Control of Once-through

Cooler

Master Thesis

of

Seyed-Hanif TABATABAEE

born on the 23rd of April 1982 in Tehran

February 22, 2007

Supervisors:

Prof. R.Longchamp

Dr. S. Hepner

Dr. A. Karimi

EPFL

Faculty of STI

Institute of Mechanical Engineering

Work realized in

Alstom Power Switzerland

Department PIT 6

Acknowledgments

I want to thank the department PIT6 of Alstom Swtitzerland, for having givenme the opportunity to live this huge experience of working in an interesting�eld which is the dynamic simulation. I have added a signi�cant load to mymechanical engineering baggage during this time, especially in thermodynamicsand heat transfer, thanks to them.I would like to give my special thanks to Jürg Ardüser, for his energy in replyingall my questions and debugging my programs. I also would like to thank Dr.Stephan Hepner, for the helpful overview he has on this �eld, an his assistanceall along this thesis.

Abstract

This thesis is about the control of a once-through cooler (heat exchanger forgas turbine). A dynamic, continuous and state space model is built, based onphysics of thermo hydraulics, to be able to simulate the system. First a counter�ow heat exchanger model has been constructed, followed by a steady-statecalibration. Then a cross-�ow model followed by a dynamic matching has beenbuilt, and satisfactory results have been obtained in terms of dynamics. Afterwards, the control system of this once-through cooler is investigated, and someimprovements have been brought to it. We showed that the use of a staticcompensator instead of a PI controller for the command control valve improvesremarkably the results. Finally some propositions for future control design aregiven as conclusion.

Contents

Acknowledgments 1

Abstract 2

List of Symbols 5

1 Introduction 71.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Task Breakdown . . . . . . . . . . . . . . . . . . . . . . . 81.3 Report plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Presentation of the system 92.1 Overview of the Plant . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Vertical Once-through cooler . . . . . . . . . . . . . . . . . . . . 112.3 Horizontal Once-through cooler . . . . . . . . . . . . . . . . . . . 13

2.3.1 Conventional, Dense and Maximum packing . . . . . . . . 14

3 Dynamic equations 173.1 Introduction to the simulator . . . . . . . . . . . . . . . . . . . . 173.2 Balance equations . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2.1 Mass balance . . . . . . . . . . . . . . . . . . . . . . . . . 193.2.2 Energy balance . . . . . . . . . . . . . . . . . . . . . . . . 193.2.3 Momentum balance . . . . . . . . . . . . . . . . . . . . . 20

3.3 Air (hy) library . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3.1 Pipe Gas (Volume) . . . . . . . . . . . . . . . . . . . . . . 213.3.2 Volume with heat exchange . . . . . . . . . . . . . . . . . 233.3.3 Control Valve . . . . . . . . . . . . . . . . . . . . . . . . . 243.3.4 Ori�ce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4 Water/Steam (ws) library . . . . . . . . . . . . . . . . . . . . . . 273.4.1 Pipe ws (Volume) . . . . . . . . . . . . . . . . . . . . . . 293.4.2 VolumeWs with heat exchange . . . . . . . . . . . . . . . 303.4.3 Control valve Ws . . . . . . . . . . . . . . . . . . . . . . . 313.4.4 Ori�ceWs . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.4.5 Mass �ow vs. pressure di�erence : Square root . . . . . . 34

3

4 Model of the Once-through cooler 354.1 Counter-�ow model . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.1.1 Use of VolumeQHE and VolumeQHEWs . . . . . . . . . . 354.1.2 The HeatExchanger component . . . . . . . . . . . . . . . 384.1.3 Calibration of the counter-�ow model . . . . . . . . . . . 424.1.4 Step response of the counter-�ow model . . . . . . . . . . 45

4.2 Cross-�ow model . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.2.1 Calibration of the cross-�ow model . . . . . . . . . . . . . 474.2.2 Step response of the cross-�ow model . . . . . . . . . . . 48

5 Control system of the OTC 515.1 The Feed-Forward mass �ow . . . . . . . . . . . . . . . . . . . . 525.2 The inner loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.2.1 Controller for the inner loop . . . . . . . . . . . . . . . . . 545.2.2 Anti-reset windup . . . . . . . . . . . . . . . . . . . . . . 56

5.3 The outer loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.4 Static compensator instead of inner loop . . . . . . . . . . . . . . 62

5.4.1 Unknown control valve characteristic . . . . . . . . . . . . 645.4.2 Uncertainty on the control valve characteristic . . . . . . 67

5.5 Comparison for further transients . . . . . . . . . . . . . . . . . . 715.5.1 Start up . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.5.2 Shut down . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.5.3 Load rejection . . . . . . . . . . . . . . . . . . . . . . . . 735.5.4 Emergency switch o� . . . . . . . . . . . . . . . . . . . . . 75

6 Conclusion and Future scope 76

Bibliography 80

List of Figures 82

List of Tables 85

A Vertical OTC 2D sketch 86

B Horizontal OTC 2D sketch 87

C Horizontal OTC HP Data sheet 88

D Horizontal OTC LP Data sheet 89

E Step response of the counter-�ow model 90

F Step response of the cross-�ow model 100

G Transients 110

List of Symbols

Parameters and constants

Symbol Description UnitA Area [m2]α Heat transfer coe�cient [W/m2.K]C Correction factor [−]cf friction factor [−]cp Speci�c heat capacity [J/kg.K]

(constant pressure)cv Speci�c heat capacity [J/kg.K]

(constant volume)d Diameter [m]ε Void fraction [−]F Force [N ]f �uid type (�ag) [−]h Speci�c Enthalpy [J/kg]H Enthalpy [J ]κ Isentropic exponent [−]Kp PID proportional gain [−]Kv volumetric mass �ow [m3/h]L Length [m]λ Thermal conductivity [W/m.K]m Mass [kg]m Mass �ow [kg/s]µ Viscosity [kg/s.m]p Pressure [Pa]Q Heat Flux [J/S] = [W ]< Universal gas constant [kJ/kmol.K]Re Reynolds Number [−]ρ Density [kg/m3]t Time [s]T Temperautre [K]Ti PID integral term gain [s]u Speci�c internal energy [J/kg]U Internal energy [J ]v Speci�c volume [m3/kg]V Volume [m3]ω Pulsation [rad/s]x Steam quality [−]

5

Subscripts

Subscript Descriptionconv Convectioncond Conductioncv control valveF Finalhy Air libraryi Insidel left hand sidelim Limitmax maximummin minimumr right hand sides steams Shell sidet Tube sidetp Two phasew Water

6

Chapter 1

Introduction

1.1 Introduction

Overall e�cieny is one of the main parameters for pro�tability of power plants.Overall e�ciency strongly depends on the e�ciency of the plant's main com-ponents (such as the steam generator and the steam turbines) as well as thecomponent cost.

For reasons of cost reduction the classical recirculation evaporators have beenreplaced by so-called "Once-Through" evaporators in recent plant projects. Re-circulation evaporators use a pressure vessel (called drum) for the evaporation ofwater. The evaporation process is strictly con�ned to this vessel by means of re-circulation. On the other hand, Once-Through evaporators perform preheatingand evaporation during a single pass of water/steam through the system. Waterenters the system on one side, heats up while it progresses through the system,then starts evaporating at a variable intermediate location, and �nally exits assteam on the opposite side. The evaporation zone within the Once-Throughcooler depends on the state of the system (pressure, temperature distribution)and his highly variable. This results in considerable stability problems that areunknown from conventional evaporators.

When using Once-Through evaporators in combined cycle plants, the steamgeneration is performed by the use of the hot exhaust gases from a gas turbine.The exhaust gas temperature may be subject to fast transients due to the op-eration concept of the gas turbine. In conjunction with the inherent stabilityproblems mentioned above, the control of Once-Through evaporators and theiradjacent components requires innovative control concepts.

1.2 Objectives

The purpose of this thesis is the investigation of the transient behaviour ofa Once-Through evaporator, and the dynamic modeling for the purposes ofcontrol. Based on the dynamic model, a control concept (possibly based ondistributed actuators/sensors) shall be developed that allows for stabilizationand control of the evaporator under various plant transients including start-upand load rejection.

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1.2.1 Task Breakdown

The task breakdown is as follows:

• Development of a dynamic model of a Once-Through Cooler.

• Model Calibration and Validation based on available measurements.

• Design of a control concept.

• Validation by simulation.

The two �rst tasks represent most of the project load (about 75% ), and thetwo remainig ones 25%.

1.3 Report plan

In chapter 2, the system itself, and its interaction with the whole plant will bedescribed. Chapter 3 will focus on the presentation of the simulator that weuse in this thesis, and will explain the basics of dynamic modeling of thermalsystems. After wards, chapter 4 will detail the steps of making our dynamicmodel, as well as the calibration. Finally in chapter 5, we will present the controlsystem of the Once-through cooler, and the modi�cations brought to it.

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Chapter 2

Presentation of the system

2.1 Overview of the Plant

The part of the air/water-steam plant of our interest is shown in the �gurebelow. The Once-through cooler is a heat exchanger, in which hot air �ows onthe shell side, and water/steam as refrigerant on the tube side.

Figure 2.1: Overview of the Plant

9

The hot air is in fact the exhaust gas coming out of the compressor of a gasturbine. It is cooled down with the help of water in the Once-through cooler.The feed water actually, is taken from the economizer of a HRSG (Heat recoverysteam generation unit) �ows through the OTC (Once-through cooler), coolsdown the hot air, and is evaporated when leaving the cooler. This vapor is thenused in a steam turbine. The cooled air �owing out of the Once-through coolerwill then be used to cool down the blades of the turbine (�lm cooling). Thereforyou can see in the �gure, that it is re-injected to the turbine. As the cooling ofthe blades on the turbine is very crucial (a too hot gas can de�nitely damagethem, a too cold gas can fail to cool them su�ciently), the outlet temperatureof the air from the Once-through cooler is very important, and as we will seelater, will be the variable we will control. It is important to emphasize that

there are two Once-through coolers:

1. Low pressure OTC

2. High presssure OTC

As a recall, the air �owing through the compressor of a gas turbo-machine,loses pressure, and is compressed. The potential energy of air obtained thanksto the compressor is then increased when ignition in the combustion chamber,and the released and transformed to mechanical energy in the Turbine. When�owing through the turbine, there is a decrease in the pressure of the air. Thismeans that pressure is variable in both Compressor and turbine.

The LP-OTC (low pressure OTC) treats the air out of the low pressurestages of the compressor, and uses it to cool down the low pressure stages of theturbine, while the HP-OTC (high pressure OTC) does the same duty on thehigh pressure stages of the turbo-machine.

The �gure below shows the location of LP and HP stages on both compressorand turbine: It is obvious that the nominal working conditions of the LP and

Figure 2.2: HP / LP stages of compressor and turbine

HP Once-through coolers are di�erent, because they are operating in di�erentpressure ranges. Therefore, the geometry and mass of these two also di�ers.Their data sheets are included in appendices C and D.

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2.2 Vertical Once-through cooler

The actual design of the OTC is the so called Vertical or Helix OTC. In this heatexchanger, as shown in the �gure below, exhaust air from the air compressorenters the cooler at the bottom, goes all the way to the top of the cooler, and iscooled when �owing back down, in interaction with the feed water. The wateractually, comes from the bottom, in 45 parallel tubes of the HP, and 58 for theLP cooler, moves to the top in a helix shape path, and leaves the cooler as steamon the top.

Figure 2.3: Vertical (Helix) OTC

The main characteristics for the HP and LP coolers are:

The Helix design is proven and robust, and has been optimized through itsyears of operation, but it is costly. An internal study has shown that the costof piping is almost equal to the cost of the cooler itself, because it's dislocatedfrom the gas turbine.

So before starting to optimize once again the Helix OTC for cost killing,Alstom engineers thought of a new design which could save money is terms ofpiping and air volume.

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Characteristic OTC HP vertical OTC LP veritcalOverall length 8.5m 8.05mVessel inside φ 1.4m 2.05mNb. of parallel Tubes 45 58Mean tube length 111.8m 139.5mTotal Mass 27′414kg 31′555kgShell side heat input 13.1MW 10.3MWShell side mass �ow 61.95kg/s 165kg/sTube side mass �ow 7.65kg/s 5.73kg/sAir volume 9.87m3 18.48m3

Piping Air volume 4.42m3 18.17m3

Table 2.1: HP and LP main characteristics for vertical OTC

We notice that the piping air volume is almost equal to the air volume insidethe vessel for the LP OTC. Figure 2.4 shows the location of the coolers and thepiping.

Figure 2.4: Location of Helix Once-through coolers

After investigating the gas turbine lay down, they found that there is anunused place below the gas turbine platform. So placing a cooler there wouldsave piping because it is very close to the gas turbine. Figure 2.5 shows thisnew envisaged arrangement.

The hight of the platform is almost 5.5m from the ground, so there was apossibility to place a �Horizontal OTC� with a vessel diameter of about 2 metersthere. Therefore the study of a new OTC generation began, and is still goingon. Next section introduces this new horizontal Once-through cooler.

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Figure 2.5: Gas Turbine lay down and OTC placement possibility

2.3 Horizontal Once-through cooler

The Horizontal OTC has a more conventional design in terms of tubes. Insteadof a helix shape, the tubes in the horizontal OTC are in so called �Serpentine�form (generally shown in �gure 2.1). The advantage with this type of piping isthat the tubes are straight, comparing to the helix design in which the tubesare curved . It is obvious that the assembly of serpentines with the vessel iseasier. One can easily imagine how complicated and costly it is to assemble thehelix tubes with the vessel.

As it is shown on the top view of appendix A, there are 3 main and 3 inter-mediate support plates through which the helix is pushed in a screw movement,and each time the movement is blocked (due to the form and the mass of thehelix) the plates are vibrated to pursuit the assembly. This assembly is called�Coiling assembly�.

The assembly of serpentines in a box to include in the vessel in easier, andalso pre-fabricated serpentines can be stacked to allow a modular assembly.

