√γRT∞ (20)

The static temperature distribution (T∞) along the edge of boundary layer is already calcu-lated.Therefore, only parameters missing are R and γ.At this point, we assumed that;

• γ and R of the external flow is constant and equal to 1.35 and 287 J/kgK, respectively.

Local Heat Transfer Coefficient Results along the midspan of NGV

Figure 40: Heat transfer coefficient distribution along pressure side

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Figure 41: Heat transfer coefficient distribution along suction side

5 Hand Calculation for Internal Flow / 1-D Flow Network Anal-ysis Results

In this section, the main goal is to describe the method and results of the 1-D coolant flow analysis.The mass flow rate of the coolant was given as 7% of the flow entering the engine which was 3.2kg/s in the previous report. However, according to CFD results, 5% mass flow rate is enough tocool down the NGV to required temperature levels. Therefore, the mass flow rate of the coolantreduced to 5% of the flow entering the engine. Inlet conditions of the coolant flow is given in table7 for one NGV.

Table 7: Parameters related to cooling flow

Parameter ValuePt [kPa] 1200Tt [K] 700m [kg/s] 8× 10−3

5.1 Details of the MethodIn this part, the steps followed in order to obtain the spanwise variations of pressure, temperature,mass flow rate and velocity of the cooling flow are presented.The NGV cooling design consists of 3different channels. The front two are connected to each other with jet holes, the flow enters fromthe back side and feeds the front channel through the jet holes. The leading edge channel has filmholes in order to eject the coolant flow. The back channel consists of trailing edge exit slot and filmholes to eject the coolant. The coolant properties inside the channels are calculated by dividing theNGV in the direction of the span. Due to the taper of the NGV, the front two and the back channelare divided into different number of sections which are shown in figure 42.

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Figure 42: Sections

Since the two channel located in the front and back channel are independent, the developmentof coolant properties through these sections are observed separately by using proper correlationsstarting from the inlet section. Each section receives the flow from the previous section and passeson to the next after the application of relevant Nusselt number and friction coefficient correlations.The details of the calculation of the mass flow rate, pressure, velocity and temperature of the coolantfor each section are described in following sections, respectively.

5.1.1 Mass Flow Rate Calculation Method and Initial Mass Flow Rates

The coolant mass flow rate is divided into two different channels which are the front channel tofeed the jet impingement holes and one at the back of the NGV. The initial mass flow rates to besupplied to the channels are decided according to results of the CFD analysis. Both the temperatureand pressure requirements are considered while deciding the mass flow rates.The final decision onmass flow rates can be seen in below table.

Table 8: Channel Initial Mass Flow Rates

mchannel1 [kg/s] 1.4× 10−3

mchannel2 [kg/s] 6.6× 10−3

Mass flow rate change through the blade is examined by using small sections defined previously.The mass flow rate of the coolant decreases from hub to tip, since it is ejected from the film coolingand jet holes or trailing edge slot.The remaining mass flow rate enters the section above. In figure53, the mass flow rates of the coolant ejecting from cooling channels by different means can beseen.

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Figure 43: Ejected Mass Flows from One Section

The mass flow discharged from the film holes and the trailing edge slot are found thanks to theCFD analysis performed. The temperature, pressure and velocity values of the coolant at the entryof the ejection holes are taken from the analysis and corresponding mass flow rates are calculatedby using below equations.

m = ρUA (21)

ρ =P

TR(22)

After finding the ejected mass flow rates, the remaining mass flow rates to enter the next sectioncan be calculated as follows for the two channels.

mchannel1,out = mchannel1,in − (mTE + mfilm) (23)

mchannel2,out = mchannel2,in − mfilm (24)

5.1.2 Coolant Flow Properties Calculation Method

The coolant flow properties change along the span due to pressure losses and heating of thecoolant.Therefore, in each section, the static pressure, temperature, mass flow rate are calculatedafter calculating the heat transfer coefficient and pressure losses.

To calculate the pressure drop and heat transfer coefficient, Gnielsky [2] correlation is usedfor the fully turbulent duct flow in smooth channel. Both the Nusselt number and friction factorcorrelations depend on the Reynolds number. Since the channels are non circular, the hydraulicdiameter is used in Reynolds number calculation. The equations used in the calculation of HTC andpressure drop of each section are shown below.

ReD =Vchannel ·Dh

ν(25)

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Dh =4AcP

where the Ac is the channel area and P is the wetted perimeter of channel.

f = (0.79 · log(ReD)− 1.64)−2 (26)

NuD =(f/8)(ReD − 1000)Pr

1 + 12.7(f/8)0.5(Pr0.66 − 1)(27)

which is valid for 0.5 < Pr < 2000 and 3000 < Re < 5× 106

NuD =(f/8)(ReD − 1000)Pr

1 + 12.7(f/8)0.5(Pr0.66 − 1)(28)

h = kNuDDh

(29)

∆P = fL

Dh

ρ

2V 2channel (30)

To calculate the pressure drop and heat transfer coefficient change along the span, one shouldfind the static properties of the flow at the inlet of the each section. Since the total pressure andtemperature values of the coolant are provided at the cooling channel inlet, the first thing that shouldbe done is to calculate Mach number and to obtain static values of temperature and pressure.Sincethe initial mass flow rate of the channels are decided, the below equation can be used to calculatethe Mach number at the inlet of the first section. After performing a few iterations in the code, theequation gives the Mach number of the flow.

M =m

A

(1 + γ−12M2)1/γ−1

ρt

1√γRTt

√1 +

γ − 1

2M2 (31)

Therefore, the static temperature and pressure can be obtained by below isentropic relations.

P =Pt

(1 + γ−12M2)γ/γ−1

(32)

T =Tt

1 + γ−12M2

(33)

The next step is to calculate velocity since velocity is needed for Reynolds number calcula-tion.The velocity of a flow with a certain amount of mass flow rate can be calculated as follows.

V =m

ρA(34)

After finding the pressure drop of coolant along the section and heat transfer coefficient betweenthe coolant and inner surface of NGV, the change in the temperature of the coolant can be determinedby establishing a 1D thermal circuit analogy as in below figure.

34

Figure 44: 1D Thermal Circuit of NGV

As seen from the figure, the heat transfer along the x direction is constant and it can be foundby using below equation. In this method, the following assumptions are done to find coolant statictemperature.

• There is no heat generation.

• Constant heat flux along one section.

• Constant properties along one section.

• There is no radiation.

• Outside surface area exposed to heat transfer and the area exposed to coolant are equal toeach other.

• Steady state condition.

qx =Taw − Tc(

1hgasA

+ LkA

+ 1hcA

) (35)

where

Taw: Averaged adiabatic wall temperature of hot gasTc: Coolant temperaturehgas: Averaged heat transfer coefficient of hot gashc: Coolant heat transfer coefficientk: Conductive heat transfer coefficient of NGV

After determining the heat transfer, one can find the outside wall temperature by using followingequation.

Tw,o = Taw − q1

Ahgas(36)

Then, in order to obtain the temperature increase of the coolant from one radius to another alongspanwise, below energy balance can be written [5].This equation relates the heat transfer to thecoolant to the temperature change of the coolant through spanwise direction.

35

mccpcdTcdR

= Ahg(Taw − Tw,o) (37)

Therefore, the below equation can be derived to estimate the temperature of the coolant at theentrance of the next section that it reaches.The subscript ’in’ indicates the temperature of the coolantentering a section and ’out’ indicates the coolant temperature after it passes through a section alongthe span.

Tc,out = Tc,in +Ahg(Taw − Tw,o)

mcpc(38)

5.2 Results for Mass Flow

Figure 45: Front Channel Mass Flow Rates along Span

Figure 46: Impingement Holes Mass Flow Rates

36

Figure 47: Front Film Holes Exit Mass Flow Rates

Figure 48: Aft Channel Mass Flow Rates along Span

Figure 49: Aft Film Holes Exit Mass Flow Rates

37

Figure 50: Trailing Edge Slot Exit Mass Flow Rates along Span

5.3 Results for Pressure, Temperature, Velocity

Figure 51: Coolant Flow Properties of Front Channel along Span

38

Figure 52: Coolant Flow Properties of Front Film Holes

39

Figure 53: Coolant Flow Properties of Aft Channel

5.4 Back Flow Margin (BFM) CalculationsBack flow margin is calculated for film hole ejections and trailing edge slot ejection. The equationused in calculation of BFM is given Equation 39. The data for calculation of BFM is directly takenfrom ANSY CFX results. For each film hole a central data point is placed at the ejection surfaces offilms. In the same manner a data line including seven data points is placed at the TE slot ejectionsurface. These ejection surfaces are described as interfaces in CFX. Total pressure value from colddomain and static pressure value from hot domain are plugged in BFM equation. The results forBFM for each ejection is shown in Table 9.

BFM =

(Pt,coolantPs,gas

− 1

)× 100 (39)

wherePt,coolant: Total pressure of ejected coolantPs,gas: Static pressure of hot gas

40

Table 9: BFM values for ejections of film holes and TE slot.

