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1 Stability Analysis of Condensate System Presented by Mitali Soni Rishikesh Bagwe

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Page 1: Stb of Condensate system

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Stability Analysis of

Condensate System

Presented by

Mitali Soni

Rishikesh Bagwe

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Acknowledgements

We would like to thank Mr. C. Saha (SME (I)) and Mr. Nilesh Bobade (SO/D), our

mentor who guided us throughout the project through his extensive knowledge of the

systems and valuable inputs wherever required. We would also like to thank Mr.

Saumitra Trivedi, ATO, Nuclear Training Center (NTC) and other members of the NTC

at TAPS 3&4 for helping and motivating us throughout the project.

We would also like to thank BITS Pilani for giving us the opportunity to attain this

exposure to industry atmosphere and functioning. Lastly we would like to thank our

Instructors In-charge, Mr. Shamuel Tharu and Mr. Pawan Poddar, as well as our

Practice School batch-mates who have been supportive and encouraging throughout

the duration of the project.

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Preface

The following project is based on the study of Programmable Logic Controllers (PLC)

and their networking at NPCIL’s Tarapur Atomic Power Station, Units 3&4 (TAPS 3&4).

It also gives information about the current profile of the nuclear energy organization,

Nuclear Corporation of India Limited (NPCIL) and its policies of managing the 20

nuclear reactors in the country. It also explains the process of conversion of nuclear

energy into electricity. There are some other benefits of nuclear energy shown in this

report.

The second part of the report emphasizes on the stability analysis of Condensate

system using MATLAB- Simulink.

This report is made in the partial fulfilment of the course Practice School -1(PS 1) of

BITS Pilani on July 9, 2014. The data in the report was gathered from various sources,

the prominent being the manuals available at TAPS 3&4 and the orientation session at

TAPS 3&4. Our mentor at TAPS 3&4, Mr. Nilesh Bobade (SO/D) guided us throughout

the project.

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Introduction

The Secondary Cycle 24

Condensate System 25

Deaerator 26

Feedwater System 27

MATLAB

Simulink 28

Simulink Real Time 29

Stability Analysis of Process Control Loops

Control System Terminology 30

Closed Loop System 32

Table of Contents

Page No.

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Company Profile: NPCIL

Introduction to Nuclear Energy

Working Principle of a Nuclear Reactor

Types of Nuclear Reactors

Pressurised Water Reactor

Boiling Water Reactor

Pressurised Heavy Water Reactor

7

Details of NPCIL plants in India 12

Need for Nuclear Energy in India 14

India’s Three Stage Nuclear Program

Stage 1 - Pressurised Heavy Water Reactors

Stage 2 - Fast Breeder Reactors

Stage 3 - Thorium Based Reactors

15

Advantages and Disadvantages of Nuclear Energy

Advantages

Disadvantages

18

General Description of TAPS 3&4 21

The Power Plant Cycle and Main Systems Involved 22

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Checking the Stability of Condensate System

Transfer Functions 32

PI controller 34

Saturator 36

E/A converter 36

Control Valve Positioner 37

Control Valve 39

Resistance of a Pipe 41

Pressure drop across Low pressure Heaters 43

Gain of Control Valve 46

Transfer function of level control process 48

Transfer function of Deaerator Storage Tank 49

References 50

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Company Profile: NPCIL

Nuclear Power Industry has developed manifold since its inception in India. Studies in

nuclear science in a systematic basis began in India during the late forties with the

establishment of Tata Institute of Fundamental research (TIFR) at Mumbai. Exploitation

of Nuclear energy for generation of electricity has supplied the country with nearly more

of electricity so far.

Keeping in mind the increasing need of industry and global competitive challenges,

Nuclear Power Corporation India Limited (NPCIL) with it’s headquarter at Vikram

Sarabhai Bhavan, Mumbai was started. NPCIL is a Public Sector Enterprise under the

Department of Atomic Energy (DAE), Government of India. It was incorporated on

September 17, 1987 as a Public Limited Company under the Companies Act 1956, with

the objective of operating the atomic power stations and implementing the atomic power

projects for the generation of electricity, in pursuance of the schemes and programmes

of Government of India under the Atomic Energy. The formation of NPCIL was

necessitated to give it operational flexibility and raise financial resources from the

capital market to finance the setting up of the projects.

Bhabha Atomic Research Center (BARC) at Mumbai is a premier institute aiming to

provide quality manpower for NPCIL’s nuclear power projects all over the country for

last 35 years. BARC encompasses fields like agriculture, medicine, computer,

electronics, R & D and other areas which are directly relevant to the development of the

nuclear resources of the country in a very efficient way.

The fundamental of the electricity generation at atomic power station is the generation

of heat by bombarding neutrons on the isotope of U-235. The heat, which is thus being

generated, is used to heat up the water to convert it into steam which is used to rotate

turbines, which further runs the turbo generator, and thus generates electricity. It is

estimate to have Nuclear Power Capacity of 20000 MW to make the country self-

sufficient in electricity production. Considering that Nuclear Power is a safe and

environmentally clean source of power generation and that India has vast thorium

reserves, NPCIL is going to play a leading role in future to meet energy demands of the

country.

With a total capacity of 1400 MW, Tarapur is the largest nuclear power station in India.

The facility is operated by the Nuclear Power Corporation of India Limited (NPCIL). It

was initially constructed with two boiling water reactor (BWR) units of 210 MWe. More

recently, an additional two pressurised heavy water reactor (PHWR) units of 540 MW

each were added

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Introduction to Nuclear Energy

Working Principle of a Nuclear Reactor

Nuclear Reactor is a source of heat, which is produced by self-sustained and controlled

chain reaction within the reactor core. The geometrical boundaries within which the

nuclear fuel, moderator, coolant and control rods are arranged to facilitate production

and control of the nuclear reaction to provide heat energy at desired rate is called the

reactor core.

The natural uranium is used as a fuel in our Pressured Heavy Water Reactors.

