33688822 Design of a Gas Turbine[1]

MEEN-646 Aerothermodynamics of Turbomachinery

Design of Gas Turbine Engine

Authors: Kapil Sharma Thomas A. Chirathadam Vishal Wadhvani Shriram Jagannathan

Presented to: Dr. T. Schobeiri

CONTENTS

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Contents 1 Introduction

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2 Compressor 2.1 Design of Compressor Blades . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Pressure distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Design of Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Steps Followed . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Radial Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Optimization of Compressor Pressure Ratio for Power Generation in Gas Turbine Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Profile Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Trailing Edge Loss . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Secondary Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4 Exit Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Results and Discussions . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Combustion Chamber

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4 Turbine 4.1 Design of Turbine Blades . . . . 4.1.1 Procedure . . . . . . . . 4.1.2 Generating Blade profile 4.2 Blade Profiles . . . . . . . . . . 4.2.1 Procedure . . . . . . . . 4.2.2 CFD Results . . . . . . 4.3 Design of Turbine . . . . . . . . 4.3.1 Assumptions . . . . . . . 4.3.2 Steps Followed . . . . . 4.3.3 Radial Equilibrium . . . 4.4 Losses . . . . . . . . . . . . . . 4.5 Results . . . . . . . . . . . . . .

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5 Stress Analysis

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CONTENTS

Project Report 6 Appendix 6.1 Compressor Codes . . . . . . . . . . . . . . . . . 6.1.1 Blade Profile Generation . . . . . . . . . . 6.1.2 Low Pressure Compressor . . . . . . . . . 6.1.3 Intermediate Pressure Compressor . . . . . 6.1.4 High Pressure Compressor . . . . . . . . . 6.1.5 Twisting the Blades in Spanwise direction 6.2 Turbine Codes . . . . . . . . . . . . . . . . . . . . 6.2.1 Design of Turbine Blades . . . . . . . . . . 6.2.2 Efficiency Calculation in Turbine . . . . .

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46 46 46 50 60 71 94 96 101 108

LIST OF FIGURES

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List of Figures 1.1 1.2 1.3 1.4 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 3.1 3.2 3.3 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16

Axial flow gas turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . Gas turbine assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cross section view of the gas turbine assembly . . . . . . . . . . . . . . Multi stage power generation gas turbine . . . . . . . . . . . . . . . . . Compressor blade profile for a lift coefficient of 0.5 . . . . . . . . . . . . Cascade Nomenclature as given in [2] . . . . . . . . . . . . . . . . . . . Pressure Distribution in the Compressor Blade . . . . . . . . . . . . . . Compressor configuration used for pressure distribution . . . . . . . . . Mesh around the compressor blade . . . . . . . . . . . . . . . . . . . . Compressor Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . Notation followed for compressor stages . . . . . . . . . . . . . . . . . . Velocity Triangle for a compressor stage. Subscripts 1 refer to rotor inlet and 2 refers to rotor exit . . . . . . . . . . . . . . . . . . . . . . . . . . h-s diagram for a compressor stage. . . . . . . . . . . . . . . . . . . . . Variation of enthalpy and static pressure in LP Compressor . . . . . . . Variation of enthalpy and static pressure in IP Compressor . . . . . . . Variation of enthalpy and static pressure in HP Compressor . . . . . . Compressor blade showing the twist in the spanwise direction . . . . . Compressor Pressure Ratio vs Thermal Efficiency for various temperature ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Trailing Edge Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . Exit of the combustion chamber . . . . . . . . . . . . . . . . . . . . . . View showing the arrangement of flame-holders . . . . . . . . . . . . . Cross section view of combustion chamber . . . . . . . . . . . . . . . . Turbine Blade Nomenclature as given in [2] . . . . . . . . . . . . . . . . Sample base profile for superposition . . . . . . . . . . . . . . . . . . . Blade Cascade . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Velocity contour around the blade . . . . . . . . . . . . . . . . . . . . . Pressure distribution around the blade . . . . . . . . . . . . . . . . . . Mesh using tetrahedral elements . . . . . . . . . . . . . . . . . . . . . . Cp plot for a turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turbine configuration used for pressure distribution . . . . . . . . . . . Section view showing the increasing cross section of the turbine . . . . Axial Turbine stage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Velocity Triangle for a Turbine stage. Subscripts 2 refer to rotor inlet and 3 refers to rotor exit . . . . . . . . . . . . . . . . . . . . . . . . . . h-s diagram for a turbine stage . . . . . . . . . . . . . . . . . . . . . . Notation for a turbine stage . . . . . . . . . . . . . . . . . . . . . . . . Turbine rotor twisted in spanwise direction. . . . . . . . . . . . . . . . Variation of density with number of stages in a turbine . . . . . . . . . Variation of Enthalpy with number of stages in a turbine . . . . . . . . iii

1 2 3 4 6 6 7 8 8 10 11 11 12 12 13 13 15 18 20 23 24 25 26 28 29 30 30 31 31 32 34 35 35 36 36 38 40 40

LIST OF FIGURES

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4.17 Variation of Mach Number with number of stages in a turbine . . . . . 4.18 Variation of static and total pressure with number of stages in a turbine 4.19 Variation of static and total temperature with number of stages in a turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.20 Variation of absolute and relative velocity with number of stages in a turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Mesh of the gas turbine casing used for simulation in COSMOS . . . . 5.2 Stress distribution on the casing of the gas turbine . . . . . . . . . . .

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LIST OF TABLES

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List of Tables 1 2 3

Loss coefficients in each section of the compressor. It includes all the four losses mentioned in this section. . . . . . . . . . . . . . . . . . . . Table that shows the variation of stage parameters and thermodynamic quantities with different stages of compressor . . . . . . . . . . . . . . . Table that shows the variation of stage parameters and thermodynamic quantities with different stages of turbine . . . . . . . . . . . . . . . . .

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Abstract This project deals with the design procedure followed for designing the 14 stage axial compressor, annular combustion chamber and a 4 stage axial turbine for a 38 MW gas turbine. The compressor section consists of 2,3 and 9 stages of low, intermediate and high pressure compressors. The design of blades, the pressure distribution along the blade is also presented. An iterative procedure is discussed that takes into account the variation of different non-dimensional stage parameters. The radial equilibrium is also considered along the spanwise direction which results in twisted blades.

1 INTRODUCTION

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1

Introduction

Typically, gas turbine engines consists of compressor section, a combustion chamber and a turbine section. Air enters the compressor where it gets compressed thereby increasing its pressure. The high pressure, high temperature gas enters the combustion chamber where it mixes with the fuel and ignites. The temperature of the gas increases tremendously as it exits the combustion chamber. It then passes over the turbine section where the gas expands and work is extracted.

Figure 1.1: Axial flow gas turbine All gas turbine component consist of a stationary component known as stator and a rotating component, rotor. The rotor blades are mounted on a shaft that connects the compressor and the turbine. This shaft is called as the hub. A set of rotor blade and a stator blade together constitute a stage. Often, calculations are made on a stage by stage basis in turbomachinery. The purpose of a compressor in a gas turbine is to increase the pressure of the air before it enters the combustion chamber. Pressure rise occurs only in the rotor since mechanical energy is added to the rotor. The power that he compressor consumes often comes from the power generated by the turbine through a generator. The stator just serves to deflect the flow at the right incidence to the rotor. Since the pressure rise per stage is minimal in compressor due to factors such as flow separation, to obtain a significant pressure rise (pressure ration close to 10), compressor section consists of more than 10˜12 stages. Once the gases enter the combustion chamber, it mixes with the fuel in the combustion chamber, ignites and produces enormous amount of energy. The hot and high pressure gas from the combustion chamber drives the rotor blades in the turbine and turns the shaft that drives the compressor.

1

Figure 1.2: Gas turbine assembly

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Figure 1.3: Cross section view of the gas turbine assembly

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Figure 1.4: Multi stage power generation gas turbine

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2 COMPRESSOR

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2

Compressor

The following sections deal with the design of compressor blades for a particular lift coefficient, and consequently a 2-D CFD simulation on the blade for obtaining the pressure distribution. The design of low, intermediate and high pressure compressors is subsequently dealt with. Particular attention is given to the twisted blades in the compressor section. The books [2] and [1] have been extensively referred to, in many of the subsequent sections.

2.1 2.1.1

Design of Compressor Blades Assumptions

• In the design of compressor blades, the lift force induced by the vortex is assumed to be linearly proportional to the blade lift. 2.1.2

Procedure

Steps followed for designing a Compressor blade : • An arbitrary chord length is taken from the user (in our case 3 units) • This chord length is divided into a number of subdivisions taken from the user. • Also the lift coefficient is varied from 0 - 1 to see the effect on the blade camber height. The important parameter for determining the shape of a compressor blade is the Mach number of the flow. The lift force in an inviscid flow is given by[2], I F = ρV∞ XΓ, where Γ = V · dc (2.1) This relationship shows the relationship between lift and circulation. Also, the circulation is related to the flow deflection from the cascade inlet to the exit. When the deviation is zero, the velocity vector is tangent to the camber line at the exit. Compressors blades are usually designed for a given lift coefficient. The lift coefficient for the camber line is given by: ρΓV1 2Γ F = ρ = CL = ρ V1 c V 2c V2 2 1 2 1

5

(2.2)

2 COMPRESSOR

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6

Cax= 10 Cl = 0.5

4

Y

2

0

-2

-4

-6

0

2

4

Cax

6

8

10

12

Figure 2.1: Compressor blade profile for a lift coefficient of 0.5

Figure 2.2: Cascade Nomenclature as given in [2] 6

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0.120

Inlet Angle (1)= 120o Exit Angle (2)= 100o

Cp

0.110

0.100

0.090

0.080 0

0.2

0.4

x/cax

0.6

0.8

1

Figure 2.3: Pressure Distribution in the Compressor Blade

2.2

Pressure distribution

A very straightforward and simple CFD analysis is performed on the compressor blade to investigate cases of separation and the amount of pressure rise per stage. The domain consists of a single blade with periodic boundary condition in the pitchwise direction. c The pitch is decided based on the assumption that the = 1, where c is the chord s and s is the pitch. A velocity inlet boundary condition is used at 1 chord upstream of the blade leading edge and pressure outlet at one chord downstream of the blade. The commercial CFD package, STAR-CCM+ was used to perform the simulation. A separation plateau could be observed from the pressure distribution along the blade when there is dip in the pressure distribution or when the pressure is constant in the streamwise direction. The figures 2.3 and 2.5 show the pressure distribution and the mesh that was used for the simulation.

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Figure 2.4: Compressor configuration used for pressure distribution

 

Figure 2.5: Mesh around the compressor blade

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2.3 2.3.1

Design of Compressor Assumptions

• α1 = 55 and β2 = 125 • A total to total isentropic efficiency of 95% for compressor and turbine. • Mean diameter varies linearly We assume that the compressor section starts with a rotor. The station numbers are such that the rotor inlets are always odd numbers and rotor exits are even numbers. 2.3.2

Steps Followed

1. The values of α1 = 55, β2 = 125, λ, ν, r are assumed to be known apriori. 2. The four non-linear equations[2] are then solved to arrive at the other 4 unknowns, namely α2 , β1 , φ and µ.

r =1+

1 − 1/ tan (β2 ) = ν/ (µ · φ) tan (α2 )

(2.3)

1 − 1/ tan (β3 ) = 1/φ tan (α3 )

(2.4)



φ2 · 1 + 1/ (tan (α3 ))2 − µ2 · 1 + 1/ (tan (α2 ))2 2·λ   ν λ=φ· µ· − 1/ tan (β3 ) − 1 tan (α2 )

(2.5) (2.6)

3. A stage by stage analysis as detailed in [2] is followed. 4. Mean diameter and tangential velocity of the rotor at all the stations is calculated. U1 = Dm1 ω

(2.7)

5. The universal gas constant, enthalpy and temperature at the exit of first stage is calculated using the following procedure : H3 = H1 − lm +

V12 − V32 2000

T3 = H3 /Cp

9

(2.8)

(2.9)

Figure 2.6: Compressor Assembly

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2 COMPRESSOR

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Notation for LP Compressor Stages 2

1 ROTOR

4

3 ROTOR

STATOR

5 STATOR

Figure 2.7: Notation followed for compressor stages

VM 1

W1 W2

U1 U2 α1

V1

β1 α2 β2

V2

VM 2

Figure 2.8: Velocity Triangle for a compressor stage. Subscripts 1 refer to rotor inlet and 2 refers to rotor exit

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h P3

P2

H03 = H02 h3

V22 /2

h2 H01

1 - Rotor Inlet 2 - Rotor Exit

V32 /2 p3 lms

p2

P1 V12 /2 p1

h1

s Figure 2.9: h-s diagram for a compressor stage.