Another advantage with the serpentine tubes is that this con�guration allowsmore space between the tubes, and as the tubes are now straight, �ns can beused to �ll these empty spaces, in order to enhance the heat transfer. Enhancingthe tubes' surface by adding �ns results in a greater heat transfer and hence areduce of the air volume is possible.

Figure 2.6 shows a conceptual sketch of how the horizontal OTC is meantto be.

The side view shows clearly the serpentine included into a box. Water isentering the OTC through a di�user at the bottom of the vessel, �ows to thetop and is extracted as steam through another di�user. The front cut viewshows the air path : air is getting into the box through a hood, �ows through

13

Figure 2.6: Horizontal OTC side view (left) and front cut (right)

the water tube bundle, is cooled down, and leaves the box at the bottom andenters the inner vessel space and �ows upwards. It leaves then the vessel throughanother hood at the top of the vessel.

The front cut view shows also that there are several serpentines of water tubein the box (in the form of parallel sheets). In each sheet, 2 parallel serpentinesperform 14 passes together. This has been made possible by the concept of�Maximum packing� which is presented in the next paragraph comparing to theconventional and dense packing.

2.3.1 Conventional, Dense and Maximum packing

Figure 2.7 shows the side view of a serpentine.

Figure 2.7: Side view of a serpentine

The support plate are visible at the right and left hand side of the serpentine.The side view of this serpentine can have on of the 3 con�gurations shown in�gure 2.8:

The conventional packing is a staggered tube arrangement pattern. Theouter circles show the �ns, and the inner one the tubes themselves. the bendsas shown on �gure 2.7 are naked tubes and in the plane of the serpentines. In

14

Figure 2.8: Conventional (left), dense (middle) and maximum (right) packing

this con�guration, the support plate is on the �nned tubes. This limits therealizable �n height and the can limit the �n e�ciency.

The dense packing (middle of �gure 2.8) is also a staggered tube arrangementpattern. Its di�erence with the conventional packing is that this time, thesupport plate is on the naked bends. This allows to move the sheets closer toeach other and increase the mass �ow density. But unfortunately, heat transfercalculations reveal low �n e�ciencies. There is still too much unused space.The mass �ow density in the gap is clearly below what is realized in the helixOTC.

The solution to this problem is the maximum packing (right side of �gure2.8). This con�guration is an inline tube arrangement pattern. The supportplate is still on bare tubes, but this time, the bend of a serpentine is out of itsplane. To allow this, the bends this time are in alternating distance from thesupport structure. This allows a higher mass �ow density. Figure 2.9 allows abetter understanding of this con�guration.

Figure 2.9: Maximum packing 3D view

The horizontal OTC is still a research project, therefore no o�cial 2D draft

15

yet exists. We have used the data sheets (appendices C and D), which them-selves come from a simulation, to draw a handmade sketch (appendix B). Therelevant characteristics of the LP and HP horizontal OTC's are nonethelessgiven here, to create a general image for the reader.

Characteristic OTC HP horizontal OTC LP horizontalOverall length 5.736m 4.855mVessel inside φ 1.957m 2.222mNb. of parallel Tubes 52 62Box dimensions (LxWxH) 4.5m× 1.2m× 1.1m 3.6m× 1.3m× 1.3mTotal Mass 22′915kg 20′151Shell side heat input 14.21MW 19.17MwShell side mass �ow 58.6kg/s 176.3kg/sTube side mass �ow 7.533kg/s 10.343kg/sAir volume 13.02m3 14.55m3

Piping Air volume 3.26m3 10.01m3

Table 2.2: HP and LP main characteristics for horizontal OTC

By summing the two last lines of table 2.1 in order to obtain the total airvolume, we obtain 50.94m3 for the vertical OTC. On table 2.2 this value is40.84m3. Hence the new horizontal design has 19.82% less air than the verticalOTC (in volume).

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Chapter 3

Dynamic equations

Our approach in this thesis is simulation based. It means that the OTC willbe modeled for simulation, and will be controlled in simulation. The simulationprogram we use is developed in Alstom Switzerland. First an introductionto the general structure of the simulator will be presented, then the dynamicequations for modeling the system will be established along with a presentationof the components of the simulator useful for the modeling of our OTC.

3.1 Introduction to the simulator

The simulation tool in use is called �Graphical modeling�. The purpose of thistool is to develop mathematical models for physical systems. The idea of gen-erating models is based on a modular approach. This tool consists of di�erentlibraries available in Simulink R©containing a set of modules, each of them rep-resenting a speci�c type of system or component to be modeled. A graphicaleditor ensures fast model generation by dragging objects from the library anddrop them into the Simulink editor. In the Simulink editor itself, the di�er-ent component objects are connected in a way to represent the system underconsideration.[4]

Having the model structure completed, the individual components can beparametrized via a calibration matrix. In order to achieve even higher accuracy,operating data of each component can additionally be entered which will triggeran automatic re-calibration of the system.

Using all the model information, simulation source code is generated, whichincludes the following features:

• The source code is either pure Fortran77 code or C++ and thereforeportable to any open simulation platform. We programed in Fortran77in this project.

• The source code is able to run under di�erent operating systems, in par-ticular Unix, Linux or Windows.

• In order to guarantee an identical behavior of the model on any simulationplatform, the numerical integration solver is part of the model.

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• Whenever source code is being generated, all the information about themodel (structure, parameters, ...) is automatically written to a documen-tation �le.

The following picture shows the entire procedure of the model generationprocess:

Figure 3.1: Tool Overview

After the assembly of di�erent components by the user in stage 2 of �gure3.1, a source code is generated. In fact, each component has its own source code,which is written on the basis of the equations that govern it. The generatedsource code in stage 3 of �gure 3.1 is an assembly of these component sourcecodes (depending on the order in which they are connected) and the solversource code.

After the generation of the source code and the documentation �le, a Simulinkblock is created, to which the user can enter the desired inputs. The inputs (in-cluding initial conditions) get to the source code through a Matlab R©gateway.The source code when executed, reads the model parameters which are writtenin the documentation �les. The generated outputs come back to the Simulinkinterface through the same gateway.

The state space modeling is our approach in this simulator. Continuoustime di�erential equations will be used to model the system. The obtainedderivatives of the states of the system will be integrated using a solver. Thesolver can be one of these di�erential equations solving algorithms: Rosenbrock,Gear, Runge-Kutta, Mid-point or forward Euler. During the modeling of oursystem, we will use the Gear algorithm with a variable step size.

As the �uids in use of a Once-through cooler are air and water, we will nowfocus on their dynamic equations, and the laws that will allow us to model asystem containing these �uids.

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3.2 Balance equations

The conservation of the extensive quantities mass, energy and momentum in aproblem domain with given volume is the physical basis for modeling of �uids.These have to be expressed for each media in our system, in order to obtainthe right dynamic behavior for our model. Let us now focus more on thesefundamental laws in the next paragraphs.

3.2.1 Mass balance

Assuming a volume with n �xed boundaries, and �ow perpendicular to its sur-face, the mass balance can be expressed as follows:

d(m(t))dt

=n∑i

mi (3.1)

This simply means that the time variation of the stored mass in the volume,is the sum of the incoming and outgoing mass �ows (by setting the conventionthat outgoing quantities from a volume are negative).

3.2.2 Energy balance

The derivative of the internal energy U(t) is equal to the sum of the incomingand outgoing enthalpy �uxes, added to the heat �ux into the volume (Assumen �ow channels and m heat inputs into the considered volume):

d(U(t))dt

=n∑i

mihi +m∑i

Qi (3.2)

Furthermore, the speci�c internal energy u is de�ned as the di�erence be-tween the speci�c enthalpy and the product of the pressure and speci�c volume[1]:

u = h− pv (3.3)

The derivation of the two sides of equation 3.3 after multiplication by themass m gives (U = mu, V = mv):

U = mh− pV d/dt−→ d(U(t))dt

=d

dt(m(t)h(t)− p(t)V ) (3.4)

The time variation of the enthalpy is the sum of the incoming and outgoingenthalpy �uxes (note that H = mh):

d(H(t))dt

=n∑i

mihi(t) (3.5)

Substituting equations 3.4 and 3.5 in 3.2 gives the general formula for theenergy balance:

d(H(t))dt

+ Q(t) = m(t)h(t) +m(t)h(t)− pV (3.6)

This energy balance has been established assuming an incompressible �uid(V = cste). The kinetic energy being only a small fraction of the total energy

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has been neglected, as well as the work of the gravitational �eld. The work ofshear stress and friction force has also been neglected.[2]

3.2.3 Momentum balance

The general form of the Momentum balance is expressed as follows:

d(I(t))dt

=∫

A

ρω(ωA − ω)ndA+ Fg + Fp + FF (3.7)

The time variation of the momentum I is equal to the sum of applied forceson the volume(gravity force, pressure force and friction force), and the term ofconvective transport of momentum (�rst term of right hand of equattion 3.7).

ρ is the density of the �uid inside the volume, ωA is the velocity of the volume,ω is the relative velocity of the �uid. n is the surface A normal on which thevelocities are projected, and dA is an in�nitesimal part of this surface.

In thermo-hydraulic systems dynamic modeling and control, the momentumbalances are replaced by a static relation of pressure drop at the out�ow, becausetheir time constants are outside the bandwidth of interest of control. In fact,the momentum balance evolves on a faster time scale than the other equations.

More information on the momentum equation can be found in [2].

After getting to know the basic equations of thermo-hydraulic modeling, wecan now focus on the components that will be useful for the modeling of ourOTC.

3.3 Air (hy) library

The �hy� library is meant to be used for single phase (like air) and two phase(mixtures: air and water) �uids. But we will need its components only to simulate theair side. This library, like others, has got two kinds of components:

1. Dynamic components

2. Quasi steady-state components

The dynamic components as their name sounds, have dynamics, so di�eren-tial equations and derivatives (like �uid volumes, compressors, heat exchangersetc.). The the Quasi steady-state components are those in which the states areconstant, and no di�erential equations govern the component.

The state vector, is a vector that contains the states of the component (dy-namic variables). The state variables for air are the pressure p(t), the temper-ature T (t) and a �ag f(t):

~x =

p(t)T (t)f(t)

(3.8)

The �ag (or the �uid types)is a help variable for the simulation. It changeswhen there is a phase switch. For the air part of the system, this �ag is notimportant, because there is no phase switch.

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3.3.1 Pipe Gas (Volume)

This component is a volume, which is thermally insulated (no heat transfer withneighboring components). The purpose of this model is to calculate the pressureand temperature derivative of the gas based on the incoming and outgoing mass�ows and temperatures (and the actual state of the component). The followingassumptions are taken for this component:

• The gas pressure and temperature distribution in the plenum is uniform.

• The volume is uniformly �lled with the �uid of type .

• All incoming and outgoing mass �ows are of �uid type .

• No heat transfer between the gas and the pipe casing.

The basic overview is shown in the following picture:

Figure 3.2: Overview of the Pipe gas

Assume nHy the number of input mass channels into the volume. The inputsto the model are :

1. Vector of incoming and outgoing mass �ows: ~mhy = [mhy,1, mhy,2, ..., mhy,nHy]

2. Vector of temperatures ~Thy = [Thy,1, Thy,2, ..., Thy,nHy]

3. Vector of �uid types ~fhy = [fhy,1, fhy,2, ..., fhy,nHy]

The outputs are:

1. Pressure derivative dp(t)dt

2. temperature derivative dT (t)dt

3. Fluid type derivative df(t)dt

4. Volume pressure p(t)

5. Volume temperature T (t)

6. Volume �uid type f(t)

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Now it is time to express the di�erential equations to obtain the derivativesof the state variables.

Pressure derivative

The universal gas constant < is de�ned as:

< = 8314.4kJ

kmolK(3.9)

The speci�c gas constant R can be calculated as follows:

R =<M

(3.10)

The expression of the ideal gas equation is :

p = ρ.R.T (3.11)

Where the de�nition of the density ρ is :

ρ =m

V(3.12)

after substitution of equation 3.12 in equation 3.11 and derivation, the pres-sure derivative for air inside the volume is obtained:

dP (t)dt

=R.

(˙m(t).T (t) +m(t). ˙T (t)

)V

(3.13)

Temperature derivative

The speci�c heat capacity of the air is de�ned as:

cp(T ) =dh(T )dT

(3.14)

at constant pressure.Substituting equation 3.13 into the energy balance, or better into equation

3.4, and using equation 3.14, we obtain the temperature derivative for the vol-ume pipe gas:

dT (t)dt

=m(t). (R.T (t)− h(t)) + dU(t)

dt

m(t). (cp(t)−R)(3.15)

In practice, �rst the temperature derivative is calculated, then the pressurederivative, because for the calculation of the pressure derivative, the tempera-ture derivative is necessary, as we can see in equation 3.13.

Fluid type derivative

As the �uid type in the pipe never changes, and remains always air, itsderivative is equal to zero:

df(t)dt

= 0 (3.16)

22

3.3.2 Volume with heat exchange

The normal volume of the last section has its own basic importance, but theproblem is that it cannot exchange heat with its neighboring components. Ourconsidered system being a heat exchanger, it is evident that we need a volumethat will exchange heat with some other volumes. So we came to create thisnew volume, that we called �VolumeQHE� (Q to say that heat was considered,and HE to say that there is heat exchange with other volumes, comparing to avolume which receives constant heat from outside.)

For this purpose, a wall has been added to the component, which exchangesheat on one hand with the �uid inside the pipe, and on the other hand withthe wall of the connected heat exchanging component. A new signal port hasbeen added to the component to allow this type of connection with neighboringcomponents. The new state variable now is the wall temperature Twall.

The overview of this component in shown in the �gure below:

Figure 3.3: Overview of volumeQHE

There is a new input into this component: Twall,neighbor. This is the walltemperature of the neighboring block which will be used for the conductive heattransfer between the two blocks.

