Energy_Audit (Sir Has Said to Use Dis File for Audit)

54Bureau of Energy Efficiency

SyllabusEnergy Management & Audit: Definition, Energy audit- need, Types of energy audit,Energy management (audit) approach-understanding energy costs, Bench marking, Energyperformance, Matching energy use to requirement, Maximizing system efficiencies,Optimizing the input energy requirements, Fuel and energy substitution, Energy auditinstruments

"The judicious and effective use of energy to maximize profits (minimizecosts) and enhance competitive positions"

(Cape Hart, Turner and Kennedy, Guide to Energy Management Fairmont press inc. 1997)

"The strategy of adjusting and optimizing energy, using systems and procedures so as toreduce energy requirements per unit of output while holding constant or reducing totalcosts of producing the output from these systems"

3.1 Definition & Objectives of Energy Management

The fundamental goal of energy management is to produce goods and provide services with theleast cost and least environmental effect.

The term energy management means many things to many people. One definition of ener-gy management is:

Another comprehensive definition is

The objective of Energy Management is to achieve and maintain optimum energy procurementand utilisation, throughout the organization and:

• To minimise energy costs / waste without affecting production & quality• To minimise environmental effects.

3.2 Energy Audit: Types And Methodology

Energy Audit is the key to a systematic approach for decision-making in the area of energy man-agement. It attempts to balance the total energy inputs with its use, and serves to identify allthe energy streams in a facility. It quantifies energy usage according to its discrete functions.Industrial energy audit is an effective tool in defining and pursuing comprehensive energy man-agement programme.

As per the Energy Conservation Act, 2001, Energy Audit is defined as "the verification, mon-

3. ENERGY MANAGEMENT AND AUDIT

3. Energy Management and Audit


itoring and analysis of use of energy including submission of technical report containing rec-ommendations for improving energy efficiency with cost benefit analysis and an action plan toreduce energy consumption".

3.2.1 Need for Energy Audit

In any industry, the three top operating expenses are often found to be energy (both electricaland thermal), labour and materials. If one were to relate to the manageability of the cost orpotential cost savings in each of the above components, energy would invariably emerge as atop ranker, and thus energy management function constitutes a strategic area for cost reduction.Energy Audit will help to understand more about the ways energy and fuel are used in anyindustry, and help in identifying the areas where waste can occur and where scope for improve-ment exists.

The Energy Audit would give a positive orientation to the energy cost reduction, preventivemaintenance and quality control programmes which are vital for production and utility activi-ties. Such an audit programme will help to keep focus on variations which occur in the energycosts, availability and reliability of supply of energy, decide on appropriate energy mix, identi-fy energy conservation technologies, retrofit for energy conservation equipment etc.

In general, Energy Audit is the translation of conservation ideas into realities, by lendingtechnically feasible solutions with economic and other organizational considerations within aspecified time frame.

The primary objective of Energy Audit is to determine ways to reduce energy consumptionper unit of product output or to lower operating costs. Energy Audit provides a " bench-mark"(Reference point) for managing energy in the organization and also provides the basis for plan-ning a more effective use of energy throughout the organization.

3.2.2 Type of Energy Audit

The type of Energy Audit to be performed depends on:- Function and type of industry- Depth to which final audit is needed, and - Potential and magnitude of cost reduction desiredThus Energy Audit can be classified into the following two types.i) Preliminary Auditii) Detailed Audit

3.2.3 Preliminary Energy Audit Methodology

Preliminary energy audit is a relatively quick exercise to:• Establish energy consumption in the organization• Estimate the scope for saving• Identify the most likely (and the easiest areas for attention• Identify immediate (especially no-/low-cost) improvements/ savings• Set a 'reference point'• Identify areas for more detailed study/measurement• Preliminary energy audit uses existing, or easily obtained data



3.2.4 Detailed Energy Audit Methodology

A comprehensive audit provides a detailed energy project implementation plan for a facility,since it evaluates all major energy using systems.

This type of audit offers the most accurate estimate of energy savings and cost. It considersthe interactive effects of all projects, accounts for the energy use of all major equipment, andincludes detailed energy cost saving calculations and project cost.

In a comprehensive audit, one of the key elements is the energy balance. This is based on aninventory of energy using systems, assumptions of current operating conditions and calculationsof energy use. This estimated use is then compared to utility bill charges.

Detailed energy auditing is carried out in three phases: Phase I, II and III.

Phase I - Pre Audit PhasePhase II - Audit PhasePhase III - Post Audit Phase

A Guide for Conducting Energy Audit at a Glance

Industry-to-industry, the methodology of Energy Audits needs to be flexible. A comprehensive ten-step methodology for conduct of Energy Audit at field level is pre-

sented below. Energy Manager and Energy Auditor may follow these steps to start with andadd/change as per their needs and industry types.



Ten Steps Methodology for Detailed Energy Audit





Phase I -Pre Audit Phase Activities

A structured methodology to carry out an energy audit is necessary for efficient working. Aninitial study of the site should always be carried out, as the planning of the procedures neces-sary for an audit is most important.

Initial Site Visit and Preparation Required for Detailed Auditing

An initial site visit may take one day and gives the Energy Auditor/Engineer an opportunity tomeet the personnel concerned, to familiarize him with the site and to assess the procedures nec-essary to carry out the energy audit.

During the initial site visit the Energy Auditor/Engineer should carry out the followingactions: -• Discuss with the site's senior management the aims of the energy audit.• Discuss economic guidelines associated with the recommendations of the audit.• Analyse the major energy consumption data with the relevant personnel.• Obtain site drawings where available - building layout, steam distribution, compressed air

distribution, electricity distribution etc.• Tour the site accompanied by engineering/production

The main aims of this visit are: -

• To finalise Energy Audit team• To identify the main energy consuming areas/plant items to be surveyed during the audit.• To identify any existing instrumentation/ additional metering required.• To decide whether any meters will have to be installed prior to the audit eg. kWh, steam,

oil or gas meters.• To identify the instrumentation required for carrying out the audit.• To plan with time frame• To collect macro data on plant energy resources, major energy consuming centers• To create awareness through meetings/ programme

Phase II- Detailed Energy Audit Activities

Depending on the nature and complexity of the site, a comprehensive audit can take from sev-eral weeks to several months to complete. Detailed studies to establish, and investigate, energyand material balances for specific plant departments or items of process equipment are carriedout. Whenever possible, checks of plant operations are carried out over extended periods oftime, at nights and at weekends as well as during normal daytime working hours, to ensure thatnothing is overlooked.

The audit report will include a description of energy inputs and product outputs by majordepartment or by major processing function, and will evaluate the efficiency of each step of themanufacturing process. Means of improving these efficiencies will be listed, and at least a pre-liminary assessment of the cost of the improvements will be made to indicate the expected pay-back on any capital investment needed. The audit report should conclude with specific recom-mendations for detailed engineering studies and feasibility analyses, which must then be per-formed to justify the implementation of those conservation measures that require investments.



The information to be collected during the detailed audit includes: -

1. Energy consumption by type of energy, by department, by major items of process equipment, by end-use

2. Material balance data (raw materials, intermediate and final products, recycledmaterials, use of scrap or waste products, production of by-products for re-use in otherindustries, etc.)

3. Energy cost and tariff data4. Process and material flow diagrams5. Generation and distribution of site services (eg.compressed air, steam).6. Sources of energy supply (e.g. electricity from the grid or self-generation)7. Potential for fuel substitution, process modifications, and the use of co-generation

systems (combined heat and power generation).8. Energy Management procedures and energy awareness training programs within the

establishment.Existing baseline information and reports are useful to get consumption pattern, production costand productivity levels in terms of product per raw material inputs. The audit team should col-lect the following baseline data:

- Technology, processes used and equipment details- Capacity utilisation - Amount & type of input materials used - Water consumption- Fuel Consumption- Electrical energy consumption- Steam consumption- Other inputs such as compressed air, cooling water etc - Quantity & type of wastes generated- Percentage rejection / reprocessing- Efficiencies / yield

DATA COLLECTION HINTS

It is important to plan additional data gathering carefully. Here are some basic tips to avoid wasting timeand effort:

• measurement systems should be easy to use and provide the information to the accuracy that is needed, not the accuracy that is technically possible

• measurement equipment can be inexpensive (flow rates using a bucket and stopwatch)• the quality of the data must be such that the correct conclusions are drawn (what grade of prod

uct is on, is the production normal etc)• define how frequent data collection should be to account for process variations. • measurement exercises over abnormal workload periods (such as startup and shutdowns)• design values can be taken where measurements are difficult (cooling water through heat exchang

er)DO NOT ESTIMATE WHEN YOU CAN CALCULATE

DO NOT CALCULATE WHEN YOU CAN MEASURE



Draw process flow diagram and list process steps; identify waste streams and obviousenergy wastage

An overview of unit operations, important process steps, areas of material and energy use andsources of waste generation should be gathered and should be represented in a flowchart asshown in the figure below. Existing drawings, records and shop floor walk through will help inmaking this flow chart. Simultaneously the team should identify the various inputs & outputstreams at each process step.

Example: A flowchart of Penicillin-G manufacturing is given in the figure3.1 below. Notethat waste stream (Mycelium) and obvious energy wastes such as condensate drained and steamleakages have been identified in this flow chart

The audit focus area depends on several issues like consumption of input resources, energyefficiency potential, impact of process step on entire process or intensity of waste generation /energy consumption. In the above process, the unit operations such as germinator, pre-fermen-tor, fermentor, and extraction are the major conservation potential areas identified.

Figure 3.1



Identification of Energy Conservation Opportunities

Fuel substitution: Identifying the appropriate fuel for efficient energy conversion

Energy generation :Identifying Efficiency opportunities in energy conversion equipment/util-ity such as captive power generation, steam generation in boilers, thermic fluid heating, optimalloading of DG sets, minimum excess air combustion with boilers/thermic fluid heating, opti-mising existing efficiencies, efficienct energy conversion equipment, biomass gasifiers,Cogeneration, high efficiency DG sets, etc.

Energy distribution: Identifying Efficiency opportunities network such as transformers,cables, switchgears and power factor improvement in electrical systems and chilled water, cool-ing water, hot water, compressed air, Etc.

Energy usage by processes: This is where the major opportunity for improvement and manyof them are hidden. Process analysis is useful tool for process integration measures.

Technical and Economic feasibility

The technical feasibility should address the following issues

• Technology availability, space, skilled manpower, reliability, service etc• The impact of energy efficiency measure on safety, quality, production or process.• The maintenance requirements and spares availability

The Economic viability often becomes the key parameter for the management acceptance. Theeconomic analysis can be conducted by using a variety of methods. Example: Pay back method,Internal Rate of Return method, Net Present Value method etc. For low investment short dura-tion measures, which have attractive economic viability, simplest of the methods, payback isusually sufficient. A sample worksheet for assessing economic feasibility is provided below:

Classification of Energy Conservation Measures

Based on energy audit and analyses of the plant, a number of potential energy saving projectsmay be identified. These may be classified into three categories:



1. Low cost - high return;2. Medium cost - medium return;3. High cost - high return

Normally the low cost - high return projects receive priority. Other projects have to be analyzed,engineered and budgeted for implementation in a phased manner. Projects relating to energycascading and process changes almost always involve high costs coupled with high returns, andmay require careful scrutiny before funds can be committed. These projects are generally com-plex and may require long lead times before they can be implemented. Refer Table 3.1 for pro-ject priority guidelines.

3.3 Energy Audit Reporting Format

After successfully carried out energy audit energy manager/energy auditor should report to thetop management for effective communication and implementation. A typical energy auditreporting contents and format are given below. The following format is applicable for most ofthe industries. However the format can be suitably modified for specific requirement applicablefor a particular type of industry.







The following Worksheets (refer Table 3.2 & Table 3.3) can be used as guidance for energyaudit assessment and reporting.