360

200

180 340

Pressure (kPa)

Enthalpy (kJ/kg)

160

320

140

120

300 100

280

0

1

2

3

4

5

80

6

Stations

1

2

3

4

5

Stations  

 

Figure 2.10: Variation of enthalpy and static pressure in LP Compressor

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440

320 300

420

Pressure (kPa)

Enthalpy (kJ/kg)

280 260

400

240

380

220 200

360 180 340

0

1

2

3

4

5

6

7

160

8

0

1

2

3

4

Stations

5

6

7

8

Stations  

 

Figure 2.11: Variation of enthalpy and static pressure in IP Compressor

650

1000

600 800

Pressure (kPa)

Enthalpy (kJ/kg)

550

500

600

450

400 400

350

0

2

4

6

8

10

12

14

16

18

200

20

Station

0

2

4

6

8

10

12

14

16

18

20

Station  

 

Figure 2.12: Variation of enthalpy and static pressure in HP Compressor

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6. Assuming an isentropic efficiency, h3s and T3s can be calculated as : H3s = η(H3 − H1 ) + H1

(2.10)

T3s = H3s /Cp

(2.11)

7. The value of gamma at T3s is calculated from the table that the instructor had given. 8. Finally,the exit pressure of first stage is calculated from the isentropic relation,  p3 = pin

Tin T3s

γ  γ−1

(2.12)

9. The properties at rotor exit (station 2) can now be calculated: H2 = r(H2 − H1 ) − H1

(2.13)

T2 = H2 /Cp

(2.14)

10. For evaluating the pressure at station 2, the polytropic constant(n) is calculated,: log



p3 pin



log



Tin T3



c=

n=

c c+1 

p2 = p 1

T1 T2

(2.15)

(2.16)

n  1−n

(2.17)

11. The above steps are repeated for the other stages as well. This would give us the exit pressure at the exit of the low pressure compressor. The percentage error between the actual pressure and the exit pressure is calculated. If the pressure error is positive, it would imply that the pressure at the outlet is greater than the exit pressure. The whole procedure is then repeated for lesser value of stage specific mechanical energy. If the error is negative, the stage specific mechanical energy is increased by 1% of its original value. This procedure is repeated until desired convergence is obtained. The convergence criteria is set as 0.2%

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Figure 2.13: Compressor blade showing the twist in the spanwise direction 2.3.3

Radial Equilibrium

The inlet angle varies in the spanwise direction and so does the exit angle. In order to comply with our assumption that the exit angle is constant, we twist the blade in spanwise direction so as to maintain a constant exit angle. The twist would account for the change in inlet angle as well. This is done by incorporating the following equations into our calculations from [2]. Rm cot α1 = (2.18) R cot α1m where the subscript m denotes the mean line values. Rm = Rh +

Bh 2

where Bh is the blade height and Rh is the Hub radius.

15

(2.19)

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2.4

Optimization of Compressor Pressure Ratio for Power Generation in Gas Turbine Engines

This section deals with the optimum pressure ratio for power generation in gas turbine engines. The pressure losses across the combustion chamber, turbine and compressor are considered while calculating the thermal efficiency of gas turbine engine. The effect of recuperator effectiveness on the overall efficiency of engine is also considered. Results of thermal efficiency and specific net power for various temperature ratios and recuperator effectiveness are plotted against compressor pressure ratio. 2.4.1

Assumptions

• The turbine and Compressor ratios are not the same due to pressure losses. • Efficiency of turbine and compressor assumed to be 0.9 • Pressure loss assumed to be 5% in both regenerator and combustion chamber. 2.4.2

Equations

Considering the T-S diagram of a gas turbine engine (with recuperator) given by the instructor, we can deduce the following efficiencies [2] : Compression Process : h2 − h1 = (h2s − h1 )/ηc

(2.20)

Regenerator : ζRA =

∆PRA P2

with ∆PRA = P2 − P5

(2.21)

Combustion Chamber Pressure Loss Coefficient : ζcc =

P5 − P 3 P2

(2.22)

Turbine Efficiency : h3 − h4 = (h3 − h4s )ηT

(2.23)

Regenerator: Gas side Pressure Loss Coefficient : ζ=

∆PRG P1

with ∆PRG = P4 − P6 16

(2.24)

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Thermal Efficiency of Gas Turbine Engine: ηin =

W˙net Q˙in

(2.25)

Considering the mass flow balance in Combustion Chamber: m˙ 1 + m˙ f = m˙ 3 β=

m˙ f m˙ 1

(2.26) (2.27)

Net Power of Gas Turbine: W˙net W˙ T m˙ 5 = (1 + β)WT − WC = m˙ 5 W˙C

(2.28)

WT = ηT c¯P T (1 + β)(T3 − T4s )

(2.29)

where,

Since the temperatures are high the , specific heat depends on temperature. h3 − h4 T3 − T4

c¯P T =

(2.30)

The relation between T3 and T4s can be obtained as they follow isentropic expansion process.  γ − 1  P3 γ m T = π T T P4 

T3 = T4s





(2.31)

From our assumption of the pressure loss coefficients, the turbine inlet and outlet pressures can be found.   P3 P2 1 − ζRA − ζcc πT = = (2.32) P4 P1 1 + ζRA We can then arrive at the following relation,   1 − ζRA − ζcc πT = πC 1 + ζRA

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(2.33)

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0.4

Thermal Efficiency

Theta=4.0 Theta=3.5 Theta=3.0

0.3

0.2

0.1

0

5

10

15

20

25

Compression Ratio

Figure 2.14: Compressor Pressure Ratio vs Thermal Efficiency for various temperature ratios where,  =

1 − ζRA − ζcc 1 + ζRA

 (2.34)

The regenerator effectiveness is defined as : ηR =

Q˙ actual h5 − h2 = ˙ h4 − h2 Qideal

(2.35)

Defining the engine temperature ratio as , ν=

T3 T1

(2.36)

Substituting the above equations into 2.25, we get the following relation for thermal efficiency,: c¯P T ηT ν[1 − (πc )−mT ](1 + β) − ηth = c¯P cc {ν(1 + β − ηR ) − [1 +

c¯P C mc π − 1)ηc ( c

πcmc − 1 ](1 − ηR ) + νηR ηT [1 − (πc )−mT ] ηc

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(2.37)

2 COMPRESSOR

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2.5

Losses

The major losses considered in the design are the following : 1. Profile Loss 2. Trailing edge loss 3. Secondary flow loss 4. Exit loss 2.5.1

Profile Loss

The profile loss is calculated from the Diffusion factor,D    sin α2 sin2 α1 (cot α2 − cot α1 ) + 1.12 , D= 0.61 sin α2 σ

(2.38)

δ2 For the value of D, corresponding value of is obtained from the graph in the [2]. c This value is used in Eqn.2.39 to calculate the profile loss coefficient.    2 δ2 σ sin α2 ζP = 2 (2.39) c sin α2 sin α2 Then the stage profile loss is given by,  2  2 Vn Wn ZP = ζP S + ζP R 2l 2l 2.5.2

(2.40)

Trailing Edge Loss

Due to the thickness of the trailing edge, there are wakes that are shed. These constitute towards trailing edge losses. As outlined in [2], the following dimensionless variables are calculated. b b D= = (2.41) s s sin α2 ∆1 = δ1 /s

(2.42)

∆2 = δ2 /s

(2.43)

The auxiliary equation are given as [2] : G1 = 1 − D − ∆1

(2.44)

G2 = 1 − D − ∆1 − ∆2

(2.45)

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Figure 2.15: Trailing Edge Thickness

 2  G21 − 2G2 + 1 2G1 − 2G2 + 1 G21 2 ζ= − cos α3 + 2 G21 G21 G2  2  2 Wn Vn + ζP R ZT = ζP S 2l 2l 2.5.3

(2.46) (2.47)

Secondary Loss

Secondary flow losses are due to the boundary layer development in the tip clearance and in the spacing between the hub abd stator. These clearances gives rise to vortices and thus constitute towards losses. α1 + α2 (2.48) cot α∞ = 2 m  2 δ − δ0 2 sin α1 ζs = 0.676 (cot α2 − cot α1 ) (2.49) sin α∞ c where δ is the actual tip clearance and δ0 is the smallest tip clearance without a tip clearance flow[2]. The above equations are repeated for stator and Eqn2.47 applies to secondary flow as well to calculate the loss coefficient. 2.5.4

Exit Loss

The exit loss is defined as the ratio of exit kinetic energy to the stage specific mechanical energy. V32 V32 ZE = = (2.50) 2l 2λU32 20

2 COMPRESSOR

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IP 0.0528

HP 0.3227

Table 1: Loss coefficients in each section of the compressor. It includes all the four losses mentioned in this section. Adding up all the loss calculations would result in the total loss coefficient which s used to find the isentropic efficiency of the compressor. Z=

n X

Zi

(2.51)

ηs = 1 − Z

(2.52)

i=1

As given in eqn.2.51, the total loss coefficient was calculated to be 0.375 and as given in 2.52 and [2], we find the isentropic efficiency to be 63%

2.6

Results and Discussions

A mean line analysis is initially performed to obtain the heights, velocities and other quantities at the mean diameter. As we move through the stages in the compressor, the blade height decreases. This is in attempt to have approximately constant axial velocity throughout the compressor section. It is important to note here that the variation of density is to be accounted for compressor. The power consumed by compressor was arrived at 40MW while that generated from turbine is 78MW. So, the power generated is close to 38MW.

21

LP

S1 S2 0.46 0.44 -0.34 -0.37 0.5 0.5 1 1 55 55 34.2 32.7 145.78 147.22 125 125 98.61 134.57 134.57 178.24 288 315 315 347 151.5 151.5 4 4

Variable

φ λ r µ α1 (◦ C) α2 (◦ C) β1 (◦ C) β2 (◦ C) pin (kP a) pout (kP a) Tin (K) Tout (K) m(kg/s) ˙ Power(MW)

S3 0.5 -0.27 0.5 1 55 38.9 141.2 125 178 213 347 365 150 2.8

IP S4 0.5 -0.28 0.5 1 55 38.3 141.7 125 213 254 365 384 150 2.8 S5 S6 0.5 0.51 -0.29 -0.28 0.5 0.5 1 1 55 55 37.9 38.7 142.2 141 125 125 254 298 298 343 384 401 401 424 150 151.3 2.8 2.67

S7 0.51 -0.28 0.5 1 55 38.4 141 125 343 394 424 441 151.3 2.67

S8 0.5 -0.29 0.5 1 55 38 142 125 394 450 441 458 151.3 2.67

S9 0.5 -0.29 0.5 1 55 38 142 125 450 512 458 475 151.3 2.67

HP S10 0.5 -0.3 0.5 1 55 37.7 142 125 512 580 475 491 151.3 2.67 S11 0.5 -0.3 0.5 1 55 37.3 143 125 580 653 491 508 151.3 2.67

S12 0.5 -0.3 0.5 1 55 37 143 125 653 732 508 524 151.3 2.67

S13 0.5 -0.3 0.5 1 55 37 143 125 732 821 524 541 151.3 2.67

S14 0.5 -0.31 0.5 1 55 36.6 143 125 821 910 541 557 151.3 2.67

Table 2: Table that shows the variation of stage parameters and thermodynamic quantities with different stages of compressor

Project Report 2 COMPRESSOR

22

3 COMBUSTION CHAMBER

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3

Combustion Chamber

The thermodynamic process in a gas turbine engine follows the Brayton cycle. The energy input to the Brayton cycle is provided by the heat generated, and hence carried by the combustion gases, in the combustor. The high pressure air from the compressor is bled into the combustion chamber, where the fuel is burned at an exorbitantly high temperature. Air flow from the compressor is mixed with the high temperature combustion gases and the resulting mixture of gases with suitably lowered temperature is fed to the turbine. The combustion chamber typically consist of a central burning zone, a swirl or mixing zone at the beginning, and a dilution zone that aids to the reduction of the overall gas temperature. The combustion chamber, however, does not completely burn the fuel, and this is represented in terms of the combustion chamber efficiency. The efficiency is computed as the ratio of the amount of heat generated to the maximum possible (theoretical) heat that can be generated using the particular type of fuel. Note that the fuel heating value is a measure of the maximum amount of heat that can be generated while burning the fuel.

Figure 3.1: Exit of the combustion chamber

23

3 COMBUSTION CHAMBER

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Figure 3.2: View showing the arrangement of flame-holders The present gas turbine layout incorporates an annular combustion chamber design, where the gas flows straight through all the combustors. The combustors periphery matches with that of the compressor and the turbine sections, and hence no hindrance to the gas flow is expected. The combustor has a baffle sort of design at the fuel inlet. This is so since the high velocity gas flow does not assist in generating a stable flame, and hence could lead to low efficiency combustion. The baffle, or swirl generators, effectively reduces the gas flow velocity in the space where the fuel is burnt by creating a turbulent flow regime. Another component inside the combustion chamber, a perforated liner, prevents excessive mass flow in the direction of the flame. The absence of a perforated liner would have made the combustion process nearly impossible as the large mass flow of gas could instantly blow away the fuel, or reduce the fuel to air ratio drastically.