Point Jet Channel Films Smooth Channel Films TE Slot1 2.49 % 1.29 % 23.04 %2 1.92 % 1.29 % 24.34 %3 1.81 % 1.34 % 26.41 %4 1.87 % 1.37 % 26.81 %5 2.16 % 1.37 % 27.79 %6 2.17 % 1.31 % 27.59 %7 2.16 % 1.21 % 29.50 %8 2.06 % 1.20 %9 1.73 % 1.32 %

10 1.72 %

6 Hand Calculation of Heat Transfer / 1-D Analysis Results

6.1 Details of the MethodThe method how we calculate the properties of internal flow in each section has already beenshown in the previous section.Thanks to the these calculations, we have the coolant temperature ateach section.Moreover, previously, external heat transfer coefficient and adiabatic wall temperaturevalues along pressure and suction sides has already been obtained in section 4. Therefore, byestablishing a thermal circuit analogy between the external and internal flow, we can easily solvefor inner (Twi) and outer (Two) surface temperatures of NGV.Below, the thermal circuit can be seen.

Figure 54: 1D Thermal Circuit

qx =Taw − Tc

Rext +Rblade +Rcoolant

(40)

41

whereRext = 1/(hextA)

Rblade = L/(kbladeA)

Rcoolant = 1/(hcoolantA)

Inner and outer surface temperatures can be calculated as follows;

qx =Taw − Tw,outer

Rext

(41)

Tw,outer = Taw − qxRext

qx =Tw,outer − Tw,in

Rblade

(42)

Tw,inner = Tw,outer − qxRblade

At this point, the resistance called Rcoolant should be calculated separately throughout the 4portions shown below.These are the front most channel where jet impingement happens, the smoothchannel supplying coolant to that channel, the smooth channel at the back and the trailing edge slotwhich we treat as a smooth channel, too. Here is an important assumption made while solving forthe inner and outer wall temperatures.It is assumed that;

• The properties of coolant in any channel do not vary along the chord of the section.

Therefore, at a spanwise section, all the points interacting with any of the 4 channels experiencethe same coolant temperature and internal heat transfer coefficient.However, the external flowcharacteristics differ from one point to another on the blade surface.In other words, the differencesbetween the circuits established for the points interacting with the same coolant channel are onlythe Taw and external heat transfer coefficient .

Figure 55: The 4 portions of the cooling system

42

We have already calculated the heat transfer coefficient distributions through the channels2,3 and 4 since they are needed to calculate the variations in coolant properties (i.e temperature,pressure, mass flow rate).However, heat transfer rate happening thanks to jet impingement in thefront most channel is not included in those equations.Jet impingement provides heat transfer onlythrough channel 1.Now, the Nusselt number correlation used in order to calculate the heat transfercoefficient between the coolant in channel 1 and the walls of that channel should be defined.

Jet Impingement Correlation

Nu = 0.44Rejet

(d

s

)0.8

exp

(−0.85

(l

d

)(d

s

)(d

D

)0.4)

(43)

This correlation is valid between the parameter ranges given in Table 5.

Table 10: Chupp et al. correlation limits [6]

Lower Limit Upper LimitRejet 3000 20000s/d 4 16l/d 4 10D/d 1.5 16

whereRejet = Vjetd/νD : target surface diameterd : jet hole diameterl : distance between the jet exit plane and target surfaces : center-to-center

In order to calculate the Reynolds number of the jet flow, velocity of the coolant flows (Vjet)ejecting through jet holes are determined by CFD results.After calculating the Nusselt number,one can determine the heat transfer coefficient between bu using below relation.Here k is theconductivity of the cooling fluid evaluated at the sectional temperature.

h = kNu

d(44)

43

Figure 56: Heat transfer coefficient (hcoolant) along the span through the 1st portion

Figure 57: Heat transfer coefficient (hcoolant) along the span through the 2nd portion

44

Figure 58: Heat transfer coefficient (hcoolant) along the span through the 3rd portion

Figure 59: Heat transfer coefficient (hcoolant) along the span through the 4th portion

6.2 Metal Temperature DistributionThe NGV wall temperature distribution in chordwise direction at different section of NGV is foundby using both hand calculation and CFD analysis. In this section, both results are given.

45

Figure 60: Blade pressure side average temperature distribution along spanwise

Figure 61: Blade suction side average temperature distribution along spanwise

Figure 62: Blade pressure side average temperature distribution (CFD)

46

Figure 63: Suction side average temperature distribution (CFD)

6.2.1 Pressure Side

Figure 64: Pressure side wall temperature distribution at midspan

47

Figure 65: Pressure side temperature distribution with CFD and hand calculation results at midspan

Figure 65 shows that there is sufficiently matching between CFD and hand calculation results. Thereare some sudden drops at temperature at hand calculations where the film holes exist. Althoughthe general behaviour is the same, there are small differences between the temperature valuescalculated by these 2 methods.This difference definitely arises due to several assumptions madewhile performing hand calculations.

6.2.2 Suction Side

Figure 66: Suction side wall temperature distribution at midspan

48

Figure 67: Suction side wall temperature distribution with CFD and hand calculation results atmidspan

At each point on the midspan section, hand calculation results are higher than the CFD results.This difference definitely arises due to the assumptions made to be able to use Nusselt numbercorrelations. However, the temperature difference occurring in suction side between the handcalculation and CFD is much greater than that is occurring in the pressure side. This is because thesuction side curvature is much higher than that of pressure side. Therefore, approaching suctionside as a flat plate does not give as accurate results as in the case of pressure side.

6.2.3 Temperature Distribution at different sections along the span of NGV

Figure 68: Pressure side wall temperature distribution at % 30 span

49

Figure 69: Pressure side wall temperature distribution at % 60 span

Figure 70: Pressure side wall temperature distribution at % 80 span

50

Figure 71: Suction side wall temperature distribution at % 30 span

Figure 72: Suction side wall temperature distribution at % 60 span

51

Figure 73: Suction side wall temperature distribution at % 80 span

Figure 74: Blade pressure side wall temperature distribution at different sections

52

Figure 75: Blade suction side wall temperature distribution at different sections

6.3 Comments on Wall Temperature ResultsWhen the figure 65 and figure 67 are examined, it is clearly seen that wall temperature result of handcalculation is higher than the CFD result. However, the location where the temperature makes apeak or decrease to minimum is the same for both result. The wall temperature makes a peak around%30 of the chord for suction side and decrease to minimum near the trailing edge. On the other hand,wall temperature increases or decreases sharply in hand calculation.In other words, the changes inwall temperatures coming from hand calculations are not as smooth as the changes occurring inCFD results. It may be resulted from ignoring the 2D conduction in the NGV. In addition, becausemany assumptions are made while calculating the temperatures by hand, we were expecting to seedifferences between the results of hand calculation and CFD analysis. In this project, the CFDresults are considered to be more reliable because assumptions of the hand calculation definitelydecrease the accuracy of the results. Therefore, the cooling effectiveness and life time calculationsare done by using CFD results.

6.3.1 Cooling Effectiveness

The cooling effectiveness of the design can be determined by the equation below [7].

φ =Tg − Tm,extTg − Tc,in

(45)

where

Tg : Outside hot gas temperatureTm,ext : Average NGV external surface temperatureTc,in : Coolant inlet temperature

After plugging the numbers, the cooling effectiveness is determined. Since the CFD resultis more reliable than 1D hand calculation, while calculating the cooling effectiveness, the NGV

53

averaged external wall temperature of CFD result is used. At the end, the cooling effectiveness ofthe NGV design is found as given below.

φ = 0.66

7 Hand Calculation for Cold Flow Test/1-D Heat Transfer Anal-ysis Results

Tamb[K] PR[-] P1[Pa] P2[Pa] MFR[g/s] Cd[-]

293.15 1.05 0.969 0.923 1.34× 10−4 0.8

293.15 1.10 1.015 0.923 1.9× 10−4 0.8

293.15 1.15 1.061 0.923 2.32× 10−4 0.8

293.15 1.20 1.108 0.923 2.68× 10−4 0.8

293.15 1.25 1.154 0.923 3× 10−4 0.8

The discharge coefficient of a slot or hole can be defined as the ratio of actual mass flow rate passingthrough the hole or slot to the mass flow rate that would pass there if there were no losses.We knowthat P1 is the inlet total pressure (Pt) and P2 is the outlet static pressure (Ps).If we assume that theflow is isentropic between the inlet and outlet section of a hole or slot, we can calculate the idealmass flow rate as follows.

Pt − Ps =1

2ρV 2 (46)

where the density is taken at 293.25 K and is equal to 1.204 kg/m3.

V =√

2(Pt − Ps)/ρ

mideal = ρA√

2(Pt − Ps)/ρ

Therefore, discharge coefficient can be calculated by the following equation.

cd =mactual

ρA√

2(Pt − Ps)/ρ(47)

Experimental results for discharge coefficient of holes oriented with 45 degrees to the mainstreamflow change depending on the pressure ratio (PR). In the literature, pressure ratios between 1 and

54

1.25, approximately yield a discharge coefficient of 0.8 [8]. Therefore, the actual mass flow ratepasses through holes are calculated as follows.

mactual = cdA√

2ρ(Pt − Ps) (48)

8 Lifetime Prediction

8.1 Stress Calculation DetailsIn this section stress calculation details are presented. As a stationary part, NGV is investigated interms of thermal stresses only since NGV do not rotate so other kind of stresses like centrifugalstresses or disk stress do not take into consideration. Thermal stress is calculated with the help ofcalculation of elongation of blade with the change of temperature.

∆L = α∆TL (49)

∆L =FL

AE(50)

σthermal =F

A= αE∆T (51)

whereα : Coefficient of thermal expansion (17.7e-6 1/C°)E : Young’s Modulus (1.35e+11 Pa)∆L : Elongation∆T : Total change in temperature between the hot and cold sides (Thotwall- Tcoldwall)A : Cross section areaF : Force due to elongationL : Lengthσthermal : Thermal stress on blade

Thermal stress is calculated in both longitudinal and lateral directions. The dimensions chosenfor calculation is shown in Figure 76.