TABLE 3.2 SUMMARY OF ENERGY SAVING RECOMMENDATIONS

S.No. Energy Saving Annual Energy Annual Capital SimpleRecommendations (Fuel & Electricity) Savings Investment Payback

Savings (kWh/MT Rs.Lakhs (Rs.Lakhs) periodor kl/MT)

1

2

3

4

Total

TABLE 3.3 TYPES AND PRIORITY OF ENERGY SAVING MEASURES

Type of Energy Annual Annual Saving Options Electricity Savings Priority

/Fuel savings

KWh/MT or kl/MT (Rs Lakhs)

A No Investment(Immediate)

- OperationalImprovement- Housekeeping

B Low Investment(Short to Medium Term)

- Controls- Equipment Modification- Process change

C High Investment(Long Term)

- Energy efficient Devices- Product modification- Technology Change



3.4 Understanding Energy Costs

Understanding energy cost is vital factor for awareness creation and saving calculation. Inmany industries sufficient meters may not be available to measure all the energy used. In suchcases, invoices for fuels and electricity will be useful. The annual company balance sheet is theother sources where fuel cost and power are given with production related information.

Energy invoices can be used for the following purposes:• They provide a record of energy purchased in a given year, which gives a base-line for

future reference• Energy invoices may indicate the potential for savings when related to production

requirements or to air conditioning requirements/space heating etc.• When electricity is purchased on the basis of maximum demand tariff• They can suggest where savings are most likely to be made.• In later years invoices can be used to quantify the energy and cost savings made through

energy conservation measures

Fuel Costs

A wide variety of fuels are available forthermal energy supply. Few are listedbelow:

• Fuel oil• Low Sulphur Heavy Stock (LSHS)• Light Diesel Oil (LDO)• Liquefied Petroleum Gas (LPG)• COAL• LIGNITE• WOOD ETC.

Understanding fuel cost is fairly simpleand it is purchased in Tons or Kiloliters.Availability, cost and quality are the mainthree factors that should be consideredwhile purchasing. The following factors should be taken into account during procurement offuels for energy efficiency and economics.

• Price at source, transport charge, type of transport• Quality of fuel (contaminations, moisture etc)• Energy content (calorific value)

Power Costs

Electricity price in India not only varies from State to State, but also city to city and consumerto consumer though it does the same work everywhere. Many factors are involved in decidingfinal cost of purchased electricity such as:

• Maximum demand charges, kVA(i.e. How fast the electricity is used? )



Figure 3.2 Annual Energy Bill



• Energy Charges, kWh(i.e., How much electricity is consumed? )

• TOD Charges, Peak/Non-peak period(i.e. When electricity is utilized ?)

• Power factor Charge, P.F(i.e., Real power use versus Apparent power use factor )

• Other incentives and penalties applied from time to time• High tension tariff and low tension tariff rate changes• Slab rate cost and its variation • Type of tariff clause and rate for various categories such as commercial, residential,

industrial, Government, agricultural, etc.• Tariff rate for developed and underdeveloped area/States • Tax holiday for new projects

Example: Purchased energy Bill

A typical summary of energy purchased in an industry based on the invoices

Unfortunately the different forms of energy are sold in different units e.g. kWh of electricity,liters of fuel oil, tonne of coal. To allow comparison of energy quantities these must be con-verted to a common unit of energy such as kWh, Giga joules, kCals etc.

Electricity (1 kWh) = 860 kCal/kWh (0.0036 GJ)Heavy fuel oil (Gross calorific value, GCV) =10000 kCal/litre ( 0.0411 GJ/litre)Coal (Gross calorific value, GCV) =4000 kCal/kg ( 28 GJ/ton)

3.5 Benchmarking and Energy Performance

Benchmarking of energy consumption internally (historical / trend analysis) and externally(across similar industries) are two powerful tools for performance assessment and logical evo-lution of avenues for improvement. Historical data well documented helps to bring out energyconsumption and cost trends month-wise / day-wise. Trend analysis of energy consumption,cost, relevant production features, specific energy consumption, help to understand effects ofcapacity utilization on energy use efficiency and costs on a broader scale.

External benchmarking relates to inter-unit comparison across a group of similar units.However, it would be important to ascertain similarities, as otherwise findings can be grossly

TABLE 3.4

Type of energy Original units Unit Cost Monthly Bill Rs.

Electricity 5,00,000 kWh Rs.4.00/kWh 20,00,000

Fuel oil 200 kL Rs.10,000/ kL 20,00,000

Coal 1000 tons Rs.2,000/ton 20,00,000

Total 60,00,000



misleading. Few comparative factors, which need to be looked into while benchmarking exter-nally are:

• Scale of operation• Vintage of technology• Raw material specifications and quality• Product specifications and quality

Benchmarking energy performance permits• Quantification of fixed and variable energy consumption trends vis-à-vis production

levels• Comparison of the industry energy performance with respect to various production

levels (capacity utilization)• Identification of best practices (based on the external benchmarking data)• Scope and margin available for energy consumption and cost reduction• Basis for monitoring and target setting exercises.

The benchmark parameters can be:• Gross production related

e.g. kWh/MT clinker or cement produced (cement plant)e.g. kWh/kg yarn produced (Textile unit)e.g. kWh/MT, kCal/kg, paper produced (Paper plant)e.g. kCal/kWh Power produced (Heat rate of a power plant)e.g. Million kilocals/MT Urea or Ammonia (Fertilizer plant)e.g. kWh/MT of liquid metal output (in a foundry)

• Equipment / utility relatede.g. kW/ton of refrigeration (on Air conditioning plant)e.g. % thermal efficiency of a boiler plante.g. % cooling tower effectiveness in a cooling towere.g. kWh/NM3 of compressed air generatede.g. kWh /litre in a diesel power generation plant.

While such benchmarks are referred to, related crucial process parameters need mentioning formeaningful comparison among peers. For instance, in the above case:

• For a cement plant - type of cement, blaine number (fineness) i.e. Portland and process used (wet/dry) are to be reported alongside kWh/MT figure.

• For a textile unit - average count, type of yarn i.e. polyester/cotton, is to be reported along side kWh/square meter.

• For a paper plant - paper type, raw material (recycling extent), GSM quality is some important factors to be reported along with kWh/MT, kCal/Kg figures.

• For a power plant / cogeneration plant - plant % loading, condenser vacuum, inlet cooling water temperature, would be important factors to be mentioned alongside heat rate (kCal/kWh).

• For a fertilizer plant - capacity utilization(%) and on-stream factor are two inputs worth comparing while mentioning specific energy consumption

• For a foundry unit - melt output, furnace type, composition (mild steel, high carbon steel/cast iron etc.) raw material mix, number or power trips could be some useful oper

ating parameters to be reported while mentioning specific energy consumption data.• For an Air conditioning (A/c) plant - Chilled water temperature level and refrigeration

load (TR) are crucial for comparing kW/TR.• For a boiler plant - fuel quality, type, steam pressure, temperature, flow, are useful com

parators alongside thermal efficiency and more importantly, whether thermal efficiency is on gross calorific value basis or net calorific value basis or whether the computation is by direct method or indirect heat loss method, may mean a lot in benchmarking exercise for meaningful comparison.

• Cooling tower effectiveness - ambient air wet/dry bulb temperature, relative humidity, air and circulating water flows are required to be reported to make meaningful sense.

• Compressed air specific power consumption - is to be compared at similar inlet air temperature and pressure of generation.

• Diesel power plant performance - is to be compared at similar loading %, steady run condition etc.

Plant Energy Performance

Plant energy performance (PEP) is the measure of whether a plant is now using more or lessenergy to manufacture its products than it did in the past: a measure of how well the energymanagement programme is doing. It compares the change in energy consumption from oneyear to the other considering production output. Plant energy performance monitoring comparesplant energy use at a reference year with the subsequent years to determine the improvementthat has been made.

However, a plant production output may vary from year to year and the output has a sig-nificant bearing on plant energy use. For a meaningful comparison, it is necessary to determinethe energy that would have been required to produce this year production output, if the planthad operated in the same way as it did during the reference year. This calculated value can thenbe compared with the actual value to determine the improvement or deterioration that has takenplace since the reference year.

Production factor

Production factor is used to determine the energy that would have been required to produce thisyear's production output if the plant had operated in the same way as it did in the reference year.It is the ratio of production in the current year to that in the reference year.

Reference Year Equivalent Energy Use

The reference year's energy use that would have been used to produce the current year's pro-duction output may be called the "reference year energy use equivalent" or "reference yearequivalent" for short. The reference year equivalent is obtained by multiplying the referenceyear energy use by the production factor (obtained above)

Reference year equivalent = Reference year energy use x Production factor

The improvement or deterioration from the reference year is called "energy performance" and



productionsyearReferencePr

'

' productionsyearCurrent=factoroduction



is a measure of the plant's energy management progress. It is the reduction or increase in thecurrent year's energy use over the reference, and is calculated by subtracting the current year'senergy use from the reference years equivalent. The result is divided by the reference yearequivalent and multiplied by 100 to obtain a percentage.

The energy performance is the percentage of energy saved at the current rate of use comparedto the reference year rate of use. The greater the improvement, the higher the number will be.

Monthly Energy Performance

Experience however, has shown that once a plant has started measuring yearly energy perfor-mance, management wants more frequent performance information in order to monitor andcontrol energy use on an on-going basis. PEP can just as easily be used for monthly reportingas yearly reporting.

3.6 Matching Energy Usage to Requirement

Mismatch between equipment capacity and user requirement often leads to inefficiencies due topart load operations, wastages etc. Worst case design, is a designer's characteristic, while opti-mization is the energy manager's mandate and many situations present themselves towards anexercise involving graceful matching of energy equipment capacity to end-use needs. Someexamples being:

• Eliminate throttling of a pump by impeller trimming, resizing pump, installing variable speed drives

• Eliminate damper operations in fans by impeller trimming, installing variable speed drives, pulley diameter modification for belt drives, fan resizing for better efficiency.

• Moderation of chilled water temperature for process chilling needs• Recovery of energy lost in control valve pressure drops by back pressure/turbine adop

tion• Adoption of task lighting in place of less effective area lighting

3.7 Maximising System Efficiency

Once the energy usage and sources are matched properly, the next step is to operate the equip-ment efficiently through best practices in operation and maintenance as well as judicious tech-nology adoption. Some illustrations in this context are:

• Eliminate steam leakages by trap improvements• Maximise condensate recovery• Adopt combustion controls for maximizing combustion efficiency• Replace pumps, fans, air compressors, refrigeration compressors, boilers, furnaces,

heaters and other energy consuming equipment, wherever significant energy efficiency margins exist.

100xequivalentyear Reference

energy syear'Current -equivalentyear Reference eperformancenergy Plant =



Optimising the Input Energy Requirements

Consequent upon fine-tuning the energy use practices, attention is accorded to considerationsfor minimizing energy input requirements. The range of measures could include:

• Shuffling of compressors to match needs.• Periodic review of insulation thickness• Identify potential for heat exchanger networking and process integration.• Optimisation of transformer operation with respect to load.

3.8 Fuel and Energy Substitution

Fuel substitution: Substituting existing fossil fuel with more efficient and less cost/less pol-luting fuel such as natural gas, biogas and locally available agro-residues.

Energy is an important input in the production. There are two ways to reduce energy depen-dency; energy conservation and substitution.

Fuel substitution has taken place in all the major sectors of the Indian economy. Keroseneand Liquefied Petroleum Gas (LPG) have substituted soft coke in residential use. Few examples of fuel substitution

• Natural gas is increasingly the fuel of choice as fuel and feedstock in the fertilizer, petrochemicals, power and sponge iron industries.

