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Energy efficiency optimization for Power PlantsTraining Program – Dr.G.G.Rajan

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ENERGY EFFICIENCY OPTIMIZATION FOR POWER

PLANTS 3 DAY TRAINING PROGRAM 19th to 21st December ‘11

Organised byNational Thermal Power Corporation

Vindhyachal

Course Director Dr.G.G.Rajan / Kochi

India


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Program objective

Existing energy scenario all over the world warrants effective control on generation,

utilization and conversion at all levels. Energy utilization must be done more judiciously than the

past, without loss of product quality / quantity and environmental factors. This could be achieved

only by optimization, a special technique involving mathematics, statistics and Operations

research.

The program will explain the concepts of energy efficiency optimization with appropriate

examples. At the end of the program, participants will be able to appreciate the role of energy

efficiency optimization and it’s practical application.

Many examples covered in the program are real life cases and hence participants may

compare their live problems and adopt a suitable problem solving approach.

I thank the program organizers Messrs NTPC,Vindhyachal,India for their efforts in

organizing this program at short notice.

Dr..G.G.Rajan


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CV OF Dr.G.G.Rajan

.

Dr.G.G.Rajan is a Chemical Engineer with qualifications in Mathematics, Statistics,

Management, Operations Research and Computer Applications. With his vast ‘hands-on’

experience and expertise in Refining, Petrochemical and Fertilizer industry, He had developed a

number of process models and had published and presented about 150 papers in national and

Inter National Seminars related to energy , environment , utility management, process

decisioneering etc. He has written a book titled ‘Optimizing Energy Efficiencies in Industry’ ,

Published by Tata McGraw Hill and McGraw Hill USA. He has also written a book titled

‘Practical Energy Efficiency optimization’ published by PennWell,US in 2006.

Rajan.G.G. the Chief Executive ,Techno Software International

Kochi, India since February 2002 was holding the position of

DGM (R&D) in Kochi Refineries Ltd.

Prior to this, he was the Chief Manager in Energy &

Environment section of KRL.He was deputed to Centre for

High Technology , Ministry of Petroleum, New Delhi, India as

an Additional Director in 1988, for six years to accomplish

certain special assignments


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He had won a number of awards for his concepts on Productivity Management. He had

been the Chairman of an Energy Audit Team in 1990s, set up by the Ministry of Petroleum

and Natural Gas for energy auditing of three major refineries ( Chennai Petroleum

Corporation, Kochi Refineries Ltd, HPCL Vizag refinery).

Besides this he had conducted energy auditing of petrochemical plants like Asian

Peroxide Ltd, HOCL,KRL-Aromatics Plant, BPC Aromatics Plant , Fine papers Dubai etc,

Based on this he had developed an auditing software called Technical Audit.

His area of specialization is Energy and Environment Auditing, Performance Evaluation,

Bench Marking , Productivity Analysis and Profit Maximization. By virtue of his contribution to

Technology, he has been selected as a member of International Who is Who Society, Gibralter.

His Software Techno Therm , Technical Audit and SCIMOD are very popular and extensively

used by many industries.

Email: [email protected]

URL: http://business.vsnl.com/ggrtech


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Program Contents

1. Energy efficiency and productivity • Fuel System

• Boiler / Steam System

• Turbine / Power system

• Thermal Efficiency / Energy transmission

• Energy Losses

2. Optimization Basics • Applications for Power Plant operation

• What to Optimize and how

• Cost / benefit Analysis – Evaluation

3. Power Plant Management • Boiler System

• Utilization of Power Management System

* Cogeneration

• Transmission System

• Waste heat recovery

• Total Power Plant

4. Performance Monitoring Techniques • Boilers

• Steam turbines

• Heat recovery steam generators

• Loss control

5. Energy Conservation through energy efficient technology & equipment • Emissivity coatings


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• Enriched air

• High efficiency air heater / burners

• On-line cleaning

• Boilers tuning control

• Turbine monitoring system

• CW/ BFW / CEP Pumps

• Fans & Blowers

• Super Heater

Summing up / Discussion / Q A Session.

Conclusion


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ENERGY EFFICIENCY AND PRODUCTIVITY

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ENERGY EFFICIENCY AND PRODUCTIVITY

Introduction:

All industries consume four resources namely men , machine , materials and money.Of

late Energy is also treated as a resource, though it forms the subset of money resource. Financial

performance of any enterprise is ultimately determined by how effectively these resources are

utilized and how much profit has been generated for the resource inputs, of which energy has a

major role to play. This evaluation is extremely complex as financial performance is linked to

productivity of labor, machine ,material and money (4 M's of management - men, machines,

materials and money) .

These resources being inter-convertible to a limited extent , an optimum resource mix

should be evolved for a given economic scenario for achieving cost-effectiveness of the industry

under consideration . Each resource has an associated cost . An example is a highly automated

industry which has less than 30% of the man-power compared to a conventional industry of

the same magnitude. and production facility .This could be achieved only because of high

degree of automation. This strategy involves heavy capital input . Hence, an economic analysis

of cost of savings in man-power against additional capital investment has to be evaluated and

justified .

Material resource ( comprising hydro carbon feed stoch, fuel etc ) ,is another cost

centre that has an impact on the financial performance of the industry. It is normally felt that

cheaper hydrocarbon feed stock shall generate higher profits for a given set of production facility

comprising primary unit and other conversion processes.. This may not be always true in view of

the fact that the yield pattern of cheaper feed stock are invariably inferior and necessitate high

degree of secondary processing ,treating and waste disposal .This tends to increase the operating

cost in terms of fuel and loss, pollution control measures, additional treatment cost etc.

Equipment performance (machine productivity) is an important factor that determines

the profitability of the industry. Organizational procurement procedures and policies necessitate


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the choice of low cost equipment during the selection stage for new / existing processes. A

scientific method need to be applied in this area as cheaper equipment normally have lower life

cycle , reliability energy efficiency and high failure rates leading to enhanced operation,

maintenance and replacement costs which in turn affect the run length of the operating units.

This reflects on production loss, wasteful expenditure on maintenance and lower profitability.

This factor that has to be evaluated to arrive at right equipment maintenance / replacement

decision. This is a crucial managerial decision as the productivity of the enterprise is linked to

equipment availability, when other parameters remain constant

Capital performance (capital productivity ) is the ultimate objective of any enterprise

. This is normally determined by the production volume ( in monetary terms) to capital input.

For example when the performance of two industries are compared, quantity of products

produced per unit of capital input is considered as a measure of performance. For achieving

effective financial performance, one has to identify all the cost and profit centres , cost-intensive

elements and develop performance models for monitoring and control.

This document covers the most modern concept of Enterprise Resource Planning using some

Linear Programming and Scientific Models which are used by modern industries for

accomplishing effective cost control and high profitability within reasonable limits of

accuracy .


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A. Productivity and Resources Utilization.

Productivity is the efficiency with which resources are used to produce goods and

services. This is generally expressed as the ratio of output to input. Most common productivity

ratios in use are labor productivity, Capital productivity, Material productivity and Machine

productivity.

Enterprise performance is the sum total of all productivity which reflects on how

effectively labor , capital, equipment and materials are deployed to achieve the output.

Hydrocarbon processing industry operations are very capital and energy intensive.

Consequently, partial productivity ratios pertaining to the efficiency of capital, energy and

feedstock play an important role in plant performance and cost effectiveness. Fig 1 shows the

concept of total and partial productivity which is used in conventional Enterprise Resources

Planning .


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B. Energy Efficiency and productivity.

Since we are more concerned with Energy productivity, energy efficiency must be high to

achieve this objective and it may be represented by a simple relationship

Energy productivity may be increased by

1. Increasing the output for the same energy input

2. By reducing the energy input for the same output or

3. By increasing the output and simultaneously reducing energy input.

Though it looks simple in the above mathematical form, this is a complex problem in the

Hydro Carbon Processing Industry as the energy input to the process are in various forms such

as fuel, steam, power, thermal energy etc.

Energy intensity (EI) is a commonly used index to estimate energy efficiency (EE) for

countries, but it neglects the specific structure of energy consumption. For micro level analysis

and energy efficiency optimization, the problem may be broken down into smaller problems and

the concept of energy supply chain must be applied.

Fig 2 explains the concept of Energy supply chain. This refers to the case of a Thermal

power plant, which uses coal , fuel oil and fuel gas as the primary fuel to generate high pressure

steam. The super heated steam is supplied to the steam turbine which generates electricity.

Electric power is then transmitted by a transformer to the consumer’s area, located at a far off

place. This is shown in the fig2. Overall efficiency of the system is the product of individual

efficiencies of elements constituting them.

Fig2. Energy supply chain – Power Generation

Ep = Output

Energy input.

Power Boiler Consumer

TransformerStation

Turbo Generator

SteamTurbine


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C. Energy efficiency vs Productivity - Fuel System.

Fuel system has an important role in maintaining energy efficiency of boilers. Different

fuels behave differently during combustion process. The most generic term used to identify

efficient fuel utilization is the fuel efficiency.

Fuel efficiency, in its basic sense, is the same as thermal efficiency, meaning the

efficiency of a process that converts chemical potential energy contained in a carrier fuel into

kinetic energy or work. Overall fuel efficiency may vary per device, which in turn may vary per

application, and this spectrum of variance is often illustrated as a continuous energy profile.

Non-transportation applications, such as industry, benefit from increased fuel efficiency,

especially fossil fuel power plants or industries dealing with combustion, such as ammonia

production during the Haber process.

Parameters affecting the fuel efficiency in the case of Process Heaters and Boilers are

Fuel type ( coal / liquid / Gaseous )

Fuel temperature

Fuel viscosity / particle size ( for solid fuels )

Air to fuel ratio ( for combustion )

Steam to Fuel ratio ( for atomization )

Burner condition etc

Fuel temperature and viscosity play an important role in combustion. Since viscosity and

temperature are interrelated, the viscosity of fuel at the burner tip is specified by the burner

manufacturers.

The production of energy by the fired heater consists of the burning of fuels that are

primarily of fossil origin. However, the variety of fuels is great when compared with other

industries. The liquid fuels burned ranges from liquid butanes and pentanes, to light fuel oils

such as diesel and #2, to heavy fuel oils such as #6, and many heavier liquids.

Gaseous fuels may come in an even wider variety. Fuels such as hydrogen, methane,

ethane, propane, butane and carbon monoxide are burned either separately or in an infinite

variety of blends which sometimes contain inert components such as nitrogen and carbon

dioxide. The heating value and specific gravity of the blends are quite variable. Lower heater


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values ranging from lower than 100 Btu per cubic foot, to in excess of 3,000 Btu per cubic foot

are not unusual.

Further complicating the firing of a process heater are requirements such as flame shape,

radiant flux rates and unique sources of oxygen for the combustion reaction. Heat transfer for

different processes and different heater designs to accomplish the heat transfer for a particular

process have engendered the design of different burners that produce a multitude of flame

shapes. The flame shapes must be compatible with both the mechanical configuration of the

radiant combustion zone of the heater, and the radiant heat flux rate required by both the process

and heater design

Burner types consist of raw gas burners that mix fuel and air in the combustion zone, and

pre-mix burners that mix the fuel and air prior to the combustion zone for gaseous fuel firing.

There is a multitude of oil burners which use many methods of atomization for the fuel oil. In

addition, there are various combinations of the raw gas, pre-mix and liquid burners.

The burning of the various fuels described produce nitrogen oxides. The production of

nitrogen oxides has become a matter of concern to industry. First, NOx has an adverse effect on

the environment. Second, there is a growing volume of regulation dealing with the emission of

NOx to the atmosphere.

Process heaters are also somewhat unique in that the oxygen for combustion can come

from a variety of sources. In the majority of cases, the source of oxygen is atmospheric air

containing 21% oxygen by volume.

However, atmospheric air can be supplied in many forms. For many years, the most

frequent method of air supply to the burner was by natural draft with the air at ambient

temperature. More recently, the combustion air has been supplied at a positive pressure to permit

better control of the flame pattern and excess air.

Further requirements to conserve energy have brought about the use of preheated

combustion air. Because of the pressure to conserve fuel, other high temperature sources of

oxygen for the combustion reaction are now being utilized. For example, the off-gas from a gas

turbine can be used. This stream contains 15% to 16% by volume of oxygen at a temperature

usually in excess of 900° F. When a stream of this type is used as a source for oxygen, the

sensible heat of the TEG stream can be recovered. Another source of oxygen is the off gases

from certain kilns.


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Steam and air assisted atomizers are widely used in oil burners in industrial furnaces or

boilers. In industrial burners, the fuel oil is delivered to the burner process by an atomizer

nozzle, also known as an “oil gun”, when referring to the fuel supply lance and oil atomizer

assembly as a whole. The quality of atomization will influence combustion performance because

combustion begins immediately downstream of the atomizer nozzle.

The combustion performance attributes that furnace operators are most interested in are

flame length, consumption of atomizing medium, turndown ratio, NOx emissions, and

particulate emissions. In the past, due to lack of a good way of measuring atomization

effectiveness, mostly empirical trial and error was used in industrial research to select the best

atomizer nozzle. Consequently the empirical approach was limited to the performance of

the atomizer nozzles that were readily available, and could not provide a method to design the

atomization specifically to achieve a given combustion performance target.

The study and comparison present findings about how the atomization parameters can

be varied to contribute improvement in specific areas of combustion performance.

A rule of thumb to determine the burner efficiency / performance is to compare the

combustion zone temperature of heater / boiler at different fuel temperatures , atomizing steam to

fuel ratio , air to fuel ratio and air temperature. This is a practical method to determine the

performance of the fuel system.

D. Case study :

Hyundai Oilbank Co., Ltd. needed to increase the processing capacity of its No. 2

Crude Distillation Unit located in Daesan, South Korea. When the operators increased the burner

heat release in the heater, by increasing oil flow to the burner, the flame length also increased

and created flame impingement.

Burner modernization in the heater by replacement of its conventional EA-type oil guns

with the new HERO oil guns having phased atomization of fuel oil to efficiently atomize the

liquid fuel with less steam consumption resulted in shorter flame lengths, reduced soot

formation, increased turndown and lower NOx emissions.

RESULTS

• Up to 15 % shorter flame lengths at design rate.


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• 36 % less atomizing steam consumption; steam-to-oil ratio reduced from 0.40 to 0.27 kg-

steam/kg-oil.

• 5 % less NOx emissions; 210 ppm NOx before change and 200 ppm NOx after change.

• Hyundai Oilbank Co. will recover its capital investment of the 64 HERO guns over 15 months

in reduced energy costs alone.

Fig 3. Burner retrofitting results in energy savings

For a fired duty of 100 million kcal/hr and the heater efficiency of 87%, the fuel

consumption shall be 11495 kg oil /hr (10000 kcal/kg LCV). The atomizing steam consumption

between the two cases will be as shown in the table below.( Ref John Zinc )

Fired duty of heater in mmkcal/h

Heaterefficiency% on LHV

Fuelconsumption in kg/hr

Steam consumption kg/hr

Annualconsumption mt/yr

100 87 11495 4598 36784

100 87 11495 3104 24832

Savings due to modernization

1494 11952

This example shows the impact of burner efficiency on the heater performance in terms

of savings in atomizing steam and fuel consumption.


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E. IMPACT OF BURNER AIR TO FUEL RATIOS

Periodic checking and resetting of air-fuel ratios is one of the simplest ways to get

maximum efficiency out of fuel-fired process heating equipment such as furnaces, ovens,

heaters, and boilers. Most high temperature direct-fired furnaces, radiant tubes, and boilers

operate with about 10 to 20 percent excess combustion air at high fire to prevent the formation

of dangerous carbon monoxide and soot deposits on heat transfer surfaces and inside radiant

tubes.

For the fuels most commonly used by the industry, including natural gas, propane, and fuel oils,

approximately one cubic foot of air is required to release about 100 British thermal units in

complete combustion. Exact amount of air required for complete combustion of commonly used

fuels can be obtained from the information given in one of the references.

Process heating efficiency is reduced considerably if the air supply is significantly higher

or lower than the theoretically required air.

Air-gas ratios can be determined by flow metering or flue gas analysis. Sometimes, a

combination of the two , works best. Fig 4 gives fuel lost as dry flue gas as % of input fuel for

various stack temperatures., which may be controlled by adjusting air-fuel ratios.

The excess air curves are labeled with corresponding oxygen percentages in flue gases.

To figure potential savings, you need to know:

• The temperature of the products of combustion as they leave the furnace

• The percentage of excess air or oxygen in flue gases, at which the furnace now operates

• The percentage of excess air or oxygen in flue gases, at which the furnace could operate.

On the chart, determine the available heat under present and desired conditions by reading up

from the flue gas temperature to the curve representing the excess air or O2 level; then, read left

to the percentage available heat (AH).

Calculate the potential fuel savings:

% Fuel Savings = 100 X ((%AH Desired - %AH Actual ) / %AH Desired)


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F. FACTORS AFFECTING EXCESS AIR LEVEL REQUIREMENTS

Combustion systems operate with different amounts of excess air between high and low fire.

Measurement of oxygen and combustibles such as carbon monoxide in flue gases can be used to

monitor changes in excess air levels.

For most systems, 2 to 3 percent oxygen with a small amount of combustibles—only 10

to 50 parts per million—indicate ideal operating conditions. Processes that evaporate moisture or

solvents need large amounts of excess air to dilute flammable solvents to noncombustible levels,

to ensure adequate drying rates, and to carry vapors out of the oven.

Lowering excess air to minimal levels can slow down the process and create an

explosion hazard.

Fig 4. % Oxygen in Stack vs Heat Loss % in dry flue gas


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G. BURNER SPECIFICATIONS

These burner specifications can improve fired heaters performance. Specifying the right

requirements for fired heater burners can improve the heater operation and reduce maintenance.

With the right type of burners, fired heater capacity can be increased by 5 to 10% and

thermal efficiency by 2 to 3 %.

Burner selection and specification should be done carefully as they have a direct impact

on heater operation. Table given below lists the major parameters for a good burner design.

TABLE —Burner selection criteria:

Burner type.

Users should specify only burner types that they have sufficient operating experience

with. Alternatively, the burner that has been proven in the industry elsewhere, or at least has

been successfully tested at the vendor's furnace test rigs under simulated conditions, should be

specified. A number of cases have been reported of severe production losses and shut downs due

to selection of unproven burners. It may appear to be costly to ask for testing of burners, as one

of the requirements, but it will prove much cheaper in the long run. This is particularly important

Burners should provide stable combustion with:

1. Ability to handle wide range of fuels

2. Provision for safe ignition

3.Easy maintenance

4.Good turn down ratio

5.Well defined flame pattern with all fuels and firing rates

6.Low excess air operation

7. Low noise and NOx levels.


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now with new low Nox burner designs that may require higher excess air and have longer flame

lengths than normal burners.

Heat release and turndown.

The burner supplier should be able to meet varying demands of the user arising from his

process design and operation philosophy. Burners are normally designed to provide 120% of

their normal heat liberation at peak heat duty.

The user should specify very clearly the normal, maximum and minimum heat release

requirements of the burner to the vendor.

Over sizing burners normally leads to over firing of the heaters. Turndown capability of

these burners depends upon the type of oil gun selected and available oil and steam pressures.

The lower limit depends upon the ability to keep a stable flame with minimum oil throughout.

Air supply.

The availability of furnace draft determines the size of air registers. In case of a natural

draft burner, draft available at the burner must be calculated accurately from the firebox

dimensions and flue gas temperature, and specified clearly. If the draft specified is lower than the

available draft, it will result in oversized burners.

On the other hand, if higher draft is specified than what is actually available, burners will

not be able to give maximum heat release. If preheated air is used for combustion then

combustion air temperature at the burner needs to be specified to apply temperature correction

for the pressure drop. If forced draft burners are also required to operate under natural draft

conditions it must he dearly specified.

The natural draft conditions are controlling in such cases. Burner air supply should be

protected against variations in wind velocity, which may cause blowback. Forced draft burners

should be provided with adjustable dampers and a wind box gauge to ensure equal distribution of

air. Uniform air distribution to the burner can also be ensured by proper duct design or flow

modeling.


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The burner wind box and front plate should be made out of at least 3-mm CS plates. This

is to prevent warping of plates and improve long-term operability. Roller bearings should be

provided for easy register movement and a positive locking device should be included to prevent

vibration from changing register position. Register air controls should be easily accessible

Excess air.

For complete combustion, it is necessary to supply enough excess air. A good burner

design calls for minimum possible excess air compatible with process requirements.

TABLE 2-Excess air levels

Normal excess air levels for different fuels are given in Table 2. As a rule of thumb every

10% extra air used in combustion translates into a loss of 0.7% in terms of efficiency. At

turndown conditions, which are mostly overlooked, the excess air requirement goes up

substantially

Fuel specifications.

The burner design is linked directly with the fuels to be fired. Some of the properties that

need to be listed very clearly for liquid fuels are fuel specific gravity, lower heating value, fuel

oil viscosity, fuel pressure, and temperature available at burner.

For gaseous fuels, molecular weight is also important. Fuel oil viscosity and pressure

available will govern the atomizer and oil tip design. If the fuel contains abrasive particles

hardened, oil tips should be specified.

Burner Type Fuel oil Fuel gas

Natural draft 15 to 25 10 to 20

Forced draft 10 to 15 5 to 10

Forced draft (preheated air) 5 to 10 5

High intensity 5 5


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COAL FIRED BOILERS

In case of coal fired boilers, quality of coal especially the ash and sulfur content plays

an important role in maintaining boiler efficiency. Following figure shows the impact of ash

content on boiler efficiency.

ASH CONTENT IN COAL vs BOILER EFFICIENCY (LHV)

Besides ash content, other parameters which affect the coal fired boiler efficiency are the

particle size of coal. Pulverization is an energy intensive process . Hence the impact of these

parameters will have to be evaluated in coal fired boilers in the optimization study.

82

83

84

85

86

87

88

'16 % ' '21 %' '30 %' '35 %'

% eff blr


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H. BOILER / STEAM SYSTEM.

Steam is consumed in process and power industries for driving compressors, turbines and

for heating process fluids ,creating vacuum by ejectors etc. Steam pressure depends on the

process configuration, type of equipments used and the choice of Combined Heat Power cycle

selected. In normal practice, cost of steam alone constitutes about 60 to 70 % of the total

utility cost.

Steam Consumption could be monitored by an overall steam balance of the set up on a

day to day basis which include the quantity of steam generated and steam consumed at various

sections. The difference between the two denotes steam loss which should be minimum.

Steam consumption is dynamic in nature and it's demand is a function of load factor of

individual unit , efficiency of various rotating equipments deployed, heat recovery level , feed

stock quality, operating severity etc of each unit / section .

Under these dynamic conditions, computer-based models could be used effectively to

monitor the steam consumption of individual unit, monitor overall steam demand and identify

deviations and problem areas for remedial action. A typical utility flow diagram of a process

unit is given in fig 5. In case of boiler system, boiler efficiency determines the quantity of steam

generated per ton of fuel burnt and the productivity parameter is the steam by fuel ratio. All

parameters remaining constant, Steam / Fuel ratio or specific fuel consumption is the key

performance indicator.

Applying Energy supply chain concept, steam generation quantity will depend on steam

and power demand of the total process. When process efficiency , power generation efficiency

and steam distribution efficiency are lower than the target, the specific consumption of fuel will

tend to go up.

Example :

A boiler generates 100 t/hr of steam at 65 kg/cm2g pressure and generates 10 mw power

in a back pressure turbine. Exhaust steam is 100 t/hr at 15 kg/cm2g pressure. Entire BP steam is

consumed in the process when the line loss is about 2.5%. After years of operation, line losses

increased to 4.0 % and process consumption increased by 5%. Under this condition steam


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generation also increases to 106.6. t/hr, assuming no change in turbine efficiency. In this case the

productivity has come down by 6.6%.