55

Figure 76: Dimensions for calculation of thermal stress.

The temperature change causes an elongation and this results in a force action on NGV. Whenforce is calculated at transverse direction length is taken as L1 and area is taken as A1. Forceat longitudinal direction is calculated with taking account of dimensions L2 as length and A2 asarea. Also NGV is divided into small elements shown in Figure 77 to increase the accuracy of thecalculation.

Figure 77: Area calculation

56

Points which seen in Figure 77, are used to calculate value. Stress is calculated with onlyvariable of temperature difference between inner and external side of blade wall. Calculation ofstress value is done by using MATLAB just as input the wall temperature at inner and external sidewhich values are taken from CFD results at the determinate points. Temperature difference taken asinner and external side of wall since the main stress occurs due to this difference in NGV [9]. As aresult maximum stress is found as 120 MPa at blade.

8.2 Lifetime Calculation DetailsLarson-Miller parameter curves are used to calculate the lifetime estimation of NGV blade. TheLarson-Miller parameter provides an estimation for the time that the material may endure atoperational temperature. Figure 78 shows the Larson-Miller parameter curve of superalloy CM247LC DS which is the material of NGV.

Figure 78: Larson-Miller Parameter Curve

LMP = 1.8× T (K)× [20 + log(t)]/1000 (52)

Temperature (T(K)) in equation 78 is taken as 1190 K which is the maximum temperaturethat the NGV experiences according to ANSYS results. To find lifetime of the NGV for bothlongitudinal and transverse, temperature distribution over the NGV is taken into consideration.Stress at all pieces is calculated with the help of MATLAB code. When we found Larson-Millerparameter from figure 78, LMP is 54 for longitudinal and 51 for lateral direction for 120 MPa stressvalue.

54 = 1.8× 1190K × [20 + log(t)]/1000 (53)

LMP = 1.8× 1190K × [20 + log(t)]/1000 (54)

Equation 53 give lifetime as 162212 hours for longitudinal direction and equation 54 givelifetime for blade at transverse direction as 6449.5 hours. These values are bigger than 5000 hours.Therefore requirement of TEKNOFEST is sustained with our design for lifetime.

57

9 Appendix

Listing 1: Algorithm which is used in Hand Calculation1 c l c ; c l e a r ;2

3 %% Flow p r o p e r t i e s4 l o a d Tab le . d a t % I d e a l gas p r o p e r t i e s wr t Tempera tu r e5 T = Tab le ( : , 1 ) ; %K e lv in6 Rho = Tab le ( : , 2 ) ;7 Cp = Tab le ( : , 3 ) ;8 Visco = Tab le ( : , 4 ) *10ˆ(−7) ;9 k = Tab le ( : , 6 ) *10ˆ(−3) ;

10 Pr = Tab le ( : , 8 ) ;11 Y c = 1 . 4 ;12 Y t = 1 . 3 5 ;13 Rc = 287 ;14 c p c = Y c*Rc / ( Y c−1) ;15 c p t = 1248 ;16 Rt = c p t * ( Y t−1) / Y t ;17

18 %% E x t e r n a l Flow19

20 %g e o m e t r i c i n p u t ( c h o r d w i s e and s p a n w i s e l o c a t i o n s )21 s s c u r v e d e l = 3 9 . 7 8 8 / 4 8 ;22 s s f l a t p l a t e =0 : s s c u r v e d e l : 3 9 . 7 8 8 ;23 p s c u r v e d e l = 3 0 . 0 9 7 / 4 9 ;24 p s f l a t p l a t e =0 : p s c u r v e d e l : 3 0 . 0 9 7 ;25 p s f l a t p l a t e ( 5 1 ) = 3 0 . 0 9 7 + 2 . 3 6 4 ;26

27

28

29 % from a n s y s30 l o a d ( ’PRESSURE SS SEC 7 . d a t ’ )31 l o a d ( ’PRESSURE SS SEC 6 . d a t ’ )32 l o a d ( ’PRESSURE SS SEC 5 . d a t ’ )33 l o a d ( ’PRESSURE SS SEC 4 . d a t ’ )34 l o a d ( ’PRESSURE SS SEC 3 . d a t ’ )35 l o a d ( ’PRESSURE SS SEC 2 . d a t ’ )36 l o a d ( ’PRESSURE SS SEC 1 . d a t ’ )37 l o a d ( ’PRESSURE PS SEC 7 . d a t ’ )38 l o a d ( ’PRESSURE PS SEC 6 . d a t ’ )39 l o a d ( ’PRESSURE PS SEC 5 . d a t ’ )40 l o a d ( ’PRESSURE PS SEC 4 . d a t ’ )41 l o a d ( ’PRESSURE PS SEC 3 . d a t ’ )

58

42 l o a d ( ’PRESSURE PS SEC 2 . d a t ’ )43 l o a d ( ’PRESSURE PS SEC 1 . d a t ’ ) % gauge p r e s s u r e44

45

46 % s t a t i c p r e s s u r e d i s t r i b u t i o n on NGV−a b s o u l u t e p r e s s u r e47 P i n f p r e s s ( : , 1 ) = PRESSURE PS SEC 1 ( : , 2 ) +101000;48 P i n f s u c t ( : , 1 ) = PRESSURE SS SEC 1 ( 5 0 : −1 : 2 , 2 ) +101000;49 P i n f p r e s s ( : , 2 ) = PRESSURE PS SEC 2 ( : , 2 ) +101000;50 P i n f s u c t ( : , 2 ) = PRESSURE SS SEC 2 ( 5 0 : −1 : 2 , 2 ) +101000;51 P i n f p r e s s ( : , 3 ) = PRESSURE PS SEC 3 ( : , 2 ) +101000;52 P i n f s u c t ( : , 3 ) = PRESSURE SS SEC 3 ( 5 0 : −1 : 2 , 2 ) +101000;53 P i n f p r e s s ( : , 4 ) = PRESSURE PS SEC 4 ( : , 2 ) +101000;54 P i n f s u c t ( : , 4 ) = PRESSURE SS SEC 4 ( 5 0 : −1 : 2 , 2 ) +101000;55 P i n f p r e s s ( : , 5 ) = PRESSURE PS SEC 5 ( : , 2 ) +101000;56 P i n f s u c t ( : , 5 ) = PRESSURE SS SEC 5 ( 5 0 : −1 : 2 , 2 ) +101000;57 P i n f p r e s s ( : , 6 ) = PRESSURE PS SEC 6 ( : , 2 ) +101000;58 P i n f s u c t ( : , 6 ) = PRESSURE SS SEC 6 ( 5 0 : −1 : 2 , 2 ) +101000;59 P i n f p r e s s ( : , 7 ) = PRESSURE PS SEC 7 ( : , 2 ) +101000;60 P i n f s u c t ( : , 7 ) = PRESSURE SS SEC 7 ( 5 0 : −1 : 2 , 2 ) +101000;61 P i n f p r e s s ( 5 1 , 1 ) = PRESSURE SS SEC 1 ( 1 , 2 ) +101000;62 P i n f p r e s s ( 5 1 , 2 ) = PRESSURE SS SEC 2 ( 1 , 2 ) +101000;63 P i n f p r e s s ( 5 1 , 3 ) = PRESSURE SS SEC 3 ( 1 , 2 ) +101000;64 P i n f p r e s s ( 5 1 , 4 ) = PRESSURE SS SEC 4 ( 1 , 2 ) +101000;65 P i n f p r e s s ( 5 1 , 5 ) = PRESSURE SS SEC 5 ( 1 , 2 ) +101000;66 P i n f p r e s s ( 5 1 , 6 ) = PRESSURE SS SEC 6 ( 1 , 2 ) +101000;67 P i n f p r e s s ( 5 1 , 7 ) = PRESSURE SS SEC 7 ( 1 , 2 ) +101000;68

69 %% LE s i l i n d i r ( l a m i n a r )70 l o a d ( ’BL STAG Vinf . d a t ’ )71 l o a d ( ’BL STAG Temp . d a t ’ )72 V i n f s t a g =BL STAG Vinf ( : , 2 ) ;73 T s LE=BL STAG Temp ( : , 2 ) ;74 d = 1 . 2 7 5 * 2 / 1 0 0 0 ; %l e d i a m e t e r75 P t i n f = 1 1 8 0 * 1 0 ˆ 3 ; %t o t a l p r e s s u r e [ Pa ]76 P r x s u c t = 0 . 7 ;77 T t i n f = 1780 ;78 % i n i t i a l i z i n g79 T w o u t s u c t = 1250* ones ( 4 9 , 7 ) ;80 T w o u t p r e s = 1250* ones ( 5 1 , 7 ) ;81 R e s u c t x ( 1 : 4 9 , 1 : 7 ) =6996000;82 R e p r e s x ( 1 : 5 1 , 1 : 7 ) =1007200;83

84

85 f o r p =1:100 % i t e r a t i o n86 f o r j =1:7%s p a n w i s e l o c a t i o n

59

87 f o r i =1:1 %c h o r d w i s e l o c a t i o n88 T w LE ( i , j ) = T w o u t s u c t ( i , j ) ; %K89 Y t ( i , j ) = 1 . 3 5 ;90 a s u c t ( i , j ) = s q r t ( Y t ( i , j ) * Rt * T s LE ( j ) ) ; %speed of sound91 Minf LE ( i , j ) = V i n f s t a g ( j ) / a s u c t ( i , j ) ;92 r LE ( i , j ) = 1 ;93 T aw LE ( i , j ) = T t i n f ;94 Tmean LE ( i , j ) = T s LE ( j ) + 0 . 5 8 * ( T w LE ( i , j )−T s LE ( j ) ) + 0 . 1 9 * (