• Replacement of coal by coconut shells, rice husk etc.• Replacement of LDO by LSHS

Few examples of energy substitution

� Replacement of electric heaters by steam heaters� Replacement of steam based hotwater by solar systems

Case Study : Example on Fuel Substitution

A textile process industry replaced old fuel oil fired thermic fluid heater with agro fuel firedheater. The economics of the project are given below:

A: Title of Recommendation : Use of Agro Fuel (coconut chips) in place of Furnaceoil in a Boiler

B: Description of Existing System and its operation : A thermic fluid heater with furnace oil currently.

In the same plant a coconut chip fired boiler is operating continuously with good performance.

C: Description of Proposed systemand its operation : It was suggested to replace the oil fired thermic

fluid heater with coconut chip fired boiler as the company has the facilities for handling coconut chip fired system.



D: Energy Saving CalculationsOld SystemType of fuel Firing : Furnace Oil fired heaterGCV : 10,200 kCal/kgAvg. Thermal Efficiency : 82%Heat Duty : 15 lakh kCal / hourOperating Hours : 25 days x 12 month x 24 hours = 7,200 hrs.Annual Fuel Cost : Rs.130 lakh (7200 x 1800 Rs./hr.)

Modified System

Type of fuel saving = Coconut chips fired HeaterGCV = 4200 kCal/kgAverage Thermal Efficiency = 72 %Heat Duty = 15 lakh kCal / hourAnnual Operating Cost = 7200 x 700 Rs./hr = 50 lakhAnnual Savings = 130 - 50 = Rs.80 lakh .Additional Auxiliary Power +Manpower Cost = Rs. 10 lakhNet Annual Saving = Rs. 70 lakh Investment for New Coconut Fired heater = Rs. 35 lakh

Simple pay back period = 6 months

3.9 Energy Audit Instruments

The requirement for an energy audit such as identification and quantification of energy neces-sitates measurements; these measurements require the use of instruments. These instrumentsmust be portable, durable, easy to operate and relatively inexpensive. The parameters generallymonitored during energy audit may include the following:

Basic Electrical Parameters in AC &DC systems - Voltage (V), Current (I), Power factor, Activepower (kW), apparent power (demand) (kVA), Reactive power (kVAr), Energy consumption(kWh), Frequency (Hz), Harmonics, etc.

Parameters of importance other than electrical such as temperature & heat flow, radiation, airand gas flow, liquid flow, revolutions per minute (RPM), air velocity, noise and vibration, dustconcentration, Total Dissolved Solids (TDS), pH, moisture content, relative humidity, flue gasanalysis - CO2, O2, CO, SOx, NOx, combustion efficiency etc.

Key instruments for energy audit are listed below.

The operating instructions for all instruments must be understood and staff should familiarizethemselves with the instruments and their operation prior to actual audit use.



Electrical Measuring Instruments:

These are instruments for measuring majorelectrical parameters such as kVA, kW, PF,Hertz, kVAr, Amps and Volts. In additionsome of these instruments also measureharmonics.

These instruments are applied on-line i.e on running motors without any need to stop the motor. Instant measurements can be taken with hand-held meters, while more advanced ones facilitates cumulative readings with print outs at specified intervals.

Combustion analyzer:

This instrument has in-built chemical cellswhich measure various gases such as O2, CO,NOX and SOX.

Fuel Efficiency Monitor:

This measures oxygen and temperature of theflue gas. Calorific values of common fuelsare fed into the microprocessor which calcu-lates the combustion efficiency.

Fyrite:

A hand bellow pump draws the flue gassample into the solution inside the fyrite. Achemical reaction changes the liquid volumerevealing the amount of gas. A separate fyritecan be used for O2 and CO2 measurement.



Contact thermometer:

These are thermocouples which measures forexample flue gas, hot air, hot water tempera-tures by insertion of probe into the stream.

For surface temperature, a leaf type probe isused with the same instrument.

Infrared Thermometer:

This is a non-contact type measurementwhich when directed at a heat source directlygives the temperature read out. This instru-ment is useful for measuring hot spots infurnaces, surface temperatures etc.

Pitot Tube and manometer:

Air velocity in ducts can be measured usinga pitot tube and inclined manometer for fur-ther calculation of flows.

Water flow meter:

This non-contact flow measuring deviceusing Doppler effect / Ultra sonic principle.There is a transmitter and receiver which arepositioned on opposite sides of the pipe. Themeter directly gives the flow. Water and otherfluid flows can be easily measured with thismeter.



Speed Measurements:

In any audit exercise speed measurements arecritical as thay may change with frequency,belt slip and loading.

A simple tachometer is a contact type instru-ment which can be used where direct accessis possible.

More sophisticated and safer ones are noncontact instruments such as stroboscopes.

Leak Detectors:

Ultrasonic instruments are available whichcan be used to detect leaks of compressed airand other gases which are normally not pos-sible to detect with human abilities.

Lux meters:

Illumination levels are measured with a luxmeter. It consists of a photo cell which sens-es the light output, converts to electricalimpulses which are calibrated as lux.

Tachometer Stroboscope



QUESTIONS

1. List down the objective of energy management..

2. What are the managerial functions involved in energy management?

3. Explain why managerial skills are as important as technical skills in energy management?

4. What are the various steps in the implementation of energy management in an organization?

5. State the importance of energy policy for industries.

6. Explain the role of training and awareness in energy management programme?

7. What is an energy audit?

8. Explain briefly the difference between preliminary and detailed energy audits?

9. What is the significance of knowing the energy costs?

10. What are the benefits of benchmarking energy consumption?

11. Explain the implications of part load operation of energy equipment with examples?

12. What do you understand by the term fuel substitution? Give examples.

13. What are the parameters that can be measured by on line power analyser?

14. Name the one instrument used to measure CO2 from boilers stack is(a) Infrared thermometer (b) Fyrite (c) Anemometer (d) Pitot tube

15. Non contact flow measurement can be carried out by(a) Orifice meter (b) Turbine flow meter (c) Ultrasonic flow meter (d) Magneticflow meter

16. Non contact speed measurements can be carried out by (a) Tachometer (b) Stroboscope (c) Oscilloscope (d) Odometer

REFERENCES

1. NPC energy audit manual and reports2. Energy management handbook, John Wiley and Sons - Wayne C. Turner3. Guide to Energy Management, Cape Hart, Turner and Kennedy4. Cleaner Production – Energy Efficiency Manual for GERIAP, UNEP, Bangkok prepared

by National Productivity Council

www.eeca.govt.nzwww.energyusernews.com/


5. ENERGY ACTION PLANNING

SyllabusEnergy Action Planning: Key elements, Force field analysis, Energy policy purpose, per-spective, Contents, Formulation, Ratification, Organizing - location of energy manage-ment, Top management support, Managerial function, Roles and responsibilities of energymanager, Accountability. Motivating-motivation of employees: Information system-designing barriers, Strategies; Marketing and communicating-training and planning.

5.1 Introduction

Energy efficiency is extremelyimportant to all organisations, espe-cially those that are energy intensive. The four vital requirements for a suc-cessful energy management is shownin Figure 5.1. Any successful energymanagement programme within anorganisation needs the total supportof top management. Hence, top man-agement support is the key require-ment for success. Top managementshould give energy efficiency equalimportance in their corporate objec-tives as manpower, raw materials, production and sales. The other important requirements area well charted strategy plan, an effective monitoring system and adequate technical ability foranalysing and implementing energy saving options.

5.2 Energy Management System

Organizations seeking financial returns from superior energy management continuously striveto improve their energy performance. Their success is based on regularly assessing energy per-formance, planning and implementing action plans to improve energy efficiency. Hence a soundenergy management system is a prerequisite for identifying and implementing energy conser-vation measures, sustaining the momentum and for effecting improvements on a continuousbasis. The various steps for energy action planning are shown in Figure 5.2.

Figure 5.1 The 4 Pillars of Successful Energy Management

5. Energy Action Planning


5.2.1 Top Management Commitment and Support

Top management shall make a commitment to allocate manpower and funds to achieve contin-uous improvement. To establish the energy management programme, leading organizationsappoint energy manager, form a dedicated energy team and institute an energy policy.

Appoint an Energy Manager

The tasks of energy manger are setting goals, tracking progress, and promoting the ener-gy management program. An Energy Manager helps an organization achieve its goals byestablishing energy performance as a core value.

The Energy Manager is not always an expert in energy and technical systems. SuccessfulEnergy Manager understands how energy management helps the organization achieve its finan-cial and environmental goals and objectives. Depending on the size of the organization, theEnergy Manager role can be a full-time position or an addition to other responsibilities.

Location of Energy Manager

The energy management function, whether vested in one "energy manager or coordinator" ordistributed among a number of middle managers, usually resides somewhere in the organiza-tion between senior management and those who control the end-use of energy. Exactly howand where that function is placed is a decision that needs to be made in view of the existingorganisational structure.

Figure 5.2 Steps in Energy Action Planning

5. Energy Action Planning


Energy Manager: Responsibilities and Duties to be Assigned Under The EnergyConservation Act, 2001.

Responsibilities

• Prepare an annual activity plan and present to management concerning financially attrac-tive investments to reduce energy costs

• Establish an energy conservation cell within the firm with management's consent aboutthe mandate and task of the cell.

• Initiate activities to improve monitoring and process control to reduce energy costs.• Analyze equipment performance with respect to energy efficiency• Ensure proper functioning and calibration of instrumentation required to assess level of

energy consumption directly or indirectly.• Prepare information material and conduct internal workshops about the topic for other

staff.• Improve disaggregating of energy consumption data down to shop level or profit center

of a firm.• Establish a methodology how to accurately calculate the specific energy consumption of

various products/services or activity of the firm.• Develop and manage training programme for energy efficiency at operating levels.• Co-ordinate nomination of management personnel to external programs.• Create knowledge bank on sectoral, national and inter-national development on energy

efficiency technology and management system and information denomination• Develop integrated system of energy efficiency and environmental up gradation.• Co-ordinate implementation of energy audit/efficiency improvement projects through

external agencies.• Establish and/or participate in information exchange with other energy managers of the

same sector through association

Duties

• Report to BEE and State level Designated Agency once a year the information withregard to the energy consumed and action taken on the recommendation of the accredit-ed energy auditor, as per BEE Format.

• Establish an improved data recording, collection and analysis system to keep track ofenergy consumption.

• Provide support to Accredited Energy Audit Firm retained by the company for the con-duct of energy audit

• Provide information to BEE as demanded in the Act, and with respect to the tasks givenby a mandate, and the job description.

• Prepare a scheme for efficient use of energy and its conservation and implement suchscheme keeping in view of the economic stability of the investment in such form andmanner as may be provided in the regulations of the Energy Conservation Act.

Form A Dedicated Energy Team

The tasks of energy team are executing energy management activities across differentparts of the organization and ensuring integration of best practices.

8. ENERGY MONITORING AND TARGETING


SyllabusEnergy Monitoring and Targeting: Defining monitoring & targeting, Elements of mon-itoring & targeting, Data and information-analysis, Techniques -energy consumption,Production, Cumulative sum of differences (CUSUM).

8.1 Definition

Energy monitoring and targeting is primarily a management technique that uses energy infor-mation as a basis to eliminate waste, reduce and control current level of energy use and improvethe existing operating procedures. It builds on the principle "you can't manage what youdon't measure". It essentially combines the principles of energy use and statistics.

While, monitoring is essentially aimed at establishing the existing pattern of energy con-sumption, targeting is the identification of energy consumption level which is desirable as amanagement goal to work towards energy conservation.

Monitoring and Targeting is a management technique in which all plant and building utili-ties such as fuel, steam, refrigeration, compressed air, water, effluent, and electricity are man-aged as controllable resources in the same way that raw materials, finished product inventory,building occupancy, personnel and capital are managed. It involves a systematic, disciplineddivision of the facility into Energy Cost Centers. The utilities used in each centre are closelymonitored, and the energy used is compared with production volume or any other suitable mea-sure of operation. Once this information is available on a regular basis, targets can be set, vari-ances can be spotted and interpreted, and remedial actions can be taken and implemented.