24

3 COMBUSTION CHAMBER

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Figure 3.3: Cross section view of combustion chamber

25

4 TURBINE

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4

Turbine

4.1

Design of Turbine Blades

The mechanical energy generated from the combustion of fuel in the combustion chamber is extracted using the turbine stage as shaft power. The turbine section consists of several alternate rows of stator and rotor blades, the number of stages dependent on the overall specific work, the blade parameters, mass flow rate, the desired power output etc. The turbine stage design follows an iterative procedure solving the nonlinear relationship between the stage parameters, and modifying the parameters during iterations, until the desired output pressure is obtained.

Figure 4.1: Turbine Blade Nomenclature as given in [2]

4.1.1

Procedure

The procedure for generating the blade profiles include generating cascade lines, followed by generating blade profiles based on the base profile chosen. 1. Depending on the chord length, draw two lines, leading and trailing edge. 26

4 TURBINE

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2. Draw a third line at a distance of 1/3rdof chord length from the leading line. 3. From an arbitrary point on the leading edge, draw a line with a slope of the inlet velocity . 4. From the point of intersection of the two line at the 1/3rd distance, draw a line with a slope of the exit velocity . 5. Now we have 3 points using which we can generate the camber line. This can be done using the BEZIER CURVE FUNCTION as follows: Bezier Curves are parametric curves used to generate smooth contours. Depending upon the number of points used to generate the curve, they can be classified as follows: Linear Bezier curve (2 points) : B(t) = P0 + t(P1 − P0 ), t ∈ [0, 1]

(4.1)

Quadratic Bezier curve (3 points) : B(t) = (1 − t2 )2 P0 + 2t(1 − t)P1 + t2 P2 , t ∈ [0, 1] 4.1.2

(4.2)

Generating Blade profile

1. Choose a base profile, for instance such as the one shown in Figure?? below. 2. Superimpose the above base profile on the camber line obtained using the above steps and the equations 4.3 for suction side, 4.4 for pressure side and 4.5 : x = xc −

t sin ν 2

(4.3a)

y = yc +

t cos ν 2

(4.3b)

t sin ν 2 t y = yc − cos ν 2 x = xc +

xc CL 1 − c ν= ln xc 4π c

27

(4.4a) (4.4b)

(4.5)

4 TURBINE

Project Report

0.5

yc

0.25

0

-0.25

-0.5 0

0.25

0.5

0.75

1

xc Figure 4.2: Sample base profile for superposition

4.2

Blade Profiles

Figure 4.3 shows a turbine blade cascade with the blade parameters displayed in the image. The blade profiles are generated using the procedure mentioned in the previous section. 4.2.1

Procedure

1. Select the cascade chord spacing as

c =1 s

2. Use the design data: Turbine pressure ratio = 1.5 and Compressor press. ratio= 1.1 p1 − px 3. Define pressure coefficient,Cp = p1 4. Mesh the blade profile using a commercial CFD package, STAR-CCM+ 5. Generate Pressure distribution along the surfaces The turbine and compressor blade parameters are as follows Turbine: α1 = 120, β2 = 16, γ = 33.4 28

(4.6)

4 TURBINE

Project Report

     

50



  

0

-50

-100

-150 -200

-150

-100

-50

0

50

100  

Figure 4.3: Blade Cascade Compressor : α1 = 104, β2 = 60, CL = 0.9 4.2.2

(4.7)

CFD Results

The pressure distribution along the blade profiles are computed using the following procedure. A pressure distribution coefficient is selected to conveniently represent the non-dimensional pressure distribution. Figure 4.6 depicts the 2D mesh surface of the fluid around the turbine blade. The general mesh contains tetrahedral elements. The fine mesh around the blade periphery contains prism elements. Figure 4.5 shows the pressure distribution around the turbine blade. Figure 4.4 displays the contour of velocity magnitude.

29

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Figure 4.4: Velocity contour around the blade

Figure 4.5: Pressure distribution around the blade

30

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Figure 4.6: Mesh using tetrahedral elements

-0.40 -0.20

Cp

0.00

Inlet Angle (1)= 76o Exit Angle (2)= 16o Stagger Angle ()= 33.4o

0.20 0.40 0.60 0

Corresponding Blade Cascade

0.2

0.4

x/cax

0.6

0.8

Figure 4.7: Cp plot for a turbine

31

1

Figure 4.8: Turbine configuration used for pressure distribution

Project Report 4 TURBINE

32

4 TURBINE

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4.3 4.3.1

Design of Turbine Assumptions

Exit Diameter = 1.12 m Mass flow rate= 151.3 kg/s Rotating speed= 469.35 rad/sec Gas constant = 1.44 Turbine inlet temperature, Tin = 1222.7F Turbine inlet pressure, Pin = 873350N/m2 Turbine outlet temperature, Tout = 806.77F Turbine outlet pressure, Pout = 102200N/m2 Stage reaction, R=0.5 Turbine efficiency = 90%

4.3.2

Steps Followed

1. The number of turbine stages required for the gas turbine is chosen as four. Mean diameters for the rotor and stator at all the stages are computed with the known data. Figure 4.13 displays the notation followed for the blades at the various turbine stages 2. The specific mechanical energy, lm for the turbine is computed as the difference in enthalpies of the outlet and inlet gas. The stage specific mechanical energy, lms , is computed from, lm (4.8) lms = 4 3. Dimensionless stage parameters are defined to solve the nonlinear relationships between the stage parameters. The dimensionless parameters are as defined in the nomenclature. 4. The introduction of these dimensionless parameters into the known relationships of the stage parameters provides the following four nonlinear equations ,2.3,2.4,2.5 and 2.6. The equations contain nine unknowns. In order to solve the four equations, at least five of the nine unknowns must be known. The diameter ratio, the stator and rotor exit angles, exit flow angle, and the stage reaction are guessed. Note that α2 = 20 , β3 = 160 is assumed. 5. The solution of the above four nonlinear equations provide λ, φ, α3 , β2 6. The meridional velocity, and the mass flow rate, along with the gas density, found using the inlet pressure and temperature aids to the estimation of the cross sectional area.

33

Figure 4.9: Section view showing the increasing cross section of the turbine

Project Report 4 TURBINE

34

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Stator

V1

Rotor

V2

W3

Figure 4.10: Axial Turbine stage

VM 3

W3

U3

V3

W2

α2 α3

U2

β2 β3

V2

VM 2

Figure 4.11: Velocity Triangle for a Turbine stage. Subscripts 2 refer to rotor inlet and 3 refers to rotor exit

35

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Po1

Po2 H1= H2 h1

P1

Enthalpy

lms P2 h2 Po3

H3 h3

P3

Entropy Figure 4.12: h-s diagram for a turbine stage

1

3

2

4

5

6

7

9

8

Stator Blade

Rotor Blade

Stage-1

Stage-2

Stage-3

Stage-4

Figure 4.13: Notation for a turbine stage UNLESS OTHERWISE SPECIFIED:

DIMENSIONS ARE IN INCHES TOLERANCES: FRACTIONAL ANGULAR: MACH BEND TWO PLACE DECIMAL THREE PLACE DECIMAL INTERPRET GEOMETRIC TOLERANCING PER:

PROPRIETARY AND CONFIDENTIAL THE INFORMATION CONTAINED IN THIS DRAWING IS THE SOLE PROPERTY OF . ANY REPRODUCTION IN PART OR AS A WHOLE WITHOUT THE WRITTEN PERMISSION OF IS PROHIBITED.

MATERIAL

USED ON

NEXT ASSY

SolidWorks Student License Academic Use 5 Only

APPLICATION

4

FINISH

NAME

DATE

DRAWN

TITLE:

CHECKED ENG APPR. MFG APPR. Q.A. COMMENTS:

SIZE DWG. NO.

A

36 2

REV

SHEET 1 OF 1

SCALE: 1:10 WEIGHT:

DO NOT SCALE DRAWING

3

Stations 1

4 TURBINE

Project Report

V m1ref = phiref ∗ U 1; Pin ρ1ref = RTin mdot Areacrosssectional = rho1ref ∗ V m1ref

(4.9a) (4.9b) (4.9c)

7. The rotor hub radius is readily found out using the blade height and the mean diameter at the first stage. The tip diameter and the hub diameter for all the stages are computed using the mean diameters at each stage. 8. The enthalpy of inlet gas is found out using the turbine inlet temperature. The computed inlet velocity aids to the estimation of the total enthalpy Areacrosssectional = π(tip2rad − hub2rad )

(4.10a)

h1 = Cp T1 (4.10b) 1 (4.10c) H1 = h1 + V12 2 9. H3 is computed using H1 and the specific stage energy, lms H3 = H1 − lms . 10. T03 is calculated using the efficiency of 90% as : T 03s(i) =

T 01(i) − ((T 01(i) − T 03(i))) η

(4.11)

11. The total pressures, P01 and P03 are computed then 1 P01 = P1 + V12 2  γ  T03 γ − 1 P03 = P01 T01

(4.12a) (4.12b)

12. Similarly, T03 and hence h3 is found out. 13. The above steps (5-12) is repeated for all the four stages. Note that α3 = 90 for the last stage. Iteration is continued until the pressure generated at the last stage is within acceptable values of the desired outlet pressure. The maximum variation in pressure allowed is 10 N/m2 . During each iteration, if the difference of the computed output pressure and the desired output pressure is greater than the convergence criterion, then the specific stage energy is either increased or decreased by 0.01%. 14. The enthalpy values are sufficiently modified to incorporate the variation in enthalpy at each stage due to the reduction in efficiency, attributed to various losses, at each stage. 37

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Project Report 4.3.3

Radial Equilibrium

The radial equilibrium as listed in section is followed for the turbine blades as well. Figure 4.14 shows a picture of the twisted blade.

Figure 4.14: Turbine rotor twisted in spanwise direction.

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4.4

Losses

The loss calculations are same as that of compressor blades. The total losses for each turbine stage comprises of the stage profile losses, the trailing edge thickness losses, the secondary flow losses and the exit losses.The total losses for each stage provides the efficiency of each stage. The design iteration for determining the exit pressure includes the determination of the efficiency of each stage, and suitably modifying the enthalpies in each stage to incorporate their corresponding losses.

39

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Results

3

Air Density

3

Air Density (kg/m )

2.5

2

1.5

1

0.5

0

1

2

3

4

5

Stage  

Figure 4.15: Variation of density with number of stages in a turbine

1.0E+06

9.5E+05

Total Enthalpy Static Enthalpy

9.0E+05

Enthalpy (J/kg)

4.5

8.5E+05

8.0E+05

7.5E+05

7.0E+05

6.5E+05

6.0E+05

5.5E+05

1

2

3

4

Stage  

Figure 4.16: Variation of Enthalpy with number of stages in a turbine

40

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Table 3: Table that shows the variation of stage parameters and thermodynamic quantities with different stages of turbine Variable Stages S1 S2 S3 S4 φ .3392 0.3426 0.3440 0.3640 λ 2.33 2.20 2.08 1.97 r 0.5 0.5 0.5 0.5 µ 0.3042 0.3092 0.3225 0.9867 α2 (◦ C) 20 20 20 20 ◦ α3 ( C) 158.5 158.5 158.5 158.5 β2 (◦ C) 40.695 40.695 40.695 40.695 β3 (◦ C) 160 160 160 160 pin (kP a) 873.35 573.91 354.91 202.21 pout (kP a) 573.91 354.91 202.21 102.19 Tin (K) 949.55 846.68 743.83 642.29 Tout (K) 846.68 743.83 642.29 546.58 m(kg/s) ˙ 151.3 151.3 151.3 151.3 Power(MW) 19.6 19.6 19.6 19.6

1 Mach Number

Mach number

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5

Stage  

Figure 4.17: Variation of Mach Number with number of stages in a turbine The iterative solving procedure provides the air density, velocity, mach no., pressure, temperature, and enthalpy of gas at each stage, and are depicted below in figures 4.18, 4.20, 4.17. 41

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Project Report

7.0E+05

6.0E+05

Total Pressure Static Pressure

Pressure (Pa)

5.0E+05

4.0E+05

3.0E+05

2.0E+05

1.0E+05

0.0E+00

1

2

3

4

Stage  

Figure 4.18: Variation of static and total pressure with number of stages in a turbine Figure 4.16 shows the enthalpy decreasing with each stage, as expected across any stage in a turbine. Similar variation would be found for temperature(fig.4.19) as well, as the relationship between them is linear.

42

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1200 Total Temperature Static Temperature

1150

1050

o

Temperature ( K)

1100

1000 950 900 850 800 750

1

2

3

4

Stage  

Figure 4.19: Variation of static and total temperature with number of stages in a turbine

900 800

Absolute Velocity Relative Velocity

Velocity (m/s)

700 600 500 400 300 200 100

1

2

3

4

Stage  

Figure 4.20: Variation of absolute and relative velocity with number of stages in a turbine Figure 4.20 shows the variation of both the absolute and relative velocity. As can be seen, the velocity increases as it progresses through the turbine stages.