Fig 5. Typical Steam , Power and Compressed Air System

If the turbine efficiency has also deteriorated during the period, the steam demand will

increase further. Assuming the steam / fuel ratio of the base case as 16.5 and the current ratio at

15.4, the increase in fuel consumption will be 106.6/15.4 – 100/16.5 = 0.8641 t/hr.

Typical energy supply chain for rhe boiler and steam system may be as shown below.


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Boiler efficiency is a function of fuel type, atomization efficiency, air / fuel ratio, air

temperature, draft profile, convection and radiation losses, boiler shell condition, water tube

condition, etc.

In case of coal fired boilers, efficiency will be determined by pulverized coal size, ash

content, coal burner efficiency etc.

Similarly for the steam system, the consumption is a function of transmission efficiency,

temperature / pressure, equipment efficiency , line losses etc.

All these parameters will have to be checked thoroughly when either the boiler or steam

system efficiency starts dropping down.

I. TURBINE AND POWER SYSTEM.

Electric Power is generated by a turbo generator in which the heat energy in steam is

converted into kinetic energy, which rotates the rotor of the Generator, which in turn generates

power. In the case of gas turbine, fuel is burnt in a combustor and the hot gases expand in the

turbine section to convert thermal energy into kinetic energy which drives the rotor of the

generator. Power generated at the terminal is linked to the efficiency of the turbine and the

generator.

Efficiency of the rotating machine is linked to

Existing machine life or years of operation

Wear and tear of various components

Condition of stator / rotor coils and

Condition of bearings

Alignment etc.

Starting from the consumer end, the energy demand at turbine terminal may be evaluated

as shown in fig 6.


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Following table shows the impact of energy efficiency of individual energy link on the

overall energy efficiency of the turbine and power system.

Consumer side demand 1000 kw

Above table shows the deterioration in overall energy efficiency of the system over a

period of operation or life cycle.

J. THERMAL EFFICIENCY AND ENERGY LOSSES

In any fuel fired equipment like heater / boiler / gas turbine etc, thermal efficiency refers

to the ratio of energy consumed for heating the process fluid or quantity of heat absorbed for

steam generation against the thermal energy input in the form of fuel or any thermal energy. This

may be represented as given below for a system

Thermal Efficiency may be mathematically expressed as

Case Turbine efficiency

Generatorefficiency

Transmissionefficiency

Consumerequipmentefficiency

Overallenergy

efficiencyBase 60.0 95.0 97.5 65.0 36.12 After 5 yrs 58.0 92.5 96.0 61.0 31.41 After 10 yrs 56.0 91.0 95.0 58.0 28.07 After 15 yrs 54.5 89.5 93.8 56.5 25.85

SYSTEM Energyinput

Energyutilized

EnergyLosses


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Thermal = Energy utilized for doing work / Energy input (1)

= ( Energy input – energy losses ) / Energy input (2)

Thermal may be increased by reducing the energy input for the same energy

utilization / output or increasing useful energy output for the same energy input. This needs a

total system analysis and an energy loss break-up.

In the case of a boiler for example, energy input ( energy supply chain ) must be analyzed

scientifically as shown by the energy supply chain diagram. From the energy system link, it may

be noted that the useful energy input is determined by fuel quality, fuel atomization, excess air,

moisture content etc.

IMPACT OF FUEL QUALITY / MIX

In Thermal Power plants, the conventional fuel used is coal. Coals used as fuel in boiler

plants have high ash content. Impact of ash content on Boiler efficiency is as shown below.

ASH CONTENT IN COAL vs EFFICIENCY (LHV)

82

83

84

85

86

87

88

'16 % ' '21 %' '30 %' '35 %'

% eff blr


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PARAMETERS USED IN EFFICIENCY CALCULATION

Fuel mix options for coal fired boilers

o Selecting low ash coal

o Blending Anthracite coal with high ash coal before crushing and pulverizing

o Blending raw petroleum coke with high ash coal and pulverizing the mix

Impact of fuel atomization / coal pulverization:

Fuel atomization / coal pulverization has direct impact on thermal efficiency of fired

heaters / boilers. Unburnt hydro carbons will be eliminated, when the degree of atomization is

high. With good atomization, air/fuel ratio requirement will be lower and the combustion zone

temperature also will be high. Hence there will be a reduction in fuel consumption and increase

in efficiency due to low heat loss from excess air.

Impact of excess air :

o As boiler excess air increases, efficiency of fired equipment like heater / boiler reduces.

o This is equipment specific and varies with the operating parameters such as fuel

composition, % Oxygen in flue gas and stack temperature.

o Figure given below shows the relationship for a specific case.

Component Wt % Wt % Wt % Wt %

Carbon 59.0 54.0 45.0 40.0

Hydrogen 3.0 3.0 3.0 3.0

Moisture 00.0 00.0 00.0 00.0

Oxygen 19.4 19.4 19.4 19.4

Sulfur 0.4 0.4 0.4 0.4

Nitrogen 2.2 2.2 2.2 2.2

Ash 16.0 21.0 30.0 35.0


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Excess air vs Efficiency

Typical energy losses from a fired heater / boiler.

a.Dry Gas Loss : 7.3617

b.Air Moisture loss : 0.0000

c.Combustion Moisture Loss : 7.5381

d.Fuel Moisture Loss : 0.7614

e.Radiation Loss : 1.2100

f.Blow Down Loss : 0.5100

g.Unaccounted Loss : 0.2000

g.Loss due to combustibles : 0.0000

Total Loss % ( dry basis) 17.5813

k. ENERGY TRANSMISSION :

In any fired equipment like a heater / boiler etc, there exists an Energy transmission chain

from the point of energy system input to the consuming end. A typical energy supply chain for a

boiler system is as shown below.

85

86

87

88

89

90

30% 25% 20%

Boiler Eff%


29

Typical Energy transmission chain

Heat of combustion between fuel and air, generates hot gases in the furnace section of

the boiler. Hot gases will be at a temperature between 900 to 1000 oC.

Energy transfer takes place between the hot gases and the circulating water in the water

wall tubes. This energy transfer produces saturated steam in the boiler drum.

The hot gases after delivering energy to the BFW in the boiler leaves at a temperature of

450 to 500 oC in the flue gas duct and enters the steam super heater as shown below.

Fuel

Air

Boiler

Air pre heater

SteamSuper heater

Economizer

Air

Flue gas to stack

Hot air

BFW


30

Flue gas temperature is now lowered to around 350 oC. Energy available in the flue gas

is utilized to pre heat the air for combustion as shown in the figure below.

Air enters the shell side of the APH while flue gas passes through tubes as shown in

the figure. This increases thermal efficiency of the boiler by 9 to 11 %

Typical Air pre heater Layout.

Energy transmission in Economizer:


31

Main objective of economizer is to increase the thermal efficiency of the boiler by pre

heating boiler feed water by exhaust flue gases as shown in the figure

From the above figures, it may be noted that energy efficiency of the total system

increases by efficient energy recovery in various sections, so that energy loss is minimized. In

the case of any fired equipment, energy losses occur in the form of exhaust flue gases, as shown

in page 28. Fig given below shows the energy loss analysis of a typical boiler.

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00

a.Dry Gas Loss :

b.Air Moisture loss :

c.Combustion Moisture Loss :

d.Fuel Moisture Loss :

e.Radiation Loss :

f.Blow Down Loss :

g.Unaccounted Loss :

g.Loss due to combustibles :

energy loss%

energy loss%


32

Exercise :

A power plant has 4 nos of coal fired boilers and 4 steam turbines to generate power.

Capacity of each boiler is is 750 t/hr and the capacity of each turbine is 120 mw with one

standby. One boiler and one turbine are standby. Coal consumption of each boiler is 135 t/hr.

Power demand is 300 mw. How will you operate the system such that the fuel consumption is

minimum ?


33

Exercise :

Following table shows the efficiency data of a steam turbo generator. What are your

observations and what are your suggestions for improvement ?

Steam turbine operating data

Month HP steam t/hr

Steam temp Power gen in mw

1 750 420 100 3 750 415 98.5 5 800 415 112.8 7 800 420 120.0 9 780 420 103.0 11 780 415 101.5


34

Notes :

35 Energy efficiency Optimization for Power plants. Faculty : Dr.G.G.Rajan

OPTIMIZATIONBASICS

2


2. OPTIMIZATION BASICS

Optimization refers to the scientific management of operating / operational

parameters, which are within the control of the organization, to maximize productivity,

profitability and performance of the organization, within the imposed internal / external

constraints. In any business activity, a number of conflicting parameters exert their

influence on the product / services provided by the organization. Hence the organization

chief has to take appropriate decision on what has to be done to maximize the

performance at a given situation. This leads to an optimization problem(s), which are

made up of three basic ingredients:

An objective function which one wants to minimize or maximize. For instance,

in a power plant where a number of boilers are operating in open cycle and co generation

mode, the objective function is to generate power to meet the demand at lowest operating

cost and maximum profit. By using experimental data, it is possible to develop a user-

defined model, which will minimize the total deviation of observed data from predicted

model output. In this problem, the variables include the boiler capacity, turn down ratio,

turbine load, turbine efficiency, fuel cost, steam cost, total operating cost at a particular

mode of operation etc. A set of constraints that allow the unknowns to take on certain

values but exclude others. In this power plant problem, the constraints are the individual

boiler loads, it’s maximum efficiency point, turbine load, it’s efficiency point etc .

The optimization problem is then narrowed down to find the values of the

variables that minimize or maximize the objective function as required, while fulfilling

all the constraints.

A. Objective Function

Almost all optimization problems have a single objective function. The two

interesting exceptions are:


B. No objective function.

In some cases (for example, design of integrated circuit layouts), the goal is to

find a set of variables that satisfies the constraints of the model. The user does not

particularly want to optimize anything so there is no reason to define an objective

function. This type of problems is usually called a feasibility problem.

Multiple objective functions.

Often, the user would actually like to optimize a number of different objectives at

once. For instance, in the power plant problem, the objective function is to minimize

operating cost and maximize power generation quantity simultaneously.

Usually, the different objectives are not compatible; the variables that optimize

one objective may be far from optimal for the others. In practice, problems with multiple

objectives are reformulated as single-objective problems by either forming a weighted

combination of the different objectives or else replacing some of the objectives by

constraints. These approaches and others are described in our section on multi-objective

optimization.

Variables

These are essential. If there are no variables, we cannot define the objective

function and the problem constraints.

Constraints

If Constraints are not present, this results in the field of unconstrained

optimization. In real life situations, all problems really do have constraints.

Example 1 Three boilers B1,B2 & B3 are in operation in a process unit, generating 100, 150

and 180 t/hr of steam respectively. Steam / Fuel ratio which is function of capacity

utilization of each boiler is given below. The design capacity of the boiler B1,B2 & B3

are 120, 180 & 210 tons/hr respectively. Determine the optimal load, the boilers should

handle to meet the steam demand of 430 t/hr at minimum fuel consumption. Fig 2.01


shows this lay out. The flow diagram shows the normal operation which was based on the

decision of the plant superintendent.

The above data will be converted into fuel / steam ratio for each boiler and a

model developed between fuel consumption and steam produced. Table 2.01 shows the

above information in terms of steam produced in tons/hr versus fuel in kg. These values

will be used in boiler loading optimization model.

Table 2.01. Steam Generation Vs Fuel consumed

Boiler 1 Boiler2 Boiler 3

steam gen fuel steam gen fuel steam gen fuel

72 6000 108 8244 126 9130

84 6720 126 9333 147 10208

96 7385 144 10286 168 11200

108 8000 162 11408 189 12194

120 8571 180 12329 210 13125

Boiler 60% 70% 80 % 90% 100%

Boiler 1 12 12.5 13.0 13.5 14.0

Boiler 2 13.1 13.5 14.0 14.2 14.6

Boiler 3 13.8 14.4 15.0 15.5 16.0

Steam / fuel ratio of boilers at various loads.

BOILER 1 Capacity 120 t/h



Fig 2.01. Boiler Load optimization problem

430 t/hr demand

100 t/hr 150 t/hr 180 t/hr


Note: Steam generation in tons/hr . Fuel Consumption in kg/hr

From the above data regression equations have been developed using MS Excel

program. Let y1,y2 and y3 be the quantities of fuel consumed for x1,x2 & x3 t/hr of

steam generated in boilers 1,2 & 3.

Then the total quantity of fuel consumed = y1 + y2 + y3 , which has to be

minimized. Regression equations for fuel consumption as a function of Boiler loads for

boilers 1,2 and 3 ( using Excel Spreadsheet ) are

y1 = 53.52381* x1 + 2196.9

y2=56.911678*x2 + 2124.82

y3 = 47.496883*x3+3191.98

Total fuel consumed at steam production rates x1,x2 & x3 are

F = 53.52381* x1+56.911678*x2+47.496883*x3+ 7513.7

The ultimate linear programming model for minimizing fuel consumption is given by

Minimize 53.52381* x1+56.911678*x2+47.496883*x3+ 7513.7

Subject to

x1+x2+x3 =430

x1<=120

x2 <=180

x3<= 210

x1>=0

x2>=0

x3>=0


Solution (From Lingo Program)

LP OPTIMUM FOUND AT STEP 2

OBJECTIVE FUNCTION VALUE = 22088.37 kg fuel

VARIABLE VALUE REDUCED COST

X1 120.000000 0.000000

X2 100.000000 0.000000

X3 210.000000 0.000000

NO. ITERATIONS = 2

As per existing operating pattern presented above, the fuel consumption is calculated as

29952.3 kg/hr which is higher than the optimum value by 7863.9 kg/hr. i.e. 35.6 % than

the optimum consumption. This example shows how operation of boilers can be

optimized to minimize fuel consumption using operations research techniques.

Case study 1

A power generation unit has five nos of power boilers B1,B2,B3,B4 & B5 whose

operating parameters are given in the following table.

Boiler

code

Steam

Generation

capacity t/hr

Efficiency at

full load in

%

Steam / fuel

weight ratio at

design load

B1

B2

B3

B4

B5

300

250

200

350

400

87.5

85.0

79.5

91.0

92.5

11.5

11.0

10.5

12.0

12.5


Steam pressure : 100 kg/cm2g : Steam temperature = 450 oC. Fuel used = FO:

Calorific value of fuel (NCV) = 10000 kcal/kg.

Total steam demand to generate power in two identical turbines is 1300 t/hr.

Determine the steam load on each boiler so that the overall efficiency of boiler is

maximum and fuel consumption is minimum.

Logical programming:

Though this is basically a linear programming problem, this can be done even by

mere logical data analysis. In this case boiler efficiencies are linear and proportional to

actual steam production. Hence, it is logical to load the boiler which gives maximum

efficiency. In this case boiler B5 shows an efficiency of 92.5 % at full load of 400 t/hr.

Hence B5 load is taken as 400. From steam demand point of view, the balance to be

generated out of other boilers is 900 ton/hr and boilers available are B1,B2,B3 and B4.

Out of these four boilers, B4 shows an efficiency of 91 % at full load. Hence B4 load is

taken as 350 t/hr. Other boilers left out are B1,B2 and B3 and the balance steam demand

is 550 tons/hr. Next boiler in the efficiency hierarchy is B1 which may be loaded to 300

t/hr. Balance demand of 250 t/hr may be met by boiler B2. B3 will be just idling.. This is

the simplest logical method followed by plant managers in real life situations .

Boilercode

Steam Generation

capacity t/hr

Efficiency at full load in %

Steam / fuel weight ratio at design load

B5

B4

B1

B2

400

350

300

250

92.5

91.0

87.5

85.0

12.5

12.0

11.5

11.0


In this case, the overall efficiency of the boiler system shall be

t = (400 x 0.925 + 350 x 0.91 + 300 x 0.875 + 250 x 0.85) / 1300

= 89.5

Fuel consumption = (400/12.5) + (350/12.0)+ (300/11.5) + (250/11.0)

= 32 + 29.17 + 26.09 + 22.73 = 110.53

Average steam / fuel = 11.76

When the steam demand changes, this exercise has to be repeated time and again .

Manual calculation being a cumbersome process, this may be programmed using

operations research techniques as shown in example 1 .

This problem may be quantitatively stated and solved as explained below.

Let x1,x2,x3,x4 and x5 be the steam loads. The main objective is to meet the steam

demand and maximize boiler efficiency and minimize fuel consumption. Hence a

mathematical relationship may be developed between steam generated quantity and fuel

consumed.

Efficiency of Boilers 1,2,3,4 & 5 at their respective loads x1,x2,x3,x4 & x5 will be

0.0029167 x1, 0.0034 x2, 0.003975 x3, 0.0026 x4 and 0.0023125 x5 respectively.

e.g. Calculation for boiler 1. : 1 = (x1/300) x 0.875 = 0.0029167

Efficiency for other boilers is calculated as shown for boiler 1.

Overall efficiency of boiler system = t = x1 * 1 + x2 * 2 + x3 * 3 + x4 * 4

x1 + x2 + x3 + x4

Substituting these values of efficiency function in the above equation, we get

t = 0.0029167 (x1)2 + 0.0034 (x2)2 + 0.003975(x3)2 + 0.0026(x4) 2 + 0.0023125 (x5)2

x1 + x2 + x3 + x4


where t is the overall efficiency of the boiler system.

The problem may now be stated as

Maximize

t = 0.0029167 (x1)2 + 0.0034 (x2)2 + 0.003975(x3)2 + 0.0026(x4) 2 + 0.0023125 (x5)2

x1 + x2 + x3 + x4

Subject to x1+x2+x3 + x4 + x5 = 1300 ( steam demand )

x1 < = 300 (boiler 1 load limit)





x1,x2,x3,x4 , x5 > = 0

This represents a non linear objective function with constraints given as above. To

solve this model using Lindo, the non linear equation must be converted into a linear

equation. Solving this problem involves a number of complicated steps and it is beyond

the scope of the book.

C. Optimization steps

This is referred to as ‘Problem Formulation’. This is the most critical aspect of

any optimization problem to achieve tangible results. Any snag in the problem

formulation, may result in non-feasible / impractical solutions. Extreme care must be

taken in ‘problem formulation’, as this is the most intelligent activity to tackle any

problem.

Following steps are involved in optimizing the performance of any system.

First step is to define the objective / goal we are aiming at.

Examples given below are typical objective functions related to production or

operation of an industry. They are


Maximizing production quantity

Maximizing operating profits

Minimizing operating costs

Minimizing energy consumption / energy costs

Minimizing emissions level

Loss reduction etc

Step 2: Identify process constraints which have an impact on the objective function.

Typical examples are

Capacity utilization or load factor in boilers, heaters, fans, blowers, turbines etc.

(As capacity utilization increases, specific energy consumption reduces and efficiency

increases up to certain capacity (i.e. design capacity ) and starts dropping down, when

the capacity utilization increases further.)

Run length .(In many equipments, wear and tear results in loss of efficiency

which will result in higher energy consumption. In the case of boilers, heaters, heat

exchangers etc fouling increases with operating period, which retard heat transfer and

decrease equipment efficiency and increase fuel consumption )

Operating severity. This refers to the reaction temperature / pressure / recycle

etc which reduce the energy efficiency of the system. In the case of conversion processes,

the objective is to increase conversion at the expense of energy / operating costs.

Feed quality / composition : In petroleum refining, petrochemical, fertilizer

plants etc feed quality / composition plays an important part in operating profit,

production rate etc.

Step 3: Establish a mathematical model between these variables and the objective

function. If these variables are linearly connected, the problem may be solved by Linear

Programming methods else it has to be solved by NLP / Parametric programming

methods.


In the boiler loading example , relationship between fuel consumption and steam

generation quantity was established using linear regression models for all the three

boilers. Using these models, total fuel consumption was established by summing up these

relationship to get an objective function equation. In the case study, where five boilers

were considered, a non linear model was developed to determine overall boiler efficiency

by individual efficiency models. In this case the objective function was to maximize

overall efficiency of the system of five operating boilers.

Step 4: Identify the constraints which impede / control the objective function. In the

example dealing with three boilers, the constraint was basically the maximum steam

generating capacity. There could be any number of constraints which will be incorporated

in the LP model.

Step 5: When once all the above steps have been applied and problem formulated

successfully, the solution may be arrived at using Operations Research tools such as

Linear / Non Linear / Parametric programming methods.

C. Common Industrial optimization problems

Single variable problem: Variables used in these problems are called

‘manipulable / endogenous ‘ variables as these may be adjusted by the user at his will.

For example, the feed rate in a heater or steam production rate from a boiler may be

varied physically by the operations staff .

What is not within his control is the heater / boiler efficiency. All the other

parameters remaining constant, heater / boiler / pump / turbine efficiency etc varies with

the load based on the load characteristic curve of the equipment which is design specific.

A typical load vs efficiency curve is shown in fig 2.02. It may be noted from the figure ,

that the equipment efficiency is maximum (87.5%) at 100 % capacity utilization.

Many designers often incorporate addition capacity of 5 to 10 % to take care of

any deficiencies. Under these conditions, the maximum efficiency point will slightly shift


to the right than what is shown in the figure. Similarly when the equipment is under

designed, the peak efficiency shall shift to the left. This aspect is shown in fig 2.03.

80

81

82

83

84

85

86

87

88

70 80 90 100 110 120

load %

effic

ienc

y

blreff %

Fig 2.02. Efficiency vs Capacity utilization – Generic Model


Above information may be represented in mathematical form to arrive at

quantitative decisions. Since the efficiency vs load shows a bell shaped curve, the model

of the form (quadratic model) ax2+bx+c or a cubic equation shall be more appropriate.

The same data may be used to develop these models. For developing these regression

models, principle of least square is used and the coefficients arrived at. When once the

quantitative models are developed, it is easy to find out the peak efficiency point.

76

78

80

82

84

86

88

90

70 80 90 100 110 120

load %

effic

ienc

y %

10% under normal 10% over

Fig 2.03. Impact of Design on peak efficiency point


Models are of the form

Eff % = -6.198286E-03 * x2 + 1.277158 * x + 20.74732 (I)

Eff % = -2.266988E-04 * x3 + 5.842239E-02 * x2 - 4.748416 * x + 204.3378 (II)

Standard error of model I and II are 0.7594 and 0.3769 respectively.

x is the load % of the equipment.

78

80

82

84

86

88

90

load %

effic

ienc

y %

actual quadratic polynomial

actual 80.5 82.1 85.0 87.6 86.5 84.3

quadratic 79.8 83.3 85.5 86.5 86.2 84.8

polynomial 80.5 82.3 84.9 87.0 87.2 84.1

70 80 90 100 110 120

Fig 2.04. Efficiency vs Load % models


D. Time dependant optimization models

Let us consider the case of a process heater whose efficiency varies with load %

as shown in the previous example. The same heater will perform differently with passage

of time say 1 yr, 1½ yrs, 2 yrs etc. Though the efficiency pattern may remain the same in

all these load conditions, observed efficiency for the same load will be somewhat lower

with passage of time. Hence necessary run length factor will have to be incorporated in

the model. A typical case is shown below and in fig 2.05.

Above table shows the behavioral pattern of the heater with passage of time

called ‘on-stream hours’. It is possible to predict what will be the behavioral pattern after

certain passage of time for taking corrective action. For this purpose, a two variable

model is built for the heater with load and operating hours as two independent variables.