The Monitoring and Targeting programs have been so effective that they show typicalreductions in annual energy costs in various industrial sectors between 5 and 20%.

8.2 Elements of Monitoring & Targeting System

The essential elements of M&T system are:• Recording -Measuring and recording energy consumption• Analysing -Correlating energy consumption to a measured output, such as production

quantity• Comparing -Comparing energy consumption to an appropriate standard or benchmark• Setting Targets -Setting targets to reduce or control energy consumption• Monitoring -Comparing energy consumption to the set target on a regular basis• Reporting -Reporting the results including any variances from the targets which have

been set• Controlling -Implementing management measures to correct any variances, which may

have occurred.Particularly M&T system will involve the following:

• Checking the accuracy of energy invoices• Allocating energy costs to specific departments (Energy Accounting Centres)

• Determining energy performance/efficiency• Recording energy use, so that projects intended to improve energy efficiency can be

checked• Highlighting performance problems in equipment or systems

8.3 A Rationale for Monitoring, Targeting and Reporting

The energy used by any business varies with production processes, volumes and input.Determining the relationship of energy use to key performance indicators will allow you todetermine:

• Whether your current energy is better or worse than before• Trends in energy consumption that reflects seasonal, weekly, and other operational para-

meters• How much your future energy use is likely to vary if you change aspects of your busi-

ness• Specific areas of wasted energy• Comparison with other business with similar characteristics - This "benchmarking"

process will provide valuable indications of effectiveness of your operations as well asenergy use

• How much your business has reacted to changes in the past• How to develop performance targets for an energy management program

Information related to energy use may be obtained from following sources:

• Plant level information can be derived from financial accounting systems-utilities costcentre

• Plant department level information can be found in comparative energy consumptiondata for a group of similar facilities, service entrance meter readings etc.

• System level (for example, boiler plant) performance data can be determined from sub-metering data

• Equipment level information can be obtained from nameplate data, run-time and sched-ule information, sub-metered data on specific energy consuming equipment.

The important point to be made here is that all of these data are useful and can be processed toyield information about facility performance.

8.4 Data and Information Analysis

Electricity bills and other fuel bills should be collected periodically and analysed as below. Atypical format for monitoring plant level information is given below in the Table 8.1.

8. Energy Monitoring and Targeting




TABLE 8.1 ANNUAL ENERGY COST SHEET

Thermal Energy Bill Electricity Bill TotalEnergy Bill

Month Fuel 1 Fuel 2 Fuel 3 Total Day Night Maximum Total Rs.LakhRs. Lakh kWh kWh Demand Rs. Lakh

1

2

3

4

5

6

7

8

9

10

11

12

Sub-Total

%

Pie Chart on Energy Consumption

All the fuels purchased by the plant should be converted into common units such as kCal. Thefollowing Table 8.2 below is for that purpose.

Figure 8.1 % Share of Fuels Based on Energy Bill

After obtaining the respective annual energy cost, a pie chart (see Figure 8.1) can be drawn asshown below:

After conversion to a common unit, a pie chart can be drawn showing the percentage dis-tribution of energy consumption as shown in Figure 8.2.

8.5 Relating Energy Consumption and Production.

Graphing the Data

A critical feature of M&T is to understand what drives energy consumption. Is it production,hours of operation or weather? Knowing this, we can then start to analyse the data to see howgood our energy management is.

After collection of energy consumption, energy cost and production data, the next stage ofthe monitoring process is to study and analyse the data to understand what is happening in theplant. It is strongly recommended that the data be presented graphically. A better appreciationof variations is almost always obtained from a visual presentation, rather than from a table ofnumbers. Graphs generally provide an effective means of developing the energy-productionrelationships, which explain what is going on in the plant.

Use of Bar Chart

The energy data is then entered into a spreadsheet. It is hard to envisage what is happening fromplain data, so we need to present the data using bar chart. The starting point is to collect andcollate 24/12 months of energy bills. The most common bar chart application used in energy



TABLE 8.2 FUEL CONVERSION DATA

Energy source Supply unit Conversion Factor to Kcal

Electricity kWh 860

HSD kg 10,500

Furnace Oil kg 10,200

LPG kg 12,000

Figure 8.2 %Share of Fuels Based on Consumption in kCals



management is one showing the energy per month for this year and last year (see Figure 8.3) -however, it does not tell us the full story about what is happening. We will also need produc-tion data for the same 24/12-month period.

Having more than twelve months of production and energy data, we can plot a moving

annual total. For this chart, each point represents the sum of the previous twelve months ofdata. In this way, each point covers a full range of the seasons, holidays, etc. The Figure 8.4shows a moving annual total for energy and production data.

This technique also smoothens out errors in the timing of meter readings. If we just plotenergy we are only seeing part of the story - so we plot both energy and production on the samechart - most likely using two y-axes. Looking at these charts, both energy and productions seemto be "tracking" each other - this suggests there is no major cause for concern. But we will needto watch for a deviation of the energy line to pick up early warning of waste or to confirm

Figure 8.3 Energy Consumption :Current Year(2000) Vs. Previous year(1999)

Figure 8.4 Moving Annual Total - Energy and Production

Prod

uctio

n

Ene

rgy



whether energy efficiency measures are making an impact.For any company, we also know that energy should directly relate to production. Knowing

this, we can calculate Specific Energy Consumption (SEC), which is energy consumption perunit of production. So we now plot a chart of SEC (see Figure 8.5).

At this point it is worth noting that the quality of your M&T system will only be as good as thequality of your data - both energy and production. The chart shows some variation - an all timelow in December 99 followed by a rising trend in SEC.

We also know that the level of production may have an effect on the specific consumption.If we add the production data to the SEC chart, it helps to explain some of the features. Forexample, the very low SEC occurred when there was a record level of production. This indi-cates that there might be fixed energy consumption - i.e. consumption that occurs regardless ofproduction levels. Refer Figure 8.6.

Figure 8.5: Monthly Specific Energy Consumption

Figure 8.6 SEC With Production

SEC

PRODUCTION

SEC



The next step is to gain more understanding of the relationship of energy and production, andto provide us with some basis for performance measurement. To do this we plot energy againstproduction - In Microsoft Excel Worksheet, this is an XY chart option. We then add a trend lineto the data set on the chart. (In practice what we have done is carried out a single variableregression analysis!). The Figure 8.7 shown is based on the data for 1999.

We can use it to derive a "standard" for the up-coming year's consumption. This chart shows alow degree of scatter indicative of a good fit. We need not worry if our data fit is not good. Ifdata fit is poor, but we know there should be a relationship, it indicates a poor level of controland hence a potential for energy savings.

In producing the production/energy relationship chart we have also obtained a relationshiprelating production and energy consumption.

Energy consumed for the period = C + M x Production for same period

Where M is the energy consumption directly related to production (variable) and C is the"fixed" energy consumption (i.e. energy consumed for lighting, heating/cooling and generalancillary services that are not affected by production levels). Using this, we can calculate theexpected or "standard" energy consumption for any level of production within the range of thedata set.

We now have the basis for implementing a factory level M&T system. We can predict stan-dard consumption, and also set targets - for example, standard less 5%. A more sophisticatedapproach might be applying different reductions to the fixed and variable energy consumption.Although, the above approach is at factory level, the same can be extended to individualprocesses as well with sub metering.

At a simplistic level we could use the chart above and plot each new month's point to seewhere it lies. Above the line is the regime of poor energy efficiency, and below the line is theregime of an improved one.

Figure 8.7: Energy vs Production

2. ELECTRIC MOTORS


Syllabus

Electric motors: Types, Losses in induction motors, Motor efficiency, Factors affectingmotor performance, Rewinding and motor replacement issues, Energy saving opportunitieswith energy efficient motors.

2.1 Introduction

Motors convert electrical energy into mechanical energy by the interaction between the mag-netic fields set up in the stator and rotor windings. Industrial electric motors can be broadly clas-sified as induction motors, direct current motors or synchronous motors. All motor types havethe same four operating components: stator (stationary windings), rotor (rotating windings),bearings, and frame (enclosure).

2.2 Motor Types

Induction Motors

Induction motors are the most commonly used prime mover forvarious equipments in industrial applications. In inductionmotors, the induced magnetic field of the stator winding inducesa current in the rotor. This induced rotor current produces a sec-ond magnetic field, which tries to oppose the stator magneticfield, and this causes the rotor to rotate.

The 3-phase squirrel cage motor is the workhorse of industry;it is rugged and reliable, and is by far the most common motortype used in industry. These motors drive pumps, blowers andfans, compressors, conveyers and production lines. The 3-phaseinduction motor has three windings each connected to a separate phase of the power supply.

Direct-Current Motors

Direct-Current motors, as the name implies, use direct-unidirectional, current. Direct currentmotors are used in special applications- where high torque starting or where smooth accelera-tion over a broad speed range is required.

Synchronous Motors

AC power is fed to the stator of the synchronous motor. The rotor is fed by DC from a separatesource. The rotor magnetic field locks onto the stator rotating magnetic field and rotates at the samespeed. The speed of the rotor is a function of the supply frequency and the number of magnetic polesin the stator. While induction motors rotate with a slip, i.e., rpm is less than the synchronous speed,the synchronous motor rotate with no slip, i.e., the RPM is same as the synchronous speed governedby supply frequency and number of poles. The slip energy is provided by the D.C. excitation power

2. Electric Motors


2.3 Motor Characteristics

Motor Speed

The speed of a motor is the number of revolutions in a given time frame, typically revolutionsper minute (RPM). The speed of an AC motor depends on the frequency of the input power andthe number of poles for which the motor is wound. The synchronous speed in RPM is given bythe following equation, where the frequency is in hertz or cycles per second:

Indian motors have synchronous speeds like 3000 / 1500 / 1000 / 750 / 600 / 500 / 375 RPMcorresponding to no. of poles being 2, 4, 6, 8, 10, 12, 16 (always even) and given the mainsfrequency of 50 cycles / sec.

The actual speed, with which the motor operates, will be less than the synchronous speed.The difference between synchronous and full load speed is called slip and is measured in per-cent. It is calculated using this equation:

As per relation stated above, the speed of an AC motor is determined by the number ofmotor poles and by the input frequency. It can also be seen that theoretically speed of an ACmotor can be varied infinitely by changing the frequency. Manufacturer's guidelines should bereferred for practical limits to speed variation. With the addition of a Variable Frequency Drive(VFD), the speed of the motor can be decreased as well as increased.

Power Factor

The power factor of the motor is given as:

As the load on the motor comes down, the magnitude of the active current reduces.However, there is no corresponding reduction in the magnetizing current, which is propor-tional to supply voltage with the result that the motor power factor reduces, with a reduction inapplied load. Induction motors, especially those operating below their rated capacity, are themain reason for low power factor in electric systems.

2.4 Motor Efficiency

Two important attributes relating to efficiency of electricity use by A.C. Induction motors areefficiency (η), defined as the ratio of the mechanical energy delivered at the rotating shaft tothe electrical energy input at its terminals, and power factor (PF). Motors, like other inductiveloads, are characterized by power factors less than one. As a result, the total current draw need-ed to deliver the same real power is higher than for a load characterized by a higher PF. Animportant effect of operating with a PF less than one is that resistance losses in wiring upstreamof the motor will be higher, since these are proportional to the square of the current. Thus, botha high value for η and a PF close to unity are desired for efficient overall operation in a plant.

Squirrel cage motors are normally more efficient than slip-ring motors, and higher-speedmotors are normally more efficient than lower-speed motors. Efficiency is also a function of

120 × FrequencySynchronous Speed (RPM) =

No. of Poles

kWPower Factor = Cos φ =

kVA

Synchronous Speed – Full Load Rated SpeedSlip (%) = × 100

Synchronous Speed

motor temperature. Totally-enclosed, fan-cooled (TEFC) motors are more efficient than screen-protected, drip-proof (SPDP) motors. Also, as with most equipment, motor efficiency increas-es with the rated capacity.