43

5 STRESS ANALYSIS

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5

Stress Analysis

This section deals with the stress analysis on the entire gas turbine assembly. This would help us in evaluating the thickness of casing for the turbine and compressor that would avoid failure due to excessive stress. The commercially available solver COSMOS was used to arrive at the VonMisses stress distribution.

Figure 5.1: Mesh of the gas turbine casing used for simulation in COSMOS

Figure 5.2: Stress distribution on the casing of the gas turbine

44

REFERENCES

Project Report

References [1] Schobeiri, M., Advanced Fluid Mechanics Springer-Verlag. [2] Schobeiri, M., Turbomachinery Flow Physics and Dynamic Performance SpringerVerlag. [3] Yunus A Cengel, Michael A.Boles Thermodynamics- An Engineering Approach [4] Wikipedia, www.wikipedia.com

45

6 APPENDIX

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6 6.1 6.1.1

Appendix Compressor Codes Blade Profile Generation

Blade Profile Generation: 1 2 3

clear all ; close all ; clc ;

4 5

C_ax = input ( ’ Input Axial Chord :

’) ;

6 7

X_by_C (: ,1) =0:0.001:1;

8 9

Cl = input ( ’ Input desired lift - coefficient ( Cl ) :

’) ;

10 11

Y_by_C = - Cl /(4* pi ) *((1 - X_by_C ) .* log (1 - X_by_C ) + X_by_C .* log ( X_by_C ) );

12 13 14

badrows = any ( isnan ( Y_by_C ) ,2) ; Y_by_C ( badrows ,:) = zeros ;

15 16 17

camber_line =[ C_ax * X_by_C C_ax * Y_by_C ];

18 19 20 21 22

23

camber_length =0; for ii =0: length ( camber_line (: ,1) ) -2 camber_length = camber_length + sqrt (( camber_line ( ii +1 ,1) camber_line ( ii +2 ,1) ) ^2+( camber_line ( ii +1 ,2) - camber_line ( ii +2 ,2) ) ^2) ; end

24 25 26

chord_length = C_ax ; s = chord_length /2;

27 28 29

Y_by_C_0 (: ,1) = Y_by_C - s ; Y_by_C_2 (: ,1) = Y_by_C + s ;

30 31 32

camber_line0 =[ X_by_C Y_by_C_0 ]; camber_line2 =[ X_by_C Y_by_C_2 ];

33 34

35

[ filename , pathname ] = uigetfile ({ ’*. dat ;*. csv ;*. txt ;*. xls ’} , ’ Pick a file ’) ; base_profile = load ( strcat ( pathname , filename ) ) ;

46

6 APPENDIX

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36

37

half_base _profi le =[ base_profile (1: ceil ( length ( base_profile ) /2) ,1) base_profile (1: ceil ( length ( base_profile ) /2) ,2) ]; half_base _profi le ( end ,2) = hal f_base _profi le (1 ,2) ;

38 39 40 41 42 43 44 45

46

47 48 49 50 51 52 53

clear T_by_C ; for ii =1: length ( X_by_C ) for jj =1: length ( h alf_ba se_pro file ) if ( X_by_C ( ii ,1) == ha lf_bas e_prof ile ( jj ,1) ) T_by_C ( ii ,1) = ha lf_bas e_prof ile ( jj ,2) ; break ; elseif ( X_by_C ( ii ,1) < ha lf_bas e_prof ile ( jj ,1) && X_by_C ( ii ,1) > half_ base_p rofile ( jj -1 ,1) ) T_by_C ( ii ,1) = ( ha lf_bas e_prof ile ( jj -1 ,2) + ( X_by_C ( ii ,1) - half_ base_p rofile ( jj ,1) ) *(( hal f_base _profi le ( jj ,2) - hal f_base _profi le ( jj -1 ,2) ) /( half_ base_p rofile ( jj ,1) - half_ base_p rofile ( jj -1 ,1) ) ) ) ; break ; end end end if ( length ( T_by_C ) < length ( X_by_C ) ) T_by_C ( ii ,1) =0; end

54 55

T = T_by_C * camber_length ;

56 57 58

%% norm_camber =[ camber_line (: ,1) / chord_length ( camber_line (: ,2) camber_line (1 ,2) ) / chord_length ];

59 60

superimp_profile = camber_length * h alf_ba se_pro file ;

61 62 63

64

%% pres_surf_x = camber_line (: ,1) -T .* sin ( atan ( camber_line (: ,2) ./ camber_line (: ,1) ) ) ; pres_surf_y = camber_line (: ,2) + T .* cos ( atan ( camber_line (: ,2) ./ camber_line (: ,1) ) ) ;

65 66

67

suct_surf_x = camber_line (: ,1) + T .* sin ( atan ( camber_line (: ,2) ./ camber_line (: ,1) ) ) ; suct_surf_y = camber_line (: ,2) -T .* cos ( atan ( camber_line (: ,2) ./ camber_line (: ,1) ) ) ;

68 69

for jj =1: length ( camber_line )

47

6 APPENDIX

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if ( pres_surf_x ( jj ,1) > camber_line ( jj ,1) || suct_surf_x ( jj ,1) < camber_line ( jj ,1) ) press ( jj ,1) = suct_surf_x ( jj ,1) ; suct ( jj ,1) = pres_surf_x ( jj ,1) ; else press ( jj ,1) = pres_surf_x ( jj ,1) ; suct ( jj ,1) = suct_surf_x ( jj ,1) ; end

70

71 72 73 74 75 76 77

end

78 79 80

press (: ,2) = pres_surf_y (: ,1) ; suct (: ,2) = suct_surf_y (: ,1) ;

81 82 83 84

blade_profile = vertcat ( press , flipud ( suct ) ) ; badrows = any ( isnan ( blade_profile ) ,2) ; blade_profile ( badrows ,:) = zeros ;

85 86

87

% rot_ang_back = pi - atan (( camber_line ( end ,2) - camber_line (1 ,2) ) /( camber_line ( end ,1) - camber_line (1 ,1) ) ) ; % rot_matrix_back =[ cos ( rot_ang_back ) - sin ( rot_ang_back ) ; sin ( rot_ang_back ) cos ( rot_ang_back ) ];

88 89 90 91 92 93 94 95 96

97

98 99

blade = blade_profile ;%* rot_matrix_back ; blade_CFD (: ,1) = blade (: ,1) ; blade_CFD (: ,2) = blade (: ,2) ; blade_CFD (: ,3) = zeros ; camber_line_CFD (: ,1) = camber_line (: ,1) ; camber_line_CFD (: ,2) = camber_line (: ,2) ; camber_line_CFD (: ,3) = zeros ; camber_line0_CFD =[ camber_line (: ,1) camber_line (: ,2) -s camber_line_CFD (: ,3) ]; camber_line2_CFD =[ camber_line (: ,1) camber_line (: ,2) + s camber_line_CFD (: ,3) ]; blade0 =[ blade (: ,1) blade (: ,2) -s ]; blade2 =[ blade (: ,1) blade (: ,2) + s ];

100 101 102 103 104

105 106 107 108

fid = fopen ( ’ Blade_Profile . txt ’ , ’w + ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , blade , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ;

48

6 APPENDIX

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109

110 111 112 113 114

115 116 117 118 119

120 121 122 123 124

125 126 127 128 129

130

dlmwrite ( ’ Blade_Profile . txt ’ , blade0 , ’- append ’ , ’ delimiter ’ , ’\ t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , blade2 , ’- append ’ , ’ delimiter ’ , ’\ t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , camber_line , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , camber_line0 , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , camber_line2 , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

131 132

133

134

135

136

dlmwrite ( ’ Blade_CFD . txt ’ , blade_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber_CFD . txt ’ , camber_line_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber0_CFD . txt ’ , camber_line0_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber2_CFD . txt ’ , camber_line2_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Blade . txt ’ , blade , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ;

49

6 APPENDIX

Project Report 6.1.2

Low Pressure Compressor

Low Pressure Compressor : 1 2

clc clear all

3 4

global a1 a2 b1 b2 nu lambda r phi mu

5 6 7 8

%%%%%%%%%%%%%%%%%%%%%%%%% % Given parameters % % % %% % % % % % % % % % % % % % % % % % % % %

9 10 11 12 13 14 15 16

% Low Pressure Section lp . mdot = 151.51; %%% lp . pin = 98.61; %%% lp . pr = 1.8048; %%% lp . Tin = 288.21; %%% lp . Tout = 347.2; %%% omega = 469.35; %%%

mass flow rate in kg / sec pressure in Kpa pressure ratio in Kelvin in Kelvin in rad / sec

17 18 19 20 21 22

% % % % %

% Intermediate Pressure Section mdot_ip = 150; pin_ip = 177.97; pr_ip = 1.6739; Tin_ip = 347.02;

23 24 25 26

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Low pressure section having 2 stages %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

27 28 29 30 31 32 33

% defining the diameters at the axial length % Dm1 = 1.2043; %%% in meters Dm5 = 1.1253; %%% in meters Dm3 = ( Dm1 + Dm5 ) /2; Dm4 = ( Dm5 + Dm3 ) /2; Dm2 = ( Dm1 + Dm3 ) /2;

34 35 36 37 38 39 40

%%%% U1 = U2 = U3 = U4 = U5 =

calculating the circumferential velocity %%% 0.5* Dm1 * omega ; 0.5* Dm2 * omega ; 0.5* Dm3 * omega ; 0.5* Dm4 * omega ; 0.5* Dm5 * omega ;

41 42 43

% Assuming the stage 1 LP angle values a1 = 55 ;

50

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44

b2 = 125;

45 46 47

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

48 49

%%%% Calculating the table

thermodynamic values from given data

50 51

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

52 53 54 55

56 57

58

59

60

61

62

63

64

Data_Compressor = importdata ( ’ Compressor . dat ’) ; % To calculate the values from the given table at given values of inlet (1) and % exit (5) temperatures lp . h1 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,2) , lp . Tin ) ; lp . Cp1 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,4) , lp . Tin ) ; lp . R1 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,6) , lp . Tin ) ; lp . Gamma1 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , lp . Tin ) ; lp . h5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,2) , lp . Tout ) ; lp . Cp5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,4) , lp . Tout ) ; lp . R5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,6) , lp . Tout ) ; lp . Gamma5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , lp . Tout ) ;

65 66 67

% Isentropic Efficiency eff = 0.95;

68 69 70

% Degree of Reaction r = 0.5;

71 72 73 74 75

% Defining velocities ( initialized as zero ) lp . V1 =0; lp . V5 =0; lp . V3 =0;

76 77

% Total Load Lm

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78 79

% lp . lm = ( lp . h5 - lp . h1 ) +0.5*( lp . V1 ^2 - lp . V5 ^2) ; lp . lm = ( lp . h1 - lp . h5 ) + 0.5 * ( lp . V1 ^2 - lp . V5 ^2) ; % Lp is negative for compressor

80 81 82 83

% Stage Load no_stages = 2; lp . lms = lp . lm / no_stages ;

84 85

Press_error = 100000;

86 87

% Start of Iteration

88 89

while abs ( Press_error ) >=0.2

90 91 92 93

lp . h3 = lp . h1 - lp . lms + 0.5*( lp . V1 ^2 - lp . V3 ^2) /1000; lp . h5 = lp . h3 - lp . lms + 0.5*( lp . V3 ^2 - lp . V5 ^2) /1000; % To interpolate the values using the enthalpy h3

94 95

96

97 98

99

100

lp . T3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , lp . h3 ) ; lp . R3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , lp . h3 ) ; lp . h3s = eff *( lp . h3 - lp . h1 ) + lp . h1 ; lp . T3s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , lp . h3s ) ; lp . Gamma3s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , lp . T3s ) ; Gamma = ( lp . Gamma1 + lp . Gamma3s ) *0.5;

101 102

% Pressure 3 is found using the adiabatic relation

103 104 105 106

lp . p3 = lp . pin *( lp . Tin / lp . T3s ) ^( Gamma /(1 - Gamma ) ) ; lp . stage1 . lambda = lp . lms *1000 / U2 ^2; lp . stage2 . lambda = lp . lms * 1000 / U4 ^2;

107 108

109

% h2 is found using the degree of reaction formula and hence the values are % found using the interpolation data

110 111 112

113

lp . h2 = r *( lp . h3 - lp . h1 ) + lp . h1 ; lp . T2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , lp . h2 ) ; lp . R2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , lp . h2 ) ;

114

52

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115

116 117 118

% Calculating the pressure 2 by assuming poltropic process from 1 to 3 c = log ( lp . p3 / lp . pin ) / log ( lp . Tin / lp . T3 ) ; n = c /( c +1) ; % Polytropic Contstant lp . p2 = lp . pin *( lp . Tin / lp . T2 ) ^( n /(1 - n ) ) ;

119 120

% To interpolate the values using the enthalpy h5s

121 122 123

124

125

126

lp . h5s = eff *( lp . h5 - lp . h3 ) + lp . h3 ; lp . T5s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , lp . h5s ) ; lp . Gamma3 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , lp . T3 ) ; lp . Gamma5s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , lp . T5s ) ; Gamma = ( lp . Gamma3 + lp . Gamma5s ) *0.5;