Uranium has a natural property to emanate radio-active particles. This element has 3

isotopes i.e. U-238, U-235 and U-234. Only the isotope U-235 which is around 0.7% in

the natural uranium is important for energy production. When thermal neutron strikes

the atom of U-235, fission of U-235 atom takes place breaking it up into two or more

fragments. During this process enormous heat energy is generated along with

production of two to three fast moving neutrons.

These fast moving neutrons are slowed down in the presence of moderator (heavy

water) and its probability to cause further fission with uranium atom increases. This

process continues and self-sustained chain reaction is maintained. This provides the

constant heat energy source. The energy produced in this process is proportional to the

neutron density in the reactor core. Thus the reactor power is regulated by controlling

the absorption of the excess neutrons in the core.

The heat produced in the reactor is used to generate light water steam at high pressure,

which drives the turbo-generator to produce electrical energy.

Types of Nuclear Reactors

Pressurised Water Reactor (PWR)

This is the most common type, with over 230 in use for power generation and several

hundred more employed for naval propulsion. The design of PWRs originated as a

submarine power plant. PWRs use ordinary water as both coolant and moderator. The

design is distinguished by having a primary cooling circuit which flows through the core

of the reactor under very high pressure, and a secondary circuit in which steam is

generated to drive the turbine. In Russia these are known as VVER types - water-

moderated and -cooled.

A PWR has fuel assemblies of 200-300 rods each, arranged vertically in the core, and a

large reactor would have about 150-250 fuel assemblies with 80-100 tonnes of uranium.

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Water in the reactor core reaches about 325°C, hence it must be kept under about 150

times atmospheric pressure to prevent it boiling. Pressure is maintained by steam in a

pressuriser (see diagram). In the primary cooling circuit the water is also the moderator,

and if any of it turned to steam the fission reaction would slow down. This negative

feedback effect is one of the safety features of the type. The secondary shutdown

system involves adding boron to the primary circuit.

A Typical Pressurized Water Reactor (PWR) (Source: http://www.world-nuclear.org/)

The secondary circuit is under less pressure and the water here boils in the heat

exchangers which are thus steam generators. The steam drives the turbine to produce

electricity, and is then condensed and returned to the heat exchangers in contact with

the primary circuit.

Boiling Water Reactor (BWR)

This design has many similarities to the PWR, except that there is only a single circuit in

which the water is at lower pressure (about 75 times atmospheric pressure) so that it

boils in the core at about 285°C. The reactor is designed to operate with 12-15% of the

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water in the top part of the core as steam, and hence with less moderating effect and

thus efficiency there. BWR units can operate in load-following mode more readily then

PWRs.

The steam passes through drier plates (steam separators) above the core and then

directly to the turbines, which are thus part of the reactor circuit. Since the water around

the core of a reactor is always contaminated with traces of radionuclides, it means that

the turbine must be shielded and radiological protection provided during maintenance.

The cost of this tends to balance the savings due to the simpler design. Most of the

radioactivity in the water is very short-lived*, so the turbine hall can be entered soon

after the reactor is shut down.

A BWR fuel assembly comprises 90-100 fuel rods, and there are up to 750 assemblies

in a reactor core, holding up to 140 tonnes of uranium. The secondary control system

involves restricting water flow through the core so that more steam in the top part

reduces moderation.

A Typical Boiling Water Reactor (BWR)

(Source: http://www.world-nuclear.org/)

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Pressurised Heavy Water Reactor (PHWR)

The PHWR reactor design has been developed since the 1950s in Canada as the

CANDU, and more recently also in India. PHWRs generally use natural uranium (0.7%

U-235) oxide as fuel, hence needs a more efficient moderator, in this case heavy water

(D2O).** The PHWR produces more energy per kg of mined uranium than other

designs.

The moderator is in a large tank called a calandria, penetrated by several hundred

horizontal pressure tubes which form channels for the fuel, cooled by a flow of heavy

water under high pressure in the primary cooling circuit, reaching 290°C. As in the

PWR, the primary coolant generates steam in a secondary circuit to drive the turbines.

The pressure tube design means that the reactor can be refuelled progressively without

shutting down, by isolating individual pressure tubes from the cooling circuit.

A Typical Pressurized Heavy Water Reactor (PHWR)

(Source: http://www.world-nuclear.org/)

A CANDU fuel assembly consists of a bundle of 37 half metre long fuel rods (ceramic

fuel pellets in zircaloy tubes) plus a support structure, with 12 bundles lying end to end

in a fuel channel. Control rods penetrate the calandria vertically, and a secondary

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shutdown system involves adding gadolinium to the moderator. The heavy water

moderator circulating through the body of the calandria vessel also yields some heat

(though this circuit is not shown on the diagram above).

Newer PHWR designs such as the Advanced CANDU Reactor (ACR) have light water

cooling and slightly-enriched fuel.

CANDU reactors can readily be run on recycled uranium from reprocessing LWR used

fuel, or a blend of this and depleted uranium left over from enrichment plants. About

4000 MWe of PWR can then fuel 1000 MWe of CANDU capacity, with addition of

depleted uranium. Thorium may also be used in fuel.

(** with the CANDU system, the moderator is enriched (i.e. water) rather than the fuel, -

a cost trade-off.)

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Details of NPCIL plants in India

At present 20 reactors are operating with an installed capacity of 4,780 MWe (including

RAPS-1 of 100 MWe owned by the Government) supplying quality electricity to

consumers in a cost effective manner.