The efficiency of a motor is determined by intrinsic losses that can be reduced only bychanges in motor design. Intrinsic losses are of two types: fixed losses - independent of motorload, and variable losses - dependent on load.

Fixed losses consist of magnetic core losses and friction and windage losses. Magnetic corelosses (sometimes called iron losses) consist of eddy current and hysteresis losses in the stator.They vary with the core material and geometry and with input voltage.

Friction and windage losses are caused by friction in the bearings of the motor and aerody-namic losses associated with the ventilation fan and other rotating parts.

Variable losses consist of resistance losses in the stator and in the rotor and miscellaneousstray losses. Resistance to current flow in the stator and rotor result in heat generation that isproportional to the resistance of the material and the square of the current (I2R). Stray lossesarise from a variety of sources and are difficult to either measure directly or to calculate, but aregenerally proportional to the square of the rotor current.

Part-load performance characteristics of a motor also depend on its design. Both η and PFfall to very low levels at low loads. The Figures 2.1 shows the effect of load on power factorand efficiency. It can be seen that power factor drops sharply at part loads. The Figure 2.2 showsthe effect of speed on power factor.

Field Tests for Determining Efficiency

No Load Test: The motor is run at rated voltage and frequency without any shaft load. Inputpower, current, frequency and voltage are noted. The no load P.F. is quite low and hence lowPF wattmeters are required. From the input power, stator I2R losses under no load are subtract-ed to give the sum of Friction and Windage (F&W) and core losses. To separate core and F &

2. Electric Motors


Figure 2.1 % Load vs. Power factor, Efficiency Figure 2.2 Speed vs. Power factor

W losses, test is repeated at variable voltages. It is useful to plot no-load input kW versusVoltage; the intercept is Friction & Windage kW loss component.

F&W and core losses = No load power (watts) - (No load current)2 × Stator resistance

Stator and Rotor I2R Losses: The stator winding resistance is directly measured by a bridgeor volt amp method. The resistance must be corrected to the operating temperature. For mod-ern motors, the operating temperature is likely to be in the range of 100°C to 120°C and nec-essary correction should be made. Correction to 75°C may be inaccurate. The correction fac-tor is given as follows :

2. Electric Motors


The rotor resistance can be determined from locked rotor test at reduced frequency, but rotorI2R losses are measured from measurement of rotor slip.

Rotor I2R losses = Slip × (Stator Input – Stator I2R Losses – Core Loss)

Accurate measurement of slip is possible by stroboscope or non-contact type tachometer.Slip also must be corrected to operating temperature.

Stray Load Losses: These losses are difficult to measure with any accuracy. IEEE Standard112 gives a complicated method, which is rarely used on shop floor. IS and IEC standards takea fixed value as 0.5 % of input. The actual value of stray losses is likely to be more. IEEE –112 specifies values from 0.9 % to 1.8 % (see Table 2.1.)

TABLE 2.1 MOTOR RATING VS. STRAY

LOSSES - IEEE

Motor Rating Stray Losses1 – 125 HP 1.8 %

125 – 500 HP 1.5 %

501 – 2499 HP 1.2 %

2500 and above 0.9 %

Pointers for Users:

It must be clear that accurate determination of efficiency is very difficult. The same motor test-ed by different methods and by same methods by different manufacturers can give a differenceof 2 %. In view of this, for selecting high efficiency motors, the following can be done:

a) When purchasing large number of small motors or a large motor, ask for a detailed test cer-tificate. If possible, try to remain present during the tests; This will add cost.

b) See that efficiency values are specified without any tolerancec) Check the actual input current and kW, if replacement is doned) For new motors, keep a record of no load input power and currente) Use values of efficiency for comparison and for confirming; rely on measured inputs for all

calculations.

R2 235 + t2= , where, t1 = ambient temperature, °C & t2 = operating temperature, °C. R1 235 +t1

Estimation of efficiency in the field can be done as follows:a) Measure stator resistance and correct to operating temperature. From rated current value ,

I2R losses are calculated.b) From rated speed and output, rotor I2R losses are calculatedc) From no load test, core and F & W losses are determined for stray loss

The method is illustrated by the following example:Example :

Motor Specifications

Rated power = 34 kW/45 HPVoltage = 415 VoltCurrent = 57 AmpsSpeed = 1475 rpmInsulation class = FFrame = LD 200 LConnection = Delta

No load test Data

Voltage, V = 415 VoltsCurrent, I = 16.1 AmpsFrequency, F = 50 HzStator phase resistance at 30°C = 0.264 OhmsNo load power, Pnl = 1063.74 Watts

2. Electric Motors


2. Electric Motors


2. Electric Motors


2.5 Motor Selection

The primary technical consideration defining the motor choice for any particular application isthe torque required by the load, especially the relationship between the maximum torque gen-erated by the motor (break-down torque) and the torque requirements for start-up (locked rotortorque) and during acceleration periods.

The duty / load cycle determines the thermal loading on the motor. One consideration withtotally enclosed fan cooled (TEFC) motors is that the cooling may be insufficient when themotor is operated at speeds below its rated value.

Ambient operating conditions affect motor choice; special motor designs are available forcorrosive or dusty atmospheres, high temperatures, restricted physical space, etc.

An estimate of the switching frequency (usually dictated by the process), whether automat-ic or manually controlled, can help in selecting the appropriate motor for the duty cycle.

The demand a motor will place on the balance of the plant electrical system is another con-sideration - if the load variations are large, for example as a result of frequent starts and stopsof large components like compressors, the resulting large voltage drops could be detrimental toother equipment.

Reliability is of prime importance - in many cases, however, designers and process engi-neers seeking reliability will grossly oversize equipment, leading to sub-optimal energy perfor-mance. Good knowledge of process parameters and a better understanding of the plant powersystem can aid in reducing oversizing with no loss of reliability.

Inventory is another consideration - Many large industries use standard equipment, whichcan be easily serviced or replaced, thereby reducing the stock of spare parts that must be main-tained and minimizing shut-down time. This practice affects the choice of motors that mightprovide better energy performance in specific applications. Shorter lead times for securingindividual motors from suppliers would help reduce the need for this practice.

Price is another issue - Many users are first-cost sensitive, leading to the purchase of lessexpensive motors that may be more costly on a lifecycle basis because of lower efficiency. Forexample, energy efficient motors or other specially designed motors typically save within a fewyears an amount of money equal to several times the incremental cost for an energy efficientmotor, over a standard-efficiency motor. Few of salient selection issues are given below:

• In the selection process, the power drawn at 75 % of loading can be a meaningful indicatorof energy efficiency.

• Reactive power drawn (kVAR) by the motor.• Indian Standard 325 for standard motors allows 15 % tolerance on efficiency for motors

upto 50 kW rating and 10 % for motors over 50 kW rating. • The Indian Standard IS 8789 addresses technical performance of Standard Motors while IS

12615 addresses the efficiency criteria of High Efficiency Motors. Both follow IEC 34-2test methodology wherein, stray losses are assumed as 0.5 % of input power. By the IECtest method, the losses are understated and if one goes by IEEE test methodology, the motorefficiency values would be further lowered.

• It would be prudent for buyers to procure motors based on test certificates rather thanlabeled values.

• The energy savings by motor replacement can be worked out by the simple relation : kWsavings = kW output × [ 1/ηold – 1/ ηnew ] where ηold and ηnew are the existing and proposedmotor efficiency values.

• The cost benefits can be worked out on the basis of premium required for high efficiencyvs. worth of annual savings.

2.6 Energy-Efficient Motors

Energy-efficient motors (EEM) are the ones in which, design improvements are incorporatedspecifically to increase operating efficiency over motors of standard design (see Figure 2.3).Design improvements focus on reducing intrinsic motor losses. Improvements include the useof lower-loss silicon steel, a longer core (to increase active material), thicker wires (to reduceresistance), thinner laminations, smaller air gap between stator and rotor, copper instead of alu-minum bars in the rotor, superior bearings and a smaller fan, etc.

Energy-efficient motors now available in India operate with efficiencies that are typically 3 to 4 percentage points higher than standard motors. In keeping with the stipulations of the BIS,energy-efficient motors are designed to operate without loss in efficiency at loads between 75 %and 100 % of rated capacity. This may result in major benefits in varying load applications. Thepower factor is about the same or may be higher than for standard motors. Furthermore, energy-

2. Electric Motors


efficient motors have lower operating temperatures and noise levels, greater ability to acceleratehigher-inertia loads, and are less affected by supply voltage fluctuations.

Measures adopted for energy efficiency address each loss specifically as under:

Stator and Rotor I2R Losses

These losses are major losses and typically account for 55% to 60% of the total losses. I2R loss-es are heating losses resulting from current passing through stator and rotor conductors. I2Rlosses are the function of a conductor resistance, the square of current. Resistance of conductoris a function of conductor material, length and cross sectional area. The suitable selection ofcopper conductor size will reduce the resistance. Reducing the motor current is most readilyaccomplished by decreasing the magnetizing component of current. This involves lowering theoperating flux density and possible shortening of air gap. Rotor I2R losses are a function of therotor conductors (usually aluminium) and the rotor slip. Utilisation of copper conductors willreduce the winding resistance. Motor operation closer to synchronous speed will also reducerotor I2R losses.

Core Losses

Core losses are those found in the stator-rotor magnetic steel and are due to hysterisis effect andeddy current effect during 50 Hz magnetization of the core material. These losses are indepen-dent of load and account for 20 – 25 % of the total losses.

The hysterisis losses which are a function of flux density, are be reduced by utilizing low-loss grade of silicon steel laminations. The reduction of flux density is achieved by suitableincrease in the core length of stator and rotor. Eddy current losses are generated by circulatingcurrent within the core steel laminations. These are reduced by using thinner laminations.

Friction and Windage Losses

Friction and windage losses results from bearing friction, windage and circulating air throughthe motor and account for 8 – 12 % of total losses. These losses are independent of load. The

2. Electric Motors


Figure 2.3 Standard vs High Efficiency Motors

reduction in heat generated by stator and rotor losses permit the use of smaller fan. The windagelosses also reduce with the diameter of fan leading to reduction in windage losses.

Stray Load-Losses

These losses vary according to square of the load current and are caused by leakage fluxinduced by load currents in the laminations and account for 4 to 5 % of total losses. These loss-es are reduced by careful selection of slot numbers, tooth/slot geometry and air gap.

Energy efficient motors cover a wide range of ratings and the full load efficiencies arehigher by 3 to 7 %. The mounting dimensions are also maintained as per IS1231 to enableeasy replacement.

As a result of the modifications to improve performance, the costs of energy-efficient motors arehigher than those of standard motors. The higher cost will often be paid back rapidly in saved oper-ating costs, particularly in new applications or end-of-life motor replacements. In cases where exist-ing motors have not reached the end of their useful life, the economics will be less clearly positive.

Because the favourable economics of energy-efficient motors are based on savings in oper-ating costs, there may be certain cases which are generally economically ill-suited to energy-efficient motors. These include highly intermittent duty or special torque applications such ashoists and cranes, traction drives, punch presses, machine tools, and centrifuges. In addition,energy, efficient designs of multi-speed motors are generally not available. Furthermore, ener-gy-efficient motors are not yet available for many special applications, e.g. for flame-proofoperation in oil-field or fire pumps or for very low speed applications (below 750 rpm). Also,most energy-efficient motors produced today are designed only for continuous duty cycle oper-ation.

Given the tendency of over sizing on the one hand and ground realities like ; voltage, fre-quency variations, efficacy of rewinding in case of a burnout, on the other hand, benefits ofEEM's can be achieved only by careful selection, implementation, operation and maintenanceefforts of energy managers.