127 128

% Pressure 5 is found using the adiabatic relation

129 130

lp . p5 = lp . p3 *( lp . T3 / lp . T5s ) ^( Gamma /(1 - Gamma ) ) ;

131 132

133

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

134 135 136

137

lp . h4 = r *( lp . h5 - lp . h3 ) + lp . h3 ; lp . T4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , lp . h4 ) ; lp . R4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , lp . h4 ) ;

138 139

140 141 142

% Calculating the pressure 4 by assuming poltropic process from 3 to 5 c = log ( lp . p5 / lp . p3 ) / log ( lp . T3 / lp . Tout ) ; n = c /( c +1) ; % Polytropic Contstant lp . p4 = lp . p3 *( lp . T3 / lp . T4 ) ^( n /(1 - n ) ) ;

143 144 145 146 147 148 149 150 151 152

% For Stage 1 lambda = lp . stage1 . lambda ; nu = U1 / U2 ; x0 = [6 0;160; 0.8;0. 9]; [ stage1_x ] = fsolve ( @LP_Stg1 , x0 ) ; lp . stage1 . a2 = stage1_x (1) ; lp . stage1 . b1 = stage1_x (2) ; lp . stage1 . phi = stage1_x (3) ; lp . stage1 . mu = stage1_x (4) ;

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153 154 155 156 157 158 159 160 161 162 163

% For Stage 2 lambda = lp . stage2 . lambda ; nu = U3 / U4 ; x0 = [6 0;160; 0.8;0. 9]; [ stage2_x ] = fsolve ( @LP_Stg1 , x0 ) ; lp . stage2 . a2 = stage2_x (1) ; lp . stage2 . b1 = stage2_x (2) ; lp . stage2 . phi = stage2_x (3) ; lp . stage2 . mu = stage2_x (4) ;

164 165 166

% Velocity and Mass flow Rates :

167 168 169 170 171

V_ax2 V_ax4 V_ax1 V_ax3

= = = =

lp . stage1 . phi * U2 ; lp . stage2 . phi * U4 ; lp . stage1 . mu * V_ax2 ; lp . stage2 . mu * V_ax4 ;

lp . V1 lp . V2 lp . V3 lp . V4

= = = =

V_ax1 V_ax2 V_ax3 V_ax4

/ / / /

lp . W1 lp . W2 lp . W3 lp . W4

= = = =

sqrt ( sqrt ( sqrt ( sqrt (

( V_ax1 ) ^2 ( V_ax2 ) ^2 ( V_ax3 ) ^2 ( V_ax4 ) ^2

172 173 174 175 176

sind ( a1 ) ; sind ( lp . stage1 . a2 ) ; sind ( a1 ) ; sind ( lp . stage2 . a2 ) ;

177 178 179 180 181

+ + + +

( U1 - lp . V1 * cos ( a1 ) ^2 ) ) ; ( U2 - lp . V2 * cos ( lp . stage1 . a2 ) ^2 ) ) ; ( U3 - lp . V3 * cos ( a1 ) ^2 ) ) ; ( U4 - lp . V4 * cos ( lp . stage2 . a2 ) ^2 ) ) ;

182 183 184

% lp . V5 = sqrt ( abs ( lp . V4 ^2 - 2000*( lp . h5 - lp . h4 ) ) ) ; % V_ax5 = lp . V5 * sind ( a1 ) ;

185 186 187 188 189 190 191 192 193 194 195 196 197 198

rho1 = lp . pin *10^3/( lp . R1 * lp . Tin ) ; rho2 = lp . p2 *10^3/ ( lp . R2 * lp . T2 ) ; rho3 = lp . p3 *10^3/ ( lp . R3 * lp . T3 ) ; rho4 = lp . p4 *10^3/ ( lp . R4 * lp . T4 ) ; rho5 = lp . p5 *10^3/ ( lp . R5 * lp . Tout ) ; lp . A1 = lp . mdot /( rho1 * V_ax1 ) ; lp . A2 = lp . mdot /( rho2 * V_ax2 ) ; lp . A3 = lp . mdot /( rho3 * V_ax3 ) ; lp . A4 = lp . mdot /( rho4 * V_ax4 ) ; % lp . A5 = lp . mdot /( rho5 * V_ax5 ) ; lp . A5 = 2* lp . A4 - lp . A3 ; % Linear variation of area V_ax5 = lp . mdot / ( rho5 * lp . A5 ) ; lp . V5 = V_ax5 / sind ( a1 ) ;

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199

lp . W5 = lp . V5 - U5 ;

200 201 202 203 204 205

lp . ht1 lp . ht2 lp . ht3 lp . ht4 lp . ht5

= = = = =

lp . A1 /( pi lp . A2 /( pi lp . A3 /( pi lp . A4 /( pi lp . A5 /( pi

* * * * *

Dm1 ) Dm2 ) Dm3 ) Dm4 ) Dm5 )

; ; ; ; ;

206 207

Press_error = ( lp . p5 - ( lp . pin * lp . pr ) ) *100/( lp . pin * lp . pr )

208 209 210 211 212 213 214

if ( Press_error >0) lp . lms = 0.99* lp . lms ; % lp . lm = 2* lp . lms ; elseif ( Press_error 0.3

91 92 93 94 95

ip . h3 = ip . h1 - ip . lms + 0.5*( ip . V1 ^2 - ip . V3 ^2) /1000; ip . h5 = ip . h3 - ip . lms + 0.5*( ip . V3 ^2 - ip . V5 ^2) /1000; ip . h7 = ip . h5 - ip . lms + 0.5*( ip . V5 ^2 - ip . V7 ^2) /1000; % To interpolate the values using the enthalpy h3

96 97

98

99 100

101

102

ip . T3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h3 ) ; ip . R3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , ip . h3 ) ; ip . h3s = eff *( ip . h3 - ip . h1 ) + ip . h1 ; ip . T3s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h3s ) ; ip . Gamma3s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , ip . T3s ) ; Gamma = ( ip . Gamma1 + ip . Gamma3s ) *0.5;

103 104

% Pressure 3 is found using the adiabatic relation

105 106 107 108 109

ip . p3 = ip . pin *( ip . Tin / ip . T3s ) ^( Gamma /(1 - Gamma ) ) ; ip . stage1 . lambda = ip . lms *1000 / U2 ^2; ip . stage2 . lambda = ip . lms * 1000 / U4 ^2; ip . stage3 . lambda = ip . lms * 1000 / U6 ^2;

110 111

112

% h2 is found using the degree of reaction formula and hence the values are % found using the interpolation data

113 114 115

ip . h2 = r *( ip . h3 - ip . h1 ) + ip . h1 ; ip . T2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h2 ) ;

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116

ip . R2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , ip . h2 ) ;

117 118

119 120 121

% Calculating the pressure 2 by assuming poltropic process from 1 to 3 c = log ( ip . p3 / ip . pin ) / log ( ip . Tin / ip . T3 ) ; n = c /( c +1) ; % Polytropic Contstant ip . p2 = ip . pin *( ip . Tin / ip . T2 ) ^( n /(1 - n ) ) ;

122 123

% To interpolate the values using the enthalpy h5s

124 125

126 127

128

129

130

131

ip . T5 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h5 ) ; ip . h5s = eff *( ip . h5 - ip . h3 ) + ip . h3 ; ip . R5 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , ip . h5 ) ; ip . T5s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h5s ) ; ip . Gamma3 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , ip . T3 ) ; ip . Gamma5s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , ip . T5s ) ; Gamma = ( ip . Gamma3 + ip . Gamma5s ) *0.5;

132 133

% Pressure 5 is found using the adiabatic relation

134 135

ip . p5 = ip . p3 *( ip . T3 / ip . T5s ) ^( Gamma /(1 - Gamma ) ) ;

136 137

138

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

139 140 141

142

ip . h4 = r *( ip . h5 - ip . h3 ) + ip . h3 ; ip . T4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h4 ) ; ip . R4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , ip . h4 ) ;

143 144

145 146 147

% Calculating the pressure 4 by assuming poltropic process from 3 to 5 c = log ( ip . p5 / ip . p3 ) / log ( ip . T3 / ip . T5 ) ; n = c /( c +1) ; % Polytropic Contstant ip . p4 = ip . p3 *( ip . T3 / ip . T4 ) ^( n /(1 - n ) ) ;

148 149 150

%---------------% To interpolate the values using the enthalpy h5s

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151

152 153

154

155

156

ip . T7 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h7 ) ; ip . h7s = eff *( ip . h7 - ip . h5 ) + ip . h5 ; ip . T7s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h7s ) ; ip . Gamma5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , ip . T5 ) ; ip . Gamma7s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , ip . T7s ) ; Gamma = ( ip . Gamma3 + ip . Gamma5s ) *0.5;

157 158

% Pressure 5 is found using the adiabatic relation

159 160

ip . p7 = ip . p5 *( ip . T5 / ip . T7s ) ^( Gamma /(1 - Gamma ) ) ;

161 162

163

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

164 165 166

167

ip . h6 = r *( ip . h7 - ip . h5 ) + ip . h5 ; ip . T6 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , ip . h6 ) ; ip . R6 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , ip . h6 ) ;

168 169

170 171 172 173

% Calculating the pressure 4 by assuming poltropic process from 3 to 5 c = log ( ip . p7 / ip . p5 ) / log ( ip . T5 / ip . T7 ) ; n = c /( c +1) ; % Polytropic Contstant ip . p6 = ip . p5 *( ip . T5 / ip . T6 ) ^( n /(1 - n ) ) ; %-------------------

174 175 176 177 178 179 180 181 182 183

% For Stage 1 lambda = ip . stage1 . lambda ; nu = U1 / U2 ; x0 = [6 0;160; 0.8;0. 9]; [ stage1_x ] = fsolve ( @LP_Stg1 , x0 ) ; ip . stage1 . a2 = stage1_x (1) ; ip . stage1 . b1 = stage1_x (2) ; ip . stage1 . phi = stage1_x (3) ; ip . stage1 . mu = stage1_x (4) ;

184 185 186 187 188

% For Stage 2 lambda = ip . stage2 . lambda ; nu = U3 / U4 ; x0 = [6 0;160; 0.8;0. 9];

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189 190 191 192 193

[ stage2_x ] = fsolve ( @LP_Stg1 , x0 ) ; ip . stage2 . a2 = stage2_x (1) ; ip . stage2 . b1 = stage2_x (2) ; ip . stage2 . phi = stage2_x (3) ; ip . stage2 . mu = stage2_x (4) ;

194 195 196 197 198 199 200 201 202 203

% For Stage 3 lambda = ip . stage3 . lambda ; nu = U5 / U6 ; x0 = [6 0;160; 0.8;0. 9]; [ stage3_x ] = fsolve ( @LP_Stg1 , x0 ) ; ip . stage3 . a2 = stage3_x (1) ; ip . stage3 . b1 = stage3_x (2) ; ip . stage3 . phi = stage3_x (3) ; ip . stage3 . mu = stage3_x (4) ;

204 205

% Velocity and Mass flow Rates :

206 207 208 209 210 211 212

V_ax2 V_ax4 V_ax6 V_ax1 V_ax3 V_ax5

= = = = = =

ip . stage1 . phi * U2 ; ip . stage2 . phi * U4 ; ip . stage3 . phi * U6 ; ip . stage1 . mu * V_ax2 ; ip . stage2 . mu * V_ax4 ; ip . stage3 . mu * V_ax6 ;

ip . V1 ip . V2 ip . V3 ip . V4 ip . V5 ip . V6

= = = = = =

V_ax1 V_ax2 V_ax3 V_ax4 V_ax5 V_ax6

213 214 215 216 217 218 219

/ / / / / /

sind ( a1 ) ; sind ( ip . stage1 . a2 ) ; sind ( a1 ) ; sind ( ip . stage2 . a2 ) ; sind ( a1 ) ; sind ( ip . stage3 . a2 ) ;

220 221 222 223 224 225 226 227

rho1 rho2 rho3 rho4 rho5 rho6 rho7

= = = = = = =

ip . pin *10^3/( ip . R1 * ip . Tin ) ; ip . p2 *10^3/ ( ip . R2 * ip . T2 ) ; ip . p3 *10^3/ ( ip . R3 * ip . T3 ) ; ip . p4 *10^3/ ( ip . R4 * ip . T4 ) ; ip . p5 *10^3/ ( ip . R5 * ip . T5 ) ; ip . p6 *10^3/ ( ip . R6 * ip . T6 ) ; ip . p7 *10^3/ ( ip . R7 * ip . Tout ) ;

228 229 230 231 232 233 234

ip . A1 ip . A2 ip . A3 ip . A4 ip . A5 ip . A6

= = = = = =

ip . mdot /( rho1 ip . mdot /( rho2 ip . mdot /( rho3 ip . mdot /( rho4 ip . mdot /( rho5 ip . mdot /( rho6