Plant Unit and Location Type Capacity in MW

Existing

TAPS Tarapur, Boisar, Maharashtra

1BWR 2BWR

160 160

RAPS Rawatbhata, Rajasthan

1 PHWR 2 PHWR 3 PHWR 4 PHWR 5 PHWR 6 PHWR

100 200 220 220 220 220

MAPS Kalpakkam, Tamil Nadu

1 PHWR 2 PHWR

220 220

KGS Kaiga, Karnataka

1 PHWR 2 PHWR 3 PHWR 4 PHWR

220 220 220 220

NAPS Narora, Uttar Pradesh

1 PHWR 2 PHWR

220 220

KAPS Kakrapar, Gujarat

1 PHWR 2 PHWR

220 220

Under Construction

Kudankulam, Tamil Nadu 1 LWR 2 LWR

1000 1000

Kakrapar, Gujarat 3 PHWR 4 PHWR

700 700

RAPS Rawatbhata, Rajasthan

7 PHWR 8 PHWR

700 700

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Source: www.npcil.nic.in

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Need for Nuclear Energy in India

Electricity is a basic input which is closely related to the economic development of a

country. In spite of the impressive strides in increasing overall installed capacity in the

country, we are still facing shortages. In a world that is increasingly demanding more

energy, resources are getting stressed and strained. Countries and governments today

have to make careful choices to generate growth

that is as much about wealth as it is about well-

being of its citizens.

Options available for commercial electricity

generation are hydro, thermal, nuclear and

renewables. In the energy planning of the country,

a judicious mix of the different types of energy

sources is an important aspect. Diversified energy

resource-base is essential to meet electricity

requirements and to ensure long-term energy

security.

Nuclear energy is a clean and sustainable source of energy. It has the potential and

capability to contribute significantly to India’s quest for long-term energy security. The

conventional fossil fuels are fast depleting in supply. Moreover, with driving global

concerns all over the world about climate change and deployment of environment

friendly power generation technologies, nuclear power has clear advantages to meet

the dual challenge of managing consumption as well as developing low-carbon

energies. India has abundant reserves of Thorium that have the potential to be utilised

to generate nuclear power – an option that is clean, sustainable and economically

viable.

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India’s Three Stage Nuclear Program

India has been following a three-stage nuclear power programme, which aims at the

development of

1. Pressurized Heavy Water Reactors, (PHWR) based on natural uranium.

2. Fast breeder reactors utilizing plutonium-uranium fuel cycle, and

3. Breeder Reactors for utilization of thorium.

The ultimate focus of the programme is on enabling the thorium reserves of India to be

utilised in meeting the country's energy requirements. Thorium is particularly attractive

for India, as it has only around 1–2% of the global uranium reserves, but one of the

largest shares of global thorium reserves at about 25% of the world's known thorium

reserves.

THE THREE-STAGE PROGRAMME

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(Source: www.conceptualphysicstoday.com )

Stage 1 – Pressurized Heavy Water Reactors

In the first stage of the programme, natural uranium fuelled pressurised heavy water

reactors (PHWR) produce electricity while generating plutonium-239 as by-product.

PHWRs was a natural choice for implementing the first stage because it had the most

efficient reactor design in terms of uranium utilisation, and the existing Indian

infrastructure in the 1960s allowed for quick adoption of the PHWR technology. India

correctly calculated that it would be easier to create heavy water production facilities

(required for PHWRs) than uranium enrichment facilities (required for LWRs).

Natural uranium contains only 0.7% of the fissile isotope uranium-235. Most of the

remaining 99.3% is uranium-238 which is not fissile but can be converted in a reactor to

the fissile isotope plutonium-239. Heavy water is used as moderator and coolant.

Indian uranium reserves are capable of generating a total power capacity of 420 GWe-

years, but in order to ensure that existing plants get a lifetime supply of uranium, it

becomes necessary to limit the number of PHWRs fuelled exclusively by indigenous

uranium reserves. US analysts calculate this limit as being slightly over 13 GW in

capacity. Several other sources estimate that the known reserves of natural uranium in

the country permit only about 10 GW of capacity to be built through indigenously fuelled

PHWRs. The three-stage programme explicitly incorporates this limit as the upper cut

off of the first stage, beyond which PHWRs are not planned to be built.

Stage 2 – Fast Breeder Reactors

In the second stage, fast breeder reactors (FBRs) would use a mixed oxide (MOX) fuel

made from plutonium-239, recovered by reprocessing spent fuel from the first stage,

and natural uranium. In FBRs, plutonium-239 undergoes fission to produce energy,

while the uranium-238 present in the mixed oxide fuel transmutes to additional

plutonium-239. Thus, the Stage II FBRs are designed to "breed" more fuel than they

consume. Once the inventory of plutonium-239 is built up thorium can be introduced as

a blanket material in the reactor and transmuted to uranium-233 for use in the third

stage.

The surplus plutonium bred in each fast reactor can be used to set up more such

reactors, and thus grow the Indian civil nuclear power capacity till the point where the

third stage reactors using thorium as fuel can be brought online, which is forecasted as

being possible once 50 GW of nuclear power capacity has been achieved.

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The uranium in the first stage PHWRs that yield 29 EJ of energy in the once-through

fuel cycle, can be made to yield between 65 and 128 times more energy through

multiple cycles in fast breeder reactor

Stage 3 – Thorium Based Reactors

Stage III Reactor or an Advanced Nuclear Power System involves a self-sustaining

series of Thorium-232 and Uranuim-233 fuelled reactors. This would be a thermal

breeder reactor, which in principle can be refuelled – after its initial fuel charge – using

only naturally occurring thorium. According to the three-stage programme, Indian

nuclear energy could grow to about 10 GW through PHWRs fuelled by domestic

uranium, and the growth above that would have to come from FBRs till about 50GW ,

The third stage is to be deployed only after this capacity has been achieved.

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Advantages and Disadvantages of Nuclear Energy

Advantages

1 .Amount of Fuel Needed

With little fuel large amounts of energy are obtained. This saves on raw materials but also in transport, handling extraction nuclear fuel. The cost of fuel is 20% of the cost of energy generated.

2. Production of electric energy is continuous.

A nuclear power plant is generating electricity for almost 90% of the hours of the year. This reduces the price volatility that exist in other fuels such as petrol. The fact that is also conducive to continuous electrical planning as no such dependency in natural aspects.