A summary of energy efficiency improvements in EEMs is given in the Table 2.2:

2. Electric Motors


TABLE 2.2 ENERGY EFFICIENT MOTORS

Power Loss Area Efficiency Improvement 1. Iron Use of thinner gauge, lower loss core steel reduces eddy current losses. Longer

core adds more steel to the design, which reduces losses due to lower operatingflux densities.

2. Stator I2R Use of more copper and larger conductors increases cross sectional area of statorwindings. This lowers resistance (R) of the windings and reduces losses due tocurrent flow (I).

3. Rotor I2R Use of larger rotor conductor bars increases size of cross section, lowering con-ductor resistance (R) and losses due to current flow (I).

4. Friction & Windage Use of low loss fan design reduces losses due to air movement.

5. Stray Load Loss Use of optimized design and strict quality control procedures minimizes strayload losses.

2.7 Factors Affecting Energy Efficiency & Minimising Motor Losses inOperation

Power Supply Quality

Motor performance is affected considerably by the quality of input power, that is the actual voltsand frequency available at motor terminals vis-à-vis rated values as well as voltage and fre-quency variations and voltage unbalance across the three phases. Motors in India must complywith standards set by the Bureau of Indian Standards (BIS) for tolerance to variations in inputpower quality. The BIS standards specify that a motor should be capable of delivering its ratedoutput with a voltage variation of +/- 6 % and frequency variation of +/- 3 %. Fluctuations muchlarger than these are quite common in utility-supplied electricity in India. Voltage fluctuationscan have detrimental impacts on motor performance. The general effects of voltage and fre-quency variation on motor performance are presented in Table 2.3:

Voltage unbalance, the condition where the voltages in the three phases are not equal, canbe still more detrimental to motor performance and motor life. Unbalance typically occurs as aresult of supplying single-phase loads disproportionately from one of the phases. It can alsoresult from the use of different sizes of cables in the distribution system. An example of theeffect of voltage unbalance on motor performance is shown in Table 2.4.

2. Electric Motors


2. Electric Motors


TA

BL

E 2

.3 G

EN

ER

AL

EF

FE

CT

S O

F V

OLTA

GE

AN

D F

RE

QU

EN

CY

VA

RIA

TIO

N O

N I

ND

UC

TIO

N M

OT

OR

CH

AR

AC

TE

RIS

TIC

S

The options that can be exercised to minimize voltage unbalance include:

i) Balancing any single phase loads equally among all the three phasesii) Segregating any single phase loads which disturb the load balance and feed them from a sep-

arate line / transformer

Motor LoadingMeasuring Load% Loading of the motor can be estimated by the following relation:% loading = Input power drawn by the motor (kW) at existing load x 100

(Name plate full load kW rating / name plate full load motor efficiency)or % loading = Input power drawn by the motor (kW) at existing load x 100

√3 x kV x I CosØ

• Never assume power factor • Loading should not be estimated as the ratio of currents.

Reducing Under-loading

Probably the most common practice contributing to sub-optimal motor efficiency is that ofunder-loading. Under-loading results in lower efficiency and power factor, and higher-than-nec-essary first cost for the motor and related control equipment. Under-loading is common for sev-eral reasons. Original equipment manufacturers tend to use a large safety factor in motors theyselect. Under-loading of the motor may also occur from under-utilisation of the equipment. Forexample, machine tool equipment manufacturers provide for a motor rated for the full capacityload of the equipment ex. depth of cut in a lathe machine. The user may need this full capacityrarely, resulting in under-loaded operation most of the time. Another common reason for under-loading is selection of a larger motor to enable the output to be maintained at the desired leveleven when input voltages are abnormally low. Finally, under-loading also results from select-ing a large motor for an application requiring high starting torque where a special motor,designed for high torque, would have been suitable.

A careful evaluation of the load would determine the capacity of the motor that should be select-ed. Another aspect to consider is the incremental gain in efficiency achievable by changing themotor. Larger motors have inherently higher rated efficiencies than smaller motors. Therefore, thereplacement of motors operating at 60 – 70 % of capacity or higher is generally not recommended.However, there are no rigid rules governing motor selection; the savings potential needs to be eval-uated on a case-to-case basis. When downsizing, it may be preferable to select an energy-efficientmotor, the efficiency of which may be higher than that of a standard motor of higher capacity.

2. Electric Motors


TABLE 2.4 EXAMPLE OF THE EFFECT OF VOLTAGE UNBALANCE ON

MOTOR PERFORMANCE

Parameter Percent unbalance in voltage*

0.30 2.30 5.40Unbalance in current (%) .................. 0.4 17.7 40.0

Increased temperature rise (°C) .................. 0 30 40

* Percent unbalance in voltage is defined as 100 (Vmax – Vavg) / Vavg, Where Vmax and Vavg are the largest andthe average of the three phase voltages, respectively.

For motors, which consistently operate at loads below 40 % of rated capacity, an inexpen-sive and effective measure might be to operate in star mode. A change from the standard deltaoperation to star operation involves re-configuring the wiring of the three phases of power inputat the terminal box.

Operating in the star mode leads to a voltage reduction by a factor of '√3'. Motor is electri-cally downsized by star mode operation, but performance characteristics as a function of loadremain unchanged. Thus, full-load operation in star mode gives higher efficiency and power fac-tor than partial load operation in the delta mode. However, motor operation in the star mode ispossible only for applications where the torque-to-speed requirement is lower at reduced load.

As speed of the motor reduces in star mode this option may be avoided in case the motor isconnected to a production facility whose output is related to the motor speed. For applicationswith high initial torque and low running torque needs, Del-Star starters are also available inmarket, which help in load following de-rating of electric motors after initial start-up.

Sizing to Variable Load

Industrial motors frequently operate under varying load conditions due to process requirements.A common practice in cases where such variable-loads are found is to select a motor based on thehighest anticipated load. In many instances, an alternative approach is typically less costly, moreefficient, and provides equally satisfactory operation. With this approach, the optimum rating forthe motor is selected on the basis of the load duration curve for the particular application. Thus,rather than selecting a motor of high rating that would operate at full capacity for only a short peri-od, a motor would be selected with a rating slightly lower than the peak anticipated load andwould operate at overload for a short period of time. Since operating within the thermal capacityof the motor insulation is of greatest concern in a motor operating at higher than its rated load, themotor rating is selected as that which would result in the same temperature rise under continuousfull-load operation as the weighted average temperature rise over the actual operating cycle.Under extreme load changes, e.g. frequent starts / stops, or high inertial loads, this method of cal-culating the motor rating is unsuitable since it would underestimate the heating that would occur.

Where loads vary substantially with time, in addition to proper motor sizing, the controlstrategy employed can have a significant impact on motor electricity use. Traditionally,mechanical means (e.g. throttle valves in piping systems) have been used when lower output isrequired. More efficient speed control mechanisms include multi-speed motors, eddy-currentcouplings, fluid couplings, and solid-state electronic variable speed drives.

Power Factor Correction

As noted earlier, induction motors are characterized by power factors less than unity, leading tolower overall efficiency (and higher overall operating cost) associated with a plant's electricalsystem. Capacitors connected in parallel (shunted) with the motor are typically used to improvethe power factor. The impacts of PF correction include reduced kVA demand (and hencereduced utility demand charges), reduced I2R losses in cables upstream of the capacitor (andhence reduced energy charges), reduced voltage drop in the cables (leading to improved volt-age regulation), and an increase in the overall efficiency of the plant electrical system.

It should be noted that PF capacitor improves power factor from the point of installation backto the generating side. It means that, if a PF capacitor is installed at the starter terminals of themotor, it won't improve the operating PF of the motor, but the PF from starter terminals to thepower generating side will improve, i.e., the benefits of PF would be only on upstream side.

2. Electric Motors


The size of capacitor required for a particular motor depends upon the no-load reactive kVA(kVAR) drawn by the motor, which can be determined only from no-load testing of the motor.In general, the capacitor is then selected to not exceed 90 % of the no-load kVAR of the motor.(Higher capacitors could result in over-voltages and motor burn-outs). Alternatively, typicalpower factors of standard motors can provide the basis for conservative estimates of capacitorratings to use for different size motors. The capacitor rating for power connection by direct con-nection to induction motors is shown in Table 2.5.

From the above table, it may be noted that required capacitive kVAr increases with decrease inspeed of the motor, as the magnetizing current requirement of a low speed motor is more in com-

2. Electric Motors


TABLE 2.5 CAPACITOR RATINGS FOR POWER FACTOR CORRECTION

BY DIRECT CONNECTION TO INDUCTION MOTORS

Motor Rating (HP) Capacitor rating (kVAr) for Motor Speed

3000 1500 1000 750 600 5005 2 2 2 3 3 3

7.5 2 2 3 3 4 4

10 3 3 4 5 5 6

15 3 4 5 7 7 7

20 5 6 7 8 9 10

25 6 7 8 9 9 12

30 7 8 9 10 10 15

40 9 10 12 15 16 20

50 10 12 15 18 20 22

60 12 14 15 20 22 25

75 15 16 20 22 25 30

100 20 22 25 26 32 35

125 25 26 30 32 35 40

150 30 32 35 40 45 50

200 40 45 45 50 55 60

250 45 50 50 60 65 70

parison to the high speed motor for the same HP of the motor. Since a reduction in line current, andassociated energy efficiency gains, are reflected backwards from the point of application of thecapacitor, the maximum improvement in overall system efficiency is achieved when the capacitoris connected across the motor terminals, as compared to somewhere further upstream in the plant'selectrical system. However, economies of scale associated with the cost of capacitors and the laborrequired to install them will place an economic limit on the lowest desirable capacitor size.

Maintenance

Inadequate maintenance of motors can significantly increase losses and lead to unreliable oper-ation. For example, improper lubrication can cause increased friction in both the motor and

associated drive transmission equipment. Resistance losses in the motor, which rise with tem-perature, would increase. Providing adequate ventilation and keeping motor cooling ductsclean can help dissipate heat to reduce excessive losses. The life of the insulation in the motorwould also be longer : for every 10°C increase in motor operating temperature over the recom-mended peak, the time before rewinding would be needed is estimated to be halved

A checklist of good maintenance practices to help insure proper motor operation would include:

• Inspecting motors regularly for wear in bearings and housings (to reduce frictional losses)and for dirt/dust in motor ventilating ducts (to ensure proper heat dissipation).

• Checking load conditions to ensure that the motor is not over or under loaded. A change inmotor load from the last test indicates a change in the driven load, the cause of which shouldbe understood.

• Lubricating appropriately. Manufacturers generally give recommendations for how and whento lubricate their motors. Inadequate lubrication can cause problems, as noted above. Over-lubrication can also create problems, e.g. excess oil or grease from the motor bearings can enterthe motor and saturate the motor insulation, causing premature failure or creating a fire risk.

• Checking periodically for proper alignment of the motor and the driven equipment.Improper alignment can cause shafts and bearings to wear quickly, resulting in damage toboth the motor and the driven equipment.

• Ensuring that supply wiring and terminal box are properly sized and installed. Inspect reg-ularly the connections at the motor and starter to be sure that they are clean and tight.

Age

Most motor cores in India are manufactured from silicon steel or de-carbonized cold-rolled steel,the electrical properties of which do not change measurably with age. However, poor maintenance(inadequate lubrication of bearings, insufficient cleaning of air cooling passages, etc.) can cause adeterioration in motor efficiency over time. Ambient conditions can also have a detrimental effecton motor performance. For example, excessively high temperatures, high dust loading, corrosiveatmosphere, and humidity can impair insulation properties; mechanical stresses due to load cyclingcan lead to misalignment. However, with adequate care, motor performance can be maintained.