Wall temperature derivative

Having chosen the wall temperature as a new state variable, we have toexpress its derivative. First, there is a convective heat transfer between the walland the �uid inside the volume. Its general equation is:

Qconv = α.A(T − Twall) (3.17)

A is the heat transfer surface and α the heat transfer coe�cient in [W/m2.K].Depending on whether or not there is quiescent �uid inside the volume, we

23

will have respectively natural or forced convection. The empirical relations tocalculate this heat transfer coe�cient can be found in [3].

Then there is conductive heat transfer between the wall of the volume, andthe wall of the neighboring volume. Its general equation is:

Qcond =λ.A

L(Twall − Twall,neighbor) (3.18)

A is the heat transfer surface, L the thickness of the wall andλ the thermalconductivity of the wall in [W/m.K].

From the �rst law of thermodynamics, the derivative of wall temperature is:

Twall(t)dt

=Qconv − Qcond

mwall.cp,wall(3.19)

3.3.3 Control Valve

The control valve is a quasi steady-state component. The purpose of this modelis to calculate the mass �ow through the valve including its temperature and�uid type. The basic overview of the model is shown the following picture: The

Figure 3.4: Overview of a control valve

inputs to the control valve are:

• Pressure, temperature and �uid type at the valve inlet (p1, T1, f1)

• Pressure at the outlet (p2)

• Valve stroke (stroke)

The outputs of the control valve are:

• Valve outlet temperature (T2)

24

• Inlet and outlet mass �ows (m1,m2)

• Valve outlet �uid type(f2)

Calculation of the control valve outputs

As there is no leakage through the control valve, the inlet and outlet mass�ows are identical:

m1 = m2 (3.20)

The �ow through the pipe is isothermic, hence the inlet temperature equalsthe outlet temperature:

T1 = T2 (3.21)

and as the �uid type remains unchanged, the inlet �uid type equals the outlet�uid type:

f1 = f2 (3.22)

The de�nition of the speci�c heat capacity cv(T ) at constant volume is [1]:

cv(T ) =du(T )dt

(3.23)

u being the internal speci�c energy. The isentropic exponent κ(T ) is de�ned as[1]:

κ(T ) =cp(T )cv(T )

(3.24)

The pressure loss (or pressure change in general) can be described as thepressure change in a convergent/divergent channel. As we want to avoid hav-ing supersonic �ows, if our calculated outlet pressure is greater than the sonicpressure (i.e. the pressure of the �uid which �ows with sound's velocity in givenconditions), we take the sonic pressure for the calculations. The sonic pressureis calculated from the inlet pressure[4]:

ps

pin=

(2

κ+ 1

) κκ−1

(3.25)

The greatest value between ps and p2 is taken for the mass �ow calculation:

p2 = max(p2, ps) (3.26)

Let pratio = p1p2, then

ψ =

√2.κκ− 1

.[(pratio)

−2κ − (pratio)

−κ−1κ

](3.27)

The mass �ow is then calculated:

m = Ap1√R.T1

ψ.C (3.28)

C is a correction factor to match the control valve to the good mass �ow atnominal conditions. The e�ective �ow area is obtained from the Kv-value usingthe following conversion:

A = Kv(stroke).√

ρ0

∆p0(3.29)

25

Kv is the volumetric �ow through the pipe as a function of the stroke anddetermines the non-linearity of the control valve. This value as well as ρ0 and∆p0 are speci�ed by the manufacturer of the control valve.

3.3.4 Ori�ce

The ori�ce is a quasi steady-state element which symbolizes the pressure loss(in a pipe or in a volume in general). The schema of an ori�ce is shown in the�gure below:

Figure 3.5: Overview of an ori�ce

The inputs to the ori�ce are:

• Pressure, temperature and �uid type at the inlet (p1, T1, f1)

• Pressure at the outlet (p2)

The outputs of the ori�ce are:

• Outlet temperature (T2)

• Inlet and outlet mass �ows (m1,m2)

• Outlet �uid type(f2)

Calculation of the ori�ce outputs

The ori�ce behaves similarly to a control valve, except that there is no strokefor the ori�ce. The equations 3.20 to 3.22 are therefore also valid for the ori�ce.

The mass �ow is also calculated with the equation 3.28, except for the e�ec-tive area A which this time is calculated based on the width of the channel (ordiameter):

26

A =π.d2

4(3.30)

3.4 Water/Steam (ws) library

This library, as its name sounds, is used to simulate the water/steam part of thesystem. It has, like the air library, dynamic and quasi steady-state components.A description of useful components for our context will follow. But �rst, let'sget to know the basics of water/steam thermodynamics, in order to chose thestate variables correctly.

The state of a water/steam �ow is characterized by three state varibles:

• The pressure p

• The enthalpy h

• The temperature T

But unlike the air, this time the enthalpy does not only depend on thetemperature as in equation 3.14, but on both temperature and pressure. Thiscan be seen on the state diagram below:

Figure 3.6: State diagram of water/steam

27

Another state variable which is important for the simulation, is the �Steamquality�. The steam quality is the ratio of steam mass to the total mass in avolume:

x(t) =ms(t)

mw(t) +ms(t)(3.31)

The subscripts w and s stand for respectively water and steam. This statevariable acts like the �uid type explained in section 3.3, and takes its impor-tance when the �ow is two phase. Let's have a look at the h − T diagram ofwater/steam to understand evaporation and condensation of water/steam:

Figure 3.7: h-T diagramm of water/steam

The blue lines are the isobars (p = cste). Let us follow the evolution ofwater/steam on an isobar. Starting on left, if we heat sub-cooled water, aftera stage, evaporation takes place in the �saturation sector�. In the saturationsector, the pressure and temperature are constant and the enthalpy increases.After total evaporation, we reach the super-heated steam area. If for di�erentpressures, we repeat the experiment and join the two limits of the obtainedsaturation sectors, the red curve is formed, which is called �saturation curve�.

In the saturation sector, the �uid is two-phase. The steam quality can nowbe de�ned as [2]:

x(t) =h(t)− hw(t)hs(t) + hw(t)

(3.32)

which is a simple lever rule on the saturation area. The value of steam qualityis between 0 and 1, but to avoid a confusion of states (for example betweensub-cooled and saturated water), we take these conventions:

• Sub-cooled water : x = −2

28

• Saturated water : x = 0

• Two phase area : x ∈ (0, 1)

• Saturated steam : x = 1

• Super-heated steam : x = 2

Now that the dependence of the steam quality on the pressure and enthalpyhas been shown, the state vector for a water/steam �ow is:

~x =(p(t)h(t)

)(3.33)

The steam quality can be calculated with the pressure and enthalpy usingsteam tables. Let's focus now on the components of this library.

3.4.1 Pipe ws (Volume)

The principle of this component is the same as the air volume (section 3.3.1).But this time the input and outputs are no more the same, because the statevector is di�erent. A schema of the so called �VolumeWs� is shown in thefollowing �gure:

Figure 3.8: Overview of VolumeWs

Assuming a number of nWs incoming water/steam channel into the volume,the inputs are the mass �ows of these channels and their enthalpies, and theoutputs are the derivative of the state vector of the volume and its boundarystate values.

Now let's express the di�erential equations to �nd the derivatives of the statevariables in a water/steam system.

Pressure and enthalpy derivative

Rearranging the energy balance expressed in equation 3.6, we obtain for thepressure derivative:

dp

dt=m(t)h(t) +m(t)h(t)− d(H(t))

dt − Q(t)V

(3.34)

From the de�nition of the density (ρ = m/V ), we have:

29

m(t) = ρ(p, h, t)V (3.35)

On the other hand, rewriting the density we obtain:

m(t) = (∂ρ

∂p

dp

dt+∂ρ

∂h

dh

dt)V (3.36)

The pressure derivative can following this equation be written as follows:

dp

dt= (

m(t)V

− ∂ρ

∂h

dh

dt).

[∂ρ

∂p

]−1

(3.37)

By putting together the two expressions for the pressure derivative, equations3.34 and 3.37, we obtain the expression for the enthalpy derivative:

dh

dt=

[m

V.

[∂ρ

∂p

]−1

+H + Q

V− mh

V

].

[m

V+∂ρ

∂h

[∂ρ

∂p

]−1]−1

(3.38)

Once the enthalpy derivative obtained, we insert it into equation 3.34 or intothe energy equation to pull out the pressure derivative. Note that ∂ρ

∂p and ∂ρ∂h

are obtained from the steam tables.

3.4.2 VolumeWs with heat exchange

The principle of this volume, called �VolumeQHEWs�, is the same as the Vol-umeQHE, explained in section 3.3.2. The new state variable is Twall. Theoverview of this component can be seen in the next �gure:

Figure 3.9: Overview of VolumeQHEWs

Assuming a number of nWs incoming water/steam channel into the volume,the inputs are the mass �ows of these channels and their enthalpies and also theneighboring component's wall temperature and the outputs are the derivativeof the state vector of the volume and its boundary state values.

The wall temperature derivative is calculated based on equations 3.17 to3.19.

30

3.4.3 Control valve Ws

The control valve ws is a quasi steady-state component. The purpose of thismodel is to calculate the mass �ow through the valve including its enthalpy andsteam quality. The basic overview of the model is shown the following picture.The inputs to the control valve are:

Figure 3.10: Overview of a control valve ws

• Pressure, enthalpy and steam quality at the valve inlet (p1, h1, x1)

• Pressure at the outlet (p2)

• Valve stroke (stroke)

The outputs of the control valve are:

• Valve outlet enthalpy (h2)

• Inlet and outlet mass �ows (m1,m2)

• Valve outlet steam quality(x2)

Calculation of the control valve ws outputs

As there is no leakage through the control valve, the inlet and outlet mass�ows are identical:

m1 = m2 (3.39)

The �ow through the pipe is isoenthalpe, hence the inlet enthalpy equals theoutlet enthalpy:

h1 = h2 (3.40)

and as the steam quality remains unchanges, the steam quality type equals thesteam quality type:

x1 = x2 (3.41)

The mass �ow of a control valve ws is calculated based on the followingformula:

31

m(t) = Kv(stroke)

√ρ2(t)cf

(p2(t)− p1(t)).C (3.42)

C is a correction factor to match the control valve to the good mass �owat nominal conditions. cf is the friction factor, which depends on the stateof the �ow. Many correlations have been established to determine this frictionfactor, as described in [6]. We will use the �Homogeneous �ow model�, especiallyto avoid jumps when phase switch happens. A homogeneous �uid is a pseudo�uid that obeys the conventional design equations for single-phase �uids and ischaracterized by suitably averaged properties of the liquid and vapor phase.

The total pressure drop of a �uid is due to the variation of kinetic and poten-tial energy of the �uid and that due to friction to the walls of the �ow channel.Thus, the total pressure drop ∆ptot is the sum of the static pressure drop (ele-vation head) ∆pstatic, the momentum pressure drop (acceleration)∆pmom, andthe frictional pressure drop ∆pfrict:

∆ptot = ∆pstatic + ∆pmom + ∆pfrict (3.43)

Ignoring the �rst two contributions, the total pressure drop is the frictionpressure loss, which can be expressed as follows:

∆pfrict =2.ftp.L.m

2

di.ρtp(3.44)

where L is the length of the pipe, di its internal diameter, ftp the two phasepressure friction factor, and ρtp the homogeneous density.

The homogeneous density is calculated as follows:

ρtp = ρw(1− εH) + ρsεH (3.45)

εH is the homogeneous void fraction and is determined from the steam qual-ity as:

εH =1

1 + ( 1−xx

ρs

ρw)

(3.46)

ftp may be expressed in terms of the Reynolds number by the Balsius equa-tion:

ftp =0.079Re0.25

(3.47)

where the Reynolds number is:

Re =mtotaldi

µtp(3.48)

µtp is the two phase viscosity and is calculated as follows:

µtp = µw(1− x) + µsx (3.49)

by substituting equation 3.48 into 3.47 and putting it into equation 3.44, wecan obtain a new expression for the mass �ow:

32

m = 7

√√√√[ρtp.d

5/4i

2.0, 079.µ1/4tp

∆Pfrict

]4

(3.50)

By equalizing equation 3.50 and 3.42 (with p2(t)−p1(t) = ∆P , and omittingKv), we can obtain the expression for the wanted friction factor cf :

cf =ρtp

∆P

[2.0, 079.µ1/4

tp

ρtp.d5/4i

]8/7

(3.51)

3.4.4 Ori�ceWs

The ori�ce ws, like the ori�ce, is a quasi steady-state element which symbolizesthe pressure loss (in a pipe or in a volume in general). The schema of an ori�cews is shown in the �gure below:

Figure 3.11: Overview of an ori�ce

The inputs to the ori�ce ws are:

• Pressure, enthalpy and steam quality at the inlet (p1, h1, x1)

• Pressure at the outlet (p2)

The outputs of the ori�ce ws are:

• Outlet enthalpy (h2)

• Inlet and outlet mass �ows (m1,m2)

• Outlet steam quality(x2)

Calculation of the ori�ce outputs

The ori�ce ws behaves similarly to a control valve ws, except that there isno stroke for the ori�ce. The equations 3.39 to 3.41 are therefore also valid forthe ori�ce.

33

The mass �ow through the ori�ce ws is calculated with equation 3.42 byomitting Kv. To characterize the diameter of the ori�ce, a supplementary cor-rection factor will be introduced when calibrating the system in the upcomingchapter.

3.4.5 Mass �ow vs. pressure di�erence : Square root

Equation 3.42 shows the function m(t) = f(∆P ) is a square root. For smallpressure di�erences, the slope of the square function is very high. This couldlead the simulation to su�er numerical problems : a step size reduction in thisregion is necessary.

To avoid this, the square function will be replaced by a third degree poly-nomial from a selectable pressure di�erence ∆Plim. This polynomial possessesa �nite upward slope in the critical region.

√∆P = Alim(∆P )3 +Blim(∆P )2 (3.52)

Both coe�cients are taken from [5] and are as follows:

Alim = −1.5∆P−5/2lim , Blim = 2.5∆P−3/2

lim

The following �gure shows the polynomial approximation comparing to thesquare root function:

Figure 3.12: Square root vs. approximation

34

Chapter 4

Model of the Once-through

cooler

In this chapter, we will present the modeling and the calibration of the OTC,based on the theory we presented in the last chapter.