T aw LE ( i , j )−T s LE ( j ) ) ;95

96 r h o d ( i , j ) = P i n f s u c t ( i , j ) / T aw LE ( i , j ) / Rt ;%P i n f s u c t ( i , j ) /Tmean LE ( i , j ) / Rt ;

97 v i s d = i n t e r p 1 ( T , Visco , T aw LE ( i , j ) ) ;%i n t e r p 1 ( T , Visco ,Tmean LE ( i , j ) ) ;

98 k d = i n t e r p 1 ( T , k , T aw LE ( i , j ) ) ;%i n t e r p 1 ( T , k , Tmean LE( i , j ) ) ;

99 Re d ( i , j ) = V i n f s t a g ( j ) * r h o d ( i , j ) *d / v i s d ; %check f o rv e l o c i t y

100 Tu = 0 . 1 ;101 Nu d ( i , j ) = 0 . 3 + ( ( 0 . 6 2 * Re d ( i , j ) ˆ ( 0 . 5 ) * 0 . 7 ˆ ( 1 / 3 ) )

/ ( 1 + ( 0 . 4 / 0 . 7 ) ˆ 2 / 3 ) ˆ 0 . 2 5 ) * ( ( 1 + ( Re d ( i , j ) / 2 8 2 0 0 0 ) ˆ ( 5 / 8 ) )ˆ ( 4 / 5 ) ) ;

102 h LE ( i , j ) = Nu d ( i , j ) * k d / d ;103 end104

105 end106

107 %% PRESSURE SIDE LOCAL HTC108 x f i r s t f i l m = 2 . 4 5 5 / 1 0 0 0 ; % f i l m h o l e l o c a t i o n s109 x s e c o n d f i l m = 9 . 8 2 6 / 1 0 0 0 ;110 % i n i t l i a z i n g111 P r x p r e s s ( 2 , 1 ) = 0 . 7 ;112 P r x p r e s s ( 2 , 2 ) = 0 . 7 ;113 P r x p r e s s ( 2 , 3 ) = 0 . 7 ;114 P r x p r e s s ( 2 , 4 ) = 0 . 7 ;115 P r x p r e s s ( 2 , 5 ) = 0 . 7 ;116 P r x p r e s s ( 2 , 6 ) = 0 . 7 ;117 P r x p r e s s ( 2 , 7 ) = 0 . 7 ;118

119 x s u c t p l a t e =( s s f l a t p l a t e ) / 1 0 0 0 ; %f l a t p l a t e l o c a t i o n s f o r l e =0 ;120 x p r e s p l a t e =( p s f l a t p l a t e ) / 1 0 0 0 ; %121 L p r e s s p l a t e = x p r e s p l a t e ( end ) ; % f l a t p l a t e l e n g t h122

123 T t i n f =1780; % t o t a l t e m p e r a t u r e o f h o t gas [K]124

60

125 f o r j =1:7126 % i n i t i l i a z i n g127 T w p r e s s ( 1 , j ) =T w LE ( 1 , j ) ; % s u r f a c e t e m p e r a t u r e128 h p r e s x ( 1 , j ) =h LE ( 1 , j ) ;129 T a w p r e s s ( 1 , j ) =T aw LE ( 1 , j ) ;130

131 f o r i =2 : s i z e ( x p r e s p l a t e , 2 )132 T w p r e s s ( i , j ) = T w o u t p r e s ( i , j ) ; %K133 Y t ( i , j ) = 1 . 3 5 ;134 M i n f p r e s s u r e ( i , j ) = s q r t ( ( ( P t i n f / P i n f p r e s s ( i , j ) ) ˆ ( ( Y t ( i , j

)−1) / Y t ( i , j ) )−1) * 2 / ( Y t ( i , j )−1) ) ;135 T s p r e s s ( i , j ) = T t i n f / ( 1 + ( Y t ( i , j )−1) / 2 * M i n f p r e s s u r e ( i

, j ) ˆ 2 ) ; %136

137 i f R e p r e s x ( i , j ) < 500000 % x c r check138 r p r e s s u r e ( i , j ) = s q r t ( P r x p r e s s ( i , j ) ) ;139 T a w p r e s s ( i , j ) = T s p r e s s ( i , j ) *(1+ r p r e s s u r e ( i , j ) * ( (

Y t ( i , j )−1) / 2 ) * M i n f p r e s s u r e ( i , j ) ˆ 2 ) ; %p r a n d t lc o r r e c t i o n

140 e l s e141 r p r e s s u r e ( i , j ) = ( P r x p r e s s ( i , j ) ) ˆ ( 1 / 3 ) ; % f u l l y

t u r b u l e n t142 T a w p r e s s ( i , j ) = T s p r e s s ( i , j ) *(1+ r p r e s s u r e ( i , j ) * ( (

Y t ( i , j )−1) / 2 ) * M i n f p r e s s u r e ( i , j ) ˆ 2 ) ; %p r a n d t lc o r r e c t i o n

143 end144 % f i l m c o o l i n g Taw c a l c u l a t i o n145 c o u n t e r =0;146 c o u n t e r 2 =0;147 i f x p r e s p l a t e ( i ) >= x f i r s t f i l m && x p r e s p l a t e ( i ) <=

x s e c o n d f i l m148 c o u n t e r = c o u n t e r +1;149 i f c o u n t e r ==1150 V e l i n f = V i n f p r e s s ( i , j ) ;151 P r e s i n f = P i n f p r e s s ( i , j ) ;152 Temp inf = T s p r e s s ( i , j ) ;153 x e j e c t i o n ( j , 1 ) = x p r e s p l a t e ( i ) ;154 P s t a t i c o u t ( j , 1 ) = P i n f p r e s s ( i , j ) ;155 end156 Tr = T a w p r e s s ( i , j ) ;157 x= x p r e s p l a t e ( i )−x f i r s t f i l m ;158 a =1;159 [ T a w p r e s s ( i , j ) , e t a f i l m ( i , j ) ] = f i l m c o o l i n g ( Tr ,

T s p r e s s ( i , j ) , x , j , a , V e l i n f , P i n f p r e s s ( i , j ) , P r e s i n f ,Temp inf ) ;

61

160

161

162 e l s e i f x p r e s p l a t e ( i ) >= x s e c o n d f i l m163 c o u n t e r 2 = c o u n t e r 2 +1;164 i f c o u n t e r 2 ==1165 V e l i n f 2 = V i n f p r e s s ( i , j ) ;166 P r e s i n f 2 = P i n f p r e s s ( i , j ) ;167 Temp inf2 = T s p r e s s ( i , j ) ;168 x e j e c t i o n ( j , 2 ) = x p r e s p l a t e ( i ) ;169 P s t a t i c o u t ( j , 2 ) = P i n f p r e s s ( i , j ) ;170 end171 Tr = T a w p r e s s ( i , j ) ;172 x= x p r e s p l a t e ( i )−x s e c o n d f i l m ;173 a =2;174 [ T a w p r e s s ( i , j ) , e t a f i l m ( i , j ) ] = f i l m c o o l i n g ( Tr ,

T s p r e s s ( i , j ) , x , j , a , V e l i n f 2 , P i n f p r e s s ( i , j ) ,P r e s i n f 2 , Temp inf2 ) ;

175

176 e l s e177 T a w p r e s s ( i , j ) = T s p r e s s ( i , j ) *(1+ r p r e s s u r e ( i , j ) * ( (

Y t ( i , j )−1) / 2 ) * M i n f p r e s s u r e ( i , j ) ˆ 2 ) ;178 end179 Tmean press ( i , j ) = T s p r e s s ( i , j ) + 0 . 5 8 * ( T w p r e s s ( i , j )−

T s p r e s s ( i , j ) ) + 0 . 1 9 * ( T a w p r e s s ( i , j )−T s p r e s s ( i , j ) ) ;180 %end of T mean c a l c u l a t i o n181

182 a p r e s s ( i , j ) = s q r t ( Y t ( i , j ) * Rt * T s p r e s s ( i , j ) ) ; %speedof sound wi th i n i t i a l gamma

183 V i n f p r e s s ( i , j ) = M i n f p r e s s u r e ( i , j ) * a p r e s s ( i , j ) ;184 P r x p r e s s ( i +1 , j ) = i n t e r p 1 ( T , Pr , Tmean press ( i , j ) ) ;185 r h o p x ( i , j ) = P i n f p r e s s ( i , j ) / Tmean press ( i , j ) / Rt ;186 D y n v i s p t ( i , j ) = i n t e r p 1 ( T , Visco , Tmean press ( i , j ) ) ;187 R e p r e s x ( i , j ) = V i n f p r e s s ( i , j ) * r h o p x ( i , j ) *

x p r e s p l a t e ( i ) / D y n v i s p t ( i , j ) ;188 k p r e s x ( i , j ) = i n t e r p 1 ( T , k , Tmean press ( i , j ) ) ;189

190 i f R e p r e s x ( i , j ) < 500000191 h p r e s x ( i , j ) = ( k p r e s x ( i , j ) / x p r e s p l a t e ( i ) ) *0 .453*