2.8 Rewinding Effects on Energy Efficiency

It is common practice in industry to rewind burnt-out motors. The population of rewoundmotors in some industries exceed 50 % of the total population. Careful rewinding can some-times maintain motor efficiency at previous levels, but in most cases, losses in efficiency result.Rewinding can affect a number of factors that contribute to deteriorated motor efficiency :winding and slot design, winding material, insulation performance, and operating temperature.For example, a common problem occurs when heat is applied to strip old windings : the insu-lation between laminations can be damaged, thereby increasing eddy current losses. A changein the air gap may affect power factor and output torque.

However, if proper measures are taken, motor efficiency can be maintained, and in somecases increased, after rewinding. Efficiency can be improved by changing the winding design,though the power factor could be affected in the process. Using wires of greater cross section,slot size permitting, would reduce stator losses thereby increasing efficiency. However, it isgenerally recommended that the original design of the motor be preserved during the rewind,unless there are specific, load-related reasons for redesign.

2. Electric Motors


The impact of rewinding on motor efficiency and power factor can be easily assessed if theno-load losses of a motor are known before and after rewinding. Maintaining documentationof no-load losses and no-load speed from the time of purchase of each motor can facilitateassessing this impact.

For example, comparison of no load current and stator resistance per phase of a rewoundmotor with the original no-load current and stator resistance at the same voltage can be one ofthe indicators to assess the efficacy of rewinding.

2.9 Speed Control of AC Induction Motors

Traditionally, DC motors have been employed when variable speed capability was desired. Bycontrolling the armature (rotor) voltage and field current of a separately excited DC motor, awide range of output speeds can be obtained. DC motors are available in a wide range of sizes,but their use is generally restricted to a few low speed, low-to-medium power applications likemachine tools and rolling mills because of problems with mechanical commutation at largesizes. Also, they are restricted for use only in clean, non-hazardous areas because of the risk ofsparking at the brushes. DC motors are also expensive relative to AC motors.

Because of the limitations of DC systems, AC motors are increasingly the focus for variablespeed applications. Both AC synchronous and induction motors are suitable for variable speedcontrol. Induction motors are generally more popular, however, because of their ruggedness andlower maintenance requirements. AC induction motors are inexpensive (half or less of the costof a DC motor) and also provide a high power to weight ratio (about twice that of a DC motor).

An induction motor is an asynchronous motor, the speed of which can be varied by chang-ing the supply frequency. The control strategy to be adopted in any particular case will dependon a number of factors including investment cost, load reliability and any special control require-ments. Thus, for any particular application, a detailed review of the load characteristics, histori-cal data on process flows, the features required of the speed control system, the electricity tariffsand the investment costs would be a prerequisite to the selection of a speed control system.

The characteristics of the load are particularly important. Load refers essentially to thetorque output and corresponding speed required. Loads can be broadly classified as either con-stant power or Constant torque. Constant torque loads are those for which the output powerrequirement may vary with the speed of operation but the torque does not vary. Conveyors,rotary kilns, and constant-displacement pumps are typical examples of constant torque loads.Variable torque loads are those for which the torque required varies with the speed of operation.Centrifugal pumps and fans are typical examples of variable torque loads (torque varies as thesquare of the speed). Constant power loads are those for which the torque requirements typi-cally change inversely with speed. Machine tools are a typical example of a constant powerload.

The largest potential for electricity savings with variable speed drives is generally in vari-able torque applications, for example centrifugal pumps and fans, where the power requirementchanges as the cube of speed. Constant torque loads are also suitable for VSD application.

Motor Speed Control Systems

Multi-speed motors

Motors can be wound such that two speeds, in the ratio of 2:1, can be obtained. Motors can also

2. Electric Motors


be wound with two separate windings, each giving 2 operating speeds, for a total of four speeds.Multi-speed motors can be designed for applications involving constant torque, variable torque,or for constant output power. Multi-speed motors are suitable for applications, which require lim-ited speed control (two or four fixed speeds instead of continuously variable speed), in whichcases they tend to be very economical. They have lower efficiency than single-speed motors

Adjustable Frequency AC Drives

Adjustable frequency drives are also commonly called inverters. They are available in a rangeof kW rating from fractional to 750 kW. They are designed to operate standard inductionmotors. This allows them to be easily added to an existing system. The inverters are often soldseparately because the motor may already be in place. If necessary, a motor can be includedwith the drive or supplied separately.

The basic drive consists of the inverter itself which coverts the 50 Hz incoming power to avariable frequency and variable voltage. The variable frequency is the actual requirement,which will control the motor speed.

There are three major types of inverters designs available today. These are known asCurrent Source Inverters (CSI), Variable Voltage Inverters (VVI), and Pulse Width ModulatedInverters (PWM).

Direct Current Drives (DC)

The DC drive technology is the oldest form of electrical speed control. The drive system con-sists of a DC motor and a controller. The motor is constructed with armature and field wind-ings. Both of these windings require a DC excitation for motor operation. Usually the fieldwinding is excited with a constant level voltage from the controller.

Then, applying a DC voltage from the controller to the armature of the motor will operatethe motor. The armature connections are made through a brush and commutator assembly. Thespeed of the motor is directly proportional to the applied voltage.

The controller is a phase controlled bridge rectifier with logic circuits to control the DCvoltage delivered to the motor armature. Speed control is achieved by regulating the armaturevoltage to the motor. Often a tacho generator is included to achieve good speed regulation. Thetacho would be mounted on the motor and produces a speed feedback signal that is used with-in the controller.

Wound Rotor AC Motor Drives (Slip Ring Induction Motors)

Wound rotor motor drives use a specially constructed motor to accomplish speed control. Themotor rotor is constructed with windings which are brought out of the motor through slip ringson the motor shaft. These windings are connected to a controller which places variable resis-tors in series with the windings. The torque performance of the motor can be controlled usingthese variable resistors. Wound rotor motors are most common in the range of 300 HP andabove.

2.10 Motor Load Survey: Methodology

Large industries have a massive population of LT motors. Load survey of LT motors can be

2. Electric Motors


taken-up methodically to identify improvement options as illustrated in following case study.

i) Sampling Criteria

Towards the objective of selecting representative LT motor drives among the motor population,for analysis, the criteria considered are:

– Utilization factor i.e., hours of operation with preference given to continuously operateddrive motors.

– Sample representative basis, where one drive motor analysis can be reasoned as representa-tive for the population. Ex : Cooling Tower Fans, Air Washer Units, etc.

– Conservation potential basis, where drive motors with inefficient capacity controls on themachine side, fluctuating load drive systems, etc., are looked into.

ii) Measurements

Studies on selected LT motors involve measurement of electrical load parameters namely volts,amperes, power factor, kW drawn.

Observations on machine side parameters such as speed, load, pressure, temperature, etc.,(as relevant) are also taken. Availability of online instruments for routine measurements, avail-ability of tail-end capacitors for PF correction, energy meters for monitoring is also looked intofor each case.

iii) Analysis

Analysis of observations on representative LT motors and connected drives is carried outtowards following outputs:

– Motor load on kW basis and estimated energy consumption.– Scope for improving monitoring systems to enable sustenance of a regular in-house Energy

Audit function.– Scope areas for energy conservation with related cost benefits and source information.

The observations are to indicate:

% loading on kW, % voltage unbalance if any, voltage, current, frequency, power factor,machine side conditions like load / unload condition, pressure, flow, temperature, damper /throttle operation, whether it is a rewound motor, idle operations, metering provisions, etc.

The findings / recommendations may include:

• Identified motors with less than 50 % loading, 50 – 75 % loading, 75 – 100 % loading, over100 % loading.

• Identified motors with low voltage / power factor / voltage imbalance for needed improve-ment measures.

• Identified motors with machine side losses / inefficiencies like idle operations, throttling /damper operations for avenues like automatic controls / interlocks, variable speed drives,etc.

Motor load survey is aimed not only as a measure to identify motor efficiency areas butequally importantly, as a means to check combined efficiency of the motor, driven machineand controller if any. The margins in motor efficiency may be less than 10 % of consumptionoften, but the load survey would help to bring out savings in driven machines / systems, whichcan give 30 – 40 % energy savings.

2. Electric Motors


2. Electric Motors


QUESTIONS

1. Name three types of motors in industrial practice.

2. What is the relation between RPM (speed) and frequency of an induction motor?

3. A 4-pole squirrel case induction motor operates with 5 % slip at full load. What isthe full load RPM you may expect, if frequency is changed by a V/F control to:(a)40 c/s (b) 45 c/s (c) 35 c/s

4. List the losses in induction motors and their expected percentage out of the total losses.

5. List the factors affecting energy efficiency of electric motors?

6. The power factor of an induction motora) increases with load b) decreases with load c) remains constant with load d) hasno relation to load

7. List factors affecting windage and friction losses while rewinding.

8. What are the factors affecting core losses while rewinding?

9. List methods by which speed control of motor can be achieved.

10. Explain the ways by which efficiencies of energy efficient motors are increased.

11. How does efficiency loss occur in a rewound motor?

12. How do you check the efficacy of rewound motor?

13. A 50 kW induction motor with 86 % present full load efficiency is being consideredfor replacement by a 89 % efficiency motor. What will be the savings in energy ifthe motor works for 6000 hours per year and cost of energy is Rs. 4.50 per kWh?

REFERENCES 1. Technology Menu (NPC)2. BEE Publications3. PCRA Publications

3. COMPRESSED AIR SYSTEM


Syllabus

Compressed air system: Types of air compressors, Compressor efficiency, Efficient com-pressor operation, Compressed air system components, Capacity assessment, Leakage test,Factors affecting the performance and efficiency

3.1 Introduction

Air compressors account for significant amount of electricity used in Indian industries. Air com-pressors are used in a variety of industries to supply process requirements, to operate pneumatictools and equipment, and to meet instrumentation needs. Only 10 – 30% of energy reaches the pointof end-use, and balance 70 – 90% of energy of the power of the prime mover being converted tounusable heat energy and to a lesser extent lost in form of friction, misuse and noise.

3.2 Compressor Types

Compressors are broadly classified as: Positive displacement compressor and Dynamiccompressor.

Positive displacement compressors increase the pressure of the gas by reducing the vol-ume. Positive displacement compressors are further classified as reciprocating and rotarycompressors.

Dynamic compressors increase the air velocity, which is then converted to increased pres-sure at the outlet. Dynamic compressors are basically centrifugal compressors and are furtherclassified as radial and axial flow types.

The flow and pressure requirements of a given application determine the suitability of a par-ticulars type of compressor.

3. Compressed Air System


Positive Displacement Compressors

Reciprocating Compressors

Reciprocating compressors are the mostwidely used type for air compression.They are characterized by a flow outputthat remains nearly constant over a rangeof discharge pressures. Also, the com-pressor capacity is directly proportionalto the speed. The output, however, is apulsating one.

Reciprocating compressors areavailable in many configurations, thefour most widely used of which arehorizontal, vertical, horizontal bal-ance-opposed and tandem. Verticaltype reciprocating compressors areused in the capacity range of 50 – 150cfm. Horizontal balance opposed compressors are used in the capacity range of 200 – 5000 cfmin multi-stage design and upto 10,000 cfm in single stage designs.

Reciprocating compressors are also available in variety of types:

• Lubricated and non-lubricated • Single or multiple cylinder

Calculate the capacity as per the formulae given below :

Actual Free air discharge



WhereP2 = Final pressure after filling (kg/cm2 a)P1 = Initial pressure (kg/cm2a) after bleedingP0 = Atmospheric Pressure (kg/cm2 a)V = Storage volume in m3 which includes receiver,

after cooler, and delivery pipingT = Time take to build up pressure to P2 in minutes

The above equation is relevant where the compressed air temperature is same as the ambi-ent air temperature, i.e., perfect isothermal compression. In case the actual compressed air tem-perature at discharge, say t2

0C is higher than ambient air temperature say t10C (as is usual case),

the FAD is to be corrected by a factor (273 + t1) / (273 + t2).