3. An alternative to fossil fuels

Thus, need not consume as much of carbon fuels like oil, so therefore the problem of global warming, which is believed to have reduced one more important influence on climate change on the planet. By reducing the consumption of fossil fuels we also improve the quality of the air we breathe with all that this implies in the decline of disease and quality of life. Interestingly, nuclear power plants these days carry a zero carbon footprint policy!

4. Cheap electricity

The cost of uranium which is used as a fuel in generating electricity is quite low. Also, set up costs of nuclear power plants is relatively high while running cost is low. The average life of nuclear reactor range from 4.-60 years depending upon its usage. These factors when combined make the cost of producing electricity very low. Even if the cost of uranium rises, the increase in cost of electricity will be much lower.

5. Low Fuel Cost

The main reason behind the low fuel cost is that it requires little amount of uranium to produce energy. When a nuclear reaction happens, it releases million times more energy as compared to traditional sources of energy

6. Nuclear power plants don't require a lot of space

They have to be built on the coast, but do not need a large plot like a wind farm

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Disadvantages

1. Efficiency

The use of nuclear energy for conversion into mechanical energy is very low.

2. Security in their use remains the responsibility of individuals.

Although there are many automated safety systems at nuclear power plants, people can make wrong or irresponsible decisions. A series of bad decisions led the worst nuclear accident in Chernobyl. Once an accident has occurred, the way how it is managed is also dependent on the decisions made by people who are in office.

3. Use that can be given to nuclear power in the defence industry.

Interestingly, nuclear debuted in front of the world as two bombs dropped on Japan at end World War II.

1. Generation of nuclear waste

The difficulty to manage the waste and it takes many years to lose its radioactivity and danger.

5. Nuclear reactors, once constructed, have an expiration date.

After this date must be dismantled, so that in the main countries producing nuclear energy to maintain constant the number of operating reactors should be built about 80 new nuclear reactors the next ten years. The investment for the construction of a nuclear plant is very high and must be recovered in no time, so this raises the cost of electricity generated. In other words, the energy generated is cheap compared to the cost of fuel, but having to repay the construction of the nuclear plants significantly more expensive.

6. Nuclear power plants are targets for terrorist organizations

7. Generation of external dependence.

Shortly countries have uranium mines and not all countries have nuclear technology, so both have to be hired abroad

8. Current nuclear reactors work by fission nuclear reactions.

These chain reactions occur so that if the control systems should fail every time more and more reactions would occur to cause a radioactive explosion that would be virtually impossible to contain

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General Description of TAPS-3&4

Tarapur Atomic Power Station (T.A.P.S.) is located in Tarapur, Maharashtra (India). It

was initially constructed with two boiling water reactor (BWR) units of 210 MWe each

initially by Bechtel and GE under the 1963 123 Agreement between India, the United

States, and the International Atomic Energy Agency. The capacity of units 1 and 2 was

reduced to 160 MWe later on due to technical difficulties. Units 1 and 2 were brought

online for commercial operation on October 28, 1969. These were the first of their kind

in Asia. More recently, an additional two pressurised heavy water reactor (PHWR) units

of 540 MW each were constructed by L &T and Gammon India, seven months ahead of

schedule and well within the original cost estimates. Unit 3 was brought online for

commercial operation on August 18, 2006, and Unit 4 on September 12, 2005.

With a total capacity of 1400 MW, Tarapur is the largest nuclear power station in India.

The facility is operated by the Nuclear Power Corporation of India Limited (NPCIL)

Tarapur nuclear plant has received the highest safety awards given to any electricity

producing plants in India.

Source: www.thehindubusinessline.com

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The Power Plant Cycle and Main Systems Involved

The conversion to electrical energy takes place indirectly, as in conventional thermal

power plants. The heat is produced by fission in a nuclear reactor (a light water reactor).

Directly or indirectly, water vapour (steam) is produced. The pressurized steam is then

usually fed to a multi-stage steam turbine. After the steam turbine has expanded and

partially condensed the steam, the remaining vapour is condensed in a condenser. The

condenser is a heat exchanger which is connected to a secondary side such as a river

or a cooling tower. The water is then pumped back into the nuclear reactor and the

cycle begins again.

Nuclear Power Plant Cycle

(Source: www.ems.psu.edu )

(The power plant cycle showing the reactor vessel, control rods, reactor, steam generator, pumps,

generator, turbine, and condenser and cooling tower.)

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Introduction

The Secondary Cycle

In nuclear power plant the heat generated by nuclear fission of uranium is used to heat

the DM (de-mineralized) water. The cycle of DM light water in which it gets heated and

the steam formed turns the turbine and again gets condensed, is called as the

secondary cycle. This cycle is common to most types power plants (like Thermal, geo-

thermal, diesel power plants).

The system on the right side in the above image is the secondary cycle. The high

pressure steam generated is first passed through a high pressure (HP) turbine, the

exhausted steam is then reheated in a moisture separator heater and send to two low

pressure (LP) turbine. The low enthalpy exhausted steam is then condensed to water in

a hot well with the help of sea-water and is fed into a steam generator(SG). In SG, the

water is converted to steam and again passed through the HP turbine. This cycle goes

on and the same water is circulated till the plant is active.

The Condensate system and Feed water system are two major systems on the

secondary cycle.

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Condensate System

The part of the secondary cycle from the hot well to the deaerator storage tank is called

the condensate system. The condensed water in the hot well is extracted by 3

Condensate Pumps and sent ahead at an increased pressure of 21 kg/cm2. This water

is passed into Gland Steam Condenser (GSC) where the remaining steam is

condensed. The pure liquid DM water at 400C is sent through a series of Low Pressure

Heaters (LPH) in a set of 2 lines. Here at TAPS 3&4 there are 3 LPH in each line. These

heaters increase the temperature of water in steps i.e. LPH1 heats the water from 40C

to 62C, LPH2 heats the water from 62C to 104C and LPH3 from 104C to 125C. This

heated water is then sent to a deaerator tank where air and other non-condensable

gases which are trapped are removed and this deaerated water is stored in deaerator

storage tank. There are control valves on both the lines to control the flow and

consequently increase the water level in deaerator storage tank.