4.1 Counter-�ow model

We can see in �gure 2.6 on page 14, the direction of the �ow in the OTChorizontal. The motion is counter-�ow if we look at the system as a whole : Airis �owing downwards when water/steam is �owing upwards.

So as a �rst approach, we decided to model the system as a counter-�owheat exchanger. For this purpose, we will use the components �VolumeQHE�and �VolumeQHEWs� detailed in sections 3.3.2 and 3.4.2 respectively.

4.1.1 Use of VolumeQHE and VolumeQHEWs

Before using these two components, we need to test them, in order to con�rmthat they respect the mass and energy balances explained in section 3.2.

For this, we will connect a VolumeQHE and a VolumeQHEWs at di�erenttemperatures and with quiescent �uids (closed valves). We expect the two blocksto get to an equilibrium after a while.

After the heat transfer is complete between the two, we will check whetheror not they have respected the energy balance equations. The blocks and theirconnection in Simulink in shown in �gure 4.1.

The energy balance in equation 3.6 will be rearranged with di�erences in-stead of di�erential for our use. For each component i, the change of internalenergy, according to equation 3.4 between the initial time tinit and the �naltime tfinal is :

∆Ui = mi(hi,final − hi,init)− Vi(pi,final − pi,init) (4.1)

in which the term ∆m.h has been ignored, because there is no mass transferin the system (closed system). To this internal energy, a ∆Q term is added,

35

Figure 4.1: Simulink assembly of VolumeQHE and VolumeQHEWs

according to the energy balance, which expresses the change of the the energyin the walls:

∆Qi = mwall,icpwall,i(Twall,i,final − Twall,i,init) (4.2)

If the whole system has respected the balance equations(Assume n compo-nents in the system), this equation should be respected:

∆E =n∑

i=1

(∆Qi + ∆Ui) = 0 (4.3)

∆E being the change of energy of the whole system.

The characteristics of the two blocks where as follows:

Characteristic VolumeQHE VolumeQHEWsVolume 1m3 1m3

Mass of wall 100kg 100kgInitial pressure 1bar 1barInitial Temp. 500 370Initial wall Temp. 510 390

Table 4.1: Volumes characteristics for test

The temperatures evolution is shown in �gure 4.2.

As expected, the system goes to a temperature equilibrium. The Wa-ter/steam side gains in temperature and hence in pressure, in the opposite ofthe air side.

The calculation of the energy balances, according to equations 4.1 to 4.3,has given the following result:

Air side:

• ∆U = −7.44.103

• ∆Q = −1.63.106

36

Figure 4.2: Temperature evolution for the test

Water/Steam side:

• ∆U = 1.37.106

• ∆Q = 2.67.105

Finally we obtained ∆E = −12.23 which is nearly negligible comparing tothe amount of exchanged energies, and we can conclude that our componentsrespect the balance equations.

Now, in order to make our counter-�ow model, we assumed an open ser-pentine (a long pipe whose length is equal to the length of the serpentine),because the geodetic pressure drops on the water/steam side are less than 1%of the operating pressure range (see appendices 2 and 3), and also becausethese pressure drops can be taken into account with the friction pressure dropsin the ori�ces. In this con�guration, the air would be �owing parallel to thewater/steam, counter direction.

The Simulink model of this con�guration is shown in the following �gure:

Figure 4.3: Simulink model of counter-�ow OTC

The air is �owing on the upper channel in VolumeQHE components, andthe water/steam in the channel below, in VolumeQHEWs components. Aircomponents (like the water/steam components) are connected to each other viaori�ces, which symbolize the pressure drop (geodetic and friction). The two �owpaths are connected to each other through heat transfer connections (verticallines in �gure 4.3).

37

But this way of modeling su�ered a major problem: calibration (or match-ing) of the model was not possible. In fact, to calibrate each component, in-put/output measurements are needed. But it is impossible to have intermediatemeasurements, on each tube fragment in the real OTC, because no sensor canbe placed inside the pipe-box. So we thought of a new component, which mod-els each path as a whole entity. This component called �HeatExchanger� is theobject of the next section.

4.1.2 The HeatExchanger component

There are two of these components: HeatExhanger and HeatExchangerWs. Thecomponent HeatExchanger models the shell side, in which air �ows, and Hea-tExchangerWs models the tube side, in which water �ows. The heat transferbetween the two is modeled by a wall (gray area in the next �gure). the following�gure shows an overview of these two elements, put together:

Figure 4.4: Overview of the heat exchanger components

The subscript s and t stand for shell and tube respectively, and the subscriptl and r stand for left and right respectively.

The inputs to the HeatExchanger component are the vector of boundarymass �ows, boundary temperatures and boundary �uid types (the length ofthese vectors depend on the number of incoming/outgoing �ow channels to thecomponent), and the output of this component is the vector of state derivatives(pressure, temperature, �uid type and metal (or wall) temperature derivatives),and the vector of boundary states, exactly as in sections 3.3.1 and 3.3.2.

The inputs to the HeatExchangerWs component are the vector of boundarymass �ows and boundary enthalpies (the length of these vectors depend on thenumber of incoming/outgoing �ow channels to the component), and the outputof this component is the vector of state derivates (pressure, enthalpy and metal(or wall) temperature derivatives), and the vector of boundary states, exactlyas in sections 3.4.1 and 3.4.2.

The heat exchanger components are parameterizable : the number of internalcells n can be selected in order to have the desired temperature pro�le accuracy.The size of the cells is equal, and the number of cells is the same on both sides.

38

A detailed overview of the heat exchanger components is provided in the �gurethat follows.

Figure 4.5: Detailed overview of the heat exchanger components

The metal blocks are in gray. The subscript m stands for �metal�. Thefollowing assumptions are taken for these components:

1. The heat transfer between the components in the x-direction is negligible.

2. The in�uence of gravity is negligible.

3. The heat capacity cp is constant for the �uids.

Internal mass �ows

The mass �ows between each two cells are calculated based on equation 3.42for the ori�ces. This equation (by omitting Kv) will also be used for the airside. Depending on the pressures of the two neighboring cells, the direction ofthe mass �ow changes. For the shell side (Air):

ms,i,i+1(t) =√ρs,i

cs(ps,i(t)− ps,i+1(t)) ps,i(t) > ps,i+1(t) (4.4)

ms,i,i+1(t) = −√ρs,i+1

cs(ps,i+1(t)− ps,i(t)) ps,i(t) < ps,i+1(t) (4.5)

The friction factor cs is constant for the air, and will be calculated duringmatching. For the tube side:

39

mt,i,i+1(t) =√ρt,i

ct(pt,i(t)− pt,i+1(t)) pt,i(t) > pt,i+1(t) (4.6)

mt,i,i+1(t) = −√ρt,i+1

ct(pt,i+1(t)− pt,i(t)) pt,i(t) < pt,i+1(t) (4.7)

The friction factor ct is calculated based on equation 3.51.

Heat Transfer

The transfered heat between a cell of the tube side Qmt,i and its correspond-ing metal part is convection, and is calculated with equation 3.17 on page 23.The heat transfer surface is the contact surface of water/steam and metal (insidearea of the tubes).

Note that depending on the state of the cell, whether or not the �uid isquiescent, whether it is simple convection, boiling or condensation, the heattransfer coe�cient varies. The equations for all these heat transfer mechanismscan be found in [3] and will not be repeated here. Just as a notice, one shouldassure avoiding jumps between the values of the heat transfer coe�cient α (thevalues are very high relatively for boiling for example) when the heat transfermechanism changes. We have done this by linearly interpolation between thecorrelations.

The transfered heat between the two metal blocks, Qmm,i is of conductionand is calculated based on equation 3.18. A is here the contact area of the twoblock metals (here: the average of inside and outside naked pipe area).

The transfered heat between a cell of the shell side Qmt,i and its corre-sponding metal part is convection, and is calculated with equation 3.17. Theheat transfer surface is the contact surface of air and metal (outside area of allpipes including �ns).

Even though the heat transfer mechanism for the air can only be naturalor forced convection (depending on whether or not the mass �ow is zero), oneshould not forget that air is �owing through a tube bundle (see �gure 2.6).In fact, the inner rows of a bundle are associated with higher heat transfercoe�cients. The tubes of the �rst few rows act as a turbulence grid, which in-creases the heat transfer for the tubes in the following rows. On the other hand,�owing through a bank of tubes increases the velocity of air, which increasesthe Reynolds number that a�ects the heat transfer coe�cient. The overallheat transfer coe�cient for bundles depends on the longitudinal and transversalpitches, diameter of tubes and the rows deep that air has to travel through.Some correlations for bundle heat transfer coe�cient calculations can be foundin [3] and [7].

Derivative of the metal temperatures

The metal temperature derivatives for both shell and tube side can be cal-culated based on equation 3.19 with the heat �uxes calculated above.

40

Mass and energy balances

To apply the mass and energy balances to each cell, the boundary cells shouldhave a di�erent treatment than the inner cells of the heat exchanger components,because they get their boundary conditions from their neighboring components.For this purpose, three di�erent kinds of cells should be distinguished regardingtheir emplacement in the component:

• Cells on the left hand side : i = 1

• Cells in the middle: 2 ≤ i ≤ n− 1

• Cells on the right hand side : i = n

According to equation 3.1, the mass �ow for each cell types above is calcu-lated. Assume nsl boundary channels on the left and nsr boundary channel onthe right of the heat exchanger component:

d(ms,1(t))dt

=nsl∑i

msl,i − ms,1,2 (4.8)

d(ms,i(t))dt

= ms,i−1,i − ms,i,i+1 2 ≤ i ≤ n− 1 (4.9)

d(ms,n(t))dt

= ms,n−1,n −nsr∑

i

msr,i (4.10)

The boundary enthalpies for the energy balance (equation 3.6) are calculatedfor each cell type as follows:

Hs,1(t) =nsl∑i

msl,ihsl,i − ms,1,2hs,1,2 (4.11)

Hs,i(t) = ms,i−1,ihs,i−1,i − ms,i,i+1hs,i,i+1 2 ≤ i ≤ n− 1 (4.12)

Hs,n(t) = ms,n−1,nhs,n−1,n −nsr∑

i

msr,ihsr,i (4.13)

Note that the equations 4.8 to 4.13 have only been written for the shell side.They are identical for the tube side.

Based on these equations, the state derivatives can be calculated for eachcell, with the given equations in section 3.3.1 for air and section 3.4.1 for wa-ter/steam.

41

4.1.3 Calibration of the counter-�ow model

The calculation for the heat transfer coe�cient in equation 3.17 has been doneby empirical relations. As well, the friction factor for pressure drop on thewater/steam side, equation 3.51, has been obtained by an empirical relation.

In order to get to the good steady-state outputs of the system, we need tocorrect the values obtained out of these relations by some correction factors(because the empirical relations never give the accurate value). We will assumethe same correction factor for the heat �ux calculations on the shell and tubeside (convection), as well as on the metal side (conduction):

Qconv = fαα.A(T − Twall) (4.14)

Qcond = fαλ.A

L(Twall − Twall,neighbor) (4.15)

for the pressure drop on the water/steam side, a correction factor for the frictionfactor will be added:

m(t) =

√ρ2(t)fc,wscf

(p2(t)− p1(t)) (4.16)

As we said, we will also determine the air side friction factor by calibration

m(t) =

√ρ2(t)cair

(p2(t)− p1(t)) (4.17)

fα, fc,ws and cair will be our optimization parameters.

Now assume a counter �ow con�guration as in the �gure that follows. Theinput and output states are determined. We want to get the correct outputs byimposing the given inputs .

Figure 4.6: Matching of two neighboring heat exchanger components

42

Outer loop

At the beginning, we initialize the optimization parameters (correction fac-tors) to some values that are not far from the reality (input/output states andnominal mass �ows are known).

First cell (right hand side)

Starting on the right hand side of the �gure (column number n), on the shellside, we know that ps,n = ps,out and Ts,n = Ts,out, because the last cell is takenas boundary value. But on the tube side, pt,n and Tt,n are the states of the�rst cell, and this has exchanged heat with the last cell of the shell side. So thestates of this �rst cell are not equal to the input states of the tube side.

Let pt,out = pt,n+1 and Tt,out = Tt,n+1 (we chose the numbeer n+ 1 for theright hand side boundary states). We initialize ht,n to some value (for exampleto the enthlpy of input water). We calculate the heat transfered from cell n+ 1to cell n and call in Qint,n:

Qint = mt(ht,n+1 − ht,n) (4.18)

Inner loop

Then we initialize Tmt,n to some value. We calculate the heat convectedfrom cell number n to metal cell number n on the tube side:

Qmt,n = fαα.A(Tt,n − Tmt,n) (4.19)

α is calculated exactly as in the simulation, and the heat transfer area A isknown.