This is a very important method for heat exchangers / coolers / condensers, as they

develop fouling with time and even after de-scaling, the original heat transfer rate is

seldom achieved. Above data when converted into a model gives,

Efficiency % = 0.1416667 * load % - 0.1101426 *op. month + 71.2722

This model may be used to predict the efficiency at any load over a reasonable

operating month. For example, the predicted heater efficiency by 35th month at 100 %

load will be of the order of 81.58 % .

Load % Base case After 1 year

After 2 years

70 80 79 78.0 80 82 80 78.0 90 85 83 81.5 100 88 87 85.5 110 85 82 81.0


Fig 2.05. Efficiency deterioration with on stream hours


E. Linear Programming (LP) Problems

We have shown some examples related to boiler efficiency optimization, which

represented a linear programming (LP) problem, in which the objective and the

constraints were linear functions of the decision variables. Typical example of a linear

function is:

75 x1 + 50 x2 + 35 x3

where x1, x2 and x3 are decision variables. The variables are multiplied by coefficients

(75, 50 and 35 above) that are constant in the optimization problem. They may be

computed in Excel worksheet. Linear programming problems are intrinsically easier to

solve than nonlinear (NLP) problems. In an NLP there may be more than one feasible

region and the optimal solution might be found at any point within any such region. In

contrast, an LP has at most one feasible region with “flat faces”. Some simpler problems

may be solved graphically. A typical example related to a heater is given below.

Case study 2

A fired heater having an operating capacity of 500 t/hr of feed uses fuel gas and

fuel oil to heat the feed to the outlet temperature of 345 oC. The calorific value of fuel oil

is 9900 kcal / kg and that of fuel gas is 10800 kcal/kg. Cost of fuel oil is 100 us$ / ton

while that of fuel gas is 60 us$/ton. Maximum Fired duty of heater is 100 mmkcal /hr.

From radiation temperature point of view, the heater should have minimum 65% oil

firing and maximum 90%. Atomizing steam is 15% of oil consumed and the cost of

steam is 45 us$/ton. Availability of fuel gas is limited to 4.0 t/hr maximum. Find the

optimum fuel mix that will minimize the total operating cost of the heater.

Formulating the problem:

Let x1 and x2 be the quantity of fuel oil and fuel gas fired in the heater to meet the heat

duty.

Total cost of fuel = 100 * x1 + 65 * x2 (I)


Cost of steam = 0.15 * x1 * 45 (II)

Total cost = 100 * x1 + 65 * x2 + 0.15 * x1 * 45

= 106.75 * x1 + 65 * x2 (III)

Objective function is to minimize fuel cost. i.e. to minimize 106.75 * x1 + 65 * x2

The total calorific value of the fuel mix supplied must meet the required fired duty. Since

the calorific value of Fuel Oil and fuel gas are 9900 and 10800 kcal/ton respectively, the

heat duty equation is formulated as

x1 * 1000 * 9900 + x2 * 1000 * 10800 = 100 * 1000000

i.e. x1 + 1.0909091 * x2 = 10.10101 (IV)

Fuel oil fired should be minimum 65 %

x1 >= 0.65 * 100 * 1000000 / (9900 * 1000) tons

i.e. x1 >= 6.565656 (V)

Fuel oil fired should be maximum 90 %

i.e. x1 <= 0.90 * 100 * 1000000 / (9900 * 1000)

i.e. x1 <= 9.09091 tons. (VI)

Fuel gas availability constraint is 4.0 t/hr.

x2 < = 4.0 (VII)

The final problem may now be written as

Minimize 106.75 * x1 + 65 * x2 (I)

Subject to

x1 + 1.0909091 * x2 = 10.10101 (II)

x1 >= 6.565656 (III)

x1 <= 9.09091 tons. (IV)

x2 < = 4.0 (V)


This is comparatively a very simple problem that may be visualized in graphical form and

may be solved by logical programming as explained below.

Since the fuel oil to be used is minimum 6.565656 tons, x2 may be calculated using

equation II.

i.e. 1.0909091 * x2 = 10.10101 – 6.565656 = 3.535354

i.e. x2 = 3.24074 tons.

This meets all the constraints.

Hence total fuel cost ( objective function) is 911.531878 us$ .

Surplus gas available due to the optimal fuel mix is = 4 –3.24074 = 0.75926 tons

Case b: The operations manager decides to use the total fuel gas in the heater by making

some operational adjustments so that the minimum oil firing reduces to 50 %. What

impact does it make on the operating cost ?

In this case constraint III may be modified as

x1 >= 0.5*100*1000000/(9800*1000)

x1 >= 5.102041 tons. (III)

Since fuel gas availability is 4.0 tons / hr and the entire fuel gas will be consumed, the oil

required to meet the heat duty may be calculated by equation (II) as given below.

x1 + 1.0909091 * x2 = 10.10101

i.e. x1 + 1.0909091 * 4.00 = 10.10101

i.e. x1 = 10.10101 - 4.3636364 = 5.7373736

This meets modified constraint III and all the other constraints.

Total fuel cost in this case is = 106.75 * x1 + 65 * x2 = 106.75 * 5.7373736 + 65 * 4.00

= 872.4646318

The reduction in fuel cost due to the revised fuel mix is

= 911.531878 -872.4646318

= 39.0672462 us$/hr = 4.2858 % reduction.


Based on 8000 operating hours, total annual savings due to the revised fuel mix works

out to 312537.9 us$ / annum for a single heater. This is fairly a simple problem where

only one heater was considered with minimum constraints.

Case study 3. Two heaters in parallel service

Two heaters having an operating capacity of 500 t/hr and 200 t/hr of feed are

operating in parallel. Both heaters use fuel gas and fuel oil to heat the feed to the outlet

temperature of 345 oC. The calorific value of fuel oil is 9900 kcal / kg and that of fuel

gas is 10800 kcal/kg. Cost of fuel oil is 100 us$ / ton while that of fuel gas is 60 us$/ton.

Maximum Fired duty of heater1 is 100 mmkcal /hr, while that of heater2 is 20

mmkcal/hr. Heater1 should have minimum 5.5 t/hr oil firing and maximum of 9.10 t/hr.

Heater2 should have minimum 1.0 t/hr oil firing. Atomizing steam is 15% of oil

consumed and the cost of steam is 45 us$/ton. Availability of fuel gas is limited to 5.0

t/hr maximum. Find the optimum fuel mix that will minimize the total operating cost of

the heater ?

Formulating the problem:

Let x1 & x2 be the quantities of fuel oil used in heaters 1 & 2 respectively. From

the fired duty, fuel gas consumed may be back calculated as shown below.

The objective function is to minimize fuel cost in the combined operation of heaters

which is stated as

Minimize x1*100 + x1*0.15*45 + x2*100 + 0.15*x2*45 + 60 * (100-9.8 x1)/10.8 + 60

* (20-9.8x2)/10.8

= 106.75 * x1 + 106.75 * x2 + 555.55 - 54.4444* x1 + 111.1111 - 54.4444 * x2

= 52.3056 * x1 + 52.3056 * x2 + 666.6611

Minimize 52.3056 * x1 + 52.3056 * x2 + 666.6611 (I)

ST

x1 >= 6.565656 (II)


x1 <= 9.09091 tons. (III)

x2 >= 1.0204 (IV)

Since the coefficients in the objective function are equal, only constraints II & IV are

important in this case.

Maximum amount of gas that may be consumed in heater 1 is = 3.3015 tons.

Remaining fuel gas available is = 6-3.3015 = 2.6985 tons

Maximum amount of gas that may be consumed in heater 2 is = 0.92592 tons

Total gas consumed = 4.2274 tons.

Gas left unutilized = 0.7726 tons

This is again based on logical programming. This problem may be solved by

using LP models considering both liquid and gas fuels individually . In this case, the

objective function will undergo a change.

Item Heater 1 Heater II Fuel oil in tons

x1 x2

Equivalentheat inmmkcal

x1*9800*1000/1000000=9.8 x1 = 9.8 x2

Balanceheat load

(100-9.8 x1) (20-9.8x2)

Eqvt fuel gas in tons

(100-9.8 x1)/10.8 (20-9.8x2)/10.8


F. LP method of solving the two heater problem :

Let x1 & x2 be the quantities of fuel oil used in heaters 1 & 2 respectively and x3

and x4 quantity of fuel gas used in the heaters 1 & 2 respectively.

The objective function is to minimize total fuel costs meeting fired duty and % fo

constraints . Considering the atomizing steam cost and other constraints the problem is

formulated as shown below.

Output from Lingo 8 program.

MIN = 106.75 * x1 + 106.75 * x2 + 60 * x3 + 60 * x4 ;

x3 + x4 = 5 ;

9.9*x1+10.8*x3 = 100;

9.9*x2+10.8*x4 = 20;

x1 >= 5.5;

x1 <= 9.09 ;

x2 >= 1.00;

x3 >= 0;

x4 >=0;

Global optimal solution found at iteration: 2

Objective value: 1011.667

Variable Value Reduced Cost

X1 5.500000 0.000000

X2 1.166667 0.000000

X3 4.217593 0.000000

X4 0.782407 0.000000

Note : In this case, entire gas has been consumed without any excess quantity. These

examples indicate, how practical optimization helps in reducing the fuel cost at micro

level. The success of this exercise is based on effective problem formulation especially in

defining the objective function and the constraints.


G. Complex Fuel Mix Problem :

In real life situations, process industries import a variety of fuel oils for their

consumption in process heaters / boilers to manufacture the product. As mentioned in

chapter 1, the energy input such as fuel, steam, power etc varies with the type of industry

and the design factors. Let us consider a complex fuel mix problem, with more number of

liquid fuels with varying sulfur levels.

Cost of sulfurous fuels are comparatively cheaper than clean fuel. Environmental

regulations limit the SO2 emission level from each heater / boiler .High sulfur fuel will

emit more SO2 than a clean fuel. There is an economic balance between use of sulfurous

fuel versus reduction in fuel cost.

Using Linear Programming models, it is possible to determine the maximum

amount of Sulfurous fuel that can be used by a fired heater / boiler such that the

emissions are contained within allowable limits at minimum fuel cost. This is a typical

environmental problem faced by process industries.

H.Impact of process modification in the boiler.

The boiler operating cost can be lowered down , by incorporating certain modifications in

the boiler flue gas sections. From the LP model, it is obvious that the constraints holding

the key to using low cost fuels are SO2 and SPM emission levels. At this stage, it is

possible to incorporate certain cost effective modifications. For removal of SO2 from the

flue gas, a Sulfur Guard may be provided before the air pre heater. This is nothing but a

Zinc wire mesh which is located in a convenient place for easy removal and replacement.

This will react with SO2 present in the flue gas by chemical reaction and is converted

into ZnSO4. After certain passage of time, the zinc mesh may be replaced, when the

reaction becomes ineffective. For removal of SPM, bag filters / or other type of filtering

material may be used. Let us visualize the scenario with S guard & filter modifications as

proposed.


Let C be the cost of modifications in US$ and project life at Y years. Incremental cost

due to the project on hourly basis ( without considering interest rate ) is given by

Incremental Cost / year = C/Y

Assuming 8000 operating hrs / year, incremental operating cost / hr = (C/8000Y)

Under these conditions, the boiler can generate the entire steam using fuel x3.

Hourly saving due to switching over to 100% of EHFO = 4226.7 – 4165.9 = 60.8

US$.

Let us assume the investment cost at 2,50,000 us$.

Extra cost incurred / hr = 10.42 US$.

Hence the effective net savings / hr = 60.8 – 10.42 US$ = 50.53 US$ /hr

Pay back period = 2,50,000 / (60.8 * 8000) = 0.51 year

This is just a hypothetical case. If the management has an investment policy for projects

with longer pay backs, the investment limit may be doubled or tripled.

It is now obvious from the above practical examples that an LP Solver needs to consider

many fewer points than an NLP Solver, and it is always possible to visualize (subject to

the limitations of finite precision computer arithmetic) that some LP problem

(i) may have no feasible solution

(ii) have an unbounded objective, or may have a globally optimal solution


I. Quadratic Programming (QP) Problems

A quadratic programming (QP) problem has an objective which is a quadratic function of

the decision variables, and constraints are all linear functions of the variables. A typical

example of a quadratic function is:

2 x12 + 3 x22 + 4 x1 x2

where x1, x2 and x3 are decision variables.

A widely used QP problem is the Markowitz mean-variance portfolio optimization

problem using a number of normal equations. The linear constraints specify a lower/

upper bound for profitability function.

QP problems, like LP problems, have only one feasible region with "flat faces" on its

surface (due to the linear constraints), but the optimal solution may be found anywhere

within the region or on its surface.

Quadratic Programming Problem – Example

Two boilers B1\and B2 are designed to operate at maximum load of 500 and 750 t/hr.

The minimum turn down ratio of B1\ & B2 are 65% and 55 % respectively. Boiler

efficiency characteristics as a function of load for the two boilers are shown in fig 2.06.

Find the optimal load on the two boilers when steam demand is 800, 900 and 1000 t/hr

such that the overall efficiency is maximum.

This may be represented by

Efficiency of boiler 1 is of the form e1 = a1* x12 + b1* x1 + c1

Efficiency of boiler 2 is of the form e2 = a2* x22 + b2* x2 + c2

Over all efficiency of the boiler = (e1*x1*c1 + e2*x2*c2) / (x1*c1+x2*c2)

e1 = -6.000395E-03 * x12 + 1.159349 * x1 + 23.45476

S.E of model : 0.3543

e2 = -6.198286E-03 * x22 + 1.277158 * x2 + 20.74732

S.E of model : 0.7594

Where x1 & x2 are % of load on design, c1& c2 are boiler capacities


The QP model may now be written as

Maximize (e1*x1*c1 + e2*x2*c2) / (x1*c1+x2*c2)

Subject to (x1*c1 + x2*c2)/100 = 800

x1 >=65

x2 >= 55

65

70

75

80

85

90

Load %

blr

eff%

LCBLR HCBLR

LCBLR 66.4 71.4 75.2 77.8 79.2 79.4 78.4HCBLR 69.1 75.1 79.8 83.3 85.5 86.5 86.2

50 60 70 80 90 100 110

Fig 2.06. Overall efficiency optimization – two boilers


Lingo Program output:

Max = ((-.006000395 * x1^3 + 1.159349 * x1 ^ 2 + 23.45476*x1)*500+750*( -

.006198286 * x2^3 + 1.277158 * x2 ^ 2 +20.74732*x2))/ (500*x1+750*x2) ;

500*x1 +750* x2 = 80000;

x1 >=65; x2 >= 55;

Local optimal solution found at iteration: 17



X1 65.00000 0.000000

X2 63.33333 0.000000

Row Slack or Surplus Dual Price

1 75.42680 1.000000

2 0.000000 0.4063477E-03

3 0.000000 -0.6137188E-01

4 8.333333 0.000000

i.e. Optimum load % x1 & x2 are = 65 and 63.3333%

i.e. 325 tons /hr and 475 tons / hr respectively. Overall boiler efficiency at this load

distribution will be 75.43 %. The same was solved by Logical Programming the output of

which is shown in the table below.

load t/hr load t/hr x1 x2 eff1 eff2 overall eff

325.0 475.0 65.0 63.3 73.46 76.77 75.43

330.0 470.0 66.0 62.7 73.83 76.44 75.37

340.0 460.0 68.0 61.3 74.54 75.76 75.25

350.0 450.0 70.0 60.0 75.21 75.06 75.13

360.0 440.0 72.0 58.7 75.82 74.34 75.01

370.0 430.0 74.0 57.3 76.39 73.60 74.89

380.0 420.0 76.0 56.0 76.91 72.83 74.77

387.5 412.5 77.5 55.0 77.26 72.24 74.67


The problem will become more complex when more and more variables are involved in

the QP. In such cases, it is a good practice to develop an Evolutionary Operations

Research Model for the required objective function incorporating all the related variables

in the model. Then the problem may be solved by any one of the solvers like Lindo,

Lingo, Excel Solver etc.

J. Evolutionary Operations Research Models for Optimization

Evolutionary Operation Research Technique or (EORT in short) involves very

systematic small changes in process variables during the operation of the process. The

results of previous small changes may be used to suggest further changes so as to

approach the optimum operating conditions by a series of small steps. Care is taken to

ensure that these individual changes in process parameters do not upset the process nor

produce any undesirable outcomes.

Evolutionary Operations Research technique may be used to identify the

combination of multi variables to enhance the response surface of any operation which in

turn improves the operational objectives and also the productivity of the complex system

considerably.

The basic concept underlying EORT is that a smooth response surface exists for a

set of variables, which ultimately tend to converge to a single optimum. This is the

principle of Advanced Control System where the impact of minor variations of process

parameters are used to vary other parameters to maximize or minimize the objective

function. ( e.g maximize energy efficiency of the equipment / total system ) .EORT

models are based on actual process variables which are dynamic in nature and fluctuate

due to exogenous or endogenous factors.

Since the effect of these manipulable variables form the inputs to the model, the

outcomes will naturally be far more precise and accurate. EORT models may be used for

carrying out the most needed sensitivity analysis of the dynamic system to identify the

output trend for a set of new variables and/or operating conditions. EORT models may


also be used to determine and monitor the system efficiency as a whole and also the sub

system constituting them. Hence corrective actions may be taken more precisely & in

time.

EORT models are real life operating models and combines the expertise of the

operations group and the system response to arrive at the right decisions compatible with

the system.

EORT models are deterministic in nature as the variables and their inter

relationship are evaluated in real life situations by observed values and performance

rather than theoretical / design mode models which suffer from many inherent

assumptions.

EORT models are system specific and cannot be generalized to all similar

situations. For developing a generalized model all variables constituting the total macro

system will have to be considered. EORT generates a set of Decision Models based on

LP / NLP algorithms and serve as a very powerful, effective and result-oriented Decision

Support System (DSS in short).

These quantitative Decision Models are logical abstraction of reality under which

the total complex system operates. While simulation models are based on certain

assumptions and hypothesis, which in practice may not be totally true or valid, EORT

models are based on observed facts and figures. All constraints experienced in real life

situations are reflected in the model output.

This type of DSS serves as a conduit for creating, revising, reviewing and

checking the real performance of the total system or subsystems which are constituents of

the main system. Also it is possible to achieve the operational objectives more easily than

one can visualize.


K. EORT Modeling Methodology

Following steps are to be followed meticulously to develop a very reliable and valid

model for the energy industry.

Defining the expected outcome/ objective of the model

Outlining the general process & data collection

Identifying the manipulable and nonmanipulable variables either by observed

data, analytical data or a combination of both.

Developing basic Model with the identified variables which affect the objective

function, and deleting the least effective variables by model revalidation.

Sensitivity analysis and model validation by manipulating the variables and

comparing the observed output with model output.

Checking model validity under observed conditions for a stipulated range of

variables.

No validation will be deemed necessary if the deviation between observed and

model outputs are within statistical limits.

When once the last stage has been reached, the model will be accepted as valid

and will form a basic DSS tool, by which the actual performance of the system may be

evaluated and compared and the reason for any discrepancy may be identified for

remedial action.

If the variation is explainable, the Decision Maker may correct the situation

immediately. If the variation is not explainable or much higher than the stipulated limits,

the situation must be taken as out of control and warrants thorough investigation. Hence

EORT models are capable of giving the warning signals at right time and could easily

avert inefficiency or even disaster in some cases.


L. Application of EORT Models

From experience it has been found that EORT Models have unlimited application

areas in the process industry.

Typical applications relate to

Cooler / Condenser performance prediction

Cooling water quality monitoring

Evaluation of inhibitor performance

Equipment deterioration study

Specific Energy consumption of a single process unit or the entire industry.

Some of the examples given here refer to actual observations of some of the industries.

Using EORT Models in the objective function.

Since EORT models combine all the independent variables in the form of an

equation and the outcome is defined, they may be used as the objective function and the

limits of the independent variables shall be the constraints.

Case Study 5 - Energy consumption model using EORT model. This example refers to the Power consumption of a Centrifugal air compressor

based on the observed operating data such as air flow rate, rpm , discharge pressure and

power consumption. Following table gives the details of parameters observed during

actual run of the air compressor. The objective is to minimize power consumption for

certain air flow and required discharge pressure.

Flow nm3/hr RPM Disch Pr. ata

PowerConsumption kwh

70000 5355 5.00 5700 75000 5355 4.95 6100 80000 5355 4.90 6300 85000 5355 4.75 6500


90000 5355 4.60 6600 95000 5355 3.80 6700 100000 5355 3.20 6750 70000 5610 5.70 6400 75000 5610 5.60 6700 80000 5610 5.45 7000 85000 5610 5.30 7300 90000 5610 5.10 7500 95000 5610 4.80 7700 100000 5610 4.55 7800 105000 5610 3.90 7900 75000 5891 6.40 7600 80000 5891 6.30 8000 85000 5891 6.15 8300 90000 5891 5.95 8700 95000 5891 5.80 8900 100000 5891 5.50 9000 105000 5891 5.25 9100

EORT Model is of the form

KW = 2.864448E-08* (Flow) 0.64983*(RPM) 2.15599*(Pr) 0.16322

The validity of the model is seen from the model output of power consumption

versus observed power consumption in KWH as shown below.

Compressor Performance Model

observed simulated error power kw/h by .Model term

5699.998 5738.24756 -38.25002

6100.002 5991.52393 108.47852

6300.002 6237.79688 62.20567

6499.999 6455.61182 44.38672

6599.998 6664.90674 -64.90868


6700.002 6691.28564 8.71616

6750.001 6726.73193 23.26927

6400.000 6480.75488 -80.75426

6700.002 6758.36816 -58.36635

7000.000 7016.66162 -16.66237

7300.002 7265.44189 34.55997

7499.998 7493.19873 6.79900

7700.000 7684.71875 15.28089

7799.998 7876.11523 -76.11685

7900.000 7927.84131 -27.84197

5699.998 5738.24756 -38.25002

6100.002 5991.52393 108.47852

6300.002 6237.79688 62.20567

6499.999 6455.61182 44.38672

6599.998 6664.90674 -64.90868

6700.002 6691.28564 8.71616

6750.001 6726.73193 23.26927

6400.000 6480.75488 -80.75426

6700.002 6758.36816 -58.36635

7000.000 7016.66162 -16.66237

7300.002 7265.44189 34.55997

7499.998 7493.19873 6.79900

7700.000 7684.71875 15.28089

7799.998 7876.11523 -76.11685

7900.000 7927.84131 -27.84197

7600.000 7674.84619 -74.84582

8000.001 7983.02832 16.97263

8299.999 8271.21289 28.78612


8700.000 8538.01855 161.98141

8899.997 8806.54102 93.45620

8999.997 9026.44629 -26.44888

9100.000 9246.72559 -146.72505

Standard Error of the estimate = 71.06937

Error level of this model being less than 1.0%, this may be used for optimizing

compressor performance. Manipulable parameter in this case is RPM. Air flow rate and

Discharge pressure of the compressor are the process requirement. Hence the problem

may be defined as minimize power consumption subject to the required air flow rate

(nm3/hr) and Discharge pressure in ATA. ( Atmosphere absolute). Using the EORT

model, the problem may now be formulated as


From the above example, the need for modeling the complex system using manipulable

variables and their impact on objective function was shown. It was also shown that this

model can be used in NLP optimization model, using the constraint equations to arrive at

the optimal solution.

Case Study 6

Three centrifugal compressors are used in an oxidation plant for various applications such

as Oxidation process, CO boiler and other process use. Compressor characteristics of the

compressors are given below. The operating range of the compressors are shown below.