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Deaerator

Deaerator is Spray cum tray type. The function of deaerator heater is to remove oxygen,

dissolved non-condensable gases and to heat boiler feed water. It works in 2 stages.

In the 1st stage it consists of a pressure vessel in which water and steam are mixed in a

controlled manner.

This steam has many sources:

A. Flash Steam from SGBD (Steam Generator Blown Down)

B. Extraction Steam from MSRs (Moisture Separator Reheater)

C. Drain Steam from HP (High pressure) heaters.

When this occurs, water temperature rises, and all non-condensable dissolved gases

are liberated and removed and the effluent water may be considered free from non-

condensable gases.

In the second stage, the heated water 1st stage is passed in between the surfaces of

trays. There are many trays inside the deaerator. Due to this even small traces of gases

in the condensate is removed.

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Feed water system

The part of the secondary cycle from feed-water suction pump (booster pumps) to

steam generator is called the feed water system. The condensed water from deaerator

storage tank is extracted by 3 booster pumps and passed to 3 boiler feed pumps (BFP)

which increases its pressure. This pressurized water is split into 2 streams and then

heated in High Pressure (HP) Feed water heaters (one on each line). Finally the high

pressure, high temperature water is fed into the Steam Generator (SG).

Between the Condenser and Feedwater Pump, the water is called condensate; between

the Feedwater Pump and the Steam Generator the water is called Feedwater.

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MATLAB

MATLAB (matrix laboratory) is a multi-paradigm numerical computing environment and

fourth-generation programming language. Developed by MathWorks, MATLAB allows

matrix manipulations, plotting of functions and data, implementation of algorithms,

creation of user interfaces, and interfacing with programs written in other languages,

including C, C++, Java, and Fortran.The language, tools, and built-in math functions

enable you to explore multiple approaches and reach a solution faster than with

spreadsheets or traditional programming languages, such as C/C++ or Java.

MATLAB can be used for a range of applications, including signal processing and

communications, image and video processing, control systems, test and measurement,

computational finance, and computational biology.Although MATLAB is intended

primarily for numerical computing, an optional toolbox uses the MuPAD symbolic

engine, allowing access to symbolic computing capabilities. An additional package,

Simulink, adds graphical multi-domain simulation and Model-Based Design for dynamic

and embedded systems.

In 2004, MATLAB had around one million users across industry and academia.

MATLAB users come from various backgrounds of engineering, science, and

economics. MATLAB is widely used in academic and research institutions as well as

industrial enterprises.

Simulink

Simulink, developed by MathWorks, is a data flow graphical programming language tool

for modeling, simulating and analyzing multidomain dynamic systems. Its primary

interface is a graphical block diagramming tool and a customizable set of block libraries.

Simulink is a block diagram environment which supports simulation, automatic code

generation, and continuous test and verification of embedded systems.

Simulink provides a graphical editor, customizable block libraries, and solvers for

modeling and simulating dynamic systems. It is integrated with MATLAB, enabling one

to incorporate MATLAB algorithms into models and export simulation results to

MATLAB for further analysis.

Capabilities:

Building the Model — Model hierarchical subsystems with predefined library

blocks.

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Simulating the Model — Simulate the dynamic behavior of your system and view

results as the simulation runs.

Analyzing Simulation Results — View simulation results and debug the

simulation.

Managing Projects — Easily manage files, components, and large amounts of

data for your project.

Connecting to Hardware — Connect your model to hardware for real-time testing

and embedded system deployment.

Applications:

Model-Based Design

Control Systems

Digital Signal Processing

Communications Systems

Image and Video Processing

Embedded Systems

Mechatronics

Simulink Real Time

Simulink Real-Time is a platform to create real-time applications from Simulink models

and run them on dedicated target computer hardware connected to a physical system. It

supports real-time simulation and testing, including rapid control prototyping, DSP and

vision system prototyping, and hardware-in-the-loop (HIL) simulation.

Simulink Real-Time can extend Simulink models with driver blocks, automatically

generate real-time applications, define instrumentation, and perform interactive or

automated runs on a dedicated target computer equipped with a real-time kernel,

multicore CPU, I/O and protocol interfaces, and FPGAs.

Simulink Real-Time and Speedgoat target computer hardware are expressly designed

to work together to create real-time systems for desktop, lab, and field environments.

Simulink Real-Time can also be used with custom target computer and I/O hardware.

Capabilities:

Setting Up the Real-Time Simulation and Testing Environment

Selecting the Target Computer Hardware

Creating and Controlling a Real-Time Application

Instrumenting a Real-Time Application

Defining Concurrent Execution for a Real-Time Application

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Creating a Standalone Real-Time Application

Using Reconfigurable FPGA I/O Modules

Stability Analysis of Process Control Loops

Control System Terminology

For understanding the stability criteria and stability analysis of process control loops,

following terminology may be used. Thus they have been briefly given in the beginning

for better understanding of the system.

Bode Diagram

A plot of log amplitude ratio and angle values on a log frequency base for a transfer

function.

Corner Frequency

In the Bode diagram, the frequency where the product wT is unity, is called as the

Corner Frequency.

Damping

Progressive reduction in the amplitude of a cycling system. Critically damped describes

a system which is damped just enough to prevent overshoot following an abrupt

stimulus.

Dead Band (Dead Zone)

The change through which the input to an instrument can be varied without initiating

instrument response.

Dead Time

Time that elapses between application of input signal to the system and the system

starts to respond.

Derivative Time

The time interval by which the effect of proportional action is advanced.

Frequency Response Analysis

A system of dynamic analysis which consists of applying sinusoidal changes to the input

and recording both input and output on the same time using an oscillograph.