R e p r e s x ( i , j ) ˆ ( 1 / 2 ) * P r x p r e s s ( i +1 , j ) ˆ ( 1 / 3 ) ;192 e l s e193 h p r e s x ( i , j ) = ( k p r e s x ( i , j ) / x p r e s p l a t e ( i ) )

*0 .0308* R e p r e s x ( i , j ) ˆ ( 4 / 5 ) * P r x p r e s s ( i +1 , j ) ˆ ( 1 / 3 ) ; %f u l l y t u r b u l e n t

194 end195

62

196

197 end198 h g b a r p r e s ( j ) = t r a p z ( x p r e s p l a t e , h p r e s x ( : , j ) ) / (

x p r e s p l a t e ( end ) ) ;199 end200

201

202

203

204 % SUCTION SIDE LOCAL HTC205 % i n i t i a l i z i n g206 P r s u c t x ( 2 , 1 ) = 0 . 7 ;207 P r s u c t x ( 2 , 2 ) = 0 . 7 ;208 P r s u c t x ( 2 , 3 ) = 0 . 7 ;209 P r s u c t x ( 2 , 4 ) = 0 . 7 ;210 P r s u c t x ( 2 , 5 ) = 0 . 7 ;211 P r s u c t x ( 2 , 6 ) = 0 . 7 ;212 P r s u c t x ( 2 , 7 ) = 0 . 7 ;213 L s u c t p l a t e = x s u c t p l a t e ( end ) ;214

215 f o r j =1:7216 % i n i t i a l i z i n g217 T w s u c t ( 1 , j ) =T w LE ( 1 , j ) ;218 h s u c t x ( 1 , j ) =h LE ( 1 , j ) ;219 T a w s u c t ( 1 , j ) =T aw LE ( 1 , j ) ;220 f o r i =2 : s i z e ( x s u c t p l a t e , 2 )221 T w s u c t ( i , j ) = T w o u t s u c t ( i , j ) ; %K222 Y t ( i , j ) = 1 . 3 5 ;223 M i n f s u c t ( i , j ) = s q r t ( ( ( P t i n f / P i n f s u c t ( i , j ) ) ˆ ( ( Y t ( i , j ) −1)

/ Y t ( i , j ) )−1) * 2 / ( Y t ( i , j ) −1) ) ;224 T s s u c t ( i , j ) = T t i n f / ( 1 + ( Y t ( i , j ) −1) / 2 * M i n f s u c t ( i , j ) ˆ 2 ) ;

%s t a t i c temp a t t h e edge of BL225

226 i f R e s u c t x ( i , j ) < 500000227

228 r s u c t ( i , j ) = s q r t ( P r s u c t x ( i , j ) ) ;229 T a w s u c t ( i , j ) = T s s u c t ( i , j ) *(1+ r s u c t ( i , j ) * ( ( Y t ( i , j )

−1) / 2 ) * M i n f s u c t ( i , j ) ˆ 2 ) ;230 e l s e231

232 r s u c t ( i , j ) = ( P r s u c t x ( i , j ) ) ˆ ( 1 / 3 ) ;233 T a w s u c t ( i , j ) = T s s u c t ( i , j ) *(1+ r s u c t ( i , j ) * ( ( Y t ( i ,

j ) −1) / 2 ) * M i n f s u c t ( i , j ) ˆ 2 ) ;234 end235

63

236 T s s u c t ( i , j ) = T t i n f / ( 1 + ( Y t ( i , j ) −1) / 2 * M i n f s u c t ( i , j) ˆ 2 ) ; %s t a t i c temp a t t h e edge of BL

237 Tmean suc t ( i , j ) = T s s u c t ( i , j ) + 0 . 5 8 * ( T w s u c t ( i , j )−T s s u c t ( i , j ) ) + 0 . 1 9 * ( T a w s u c t ( i , j )−T s s u c t ( i , j ) ) ;

238

239 %end of T mean c a l c u l a t i o n240 a s u c t ( i , j ) = s q r t ( Y t ( i , j ) * Rt * T s s u c t ( i , j ) ) ; %speed

of sound wi th i n i t i a l gamma241 V i n f s u c t ( i , j ) = M i n f s u c t ( i , j ) * a s u c t ( i , j ) ;242 P r s u c t x ( i +1 , j ) = 0 . 7 ; %i n t e r p 1 ( T , Pr , Tmean suc t ( i , j ) ) ;243 r h o s x ( i , j ) = P i n f s u c t ( i , j ) / Tmean suc t ( i , j ) / Rt ;244 D y n v i s s t ( i , j ) = i n t e r p 1 ( T , Visco , Tmean suc t ( i , j ) ) ;245 R e s u c t x ( i , j ) = V i n f s u c t ( i , j ) * r h o s x ( i , j ) * x s u c t p l a t e ( i )

/ D y n v i s s t ( i , j ) ;246 k s u c t x ( i , j ) = i n t e r p 1 ( T , k , Tmean suc t ( i , j ) ) ;247

248 i f R e s u c t x ( i , j ) < 500000249 h s u c t x ( i , j ) = ( k s u c t x ( i , j ) / x s u c t p l a t e ( i ) ) *0 .332*

R e s u c t x ( i , j ) ˆ ( 1 / 2 ) * P r s u c t x ( i , j ) ˆ ( 1 / 3 ) ;250 e l s e251 c r p l a c e ( j ) = i ;252 h s u c t x ( i , j ) = ( k s u c t x ( i , j ) / x s u c t p l a t e ( i ) ) *0 .0296*

R e s u c t x ( i , j ) ˆ ( 4 / 5 ) * P r s u c t x ( i , j ) ˆ ( 1 / 3 ) ;253 end254 % a v a r e g e d HTC255 end256 k b a r = t r a p z ( x s u c t p l a t e , k s u c t x ( : , j ) ) / x s u c t p l a t e ( end ) ;257 U bar = t r a p z ( x s u c t p l a t e , V i n f s u c t ( : , j ) ) / x s u c t p l a t e ( end ) ;258 d y v i s b a r = t r a p z ( x s u c t p l a t e , D y n v i s s t ( : , j ) ) / x s u c t p l a t e ( end )

;259 P r a v e = t r a p z ( x s u c t p l a t e , P r s u c t x ( 1 : i , j ) ) / x s u c t p l a t e ( end ) ;260 r h o b a r = t r a p z ( x s u c t p l a t e , r h o s x ( : , j ) ) / x s u c t p l a t e ( end ) ;261 Re L = r h o b a r * U bar * x s u c t p l a t e ( end ) / d y v i s b a r ;262 Re xc = R e s u c t x ( c r p l a c e ( j ) , j ) ;263 h g b a r s u c t ( j ) = ( k b a r / x s u c t p l a t e ( end ) ) * ( 0 . 0 3 8 5 * Re L ˆ ( 4 / 5 )

−0.0385* Re xc ˆ ( 4 / 5 ) +0.906* Re xc ˆ ( 1 / 2 ) ) * P r a v e ˆ ( 1 / 3 ) ;264

265

266 end267

268 %−−−−−− end of t h e e x t e r n a l f low −−−−−−−−−−−−−269 %% I n t e r n a l f low270 % back smooth c h a n n e l271 % geo i n p u t s272 l o a d ( ’ S m o o t h a r e a s . t x t ’ )

64

273 A= S m o o t h a r e a s . / 1 0 0 0 0 0 0 ;274 % A c h a n n e l (A( i ) )275 l o a d ( ’ S m o o t h p e r i m e t e r . t x t ’ )276 P e r i m e t e r s u c t = S m o o t h p e r i m e t e r ( : , 1 ) . / 1 0 0 0 ;277 P e r i m e t e r p r e s = S m o o t h p e r i m e t e r ( : , 2 ) . / 1 0 0 0 ;278 P e r i m e t e r s i d e = S m o o t h p e r i m e t e r ( : , 3 ) . / 1 0 0 0 ;279 % t o t a l p e r i m e t e r ( P t o t ( i )280 P t o t = P e r i m e t e r s u c t + P e r i m e t e r p r e s + P e r i m e t e r s i d e ;281 P s i d e = P e r i m e t e r s u c t + P e r i m e t e r p r e s ;282 % s i d e p e r i m e t e r P s d e ( i )283 l o a d ( ’ S m o o t h l e n g t h . t x t ’ )284 L= S m o o t h l e n g t h . / 1 0 0 0 ;285 A te =L . * ( 0 . 3 / 1 0 0 0 ) ;286 mc ( 1 ) = 0 . 0 0 6 6 ; %f i r s t s e c t i o n mass f low r a t e287 % I n i t i a l Mach c a l c u l a t i o n288 T c t =700;289 P c t =1200000;290 r h o t 1 = P c t / T c t / Rc ;291 m= 0 . 1 ;292

293 f o r j =1:100294 m new =(mc ( 1 ) /A( 1 ) ) * ( ( 1 + ( Y c−1) / 2 *mˆ 2 ) ˆ ( 1 / ( Y c−1) ) / r h o t 1 ) * ( 1 /

s q r t ( Y c*Rc* T c t ) ) * s q r t ( 1 + ( Y c−1) / 2 *mˆ 2 ) ;295 e=m−m new ;296 i f e<0.00000001297 M=m new ;298 e l s e299 m=m new ;300 end301 end302 t o t d e l P =0;303 Pc ( 1 ) = P c t / ( ( 1 + ( Y c−1) / 2 * (Mˆ 2 ) ) ˆ ( Y c / ( Y c−1) ) ) ;304 Tc ( 1 ) = T c t / ( 1 + ( Y c−1) / 2 *Mˆ 2 ) ;305 r h o c ( 1 ) = r h o t 1 / ( ( 1 + ( Y c−1) / 2 *Mˆ 2 ) ˆ ( 1 / ( Y c−1) ) ) ;306 Vc ( 1 ) =mc ( 1 ) / r h o c ( 1 ) /A( 1 ) ;307 h g b a r p l a t e =( h g b a r s u c t + h g b a r p r e s ) . / 2 ;308 m i n l e t ( 1 ) = 0 . 0 0 6 6 ;309