KW1 = 2.86E-08* (F1)^0.64983*(RPM1)^ 2.15599*(Pr1)^0.16322

KW2 = 2.89E-08* (F2)^0.65*(RPM2)^ 2.45599*(Pr2)^0.19322

KW3 = 2.88E-08* (F3)^0.65*(RPM3)^ 2.45599*(Pr3)^0.175

EORT Model application in optimization

Min = 2.864448E-08* (Flow)^0.64983*(RPM)^ 2.15599*(Pr)^0.16322; Flow = 95000; Pr = 5.85; RPM <=5800; RPM >= 5300;




FLOW 95000.00 0.000000

RPM 5300.000 0.000000

PR 5.850000 0.000000

Row Slack or Surplus Dual Price

1 7021.530 -1.000000

2 0.000000 -0.4802949E-01

3 0.000000 -195.9033

4 500.0000 0.000000

5 0.000000 -2.856293

Since the minimum RPM of the compressor is 5355, we shall stipulate this condition in the model , by

modifying the last line in the model as RPM >= 5500. The modified output is given below.




FLOW 95000.00 0.000000

RPM 5500.000 0.000000

PR 5.850000 0.000000

Note the increase in power consumption by 584 KWH for the revised RPM level. The model shows that the

RPM must be as low as possible for low energy consumption.


Operating Parameters Range

Flow RPM Pressure (kg/cm2g)

Compressor 1 50000-100000 5000-7500 5.5-8.0

Compressor 2 65000-120000 5000-7500 5.5-8.0

Compressor 3 45000 – 95000 5000-7500 5.5-8.0

Total requirement of air = 195000 nm3/hr

Minimum required Discharge pressure = 7.5 kg/cm2g

What is the optimal operation of these compressors that will minimize the total power

consumption ?

Formulation of the problem:

Min = 2.86E-08* (F1)^0.64983*(RPM1)^ 2.15599*(Pr1)^0.16322+2.89E-08*

(F2)^0.65*(RPM2)^ 2.45599*(Pr2)^0.19322+ 2.88E-08* (F3)^0.65*(RPM3)^

2.45599*(Pr3)^0.175;

F1 + F2 + F3 = 195000;

F1>=50000;

F2>=65000;

F3>=45000;

F1<=100000;

F2<=120000;

F3<=95000;

Pr1 = 7.5;

Pr2 = 7.5;

Pr3 = 7.5;

RPM1 >=5000;

RPM1 <=7500;

RPM2 >=5000;

RPM2 <=7500;

RPM3 >=5000;


RPM3 <=7500;

Program Output.


Objective value: 128345.8 KWH


F1 85000.00 0.000000

RPM1 5000.000 0.000000

PR1 7.500000 0.000000

F2 65000.00 0.000000

RPM2 5000.000 0.000000

PR2 7.500000 0.000000

F3 45000.00 0.000000

RPM3 5000.000 0.000000

PR3 7.500000 0.000000

In the optimization model , EORT model of the three compressors were taken in

the objective function and the operational limits for RPM, Flow and Pressure were used

in the constraint equations. The output shows that the minimum power consumption will

be 128345.8 KWH.

In this problem, it is formulated in such a way that all the three compressors will

be running on load. This is to take care of any eventualities of compressor failure which

may jeopardize the process. For a two compressor operation, the formulation has to be

modified and program rerun to get the optimum results. Since the compressed air demand

is just 195000 nm3/hr, the same can be met by a combination of compressor 1 & 2 or

compressor 2&3 keeping the other idle. Compressor 1 & 3 option is left out as there is no

cushion in air capacity. The output for these 2 combinations is shown below.


Compressor 1 & 2 in operation. Compressor 3 idle:


Objective value: 95809.07 kwh


F1 100000.0 0.000000

RPM1 5000.000 0.000000

PR1 7.500000 0.000000

F2 95000.00 0.000000

RPM2 5000.000 0.000000

PR2 7.500000 0.000000

F3 0.000000 0.000000

RPM3 5000.000 0.000000

PR3 7.500000 0.000000

Note this operation mode results in a power saving of 32536.73 kwh i.e. equivalent to

25.35 % on base case operation. This operation has a cushion of 11.4 % on air supply. If

the demand exceeds this supply, the process will have a set back.

Let us now consider the next option of running compressor 2 & 3 .


Objective value: 177214.7 kwh


F1 0.000000 0.000000

RPM1 5000.000 0.000000

PR1 7.500000 0.000000

F2 120000.0 0.000000

RPM2 5000.000 0.000000

PR2 7.500000 0.000000

F3 75000.00 0.000000

RPM3 5000.000 0.000000


PR3 7.500000 0.000000

This option shows an increase of 48868.9 kwh for the same demand of 195000

nm3/hr. i.e 38% higher than the three compressor operation.

Hence this is not an economically viable option. Consolidated operational

analysis of these three options related to air compressor is given below. From this it is

obvious that Optimization of operational parameters can yield energy savings with

substantial reduction in operating costs. In complex situations, there is no substitute for

quantitative operational analysis involving LP/NLP/EORT/QP techniques.

The most important property of QP's quadratic objective is that it has an

algebraic property of definiteness. The "best" quadratics are positive definite (in a

minimization problem) or negative definite (in a maximization problem). You can picture

the graph of these functions as having a "round bowl" shape with a single bottom (or

top). Refer to figs 2.05 to 2.06. ( maximization problem)

In minimization problems, the curve will be bell shaped. As the value of

independent variable increases, the vale of objective function will become lower, reach a

minimum point and start raising again.

Option Power Consumption

Kwh

Remarks

All 3 compressors in operation

128345.8 Base case

Compressors 1 & 2 only

95809.07 Lowest energy

Compressors 2 & 3 only

177214.7 Highest energy


Solving LP and QP Problems

LP problems are usually solved via the Simplex method. This method, originally

developed by Dantzig in 1948, has been dramatically enhanced in the last decade, using

advanced methods from numerical linear algebra. This has made it possible to solve LP

problems with up to hundreds of thousands -- sometimes millions -- of decision variables

and constraints.

An alternative to the Simplex method, called the Interior Point or Newton-Barrier

method, was developed by Karmarkar in 1984. Also in the last decade, this method has

been dramatically enhanced with advanced linear algebra methods so that it is often

competitive with the Simplex method, especially on very large problems.

A faster and more reliable way to solve a QP problem is to use an extension of the

Simplex method or an extension of the Interior Point or Barrier method.

Nonsmooth Optimization (NSP)

The most difficult type of optimization problem is a non smooth problem (NSP)

which has multiple feasible regions and multiple optimal points within each region. In

most of these problems, it is impractical to enumerate all the possible solutions and pick

the best one. These techniques rely on some sort of random sampling of possible

solutions. Such methods are non deterministic or stochastic -- they may yield different

solutions on different runs, even when started from the same point on the same model,

depending on which points are randomly sampled.

Solving NSP Problems

Genetic or Evolutionary Algorithms as shown in the previous examples a way to

find "good" solutions to non smooth optimization problems. (In a genetic algorithm the

problem is encoded in a series of bit strings that are manipulated by the algorithm; in an

"evolutionary algorithm," the decision variables and problem functions are used directly.

Most commercial Solver products are based on evolutionary algorithms.) These

algorithms maintain a population of candidate solutions, rather than a single best solution


so far. From existing candidate solutions, they generate new solutions through either

random mutation of single points or crossover or recombination of two or more existing

points. The population is then subject to selection that tends to eliminate the worst

candidate solutions and keep the best ones. This process is repeated, generating better and

better solutions; however, there is no way for these methods to determine that a given

solution is truly optimal. This method uses memory of past search results to guide the

direction and intensity of future searches. These methods generate successively better

solutions, but as with genetic and evolutionary algorithms, there is no way for these

methods to determine that a given solution is truly optimal.


L. Simplex method :

Simplex method is the most basic and logical method of solving an LP problem.

In this method all the relationship are converted into equations as explained below.

Linear programming is concerned with solutions to simultaneous linear equations. These

equations are developed on the basis of restrictions on the variables. These restrictions

are always expressed as inequalities. The first step is to convert them into equations.This

is done by incorporating a new variable known as slack variable. For example if the

inequality is of the form a1x1+a2x2+a3x3 < = b , this may be converted into

a1x1+a2x2+a3x3+S1 = b. The slack variable will take the value , which will satisfy the

equation. If the inequality is of the form a1x1+a2x2+a3x3 > = b, this will be converted

into an equation by subtracting a slack variable as given by a1x1+a2x2+a3x3 – S1= b.

This step is carried out for all the constraints, so that only linear equations shall

appear in the problem. For example let us consider a coal fired boiler problem as

explained below.

Case study:

A coal boiler produces 750 t/hr of steam at 100 kg/cm2 pressure and 500 oC. The

boiler uses three grades of coal c1,c2 and c3 with the following specifications.

Coal Sulfur Phosphorous Ash Net CV Wt % Wt % Wt % kcal/kg

c1 3.8 0.02 3.0 4000

c2 1.9 0.01 2.0 4500

c3 1.0 0.005 5.0 4900

Presence of excess sulfur SO2 emissions and based on the environmental

regulations, S in the coal mix should not exceed 1.8% by weight. Phosphorous % is

restricted to 0.015 % by weight. Ash content should not exceed 3.0 %. Fuel efficiency of

the boiler is 86.0, 87 and 89 % respectively. The cost of coal is 25, 28 and 31 US$/ton

respectively. Determine the coal mix that will minimize the fuel cost meeting all the


constraints , at a steam generation rate of 750 t/hr of steam. In this example, we shall

develop the Linear equations for the above mentioned constraints.

Formulation of the problem.

Let x1,x2 and x3 be the quantity of coal used in the process.

Heat output by x1 tons of coal / hr = x1 * 1000 * 4000 kcal/hr.

Heat absorbed by boiler = x1 * 1000 * 4000 * 0.86 kcal /hr.

Steam Generated by x1 t/hr of fuel = (x1 * 1000 * 4000 * 0.86) / (780*1000)

(Enthalpy of steam is 780 kcal/kg) = 4.410256 x1 t/hr

Steam Generated by x2 t/hr of fuel = (x2 * 4500 *0.87 ) /780 = 5.01923 x2 t/hr

Steam Generated by x3 t/hr of fuel =(x3 * 4900*0.89) / 780 = 5.591025 x3

The sum of these 3 equations should be equal to 750 t/hr.

Steam Generation constraint:

i.e. 4.410256 x1 + 5.01923 x2 + 5.591025 x3 >= 750 (I)

Sulfur Content Constraint.

(3.8 x1 + 1.9 x2 + 1.0 x3) / (x1+x2+x3) < = 1.8

i.e. 2.0 x1 + 0.1 x2 –0.8 x3 <= 0 (II)

Phosphorous Content constraint.

0.02 x1 + 0.01 x2 + 0.005 x3 / (x1 + x2 + x3) <= 0.015

i.e. 0.005 x1 – 0.005 x2 – 0.010 x3 <= 0 (III)

Ash content constraint:

3.0 x1 + 2.0 x2 + 5.0 x3 / (x1 + x2 + x3) <= 3.0

i.e. 0 x1 –1.0 x2 + 2.0 x3 <= 0 (IV)

Objective function :

Minimize 25 x1 + 28 x2 +31 x3

Let s1,s2,s3 and s4 be the slack variables added to equations I to IV. The final LP

problem is represented by

Minimize 25 x1 + 28 x2 +31 x3

ST


4.410256 x1 + 5.01923 x2 + 5.591025 x3 + S1 = 750 (Ia)

2.0 x1 + 0.1 x2 –0.8 x3 –s2 = 0 (IIa)

0.005 x1 – 0.005 x2 – 0.010 x3 – s3 = 0 (IIIa)

–1.0 x2 + 2.0 x3 – s4 = 0 (IVa)

To get the initial feasible solution from the objective function, x1 should be maximized to

minimize the fuel cost.

LP program output

Min = 25 * x1 + 28 * x2 +31* x3;

4.410256* x1 + 5.01923 * x2 + 5.591025 * x3 >= 750;

2.0*x1 + 0.1* x2 -0.8* x3 < = 0;

0.005 * x1 - 0.005* x2 - 0.010* x3 <= 0;

-1.0 * x2 + 2.0 * x3 <=0;

Global optimal solution found at iteration: 3



x1 0.000000 0.4507412

x2 95.97245 0.000000

x3 47.98623 0.000000

Coal Wt tons Wt fraction

Scontent

Phosporous Ash % Steam gen t/hr

x1 0 0 x2 95.97245 0.66667 1.9 0.01 2.0 5.01923 x3 47.98623 0.33333 1.0 0.005 5.0 5.591025Total 143.95868 1.0000 1.6 0.0083 2.99999 750 Targets 1.80 0.015 3.000 750

Check for constraints fulfillment .


For solving the above equations Ia through IVa and getting the optimal values,

readers are advised to refer to any standard Operations Research / Linear Programming

books listed in the reference. Above solution has been obtained by using Lingo 8

program, which may be downloaded from www.lindo.com. EORT Models have been

developed using SCIMOD software or Excess spread sheet programming.

Energy Efficiency Optimization Techniques - Overall System

Any complex system comprises many subsystems, some energy intensive, some

not so intensive and others least intensive. Energy efficiency optimization of the overall

system may be achieved by two methods.

By overall optimization of the total energy system data or by synthesizing the

individual sub system data into one consolidated data.

A typical example is the energy consumption of a complex refinery which

comprises many sections or sub processes which has certain process objective. Each unit

consumes energy resources such as fuel, steam and power.

Using the operating data, specific energy consumption of each unit may be

determined. This may be synthesized to arrive at overall energy consumption of the

process. In case of overall optimization method, all the energy cost centers will be taken

into the data as input parameters and a multivariable model developed to determine the

total energy consumption of the process. The observation is given in the table below. The

advantage with individual unit specific consumption is that it is possible to identify

problem areas / units more precisely for remedial action.

The overall energy consumption is the sum of individual unit consumptions. This

may vary with plant capacity utilization, equipment efficiency and other operating

parameters. In the example given below there are sixteen process units including steam

generation units in the refinery. Assuming 5 critical parameters per unit, total number of

variables to be used in the overall model shall be eighty. To arrive at a fairly accurate


model, a minimum of 80 to 90 observations are required. This is a very time consuming

process and requires extensive computer programs to solve.

The approach to modeling and optimization application is similar to EORT modeling

discussed earlier. For successful overall system energy optimization, following steps are

to be followed.

UNIT T'PUT STEAM FUEL POWER EXPORT EXPORT SP.NRG Name t/hr t/hr t/h mw STM t/h PWR mw '000/t crude 500 28.5 9.55 1275 0 0 241.5

vb 100 12.8 5.65 1585 1.2 0 686.5 cru 60 35 2.05 1200 12.5 0 643.5

hds1 100 6.5 3.21 1185 0 0 391.4 hds2 120 8.5 3.55 1275 0 0 367.9 hds3 100 6.9 3.45 1285 0 0 420.2 crude 400 25 8.05 1495 0 0 257.0 vac1 100 12.8 5.65 1585 0 0 693.9 vac2 500 28.5 9.55 1275 0 0 241.5 bitum 100 12.8 5.65 1585 0 0 693.9 treatr1 50 8.5 1.55 1075 0 0 459.6 treatr2 75 10.8 1.65 1385 0 0 344.8

lpg 35 10.5 1.55 1175 0 0 697.9 blr1 100 12.8 6.65 1585 90 0 240.9 blr2 100 15.5 6.15 1275 85 6000 183.5 blr3 100 13.8 6.05 1385 85 0 214.0

Note: Total Energy Consumed kcal/h : 8.527796E+08 Equivalent Fuel in kg/h : 81217.11 Fuel Calorific Value in kcal/kg 10500 Steam Enthalpy for consumption 680 Steam Enthalpy for generation 620 Specific Energy is given as : '000 kcal/ton


Data collection / validation / reconciliation / transfer to / modeling input files

Error detection and elimination

Parameter Estimation

Performance Monitoring - generation of Key Performance Indicators (KPI)

Optimization - improvement of efficiency and profitability

Modeling - representation of plant using mathematical models

This may be automated or manually done based on the complexity of the process and

requirement. When more number of parameters are to be used in the optimization model,

the data may be transferred from DIDC console using suitable interface modules. This

solves the problem of manual data entry. When the parameters are less, manual data entry

and processing may be done.

Since the objective of the whole exercise is energy efficiency optimization, only the

energy related parameters need to be involved in the model. To identify performance

deterioration of a particular unit, the specific energy consumption value given in the last

column may be compared to the standard / basic case.

If the deterioration is very critical, corrective actions may be taken to restore the

energy efficiency. A time dependant model may be used to indicate energy efficiency

trend of all energy intensive sub systems as explained earlier.

Energy efficiency optimization - benefits.

Optimization of equipment performance against “base case ‘ or ‘zero date ‘ or design

conditions allows the operator to assess the current and future operation against the

energy demands imposed by the site-wide balances. Proper evaluation of current

equipment performance is a vital component in the optimization strategy, providing a

more accurate energy efficiency of the equipment due to potential changes in operating

variables.


M. Optimal Operation.

The primary concern of any utilities optimization system is to satisfy the site-wide

energy balances. In simple terms, all process unit and internal utilities plant requirements

must be achieved. Utilities optimization is not generally concerned with minimizing

process unit energy requirements; these are essentially fixed by operating conditions.

Process unit energy consumption should, however, be considered within any site-wide LP

or rigorous optimization strategy which defines the current operating conditions.

Optimization aims at meeting the requisite energy at the minimum cost within the

constraints imposed on the process by energy balances, environmental considerations and

utilities equipment limitations. These have been shown in a number of examples and case

studies.

Once these energy demands are satisfied, benefits are delivered by operating the

utilities process in the most cost-effective manner. This involves determining the

optimum loading for individual items of equipment.

Typical examples are

Power generation on each turbine set.

Boiler steam generation rates.

Secondary set-points such as turbine pass-out rates.

Allocation of fuel across the available fired boilers.

Given the constantly changing nature of the energy balances, this problem may be

typically solved by an automated system .

Optimal Selection

In addition to the benefits generated by optimizing the existing utilities

configuration, the operator can consider an alternative scenario whereby the operating

set-up is changed. For example:

Determine the optimum number of operating turbine sets and line-up.


Choose between steam and electricity to drive auxiliary devices.

Make a decision on the best choice of fuel types.

This problem is typically solved less frequently and must consider equipment

availabilities together with any costs associated with altering the current configuration.

Optimization of Maintenance and Planning activities.

A properly applied process model will reflect current equipment performance.

This information may be used to compute the cost of degraded performance. A

comparison is made between the current degraded performance and the “day-zero” or

base case / clean performance. (e.g. performance of condensers / coolers etc)

The incremental cost associated with degraded performance is used together with

the anticipated costs of equipment cleaning .

These costs include:

Maintenance charges, which are generally well known.

Outage cost of the maintenance period, which is more complex and can be

computed by the optimizer.

Equipment maintenance strategies can be better planned if trends are developed and used

as a basis for evaluating future performance and costs against anticipated energy

requirement profiles. From these figures, the operator can determine the correct time for

equipment cleaning or replacement.

Another challenge facing the process industry is how to effectively manage

planned shut-down of equipment for statutory inspections or plant turnarounds.

Extending the optimization to include “what-if” scenario evaluation allows the user to

develop appropriate strategies to deal with the substantial changes in energy balances that

result from the shutting down of large-scale producers or consumers.

The Integrated Approach

Fig 2.07 shows this approach of energy efficiency optimization of the total

system. This was explained in chapter 1 with specific examples.


Process Units Optimization. This level is concerned with individual process areas within the Refinery /

Petrochemical complex and is an activity carried out frequently , depending on the

process being considered. The primary activity is Performance Monitoring of the major

steam and electricity generators and users, on an individual basis. The individual units are

modeled in a sophisticated and rigorous manner. Each equipment model contains one or

more parameters with explicit engineering meanings that can be updated to reflect the

current performance of the units. This section may be considered the base level, on top of

which higher-level functionality is built. These parameters are used in the higher-level

Optimization calculations. Using these parameter values , it is possible to evaluate and

predict equipment performance / deterioration and plan for maintenance. Hence it is

possible to maintain the energy efficiency of the total system at it’s peak.

Fig 2.07. Optimizing total energy of the system


A very important feature of optimization is the calculation and prediction of the

various restricted emissions, such as NOx, SOx and Cox which were covered in some of

the case studies. This may be evaluated for the total system. Simplified models, based on

regressed plant data relate process unit heat and power usage against production rate.

When this information is integrated for the entire units, the model output shall give the

total steam, power, water, air demand etc. This output shall be used in the utility

optimization model. These relationships are reduced to comparatively simple “specifics”

– equations used by the production planning department and expressed as “MW/ te of

product”, or “te of HP steam / te of product”.

Utilities Requirements

If an online system is used, optimization model coefficients may continuously be

upgraded to provide better level of accuracy required for optimization. These models

should be flexible to incorporate different operating modes, which can significantly alter

the quantities of energy required by a process unit. Indeed, certain process units can

switch between net export and net import depending on the operating mode.

Optimization calculations, operating around the updated process models,

determine the optimum operating conditions for the equipment currently online using the

fuels presently in use.The optimizer operates within a feasible region bounded by

operating constraints such as

Site-wide energy requirements

Equipment limits and environmental targets.

Simultaneous balancing of all steam, fuel and power demands is performed at this level,

with minimal operating cost objective for the optimizer. Operating cost typically includes

cost of different fuels, electricity import / export, cost of de-mineralized water. The

technique may be used for equipment selection / maintenance decisions. Here, the

optimum combination of all available units may be evaluated to meet the current site


demands. The operating cost function may be extended to incorporate penalties for

starting up and shutting down equipment units. This is designed to establish a realistic

pattern of equipment selection that can be effectively implemented by the utilities

operators.

Power Balance

Optimization can be extended to consider the possibilities of power import and export.

Performance of electrical equipment, balancing the load etc may also be carried out by

optimization.


Case Study – Total Energy Optimization

This explains the application of simple optimization exercise on a micro level

operation of a petrochemicals plant using a single variable throughput. (load factor). All

the process parameters such as the feed quality, pressures, temperatures etc are taken as

constant. A series of models are developed to determine the fuel, steam and power

demand including loss. Fig 2.08, 2.09 and 2.10 show the output of models in graphical

form for fuel, steam and power demand as a function of unit throughput.

Fig 2.07 shows the consumption of fuel in ‘000 tons per ‘000 tons feed to the unit.

These values represent the actual consumption of fuel based on the operating data of the

Fig 2.08. Fuel Consumption model for an aromatics unit

10.0

12.0

14.0

16.0

18.0

20.0

22.0

t'put 000 tons

fuel

'000

tons

non linear 11.80 12.00 15.33 18.23 19.79

polynomial 11.39 12.69 14.95 17.96 20.15

actual 11.36 12.75 14.79 18.24 20.00

598 675 870 960 1000


unit. The calorific value of the fuel is taken as 10000 kcal /kg and is known as Standard

Refinery Fuel. This considers the effect of fuel mix and its calorific value as an imaginary

fuel with a net calorific value of 10000 kcal/kg. This is the standard practice in Refinery

Engineering evaluations.

Fig 2.09 shows the consumption of steam in ‘000 tons per ‘000 tons feed to the unit.

These values represent the actual consumption of steam based on the operating data of

150

170

190

210

230

250

270

t'put '000 t

stea

m '0

00 t

non linear polynomial actual

non linear 169.7 171.6 207.2 238.7 255.7

polynomial 169.2 172.5 206.7 238.3 256.2

actual 167.8 175.5 200.0 249.6 250.0

598 675 870 960 1000

Fig 2.09. Steam Consumption model for an aromatics unit


the unit. Steam pressure to the unit was 15 kg/cm2 g at 320 oC. Condensate recovery

from the unit was 40 %. This information may be used to determine the specific energy

consumption of the unit at various throughput levels.