Gain

The ration of change in output to input which caused it.

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Gain Margin

It is the factor by which the gain of a system can be increased to drive it to the verge of

instability. It can also be defined as the reciprocal of the gain at which phase angle

becomes -180 in the bode plot.

Gain Crossover Frequency

The frequency at which amplitude ratio is unity is called gain crossover frequency.

Hysteresis

The maximum difference in output, at any measured value within the specified range,

when the value is approached first increasing and then with decreasing measurement.

Maximum Overshoot

Largest deviation of the output over the step input during the transient cycle.

Phase Crossover Frequency

The frequency at which the phase angle is -180 is called the phase crossover

frequency.

Phase Margin

The amount of additional phase lag which can be increased to drive it to the verge of

instability.

Time Constant

The product of resistance and capacitance T=RC which becomes the time required for

the first order system to reach 63.2% of a total change when forced by a step input.

Time constant defines the speed of response of a system. In higher order systems there

is a time constant for each of the first order components.

Transportation Lag

A delay caused by the time required for the material to travel from one point to another.

E.g. water flowing at the rate 10feet/sec requires 10 sec to travel 100feet and if this

100feet exists between manipulation and measurement, it would constitute a 10sec lag.

Open Loop System

Open-loop system, also referred to as non-feedback system, is a type of continuous

control system in which the output has no influence or effect on the control action of the

input signal. In other words, in an open-loop control system the output is neither

measured nor ―fed back‖ for comparison with the input. Therefore, an open-loop system

is expected to faithfully follow its input command or set point regardless of the final

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result.

Closed Loop System

A Closed-loop Control System, also known as a feedback control system is a control

system which uses the concept of an open loop system as its forward path but has one

or more feedback loops (hence its name) or paths between its output and its input. The

reference to ―feedback‖, simply means that some portion of the output is returned ―back‖

to the input to form part of the systems excitation.

Transfer Function

A Transfer Function is the ratio of the output of a system to the input of a system, in the

Laplace domain considering its initial conditions and equilibrium point to be zero. If we

have an input function of X(s), and an output function Y(s), we define the transfer

function H(s) to be:

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Checking the stability of the Condensate System in Matlab-Simulink

Transfer Functions

Transfer Function of a Controller

Assuming no reset and derivative action, the transfer function of the P-Controller with Time constant τc and dead time τd can be written as

𝐺𝑐 =𝐾𝑝

1+sτ𝑐e-td*s

Assuming no reset action, the transfer function of PD Controller can be written as

Gc= Bagwe Eqn

Where Td is the derivative time in seconds.

As the gain of the controller is unity and being a digital controller it has a cycle time of

175 msec which can be treated as a pure dead time.In the design of DPHS-PCS, the

output rate of rise is in ramp fashion. Hence when any step change is expected in the

output, the final value is achieved through a ramp function. The slope of the ramp is

adjustable. For TAPS 3 & 4 purpose this rate has been selected as 5% of Span per sec.

Thus it takes 20sec to reach the final value. Therefore the time constant can be taken

as

τc =20

5= 4 sec

Hence, proportional controller transfer function can be written as

𝐺𝑐 =𝐾𝑝𝑒

−0.175𝑠

4𝑠 + 1

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1. PI without dead time

𝑇𝐹 =66.4𝑠 + 1.66

40𝑠

MATLAB Code:

>>num=[66.4 1.66] >>den=[40 0] >>sys=tf(num,den) sys = 66.4 s + 1.66 ------------- 40 s >>bode(sys) >>margin(sys) >> [Gm, Pm, Wgm, Wpm]=margin(sys)

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2. PI with Dead time

𝑇𝐹 =−5.81𝑠2 + 66.25𝑠 + 1.66

3.5𝑠2 + 40𝑠

MATLAB Code: >>num=[-5.81 66.25 1.66] >>den=[3.5 40 0] >>sys=tf(num,den) sys = -5.81 s^2 + 66.25 s + 1.66 -------------------------- 3.5 s^2 + 40 s >>bode(sys) >>margin(sys) >> [Gm, Pm, Wgm, Wpm]=margin(sys)

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Results: Gm = 0.6024 Pm = Inf Wgm= Inf Wpm = NaN

3. Gain (Saturator)

Converts 1-100 output of PI Controller to 4-20mA current output

𝐺1 =20 − 4

100 − 0= 0.16

𝑇𝐹1 =−0.9296𝑠2 + 10.6𝑠 + 0.2656

3.5𝑠2 + 40𝑠

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Results: Gm= 3.7651 Pm = 105.3324 Wgm= Inf Wpm = 0.0069

4. E/A Converter E/A converter receive current output from the (4-20A) from PI controller and sends pneumatic signal (3-15psi) to the control valve positioner. Its transfer function can be computed in terms of psi per mA input signal from the controller. Hence for feed valve E/A converter, the gain can be computed as

𝐺𝑎𝑖𝑛 = 𝐴1 =15 − 3

20 − 4= 0.75 𝑝𝑠𝑖/𝑚𝐴

Time constant τ2 of the E/A converter (as obtained from them manufacturer’s data sheet) is computed as follows: As per specification the corner frequency (fc) for E/A converter is 2Hz. At the corner frequency, the product of corner frequency and time constant is unity.

ωc * τ2= 1 Where, ωc is corner frequency of E/A converter and τ2 is a time constant of E/A converter ωc=2πfc

ωc=2π*2 ωc=4π

We have, ωc * τ2= 1 Substituting ωc=4π in the above equation, we get,

τ2 =1

4𝜋 𝑠𝑒𝑐

τ2 = 0.078 seconds

Hence, transfer function of E/A converter of feed control valves is,

𝑇𝑓 =0.75

0.078𝑠 + 1

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Results: Pm =Gm=Inf

Combined Transfer Function:

𝑇𝐹2 =−0.6972𝑠2 + 7.95𝑠 + 0.1992

0.273𝑠3 + 6.62𝑠2 + 40𝑠

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Results: Gm = 9.4890 Pm = 101.4157 Wgm = 20.5641 Wpm = 0.0051

Control Valve Positioner

A position controller (servomechanism) that is mechanically connected to a moving part

of a control valve or its actuator and that automatically adjusts its output to the actuator

to maintain a desired position in proportion to the input signal.