As this is a steady-state matching, there heat storage in the cells is con-stant, hence the equality Qint,n = Qmt,n should be respected. The internaloptimization loop (least-square problem) is:

minTmt,n

(Qint − Qmt,n)2 (4.20)

End of inner loop

This will give us the value of Tmt,n. The shell side metal temperature, Tms,n,can now be calculated by conduction (equation 4.15):

Tms,n =Qmm,nL

λAfα+ Tmt,n (4.21)

with Qmm,n = Qmt,n again because of constant mass storage of the cells. Havingobtained Tms,n, we can calculate the convected heat on the shell side:

Qms,n = fαα.A(Ts,n − Tms,n) (4.22)

This completes the middle loop (with ht,n as parameter) : if Qms,n and Qint

are not equal, ht,n should be changed to equalize them. This is again a leastsquare optimization problem:

43

minht,n

(Qint − Qms,n)2 (4.23)

Next cells (n→ n− 1)

Doing an energy balance on cell n will give us the enthalpy of cell n− 1:

mtht,n−1 = mtht,n − Qmt,n ⇒ ht,n−1 = ht,n −Qmt,n

mt(4.24)

The same calculation can be done to obtain the enthalpy (and then the tem-perature) for the shell side. To obtain the pressure on the tube side, we useequation 4.16:

pt,n−1 =m(t)2fc,wscf

ρn+ pn,t (4.25)

On the shell side, because of the direction of the �ow, to calculate ps,n−1,we will also need ρs,n−1. This is another least-square optimization problem:

Inner loop

We initialize ps,n−1 and having Ts,n−1 from the energy balance, we cancalculate ρs,n−1. we can also calculate a mass �ow based on these estimations:

mestim,n =√ρs,n−1

cair(ps,n−1 − ps,n) (4.26)

The least square optimization problem for this inner loop is :

minps,n−1

(mestim,n − ms)2 (4.27)

End of inner loop

At this stage, the states of the cells on column n− 1 are known. The samecalculations can be repeated to get to the boundary on the left hand side. Thevalues obtained in this way will be compared to the given nominal values, andthe correction factors will be changed and the outer loop launched again untilreaching an error smaller to a speci�ed error. The outer least-square problemis:

minfα,fc,ws,cair

[(ps,in − ps,1)2 + (Ts,in − Ts,1)2 + (pt,out − pt,1)2 + (Tt,out − Tt,1)2

](4.28)

End of outer loop

Implementing this for the HP Once-through cooler (which will be the systemthat we will study from now on), we needed 9 cells on each side to converge (aminimum number of cells is required to converge for each problem, becausethe dynamics of the system requires a speci�c pressure and enthalpy gradient).for the nominal input/output states, see appendix C. The obtained correctionfactors were:

44

• fα = 1.329

• fc,ws = 0.97

• cair = 0.176

Except for cair which is a friction factor, the correction factors show thatour heat transfer and water friction factor predictions were not far away fromthe physics.

4.1.4 Step response of the counter-�ow model

To verify the results and have an idea of the dynamics of the system, we imposeda step to the command of the system (the water inlet control valve stroke), andleft the system to get to equilibrium (steady-state). Let us have a look atappendix E. It is important to mention that, our calculation times showed thatwe are about 100 times faster than real time (a simulation time of 100 secondsfor a system simulated for 10000 seconds in real time).

Note that water/steam �ows from cells 9 to 1, and air �ows from cells 1 to9, in counter direction of water/steam (the water enthalpy increases from cell 9to 1 in �gure E.2, and air temperature decreases from cell 1 to 9 in �gure E.7).

To verify if the matching has given the expected outputs, let us check theoutput states of the heat exchanger components on both sides:

• Figure E.1 shows an output pressure of 137.30bars for cell 1, which iscorrect according to appendix C

• Figure E.2 shows an output enthalpy of 3.1182.106J/Kg for cell 1, whichis correct according to appendix C

• Figure E.6 shows an output pressure of 37.087bars for cell 9, which iscorrect according to appendix C

• Figure E.7 shows an output Temperature of 613.15K for cell 9, which iscorrect according to appendix C

Moving from cell 1 to 9 on water/steam side, we go from superheated steamto sub-cooled water. As we can see on �gure E.5, when steady-state, cells 1and 2 are superheated steam, cells 3 to 7 are two-phase, and cells 8 and 9 aresub-cooled water. This is con�rmed of �gure E.3, when we can see that the two-phase cells, in which evaporation is taking place, have the same temperature.

A snap-shot of the steam qualities at equilibrium con�rms this in �gure E.9,and the temperature snap-shot at equilibrium recon�rms the fact that two-phasecells have constant temperatures (�gure E.10).

On the air side, we can see in �gure E.7, that at the beginning, when theinlet water/steam control valve is closed, air is not cooled down and hence, allair cells are at the same temperature. All of the water/steam cells in this timeperiod have steam, as we can see in �gure E.5, because hot air all along the

45

path has evaporated the initial water. After opening the valve on water/steamside, we can see in �gure E.7 that all the air side cells are cooled down : cellnumber 9 has a smaller time constant and goes to a cooler temperature thancell number 1, because cell 9 is in the neighborhood of a water cell (cell 9 ofwater/steam side), while cell 1 is in the neighborhood of a steam cell.

Looking at the output temperature on air side, namely the output of thecontrolled system (cell 9 of �gure E.7), we have calculated a time constantof about 40 seconds. This dynamic of the system, is much quicker than theexperience that exists of these kinds of coolers: The time constant should bearound 60 seconds.

This error comes from the conservatism we accepted in making the heatexchanger components: We did not take into account the air �owing outsidethe box, which has an important volume. Even if we had considered this air byadding it to the inside box air, we still have ignored the mass of the vessel. Thisinertia is very important especially at start-up (and other transients), when itis cold, and has to be heated up to get to the equilibrium. The heat exchangercomponents do not allow us to insert intermediate components in order to modelthe air outside of the box and the vessel mass.

Another problem that exists with this counter-�ow model, is the temperaturedistribution. Firstly, going from inlet to outlet air, the next air cell is exchangingheat with the next water/steam cell, which in a cross-�ow con�guration is notright: the next air cell exchanges heat with the water/steam cell in a tube rowbelow. This leads to completely di�erent dynamics.

For these reasons, we decided to build a counter �ow model, which is theobject of the next section

4.2 Cross-�ow model

In the cross �ow model, we have assembled VolumeQHE and VolumeQHEWscomponents in order to represent the system. Figure 4.7 shows the con�gurationof the components.

Water penetrates the system from bottom in the control valve �WS IN� and�ows all the way to the top, in a serpentine path, in the blue cells, which repre-sent the VolumeQHEWs components, from WS1 to WS9. Steam leaves the sys-tem from the control valve �WS OUT�. Between each two cells of water/steam,there is a pressure drop represented by an ori�ce.

Air enters the system from the top, in the control valve �AIR IN� (Air cellsare in green). Then �ows in the Volume components �MIX IN�, which canbe seen as the input hood of air (see �gure 2.6). Air is then separated in threechannels : one boundary channels on the left (cells GT1, GT10, GT2 and GT3),one middle channel (cells GT4 to GT6) and another boundary channel on theright (cells GT7, GT11, GT8 and GT9). Between each two cells of air, there isa pressure drop represented by an ori�ce.

Through their path, the air cells exchange heat with the water/steam cells.This heat exchange is represented by the red connection lines in �gure 4.7. The

46

Figure 4.7: Discretization of the OTC HP

air cells presented until now (GT1 to GT11) are inside the tube box. Air leavesthe tube box at the bottom, from cells GT3 and GT9, and goes to the com-ponents GT12 and GT13, which are outside the box. These volumes exchangeheat with the air inside the box, with components GT10 and GT11 respectively.

As we have decided to take a �ner discretization for the air inside the box,the components GT12 and GT13 outside the box have a bigger volume. Theyrepresent the volume of air outside the box (inside the vessel) and the vesselinertia (mass). The heat transfer mechanism is also di�erent in these two cells,because their air is not �owing through a pipe bundle (it is natural or forcedconvection). Note that the middle air channel does not have an outside boxvolume, in order to have a di�erent dynamic from the two boundary channels.

The three air channels then are mixed in component �MIX OUT� and leavethe system on the top (as in the real system, �gure 2.6) through the air controlvalve �AIR OUT�.

4.2.1 Calibration of the cross-�ow model

The calibration problematic is exactly the same as in section 4.1.3. The em-pirical relations 4.14 to 4.17 have to be corrected through correction factors toget to the exact steady-state conditions of the system. But here, a steady-statecalibration as in section 4.1.3 is not possible, because a column-wise organi-zation like in the counter-�ow model (HeatExchanger components assembled)

47

does not exist. Also it is not possible to calibrate each component separately,because no intermediate data exists (also in reality, one does not have access tothe space inside the tube box). The solution here is what we call the Dynamicmatching.

Assume a system whose dynamics are presented by a set of di�erential equa-tions (x being the state array, u the inputs and z the matching parameters asin section 4.1.3):

x(t) = f(x(t), u(t), z) (4.29)

A steady state calibration would mean to pose x = 0 and solve the steadystate equations until getting to the correct outputs:

f(xF , uF , z = z0) = 0 (4.30)

This set of equations will give xf which is the steady-state state array, as afunction of uN the input array and z0 the set of optimization parameters (cor-rection factors) that gives the correct outputs (solves the steady-state matchingproblem). xf will then be compared to the output state array to establish theleast-square problem (4.28).

A dynamic matching, would be to initialize the z array at some value, solvethe set of di�erential equations (integrate equation on a su�cient time (TF ) toreach the steady-state conditions), and then minimize the least square problemfor the outputs (equation 4.28):

x(t) = f(x(t), u(t), z)→∫ TF

0

x→ XF (z, TF ) (4.31)

The obtained state array XF will be compared to the output state array inthe least square problem (equation 4.28).

Implementing this on the HP Once-through cooler (see nominal input/outputstates in appendix C), we obtained the following correction factors:

• fα = 1.362

• fc,ws = 0.887

• cair = 2.631

which are of course di�erent from those of the counter-�ow model in section4.1.3, because the �ow con�guration, masses and volumes are di�erent.

4.2.2 Step response of the cross-�ow model

To verify the results and have an idea of the dynamics of the system, we imposeda step to the command of the system (the water inlet control valve stroke), andleft the system to get to equilibrium (steady-state). Let us have a look atappendix F. The simulation times here were in the same order that those of thecounter-�ow model (100 times faster than real time).

To verify if the matching has given the expected outputs, let us check theoutput states of the heat exchanger components on both sides:

48

• Figure F.2 shows an output pressure of 137.30bars for cell 9, which iscorrect according to appendix C

• Figure F.3 shows an output enthalpy of 3.1182.106J/Kg for cell 9, whichis correct according to appendix C

• Figure F.7 shows an output pressure of 37.087bars for MIX OUT, whichis correct according to appendix C

• Figure F.8 shows an output Temperature of 613.15K for MIX OUT, whichis correct according to appendix C

For the water/steam side, we can see on �gure F.6 that the three �rst cellsare �lled with water, cells 4 to 7 are two-phase and cell 8 and 9 have super-heatedsteam at steady state conditions. This is also con�rmed in �gure F.4 where wecan see that in cells 4 to 7 where evaporation takes place, the temperature isidentical. The snapshot of the steam quality (�gure F.10) and the temperature(�gure F.11) at steady-state conditions con�rms this.

On the air side, the temperature distribution is interesting. Let's focus on�gure F.8. At steady state conditions, there is a grouping of cells according totheir temperatures:

• Cells GT1, 4, 7, 10 and 11 : the upper row

• Cells GT2, 5, 8 : the middle row

• Cells GT3, 6, 9 : the lower row

At time t=5100s, the temperature pro�le of cells 2, 5 and 8 cross each other.At the beginning (t= 5000s to 5100s), the cell order from warm to cold is 2,5 and 8. At around 5100s, the order becomes 8, 5 and 2. This is due to thecondensation of water: water/steam cell WS4 begins �rst to condensate, whichin the neighborhood of GT8, so this cell is cooled down faster. The WS5 andWS6 condensate after wards, which produces this temperature order on the airside. The same phenomena happens for the lower air row cells.

We can also see cells GT1 and GT10 crossing themselves. Cell GT10 shouldbe colder than GT1, but is warmer at the beginning, and then becomes coolerthan GT1 after a while. This is because cell GT10 is in the neighborhood of theboundary cell GT12 which has a higher inertia (more air volume, bigger mass),and needs more time to get to its equilibrium. So the heat exchange betweenGT10 and GT12 is slow, which leads to this behavior. The same phenomenahappens for cells GT7 and GT11.

Now if we look at the air cell columns (left column GT1, 2, 3 - middlecolumn GT4, 5, 6 - right column GT7, 8, 9), the right column is warmer thanthe middle column, the left column is the coldest one (if we compare element-wise each row). This is why, we see that the temperatures of cells GT3 andGT12 are colder than the cell MIX OUT, cell GT9 is warmer than MIX OUTand cell GT6 has about the MIX OUT temperature. Now, in each row, themiddle column element is closer to the right column element. This is becausethe middle row does not have any heat exchange with the air outside the boxand hence remains warmer.

49

The metal temperatures for the air side in �gure F.9 show an importantissue : The metal temperature evolution of cells GT10, 11, 12 and 13 go muchslower to equilibrium. This is mainly because of the important metal massesof cells GT12 and GT13, and also because of their bigger volume. This is thefactor which increases the time constant of this system.

Finally we can see from the temperature pro�le of MIX OUT (the outputair temperature), a time constant of about 65 seconds, which is more in linewith our expectations.

The model is now valid, and we can start the control design in the upcomingchapter.

50

Chapter 5

Control system of the OTC

The control system of the OTC is presented in the �gure that follows. Theoutput of the system to be controlled, is as mentioned before, the outlet airtemperature. This temperature has to be monitored, because this outgoing airserves the cooling of the turbine blades. A too high temperature would lead tothe waste of these blades and a too low temperature would not cool down theblades su�ciently.

Figure 5.1: Control system of the OTC

The outlet air temperature of the OTC is compared to the set temperature,which is for the HP OTC, 340◦C. A �rst controller, K1, which we call �the outerloop controller�, gives a water mass �ow as output. This mass �ow, is the changeof mass �ow brought to the feed water, in order to annul the temperature error.

But as the changes of temperatures are very slow when responding to a feedwater mass �ow command (as explained in previous page, the time constant isabout 1 minute), a feed forward command is added to this mass �ow : mFF orthe feed forward mass �ow. This water mass �ow is calculated using the inletair conditions (relating to the transient applied), and gives a prediction of thewater mass �ow which should be injected to the system. This feed forward mass�ow will be explained in the upcoming section.

This mass �ow (the addition of ∆m and mFF ) will be requested from the

51

inlet control valve of water. The �inner control loop� controlling the controlvalve, based on this reference mass �ow will give a stroke, which is the commandof the whole control system.

5.1 The Feed-Forward mass �ow

This section details the stages of the feed-forward mass �ow calculation. Let'ssee the following �gure to de�ne the input/output states that will be used here.

Figure 5.2: Feed forward mass �ow calculation conventions

We Assume an e�ciency of 100% for the OTC. In reality, this happens atpart load. This would mean all the heat out of air, is transfered to the water.