Fig 2.10 shows the electrical power consumption of the aromatics unit at various

throughput levels. This represents the actual data of an aromatics plant. This may be used

in the specific energy consumption model, using appropriate conversion factors.

80.0

90.0

100.0

110.0

120.0

130.0

140.0

150.0

160.0

t'put '000t

pow

er '0

00 m

w

nonlin polynomial actual

nonlin 93.4 105.6 133.7 145.3 150.2polynomial 92.0 107.9 132.4 144.4 151.4actual 91.7 108.5 130.9 147.0 150.0

598 675 870 960 1000

Fig 2.10. Power Consumption model for an aromatics unit


During the operation of any process plant, consuming liquid and gaseous fuels,

some quantity of fuel will always be lost in the form of flare loss or vent loss. In

aromatics production plant, additional losses encountered are hydrogen and solvent used

for extraction.

There are two methods of evaluating the specific energy consumption of the

plant. They are given by

Specific fuel and loss method and

Specific fuel consumption method.

Difference between the two values is denoted by specific energy loss of the plant,

a potion of which is controllable and the balance uncontrollable.

For reducing this energy loss, certain technological / operational changes may be

required and must be determined by cost-benefit analysis of the system on a case to case

basis. Fig 2.11 shows the hydrocarbon loss from the aromatics unit example. For this

purpose, actual plant data from a running unit has been considered.

Advantages of ARU energy consumption models:

These models shown in the Aromatics Unit example are the simplest and does not require

any rigorous process calculations and only the minimum plant data will be used for

developing these. Since this approach considers only the feed rate and other energy input

parameters independently, the plant energy consumption may be obtained by

synthesizing these values by incorporating an additional conversion program.

From these models, it is easy to identify areas of higher energy consumption

energy wise or cost wise from the operating parameters. Area of deterioration may be

pinpointed for corrective action at the right time.


A total energy consumption model for various units using energy synthesis approach is

given in the table in previous pages.

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

t'put '000 tons

hc lo

ss '0

00 to

ns

actual nl model poly model

actual 5.21 5.31 5.40 6.09 7.20

nl model 5.39 5.02 5.49 6.39 6.93

poly model 5.19 5.36 5.30 6.26 7.11

598 675 870 960 1000

Fig 2.11. Hydrocarbon Loss model for an aromatics unit


Methodology for conversion of energy inputs into specific energy consumption is

shown in the table below.

470.0

480.0

490.0

500.0

510.0

520.0

530.0

540.0

550.0

t'put 000 tons

sp.e

nerg

y '0

00 k

cal/t

sp.energy

sp.energy 540.7 529.9 478.7 524.5 524.0

598 675 870 960 1000

Fig 2.12. Actual Specific energy consumption for an aromatics unit

Specific total energy consumption - Aromatics Plantt'put sp.fuel sp.steam sp.power sp.energy 000t kg/ton kg/ton kw/ton 000 kcal/t 598 19.0 280.6 153.3 540.7 675 18.9 260.0 160.7 529.9 870 17.0 229.9 150.5 478.7 960 19.0 260.0 153.1 524.5

1000 20.0 250.0 150.0 524.0


It may be seen from fig 2.12, that specific energy consumption of the unit is

minimum at the throughput of 870000 tons capacity. Using the energy cost details, it is

possible to determine the optimum plant load that will minimize the total energy cost.

Probably, that may show a different scenario as the energy cost of each input is likely to

be different depending on the energy cost at the generation source. For example, if the

steam and power are supplied from waste heat cogeneration system, the cost shall be

lower, than if they were generated by conventional fuel fired cogeneration system.

Solvent Loss Model.

In the Aromatics production unit, a solvent such as NMP or Sulfoline is used to

extract the aromatics from the aromatics / non aromatics solution after the conversion

stage. These solvents are volatile and are recovered in a recovery column. Loss of solvent

is minimized by maintaining proper bottom temperature in a steam heated reboiler. Loss

of solvent is also a form of energy loss in an aromatics unit. This also tends to vary with

the capacity utilization of the plant. Fig 2.13 shows the solvent loss as a function of

throughput of the Aromatics plant under consideration. Fig 2.14 shows the specific

solvent loss actual for the aromatics plant under study.

From the data given in the label, it is possible to determine the specific solvent

loss for the aromatics plant. Once this is done, it becomes easier to determine the

optimum plant capacity at which the operating profit will be maximum.


Using the above specific consumption details, the unit operating capacity may be

optimized as shown below.

2.50

3.00

3.50

4.00

4.50

5.00

5.50

t'put '000 t

solv

ent '

000

t

actual nl model poly model

actual 2.93 2.77 3.48 4.32 5.00

nl model 2.95 2.74 3.47 4.40 4.94

poly model 2.92 2.79 3.44 4.39 4.96

598 675 870 960 1000

Fig 2.13. Solvent Loss model for an aromatics unit


3.80

4.00

4.20

4.40

4.60

4.80

5.00

5.20

t'put 000 tons

solv

ent l

oss k

g/to

n

actual

actual 4.90 4.10 4.00 4.50 5.00

598 675 870 960 1000

Fig 2.14. Specific Solvent Loss model for an aromatics unit


Total operating cost model

From the specific consumption models for fuel, steam, power, hydrocarbon loss

and solvent loss it is possible to develop a costing model which may be used to develop a

profitability model for the unit which may be optimized using the techniques mentioned

earlier .

If Cf, Cs,Cp,Chc & Cs are the cost of fuel, Steam, power, hydrocarbon and

solvent per ton respectively and specific consumption functions are f(x),s(x),p(x),h(x)

and l(x) respectively, then the total costs incurred are

Ct = Cf*f(x)+Cs*s(x)+Cp*p(x)+Chc*h(x)+Cs*l(x)

This is called ‘cost function’ which may be optimized to minimize the total cost.

As could be seen from the cost function equation, total cost is determined by the

individual cost of each energy component and the consumption pattern. For example let

us take just three energy components given by the equation

Ct = Cf*(a1*x2+b1*x +c1 ) + Cs*(a2*x2+b2*x+c2)+Cp*(a3x2+b3*x+c3)

Differentiating the above equation with respect to x and equating to zero, we get

dct /dx = (2 *Cf *a1 * x + b1) + (2 *Cs *a2 * x + b2) + (2 *Cp *a3 * x + b3)

= 2 x* (Cf *a1 + Cs *a2 + Cp *a3) + (b1+b2+b3) = 0

i.e. x = -(b1+b2+b3) / 2* (Cf *a1 + Cs *a2 + Cp *a3)

Differentiating the above differential w.r.t ‘x’ again, and substituting the value of

x, if the second differential value is negative, then x is maximum. If the value is positive,

then the value of x is minimum.

If the value of x fulfills the constraints stipulated, then x represents a feasible

solution. Else another value of x nearest to the lowest value of x in the constraint will

become the optimal solution.

Table given below shows the specific consumption models for fuel, steam, power, hc loss

and solvent loss for the Aromatics plant example.


Cost of fuel, steam, power , hydro carbon mixture and solvent are used to develop

a cost function model as shown below.

Sp cost = 0.15*(5.900398E-05 * (x)^2 - 9.345668E-02 * x + 55.790424) + 0.02 *

(7.923447E-04 * (x)^2 - 1.328423 * x + 813.914264)+ 0.3* (1.934422E-04 * (x)^2 -

.3310468 * x + 298.171512) + 0.3 * (3.645079E-05 * (x)^2 - 6.303739E-02 * x +

34.19038)+ 2.50 * (2.872752E-05 * (x)^2 - 4.560455E-02 * x + 22.01971).

On simplification this is reduced to

Sp.total cost = 0.000165484188* x^2 - 0.272823594 * x + 55.049275

Following table shows the workings of arriving at the final coefficients of the Cost

function model.

Specific Consumption Models - Aromatics Unit

spfmax = (5.900398E-05 * (x)^2 - 9.345668E-02 * x + 55.790424)

spstmmax = (7.923447E-04 * (x)^2 - 1.328423 * x + 813.914264)

Sppwrmax = (1.934422E-04 * (x)^2 - .3310468 * x + 298.171512)

Hclossmax = (3.645079E-05 * (x)^2 - 6.303739E-02 * x + 34.19038)

Sp.solvmax = (2.872752E-05 * (x)^2 - 4.560455E-02 * x + 22.01971)

Cost of fuel = 150 us$/t = 0.15 us$/kg

Cost of steam = 20 us$/t = 0.02 us$/kg

Cost of power = 300 us$/mw = 0.30 us$/kw

Cost of hc = 300 us$/t = 0.30 us$/kg

Cost of Solvent = 2500 us$/t = 2.50 us$/kg


This on differentiated with respect to x and equating to zero gives

0.000330968 x = 0.272823594.

i.e. x = 0.272823594 / 0.000330968 = 824.3 (‘000 tons)

Second order differential gives a positive value. Hence the value of x is

minimum. This shows that at 824.3 (‘000) tons of throughput level, the total cost of

energy components will be minimum.

This is a deterministic model where the outcome is given by a cost function

equation. It may be noted, that in arriving at this cost function, all the cost parameters

have been given a firm value.

Since the cost of these energy components are dynamic, the optimum level of

operation will shift from time to time. Hence developing a cost function program will be

very useful for taking care of cost dynamics.

Fig 2.15 shows the specific total cost model derived from these observations.

cost fn x^2 fn x fn const

0.15 8.8506E-06 -0.0140185 8.3685636

0.02 1.58469E-05 -0.0265685 16.278285

0.30 5.80327E-05 -0.0993140 89.451454

0.30 1.09352E-05 -0.0189112 10.257114

2.50 7.18188E-05 -0.1140114 55.049275

total 0.000165484 -0.2728236 179.40469


This example shows the importance of optimization in plant operation which will

minimize the total energy cost and maximize net operating profit of any process. Most

important part of the whole exercise is the data must be consistent and reliable. In many

cases, data reconciliation must be applied. However reliability of the model depends on

the number of observations used in the model.

67

68

69

70

71

72

73

t'put '000 MT

sp.to

tal c

ost u

s$

sp.total cost

sp.total cost 67.87 67.06 67.07 67.91 69.57 72.07

750 800 850 900 950 1000

Fig 2.15. Specific total cost vs throughput – Aromatics Unit.


Impact of process variables:

Accuracy of the models may be improved , by incorporating additional process

parameters which have an impact on energy consumption of the process. In the case of

Aromatics Unit, feed quality represented by PONA analysis ( Paraffin, Olefin, Naphthene

and Aromatic content) plays an important role in energy consumption.

Hence the models may be modified into a two variable model of the form

E = a * (t’put)b * (N+A)c

where E is the energy consumption , N and A are volume % of naphthenes and aromatics

present in the feed. In the single variable model, all these parameters are assumed to be

constant and only the capacity utilization has been considered. Other parameter which

affects energy consumption in the Aromatics Unit is the hydrogen/hydro carbon ratio of

the recycle gas ( hydrogen purity), Severity of reforming operation and catalyst activity.

An EORT model covering these parameters may be developed to determine

energy consumption as a function of conversion.

Case study – optimum insulation thickness

This is a very common optimization problem encountered in insulation technology. In hot

product lines such as steam pipelines, hot oils etc thickness of insulation has to be

optimized. If x is the thickness of insulation, fixed increases with thickness. But cost of

heat loss reduces.

Hence the total cost which is the sum of fixed cost and variable cost reaches an

optimum level, at which the total cost must be minimum. Equations may be written for

these costs as given below.

Cf = a * x + b

Where Cf is the fixed cost, x is insulation thickness and a & b are constants.

Cost of heat loss may be written as

Chl = c/x + d


Where Chl is the cost of heat loss, c and d are constants.

Total cost Ct = Cf + Chl = a * x + b + c/x + d

Differentiating this function w.r.t x we get

dCt / dx = a – c/x2 = 0

x2 = c/a

x = c/a

Differentiating the function again, we get d2Ct / dx2 = + 2*c/3*x3 = +ve

Hence x value represent the minimum point in the total cost function.

Optimization procedure – single variable.

The first step in optimization is to determine which parameter has to be

optimized. Typical factors could be total cost per unit of production (e.g. Aromatics Unit

case discussed above), profit, % conversion etc. Once the basis is determined, relations

between parameters as a function of objective variable must be developed by regression

methods or process models. Finally these relationships are combined to give the desired

optimum conditions. This has been shown in the insulation thickness. In this the objective

function is to determine optimum insulation thickness which will minimize the total cost .

The variables are the fixed costs, variable costs which are combined to give the total cost.

The problem was solved by analytical method using differential calculus.

Optimization procedure – two or more variables.

When two or more variables are involved in optimizing an objective function, the

procedure for optimization is tedious. However the procedure is similar to a single

variable problem. Out of the two independent variables, one variable is kept constant and

the other optimal variable is determined.

Let us consider the total cost function given by

Ct = ax + b/xy + cy + d


Where a,b,c & d are positive constants.

The objective function is to find out the values of x and y which will minimize the total

cost given by the above equation. For this purpose, partial derivatives of x & y keeping

one variable constant is carried out as shown.

( Ct / x) = a – b/x2y = 0

( Ct / y) = -b/xy2 + c = 0

Solving these two equations, we get x = (cb/a2)1/3 and y = (ab/c2)1/3

When more than two variables are involved, the same procedure is adopted to determine

the value of independent variables.

Let us assume that the range of x variable is 10 to 20 and the range of y is 12 to 36 and

the coefficients in the cost function are 2.33, 11900,1.86 and 10. When these values are

substituted in the equation, the distribution of x, y and Ct will be as shown in the

following table.

Table. Optimizing two variables ( x and y) in the above example

This gives an optimum value of x = 16 and y =20. Total Cost is 121.7

y values

12 18 24 30 36 20

x

10 154.8 132.9 127.5 128.8 133.3 130.0

11 148.1 129.2 125.3 127.5 132.6 126.9

12 142.9 126.5 123.9 126.8 132.5 124.7

13 138.9 124.6 123.1 126.6 132.7 123.3

14 135.8 123.3 122.7 126.8 133.2 122.3

15 133.4 122.5 122.6 127.2 133.9 121.8

16 131.6 122.1 122.9 127.9 134.9 121.7

17 130.3 122.0 123.4 128.7 136.0 121.8

18 129.4 122.1 124.1 129.8 137.3 122.2

19 128.8 122.5 125.0 130.9 138.6 122.8

20 128.5 123.1 126.0 132.2 140.1 123.6


Analytical method gives x = 15.97 and y = 20.01.

This problem may be solved by any NLP solvers like Lingo. The problem may be

formulated as

Minimize 2.33 * x + 11900 /x *y + 1.86 * y + 10

Subject to

x >= 10

x < 20

y >= 12

y < 20

Table given above shows the methodology adopted analytically, using the same

programming steps.

Practical Energy Efficiency constraints in optimization

This is an important factor that must be considered in energy efficiency

optimization problems. These constraints should be fixed with thorough knowledge of the

energy generating equipments like turbines, boilers, heaters etc. Impact of process

parameters on energy efficiency must be understood clearly.

In the case of gas turbine for example, the limitations are fuel input, heat release,

air/fuel ratio, power output at terminals and exhaust gas temperature. From pollution

point of view, Sulfur level in the fuel also forms a constraint. In the case of Heat

Recovery Steam Generator, the constraints are waste gas flow rate, its temperature, stack

temperature, steam generation rate etc. An EORT model may be developed using these

operating parameters and this will form the objective function. Since most of the energy

intensive equipments deteriorate with use, these constraints also will undergo a change

depending on the existing condition of the hardware configuration, which will ultimately

affect the optimal operating parameters. This is valid for almost all equipments in

operation and optimization is a dynamic exercise that must be conducted regularly to

minimize the overall energy cost.


Impact of energy efficiency on production cost and profitability.

It was proved in the examples and case studies that energy efficiency optimization

does reduce the production cost and increases profitability. Tangible results could be

obtained only when optimization is carried out on class A equipment series, as they

consume mote than 70 to 75% of total energy input. Besides this, waste heat recovery

must be optimized wherever possible, as this amounts to energy recycling and the

specific energy consumption and energy cost will be minimum at optimum energy

recycling. This is a dynamic exercise and the optimum point is determined by the energy

cost equivalent of fuel.

Impact of energy efficiency on environment

Efficient heater / boiler operation conserves fuel and simultaneously reduce

pollution . However, for every fired heater / boiler an optimum excess air has to be

maintained, below which energy conservation may tilt the balance to increase pollutants (

in the form of Carbon monoxide , Soot, SPM etc) as given in the fuel mix example. When

the excess air is reduced, heater / boiler efficiency starts improving and SO2 emissions

will start coming down as shown in figs 2.16. After certain point, fuel input will start

increasing, as the entire fuel will face air deficiency and will result in incomplete

combustion Fig 2.16. SO2 emission in kg/hr vs Heater efficiency.


The flue gas will indicate presence of Carbon monoxide, soot etc as shown in fig

2.17. In this typical case, the flue gas does not contain any CO or soot at 10% excess air.

When an attempt is made to reduce the excess air to 5,2.5 & 1 %, CO in the flue gas

starts increasing to 3000, 5000 & 12000 ppm respectively. This indicates incomplete

combustion of fuel. Under these conditions, smoke could be seen from the stack.

It is obvious from this example, that each heater has an optimum excess air, below

which pollution starts increasing in the form of soot, CO etc. Similarly at high excess air

rates also pollution will start increasing due to lower efficiency and higher fuel

consumption for the same heat duty. Hence efficient heater operation could be achieved

by operating the heaters / boilers at optimum excess air level. Optimum excess air is a

dynamic parameter that varies with the design, life, burner condition etc. These

parameters need to be monitored on a continuous basis to achieve high efficiency and low

Fig 2.17. CO emission in ppm vs % Excess air


emissions. Carbon monoxide is a toxic gas and could endanger the environment if

adequate care is not taken. In the example given above , while excess air reduction from

10% to 1% reduces emission from 37.2 ton/h to 34.32, CO pollution starts increasing

from 0 ppm to 12000 ppm which is higher than the allowable limit of 1200 ppm. Above

is a clear case of sub optimal operation. These factors must be taken into consideration in

fixing the constraints while optimization exercise is carried out.

Optimal fuel mix by LP Model

Fuel mix plays an important role in controlling atmospheric emissions. Cleaner fuels cost

more than the dirty stocks, but reduce pollution control costs and offer longer equipment

life and better efficiency. An optimum mix may be arrived at using LP techniques and

NLP models as shown in the fuel mix examples. In general, impact of energy efficiency

on environment should be considered in the constraints.

Impact of energy efficiency on safety

Safety is of utmost importance in optimization exercise. There are many parameters,

which are likely to be unnoticed / sacrificed during energy efficiency optimization. It is a

good practice, if the constraint relationships include safety parameters quantitatively

which may be judged by past experience and historical accident data.

Practical constraints in equipment, sub systems and systems.

A number of constraints may be faced in real life situations during operation.

There are occasions in which the energy efficiency has to be sacrificed for meeting the

production targets, at a lower overall efficiency. A less efficient system of heaters /

boilers will have to be necessarily operated to meet the production target. It is the judicial

choice of the operations chief that will determine the implementation of optimal

parameters.

EORT models will be an answer to fixing the operational constraint limits, as his is built

on actual operating parameters under various conditions. When once the model is

developed, it is possible to determine the feasibility of meeting the constraints by a trial


run at the stipulated parameters. Hence, plant parameter optimization exercise shall be

carried out successfully after getting the concurred values of the various constraints.

Another problem experienced in the complex plant is the down stream unit capacity /

equipment sizing, which may not match the upstream process flows / conditions leading

to a non optimal condition. This is a challenge to be tackled by the optimization group

which may warrant a rigorous process study.

Conclusions

In this chapter we have covered the basic principles of optimization that may be

used in real life situations. A number of examples such as boiler loading, heater

efficiency optimization, fuel mix optimization, EORT modeling etc were explained with

live cases. A typical Aromatics Unit example was shown in detail as to how the capacity

of the plant may be optimized for maximum operating profit. These methods may be

applied in real life situations to get the best out of competing resources.


1. Rajan.G.G., "Computer-aided Utility Management', Paper, International

Conference on Computer Simulation, Oregon, May.1996

2. W.L.Nelson, 'Petroleum Refinery Engineering ', McGraw-Hill Book Company,

Inc. New York

3. David M.Himmelblau, 'Process Analysis by Statistical Methods', John Wiley &

Sons, Inc. New York

4. Dr.Rajan.G.G., 'Optimal Management of Resources in Process Industries ', Indian

Society for Technical Education, April 1982

5. Technical Audit - User's Mnual ', Techno Software International, India.

6. Arnold H.Boas, 'Modern Mathematical Tools for Optimisation', Reprint Chemical

Engineering, McGraw-Hill Book Company, Inc. New York

7. Dr.Rajan.G.G., 'Energy Efficiency Improvement - Decision Models for selection

of alternates and cost effective solutions', International Conference on Energy,

WEC, New Delhi. 12th Jan '96.

8. Dr.Rajan.G.G.,'Operational Analysis of Process Industries for Productivity

Improvement', Productivity, NPC,New Delhi, July-Sep '89

9. Dr.Rajan.G.G., 'Computer-aided Process Decisions', Summer Computer

Simulation Conference, Ottawa,Canada, May 1995

10. Dr.Rajan.G.G., 'ERP Models for the Hydro Carbon Processing Industry', MRL-

CHT Oil Industry Computer Meet, Chennai, April '99..

11. Samuel E.Bodily, 'Modern Decision Making', McGraw-Hill Book Company, Inc.

New York

12. Max S.Peters, Klaus D.Timmerhaus, Plant Design and Economics for Chemical

Engineers, McGraw Hill Kogakusha Ltd, Toranto

13. Gupta P.K , Hira D.S., Operations Research , S.Chand and Company, New Delhi,

India

14. Loomba M.P., Linear Programming , Tata McGraw Hill Publishing Company

Ltd, New Delhi, India.

15. Rajan.G.G., 'Computer-applications and Decision Support System for Efficient

Management of Industries',PhD. Thesis, BITS,Pilani, 1996

16. Owen L.Davies,Peter L.Goldsmith, ' Statistical Methods in Research and

Production', Longman,New York

References

Energy Efficiency Optimization for power Plants Training Program – Dr.G.G.Rajan

109

POWER PLANT MANAGEMENT

3


110

3. POWER PLANT MANAGEMENT

Introduction :

A modern power plant has a typical layout as shown in fig 3.01. Power plant design has

a number of options .The design criteria , based on process economics could be

A coal fired boiler ( pulverized coal firing type, Fluidized bed combustion

type )

An oil fired boiler or

A gas fired boiler

A waste heat boiler.

Thermal Power Plant

In a thermal power plant, steam is produced and used to spin a turbine that operates a

generator. Shown here is a diagram of a conventional thermal power plant, which uses coal, oil,

or natural gas as fuel to boil water to produce the steam. The electricity generated at the plant is

sent to consumers through high-voltage power lines.


111

In thermal power plant operation, each equipment contributes to the overall energy

efficiency of the plant. Since efficiency of each equipment contributes to the overall efficiency of

the power plant, the total system has to be optimized for achieving the maximum efficiency at

required power demand, which is dynamic.

Boiler system :

Factors which affect the performance of boiler may be denoted by an energy supply chain

concept. Starting from the fuel, till steam generation a number of parameters influence boiler

efficiency.

Energy Resources

All Energy Consuming Systems / Devices / equipment consume Energy in one of the

following forms or a combination depending on the technology and / or requirement.