The positioner converts the pressure applied on the valves actuator(3 – 15 psi) into the

percent opening of closing member of the control valve.

𝐺𝑎𝑖𝑛 = 𝐴3 =100 − 0

15 − 3= 8.33%

To fully open i.e. from 0 to 100% it has 20 sec, so the time constant of the positioned is

20/4 = 5 sec.

So the transfer function of the positioner becomes ;

Tf = 8.333

5𝑠+1

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Results: Gm = Inf Pm = 96.8969 Wgm = NaN Wpm =1.6535 Combined transfer function:

𝑇𝐹3 =−5.807𝑠2 + 66.223𝑠 + 1.659

1.365𝑠4 + 33.37𝑠3 + 206.62𝑠2 + 40𝑠

Results: Gm= 23.9750 Pm = 117.6771 Wgm = 6.9841 Wpm = 0.2664

Transfer function of a Control Valve

Control Valve

Process plants consist of hundreds, or even thousands, of control loops all networked

together to produce a product to be offered for sale. Each of these control loops is

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39

designed to keep some important process variable such as pressure, flow, level,

temperature, etc. within a required operating range to ensure the quality of the end

product. Each of these loops receives and internally creates disturbances that

detrimentally affect the process variable, and interaction from other loops in the network

provides disturbances that influence the process variable. To reduce the effect of these

load disturbances, sensors and transmitters collect information about the process

variable and its relationship to some desired set point. A controller then processes this

information and decides what must be done to get the process variable back to where it

should be after a load disturbance occurs. When all the measuring, comparing, and

calculating are done, some type of final control element must implement the strategy

selected by the controller. The most common final control element in the process control

industries is the control valve. The control valve manipulates a flowing fluid, such as

gas, steam, water, or chemical compounds, to compensate for the load disturbance and

keep the regulated process variable as close as possible to the desired set point.

For a control valve :

Cv = 𝑄𝑐𝑣 ×√𝑆𝐺

√∆𝑃𝑐𝑣

Where,

Cv = Flow coefficient

Qcv = flow through the valve

∆PCV = Pressure drop across the valve

SG = specific gravity

Since we have water SG= 0.992 and √(0.992) = 0.9959

Cv = 𝑄𝑐𝑣 ×0.9959

√∆𝑃𝑐𝑣

The inherent characteristics links the Cv with the % opening of the valve. From this we can link % opening with flow from above equation.

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The inherent characteristics of the control from the manual.

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Calculation of Pressure Drop across Control Valve For calculating pressure drop across the control valve we should first calculate its upstream pressure and downstream pressure. And for this we should calculate the pressure drops across each pipes and low pressure heaters in the condensate system.

Resistance of a pipe of flow

The connecting pipes in the condensate system offer resistance due to which there is a pressure drop across them. The pressure depends on the resistance and flow; and the resistance depends on pipe geometric and fluid characteristics. ∆P = R × w2 Where, ∆P = pressure drop across the pipe R = Resistance of the pipe w = velocity of flow

R = λ × 𝐿

𝐷 ×

ρ

2

Where, λ = pipe friction coefficient L = length of pipe D = diameter of pipe ρ = density of fluid

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Reynolds number(Re) plays a role in determining the pipe friction coefficient

Re = 64

λ

If Re < 2320, then it’s a laminar flow

Re > 2320, then it’s a turbulent flow

λ = 64

𝑅𝑒; but Re =

𝑤×𝐷

𝑣 ; where v = kinematic viscosity.

∆Ppipe = 64×𝑣×𝐿×ρ

(𝐷^2)×2 × w

∆Ppipe = 81.536×𝑣×𝐿×ρ

𝐷4 ×2 × Q ; since w =

1.274

(𝐷^2) × Q

Q = flow through the pipe.

Using the above formula, and calculating the actual lengths and diameters of

pipes from the field, we calculated the pressure drops across various pipes.

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Calculation of Resistances of Low Pressure Heaters(LPH) : LPH1

Calculation of Frictional Head Loss through Tubes

for given flow rate, Q, pipe diam., D, pipe length, L, number of tubes, N,

pipe roughness, e, and fluid properties, r & m.

1. Determ. Frict. Factor, f, assuming completely turbulent flow { f = [1.14 + 2 log10(D/e )]-2

}

Inputs No. of tubes, N = 589.0

Pipe Diameter, D 18.05 mm Calculations

Pipe Roughness, e 0.15 mm Pipe Diameter, D 0.018 m

Pipe Length, L 9.677 m Friction Factor, f 0.03559

Pipe Flow Rate, Q 0.83333 m3/s Cross-Sect. Area, A = 0.150716 m2

Fluid Density, r 988.039 kg/m3 Ave. Velocity, V 5.529 m/s

(tubeside fluid)

Fluid Viscosity, m 0.000547 N-s/m2 Reynolds number, Re 180,270

(tubeside fluid)

2. Check on whether the given flow is "completely turbulent flow"

(Calculate f with the transition region equation and see if it differs from the one calculated above.)

f = {-2*log10[((e /D)/3.7)+(2.51/(Re*(f1/2

))]}-2

Transistion Region Friction Factor, f: f = 0.0360

Repeat calc of f using new value of f: f = 0.0360

Repeat again if necessary: f = 0.0360

3. Calculate hL and DPf, using the final value for f calculated in step 2

(hL = f(L/D)(V2/2g) and DPf = rghL)

Frictional Head Loss, hL 30.08737 m

Frictional Pressure

Drop, DPf 291627 N/m2

Frictional Pressure

Drop, DPf 291.6 kN/m2

Pressure Drop 2.974593

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LPH2

Calculation of Frictional Head Loss through Tubes

for given flow rate, Q, pipe diam., D, pipe length, L, number of tubes, N,

pipe roughness, e, and fluid properties, r & m.