The feed forward mass �ow, is actually the water mass �ow needed in caseof 100% e�ciency for the OTC to give the wanted air outlet temperature. Inthese conditions, the outlet steam temperature would be equal to the inlet airtemperature:

Tsteam,out = Tair,in (5.1)

The outlet enthalpy of steam will be calculated based on this temperature, andthe outlet pressure:

hsteam,out = f(Tair,in, psteam,out) (5.2)

The inlet enthalpy of water will be calculated based on inlet water temperature,and its pressure:

hwater,in = f(Twater,in, pwater,in) (5.3)

The inlet enthalpy of air will be calculated based on inlet air temperature:

hair,in = g(Tair,in) (5.4)

The outlet enthalpy of air will be calculated based on outlet air set temperature(because it's the air outlet temperature we want to reach ):

52

hair,out = g(Tair,out,set) (5.5)

The heat transfered from air, and the heat transfered to water are respectively:

Qair = mair(hair,in − hair,out) (5.6)

Qwater = mFF (hsteam,out − hwater,in) (5.7)

Assuming a 100% e�ciency, equations 5.6 and 5.7 would be equal, and we cannow calculate the feed forward water mass �ow:

mFF = mair(hair,in − hair,out)

(hsteam,out − hwater,in)(5.8)

This is a prediction (or estimation) of the feed water mass �ow needed,depending on the transient imposed to the system (the inputs), if we wanted toreach the air outlet set temperature by assuring a 100% e�ciency.

5.2 The inner loop

Equation 3.42 on page 32 shows the relation between input and the output ofthe control valve (stroke and mass �ow). As mentioned in section 3.3.3, Kv isthe volumetric mass �ow of the control valve as a function of the stroke. Thisis called the characteristic of the control valve, and determines its non-linearity.For our inlet water control valve, we have chosen a standard characteristic,which can be seen in the �gure that follows.

Figure 5.3: Characteristic of the inlet water control valve

53

The square root in equation 3.42 has also time dependent variables : theinlet and outlet pressures, and the density upstream of the control valve. So thenon linearity of the control valve is also a consequence of this.

5.2.1 Controller for the inner loop

To design a controller for the control valve, we linearize it in its nominal oper-ating point (when the system is steady-state). In this conditions, the pressuresand the density upstream of the control valve are constant, so the mass �ow-stroke function is a constant gain times Kv-stroke function. We can verify thisin the �gure below.

Figure 5.4: Mass �ow vs. stroke at steady-state conditions

We can see, according to appendix C, that the water mass �ow at steady-state conditions is 7.533[kg/s], when the stroke is 1. Linearizing the mass �ow-stroke function in this area will give us a constant gain as transfer function:

Gcv(s) = 7.533 (5.9)

The bode diagram of this transfer function is a constant at 20log(7.533) =17.54dB for the magnitude, and has a constant phase of 0◦.

We chose a closed-loop time constant of τ = 2s. According to [8] page 265,the open-loop cut-o� frequency is equal to 1/τ . The cut-o� pulsation is thenωx = π[Hz].

We have decided to design a PI controller for the control valve. The transferfunction of such a controller in continuous time is :

54

Kcv(s) = Kp(1 +1Tis

) (5.10)

According to [8], the open loop magnitude bode diagram of the controlledsystem, Kcv(s)Gcv(s) should cut the frequency axis with a slope of −20dB/decin a large enough neighborhood, to assure a good behavior of the closed loopand enough margins (robustness). As the magnitude on the bode diagram isconstant, the choice of Ti is free : we have chosen Ti = 1/5.

The open loop transfer function is:

GcvKcv(s) = 7.533Kp(s+ 5s

) (5.11)

The cut-o� pulsation is when the magnitude is 1 dB (or the amplitude is1.12):

1.12 = 7.533Kp

∣∣∣∣jω + 5jω

∣∣∣∣ (5.12)

which gives Kp = 0.079.

The open loop bode diagram is shown in the �gure that follows:

Figure 5.5: Open loop bode diagram of the for the control valve

55

We can see that the cut-o� pulsation is about π[rad/s], and that the bodemagnitude diagram has a slope of −20dB/dec near this pulsation. We obtain again margin of in�nite, and a phase margin of 126.5◦ for the controlled system.

5.2.2 Anti-reset windup

An anti-reset windup module should almost always be used when the controllerhas an integral term. According to [8], when the gap between the set point forthe output and the output stays large over a long time period, the accumulationof these gaps in the integral term results in an increased command signal, whichcan overstep the saturation limits of the system's input. This is called resetwindup or integral windup. The use of an anti reset-windup (ARW) strategyprevents this to happen.

In this project, we use the so-called �Standard anti-reset windup�. Assumea command signal whose saturation limits are umax = µ and umin = −µ. Thestrategy here is, when the command signal is in saturation, to recalculate thesignal coming out of the integral term, producing a command signal equal to µor −µ. This can lead the integral term do diminish although the gap betweenthe set point and the output is increasing.

The command signal is the addition of two terms, term proportional and theintegral term (for the PI controller):

u(t) = Kp(e(t) + ui(t)) (5.13)

e(t) being the gap between the set point and the output, and ui(t) the signalfrom the integral term. If the the command signal is bigger than umax, anintegral term will be calculated in order to make in equal to umax, if ui(t) > µ:

µ = Kp(e(t) + ui(t))⇒ ui(t) =µ

Kp− e(t) (5.14)

and the same way for the lower saturation boundary, if ui(t) < −µ:

−µ = Kp(e(t) + ui(t))⇒ ui(t) = − µ

Kp− e(t) (5.15)

Figure 5.6 shows how the PI controller for the control valve accompanied bya ARW has been realized in Simulink.

Figure 5.6: PI controller for the control valve with ARW

56

The lower saturation limit for the control valve input (stroke) is 0 and itsupper saturation limit is 1. The green components are part of the PI controlleritself, and the yellow ones are part of the anti-reset windup module.

5.3 The outer loop

The outer loop is the loop containing the controller which controls the wholesystem : the control valve loop together with the process. We did not linearizethe process, for the following reasons:

1. Hydraulic systems are highly non-linear, because of the properties of waterand steam are far from each other, and when evaporation or condensationhappens, it is not possible to linearize the system around one operatingpoint.

2. The transients that the system su�ers, see appendix G, involve highchanges in the inlet air temperature and mass �ow : the disturbance tothe process is far from being constant. Therefore, it is impossible, even ifwe had only one case to handle, to chose an operating point.

3. Figure F.8, shows a step response of the system output which is similarto a �rst order �lter. Also, the open loop system is stable, so linearizingwould not have any advantage as a �rst approach.

4. Because of lack of time for the control part of this project, we decided to�rst realize a �rst sketch of the control system. So we left the system nonlinear and tried to tune the controller manually.

Using the �rst Ziegler-Nichols method on the step response of the system,�gureF.8, to tune a controller manually (not that the controller is also tunedmanually in reality), we obtained the following parameters for the PI controller:

• Kp = −0.05

• Ti = 0.01

Applying an full load trip (see appendix G) as disturbance to the system, andusing this controller with the ARW explained in section 5.2.2, we obtained theresults in �gures 5.7 and 5.8.

As we can see in �gure 5.7, the output of the system takes a long time beforegetting back to the set point (which is 613.15 K here). Let's check the input ofthe system (inlet water/steam control valve) in �gure 5.8.

We can see a dead time of 50s. The disturbance having started at t=4000s,the system sees only after 50 seconds that it should close the water controlvalve (because the outlet air temperature is colder than what it should be, sothe logical reaction of the system should be to reduce the amount of water).

Let's focus on the control loop schema of the system, �gure 5.1. The inputto the control valve controller (K2, inner loop), mref , is the limitation for the

57

Figure 5.7: Temperature at full load trip

Figure 5.8: Stroke at full load trip

output signal of the controller K1. As the controller K2 should not produce astroke higher than 1 and lower than 0 :

• mref,max corresponds to the mass �ow delivered by the control valve, atstroke = 1. This value is not constant and changes during the simulation,because it depends on the state of the system as in equation 3.42.

• mref,min corresponds to the mass �ow delivered by the control valve, atstroke = 0, and is equal to 0.

The maximum and minimum boundaries for ∆m are then:

∆mmax(t) = mref,max(t)− mFF (t) (5.16)

∆mmin(t) = mref,min − mFF (t) (5.17)

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So both boundaries of the output of the outer loop controller K1 are variablein time. This means that the capacity of the integrator should change in time,to comply with the new output limitations. Assume an error e at time t asinput of the controller K1. Assume the upper limit of the output reducing fromµ1 to µ2. Then according to equation 5.14:

ui,1 =µ1

Kp− e (5.18)

ui,2 =µ2

Kp− e (5.19)

Then ui,1 > ui,2 . As the integral cumulates the past errors, the error due tothis boundary change will be saved in the integrator and accumulated: this isagain a windup problem. This means That the output of the integral should bereduced if the saturation boundary changes. Hence, integral should be emptied.

If that maximum allowed output of the integral term (ui,max)is smaller thanthe actual integrator signal (ui), it means that such an error has been saved: the capacity of the integrator has been reduced. Then the integrator shouldbe emptied by ui − ui,max. The same operation should be done on the lowerboundary : if ui,min is bigger than ui, the integrator should be emptied byui,min − ui.

This has been realized on our K1 controller. This can be seen in the �gurebelow. In this �gure, we can also see the the block on the left (add block) whichis the point where the feed back and the set point come toghether.

Figure 5.9: Simulink diagramm of the moving saturation ARW

The green blocks are the PI controller components, the yellow ones are thestandard ARW components, and the light blue ones are the components addedfor the compensation of saturation limits changes.

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The corrections brought to the ARW strategy improved the results. Theseresults are shown for the full load trip:

Figure 5.10: Temperature at full load trip (ARW comarison)

Figure 5.11: Stroke at full load trip (ARW comarison)

We can see on �gure 5.11 that the dead time for the command signal (inputwater stroke) has been removed, and that the command reacts much faster tothe disturbance (�gure 5.10), which leads to a faster rejected disturbance, andreduces the overshoot on the output air temperature.

We have also performed a test for the designed K1 controller, in case ofStart-up (see appendix G). The results are shown in the �gures that follows.

As we can see on �gure 5.12, at the beginning, the inlet temperature is colderthan the outlet set temperature. This leads to a colder outlet temperature. Inthis case, the reaction of the outer loop controller is to close the water controlvalve, to allow a less cooled air �ow. The control valve is closed to the maximumto minimize the output error. It is not possible to further minimize the error

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Figure 5.12: Temperature pro�le in start up (inner loop PI)

by the action of the stroke (The only way is to increase the air inlet tempera-ture), the stroke remains at zero (see �gure 5.13 ), until an increase of the inlettemperature at t=4000s.

Figure 5.13: Stroke for startup (inner loop PI)

The outlet temperature starts to increase under the e�ect of the inlet tem-perature at t=4000s and the stroke is still closed. When the outlet temperatureexceeds its set point (613.15 K) at t=4306s (the error changes sign), the strokeis logically opened to cool down the outlet air temperature, and goes toward 1,because of the increase of inlet temperature. The fact that the stroke starts toopen exactly when the temperature exceeds its set point, shows that there is nowindup problems anymore.

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Figure 5.13 shows a peak in the command at around t=4040s which disap-pears after 20 seconds. This is the e�ect of the feed forward mass �ow. In fact,at this time, the outlet enthalpy of water (calculated based on the inlet temper-ature of air, and the outlet pressure of steam, see equation 5.2)diminishes tillequaling the inlet enthalpy of water, and the denominator of equation 5.8 tendsto in�nite (see �gure 5.14).

Figure 5.14: Feed forward mass �ow startup (inner loop PI)

The feed forward mass �ow tends to −∞ right after the outlet steam en-thalpy has crossed the inlet water enthalpy, which is not re�ected on the strokecommand, because it would lead to a negative stroke (it is prevented by theouter loop controller K1's ARW).

This peak of the command is not �seen� by the OTC because of its slow timeconstants (the change of the water stroke is such a short time does not a�ectthe outlet temperature of air).

5.4 Static compensator instead of inner loop

Reconsidering the inner loop to control the inlet led us to the following argu-mentation:

1. The input/output relation of the control valve is not a constant transferfunction as we assumed. In fact, the mass �ow out of a control valve,according to equation 3.42, depends on the conditions upstream of thevalve, and consequently on the inputs (or disturbances) that are imposed

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to the system (OTC), and the state of the system itself. So if the controlleris designed in conditions where the control valve has a high gain, thecontrol valve loop will reduce the bandwidth of the whole system whenthe control valve operates at low gains. In opposite, designing the innerloop controller in conditions where the control valve has a low gain, canmake the output (mass �ow) oscillate if the control valve operates at highgains.

2. A static compensator won't need a feed back loop, so the cost is reducedby the suppression of the instrumentation for the closed loop. Also thedelays due to the information coming back all along the feedback pathwon't exist anymore.

This way, if we know exactly the characteristic of the control valve (see �gure5.3), the mass �ow-stroke function for the compensator is exactly the inversefunction of the stroke-mass �ow function for the control valve. So when theouter loop controller determines which water mass �ow the system needs, thecompensator would determine the stroke that the control valve should have inorder to deliver that mass �ow. Knowing the mass �ow, the Kv value would bedetermined from equation 3.42, and the stroke would be determined from thevalve characteristic (�gure 5.3).

As the controller K2 has been designed in the nominal operating point, wherethe control valve has a high gain, we expect if we run the system where a lowwater mass �ow is needed (for example full load trip), the closed loop containinga compensator will have a faster action. Let's see the results. Note that now theouter loop controller K1 should have di�erent parameters. We have manuallytuned the outer loop controller K1 to these values (in order to make it controlall the transients): Kp = −0.1 and Ti = 0.004.

Figure 5.15: Output temperature comaprison full load trip

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Figure 5.16: Stroke comparison full load trip

We can see on �gure 5.16 that the stroke command for the case when a PIcontroller is used for the inner loop, is slower in the region where a low stroke isdemanded. Also that the output temperature is logically reacting slower in thecase of a PI controller (�gure 5.15). As predicted, both the compensator andthe PI controller have the same behavior where a higher stroke is demanded.