They are

Fuel Energy

Electrical energy

Steam energy

Thermal energy

Total boiler system may be split into sub systems such as fuel system, Air system, BFW

system, boiler main , steam system, Flue gas system and heat recovery system. Each subsystem

performance has an impact on overall boiler efficiency. Parameters that affect the performance

of each subsystem is given in the following table.


112

Processparameter

Associatedattribute

Parameters Key indicators

Fuel Solid

Liquid

Gas

Type, c/h ratio, particle size, ash content, burner efficiency., temperature

Density, c/h ratio, viscosity, ash content, atomization efficiency., temperature

Gas composition , burner efficiency, temperature

Good quality and particle size improve combustion efficiency.Combustion zone temperature, air / fuel ratio, specific fuel consumption under identical conditions.

,, ,,

,, ,,

Air Air purity, humidity and temperature, excess air quantity

Air contamination and humidity reduces efficiency . High air temperature increases efficiency. Pre heated air saves fuel burnt in the boiler for the same load.

Key indicators : Combustion zone temperature, excess air for identical process conditions. Reduces FD fan load and energy consumption.

Burner system Burner efficiency

Fuel temperature / viscosity, Atomizing steam to fuel ratio, Fuel pressure, nozzle diameter

High fuel temperature and low viscosity improves burner efficiency. Steam to fuel ratio must be optimum. Flame length, color and absence of soot are visible parameters. Combustion zone temperature is an indication of good combustion.

Combustion system

Combustion efficiency

Flame temperature and good flame pattern

Proper combustion increases boiler efficiency all other parameters remaining constant.


113

Boiler Main BFW quality

TDS should be within allowable limits. pH of BFW should be between 7 to 7.2.

If TDS becomes excessive, water tubes will be blocked with Ca / Mg salts and may lead to tube failure. If pH is very low tube failure will be caused by corrosion.

Economizer Economizer efficiency. Higher the temperature pick up of temperature by BFW, higher will be the efficiency of boiler. and lower the flue gas temperature

Any fouling on economizer tubes, will retard heat transfer and increase fuel consumption in boiler. Economizer surface must be as clean as possible.


114

BOILER THEORY AND OPERATING PRINCIPLES

Steam generator is a pressure vessel into which liquid water is pumped at the operating

pressure . After the heat has vaporized the liquid , resulting steam is then ready for use or for

further heating in a super heater . A typical section of a water tube boiler ( end view from drum

side ) is shown in fig 3.01 .Hot boiler feed water (bfw) enters the boiler drum at a pressure

slightly higher than the drum pressure .The water gets circulated in the water tubes which are

heated by the hot gases passing through the combustion chamber .

During circulation, water is converted into steam and flashes in the upper portion of the

drum.This steam will be at it's saturated temperature corresponding to the operating pressure

of the drum .This steam passes through a set of coils into the combustion chamber, where it gets

super heated to the required temperature by contact with hot gases produced in the combustion

chamber due to combustion of fuel .

Super heater is a steam header which is placed in the combustion chamber as shown in

the figure . After getting super heated , the steam may be used in turbine for generating

power / work or may be used in process as per the heat requirement . For maintaining the

required super heated steam temperature , a de- super heater is placed outside the boiler


115

chamber , where boiler feed water is added at a lower temperature and required quantity , by

a flow control device.

During boiler feed water circulation from the drum to lower pot and vaporization , the

concentration of hardness producing solids increase and tend to produce severe scales in the

tubes , drum walls etc . This tendency retards heat transfer and thermal efficiency of the

boiler . These scales reduce flow rates in the coil section and cause tube failure also. To over

come this problem , a part of the water is blown down periodically and in some cases

continuously from the drum , based on the boiler feed water hardness and quality. This is a

source of energy loss , as excess of ' blow down ' corresponds to excess fuel input to the boiler.

From energy management point of view, thermal efficiency of the boiler which is

defined as the ratio of steam generated to the thermal energy input may be evaluated by

direct method or in-direct method as in the case of a heater. Fig 3.02 shows the energy input

and output streams for a boiler, used for thermal efficiency calculation.

Energy Input to boiler ( fuel side) :

Enthalpy + Enthalpy + Heat ofin fuel in air combustion

Energy losses ( fuel side ) :

Enthalpy in + heatflue gases losses ( convection & radiation )


116

Thermal Efficiency of a boiler is determined by one of the following methods.

Method A : from heat balance

Method B : from heat losses

In either case , the thermal efficiency may be expressed in terms of either net or

gross heating value of the fuel . Heat output in the case of a boiler is the heat value of steam

generated by the boiler less the heat value of feed water and steam returned to the unit.

Heat input includes the heat value of fuel used by the unit ( based on net or gross

calorific value) plus any other waste heat that may be supplied to the boiler from an external

source.

Energy Efficiency of boilers

A heat balance of boiler indicates the type of heat losses which could be used for

identifying areas of improvement . A number of parameters affect the thermal efficiency of

the boiler as in the case of a heater. They are

* Sensible heat in flue gases

* Convection and Radiation losses.

* Unburnt Combustibles in flue gases.

* Unburnt combustibles in refuse

* Blow Down losses etc

3..2.1.Sensible heat in flue gases:

This loss is the largest in a boiler ( refer box 4.10 ) and represents the heat carried

away by the flue gases and released to the atmosphere without doing any useful work .

Hence it is obvious that if more than the required quantity of air is used in a boiler ,

more will be the loss in flue gases and the thermal efficiency of the boiler will be reduced

correspondingly.


117

3.2.2 Convection and Radiation losses :

This depends on the temperature of boiler's external surface . Quantity of heat lost by

convection and radiation is a function of shell temperature and wind velocity. This loss occurs

basically due to poor insulation and poor design characteristics . If the refractory lining and

other insulating materials are not in good condition , these losses will tend to increase and

reduce the thermal efficiency of the boiler. Hence it is imperative that the surface temperature

at various sections should be monitored periodically to minimise this loss for timely action.

3.2.3 Unburnt matter in flue gas:

When combustion is incomplete , part of the carbon present in the hydro carbon fuel

may be converted into carbon monoxide instead of carbon-di-oxide.When the carbon content

of the fuel is not fully burnt to carbon-di-oxide, there will be substantial energy loss to the

atmosphere from the flue gases ( besides atmospheric pollution) , the loss being proportional to

the amount of carbon monoxide produced .This could be estimated by a carbon balance across

the boiler taking into consideration carbon content in fuel , carbon-di-oxide and carbon

monoxide in the flue gases and carbon content in the refuse.

Unburnt matter may show itself in the flue gas in the form of black smoke which

represents presence of carbon particles in the flue gas. It may also show itself as Carbon

monoxide in the flue gas . Heat loss due to incomplete combustion may be estimated from the

heat of combustion data as given below.

C + O2 = CO2 + 8084 kcal / kg of carbon

2C + O2 = 2CO + 2430 kcal / kg of carbon

While every kg of carbon present in the flue gas represents a loss of 8084 kcals , every kg of

Carbon partially oxidized to CO results in a loss of 5654 kcals . This situation could be

controlled with optimum excess air input to the boiler.


118

Another source of loss is from the refuse formed which is applicable to solid fuels .

This is estimated by evaluation of refuse formed per unit weight of fuel burnt and from the

analysis of combustibles in the refuse .

Unburnt matter for normal efficiency calculation is taken as Carbon. For example if

Wr is the quantity of refuse produced per unit weight of fuel and if C is the weight fraction of

carbon in refuse, then heat lost in the refuse is given by

Hr = Wr x C x 8084 kcal / kg fuel burnt

3.2.4 Blow Down Losses

Dissolved salts find entry to the boiler through make-up water which is continuously fed

by the Boiler Feed Water pump ( bfw) . In the boiler , there is continuous evaporation of water

into steam . \

This leaves behind the salts in the boiler. Concentration of these salts , tend to increase

in the boiler drum and starts precipitation after certain concentration level .

Water from the drum should be blown down to prevent concentration of salts beyond

certain limits . Since the water in the boiler drum is at a high temperature ( equivalent to it's

saturation temperature at boiler drum pressure ) , excess blow-down will lead to loss of energy

known as 'blow-down losses ' .

Blow-down rate reduces the boiler efficiency considerably as could be seen from

fig 3.-03.

Hence it is imperative that blow-down rates are optimized ,based on the hardness levels

of boiler drum water which is a function of the operating pressure.


119

Example 1 – Power Generation

In power generation a number of combinations of energy resource may be

deployed.

Conventional Power Generation Route is given below.

Factors that affect power generation efficiency are

Combustion Efficiency

Boiler Efficiency / Steam transmission efficiency

Turbine Efficiency and

Generator efficiency.

In this case optimization refers to fuel, Steam and Power system

Mathematical Model

Mathematically, the overall efficiency of the micro power Generation system may be

represented by

E overall = E comb * E boiler * E turbine * E generator

In a typical case following

conditions were observed as

against design values .

Observed Design

Combustion efficiency 1.00 1.00

Boiler Efficiency 0.85 0.91

Turbine Efficiency 0.40 0.45

Generator Efficiency 0.95 0.975

Overall System efficiency 0.3230 0.39926

Productivity as % of base 0.8089 1.0000


120

Efficiency Deviations

In the example

Actual boiler efficiency is lower by 16%

Turbine efficiency is lower by 5%

Generator Efficiency is lower by 2.5 %

Overall system efficiency is lower by 6.626 %

Productivity is lower by 19. 11 %

This aspect is seldom noticed by energy users.

In power plant operation , this will be observed in terms of excessive fuel consumption

and high operating cost.

Power plant management :

An effective power plant management system covers efficiency optimization of

o Fuel system

o Boiler feed water system

o Combustion control system

o Boiler furnace

o Air pre heater / Flue gas control system

o Super heater system

o Economizer system etc.

Following formats show the amount of information required for evaluating the boiler

efficiency by indirect method.

When boiler efficiency is calculated periodically, it is possible to identify efficiency

deterioration of a particular boiler in a power plant for corrective action.

Power plant operating economics depends of the fuel cost, steam / fuel ratio and power

generation quantity. Hence turbine efficiency also has to be monitored periodically. A rule of

thumb is the specific steam consumption / mw power generated.


121

1. Boiler Efficiency evaluation by indirect method

Design case A User : ABC Power plant B Unit : TT Power Division C Equipment : Boiler no 6

Date of Observation Stream Day : Run Date : Run No from start

D I.Fuel Data (coal) Weight Liqfuel

fraction Carbon Hydrogen Moisture Oxygen Sulfur Nitrogen Ash

E II.Observed Process parameters

Boiler Duty mmkcals/h % Unburnt matter in refuse Amb.Temperature oC Flue Gas Temp o oC Relative Humidity %

F II a. Flue gas data

O2 vol % CO2 vol % CO vol % SO2 vol % Excess Air


122

Fuel High Heating value Kcal/Kgm Fuel Low Heating value Kcal/kgm

G III.Energy Loss data

e.Radiation Loss : f.Blow Down Loss : g.Unaccounted Losses h.Loss due to combustibles

Total Loss % (on dry basis)

H IV.Boiler Efficiency design a.EFFICIENCY HHV basis b.EFFICIENCY LHV basis It is preferable that observed data is provided for various load conditions.

II. Economizer performance evaluation design Observed data and date

1 Flue gas flow t/hr * to be calculated

2 inlet temperature oC 553 3 outlet temperature 362 4 BFW flow t/hr 290 5 BFW inlet temp oC 230 6 BFW outlet temp oC 292 7 Area of heat transfer m2

III. Air preheater performance evaluation design Observed data and date

1 Flue gas flow t/hr

2Flue gas inlet temperature oC 362

3 outlet temperature 147 4 Total Air flow t/hr 313.75 5 Primary air flow t/hr 128.6 6 Secondary air flow t/hr 185.15


123

7 Air inlet temp oC 30 8 Air outlet temp oC (PRI) 303 9 Air outlet temp oC (sec) 317

10 Area of heat transfer m2

IV. Superheater performance evaluation design Observed data and date

1 Flue gas flow t/hr * to be calculated

2 SH inlet temp oC (stage I) 1140 3 SH stage II inlet temp oC 997 4 SH outlet oC 743 5 Steam ( SH outlet ) t/hr 290 6 Steam inlet temp oC 7 SH steam outlet temp oC 8 Area of heat transfer m2


124

V. Steam turbine performance evaluation design Observed data and date

1 Steam inlet to turbine t/hr 288.61

2 I stage Steam flow t/hr 18.91 ,, pressure kg/cm2 31.76 ,, temp oC 372.6

2 II stage Steam flow t/hr 17.47 ,, pressure kg/cm2 17.46 ,, temp oC 296.1

3 III stage Steam flow t/hr 19.92 ,, pressure kg/cm2 7.781 ,, temp oC 257

4 IV stage Steam flow t/hr 19.09 ,, pressure kg/cm2 2.199 ,, temp oC 123.3

5 V stage Steam flow t/hr 11.18 ,, pressure kg/cm2 0.511 ,, temp oC 81.9

6 Final stage Steam flow t/hr 201.96 ,, pressure kg/cm2 0.107 ,, temp oC 47.2

7 Power Generated in MW 74

8 Heat rate KJoule/KWH 9359.5 Kcal / KWH 2235.5 Efficiency % 38.47 Weighted TD efficiency %


125

Typical boiler efficiency calculation

Date : 12.10.07 Stream Day : 200

Input Data

item unit quantity

Fuel Data

1.Fuel t/hr 5.0000

2.Type Liquid

3.Density g/ml 0.9976

4.Gross CV kcal/kg 10800

5.Fuel temp oC 98.5

Flue Gas

1.CO2 % 12.7

2.O2 % 3.65

3.CO % 0

4.N2 % 83.65

5.Temperature oC 215

6.Air temp oC 30

Atomising Steam

1.Steam Flow t/hr 0.75

2.Pressure kg/cm2 10


Boiler Feed Water

1.BFW Flow t/hr 78.5



Steam Generation

1.Steam Flow t/hr 74.5


126



Blow Down

1.Blow Down t/hr 3



Design Duty

1.Heat Duty mmkcal/h 60

Material Balance

input output

1.Fuel t/h 5.0000

2.A.Steam t/h 0.7500

3.Dry Air t/h 81.7923

4.moisture t/h 0.0009

Total t/h 87.5432

Flue Gases

CO2 t/h 16.3443

CO t/h 0.0000

O2 t/h 3.7276

N2 t/h 62.5171

SO2 t/h 0.1000

H2O t/h 4.4332

error term t/h -0.4211

Total t/h 86.7011


127

ENERGY BALANCE

Basis: 1 kg fuel/hr

input

1.Heat of combustion kcal/hr 10800.0

2.Heat in pre heat kcal/hr 36.5

3.Pre Heat in Air kcal/hr 58.9

4.Heat in atomising

steam kcal/hr 103.8

Total Heat input kcal/hr 10999.2

output

1.Heat Gain in BFW kcal/hr 1617.1

2.Steam Generation kcal/hr 8344.9

3.Dry Gas Loss kcal/hr 802.0

4.Water formed kcal/hr 483.1

5.Air moisture kcal/hr 1.6

6.Blow Down kcal/hr 179.1

7.Incomplete Combn kcal/hr 0.0

8.Others kcal/hr 0.0

9.Error term kcal/hr -428.6

Total Heat output kcal/hr 10999.2

Reconciled output

%

1.Heat Gain in BFW kcal/hr 1556.4 14.15

2.Steam Generation kcal/hr 8031.9 73.02

3.Dry Gas Loss kcal/hr 771.9 7.02

4.Water formed kcal/hr 465.0 4.23

5.Air moisture kcal/hr 1.5 0.01

6.Blow Down kcal/hr 172.4 1.57

7.Incomplete Combn kcal/hr 0.0 0.00

8.Others kcal/hr 0.0 0.00

Total Heat output kcal/hr 10999.2 100.00


128

Boiler efficiency by direct

method = (8344.9+1617.1)/10999.2) x 100

= 90.57%


129

Power Generation System

TURBINES

All process and power industries require power and / or steam for driving various types

of equipments which are design specific. These include steam turbines, gas turbines,

electric motors, hydraulic turbines, turbo-expanders, and engines. However steam and gas

turbines are the prime movers for most applications.

Turbines may be classified as in Table 3.1.

Classification of turbines

Table 3.1 Classification of turbines

Name Type Application a.Steam urbine

back-Pressuretype

power generation

Extraction cum condensing

power and steam

totalcondensingturbine

Power generation or as prime mover for compressors, pumps etc.

b.Gas turbine

For power generation and / or as prime mover

Steam Turbines

Dependability, variable speed-operation and possibility of energy savings are the

basic factors that are considered for choice of steam turbines for many process

applications . This is also determined by Heat-Power requirement of the process.

The steam turbine is a very satisfactory and dependable prime-mover for many

process machines such as pumps ,fans , blowers and compressors . It is often used as a

driver for electric generator to provide main power to the process plant motor drives .

Steam turbine's feature of variable speed is useful for saving energy on pump , blower,

compressor drives etc.


130

Where process load fluctuations are frequently encountered, steam turbines offer

the best solution for energy saving. Steam turbines are more dependable and no process

interruptions, power failures, transmission problems etc are ever faced in case of

steam turbines. It is a common practice to keep steam driven equipment running in

critical services, where power interruptions may cause a serious problem .

Types of steam turbines

All conventional process - plant steam turbines are axial-flow turbo machines in

which steam flows parallel to the shaft axis. These turbines may be of single stage or

multi stages. In single stage turbines, steam expands through a set of nozzles only once.

They are most suited for smaller applications where a few horse-power of power (say 50

to 150 HP) is required. On the other hand, multi stage turbines are used where the

requirement of power is very high.(say 1000 to 2000 hp ). Multi stage turbines have

two or more expansions through a set of nozzles in each stage.

When the exhaust steam from any steam turbine is above atmospheric pressure,

the turbine is called non-condensing. When the exhaust steam is below atmospheric

pressure, this is labeled condensing turbine. Using second law of thermodynamics and

Bernoulli's Equation, it may be seen that

V 12 v 2 2

------- + h1 = ------- + h2 + W 2g J 2g J

where

v 1 and v 2 are velocity at inlet and outlet in ft /sec respectively

h 1 and h 2 are enthalpy at inlet and outlet conditions of steam in

btu / lb respectively

g = gravitational constant ( 32.2 ft/s2)

J = Joule's constant ( 778 ft-lb / btu)

The velocity of steam flowing at inlet and outlet conditions is approximately

equal. Hence v 1 may be taken as equal to v 2 in the above equation.This reduces the

above equation to


131

h 1 - h 2 = W

i.e. Work done by the steam turbine is directly proportional to the enthalpy difference at

inlet and outlet conditions. It may be noted that inlet and outlet steam conditions are

fixed. Hence enthalpy values may be found from the steam - tables or Mollier diagram

for the inlet and outlet conditions. However, it may be found that in actual practice

isentropic expansion is never achieved due to energy losses in friction and inefficiencies

of the turbine. The actual enthalpy will be slightly higher than the constant entropy value

but in the same pressure line.

Hence overall turbine efficiency per stage may be defined as

E stage = ( h 1 - h 2' ) / (h 1 - h 2 )

Where

h 2' = actual enthalpy observed at the outlet conditions.

A typical single stage back pressure steam turbine is given in fig 3.4 .

Performance of steam turbine is represented by Theoretical Steam Rates (TSR)

and Actual Steam Rates (ASR) which is the heat quantity in BTUs or KCALs required

to generate one kwh of power.

HP STEAM INLET

MP / LP STEAM OUTLET

~


132

fig 3.4 . Back Pressure Turbine

Since 1 kwh = 3413 btu / hr or

= 860 kcal / hr

TSR = 3413 / ( h 1 - h 2 ) lb / kwh

ASR = 3413 / (h 1 - h 2 ' ) lb / kwh

Where

h 1 = enthalpy at inlet of the turbine in Btu / lb

h 2 = enthalpy at outlet of the turbine in Btu / lb (isentropic)

h 2 ' = enthalpy at outlet of the turbine in Btu / lb (actual)

In metric units, the same is represented by

TSR = 860 / ( h 1 - h 2 ) kg / kwh

ASR = 860 / (h 1 - h 2 ' ) kg / kwh

E stage = TSR / ASR

Above relationship could be monitored from the power generation and steam input,

output data. As can be seen from the above efficiency relationship, greater the steam

pressure drop through the turbine, greater will be the power output. A reduction in

exhaust pressure causes a greater power generation than an increase in the inlet steam

pressure. It may be noted that specific steam consumption depends on the absolute

pressure ratio of the turbine.

Back Pressure turbines are thermally efficient having cycle thermal efficiencies in the

range of 75 to 85 % (ref fig 3.4)

Condensing turbines tend to be high in cost and thermally inefficient - giving around 10

to 25 % efficiency.

This could be economically exploited, only in processes where large amounts of waste

heat is available and where no other recovery options are found lucrative. It is found

more energy efficient to exploit this waste heat from low pressure steam sources such as

evaporators to drive the local devices, than to generate power.


133

Fig 3.5 is a typical condensing turbine . In a total condensing turbine, all the

steam entering the turbine is condensed in a surface condenser, which is located at the

exhaust section of the turbine.

Last stage of the turbine (surface condenser) is under vacuum and the

condensation takes place only in the surface condenser. Steam should not be allowed to

condense in the impeller blades, as this may cause rotor imbalances, Dew point corrosion,

and silica deposits in the rotor.

fig 3.5 total condensing turbine

Extraction cum condensing turbines is the most common type for total generation

schemes. This offers excellent flexibility to meet the steam and power demand of the

process and is energy efficient also. Cycle thermal efficiencies of this type is around 50 to

70 %.

~

HP STEAM

Surface

Condenser

Condensate


134

Fig 3.6 shows an Extraction cum condensing turbine . This turbine combines

the function of both back pressure and condensing turbine. Part of the steam after doing

work is taken out at medium pressure for process use, while the balance quantity goes to

the surface condenser for condensation. Provision exists to vary the extraction steam to

condensing steam, by a pass out valve. Power generated or work done varies with the

extraction steam and condensed steam.

fig 3.6. Extraction cum condensing turbine

~

HP STEAM

EXTRACTIONSTEAM - MP

SURFACE

CONDENSER


135

Gas Turbines:

Gas turbines are self-contained power plants. A gas turbine as the name implies uses gas

as the fuel , which burns with compressed air in the combustor and is expanded through the

turbine which drives the air compressor and delivers the excess of power generated as output

. In the last step, the products of combustion are exhausted to the atmosphere.

As soon as the fuel burns in the combustion chamber, the temperature of the hot gases reaches

a maximum. If t 1 is the temperature achieved by combustion and to is the ambient

temperature and q is the energy input to the turbine, then the maximum recoverable work is

given by

W max = (( t 1 - t 0) / t 1 ) * q

W loss = ( t 0 / t 1 ) * q

Eff = (t 1 - t o) / t 1

fig 3.7. Simple Cycle Gas Turbine.

Above relationship is based on 'Carnot Cycle ' which gives the highest possible

conversion rate of heat into work. Actual efficiency will be much lower than the Carnot

Efficiency for the same temperature and flow conditions. Thermal Efficiency of a gas

HOT GAS T1 oC

TURBINE

EXHAUST To C

WORK


136

turbine depends on pressure ratio and turbine inlet temperature. Fig 7.4 shows the

typical arrangement of a simple cycle gas turbine.

Turbine efficiency

Turbine efficiency is the ratio of energy output to energy input irrespective of whether it

is steam turbine or gas turbine. This is calculated by an energy balance across the turbine. A

typical enthalpy balance for a multi stage extraction cum condensing steam turbine is similar to

fig 7.3, but the number of extraction may be two to three at different extraction pressures.