* * * * * *

V_ax1 ) V_ax2 ) V_ax3 ) V_ax4 ) V_ax5 ) V_ax6 )

; ; ; ; ; ;

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235 236 237

ip . A7 = 2* ip . A6 - ip . A5 ; % Linear variation of area V_ax7 = ip . mdot / ( rho7 * ip . A7 ) ; ip . V7 = V_ax7 / sind ( a1 ) ;

238 239 240 241 242 243 244 245

ip . ht1 ip . ht2 ip . ht3 ip . ht4 ip . ht5 ip . ht6 ip . ht7

= = = = = = =

ip . A1 /( pi ip . A2 /( pi ip . A3 /( pi ip . A4 /( pi ip . A5 /( pi ip . A6 /( pi ip . A7 /( pi

* * * * * * *

Dm1 ) Dm2 ) Dm3 ) Dm4 ) Dm5 ) Dm6 ) Dm7 )

; ; ; ; ; ; ;

246 247

Press_error = ( ip . p7 - ( ip . pin * ip . pr ) ) *100/( ip . pin * ip . pr )

248 249 250 251 252 253

if ( Press_error >0) ip . lms = 0.99* ip . lms ; elseif ( Press_error 0.1

122 123 124 125 126 127 128 129 130 131

hp . h3 = hp . h1 - hp . lms + 0.5*( hp . V1 ^2 - hp . V3 ^2) /1000; hp . h5 = hp . h3 - hp . lms + 0.5*( hp . V3 ^2 - hp . V5 ^2) /1000; hp . h7 = hp . h5 - hp . lms + 0.5*( hp . V5 ^2 - hp . V7 ^2) /1000; hp . h9 = hp . h7 - hp . lms + 0.5*( hp . V7 ^2 - hp . V9 ^2) /1000; hp . h11 = hp . h9 - hp . lms + 0.5*( hp . V9 ^2 - hp . V11 ^2) /1000; hp . h13 = hp . h11 - hp . lms + 0.5*( hp . V11 ^2 - hp . V13 ^2) /1000; hp . h15 = hp . h13 - hp . lms + 0.5*( hp . V13 ^2 - hp . V15 ^2) /1000; hp . h17 = hp . h15 - hp . lms + 0.5*( hp . V15 ^2 - hp . V17 ^2) /1000; hp . h19 = hp . h17 - hp . lms + 0.5*( hp . V17 ^2 - hp . V19 ^2) /1000;

132 133 134 135 136 137 138 139 140 141 142

hp . stage1 . lambda hp . stage2 . lambda hp . stage3 . lambda hp . stage4 . lambda hp . stage5 . lambda hp . stage6 . lambda hp . stage7 . lambda hp . stage8 . lambda hp . stage9 . lambda

= = = = = = = = =

hp . lms hp . lms hp . lms hp . lms hp . lms hp . lms hp . lms hp . lms hp . lms

*1000 / U2 ^2; * 1000 / U4 ^2; * 1000 / U6 ^2; * 1000 / U8 ^2; * 1000 / U10 ^2; * 1000 / U12 ^2; * 1000 / U14 ^2; * 1000 / U16 ^2; * 1000 / U18 ^2;

143 144

%%%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

145

% STAGE 1

146 147

% To interpolate the values using the enthalpy h3

148 149

150

151 152

153

154

hp . T3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h3 ) ; hp . R3 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h3 ) ; hp . h3s = eff *( hp . h3 - hp . h1 ) + hp . h1 ; hp . T3s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h3s ) ; hp . Gamma3s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T3s ) ; Gamma = ( hp . Gamma1 + hp . Gamma3s ) *0.5;

155 156

% Pressure 3 is found using the adiabatic relation

157 158

hp . p3 = hp . pin *( hp . Tin / hp . T3s ) ^( Gamma /(1 - Gamma ) ) ;

159 160

74

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161

162

% h2 is found using the degree of reaction formula and hence the values are % found using the interpolation data

163 164 165

166

hp . h2 = r *( hp . h3 - hp . h1 ) + hp . h1 ; hp . T2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h2 ) ; hp . R2 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h2 ) ;

167 168

169 170 171

% Calculating the pressure 2 by assuming poltropic process from 1 to 3 c = log ( hp . p3 / hp . pin ) / log ( hp . Tin / hp . T3 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p2 = hp . pin *( hp . Tin / hp . T2 ) ^( n /(1 - n ) ) ;

172 173

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

174

% STAGE 2

175 176

% To interpolate the values using the enthalpy h5s

177 178

179 180

181

182

183

184

hp . T5 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h5 ) ; hp . h5s = eff *( hp . h5 - hp . h3 ) + hp . h3 ; hp . R5 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h5 ) ; hp . T5s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h5s ) ; hp . Gamma3 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T3 ) ; hp . Gamma5s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T5s ) ; Gamma = ( hp . Gamma3 + hp . Gamma5s ) *0.5;

185 186

% Pressure 5 is found using the adiabatic relation

187 188

hp . p5 = hp . p3 *( hp . T3 / hp . T5s ) ^( Gamma /(1 - Gamma ) ) ;

189 190

191

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

192 193 194

hp . h4 = r *( hp . h5 - hp . h3 ) + hp . h3 ; hp . T4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h4 ) ;

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195

hp . R4 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h4 ) ;

196 197

198 199 200

% Calculating the pressure 4 by assuming poltropic process from 3 to 5 c = log ( hp . p5 / hp . p3 ) / log ( hp . T3 / hp . T5 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p4 = hp . p3 *( hp . T3 / hp . T4 ) ^( n /(1 - n ) ) ;

201 202

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

203

% STAGE 3

204 205 206

207

208 209

210

211

212

% To interpolate the values using the enthalpy h5s hp . T7 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h7 ) ; hp . R7 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h7 ) ; hp . h7s = eff *( hp . h7 - hp . h5 ) + hp . h5 ; hp . T7s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h7s ) ; hp . Gamma5 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T5 ) ; hp . Gamma7s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T7s ) ; Gamma = ( hp . Gamma3 + hp . Gamma5s ) *0.5;

213 214

% Pressure 7 is found using the adiabatic relation

215 216

hp . p7 = hp . p5 *( hp . T5 / hp . T7s ) ^( Gamma /(1 - Gamma ) ) ;

217 218

219

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

220 221 222

223

hp . h6 = r *( hp . h7 - hp . h5 ) + hp . h5 ; hp . T6 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h6 ) ; hp . R6 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h6 ) ;

224 225

226 227 228

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p7 / hp . p5 ) / log ( hp . T5 / hp . T7 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p6 = hp . p5 *( hp . T5 / hp . T6 ) ^( n /(1 - n ) ) ;

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229 230

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

231 232

% STAGE 4

233 234 235

236

237 238

239

240

241

% To interpolate the values using the enthalpy h5s hp . T9 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h9 ) ; hp . R9 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h9 ) ; hp . h9s = eff *( hp . h9 - hp . h7 ) + hp . h7 ; hp . T9s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h9s ) ; hp . Gamma7 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T7 ) ; hp . Gamma9s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T9s ) ; Gamma = ( hp . Gamma5 + hp . Gamma7s ) *0.5;

242 243

% Pressure 7 is found using the adiabatic relation

244 245

hp . p9 = hp . p7 *( hp . T7 / hp . T9s ) ^( Gamma /(1 - Gamma ) ) ;

246 247

248

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

249 250 251

252

hp . h8 = r *( hp . h9 - hp . h7 ) + hp . h7 ; hp . T8 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h8 ) ; hp . R8 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h8 ) ;

253 254

255 256 257

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p9 / hp . p7 ) / log ( hp . T7 / hp . T9 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p8 = hp . p7 *( hp . T7 / hp . T8 ) ^( n /(1 - n ) ) ;

258 259

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

260 261

% STAGE 5

262 263

% To interpolate the values using the enthalpy h5s

77

6 APPENDIX

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264

265

266 267

268

269

270

hp . T11 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h11 ) ; hp . R11 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h11 ) ; hp . h11s = eff *( hp . h11 - hp . h9 ) + hp . h9 ; hp . T11s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h11s ) ; hp . Gamma9 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T9 ) ; hp . Gamma11s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T11s ) ; Gamma = ( hp . Gamma7 + hp . Gamma9s ) *0.5;

271 272

% Pressure 7 is found using the adiabatic relation

273 274

hp . p11 = hp . p9 *( hp . T9 / hp . T11s ) ^( Gamma /(1 - Gamma ) ) ;

275 276

277

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

278 279 280

281

hp . h10 = r *( hp . h11 - hp . h9 ) + hp . h9 ; hp . T10 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h10 ) ; hp . R10 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h10 ) ;

282 283

284 285 286

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p11 / hp . p9 ) / log ( hp . T9 / hp . T11 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p10 = hp . p9 *( hp . T9 / hp . T10 ) ^( n /(1 - n ) ) ;

287 288

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

289 290 291

% STAGE 6

292 293 294

295

296

% To interpolate the values using the enthalpy h5s hp . T13 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h13 ) ; hp . R13 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h13 ) ; hp . h13s = eff *( hp . h13 - hp . h11 ) + hp . h11 ;

78

6 APPENDIX

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297

298

299

300

hp . T13s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h13s ) ; hp . Gamma9 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T11 ) ; hp . Gamma11s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T13s ) ; Gamma = ( hp . Gamma9 + hp . Gamma11s ) *0.5;

301 302

% Pressure 7 is found using the adiabatic relation

303 304

hp . p13 = hp . p11 *( hp . T11 / hp . T13s ) ^( Gamma /(1 - Gamma ) ) ;

305 306

307

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

308 309 310

311

hp . h12 = r *( hp . h13 - hp . h11 ) + hp . h11 ; hp . T12 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h12 ) ; hp . R12 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h12 ) ;

312 313

314 315 316

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p13 / hp . p11 ) / log ( hp . T11 / hp . T13 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p12 = hp . p11 *( hp . T11 / hp . T12 ) ^( n /(1 - n ) ) ;

317 318

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

319 320

% STAGE 7

321 322 323

324

325 326

327

328

329

% To interpolate the values using the enthalpy h5s hp . T15 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h15 ) ; hp . R15 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h15 ) ; hp . h15s = eff *( hp . h15 - hp . h13 ) + hp . h13 ; hp . T15s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h15s ) ; hp . Gamma11 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T13 ) ; hp . Gamma13s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T15s ) ; Gamma = ( hp . Gamma11 + hp . Gamma13s ) *0.5;

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330 331

% Pressure 7 is found using the adiabatic relation

332 333

hp . p15 = hp . p13 *( hp . T13 / hp . T15s ) ^( Gamma /(1 - Gamma ) ) ;

334 335

336

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

337 338 339

340

hp . h14 = r *( hp . h15 - hp . h13 ) + hp . h13 ; hp . T14 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h14 ) ; hp . R14 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h14 ) ;

341 342

343 344 345

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p15 / hp . p13 ) / log ( hp . T13 / hp . T15 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p14 = hp . p13 *( hp . T13 / hp . T14 ) ^( n /(1 - n ) ) ;

346 347

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

348 349

% STAGE 8

350 351 352

353

354 355

356

357

358

% To interpolate the values using the enthalpy h5s hp . T17 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h17 ) ; hp . R17 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h17 ) ; hp . h17s = eff *( hp . h17 - hp . h15 ) + hp . h15 ; hp . T17s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h17s ) ; hp . Gamma13 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T15 ) ; hp . Gamma15s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T17s ) ; Gamma = ( hp . Gamma13 + hp . Gamma15s ) *0.5;

359 360

% Pressure 7 is found using the adiabatic relation

361 362

hp . p17 = hp . p15 *( hp . T15 / hp . T17s ) ^( Gamma /(1 - Gamma ) ) ;

363 364

% h4 is found using the degree of reaction formula and hence the values are

80

6 APPENDIX

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365

% found using the interpolation data

366 367 368

369

hp . h16 = r *( hp . h17 - hp . h15 ) + hp . h15 ; hp . T16 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h16 ) ; hp . R16 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h16 ) ;

370 371

372 373 374

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p17 / hp . p15 ) / log ( hp . T15 / hp . T17 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p16 = hp . p15 *( hp . T15 / hp . T16 ) ^( n /(1 - n ) ) ;

375 376

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

377 378 379

% STAGE 9

380 381 382

383

384 385

386

387

388

% To interpolate the values using the enthalpy h5s hp . T19 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h19 ) ; hp . R19 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h19 ) ; hp . h19s = eff *( hp . h19 - hp . h17 ) + hp . h17 ; hp . T19s = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h19s ) ; hp . Gamma15 = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T17 ) ; hp . Gamma17s = interp1 ( Data_Compressor . data (: ,1) , Data_Compressor . data (: ,5) , hp . T19s ) ; Gamma = ( hp . Gamma15 + hp . Gamma17s ) *0.5;