1. Determ. Frict. Factor, f, assuming completely turbulent flow { f = [1.14 + 2 log10(D/e )]-2

}

Inputs No. of tubes, N = 958.0

Pipe Diameter, D 15 mm Calculations

Pipe Roughness, e 0.15 mm Pipe Diameter, D 0.015 m

Pipe Length, L 11.696 m Friction Factor, f 0.03785

Pipe Flow Rate, Q 0.32430 m3/s Cross-Sect. Area, A = 0.169293 m2

Fluid Density, r 971.8007 kg/m3 Ave. Velocity, V 1.916 m/s

(tubeside fluid)

Fluid Viscosity, m 0.000355 N-s/m2 Reynolds number, Re 78,659

(tubeside fluid)

2. Check on whether the given flow is "completely turbulent flow"

(Calculate f with the transition region equation and see if it differs from the one calculated above.)

f = {-2*log10[((e /D)/3.7)+(2.51/(Re*(f1/2

))]}-2

Transistion Region Friction Factor, f: f = 0.0387

Repeat calc of f using new value of f: f = 0.0387

Repeat again if necessary: f = 0.0387

3. Calculate hL and DPf, using the final value for f calculated in step 2

(hL = f(L/D)(V2/2g) and DPf = rghL)

Frictional Head Loss, hL 5.63838 m

Frictional Pressure

Drop, DPf 53753 N/m2

Frictional Pressure

Drop, DPf 53.8 kN/m2

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LPH3

Calculation of Frictional Head Loss through Tubes

for given flow rate, Q, pipe diam., D, pipe length, L, number of tubes, N,

pipe roughness, e, and fluid properties, r & m.

1. Determ. Frict. Factor, f, assuming completely turbulent flow { f = [1.14 + 2 log10(D/e )]-2

}

Inputs No. of tubes, N = 630.0

Pipe Diameter, D 18.05 mm Calculations

Pipe Roughness, e 0.15 mm Pipe Diameter, D 0.018 m

Pipe Length, L 9.573 m Friction Factor, f 0.03559

Pipe Flow Rate, Q 0.32430 m3/s Cross-Sect. Area, A = 0.161207 m2

Fluid Density, r 958.3665 kg/m3 Ave. Velocity, V 2.012 m/s

(tubeside fluid)

Fluid Viscosity, m 0.000282 N-s/m2 Reynolds number, Re 123,402

(tubeside fluid)

2. Check on whether the given flow is "completely turbulent flow"

(Calculate f with the transition region equation and see if it differs from the one calculated above.)

f = {-2*log10[((e /D)/3.7)+(2.51/(Re*(f1/2

))]}-2

Transistion Region Friction Factor, f: f = 0.0362

Repeat calc of f using new value of f: f = 0.0362

Repeat again if necessary: f = 0.0362

3. Calculate hL and DPf, using the final value for f calculated in step 2

(hL = f(L/D)(V2/2g) and DPf = rghL)

Frictional Head Loss, hL 3.95874 m

Frictional Pressure

Drop, DPf 37218 N/m2

Frictional Pressure

Drop, DPf 37.2 kN/m2

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Gain of a control valve From the pressure drops calculated from the above equations we could calculate the total pressure on the upstream and downstream of control valve.

The installed characteristics from the above calculations using ‘cftool’ command in Matlab in its Curve Fitting Tool Box is.

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We have to analyze the stability of the system at maximum change in slope (i.e. where it is mostly likely to be unstable)

The maximum change in slope occurs at flow of 1000 m3/hr. So take the gain of the control valve as 47.16. Dead time = 100 msec So the combined transfer function (with control valve )

TF = 12𝑠3−396.04𝑠2+3119.04𝑠+77.81

0.0682𝑠5+3.033𝑠4+ 43.7𝑠3+ 208.62𝑠2+ 40𝑠

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So bode plot of the system till control valve is

Transfer Function of a Level Control Process Time constant τ =

𝑐

𝑎2

Where,

c=𝑑𝑉

𝑑𝐿 ; unit change in volume of liquid produced by unit change in level at normal

operating conditions.

a2=𝜕𝑄

𝜕𝐿; unit change in flow of liquid through the valve produced by unit change in tank

level keeping valve stroke constant.

Gain Kp = 𝑎1

𝑎2;

Where,

a1=𝜕𝑄

𝜕𝑋 ; unit change in flow of liquid through the valve opening keeping tank level

constant.

TF = 𝐾𝑝

𝜏𝑠+1

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Transfer function of Deaerator Storage Tank Water from deaeartor is stored into the deaerator storage tank (DST). DST is a horizontal cylindrical tank with semi ellipsoidal ends. Dimensions: length=2783mm and height=3980mm and radius=1990mm and width of ellipsoidal end=995mm Volume as a function of height can be calculated for horizontal cylindrical tank with ellipsoidal ends (2:1) V=L (r2acos (1-h/r)-(r-h)(2rh-h2)1/2)+πah2(1-h/3r) Where, r= radius of cylinder V= volume occupied at height h h = height from the bottom of tank a = width of ellipsoidal length Since the the tank is ellipsoidal, the change in height with respect to change in flow will be different at each height. So we are calculating the gain function at the operating point i. e. change in the level per unit change in volume. Operating point: Volume corresponding to 2614mm is 240m3.

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Refrences Websites

www.engineeringtoolbox.com

www.wikipedia.com

www.mathforum.org

www.valvias.com

www.mathworks.com Literature

Condensate System Design Manual

Manufacturing manual of Condensate Extraction Pump(CEP)

Manufacturing manual of Control Valve

Flow sheets of Condensate System

Stability Analysis Paper