Material balance

w 1 = e 1 + e 2 + c 1 + l 1

where

w 1 = steam input to turbine in t/hr

e 1 = extraction rate at stage 1 in t/hr

e 2 = extraction rate at stage 2 in t/hr

c 1 = condensed steam in t/hr.

l 1 = steam losses in t/hr.

Energy Balance

Input = steam extraction I extraction II condensing loss input - steam - steam - steam - steam energy energy energy energy energy

Energy content of each stream is calculated from the enthalpy values at it's actual pressure and

temperature conditions. Hence the above relationship may be given by

E i = w1 * h 1 - ( w 2 * (h 1 - h 2) + w 3 * (h 1 - h 3) + c1 * (h 1 - h 4) )

where

h 1 = steam enthalpy at pressure p1 and temperature t1

h 2 = steam enthalpy at pressure p2 and temperature t2h 3 = steam enthalpy at pressure p3 and temperature t3


137

h 4 = steam enthalpy at pressure p4 and temperature t4respectively

Output

E o = 860 * p

where

p is the power output in kw/h

860 conversion factor

E o is kcal/hr.

Turbine Efficiency (Actual)

Eff = ( E o / E i ) * 100

As a thermal prime mover, the efficiency of a turbine is the useful work energy that

appears as shaft power and is presented as a percentage of the chargeable heat energy .

The over-all thermal efficiency of a steam turbine is given by W / JQ where W is the

shaft work in ft lb or kg metre and Q is the btu or kcal chargeable. This leads to several

expressions for thermal efficiency as follows.

Where a number of extraction rates are carried out in a steam turbine , the efficiency is

given by

2545 Eff ' = ----------------------------------------------------------------------------------- w x h1 - ( w ' a x h a + w ' b x h b + w 'c x h c + ....) - w 2 x h f 2

where

w = inlet steam in t/hr

w 'a = Extraction steam stage I in kg/hr

w 'b = Extraction steam stage II in kg/hr

w 'c = Extraction steam stage III in kg/hr

w2 = Exhaust steam in kg/hr

h1 = Enthalpy of steam at inlet conditions kcal/kg

ha = Enthalpy at stage I extraction conditions kcal/kg


138

hb = Enthalpy at stage II extraction conditions kcal/kg

hc = Enthalpy at stage III extraction conditions kcal/kg

hf2 = Enthalpy at exhaust conditions kcal/kg

Gas Turbines

Energy distribution in simple GT

A look at simple gas turbine process shows that most of the added fuel energy disappears

with the exhaust gas as shown below

Fuel Energy 100 %

Energy distribution pattern in a simple Gas Turbine

Only about 29% of the fuel energy is converted into useful power, since the exhaust gases

still have high temperature. For minimizing the energy losses, it is imperative to use the fuel with

the highest possible efficiency. This calls for utilization of the heat in the exhaust turbine gases

in the most economical and feasible way.

Combustion Chamber

TurbineElectrical

Electrical and mechanical

Exhaust Gas Losses


139

Turbine Exhaust Temperature:

This may be calculated using the gas law and is given by

T2/T1 = ( P2/P1) ( -1)/ )

Where

o P2 & P1 are turbine exhaust and inlet pressures in absolute units

o T2 & T1 are exhaust and inlet temperatures in absolute units.

o is the adiabatic exponent.

e.g:

P1=4.5

P2=0.5(kg/cm2g)

T1 = 850 oC

y = 1.25

i.e T2/(850+273) = (1.523/5.523)^(1.25-1)/1.25): T2 = 594.9 oC

Carnot Cycle Method

When a gas expands adiabatically from a Pressure P1 to P2 and at inlet temperature of T1 oC it

may be noted that

1.Turbine exhaust temperature is a function of pressure ratio

2.For the same inlet temperature, turbine efficiency is a function of exhaust temperature.

3.As inlet temperature increases, turbine efficiency also increases for the same exhaust

temperature. This concept is shown in the next slide.

4.For the 4 inlet temperatures (i.e. 800,900,1000,1100 oC), efficiency is highest at 1100 oC,

followed by 1000 oC and so on for the same exhaust temperature. High inlet temperature and

pressure demands special metallurgy , which is cost intensive.


140

THERMO DYNAMIC EFFICIENCY OF TURBINES ( Carnot Cycle )

10

15

20

25

30

35

40

45

50

650 600 550 500 450

exhaust oC

ther

mo

dyna

mic

eff

icie

ncy

800 oC 900 oC 1000 oC 1100 oC

Turbine Efficiency is the ratio of actual work output of the turbine to the net

input energy supplied in the form of fuel.

For stand alone Gas Turbines, without any heat recovery system the

efficiency will be as low as 35 to 40%.

This is attributed to the blade efficiency of the rotor, leakage through clearance

spaces, friction, irreversible turbulence etc.


141

Increasing Overall efficiency:

Since Exhaust gases from the Gas Turbine is high, it is possible to recover

energy from the hot gas by a Heat Recovery Steam Generator.(HRSG in short) and use the steam for process.Various options are given in the next few slides.

Turbine Efficiency is a function of

* Running Hours

* Load factor

* Fuel type

* Combustor efficiency

* System Energy Loss etc.

It is possible to predict turbine efficiency using the models.

Part load operating efficiencies of gas turbine power plant is given in the next slide for various

hard ware configurations. Refer the book for more information


142

Cogeneration

Cogeneration (also called as combined heat and power, CHP) is the use of a heat engine

or a power station to simultaneously generate both electricity and useful heat.

Conventional power plants emit the heat created as a by-product of electricity generation

into the environment through cooling towers, flue gas, or by other means. CHP or a bottoming

cycle captures the by-product heat for domestic or industrial heating purposes, either very close

to the plant, or —especially in Scandinavia and eastern Europe—for distribution through pipes to

heat local housing. This is also called decentralized energy.[1]

In the United States, Con Edison produces 30 billion pounds of steam each year through

its seven cogeneration plants (which boil water to 1,000°F/538°C before pumping it to 100,000

buildings in Manhattan—the biggest commercial steam system in the world.

By-product heat at moderate temperatures (212-356°F/100-180°C) can also be used in

absorption chillers for cooling. A plant producing electricity, heat and cold is sometimes called

tri generation or more generally: poly generation plant.

Cogeneration is a thermodynamically efficient use of fuel. In separate production of

electricity some energy must be rejected as waste heat, but in cogeneration this thermal energy is

put to good use.


143

Masnedø CHP power station in Denmark. This station burns straw as fuel. The adjacent

greenhouses are heated by district heating from the plant.

Thermal power plants (including those that use fissile elements or burn coal, petroleum,

or natural gas), and heat engines in general, do not convert all of their available energy into

electricity. In most heat engines, a bit more than half is wasted as excess heat . By capturing the

excess heat, CHP uses heat that would be wasted in a conventional power plant, potentially

reaching an efficiency of up to 89%, compared with 55% for the best conventional plants. This

means that less fuel needs to be consumed to produce the same amount of useful energy. Also,

less pollution is produced for a given economic benefit.

Some tri-cycle plants have utilized a combined cycle in which several thermodynamic

cycles produced electricity, and then a heating system was used as a condenser of the power

plant's bottoming cycle. For example, the RU-25 MHD generator in Moscow heated a boiler

for a conventional steam power plant, whose condensate was then used for space heat. A more

modern system might use a gas turbine powered by natural gas, whose exhaust powers a steam

plant, whose condensate provides heat. Tri-cycle plants can have thermal efficiencies above

80%.

An exact match between the heat and electricity needs rarely exists. A CHP plant can

either meet the need for heat (heat driven operation) or be run as a power plant with some use of

its waste heat.

CHP is most efficient when the heat can be used on site or very close to it. Overall

efficiency is reduced when the heat must be transported over longer distances. This requires

heavily insulated pipes, which are expensive and inefficient; whereas electricity can be

transmitted along a comparatively simple wire, and over much longer distances for the same

energy loss.


144

A car engine becomes a CHP plant in winter, when the reject heat is useful for warming

the interior of the vehicle. This example illustrates the point that deployment of CHP depends on

heat uses in the vicinity of the heat engine.

Cogeneration plants are commonly found in district heating systems of big towns,

hospitals, prisons, oil refineries, paper mills, wastewater treatment plants, thermal enhanced oil

recovery wells and industrial plants with large heating needs.

Thermally enhanced oil recovery (TEOR) plants often produce a substantial amount of

excess electricity. After generating electricity, these plants pump leftover steam into heavy oil

wells so that the oil will flow more easily, increasing production. TEOR cogeneration plants in

Kern County, California produce so much electricity that it cannot all be used locally and is

transmitted to Los Angeles[citation needed].

Types of plants

Topping cycle plants primarily produce electricity from a steam turbine. The exhausted

steam is then condensed, and the low temperature heat released from this condensation is utilized

for e.g. district heating.

Bottoming cycle plants produce high temperature heat for industrial processes, then a

waste heat recovery boiler feeds an electrical plant. Bottoming cycle plants are only used when

the industrial process requires very high temperatures, such as furnaces for glass and metal

manufacturing, so they are less common.

Large cogeneration systems provide heating water and power for an industrial site or an

entire town. Common CHP plant types are:

o Gas turbine CHP plants using the waste heat in the flue gas of gas turbines

o Combined cycle power plants adapted for CHP

o Steam turbine CHP plants that use the heating system as the steam condenser for the

steam turbine.

o Molten-carbonate fuel cells have a hot exhaust, very suitable for heating.

o Smaller cogeneration units may use a reciprocating engine or Stirling engine. The heat is

removed from the exhaust and the radiator. These systems are popular in small sizes

because small gas and diesel engines are less expensive than small gas- or oil-fired

steam-electric plants.


145

o Some cogeneration plants are fired by biomass [5], or industrial and municipal waste (see

incineration).

Combined Heat Power Cycle / cogeneration:

Combined heat and power refers to the industrial units which generate their own

electricity and heat in the form of steam for meeting their heat and power demand of the

process

In most of the CHP systems, total heat and power is generated at an efficiency of

over 75 %, compared to the power generation efficiency of mere 30 to 35 %.

The optimum output of CHP is based on the total energy concept, in which the

primary fuel is used to generate both electricity and heat in one place.

Base case


146


147

Fuel Savings – CHP


148

INTEGRATING TURBINE EXHAUST TO FURNACES

FUEL

AIRTURBINE

HEATER

BASIS: 1 TON FUEL AIR TH : 16.5 AIR ACT 28.0 EXH.TEMP 550 PRE-HEAT AVAILABILITY29*0.28*1000*(550-30)=4.22 MILLION KCAL/HR


149


150


151

Waste Heat Recovery

o Refers to

o Recovery of Heat from Flue Gases

o Steam Condensates

o Waste Steam Recovery / Utilization

o Hot Waste Water & Process Streams

Waste Heat is the Heat that is rejected from a process.

Sources of Waste Heat are classified according to temperature in three ranges.

* High temperature range above 650 oC

* Medium temperature range between 250 to 650 oC

* Low temperature heat i.e. below 250 oC

Parameters to be evaluated for Waste Heat Recovery

* Temperature of Waste Heat Fluid

* Flow rate of waste heat fluid

* Chemical Composition of waste heat fluid

* Minimum allowable temperature of Waste Heat fluid

* Temperature / Chemical composition of cold fluid

* Maximum allowable temperature of Cold fluid.

* Control temperature if required for cold stream.


152

Examples :

Waste Heat Recovery from CO Boiler


153


154

Air Pre Heater Lay Out for Waste Heat Recovery

Air-Pre Heating - Advantages

Air Pre-heating reduces fuel consumption by 25 %

Typical Case :

* Fired Heater Efficiency before APH installation: 70 - 71 %

* ,, ,, ,, after APH installation 85 - 89 %

* ,, ,, ,, with high Efficiency APH 91 %

* Reduction of 20 - 23 oC of Flue Gas temperature results in 1.% of fuel saving.

* Throughput can be increased by 20 %


155


156

TOTAL POWER PLANT OPTIMIZATION

Total power plant optimization involves optimization of

o BFW system including economizer

o Air preheating system

o Fuel / burner system

o Combustion control

o Boiler furnace control

o Steam super heating system

o Flue gas system

o Steam distribution system

o Turbine system

o Generator system and

o Transmission loss control.


157

BFW Quality:

BFW quality plays an important role in boiler efficiency maintenance.

Dissolved salts find entry to the boiler through make-up water which is

continuously fed by the Boiler Feed Water pump ( bfw) . In the boiler , there is

continuous evaporation of water into steam . This leaves behind the salts in the

boiler. Concentration of these salts , tend to increase in the boiler drum and

starts precipitation after certain concentration level .

Water from the drum should be blown down to prevent concentration of salts

beyond certain limits . Since the water in the boiler drum is at a high temperature

( equivalent to it's saturation temperature at boiler drum pressure ) , excess blow-

down will lead to loss of energy known as 'blow-down losses ' . Blow-down rate

reduces the boiler efficiency considerably as could be seen from fig 3.8. Hence it

is imperative that blow-down rates are optimised ,based on the hardness levels of

boiler drum water which is a function of the operating pressure.


158

In boiler operation practice , rate of blow down increases with steam pressure as the

scaling tendency increases with high temperature because the hardness limits are very

stringent .While fig 3.8 gives an estimate of % blow down on losses ,the same may be

calculated from the hardness levels of make-up water , flow rate ,steam generation rate and

the hardness level of drum water ( observed) .

Polynomial model given below could be used to determine the maximum limits of

TDS (total dissolved solids) that could be tolerated in the boiler drum operating at various

pressures .This is based on American Boiler Manufacturers' Association code of practice.

However, if the limits stipulated by the Boiler Designer is less than this value , the

lower of the two must be taken as the tolerance limit.

Fig 3.8


159

Energy transmission system :

Total power plant management involves monitoring the performance of each sub system

by appropriate methods which is covered in the next few chapters.


160

Calculating Energy Efficiency of Steam Turbines.

Steam turbines are devices which convert the energy stored in steam into rotational

mechanical energy. These machines are widely used for the generation of electricity in a number

of different cycles, such as:

Rankine cycle

Reheat cycle

Regenerative cycle

Combined cycle

The steam turbine may consists of several stages. Each stage can be described by

analyzing the expansion of steam from a higher pressure to a lower pressure. The steam may be

wet, dry saturated or superheated.

.

Consider the steam turbine shown in the cycle above. The output power of the turbine at

steady flow condition is:


161

P = m (h1-h2)

where m is the mass flow of the steam through the turbine and h1 and h2 are specific enthalpy

of the steam at inlet respective outlet of the turbine.

The efficiency of the steam turbines are often described by the isentropic efficiency for

expansion process.

The presence of water droplets in the steam will reduce the efficiency of the turbine and

cause physical erosion of the blades. Therefore the dryness fraction of the steam at the outlet of

the turbine should not be less than 0.9.


162

Isentropic Efficiency

The isentropic efficiency for an expansion process is defined as:


163

Typical data for a condensing turbine


164

EFFECT OF STEAM PRESSURE – CONDENSING TURBINE

From the figure given above, it may be noted that steam pressure has an impact on

specific steam consumption of the turbine. In this particular case, increasing steam

pressure from 35 to 45 kg/cm2, reduces steam consumption by 2.1%.

Hence steam boiler pressure is an important parameter in improving the energy efficiency

of the turbine.

Above information may be converted into turbine efficiency vs steam pressure,

taking the base case efficiency as given .


165

EFFECT OF STEAM PRESSURE ON EFFICIENCY – CONDENSING TURBINE

As inlet steam pressure increases, turbine efficiency also increases. For an increase of

steam pressure from 35 to 45 kg cm2, turbine efficiency increases from 33.1 to 34.2 %.

Steam inlet temperature is also one of the parameters that determines the efficiency of

condensing steam turbine. Next figure explains this phenomena .


166

STEAM INLET TEMPERATURE vs CONSUMPTION – CONDENSING TURBINE

As the steam inlet temperature increases from 300 to 400 oC, steam consumption %

drops from 100 to 88%.

In other words, steam turbine efficiency increases substantially as given in the next

diagram. Steam turbine efficiency increases from 33.1 to 34.4 % when inlet temperature

increases from 300 to 400 oC.

In Power plant management, these parameters may be optimized to get the best efficiency

possible, within the operational constraints..


167

Exercise :

Following tables show the impact turbine operating parameters on efficiency. What are

the optimal conditions that can give the best efficiency of the turbine. What are the tools that

may be used in the evaluation process ? Maximum operating pressure of boiler 50 kg/cm2.

No Steam pressureKg/cm2

Turbineefficiency

in % 1 35 33.10

2 37 33.33

3 39 33.55

4 41 33.78

5 43 34.01

6 45 34.20

7 47 34.45

8 49 34.90

No Steam temperature

oC

Turbineefficiency

in % 1 300 33.10 2 320 33.35 3 340 33.55 4 360 33.80 5 380 34.10

6 400 34.35 7 420 34.61 8 440 34.85

Energy Efficiency Optimization Training Program – Dr.G.G.Rajan

160

4

PERFORMANCE MONITORING TECHNIQUES


161

4. Performance Monitoring Techniques

1. Monitoring Boiler Efficiency / Performance :

Boiler efficiency is determined by operation and maintenance practices of the total

system. Efficiency is also an indicator of mechanical condition of the boiler system as a whole.

Hence if the boiler efficiency is regularly monitored, it is possible to identify deterioration trend

and take corrective action. Using the deterioration trend, it is possible to predict the boiler

efficiency at a future period, by using historical models.

For boiler efficiency monitoring purposes either direct of indirect method or the

average may be selected and consistently used. A typical output of the boiler efficiency

program was given in the earlier chapter.

Precautions :Certain precautions are to be followed when boiler efficiency is determined as

listed below.

o Boiler load must be consistent.

o All flow meters must be calibrated and errors minimized.

o Pressure , temperature and draft profiles in the boiler circuit must be kept consistent

during the run.

While fig 4.01 gives a typical boiler efficiency curve as a function of load , impact of

running hrs on the efficiency is also given in the figure for various load conditions . Box 4.02

gives the performance models at base case and after ten months of operating time which may

be used for evaluation of performance and timely maintenance action.

Differentiating the above functions w.r.t load % and equating to 0 it may be found

that the maximum achievabe efficiency in the base case is 89.80 at 103.3% of design load ,

whereas the maximum achievable efficiency at the end of ten months of operation is 84.00 % at

a load of 99.16% .This clearly indicates , there is a drop in the efficiency of boiler with

the passage of time and load% .

This tends to affect the operating cost of the boiler and it is possible to determine

the optimum cycle length at which the boiler has to be shut down and cleaned so that the

efficiency may be restored to the level of base case .


162

Fig 4.01. Boiler Efficiency vs Load %


163

Performance monitoring technique:

Efficiency of boilers must be calculated for each boiler periodically and recorded

stream day wise. All the input parameters must be as accurate as possible and may be used as an

effective management tool to determine

o What is the current efficiency of the boiler ?

o What should be the efficiency at the current stream day and load %

o Whether the efficiency is within tolerable limits ?

o If not what are the reasons for lower efficiency ?

o What are the actual losses against base case ?

o Should the boiler be operated in the existing condition or should it be repaired ?

o Whether the boiler can be maintained or replaced etc.

A typical decision flow diagram to arrive at the right decision is given in fig 4.05 .

This methodology could be arrived at for any number of boilers when once the models are

developed for each controllable parameter .This uses a number of models for arriving at the

right decision.

Fig 4.05. Boiler Efficiency

management


164

Typical boiler efficiency program output

=================================================================

RESULTS - BOILER EFFICIENCY

=================================================================

User : ABC Refinery

Unit : India

Equipment : blr1

Input file : c:\tthermw\blr1.in

Output files: c:\tthermw\blr1.out

Output file2: c:\tthermw\blr1.mv

Archive file: c:\tthermw\boiler.arc

Date of Obs : 12.10.07

Stream Day : 100

Run Date : Run No : 1

I.Fuel Data

Carbon .87000

Hydrogen .12000

Moisture .00000

Oxygen .00000

Sulfur .01000

Nitrogen .00000

Ash .00000

II.Observed Process parameters

Fired Duty mmkcal/h 65.0000

% Heat load 100.0000

% Unburnt matter in refuse 0.0000

Amb.Temperature oC 30.0000

Flue Gas Temp oC 180.0000

Relative Humidity 0.0000


165

Excess Air 15.6454

Fuel High Heating Value - Kcal/Kgm 11163.8896

Fuel Low Heating Value - Kcal/kgm 10515.8887

III.Energy Losses










IV.Boiler Efficiency

a.EFFICIENCY HHV basis : 84.9498

b.EFFICIENCY LHV basis : 90.1845

NOTE:

a.Air misture loss is due to moisture present in combustion air.

b.Combustion moisture is due to combustion of H2 in fuel to water.

c.Fuel moisture is due to presence of water in the fuel fired.

d.Radiation loss is due to heat loss from the exposed boiler surface.

e.Blow down is energy equivalent as % of fuel calorifiv value.

*** end of boiler efficiency program (indirect method) ***

Output information may be used to analyze performance of each boiler for efficiency

improvement.


166

A typical energy loss break-up for the boiler is given in fig 4.06. From this it is

obvious that all losses except wet losses show an increasing trend with respect to time.

Controllable losses in this case are dry gas loss , convection and radiation losses and blow

down loss .

Fig 4.06. Energy loss break up from boiler efficiency program output.


167

Observations:

o Increase in dry gas loss indicates gradual increase in excess air which could be due to

air leaks in the shell section . This is normally diagnosed by oxygen analysis at

combustion , pre-convection and post convection zones.

o Increasing trend in convection and radiation losses is indicated by high skin

temperature of the shell. This situation occurs due to thinning of the insulation brick. It

is a good practice to monitor this loss known as setting loss periodically, for deciding

the optimal replacement time.

o High blow-down rate could be due to poor operation practice or poor boiler feed

water quality . This may be checked from the TDS levels of make-up water , drum

water and allowable TDS limits.

Systematic recording and review of boiler parameters is very imperative to determine the

energy efficiency of a boiler.

A typical analysis data for a set of boilers is given in box 4.03. This is actually the

program output of boiler efficiency monitoring system program using a popular program,

which utilizes indirect method of determining boiler efficiency.

Blow down rate as % of energy input must be added to the last column to determine the

efficiency of boilers listed below.

Boiler Efficiency Models

Since boiler efficiency is determined by a number of parameters , it is imperative to

identify the impact of all parameters by suitable scientific methods. Boiler efficiency is

invariably affected by running days and load factor.

Scaling increases with the passage of time on both the water side and tube side. This

is reflected by rise in water tube skin temperature, flue gas temperature etc .

Since the boiler load is always dynamic depending on steam demand, the efficiency

could be estimated as a function of operating period and load % involving two variables .

Even a linear multi variable model is good enough for monitoring purpose. For this

purpose, common programs like MS Excel may be used .


168

Performance Monitoring of boilers

Performance monitoring of boilers is very imperative in power and process industries

as the major chunk of cost is governed by the fuel consumed in the boilers .

Since fuels in boilers are not in their pure form , they tend to deposit soot , scales etc in

the economizer , super heater , water tubes etc which retard heat transfer efficiency with the

passage of time .

In addition to this , steady loss in evaporating capacity is also experienced due to

accumulation of scales on the water side with the passage of time . The decrease in heat-

transfer rate is a function of time and is given by

where

U is the heat transfer coefficient in kcal /hr / m2 /oC

t is the time lapsed since last cleaning.

For minimizing the scale deposition on the convective section of the boiler, soot blowers

are used periodically to remove the loose deposits .However, hard scales tend to deposit with the

passage of time .

Hence performance monitoring of the boiler is essential to decide at what point of time

the boiler has to be taken out of service for cleaning and other maintenance jobs.