389 390

% Pressure 7 is found using the adiabatic relation

391 392

hp . p19 = hp . p17 *( hp . T17 / hp . T19s ) ^( Gamma /(1 - Gamma ) ) ;

393 394

395

% h4 is found using the degree of reaction formula and hence the values are % found using the interpolation data

396 397 398

hp . h18 = r *( hp . h19 - hp . h17 ) + hp . h17 ; hp . T18 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,1) , hp . h18 ) ;

81

6 APPENDIX

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399

hp . R18 = interp1 ( Data_Compressor . data (: ,2) , Data_Compressor . data (: ,6) , hp . h18 ) ;

400 401

402 403 404

% Calculating the pressure 6 by assuming poltropic process from 4 to 7 c = log ( hp . p19 / hp . p17 ) / log ( hp . T17 / hp . T19 ) ; n = c /( c +1) ; % Polytropic Contstant hp . p18 = hp . p17 *( hp . T17 / hp . T18 ) ^( n /(1 - n ) ) ;

405 406

%%%%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

407 408 409 410 411

%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

412

% For Stage 1 lambda = hp . stage1 . lambda ; nu = U1 / U2 ; x0 = [6 0;160; 0.8;0. 9]; [ stage1_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage1 . a2 = stage1_x (1) ; hp . stage1 . b1 = stage1_x (2) ; hp . stage1 . phi = stage1_x (3) ; hp . stage1 . mu = stage1_x (4) ;

413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430

% For Stage 2 lambda = hp . stage2 . lambda ; nu = U3 / U4 ; x0 = [6 0;160; 0.8;0. 9]; [ stage2_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage2 . a2 = stage2_x (1) ; hp . stage2 . b1 = stage2_x (2) ; hp . stage2 . phi = stage2_x (3) ; hp . stage2 . mu = stage2_x (4) ;

431 432 433 434 435 436 437 438 439 440

% For Stage 3 lambda = hp . stage3 . lambda ; nu = U5 / U6 ; x0 = [6 0;160; 0.8;0. 9]; [ stage3_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage3 . a2 = stage3_x (1) ; hp . stage3 . b1 = stage3_x (2) ; hp . stage3 . phi = stage3_x (3) ; hp . stage3 . mu = stage3_x (4) ;

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441 442 443 444 445 446 447 448 449 450

% For Stage 4 lambda = hp . stage4 . lambda ; nu = U7 / U8 ; x0 = [6 0;160; 0.8;0. 9]; [ stage4_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage4 . a2 = stage4_x (1) ; hp . stage4 . b1 = stage4_x (2) ; hp . stage4 . phi = stage4_x (3) ; hp . stage4 . mu = stage4_x (4) ;

451 452 453 454 455 456 457 458 459 460

% For Stage 5 lambda = hp . stage5 . lambda ; nu = U9 / U10 ; x0 = [6 0;160; 0.8;0. 9]; [ stage5_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage5 . a2 = stage5_x (1) ; hp . stage5 . b1 = stage5_x (2) ; hp . stage5 . phi = stage5_x (3) ; hp . stage5 . mu = stage5_x (4) ;

461 462 463 464 465 466 467 468 469 470

% For Stage 6 lambda = hp . stage6 . lambda ; nu = U11 / U13 ; x0 = [6 0;160; 0.8;0. 9]; [ stage6_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage6 . a2 = stage6_x (1) ; hp . stage6 . b1 = stage6_x (2) ; hp . stage6 . phi = stage6_x (3) ; hp . stage6 . mu = stage6_x (4) ;

471 472 473 474 475 476 477 478 479 480

% For Stage 7 lambda = hp . stage7 . lambda ; nu = U14 / U15 ; x0 = [6 0;160; 0.8;0. 9]; [ stage7_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage7 . a2 = stage7_x (1) ; hp . stage7 . b1 = stage7_x (2) ; hp . stage7 . phi = stage7_x (3) ; hp . stage7 . mu = stage7_x (4) ;

481 482 483 484 485 486

% For Stage 8 lambda = hp . stage8 . lambda ; nu = U16 / U17 ; x0 = [6 0;160; 0.8;0. 9];

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487 488 489 490 491

[ stage8_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage8 . a2 = stage8_x (1) ; hp . stage8 . b1 = stage8_x (2) ; hp . stage8 . phi = stage8_x (3) ; hp . stage8 . mu = stage8_x (4) ;

492 493 494 495 496 497 498 499 500 501

% For Stage 9 lambda = hp . stage9 . lambda ; nu = U18 / U19 ; x0 = [6 0;160; 0.8;0. 9]; [ stage9_x ] = fsolve ( @LP_Stg1 , x0 ) ; hp . stage9 . a2 = stage9_x (1) ; hp . stage9 . b1 = stage9_x (2) ; hp . stage9 . phi = stage9_x (3) ; hp . stage9 . mu = stage9_x (4) ;

502 503 504 505 506

% Velocity and Mass flow Rates :

507 508 509 510 511 512 513 514 515 516

V_ax2 = hp . stage1 . phi * U2 ; V_ax4 = hp . stage2 . phi * U4 ; V_ax6 = hp . stage3 . phi * U6 ; V_ax8 = hp . stage4 . phi * U8 ; V_ax10 = hp . stage5 . phi * U10 V_ax12 = hp . stage6 . phi * U12 V_ax14 = hp . stage7 . phi * U14 V_ax16 = hp . stage8 . phi * U16 V_ax18 = hp . stage9 . phi * U18

; ; ; ; ;

517 518 519 520 521 522 523 524 525 526

V_ax1 = hp . stage1 . mu * V_ax2 ; V_ax3 = hp . stage2 . mu * V_ax4 ; V_ax5 = hp . stage3 . mu * V_ax6 ; V_ax7 = hp . stage4 . mu * V_ax8 ; V_ax9 = hp . stage5 . mu * V_ax10 ; V_ax11 = hp . stage6 . mu * V_ax12 ; V_ax13 = hp . stage7 . mu * V_ax14 ; V_ax15 = hp . stage8 . mu * V_ax16 ; V_ax17 = hp . stage9 . mu * V_ax18 ;

527 528 529 530 531 532

hp . V1 hp . V2 hp . V3 hp . V4

= = = =

V_ax1 V_ax2 V_ax3 V_ax4

/ / / /

sind ( a1 ) ; sind ( hp . stage1 . a2 ) ; sind ( a1 ) ; sind ( hp . stage2 . a2 ) ;

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6 APPENDIX

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533 534 535 536 537 538 539 540 541 542 543 544 545 546

hp . V5 = V_ax5 / hp . V6 = V_ax6 / hp . V7 = V_ax7 / hp . V8 = V_ax8 / hp . V9 = V_ax9 / hp . V10 = V_ax10 hp . V11 = V_ax11 hp . V12 = V_ax12 hp . V13 = V_ax13 hp . V14 = V_ax14 hp . V15 = V_ax14 hp . V16 = V_ax16 hp . V17 = V_ax17 hp . V18 = V_ax18

sind ( a1 ) ; sind ( hp . stage3 . a2 ) ; sind ( a1 ) ; sind ( hp . stage4 . a2 ) ; sind ( a1 ) ; / sind ( hp . stage5 . a2 ) ; / sind ( a1 ) ; / sind ( hp . stage6 . a2 ) ; / sind ( a1 ) ; / sind ( hp . stage7 . a2 ) ; / sind ( a1 ) ; / sind ( hp . stage8 . a2 ) ; / sind ( a1 ) ; / sind ( hp . stage9 . a2 ) ;

547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566

rho1 = hp . pin *10^3/( hp . R1 * hp . Tin ) ; rho2 = hp . p2 *10^3/ ( hp . R2 * hp . T2 ) ; rho3 = hp . p3 *10^3/ ( hp . R3 * hp . T3 ) ; rho4 = hp . p4 *10^3/ ( hp . R4 * hp . T4 ) ; rho5 = hp . p5 *10^3/ ( hp . R5 * hp . T5 ) ; rho6 = hp . p6 *10^3/ ( hp . R6 * hp . T6 ) ; rho7 = hp . p7 *10^3/ ( hp . R7 * hp . T7 ) ; rho8 = hp . p8 *10^3/ ( hp . R8 * hp . T8 ) ; rho9 = hp . p9 *10^3/ ( hp . R9 * hp . T9 ) ; rho10 = hp . p10 *10^3/ ( hp . R10 * hp . T10 ) ; rho11 = hp . p11 *10^3/ ( hp . R11 * hp . T11 ) ; rho12 = hp . p12 *10^3/ ( hp . R12 * hp . T12 ) ; rho13 = hp . p13 *10^3/ ( hp . R13 * hp . T13 ) ; rho14 = hp . p14 *10^3/ ( hp . R14 * hp . T14 ) ; rho15 = hp . p15 *10^3/ ( hp . R15 * hp . T15 ) ; rho16 = hp . p16 *10^3/ ( hp . R16 * hp . T16 ) ; rho17 = hp . p17 *10^3/ ( hp . R17 * hp . T17 ) ; rho18 = hp . p18 *10^3/ ( hp . R18 * hp . T18 ) ; rho19 = hp . p19 *10^3/ ( hp . R19 * hp . Tout ) ;

567 568 569 570 571 572 573 574 575 576 577 578

hp . A1 = hp . mdot /( rho1 * hp . A2 = hp . mdot /( rho2 * hp . A3 = hp . mdot /( rho3 * hp . A4 = hp . mdot /( rho4 * hp . A5 = hp . mdot /( rho5 * hp . A6 = hp . mdot /( rho6 * hp . A7 = hp . mdot /( rho7 * hp . A8 = hp . mdot /( rho8 * hp . A9 = hp . mdot /( rho9 * hp . A10 = hp . mdot /( rho10 hp . A11 = hp . mdot /( rho11

V_ax1 ) ; V_ax2 ) ; V_ax3 ) ; V_ax4 ) ; V_ax5 ) ; V_ax6 ) ; V_ax7 ) ; V_ax8 ) ; V_ax9 ) ; * V_ax10 ) ; * V_ax11 ) ;

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579 580 581 582 583 584 585

hp . A12 hp . A13 hp . A14 hp . A15 hp . A16 hp . A17 hp . A18

= = = = = = =

hp . mdot /( rho12 hp . mdot /( rho13 hp . mdot /( rho14 hp . mdot /( rho15 hp . mdot /( rho16 hp . mdot /( rho17 hp . mdot /( rho18

* * * * * * *

V_ax12 ) V_ax13 ) V_ax14 ) V_ax15 ) V_ax16 ) V_ax17 ) V_ax18 )

; ; ; ; ; ; ;

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hp . A19 = 2* hp . A18 - hp . A17 ; % Linear variation of area V_ax19 = hp . mdot / ( rho19 * hp . A19 ) ; hp . V19 = V_ax19 / sind ( a1 ) ;

590 591 592 593 594 595 596 597

hp . ht1 hp . ht2 hp . ht3 hp . ht4 hp . ht5 hp . ht6 hp . ht7

= = = = = = =

hp . A1 /( pi hp . A2 /( pi hp . A3 /( pi hp . A4 /( pi hp . A5 /( pi hp . A6 /( pi hp . A7 /( pi

* * * * * * *

Dm1 ) Dm2 ) Dm3 ) Dm4 ) Dm5 ) Dm6 ) Dm7 )

; ; ; ; ; ; ;

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Press_error = ( hp . p19 - ( hp . pin * hp . pr ) ) *100/( hp . pin * hp . pr )

600 601 602 603 604 605

if ( Press_error >0) hp . lms = 0.99* hp . lms ; elseif ( Press_error half_ base_p rofile ( jj -1 ,1) ) T_by_C ( ii ,1) = ( ha lf_bas e_prof ile ( jj -1 ,2) + ( X_by_C ( ii ,1) - half_ base_p rofile ( jj ,1) ) *(( hal f_base _profi le ( jj ,2) - hal f_base _profi le ( jj -1 ,2) ) /( half_ base_p rofile ( jj ,1) - half_ base_p rofile ( jj -1 ,1) ) ) ) ; break ; end

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84

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end end if ( length ( T_by_C ) < length ( X_by_C ) ) T_by_C ( ii ,1) =0; end T = T_by_C * camber_length ;

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%% norm_camber =[ new_camber1 (: ,1) / chord_length ( new_camber1 (: ,2) new_camber1 (1 ,2) ) / chord_length ];

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superimp_profile = camber_length * h alf_ba se_pro file ;

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%% pres_surf_x = new_camber1 (: ,1) -T .* sin ( atan ( new_camber1 (: ,2) ./ new_camber1 (: ,1) ) ) ; pres_surf_y = new_camber1 (: ,2) + T .* cos ( atan ( new_camber1 (: ,2) ./ new_camber1 (: ,1) ) ) ;

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suct_surf_x = new_camber1 (: ,1) + T .* sin ( atan ( new_camber1 (: ,2) ./ new_camber1 (: ,1) ) ) ; suct_surf_y = new_camber1 (: ,2) -T .* cos ( atan ( new_camber1 (: ,2) ./ new_camber1 (: ,1) ) ) ;