310 %smooth c h a n n e l c o r r e l a t i o n − G n i e l s k i311 % a n s y s r e s u l t s312 l o a d ( ’TE ENTER PRESSURE . d a t ’ )313 P o u t t e =TE ENTER PRESSURE ( : , 2 ) +101000;314 l o a d ( ’TE ENTER TEMPERATURE . d a t ’ )315 T t e o u t =TE ENTER TEMPERATURE ( : , 2 ) ;316 l o a d ( ’TE ENTER VELOCITY . d a t ’ )

65

317 V t e o u t =TE ENTER VELOCITY ( : , 2 ) ;318 l o a d ( ’ SMOOTH FILM IN Velocity . d a t ’ )319 V f i lm =SMOOTH FILM IN Velocity ( : , 2 ) ;320 f o r i =1:8 %s p a n w i s e l o c a t i o n321 i f i ==8322 j =7 ;323 e l s e324 j = i ;325 end326 T r a v e p ( i ) = t r a p z ( x p r e s p l a t e , T a w p r e s s ( : , j ) ) / (

x p r e s p l a t e ( end ) ) ;327 T r a v e s ( i ) = t r a p z ( x s u c t p l a t e , T a w s u c t ( : , j ) / x s u c t p l a t e (

end ) ) ;328 T r a v e ( i ) = ( T r a v e s ( i ) + T r a v e p ( i ) ) / 2 ;329 Taw ave ( i ) = T r a v e ( i ) ;330

331 [ Tc ( i +1) , Pc ( i +1) , r h o c ( i +1) , Vc ( i +1) , m i n l e t ( i +1) , hc ( i ) ,mTe( i ), k c ( i ) , m f i lm ( i ) , Re D ( i ) , check ( i ) , P t f i l m s m o o t h ( i ) ] =s m o o t h d u c t t e ( j ,A( i ) , A te ( i ) ,A( i +1) , P t o t ( i ) , P s i d e ( i ) . . .

332 , Tc ( i ) , Pc ( i ) , r h o c ( i ) , Vc ( i ) ,L ( i ) , h g b a r p l a t e ( j ) , Taw ave (i ) , m i n l e t ( i ) , P o u t t e ( j ) , T t e o u t ( j ) , V t e o u t ( j ) ,V f i lm ( i ) ) ;

333

334

335 % f i l m backf low margin336 BFM smooth fi lm ( i ) = ( Pc ( i +1) / P s t a t i c o u t ( j , 2 ) −1) *100 ;337 end338

339

340

341

342

343 %% j e t c h a n n e l344

345 %J e t Channel Dimension346

347 m j c ( 1 ) = 0 . 0 0 1 4 ; %% mass f low r a t e348 l o a d ( ’ J e t a r e a s . t x t ’ )349 A j e t = J e t a r e a s / 1 0 0 0 0 0 0 ;350 l o a d ( ’ J e t p e r i m e t e r . t x t ’ )351 P j e t t o t = ( J e t p e r i m e t e r ( : , 1 ) + J e t p e r i m e t e r ( : , 2 ) + J e t p e r i m e t e r

( : , 3 ) + J e t p e r i m e t e r ( : , 4 ) ) / 1 0 0 0 ;352 P j e t s i d e = ( J e t p e r i m e t e r ( : , 1 ) + J e t p e r i m e t e r ( : , 2 ) ) / 1 0 0 0 ;353 l o a d ( ’ J e t l e n g t h . t x t ’ )354 L j e t = J e t l e n g t h / 1 0 0 0 ;

66

355 % a n s y s i n p u t s356 l o a d ( ’ JET IMP IN Veloc i ty . d a t ’ )357 V j e t = JET IMP IN Veloc i ty ( 8 : −1 : 1 , 2 ) ;358 m= 0 . 1 ;359 f o r j =1:100360 m new =( m j c / A j e t ( 1 ) ) * ( ( 1 + ( Y c−1) / 2 *mˆ 2 ) ˆ ( 1 / ( Y c−1) ) / r h o t 1 )

* ( 1 / s q r t ( Y c*Rc* T c t ) ) * s q r t ( 1 + ( Y c−1) / 2 *mˆ 2 ) ;361 e=m−m new ;362 i f e<0.00000001363 M je t =m new ;364 e l s e365 m=m new ;366 end367 end368 P c t = 1300000;369 P c j e t ( 1 ) = P c t / ( ( 1 + ( Y c−1) / 2 * ( M je t ˆ 2 ) ) ˆ ( Y c / ( Y c−1) ) ) ;370 T c j e t ( 1 ) = T c t / ( 1 + ( Y c−1) / 2 * M je t ˆ 2 ) ;371 r h o c j e t ( 1 ) = r h o t 1 / ( ( 1 + ( Y c−1) / 2 * M je t ˆ 2 ) ˆ ( 1 / ( Y c−1) ) ) ;372 V c j e t ( 1 ) = m j c ( 1 ) / r h o c j e t ( 1 ) / A j e t ( 1 ) ;373 m i n l e t j e t ( 1 ) = 0 . 0 0 2 6 ;374 f o r i =1:8375 i f i ==1 | | i ==2376 j =1 ;377 e l s e378 j = i −1;379 end380

381 T r a v e p ( i ) = t r a p z ( x p r e s p l a t e , T a w p r e s s ( : , j ) ) / (x p r e s p l a t e ( end ) ) ;

382 T r a v e s ( i ) = t r a p z ( x s u c t p l a t e , T a w s u c t ( : , j ) / x s u c t p l a t e (end ) ) ;

383 T r a v e ( i ) = ( T r a v e s ( i ) + T r a v e p ( i ) ) / 2 ;384

385 Taw ave ( i ) = T r a v e ( i ) ;386 [ T c j e t ( i +1) , P c j e t ( i +1) , r h o c j e t ( i +1) , V c j e t ( i +1) ,

m i n l e t j e t ( i +1) , h c j e t ( i ) , m j e t ( i ) , k c j e t ( i ) , P t f i l m j e t( i ) ] = s m o o t h d u c t l e ( A j e t ( i ) , A j e t ( i +1) , P j e t t o t ( i ) ,P j e t s i d e ( i ) . . .

387 , T c j e t ( i ) , P c j e t ( i ) , r h o c j e t ( i ) , V c j e t ( i ) , L j e t ( i ) ,h g b a r p l a t e ( j ) , Taw ave ( i ) , m i n l e t j e t ( i ) , V j e t ( i ) ) ;

388

389 B F M j e t f i l m ( i ) = ( P c j e t ( i +1) / P s t a t i c o u t ( j , 1 ) −1) *100 ;390 end391

392 %% JET IMPINGEMENT

67

393 d i a H o l e = 0 . 0 0 0 8 ;394 c e n t d i s = 0 . 0 0 2 4 5 9 ;395 t a r g d i s t = 0 . 0 0 3 2 ;396 d i a T a r g e t = 1 . 2 7 5 / 1 0 0 0 ;397 f o r i =1:8398 V j e t ( i ) = m j e t ( i ) / ( r h o c j e t ( i ) * p i * ( d i a H o l e / 2 ) ˆ 2 ) ;399 D y n v i s c j e t ( i ) = i n t e r p 1 ( T , Visco , T c j e t ( i ) ) ;400 R e j e t ( i ) = r h o c j e t ( i ) * V j e t ( i ) * d i a H o l e / D y n v i s c j e t ( i ) ;401 N u s s e l t N u m b e r j e t ( i ) = 0 .44 * R e j e t ( i ) ˆ 0 . 7 * ( d i a H o l e / c e n t d i s )

ˆ 0 . 8 * exp ( −0 . 8 5 . * ( t a r g d i s t / d i a H o l e ) * ( d i a H o l e / c e n t d i s ) * (d i a H o l e / d i a T a r g e t ) ˆ 0 . 4 ) ;

402 h j e t ( i ) = k c j e t ( i ) * N u s s e l t N u m b e r j e t ( i ) / d i a H o l e ;403 end404

405 %% T r a i l i n g Edge E x i t C o n v e c t i o n406 % geo i n p u t s407 A e x i t T e = A te ;408 L TE = 0 . 0 0 6 9 2 ;409 Aside TE = L . * L TE ;410 P t e = 2*L + 2 * ( 0 . 3 / 1 0 0 0 ) ;411

412

413 f o r i =1:8414 Dh te ( i ) = 4* A e x i t T e ( i ) / ( P t e ( i ) ) ;415 Vc te ( i ) = mTe( i ) / r h o c ( i ) / A e x i t T e ( i ) ;416 D y n v i s c ( i ) = i n t e r p 1 ( T , Visco , Tc ( i ) ) ;417 Re D te ( i ) = Vc te ( i ) * Dh te ( i ) * r h o c ( i ) / D y n v i s c ( i ) ;418 f t e ( i ) = ( 0 . 7 9 * l o g ( Re D te ( i ) ) −1.64) ˆ(−2) ;419 N u c t e ( i ) = 0 .023* Re D te ( i ) ˆ ( 4 / 5 ) * 0 . 7 ˆ 0 . 4 ;420 h c t e ( i ) = N u c t e ( i ) * k c ( i ) / Dh te ( i ) ;421 d e l P t e ( i ) = f t e ( i ) * ( L TE / Dh te ( i ) ) * r h o c ( i ) / 2 * Vc te ( i ) ˆ 2 ;422 P t e e x i t ( i ) =Pc ( i )−d e l P t e ( i ) ;423 M te = Vc te ( i ) / s q r t ( Y c*Rc*Tc ( i ) ) ;424 P t t e e x i t ( i ) = P t e e x i t ( i ) / ( 1 + 0 . 5 * ( Y c−1)* M te ˆ 2 ) ˆ ( ( Y c−1) / Y c