Like the case of heaters, boiler efficiency also varies with the passage of time due to the

above factors. In addition to this , boiler capacity utilization also affects the efficiency and

boiler manufacturers provide the boiler operating characteristics data.

Hence the performance of the boiler may be evaluated by the observed efficiency and

compared to the basic characteristics for the same operating conditions. Any abnormal

deviations observed could therefore be investigated to pin-point the problem area for

necessary action.

It is obvious from the above the boiler efficiency can be managed by monitoring the

efficiency of economizer, air pre heater and super heater by appropriate techniques.

1 / U2 = A + B * t


169

Specific Fuel Consumption

Specific fuel consumption is normally assumed to be a performance Indicator of energy

efficiency of the system. The system could be a heater or boiler or turbine. In conventional

Energy Management, Specific Energy Consumption is defined as the quantity of fuel consumed

in kgs or Lbs as given below.

Equipment Specific Fuel Consumption

1.Heater Kg fuel / Ton of feed processed

2.Boiler ton fuel / Ton steam produced.

3.Gas Turbine ton fuel / MW power generated.

4.Steam Turbine ton fuel eqvt steam / MW of Power

generated.

Examples given below are indicative of the concept of specific fuel consumption.

Specific fuel consumption may be used as a thumb rule to determine the equipment performance

/ efficiency by a quick glance. When all the parameters remain constant, specific fuel

consumption is definitely a simple and good performance indicator of the system under

consideration.

Item Feed tons Fuel tons Sp.Fuel

Consumption Unit

1.Heater 500 7.5 15 Kg/ton of

feed

2.Boiler 150 10 0.06667 Ton/ton of

steam prodn


170

3.Gas Turbine 25 MW 4.50 0.180 Tons of fuel /

MW

4.Steam

Turbine

10 MW 100 tons

steam

10 Tons steam /

mw

Historical data of specific energy consumption of system / systems in an indicator of

efficiency. With the passage of time efficiency starts dropping down, thereby increasing the

specific consumption of fuels.

Factors Affecting Specific Fuel Consumprion :

Fuel is consumed only in process heaters, boilers and Gas Turbines in the refining

industry. A number of process parameters affect specific fuel consumption in the case of heater /

boiler. They are

Capacity utilization and

Heater operating parameters such as

Draft

Transfer Temperature

Fuel type and it's Calorific Value

Fuel Mix ( % Oil , % Gas ) and

Combustion Efficiency of Fuel

Impact of capacity utilization on a fired heater efficiency was shown in earlier chapter when

all the other parameters are constant. Combustion Efficiency is a function of fuel mix used in a

heater / boiler.


171

Performance Monitoring of turbines.

Performance and condition monitoring of turbines are two different areas of energy

efficiency management. While condition monitoring refers to the mechanical condition of the

rotating equipment and is concerned with vibration level, bearing conditions , mechanical

condition etc. performance monitoring refers to the energy efficiency of the turbine under

consideration.

But it is necessary to know that condition of the rotating equipment has direct impact on

energy consumption and efficiency of the machine.The physical layout of various turbines were

presented in earlier chapters.

Performance Monitoring of steam turbines

While turbine efficiency is calculated based on the input/output principles, performance

of turbines vary with the load factor and aging / on -stream hours / days. Turbine

manufacturers often supply the performance characteristics of each machine in the form of

a curve for various energy input conditions.

Actual performance of the turbine may be compared with the base case to determine

whether or not the efficiency is within the acceptable limits. A typical steam turbine

characteristic is given in fig 4.5(a).

This characteristic curve gives the input steam rate vs output in kw for any chosen

extraction steam rate.

The graph is developed from the actual performance data for a small machine. For

performance monitoring of the turbine , the output observed at any load is compared against

the base case ( in this case characteristics curve) and checked whether the performance is

normal or not.

Case given in fig 4.5(a) has extraction rates of 0 ,5 ,10 and 15 t/hr and any intermediate

value is interpolated.

An example of how the turbine characteristics may be used for performance monitoring

is given in the section.


172

fig 4.5(a) .Typical Steam turbine characteristics.

observed data: steam inlet 20.8 t/h

pass - out 10.0 t/h

power output 2100 kw

From the characteristics, power output should be 2750 kw.

Observed Deviation in performance is -650 kw

% deviation from base value = (-650/2750)* 100

= 23.60%

The deviation is very high and needs a thorough investigation. If the turbine operation

continues, excess steam consumption would have been about 3.0 t/hr for the same output. (

given by y-x )


173

Increase in annual steam consumption based on 8000 hrs of operation works out to

24000 tons. For applying this monitoring technique, user must have the complete turbine

characteristic for each machine and refer to the same every time the checking is required.

Computer-aided performance monitoring involves mainly the collection of performance

data for the turbines operating under varying loads and other parameters and using these data to

develop time-dependent and non-linear / linear multi variable models for performance

prediction.

Typical steam turbo generator data

It may be noted that turbine efficiency varies with the connected load as in the example

given above and refers to the tests carried out on a 75-kw condensing steam turbine.

Computer-aided performance monitoring of gas / steam turbine involves the following steps .

1.Turbine data collection. ( energy input / output information )

2.Evaluation of enthalpies of various streams

3.Developing material balance

4.Developing energy balance.

5.Efficiency calculation for the observed conditions.

Load

item 25% 50% 75% 100%

a.Design load kw 75.00

b.Connected load kw 18.75 37.50 55.25 75.00

c.Steam kg/kw 21.61 15.52 12.21 11.39

d.Efficiency % 21.10 30.70 38.20 40.00

e.Steam press kg/cm2 ------------------13.60 ---------------

f.Steam temp oC ------------------ 225 ----------------

g.exhaust press hg mm ---------------- 68.00----------------

h.exhaust temp oC ----------------- 42.00----------------


174

6.Deviation Analysis with base data.

7.Identifying causes for corrective action.

Performance of turbines vary with time and connected load as shown in the above

table. As the shaft load increases , efficiency of the turbine increases and the specific

consumption of energy decreases .

It is possible to develop a suitable performance model based on the actual observations.

Necessary flow rate corrections are to be applied for the steam pressure, temperature and flow

rates. Typical output of steam turbine models using the above data is given in box no 4.1. Box

no 4.2 gives the simulated turbine efficiency for new conditions, based on the performance

model developed by this method.

Box 4.1 Steam Turbine Performance Model (load % vs efficiency)

S.E of Model : 0.0001

Box 4.2 Simulated Performance for new conditions.

load% Z1 simulated efficiency

20 20.0000 19.2446

35 35.0000 24.9617

65 65.0000 35.6824

85 85.0000 39.8055

105 105.0000 39.3594

load% X1 efficiency %

observed simulated

25 25.0000 21.1000 21.1000

50 50.0000 30.7000 30.7001

75 75.0000 38.2000 38.1999

100 100.0000 40.0000 40.0000


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Steam Turbine Performance Model 2 ( load vs sp.cons)

load X1 observed simulated sp.con sp.con

25 25.0000 21.6100 21.6100

50 50.0000 15.5200 15.5199

75 75.0000 12.2100 12.2101

100 100.0000 11.3900 11.3900

S I M U L A T E D O U T P U T F O R N E W D A T A

When once the models are developed it is possible to determine the performance level of

each machine with reasonable accuracy at any point of time for various connected load

conditions.

In the case of gas turbines , the above models may be modified to give specific energy

input / kw of net shaft power and efficiency vs load. This technique is superior to the

conventional static performance monitoring method, which does not account for random /

variable parameters.

load Z1 simulated load Z1 simulated sp.cons sp.cons

20 20.0000 23.1873 65 65.0000 13.2190

35 35.0000 18.8217 85 85.0000 11.5996


176

Trouble shooting Turbine problems

Turbine problems are similar to compressor problems and may be divided into

mechanical and operational problems.

Mechanical problems include

� Vibration

� Rotor shaft displacement

� Over heating of bearings

� Excessive noise during running

� Impeller failure

� Shaft failure

Operational problems include

� High steam consumption

� Inadequate power generation at full load

� High pressure drop in last stage

� Improper functioning of surface condenser

� Frequent tripping of turbine

� Auxiliary pump failures

Vibration:

Vibration in turbines may be caused not necessarily from turbine section, but it may be due to a

coupled generator.

The turbine has to be decoupled and vibration checked.

In many cases, vibration may be due to poor bearing conditions and / or damaged bearings.

This may also happen due to crevices formed in the rotor shaft at bearing contact sections.

Another reason could be due to rotor imbalance, caused by damage to turbine blades or silica

deposits or products of corrosion.

A systematic analysis must be carried out to determine the cause for vibration.

Vibration trips:

All rotating equipments like turbines, compressors etc are provided with a vibration monitor cum

trip device. When the vibration exceeds a certain allowable limit, the machine will trip

automatically. By analyzing the vibration readings at various sections, it is possible to identify

the section causing vibration and rectify the problem.


177

A regular vibration monitoring program may even predict the likely failure of a rotating machine

beforehand.

Rotor shaft displacement:

Reasons for this problem may be due to worn out thrust bearings or crevices in the shaft section

of the bearing housing. When this problem occurs, the bearing temperature will be higher by 40

to 50 oC more than normal. This may be checked in the usual manner.

Over heating of bearings:

Over heating of bearings may be caused by poor / improper lubrication and / or the presence of

wear particles removed from the shaft or bearing. Continuous monitoring of bearing

temperatures, lubrication and lube oil analysis is imperative to check the problem.

Excessive noise during running :

Normally this problem shall be due to rough bearings / shafts or presence of hard materials in the

rotating section. Even if a section of impeller is damaged, this noise may be noticed. Turbine

shut down is necessary to check the problem.

Failure of impellers / shafts :

Thermal shocks, frequent trips, poor material of construction, aging and creep stresses may be

attributed to these failures.

OPERATIONAL PROBLEMS:

High steam consumption :

This could be due to some hindrances in the steam passage. Scales, silica and bacterial / fungus

growth may cause this problem. Besides, clogged impellers, excess friction etc may also cause

this problem.

Inadequate power generation at full load :

For a given steam flow rate at the design temperature and pressure, the power generated must

tally with the turbine characteristic. If the deviation is excessive check the steam quality, flow

rate and the power meter. Also check the RPM of turbine. When all these parameters are normal,

the reason could be with the generator section. Check all the generator parameters like coil

insulation resistance, temperature etc.


178

High pressure drop in last stage :

Normally, this happens due to the deposition of silica in the last impeller stage, where the

temperature is low. Silica is converted into silicic acid, which forms iron silicate. This impedes

the flow of steam, besides imbalances in the rotor.

Improper functioning of surface condenser :

This occurs due to poor vacuum in the surface condenser and / of fouling up of the water tubes,

flowing through the condenser for condensing the steam. Check for fouling and poor vacuum in

the SC section by vacuum holding test and tube side pressure drop measurement.

It may be noted that for the same shaft load % , specific consumption of steam or energy

increases with the passage of operating time. This is very reasonable, as turbine blades are

subjected to wear and tear like attrition / erosion , silica deposition , corrosion etc and cause an

imbalance of the rotor. Besides this, the surface condenser also gets fouled up with passage of

time and affects the turbine performance.

box 4.3 Turbine Performance with time

.No of data sets used in the model : 5

Independant variables used in the model : 2

Variables used in the model are ......

variable 1 is load%

variable 2 is year

variable 3 is sp.con kg/kw

load% year sp.con

kg/kw

25.0 1.0 21.61

50.0 1.0 15.52

75.0 1.0 12.21

100.0 1.0 11.39

25.0 2.0 22.55

50.0 2.0 16.52

75.0 2.0 13.35


179

100.0 2.0 12.51

25.0 4.0 23.00

50.0 4.0 17.20

75.0 4.0 14.10

100.0 4.0 13.30

Model is of the form

sp.con kg/kw = 100.3775 * (load %) -0.47874 * (op.year) 0.06896

Standard Error of the model = 0.30353

SIMULATED OUTPUT FOR NEW INPUT PARAMETERS

LOAD % OPTG SP.CONS YR KG/KW

90.0 4.5 12.91545

70.0 4.5 14.56670

65.0 4.5 15.09278

55.0 4.5 16.34941

40.0 4.5 19.04203

35.0 4.5 20.29907


180

Loss Control Related to Power Plants

Introduction:In power plant operation physical losses occur in fuel section and steam circuit during

various stages of operation. The first step is to identify the loss areas and quantify the losses for

control action. In power plant operation, factors attributing to the losses can be classified as :

Coal handling losses during transportation

Losses due to coal dust generation

Spillage

Day to day quality variation

Measurement

Steam transportation losses

Safety valve popping

ID / FD fan losses etc.

Plant Losses Break-up:

Table 3.01 shows a typical break-up of various losses starting from coal transportation.

This will vary from plant to plant and the type of coal handling and transportation methods, coal

conveyor maintenance efficiency of the o&m personnel etc.

This type of loss break up may be carried out for any coal based power plant for loss

evaluation and control.

This exercise will help the power plant personnel refiners in taking appropriate

corrective action for loss control. As could be seen from the loss analysis table shown above,

maximum loss iin the refinery occurs in the form of process loss amounting to 69 % of the total

loss. For this particular system under consideration, loss control priorities must be set in the

following order.

Coal Handling Area

Coal receipt / dispatch / transfer

Coal crushing section

Pulverization section


181

Coal fluidization

Combustion

Carry over in fly ash

Carbon in refuse

Carbon Losses in boiler operation :

Wagons

As may be seen from the typical loss figures in various section, the quantity of coal

reaching the boiler as pulverized coal = 0.99 x 0. 0.995 x 0.995 x 0.995= 0.9752 / unit weight of

raw coal.

Entire pulverized fuel is not burnt in the boiler to complete combustion. When

investigating steam systems, the boiler is one of the primary targets for energy-efficiency

improvement. There are many tools used in the evaluation and management of boiler

performance. One of the most useful tools is boiler efficiency. Boiler efficiency describes the

fraction of fuel energy that is converted into useful steam energy. Of course, the fuel input

energy that is not converted into useful steam energy represents the losses of the boiler operation.

Coal receipt areaSpillage during receipt 0.5 to 2.0 % on billed Quantity

Transfer to crusher house by conveyors

Transfer loss 0.2 to 0.5%

Crushingoperation

Losses during crushing0.5 to 0.6 %

PulverizationLosses 0.5 to 1.0 %


182

Boiler investigations generally examine the losses by identifying the avenues of loss, measuring

the individual loss and developing a strategy for loss reduction.

There are many avenues of loss encountered in boiler operations. Typically, the dominant

loss is associated with energy leaving the boiler with combustion gases. The temperature of

exhaust gases is an indication of their energy content.

Ensuring that the heat-transfer surfaces of the boiler are clean is a major point of focus

for managing the thermal energy in exhaust gases. Energy can be recovered from exhaust gases

by transferring thermal energy from the high-temperature gases to boiler feed water, or to the

combustion air entering the boiler.

Another aspect of exhaust-gas energy management, which is the focus here, is

combustion management. It should be noted that the temperature- and combustion-related

attributes of exhaust gases are interrelated – they combine to represent the stack loss of the

boiler. Again, this is typically the dominant loss for the boiler. Stack loss is dependent on the

operating characteristics of the boiler, the equipment installed and the type of fuel burned in the

boiler. Stack loss generally ranges from as much as 30 percent for a green-wood-fired boiler, to

18 percent for a typical natural-gas-fired boiler, to 12 percent for an oil-fired boiler, to as low as

9 percent for a coal-fired boiler.

It must be pointed out that the stack-loss range is wide for any given fuel. To address

your question, we will examine the combustion of a simple fuel – methane (CH4). The chemical

equation for the reaction of methane with oxygen (O2) is presented below:

CH4 + 2O2 CO2 + 2H2O

The combustion process does not proceed in a perfect manner. A fuel molecule may

encounter less oxygen than is required for complete combustion. The result will be partial

combustion; the exhaust gases will then contain some unreacted fuel and some partially reacted

fuel. Generally, these unburned fuel components are in the form of carbon monoxide (CO),

hydrogen (H2) and other fuel components that may include the fully unreacted fuel source,

which in this case is methane.

When unburned fuel is found to be part of the combustion products, a portion of the fuel

that was purchased is consequently discharged from the system, unused. It is also important to

note that unburned fuel can accumulate to a point where a safety hazard could result. Unburned


183

fuel can burn in a part of the boiler not designed for combustion – under certain conditions, the

materials can even explode.

Pump and Compressor seal leakage

Normally gland packing or mechanical seals are used but leakage through packing is inevitable

while the leakage through Mechanical seal is almost nil. Seal leakage will vary significantly

with operating pressure, gas properties and condition of the seals. To estimate the gland leak,

sample from the leakage source is collected in a graduated cylinder for a known period of time

and oil content estimated by laboratory analysis or visually. Then the leakage rate is calculated

over a period on one hr. When this exercise is carried out periodically, it is possible to determine

the process loss from leakage. A typical Format is given below for determination of leakage

loss.

Format 1. Leakage Determination from Plant Area

No Unit Pump /

Comp

time

mt /sec

Volum

e in cc

Density

in

gm/ml

Loss

kg/hr

Loss is calculated by the following formula. Loss in kg/hr = ( Vol in cc / time in mt ) * 60 * density * 1/1000Based on the loss data, maintenance priorities could be accorded. From a historical data of leaks, it is possible to identify the source of frequent leaks and causes for remedial action.Example:

1 BFW BFW! 30 325 1.0000 22.4246


184

Loss from BFW1 (325/30)*3600*1.0000*1/1000 = 39.0 kg/h Yearly loss @ 8000 hrs of operation = 312.0 tpy

Loss Monitoring by Models:

From the historic data maintained above, it is possible to develop a hydrocarbon loss

scenario and it's impact on profitability if the situation remains unattended. This approach

could be used for economic justification of equipment replacement / maintenance. A

typical data input and hydrocarbon loss prediction is given in table 3.03 and fig 3.02 &

3.03.

Table 3.03 Leakage Determination by NL Model

month X1 ACTUAL ESTIMATED leak kg/h leak kg/h

1 1.0000 0.3500 0.6975 2 2.0000 1.8900 2.8464 3 3.0000 5.4700 4.9154 4 4.0000 8.8600 6.9043 6 6.0000 10.2100 10.6422 8 8.0000 13.1200 14.0601 10 10.0000 16.7100 17.1580 12 12.0000 20.5500 19.9360

STANDARD ERROR OF ESTIMATE : 0.9231

Leakage forecast using NL Model

month Z1 ESTIMATED leak kg/h

14 14.0000 22.3939 16 16.0000 24.5318


185

18 18.0000 26.3498

Loss during strainer cleaning

Strainer cleaning is a periodical job and during this operation a sizeable amount is being

drained either to ETP or open drain. Although theoretically it should go to ETP/CBD, in

practice it goes to surface drain. Hence, care should be taken during cleaning of the

strainer so that drained material should go to ETP/CBD and ensure no material escapes to

the surface drain. This leads to environmental pollution, safety hazard and also loss of

hydrocarbons by evaporation. This could be ensured by the analysis of surface drain

water for oil content. A format could be developed in the same lines as above to

determine losses due to draining. A historical data on oil content will reveal the operation

and maintenance efficiency of the refinery.

Typical values for well maintained seals

100 SCF/ W per rod for reciprocating comp.

8.5 1b/day per seal for centrifugal comp.

4.0 1b/day per seal for pump

These are representative average numbers only and actual rates at a given location for a

given pump or compressor will vary considerably. Replacement of packing by good

quality mechanical seal will reduce the losses. It will also reduce the power consumption

of the drive and hence energy savings. Failure of seals contribute to losses which can be

minimized by prompt rectification / replacement of seals.

Drip Losses:

Drips and leaks add to high losses. A hydrocarbon leak is easily detectable. Necessary

steps should therefore, be taken to stop the same without any loss of time. The table 3.03

shows the amount of leakage loss in litres (approx.)

Table 3.03 . Estimation of Leakage Loss due to drips & leaks

Leakage In one

minute

In one

hour (Lts)

In one

day (Lts)

In one

week

In one

month (Lts)


186

(Lts.) (Lts)

1. Drop/sec. 0.003 0.17 4.4 31.0 132

2. Drop/sec. 0.009 0.57 14.6 101.0 398

A drop breaking

into

a stream

0.057 3.89 93.4 634 2705

1/8" Stream 0.65 43.0 1011 6842 29594

3/16" Stream 1.11 70.0 1591 11456 49165

1/4" Stream 2.36 152 3500 248211 106445

Leakage of product not only cost excise duty & product loss but also is a fire hazard.

Leakage in a refinery usually occur at :

Pipe joints

Flanges

Valve glands

Pump glands

Storage tanks

Loading rack

PREVENTION AND REDUCTION OF LEAKAGE LOSSES

a) Leakage through pipe joints

As and when a leakage is detected through a threaded pipe joint, the pipe line should be

immediately repaired. Expansion joint should be provided on long pipeline. Thermal

expansion relief provision must be made. This will take the undue stress and strain of the

pipelines due to variation of day's temperature.

b) Leakage through Flanges


187

Flanges should be of standard thickness. The nuts should be tightened crosswise and

diagonally opposite. All nuts should be tightened little by little cross-wise to ensure

equal tightening. Proper gasket packing with liberal use of grease is to be used. Packing

should not extend beyond the flange collar. This helps in tightening the holes evenly all

round.

c) Leakage through valve glands

For globe valves the gland packing should be asbestos rope packing.

For gate valves and valves of bigger diameter where the gap between the valve

stem and the body is more than 3 mm, graphite asbestos rope packing of square

cross section should be used.

The old packing must be removed from the valve and then replace with the new

packing adequately. The gland nuts should be tightened evenly.


188

Controlling Steam losses by survey Steam losses occur due to leaks from gaskets, pipe flanges, pin holes, steam trap passing

etc. This may be quantified and exact reason identified for correcting the situation and

minimizing steam leaks. A typical output is given below.

=======================================================

S T E A M L E A K E S T I M A T I O N

ABC Oil Corporation Ltd

=======================================================

User : xxx Refinery

Unit : India

Input file : C:\tthermw\stmleak.in

Output file : C:\tthermw\stmleak.out

Archive ,, : c:\tthermw\stmleak.arc

Date : 01-22-2008\05:54:44

1. P R O G R A M

O U T P U T UNIT TAG LEAKAGE PLUME SIZE STEAM Loss CODE NO CATEGORY (metre) crude 1 a 1.55 44.826 crude 2 b 1.72 61.341 crude 3 a 1.55 44.826 crude 4 b 1.72 61.341 crude 5 a 1.5 40.876 crude 6 b 1.02 16.860 crude 7 a 1.34 30.428 crude2 2 b 0.72 9.693 crude2 1 a 1.25 25.772 crude2 2 b 1.22 24.385 crude2 1 a 1.15 21.430 crude2 2 b 1.02 16.860 crude2 1 a 1.15 21.430 VBU 2 b 1.22 24.385 VBU 1 a 1.15 21.430 VBU 2 b 1.42 35.267


189

2.TOTAL STEAM LEAK IDENTIFIED

Total Steam Leaks in kg/h 501.1516

Equivalent Fuel in kg/h 7099.648

Steam Enthalpy for estimation (kcal/kg) : 10200

Fuel Calorific Value in kcal/kg : 720

3. LEAK CATEGORISATION

Code 'a' denotes leaks from pin holes & weld joints

Code 'b' denotes leaks from valve glands.

Code 'c' denotes gasket leak from joints.

Code 'd' denotes gasket miscellaneous leaks

***** end of steam leak program *****


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4

PERFORMANCE MONITORING TECHNIQUES