5.4.1 Unknown control valve characteristic

If the characteristic of the control valve is unknown, we might establish itthrough some measurement. In this case, we may draw a characteristic, whichis not exactly re�ecting the reality, and our compensator design based on thischaracteristic, will not give the right stroke for the control valve. We want tosee if the outer PI controller will take care of this error. Let's say that our errorlays in a threshold of 15% from reality. In this case, we will multiply our controlvalve characteristc Kv by 0.85 and 1.15 respectively.

It is important to mention that, we should be consistent in choosing theupper saturation limit for the outer loop's ARW. Once the characteristic chosen,although we might be wrong, if the upper saturation limit for the K1 controlleris �xed with the same characteristic (and relation) that has been used to buildthe compensator, then there would be no windup problems.

So let's test to see if by inducing 15% of error in the characteristic of thevalve, we are still faster with the compensator. For this purpose, we havemuliplied the Kv of the valve by a factor of 0.85 and 1.15 (�gure 5.17).

We can see in �gure 5.18, that on one hand, as the time constant of theclosed loop has been decreased by the presence of the inner loop PI controller(in low strokes), the green line corresponding to the stroke command of the PIcontroller in the slowest one.

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Figure 5.17: Uncertain valve characteristic

Also we can see the the command to increase the stroke appears the latestfor the PI controller and the most important one, the slope of the stroke in thisregion (around t=4400 s) is less steep than the curves for the compensator.

Figure 5.19 shows that the outlet air temperature response is also the slowestfor the PI controller, and that the overshoot is the most important for the PIcontroller.

Figure 5.18: Stroke for uncertain case (full load trip)

Logically thinking, for the same reference mass �ow into the compensator(same expected Kv), the Kv ∗0.85 compensator should give a higher stroke, andhence there would be more water in the system, and the output temperature

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Figure 5.19: Temperature for uncertain case (full load trip)

would be colder. But this is not what we see in �gures , 5.19, and 5.20.

Figure 5.20: Zoom of temperature in case of uncertainty (full load trip)

To understand this, let remove the feedback loop, and reason in open loop(same reference mass �ow : mFF ). Figure 5.21 shows that our theory was right: the output temperature is colder for the case Kv ∗ 0.85. But as the inletair temperature decreases, the steepest slope for the outlet temperature is forthe case of Kv ∗ 1.15, and this is independent from the reference mass �ow,because it is the same for all. This shows that at this point, the same changein the reference mass �ow, ends up in a bigger change in the stroke for thecase Kv ∗ 1.15 than respectively the cases Kv and Kv ∗ 0.85. It is important

66

to mention that, right after t=4000s, these slopes are not the consequence ofthe amount of water into the system, because of the high time constant of thesystem. So these are simply the di�erent ways of outlet air cooling down underthe e�ect of inlet air cooling down, regarding the compensator which has beenembedded. This justi�es why the air outlet temperature is warmer for the caseKv ∗ 0.85 in �gure 5.20.

Figure 5.21: Temperature pro�le in case of no feedback

The error then resulting of the behavior of the outlet air temperature, willde�ne a stroke for inlet water: so we understand that �gure 5.18 is a consequenceof �gure 5.19 rather than vice versa.

So we can conclude that even if the characteristic used to build the com-pensator is wrong by 15%, we are still better than the PI controller in the lowstroke zone.

5.4.2 Uncertainty on the control valve characteristic

Most of the time, the valve characteristic is given by the manufacturer. But Evenif we knew exactly the valve characteristic, it may change when the control valveis installed on the system. The length of the pipes connecting the valve to therest of the system, and their con�guration, can induce more or less friction tothe �uid going through the valve.

The risk of using a compensator is that for its construction, we use char-acteristics given by the manufacturer (Kv), which are not the reality in oursystem. Then the compensator might not give the good signal to the stroke ofthe valve, and the choice of a PI controller might be more justi�ed in this case.

But also in this case, the outer loop will take care of the inaccuracy. Thecrucial constraint here, is again the upper saturation limit for the ARW of the

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outer loop's controller K1 (the lower limit is always 0). If this limit is chosenwith the same characteristic that is used to build the compensator, there willbe no windup problems (and this is possible, because if we are the ones whobuild the compensator, we can also determine its upper input saturation limit).

So then, according to its reference mass �ow (which will not saturate itsinput, because of what has been said in the previous paragraph), the compen-sator will generate a stroke signal between 0 and 1, which will be the inputfor the water control valve (Note that the control valve is always calibrated tohave the right stroke at the nominal mass �ow). The K1 controller will makethe necessary changes to the reference mass �ow to have the good outlet airtemperature.

For changes of the control valve characteristic by 15%, we can expect thesame results as in �gures 5.16 and 5.15 to happen, because a direct comparisonbetween a PI controller and a compensator designed for the nominal valve char-acteristic (given by the manufacturer), used to control the same control valvewhich does not have the nominal characteristic will be performed. Of course,a comparison between the cases when the control valve has the characteristicsKv, Kv ∗ 0.85 and Kv ∗ 1.15 is meaningless (because we want to compare thetwo controllers for a given control valve).

The comparison has already been done for the case of the normal controlvalve (with Kv as characteristic) in �gures 5.15 and 5.16. Now let's see ifthe characteristic of the control valve in reality is Kv ∗ 0.85, whether or not acompensator is still better than a PI controller :

Figure 5.22: Comparison PI / Compensator for the case Kv*0.85

We can see that the compensator is still much faster at low strokes. Ataround time t=4400s, we can clearly see that the slope of the stroke produced bythe compensator is steeper than the one for the PI controller. At time t=4400s,a higher stroke for the compensator leads to a colder air outlet temperature.This is why we see an overshoot for the PI controller.

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For the case in which the real characteristic of the valve is Kv ∗ 1.15, weobtained these results:

Figure 5.23: Comparison PI / Compensator for the case Kv*1.15

Also here, the same arguments are valid and the compensator behaves better.It is important to mention that for the two cases, the PI controller has been ableto control the system. This means that the cascade combination (controllers K1and K2) is robust for these approximations on the water inlet control valve.

Figure 5.24: Outlet temperature comparison PI on uncertain Control valve

We can also see that the K1 controller combined to the compensator is alsorobust for these uncertainties. Next �gure shows the way the K1 / compensatorcombination can control the system, in case of di�erent characteristics of the

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control valve.

Figure 5.25: Output temperature in case of incertain Kv for compensator

The stroke has been shown on �gure 5.26. Of course, if the in reality thecharacteristic is bigger than thought, we will need less stroke to achieve thedesired mass �ow into the system.

Figure 5.26: Stroke in case of incertain Kv for compensator

As the all the control valves have been matched to have the nominal mass�ow at the maximum stroke, at the beginning, when the maximum water mass�ow (7.533kg/s) is needed to have the speci�ed outlet air temperature, thestroke given by the compensator is the same (1) for all the control valves. Thiscan be seen in next �gure, which shows the commanded mass �ows (referencefor the compensator) and the mass �ows produced by the control valve whichgoes into the OTC.

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Figure 5.27: Refernce mass �ows for the compensator

As expected, on the right subplot of �gure 5.27, we can see that the com-manded mass �ow for the case the characteristic of the valve is Kv ∗ 1.15 is thelowest one. At the beginning, when a maximum mass �ow is needed, the strokeis equal to 1 as predicted before.

We can now conclude, that even if the characteristic of the control valve isunknown and we establish it with 15% of error, or if it is known, and changes inthe reality (and our compensator is built based on the original characteristic),using a static compensator will always give better results in regions where a lowstroke is needed.

5.5 Comparison for further transients

Now that we know the bene�t of using a compensator instead of the K2 con-troller, we will see if the K1 / Compensator combination, with the chosen pa-rameters for the K1 controller, is robust enough to survive the other transients(we will not do the tests for cases of uncertainty, the results of last section showthat there is no need for this) We will do the same tests for the K1 / K2 cascadecombination, and we will compare the results.

5.5.1 Start up

At start up, the power of the compressor increases gradually. The mass �ow andtemperature outputs of the compressor are shown in appendix G. We alreadydid the test for the K1/K2 combination (PI) in section 5.3. Let's do the testfor the K1/Compensator combination.

Although the outlet air temperature for the case where a compensator isused shows some oscillations, the overshoot is smaller of about 5[K].

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Figure 5.28: Air Temperature output comparison at start up PI/compensator

In �gure 5.29, we can see on the zoom, that again, at low strokes, thecompensator acts much faster than the PI controller. This behaviour leads thecompensator to produce even a stroke of 1 at around t=4400s. This injects morewater into the system, and cools down the air outlet. This is why the overshootis much smaller for the K1/Compensator combination.

Figure 5.29: Stroke comparison at start up PI/compensator

5.5.2 Shut down

In this transient, the air temperature and mass �ow out of the compressordecrease, as the full load trip, but in a longer time. The compressor �rst deloads,then stays idle for about 8 minutes, and runs down after wards.

We can see in �gure 5.30, that the air outlet temperature in case a compen-sator is used, stays closer to the set temperature (613.15 [K]).

Figure 5.31 shows again that the compensator reacts faster to the distur-

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Figure 5.30: Air Temperature output comparison at shut down PI/compensator

bances than the PI controller.

Figure 5.31: Stroke comparison at start up PI/compensator

5.5.3 Load rejection

The load rejection transient pro�le, in terms of outlet air temperature and mass�ow of the compressor, is similar to the full load trip, except that the changesin temperatures and mass �ows are smaller. In this transients, the compressordeloads for about 150 seconds, and then remains idle.

The temperature pro�le again shows a smaller over shoot for the outlet airtemperature in the case of K1/Compensator combination : at t=4600s, theovershoot for this combination is about 2[K], while for the K1/K2 combination(PI), it is about 7[K] (see �gure 5.32, zoom).

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Figure 5.32: Air Temperature output comparison at load rejection PI / com-pensator

The stroke also shows a faster reaction for the K1/Compensator combina-tion. The valve is opened earlier for this combination (at about t=4370s) whichleads to a smaller overshoot for the outlet air temperature on �gure 5.32. Forthe other combination, the stroke is increased at t=4440s, which leads to thementioned overshoot. But this is also a consequence of the fast reaction of thecompensator at t=4000s (stroke=1).

Figure 5.33: Stroke comparison at load rejection PI/compensator

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5.5.4 Emergency switch o�

In case of emergency shut o�, the changes of temperature and mass �ow of thecompressor outlet air are smaller than the load rejection or the full load trip,but happen much faster (in 30 seconds). The compressor goes from base loadto 25% load in this time interval.

Figure 5.34: Air Temperature output comparison at emergency shut o� PI /compensator

The results here are similar to those of load rejection. We can see the over-shoot of air outlet temperature for the K1/K2 combination (PI) at time t=4700s.The slow reaction of this combination on �gure 5.35 is again responsible for thisovershoot.

Figure 5.35: Stroke comparison at emergency shut o� PI/compensator

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Chapter 6

Conclusion and Future scope

In last chapter, we the control system of the OTC, and the improvementsbrought to it. Because of lack of time, we were not able to �nd a good so-lution to linearize the system, and apply the theories of linear system control.The controllers were tuned manually and we tried to obtain the best possibleperformances for all the transients. So what presented in the last chapter wasonly a �rst sketch of what the control system can be.

The biggest challenge was to �nd a controller (or combination of controllers),which would control the system, in cases of full load trip and startup. Thisbecause, the transients shut down, load rejection and emergency shut o� havea similar pro�le to the full load trip, in terms of inlet air evolution (mass �owand temperature). We experienced that a controller tuned for the case of fullload trip, if it has a minimum robustness, would suit the cases of shut down,load rejection and emergency shut o�, but not necessarily the start up.

As in the reality, the controllers are also tuned manually (on the existingsystem, the vertical OTC), the risk is big. For example, we tried to get a goodbehavior for the start up (small overshoot) in the case of K1/K2 controllerscombination, but the result for the full load trip case was not acceptable:

Figure 6.1: Air temperature for startup (left) and full load trip(right)

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Comparing to �gure 5.12, the overshoot is much smaller at startup, but highoscillations are seen on the air outlet temperature in case of full load trip.

The problem of designing a controller which would suit the startup and loadfull load trip transients, lays in the fact that the time constant of the OTC isvariable according to the conditions the OTC is in.

In a case similar to the startup, where at the beginning, the water stroke isclosed, and then the air inlet increases in mass �ow and temperature. This leadsto the opening of the water control valve. In this case, the time constant of theOTC is at its minimum. But is a case like the full load trip, the system is hot atthe beginning, and then the air inlet mass �ow and temperature decreases. Thecontrol system sees a lower outlet temperature and reduces the water stroke. Inthese kind of cases, the time constant of the system is at its maximum.

To understand this, assume the OTC is a 100% e�cient heat exchanger asin 5.1. The equation 5.6, will help us understand this. Writing this equationand substituting the enthalpies by temperatures gives:

Qair = maircp,air(Tair,in − Tair,out) (6.1)

In the case of startup, as �gure 5.12 shows, the temperature di�erence ∆Tbetween the inlet and outlet air is zero at the beginning, and increases gradually.Also the inlet air pro�le for the startup (see appendix G) shows that the airinlet mass �ow increases gradually to base load. So mair also increases. Thesetwo changes mean that the heat transfer rate Qair[(J/s)] increases in this case.

But in the case of full load trip, as for example �gure 5.15 shows , thetemperature di�erence ∆T between the inlet and outlet air is high at the be-ginning, and decreases gradually. Also the inlet air pro�le for the full load trip(see appendix G) shows that the air inlet mass �ow decreases gradually frombase load to nearly zero. So mair also decreases. These two changes mean thatthe heat transfer rate Qair[(J/s)] decreases in this case, and leads to a slowerheat transfer, and hence a slower equilibrium achievement.