) ;425 i f i ==8426 j =7 ;427 e l s e428 j = i ;429 end430 B F M t e s l o t ( i ) = ( P t t e e x i t ( i ) / P i n f p r e s s ( 4 9 , j )−1) *100 ;431 end432

433 %% Wall Tempera tu re−c h o r d w i s e434

68

435 % s u c t i o n s i d e436 k cond = 1 7 . 7 ;437 f o r j =1:7438

439 f o r i =1 : s i z e ( T aw suc t , 1 )440

441 i f i < 2 % s t a g n a t i o n p o i n t442 hg ( i ) =h LE ( 1 , j ) ;443 HTC c ( i ) = h j e t ( j ) ;444 T c o o l a n t ( i ) = T c j e t ( j ) ;445 e l s e446 hg ( i ) = h s u c t x ( i , j ) ;447

448 i f x s u c t p l a t e ( i ) < 5 . 5 0 9 / 1 0 0 0% end of j e t449 HTC c ( i ) = h j e t ( j ) ;450 T c o o l a n t ( i ) = T c j e t ( j ) ;451 e l s e i f x s u c t p l a t e ( i ) < 8 . 8 8 1 / 1 0 0 0 % end of j e t

c h a n n e l452 HTC c ( i ) = h c j e t ( j ) ;453 T c o o l a n t ( i ) = T c j e t ( j ) ;454 e l s e i f x s u c t p l a t e ( i ) < 3 0 . 2 6 8 / 1 0 0 0 %end of smooth

c h a n n e l455 HTC c ( i ) =hc ( j ) ;456 T c o o l a n t ( i ) =Tc ( j ) ;457 e l s e458 HTC c ( i ) = h c t e ( j ) ;459 T c o o l a n t ( i ) =Tc ( j ) ;460 end461 end462 % h e a t f l u x463 T r a v e p ( i ) = t r a p z ( x p r e s p l a t e , T a w p r e s s ( : , j ) ) / (

x p r e s p l a t e ( end ) ) ;464 T r a v e s ( i ) = t r a p z ( x s u c t p l a t e , T a w s u c t ( : , j ) /

x s u c t p l a t e ( end ) ) ;465 R hot = 1 / hg ( i ) ;466 R b l a d e = 0 . 0 0 1 / k cond ;467 R co ld = 1 / HTC c ( i ) ;468 R = R hot + R b l a d e + R co ld ;469 q = ( T a w s u c t ( i , j )−T c o o l a n t ( i ) ) / R ;470 T w o u t s u c t ( i , j ) = T a w s u c t ( i , j )−q* R hot ;471 T w i n s u c t ( i , j ) = T w o u t s u c t ( i , j )−q* R b l a d e ;472

473

474 end

69

475 T w o u t a v e s u c t ( j ) = t r a p z ( x s u c t p l a t e , T w o u t s u c t ( : , j ) ) / (x s u c t p l a t e ( end ) ) ;

476 T w o u t a v e s u c t ( j ) = t r a p z ( x s u c t p l a t e , T w i n s u c t ( : , j ) ) / (x s u c t p l a t e ( end ) ) ;

477 T w s u c t a v e ( j ) =( T w o u t a v e s u c t ( j ) + T w o u t a v e s u c t ( j ) ) / 2 ;478 f o r i =1 : s i z e ( T aw pres s , 1 )479

480 i f i < 2 % s t a g n a t i o n p o i n t481 hg ( i ) =h LE ( 1 , j ) ;482 HTC c ( i ) = h j e t ( j ) ;483 T c o o l a n t ( i ) = T c j e t ( j ) ;484 e l s e485 hg ( i ) = h p r e s x ( i , j ) ;486

487 i f x p r e s p l a t e ( i ) < 5 . 4 3 3 / 1 0 0 0 % end of j e t488 HTC c ( i ) = h j e t ( j ) ;489 T c o o l a n t ( i ) = T c j e t ( j ) ;490 e l s e i f x p r e s p l a t e ( i ) < 6 . 6 2 9 / 1 0 0 0 % end of j e t

c h a n n e l491 HTC c ( i ) = h c j e t ( j ) ;492 T c o o l a n t ( i ) = T c j e t ( j ) ;493 e l s e i f x p r e s p l a t e ( i ) < 2 3 . 8 6 9 / 1 0 0 0 %end of smooth

c h a n n e l494 HTC c ( i ) =hc ( j ) ;495 T c o o l a n t ( i ) =Tc ( j ) ;496 e l s e497 HTC c ( i ) = h c t e ( j ) ;498 T c o o l a n t ( i ) =Tc ( j ) ;499 end500 end501 % h e a t f l u x502 R hot =1/ hg ( i ) ;503 R b l a d e = 0 . 0 0 1 / k cond ;504 R co ld =1/ HTC c ( i ) ;505 R= R hot + R b l a d e + R co ld ;506 q =( T a w p r e s s ( i , j )−T c o o l a n t ( i ) ) / R ;507 T w o u t p r e s ( i , j ) = T a w p r e s s ( i , j )−q* R hot ;508 T w i n p r e s ( i , j ) = T w o u t p r e s ( i , j )−q* R b l a de ;509 end510 T w o u t a v e p r e s ( j ) = t r a p z ( x p r e s p l a t e , T w o u t p r e s ( : , j ) ) / (

x p r e s p l a t e ( end ) ) ;511 T w o u t a v e p r e s ( j ) = t r a p z ( x p r e s p l a t e , T w i n p r e s ( : , j ) ) / (

x p r e s p l a t e ( end ) ) ;512 Tw pres ave ( j ) =( T w o u t a v e p r e s ( j ) + T w o u t a v e p r e s ( j ) ) / 2 ;513 end

70

514

515 end

1 f u n c t i o n [ Tc nex t , Pc nex t , r h o c n e x t , Vc out , m out , hc , m ex i t , k c ,P t f i l m ] = s m o o t h d u c t l e (A, A next , P t o t , P s i d e , Tc , Pc , rho c , Vc, L , h g b a r p l a t e , Taw ave , m i n l e t , V j e t )

2 %% Air p r o p e r t i e s3 l o a d Tab le . d a t % I d e a l gas p r o p e r t i e s wr t Tempera tu r e4 T = Tab le ( : , 1 ) ; %Ke lv in5 Rho = Tab le ( : , 2 ) ;6 Cp = Tab le ( : , 3 ) ;7 Visco = Tab le ( : , 4 ) *10ˆ(−7) ;8 k g a s = Tab le ( : , 6 ) *10ˆ(−3) ;9 Pr = Tab le ( : , 8 ) ;

10 Y c = 1 . 4 ;11 Rc = 287 ;12 c p c = Y c*Rc / ( Y c−1) ;13

14 k = 1 7 . 7 ;15

16 Dh = 4*A / ( P t o t ) ;17 Length = 2 3 . 3 / 1 0 0 0 ;18 D y n v i s c = i n t e r p 1 ( T , Visco , Tc ) ;19 Re D = Vc*Dh* r h o c / D y n v i s c ;20 f = ( 0 . 7 9 * l o g ( Re D ) −1.64) ˆ(−2) ;21 P r c = i n t e r p 1 ( T , Pr , Tc ) ;22 k c = i n t e r p 1 ( T , k gas , Tc ) ;23 Nu c = ( f / 8 ) * ( Re D−1000)* P r c / ( 1 . 0 + 1 2 . 7 * ( ( f / 8 ) ˆ 0 . 5 ) * ( P r c

ˆ ( 0 . 6 6 ) −1) ) ;24 hc = Nu c* k c / Dh ;25 de lP = f * (L / Dh ) * r h o c / 2 * Vc ˆ 2 ;26 % h e a t f l u x27 Aside = P s i d e *L ;28 R hot = 1 / h g b a r p l a t e / As ide ;29 R b l a d e = 0 . 0 0 1 / k / Aside ;30 R co ld = 1 / hc / Aside ;31 R = R hot + R b l a de + R co ld ;32 q = ( Taw ave−Tc ) / R ;33 Tw out = Taw ave−q* R hot ;34 Tw in = Tw out−q* R b l a de ;35

36

37 T c n e x t =Tc +( Aside * h g b a r p l a t e * ( Taw ave−Tw out ) ) / ( m i n l e t *c p c ) ;

38 P c n e x t =Pc−de lP ;

71

39

40

41 A j e t = ( ( 0 . 4 / 1 0 0 0 ) ˆ2* p i ) ;42 r h o c n e x t = P c n e x t / Rc / T c n e x t ;43 m e x i t = r h o c n e x t * A j e t * V j e t ;44 m out = m i n l e t−m e x i t ;45 Vc out = m out / r h o c / A next ;46