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for jj =1: length ( new_camber1 ) if ( pres_surf_x ( jj ,1) > new_camber1 ( jj ,1) || suct_surf_x ( jj ,1) < new_camber1 ( jj ,1) ) press ( jj ,1) = suct_surf_x ( jj ,1) ; suct ( jj ,1) = pres_surf_x ( jj ,1) ; else press ( jj ,1) = pres_surf_x ( jj ,1) ; suct ( jj ,1) = suct_surf_x ( jj ,1) ; end end

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press (: ,2) = pres_surf_y (: ,1) ;

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117

suct (: ,2) = suct_surf_y (: ,1) ;

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blade_profile = vertcat ( press , flipud ( suct ) ) ; badrows = any ( isnan ( blade_profile ) ,2) ; blade_profile ( badrows ,:) = zeros ;

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rot_ang_back = pi - atan (( camber_line ( end ,2) - camber_line (1 ,2) ) /( camber_line ( end ,1) - camber_line (1 ,1) ) ) ; rot_matrix_back =[ cos ( rot_ang_back ) - sin ( rot_ang_back ) ; sin ( rot_ang_back ) cos ( rot_ang_back ) ];

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134

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blade = blade_profile * rot_matrix_back ; blade_CFD (: ,1) = blade (: ,1) ; blade_CFD (: ,2) = blade (: ,2) ; blade_CFD (: ,3) = zeros ; camber_line_CFD (: ,1) = camber_line (: ,1) ; camber_line_CFD (: ,2) = camber_line (: ,2) ; camber_line_CFD (: ,3) = zeros ; camber_line0_CFD =[ camber_line (: ,1) - chord_length camber_line (: ,2) camber_line_CFD (: ,3) ]; camber_line2_CFD =[ camber_line (: ,1) + chord_length camber_line (: ,2) camber_line_CFD (: ,3) ]; blade0 =[ blade (: ,1) - chord_length blade (: ,2) ]; blade2 =[ blade (: ,1) + chord_length blade (: ,2) ];

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147 148 149 150 151

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fid = fopen ( ’ Blade_Profile . txt ’ , ’w + ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , blade , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , blade0 , ’- append ’ , ’ delimiter ’ , ’\ t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , blade2 , ’- append ’ , ’ delimiter ’ , ’\ t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ;

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156

157 158 159 160 161

162 163 164 165 166

167

dlmwrite ( ’ Blade_Profile . txt ’ , camber_line , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , camber_line0 , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fid = fopen ( ’ Blade_Profile . txt ’ , ’a ’) ; fprintf ( fid , ’ ZONE \n ’) ; fclose ( fid ) ; dlmwrite ( ’ Blade_Profile . txt ’ , camber_line2 , ’- append ’ , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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170

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172

dlmwrite ( ’ Blade_CFD . txt ’ , blade_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber_CFD . txt ’ , camber_line_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber0_CFD . txt ’ , camber_line0_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ; dlmwrite ( ’ Camber2_CFD . txt ’ , camber_line2_CFD , ’ delimiter ’ , ’\t ’ , ’ precision ’ , 20) ;

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Design of Turbine Blades

Design of Turbine Blades: 1 2

clear all ; clc ;

3 4

global a2 b3 nu lambda r a2_ref b3_ref nu_ref lambda_ref i

5 6 7 8

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Dm_exit =1.120; Dm1 =1.062; Dm2 =1.0649; Dm3 =1.07215; Dm4 =1.0794; Dm5 =1.08665; Dm6 =1.0939; Dm7 =1.10115; Dm8 =1.1084; Dm9 =1.11565; m_dot =151.3; omega =469.35; gm =1.44; R =288.15; cp_1 = gm /( gm -1) * R ; r =0.5; T_in =1222.7 -273.15; P_in =873350; T_out =806.77 -273.15; P_out =102200; eff =0.9; U1 = Dm1 /2* omega ; U2 = Dm2 /2* omega ; U3 = Dm3 /2* omega ; U4 = Dm4 /2* omega ; U5 = Dm5 /2* omega ; U6 = Dm6 /2* omega ; U7 = Dm7 /2* omega ; U8 = Dm8 /2* omega ; U9 = Dm9 /2* omega ;

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data = importdata ( ’ D A T A _ T e m p _ V s _ E n t h a l p y . txt ’) ; T = data (: ,1) ; h = 1000* data (: ,2) ; cp = data (: ,4) ; ka = data (: ,5) ;

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h_in = interp1 (T ,h , T_in ) ; h_out = interp1 (T ,h , T_out ) ; lm = h_in - h_out ; lms = lm /4;

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press_err =10000;

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while abs ( press_err ) >=20

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lambda_ref = lms / U3 (1) ^2; a2_ref =20* pi /180; b3_ref =160* pi /180; nu_ref = U2 / U3 ; x0 = [4 0;160; 0.8;0. 5]; [ X ] = fsolve ( @myfun1 , x0 ) ;

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b2_ref = X (1) ; a3_ref = X (2) ; mu_ref = X (3) ; phi_ref = X (4) ;

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Vm1_ref = phi_ref * U1 ; rho1_ref = P_in /( R * T_in ) ; xsec1_ref = m_dot /( rho1_ref * Vm1_ref ) ; BH = xsec1_ref /( Dm1 * pi ) ;

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hub_rad =( Dm1 - BH /2) /2; Dm_slope = atan ((1.12 -1.062) /.800) ;

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for i =1:4 if i ==1 P1 ( i ) = P_in ; T1 ( i ) = T_in ; else P1 ( i ) = P3 (i -1) ; T1 ( i ) = T3 (i -1) ; end if i ==1 tip ( i ) = hub_rad +2*( Dm1 /2 - hub_rad ) ;

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%

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%

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%

elseif i ==2 tip ( i ) = hub_rad +2*( Dm3 /2 - hub_rad ) ; elseif i ==3 tip ( i ) = hub_rad +2*( Dm5 /2 - hub_rad ) ; elseif i ==4 tip ( i ) = hub_rad +2*( Dm7 /2 - hub_rad ) ; end xsec ( i ) = pi *( tip ( i ) ^2 - hub_rad ^2) ; rho ( i ) = P1 ( i ) /( R *( T1 ( i ) +273.15) ) ; V1 ( i ) = m_dot /( rho ( i ) * xsec ( i ) ) ; h1 ( i ) = cp_1 * T1 ( i ) ; h1 ( i ) = interp1 (T ,h , T1 ( i ) ) ; H1 ( i ) = h1 ( i ) +1/2* V1 ( i ) ^2; H3 ( i ) = H1 ( i ) - lms ; T01 ( i ) = H1 ( i ) / cp_1 ; T01 ( i ) = interp1 (h ,T , H1 ( i ) ) ; T03 ( i ) = H3 ( i ) / cp_1 ; T03 ( i ) = interp1 (h ,T , H3 ( i ) ) ; T03s ( i ) = T01 ( i ) -( T01 ( i ) - T03 ( i ) ) / eff ; P01 ( i ) = P1 ( i ) +1/2* rho ( i ) * V1 ( i ) ^2; P03 ( i ) = P01 ( i ) *( T03s ( i ) / T01 ( i ) ) ^( gm /( gm -1) ) ; P3 ( i ) = P03 ( i ) *( T1 ( i ) / T01 ( i ) ) ^( gm /( gm -1) ) ;

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%

T3 ( i ) =( P3 ( i ) / P03 ( i ) ) ^(( gm -1) / gm ) * T03 ( i ) ; h3 ( i ) = cp_1 * T3 ( i ) ; h3 ( i ) = interp1 (T ,h , T3 ( i ) ) ; if i ==1 U2 ( i ) =( tan ( Dm_slope ) *0.080+ hub_rad ) * omega ; U3 ( i ) =( tan ( Dm_slope ) *0.180+ hub_rad ) * omega ; elseif i ==2 U2 ( i ) =( tan ( Dm_slope ) *0.280+ hub_rad ) * omega ; U3 ( i ) =( tan ( Dm_slope ) *0.380+ hub_rad ) * omega ; elseif i ==3 U2 ( i ) =( tan ( Dm_slope ) *0.480+ hub_rad ) * omega ; U3 ( i ) =( tan ( Dm_slope ) *0.580+ hub_rad ) * omega ; elseif i ==4 U2 ( i ) =( tan ( Dm_slope ) *0.680+ hub_rad ) * omega ; U3 ( i ) =( tan ( Dm_slope ) *0.780+ hub_rad ) * omega ; end lambda ( i ) = lms / U3 ( i ) ^2; a2 ( i ) =20; b3 ( i ) =160; nu ( i ) = U2 ( i ) / U3 ( i ) ; x0 = [40 ;160; 0.8;0. 5]; if i ~=4 [ X1 ] = lsqnonlin ( @GT_solve , x0 ) ; elseif i ==4 a3 (4) =90; phi (4) =1/( cot ( a3 (4) * pi /180) - cot ( b3 (4) * pi /180) ) ; nu (4) = U2 (4) / U3 (4) ; Vm3 (4) = phi (4) * U3 (4) ; Vm2 (4) = phi (4) * U2 (4) ; mu (4) = Vm2 (4) / Vm3 (4) ; b2 (4) = acotd ( cot ( a2 (4) * pi /180) - nu (4) /( mu (4) * phi (4) ) ) ; V3 (4) = sqrt (2*( H3 (4) - h3 (4) ) ) ; Vu3 (4) = sqrt ( V3 (4) ^2 - Vm3 (4) ^2) ; W3 (4) = sqrt ( Vm3 (4) ^2+( U3 (4) + Vu3 (4) ) ^2) ; V2 (4) = Vm2 (4) / sin ( a2 (4) * pi /180) ; Vu2 (4) = V2 (4) * cos ( a2 (4) * pi /180) ; W2 (4) = sqrt ( Vm2 (4) ^2+( Vu2 (4) - U2 (4) ) ^2) ; M3 (4) = V3 (4) / sqrt (( gm * R * T3 (4) ) ) ; break ; end b2 ( i ) = X1 (1) ; a3 ( i ) = X1 (2) ; mu ( i ) = X1 (3) ; phi ( i ) = X1 (4) ; V3 ( i ) = sqrt (2*( H3 ( i ) - h3 ( i ) ) ) ; M3 ( i ) = V3 ( i ) / sqrt (( gm * R * T3 ( i ) ) ) ;

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Vm3 ( i ) = V3 ( i ) * sin ( a3 ( i ) * pi /180) ; Vu3 ( i ) = sqrt ( V3 ( i ) ^2 - Vm3 ( i ) ^2) ; W3 ( i ) = sqrt ( Vm3 ( i ) ^2+( U3 ( i ) + Vu3 ( i ) ) ^2) ; Vm2 ( i ) = mu ( i ) * Vm3 ( i ) ; Vu2 ( i ) =( lms - U2 ( i ) * Vu3 ( i ) ) / U2 ( i ) ; W2 ( i ) = sqrt ( Vm2 ( i ) ^2+( Vu2 ( i ) - U2 ( i ) ) ^2) ; V2 ( i ) = sqrt ( Vm2 ( i ) ^2+( Vu2 ( i ) ) ^2) ;

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end

140 141

lm = H1 (1) - H3 (4) ; press_err = P3 (4) - P_out

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% % % % %

if i ==1 a1 =90; elseif i >0 a1 = a3 (i -1) ; end

150

% ***************************************************** % ***************************************************** % Define function using the following parameters

151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171

172 173

174 175

% % % % % % % % % % % % % % % % % % % % %

CS =0.022; % temporarily assumed CHECK , CHECK , CHECK if i ==1 Dh2 ( i ) = Dm2 - CS / ( pi * Dm2 ) ; Dh3 ( i ) = Dm3 - CS / ( pi * Dm3 ) ; elseif i ==2 Dh2 ( i ) = Dm4 - CS / ( pi * Dm4 ) ; Dh3 ( i ) = Dm5 - CS / ( pi * Dm5 ) ; elseif i ==3 Dh2 ( i ) = Dm6 - CS / ( pi * Dm6 ) ; Dh3 ( i ) = Dm7 - CS / ( pi * Dm7 ) ; elseif i ==4 Dh2 ( i ) = Dm8 - CS / ( pi * Dm8 ) ; Dh3 ( i ) = Dm9 - CS / ( pi * Dm9 ) ; end

% F = GT_efficiency ( a1 , a2 , a3 , b2 , b3 , Dh2 , Dh3 , V2 , W3 , lm , phi , lambda , mu , nu ) Efficiency = GT_efficiency ( a1 , a2 ( i ) , a3 ( i ) , b2 ( i ) , b3 ( i ) , Dh2 ( i ) , Dh3 ( i ) , V2 ( i ) , W3 ( i ) ,lm , phi ( i ) , lambda ( i ) , mu ( i ) , nu ( i ) ) ; % TO BE DONE :: Modify enthalpy using efficiency

176

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177 178 179 180

if press_err >0 lms =1.001* lms ; elseif press_err