It is not possible to impose a step to the command of the system when theboundary conditions are changing (the system is not steady state). So to havea rough idea of what the time constant can be for cases like the startup, we willkeep a constant mass �ow, and also a constant high temperature for the inletair, and we will increase the stroke of water from 0 to 1. In this case the ∆T forair increases gradually. This has been already done in appendix F, �gure F.8.The time constant was about 60 seconds.

To have an idea of what the time constant can be in cases like the full loadtrip, we have initialized the OTC at the same conditions, but this time the waterstroke is open at the beginning, at a step is done to 0.6 (and not to 0, because ifwe completely close the control valve, there is no more interaction between airand water, and only the inertia of the whole system (metal masses etc.) playsits role).

The step on inlet water stroke has been shown in �gure 6.2.

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Figure 6.2: Step of inlet water control valve stroke

The inlet and outlet air temperatures under the e�ect of this step commandare in �gure 6.3, left subplot. The comparison is made with the previous casewhich in in the right subplot.

Figure 6.3: outlet Temperature response comparison

We can see that the time constant is much lower for the case of startup likes.In the case which is referred to the full load trip, the time constant is about 160seconds, which is almost 3 times higher than the case which is referred to thestartup.

This can also be explained by the amount of water and/or steam that is inthe system during the transients. In cases like the full load trip, by reducing themass �ow of water into the system, the residence time and the amount of waterin tubes decreases. In this case, after a while, a big part of water is evaporated.We can see in �gure 6.4, that shortly after reducing the water stroke (t=4000s),all the water in cells 2 to 9 is evaporated, and only cell 1 remains two phase. Asthe heat transfer coe�cient for steam is much smaller than water, the response

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of the system is slower. Figure F.6 shows on the other hand, that for cases likethe startup, that only cell 9 on water/steam side contains super-heated steam,and all other cells contain sub-cooled water or a two phase mixture.

So as mentioned in the introduction, the evaporation point in the tubes,plays a major role in determining the dynamics of the system.

Figure 6.4: Evolution of steam qualities during step (full load trip likes)

So in general, the cases where the air inlet mass �ow and temperature in-crease (like startup), should be separated from the cases where the air inlet mass�ow and temperature decrease (like full load trip). Then, after having found ane�cient way to linearize the system for the two cases, for example assuming con-stant boundary conditions, once at low air inlet mass �ow and temperature,inwhich a low water stroke will be needed, and once for the opposite case, anadaptive control design can be considered.

It means that we will design a controller for the operating ranges of start up,and another controller will be designed by a robust control design method (forexample an RST controller via convex optimization) for the cases in the rangeof full load trip (shut down, load rejection, emergency shut o�).

Another solution would be to use the metal temperatures as indicator. Itis actually possible to measure the temperature of the tubes at their bend. Iffor example we have 14 passes of the tubes, we will have 7 bends on whichto measure the metal temperature. Assume the black line to be the metaltemperatures of the bends at nominal conditions in �gure 6.5.

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Figure 6.5: Metal temperature pro�le for bends

The red line shows the actual metal temperatures at the bends. In this case,the outlet air temperature would be too high. A state feed back controller couldbe designed, which according to the temperature deviation at each bend, willproduce a signal for the command (water stroke). This is more in coordinationwith our theory that the evaporation point plays a big role in the dynamics,because the straight lines in �gure 6.5 give us the important information aboutwhere water is evaporating. But the major problem here is, that the systemwill not be commendable anymore. So at the same time, we will have to thinkabout new commands that can a�ect the outlet air temperature.

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Bibliography

[1] Lucien BOREL, Daniel FAVRAT : Thermodynamique et énergétique.Presses Polytechniques et universitaires romandes, 2005

[2] Hubertus TUMMESCHEIT: Design and Implementation of Object-

Oriented Model Libraries using Modelica. Department of AutomaticControl, Lund Institute of Technology, 2002

[3] Frank P. INCROPERA, David P. DEWITT : Fundamentals of Heatand Mass Transfer. 5th edition, John Wiley and Sons, 2002

[4] Fabian HUESSER : GT Dynamic Modeling: Component Models

Documentation. ALSTOM (Switzerland), 2005.

[5] Christoph VOSER : Dynamisches Modell eines Wärmetaschers. AL-STOM (Switzerland), 2005

[6] John R. THOME : Engineering Data BOOK III. Wolverine TubeInc., 2004-2006

[7] K.-H. GROTE, J. FELDHUSEN : Dubbel, Taschenbuch für den

Maschinenbau. 21st Edition, Springer-Verlag Berlin, 2005

[8] Roland LONGCHAMP : Commande Numérique de systèmes dy-

namiques. Preses Polytechniques et Universitaires Romandes, 2006

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List of Figures

2.1 Overview of the Plant . . . . . . . . . . . . . . . . . . . . . . . . 92.2 HP / LP stages of compressor and turbine . . . . . . . . . . . . . 102.3 Vertical (Helix) OTC . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Location of Helix Once-through coolers . . . . . . . . . . . . . . 122.5 Gas Turbine lay down and OTC placement possibility . . . . . . 132.6 Horizontal OTC side view (left) and front cut (right) . . . . . . . 142.7 Side view of a serpentine . . . . . . . . . . . . . . . . . . . . . . . 142.8 Conventional (left), dense (middle) and maximum (right) packing 152.9 Maximum packing 3D view . . . . . . . . . . . . . . . . . . . . . 15

3.1 Tool Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Overview of the Pipe gas . . . . . . . . . . . . . . . . . . . . . . 213.3 Overview of volumeQHE . . . . . . . . . . . . . . . . . . . . . . . 233.4 Overview of a control valve . . . . . . . . . . . . . . . . . . . . . 243.5 Overview of an ori�ce . . . . . . . . . . . . . . . . . . . . . . . . 263.6 State diagram of water/steam . . . . . . . . . . . . . . . . . . . . 273.7 h-T diagramm of water/steam . . . . . . . . . . . . . . . . . . . . 283.8 Overview of VolumeWs . . . . . . . . . . . . . . . . . . . . . . . . 293.9 Overview of VolumeQHEWs . . . . . . . . . . . . . . . . . . . . . 303.10 Overview of a control valve ws . . . . . . . . . . . . . . . . . . . 313.11 Overview of an ori�ce . . . . . . . . . . . . . . . . . . . . . . . . 333.12 Square root vs. approximation . . . . . . . . . . . . . . . . . . . 34

4.1 Simulink assembly of VolumeQHE and VolumeQHEWs . . . . . . 364.2 Temperature evolution for the test . . . . . . . . . . . . . . . . . 374.3 Simulink model of counter-�ow OTC . . . . . . . . . . . . . . . . 374.4 Overview of the heat exchanger components . . . . . . . . . . . . 384.5 Detailed overview of the heat exchanger components . . . . . . . 394.6 Matching of two neighboring heat exchanger components . . . . 424.7 Discretization of the OTC HP . . . . . . . . . . . . . . . . . . . . 47

5.1 Control system of the OTC . . . . . . . . . . . . . . . . . . . . . 515.2 Feed forward mass �ow calculation conventions . . . . . . . . . . 525.3 Characteristic of the inlet water control valve . . . . . . . . . . . 535.4 Mass �ow vs. stroke at steady-state conditions . . . . . . . . . . 545.5 Open loop bode diagram of the for the control valve . . . . . . . 555.6 PI controller for the control valve with ARW . . . . . . . . . . . 565.7 Temperature at full load trip . . . . . . . . . . . . . . . . . . . . 585.8 Stroke at full load trip . . . . . . . . . . . . . . . . . . . . . . . . 58

82

5.9 Simulink diagramm of the moving saturation ARW . . . . . . . . 595.10 Temperature at full load trip (ARW comarison) . . . . . . . . . . 605.11 Stroke at full load trip (ARW comarison) . . . . . . . . . . . . . 605.12 Temperature pro�le in start up (inner loop PI) . . . . . . . . . . 615.13 Stroke for startup (inner loop PI) . . . . . . . . . . . . . . . . . . 615.14 Feed forward mass �ow startup (inner loop PI) . . . . . . . . . . 625.15 Output temperature comaprison full load trip . . . . . . . . . . . 635.16 Stroke comparison full load trip . . . . . . . . . . . . . . . . . . . 645.17 Uncertain valve characteristic . . . . . . . . . . . . . . . . . . . . 655.18 Stroke for uncertain case (full load trip) . . . . . . . . . . . . . . 655.19 Temperature for uncertain case (full load trip) . . . . . . . . . . 665.20 Zoom of temperature in case of uncertainty (full load trip) . . . . 665.21 Temperature pro�le in case of no feedback . . . . . . . . . . . . . 675.22 Comparison PI / Compensator for the case Kv*0.85 . . . . . . . 685.23 Comparison PI / Compensator for the case Kv*1.15 . . . . . . . 695.24 Outlet temperature comparison PI on uncertain Control valve . . 695.25 Output temperature in case of incertain Kv for compensator . . . 705.26 Stroke in case of incertain Kv for compensator . . . . . . . . . . 705.27 Refernce mass �ows for the compensator . . . . . . . . . . . . . . 715.28 Air Temperature output comparison at start up PI/compensator 725.29 Stroke comparison at start up PI/compensator . . . . . . . . . . 725.30 Air Temperature output comparison at shut down PI/compensator 735.31 Stroke comparison at start up PI/compensator . . . . . . . . . . 735.32 Air Temperature output comparison at load rejection PI / com-

pensator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.33 Stroke comparison at load rejection PI/compensator . . . . . . . 745.34 Air Temperature output comparison at emergency shut o� PI /

compensator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.35 Stroke comparison at emergency shut o� PI/compensator . . . . 75

6.1 Air temperature for startup (left) and full load trip(right) . . . . 766.2 Step of inlet water control valve stroke . . . . . . . . . . . . . . . 786.3 outlet Temperature response comparison . . . . . . . . . . . . . . 786.4 Evolution of steam qualities during step (full load trip likes) . . . 796.5 Metal temperature pro�le for bends . . . . . . . . . . . . . . . . 80

E.1 Pressures on water/steam side . . . . . . . . . . . . . . . . . . . . 91E.2 Enthlpies on water/steam side . . . . . . . . . . . . . . . . . . . . 92E.3 Temperatures on water/steam side . . . . . . . . . . . . . . . . . 93E.4 Metal temperatures on water/steam side . . . . . . . . . . . . . . 94E.5 Steam Qualities on water/steam side . . . . . . . . . . . . . . . . 95E.6 Pressures on air side . . . . . . . . . . . . . . . . . . . . . . . . . 96E.7 Temperatures on air side . . . . . . . . . . . . . . . . . . . . . . . 97E.8 Metal Temperatures on air side . . . . . . . . . . . . . . . . . . . 98E.9 Steam Quality vs. Cell number for heat exchanger model at

steady-state conditions . . . . . . . . . . . . . . . . . . . . . . . . 99E.10 Temperature vs. Cell number for heat exchanger model at steady-

state conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

F.1 input step stroke . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

83

F.2 Pressures on water/steam side . . . . . . . . . . . . . . . . . . . . 101F.3 Enthlpies on water/steam side . . . . . . . . . . . . . . . . . . . . 102F.4 Temperatures on water/steam side . . . . . . . . . . . . . . . . . 103F.5 Metal temperatures on water/steam side . . . . . . . . . . . . . . 104F.6 Steam Qualities on water/steam side . . . . . . . . . . . . . . . . 105F.7 Pressures on air side . . . . . . . . . . . . . . . . . . . . . . . . . 106F.8 Temperatures on air side . . . . . . . . . . . . . . . . . . . . . . . 107F.9 Metal temperatures on air side . . . . . . . . . . . . . . . . . . . 108F.10 Steam Quality vs. Cell number for heat exchanger model at

steady-state conditions . . . . . . . . . . . . . . . . . . . . . . . . 109F.11 Temperature vs. Cell number for heat exchanger model at steady-

state conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

84

List of Tables

2.1 HP and LP main characteristics for vertical OTC . . . . . . . . . 122.2 HP and LP main characteristics for horizontal OTC . . . . . . . 16

4.1 Volumes characteristics for test . . . . . . . . . . . . . . . . . . . 36

85

Appendix A

Vertical OTC 2D sketch

86

Appendix B

Horizontal OTC 2D sketch

87

Appendix C

Horizontal OTC HP Data

sheet

88

Appendix D

Horizontal OTC LP Data

sheet

89

Appendix E

Step response of the

counter-�ow model

The imposed step command to the inlet control valve stroke of the water/steamside on the OTC horizontal HP is shown in the �gure which follows:

The input/output responses of the heat exchanger states will be presentedin the �gures that follow.

90

Figure E.1: Pressures on water/steam side

91

Figure E.2: Enthlpies on water/steam side

92

Figure E.3: Temperatures on water/steam side

93

Figure E.4: Metal temperatures on water/steam side

94

Figure E.5: Steam Qualities on water/steam side

95

Figure E.6: Pressures on air side

96

Figure E.7: Temperatures on air side

97

Figure E.8: Metal Temperatures on air side

98

Figure E.9: Steam Quality vs. Cell number for heat exchanger model at steady-state conditions

Figure E.10: Temperature vs. Cell number for heat exchanger model at steady-state conditions

99

Appendix F

Step response of the

cross-�ow model

The imposed step command to the inlet control valve stroke of the water/steamside on the OTC horizontal HP is shown in the �gure which follows:

Figure F.1: input step stroke

The input/output responses of the heat exchanger states will be presentedin the �gures that follow.

100

Figure F.2: Pressures on water/steam side

101

Figure F.3: Enthlpies on water/steam side

102

Figure F.4: Temperatures on water/steam side

103

Figure F.5: Metal temperatures on water/steam side

104

Figure F.6: Steam Qualities on water/steam side

105

Figure F.7: Pressures on air side

106

Figure F.8: Temperatures on air side

107

Figure F.9: Metal temperatures on air side

108

Figure F.10: Steam Quality vs. Cell number for heat exchanger model at steady-state conditions

Figure F.11: Temperature vs. Cell number for heat exchanger model at steady-state conditions

109

Appendix G

Transients

110