47 M c = Vc out / s q r t ( Y c*Rc* T c n e x t ) ;48 P t f i l m = P c n e x t / ( 1 + 0 . 5 * ( Y c−1)*M c ˆ 2 ) ˆ ( ( Y c−1) / Y c ) ;49 end

1 f u n c t i o n [ Tc nex t , Pc nex t , r h o c n e x t , Vc out , m out , hc , m te , k c ,m fi lm , Re D , Dh , P t f i l m ] = s m o o t h d u c t t e ( j , A, A te , A next , P t o t, P s i d e , Tc , Pc , rho c , Vc , L , h g b a r p l a t e , Taw ave , m i n l e t , P o u t t e, T o u t t e , V o u t t e , V f i lm )

2 %% Air p r o p e r t i e s3 l o a d Tab le . d a t % I d e a l gas p r o p e r t i e s wr t Tempera tu r e4 T = Tab le ( : , 1 ) ; %K e lv in5 Rho = Tab le ( : , 2 ) ;6 Cp = Tab le ( : , 3 ) ;7 Visco = Tab le ( : , 4 ) *10ˆ(−7) ;8 k g a s = Tab le ( : , 6 ) *10ˆ(−3) ;9 Pr = Tab le ( : , 8 ) ;

10 Y c = 1 . 4 ;11 Rc = 287 ;12 c p c = Y c*Rc / ( Y c−1) ;13 %−−−−−−−−−−−−−−−−−−−−−−−−−14 k = 1 7 . 7 ; %NGV c o n d u c t i v e h t p a r a m e t e r15 %−−−−−−−−−16

17 Length = 1 6 . 9 / 1 0 0 0 ;18 Dh = 4*A / ( P t o t ) ; % h y d r a u l i c d i a19 D y n v i s c = i n t e r p 1 ( T , Visco , Tc ) ;20 Re D = Vc*Dh* r h o c / D y n v i s c ;21 f = ( 0 . 7 9 * l o g ( Re D ) −1.64) ˆ(−2) ;22 P r c = i n t e r p 1 ( T , Pr , Tc ) ;23 k c = i n t e r p 1 ( T , k gas , Tc ) ;24 Nu c = ( f / 8 ) * ( Re D−1000)* P r c / ( 1 . 0 + 1 2 . 7 * ( ( f / 8 ) ˆ 0 . 5 ) * ( P r c

ˆ ( 0 . 6 6 ) −1) ) ;25 hc = ( Nu c* k c / Dh ) ;26 de lP = f * (L / Dh ) * r h o c / 2 * Vc ˆ 2 ;27

28 % h e a t f l u x29 Aside = P s i d e *L ;

72

30 R hot = 1 / h g b a r p l a t e / As ide ;31 R b l a d e = 0 . 0 0 1 / k / Aside ;32 R co ld = 1 / hc / Aside ;33 R = R hot + R b l a de + R co ld ;34 q = ( Taw ave−Tc ) / R ;35 Tw out = Taw ave−q* R hot ;36 Tw in = Tw out−q* R b l a d e ;37

38

39 T c n e x t = Tc +( Aside * h g b a r p l a t e * ( Taw ave−Tw out ) ) / ( m i n l e t *c p c ) ;

40 P c n e x t = Pc−de lP ;41

42 l o a d ( ’ SMOOTH FILM IN Velocity . d a t ’ )43

44 v e l = SMOOTH FILM IN Velocity ( 9 : −1 : 1 , 2 ) ;45

46 i f j ==147 V fi lm2 = v e l ( j +1) ;48 e l s e49 V f i lm = v e l ( j +1) ;50 end51

52 A f i lm = ( ( 0 . 4 / 1 0 0 0 ) ˆ2* p i ) ;53 r h o c n e x t = P c n e x t / Rc / T c n e x t ;54 r h o t e = P o u t t e / ( Rc* T o u t t e ) ;55 m te = r h o t e * A te * V o u t t e ;56

57 i f j ==158 m fi lm = r h o c n e x t * A f i lm * V f i lm + r h o c n e x t * A f i lm *

V f i lm2 ;59 e l s e60 m fi lm = r h o c n e x t * A f i lm * V f i lm ;61 end62

63 m e x i t = m te + m fi lm ;64 m out = m i n l e t−m e x i t ;65 Vc out = m out / r h o c / A next ; %A i s t h e a r e a o f n e x t s e c t i o n66

67

68

69 M c = Vc out / s q r t ( Y c*Rc* T c n e x t ) ;70 P t f i l m = P c n e x t / ( 1 + 0 . 5 * ( Y c−1)*M c ˆ 2 ) ˆ ( ( Y c−1) / Y c ) ;71

72

73

73

74

75

76

77 end

1 f u n c t i o n [ Taw , e t a ] = f i l m c o o l i n g ( T r i n f , T i n f , x , j , a , Uinf , P in f ,P i n f t , T i n f t )

2 %% Air p r o p e r t i e s3 l o a d ( ’ Tab le . d a t ’ )4 Tempera tu r e = Tab le ( : , 1 ) ; %K e l v in5 D e n s i t y = Tab le ( : , 2 ) ;6 Cp = Tab le ( : , 3 ) ;7 Dyn Visco = Tab le ( : , 4 ) *10ˆ(−7) ;8 k = Tab le ( : , 6 ) *10ˆ(−3) ;9 Pr = Tab le ( : , 8 ) ;

10 c p i n f = i n t e r p 1 ( Tempera tu re , Cp , T i n f ) ;11 n u i n f = i n t e r p 1 ( Tempera tu re , Dyn Visco , T i n f ) ;12

13 Yt = 1 . 3 5 ;14 Rt = c p i n f * ( Yt−1) / Yt ;15 r h o i n f = P i n f / T i n f / Rt ;16 Taw = T r i n f −1;17 e r r o r =10;18 s = 1 . 1 1 6 / 1 0 0 0 ;19 %% c o o l a n t Tr20 i f a==121 l o a d ( ’ JET FILM IN Pressu re . d a t ’ )22 l o a d ( ’ JET FILM IN Tempera ture . d a t ’ )23 l o a d ( ’ JET FILM IN Veloc i ty . d a t ’ )24 T2=JET FILM IN Tempera ture (11− j −2 ,2) ;25 s p e e d s o u n d = s q r t ( 1 . 4 * 2 8 7 * T2 ) ;26 V2= JET FILM IN Veloc i ty (11− j −2 ,2) ;27 M2=V2 / s p e e d s o u n d ;28 P2= JET FILM IN Pressu re (11− j −2 ,2) +101000;29

30 e l s e31 l o a d ( ’ SMOOTH FILM IN Pressure . d a t ’ )32 l o a d ( ’ SMOOTH FILM IN Temperature . d a t ’ )33 l o a d ( ’ SMOOTH FILM IN Velocity . d a t ’ )34 temp=SMOOTH FILM IN Temperature ( 9 : −1 : 1 , 2 ) ;35 p r e s =SMOOTH FILM IN Pressure ( 9 : −1 : 1 , 2 ) +101000;36 v e l =SMOOTH FILM IN Velocity ( 9 : −1 : 1 , 2 ) ;37

38

74

39 i f j ==140 T2 f =temp ( j ) ;41 V2 f= v e l ( j ) ;42 P 2 f = p r e s ( j ) ;43 T2=temp ( j +1) ;44 V2= v e l ( j +1) ;45 P2= p r e s ( j +1) ;46

47 e l s e48 T2=temp ( j +1) ;49 V2= v e l ( j +1) ;50 P2= p r e s ( j +1) ;51 end52 s p e e d s o u n d = s q r t ( 1 . 4 * 2 8 7 * T2 ) ;53 M2=V2 / s p e e d s o u n d ;54

55 end56 R=287;57 r h o 2 = P2 / T2 / R ;58 r h o i n f t = P i n f t / T i n f t / Rt ;59 M=V2* r h o 2 / Uinf / r h o i n f t ;60

61

62 Y t = 1 . 4 ;63 r p r e s s u r e = s q r t ( 0 . 7 ) ;64 T r2 = T2 *(1+ r p r e s s u r e * ( ( Y t−1) / 2 ) *M2ˆ 2 ) ; %p r a n d t l

c o r r e c t i o n65

66 w h i l e e r r o r > 0 . 0 167 T s t a r = T i n f + 0 . 7 2 * ( Taw−T i n f ) ;68 n u s t a r = i n t e r p 1 ( Tempera tu re , Dyn Visco , T s t a r ) ;69 nu 2 = i n t e r p 1 ( Tempera tu re , Dyn Visco , T2 ) ;70 r h o s t a r = P2 / T s t a r / R ;71

72 cp 2 = i n t e r p 1 ( Tempera ture , Cp , T2 ) ;73 Dynvis 2 = i n t e r p 1 ( Tempera tu re , Dyn Visco , T2 ) ;74 Re c = V2* r h o 2 *x / Dynvis 2 ;75 k s i s t a r = ( x /M/ s ) * ( Re c * nu 2 / n u s t a r ) ˆ ( −0 . 2 5 ) * ( r h o s t a r / r h o i n f

) ;76

77 e t a = ( 1 + ( c p i n f / cp 2 ) * ( 0 . 3 3 * ( 4 + k s i s t a r ) ˆ0 .8−1) ) ˆ(−1) ;78 Taw next = e t a * ( T r2−T r i n f ) + T r i n f ;79 e r r o r = abs ( Taw next−Taw ) ;80 Taw = Taw next ;81 end

75

82

83

84 end

References[1] Loic Calba Laura Maestro Efrain Carreno-Morelli Mehdi Rahimian Srdjan Milenkovic Ilchat

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