EXAMPLE

An instrument air compressor capacity test gave the following results (assume the final com-pressed air temperature is same as the ambient temperature) - Comment?

Time taken to build up pressure : 4.021 minutes

8.287= 13.12 m3/minute

7.79 + 0.4974 = 8.287m3

Capacity shortfall with respect to 14.75 m3/minute rating is 1.63 m3/minute i.e., 11.05%,which indicates compressor performance needs to be investigated further.

3.7 Checklist for Energy Efficiency in Compressed Air System

• Ensure air intake to compressor is not warm and humid by locating compressors in well-ventilated area or by drawing cold air from outside. Every 4°C rise in air inlet temperaturewill increase power consumption by 1 percent.

• Clean air-inlet filters regularly. Compressor efficiency will be reduced by 2 percent forevery 250 mm WC pressure drop across the filter.

• Keep compressor valves in good condition by removing and inspecting once every sixmonths. Worn-out valves can reduce compressor efficiency by as much as 50 percent.

• Install manometers across the filter and monitor the pressure drop as a guide to replacementof element.

• Minimize low-load compressor operation; if air demand is less than 50 percent of compres-sor capacity, consider change over to a smaller compressor or reduce compressor speedappropriately (by reducing motor pulley size) in case of belt driven compressors.

• Consider the use of regenerative air dryers, which uses the heat of compressed air to removemoisture.

• Fouled inter-coolers reduce compressor efficiency and cause more water condensation in airreceivers and distribution lines resulting in increased corrosion. Periodic cleaning of inter-coolers must be ensured.

• Compressor free air delivery test (FAD) must be done periodically to check the presentoperating capacity against its design capacity and corrective steps must be taken if required.

• If more than one compressor is feeding to a common header, compressors must be operat-ed in such a way that only one small compressor should handle the load variations whereasother compressors will operate at full load.

• The possibility of heat recovery from hot compressed air to generate hot air or water forprocess application must be economically analyzed in case of large compressors.

• Consideration should be given to two-stage or multistage compressor as it consumes lesspower for the same air output than a single stage compressor.

• If pressure requirements for processes are widely different (e.g. 3 bar to 7 bar), it is advis-able to have two separate compressed air systems.

• Reduce compressor delivery pressure, wherever possible, to save energy.• Provide extra air receivers at points of high cyclic-air demand which permits operation

without extra compressor capacity.• Retrofit with variable speed drives in big compressors, say over 100 kW, to eliminate the

`unloaded' running condition altogether.• Keep the minimum possible range between load and unload pressure settings.• Automatic timer controlled drain traps wastes compressed air every time the valve opens.

So frequency of drainage should be optimized.• Check air compressor logs regularly for abnormal readings, especially motor current cool-

ing water flow and temperature, inter-stage and discharge pressures and temperatures andcompressor load-cycle.

• Compressed air leakage of 40 – 50 percent is not uncommon. Carry out periodic leak teststo estimate the quantity of leakage.

• Install equipment interlocked solenoid cut-off valves in the air system so that air supply toa machine can be switched off when not in use.

• Present energy prices justify liberal designs of pipeline sizes to reduce pressure drops. • Compressed air piping layout should be made preferably as a ring main to provide desired

pressures for all users.• A smaller dedicated compressor can be installed at load point, located far off from the cen-

tral compressor house, instead of supplying air through lengthy pipelines.



• All pneumatic equipment should be properly lubricated, which will reduce friction, preventwear of seals and other rubber parts thus preventing energy wastage due to excessive airconsumption or leakage.

• Misuse of compressed air such as for body cleaning, agitation, general floor cleaning, andother similar applications must be discouraged in order to save compressed air and energy.

• Pneumatic equipment should not be operated above the recommended operating pressure asthis not only wastes energy bus can also lead to excessive wear of equipment's componentswhich leads to further energy wastage.

• Pneumatic transport can be replaced by mechanical system as the former consumed about 8times more energy. Highest possibility of energy savings is by reducing compressed air use.

• Pneumatic tools such as drill and grinders consume about 20 times more energy than motordriven tools. Hence they have to be used efficiently. Wherever possible, they should bereplaced with electrically operated tools.

• Where possible welding is a good practice and should be preferred over threaded connec-tions.

• On account of high pressure drop, ball or plug or gate valves are preferable over globevalves in compressed air lines.



4. HVAC AND REFRIGERATION SYSTEM


SyllabusHVAC and Refrigeration System: Vapor compression refrigeration cycle, Refrigerants,Coefficient of performance, Capacity, Factors affecting Refrigeration and Air conditioningsystem performance and savings opportunities.Vapor absorption refrigeration system: Working principle, Types and comparison withvapor compression system, Saving potential

4.1 Introduction

The Heating, Ventilation and Air Conditioning (HVAC) and refrigeration system transfers theheat energy from or to the products, or building environment. Energy in form of electricity orheat is used to power mechanical equipment designed to transfer heat from a colder, low-ener-gy level to a warmer, high-energy level.

Refrigeration deals with the transfer of heat from a low temperature level at the heatsource to a high temperature level at the heat sink by using a low boiling refrigerant.

There are several heat transfer loops in refrigeration system as described below:

Figure 4.1 Heat Transfer Loops In Refrigeration System

In the Figure 4.1, thermal energy moves from left to right as it is extracted from the space andexpelled into the outdoors through five loops of heat transfer:

– Indoor air loop. In the leftmost loop, indoor air is driven by the supply air fan through a cool-ing coil, where it transfers its heat to chilled water. The cool air then cools the building space.

– Chilled water loop. Driven by the chilled water pump, water returns from the cooling coilto the chiller’s evaporator to be re-cooled.

– Refrigerant loop. Using a phase-change refrigerant, the chiller’s compressor pumps heatfrom the chilled water to the condenser water.

– Condenser water loop. Water absorbs heat from the chiller’s condenser, and the con-denser water pump sends it to the cooling tower.

– Cooling tower loop. The cooling tower’s fan drives air across an open flow of the hotcondenser water, transferring the heat to the outdoors.

Air-Conditioning Systems

Depending on applications, there are several options / combinations, which are available for useas given below:

� Air Conditioning (for comfort / machine) � Split air conditioners� Fan coil units in a larger system� Air handling units in a larger system

Refrigeration Systems (for processes)

� Small capacity modular units of direct expansion type similar to domestic refrigerators,small capacity refrigeration units.

� Centralized chilled water plants with chilled water as a secondary coolant for temperaturerange over 5°C typically. They can also be used for ice bank formation.

� Brine plants, which use brines as lower temperature, secondary coolant, for typically subzero temperature applications, which come as modular unit capacities as well as large cen-tralized plant capacities.

� The plant capacities upto 50 TR are usually considered as small capacity, 50 – 250 TR asmedium capacity and over 250 TR as large capacity units.

A large industry may have a bank of such units, often with common chilled water pumps, con-denser water pumps, cooling towers, as an off site utility.

The same industry may also have two or three levels of refrigeration & air conditioning such as:

� Comfort air conditioning (20° – 25° C)� Chilled water system (8° – 10° C)� Brine system (sub-zero applications)

Two principle types of refrigeration plants found in industrial use are: Vapour CompressionRefrigeration (VCR) and Vapour Absorption Refrigeration (VAR). VCR uses mechanical ener-gy as the driving force for refrigeration, while VAR uses thermal energy as the driving force forrefrigeration.

4.2 Types of Refrigeration System

Vapour Compression Refrigeration

Heat flows naturally from a hot to a colder body. In refrigeration system the opposite must occuri.e. heat flows from a cold to a hotter body. This is achieved by using a substance called a refrig-erant, which absorbs heat and hence boils or evaporates at a low pressure to form a gas. Thisgas is then compressed to a higher pressure, such that it transfers the heat it has gained to ambi-ent air or water and turns back (condenses) into a liquid. In this way heat is absorbed, orremoved, from a low temperature source and transferred to a higher temperature source. The refrigeration cycle can be broken down into the following stages (see Figure 4.2):

1 – 2 Low pressure liquid refrigerant in the evaporator absorbs heat from its surroundings,usually air, water or some other process liquid. During this process it changes its state from aliquid to a gas, and at the evaporator exit is slightly superheated.

4. HVAC and Refrigeration System


4.8 Energy Saving Opportunities

a) Cold Insulation

Insulate all cold lines / vessels using economic insulation thickness to minimize heat gains; andchoose appropriate (correct) insulation.

b) Building Envelope

Optimise air conditioning volumes by measures such as use of false ceiling and segregation ofcritical areas for air conditioning by air curtains.

c) Building Heat Loads Minimisation

Minimise the air conditioning loads by measures such as roof cooling, roof painting, efficientlighting, pre-cooling of fresh air by air- to-air heat exchangers, variable volume air system, otpi-mal thermo-static setting of temperature of air conditioned spaces, sun film applications, etc.

e) Process Heat Loads Minimisation

Minimize process heat loads in terms of TR capacity as well as refrigeration level, i.e., tem-perature required, by way of:i) Flow optimization ii) Heat transfer area increase to accept higher temperature coolantiii) Avoiding wastages like heat gains, loss of chilled water, idle flows.iv) Frequent cleaning / de-scaling of all heat exchangers

f) At the Refrigeration A/C Plant Area

i) Ensure regular maintenance of all A/C plant components as per manufacturer guide-lines.

ii) Ensure adequate quantity of chilled water and cooling water flows, avoid bypass flowsby closing valves of idle equipment.

iii) Minimize part load operations by matching loads and plant capacity on line; adopt vari-able speed drives for varying process load.

iv) Make efforts to continuously optimize condenser and evaporator parameters for mini-mizing specific energy consumption and maximizing capacity.

v) Adopt VAR system where economics permit as a non-CFC solution.

4. HVAC and Refrigeration System


6. PUMPS AND PUMPING SYSTEM


Syllabus

Pumps and Pumping System: Types, Performance evaluation, Efficient system opera-tion, Flow control strategies and energy conservation opportunities

6.1 Pump Types

Pumps come in a variety of sizes for a wide range of applications. They can be classifiedaccording to their basic operating principle as dynamic or displacement pumps. Dynamicpumps can be sub-classified as centrifugal and special effect pumps. Displacement pumps canbe sub-classified as rotary or reciprocating pumps.

In principle, any liquid can be handled by any of the pump designs. Where different pumpdesigns could be used, the centrifugal pump is generally the most economical followed byrotary and reciprocating pumps. Although, positive displacement pumps are generally moreefficient than centrifugal pumps, the benefit of higher efficiency tends to be offset by increasedmaintenance costs.

Since, worldwide, centrifugal pumps account for the majority of electricity used by pumps,the focus of this chapter is on centrifugal pump.

Centrifugal Pumps

A centrifugal pump is of a very simple design. The two main parts of the pump are the impellerand the diffuser. Impeller, which is the only moving part, is attached to a shaft and driven by amotor. Impellers are generally made of bronze, polycarbonate, cast iron, stainless steel as wellas other materials. The diffuser (also called as volute)houses the impeller and captures and directs the wateroff the impeller.

Water enters the center (eye) of the impeller and exitsthe impeller with the help of centrifugal force. As waterleaves the eye of the impeller a low-pressure area is cre-ated, causing more water to flow into the eye.Atmospheric pressure and centrifugal force cause this tohappen. Velocity is developed as the water flows throughthe impeller spinning at high speed. The water velocity iscollected by the diffuser and converted to pressure byspecially designed passageways that direct the flow tothe discharge of the pump, or to the next impeller shouldthe pump have a multi-stage configuration.

The pressure (head) that a pump will develop is indirect relationship to the impeller diameter, the numberof impellers, the size of impeller eye, and shaft speed. Capacity is determined by the exit widthof the impeller. The head and capacity are the main factors, which affect the horsepower size ofthe motor to be used. The more the quantity of water to be pumped, the more energy is required.

Figure 6.1 Centrifugal pump

A centrifugal pump is not positive acting; it will not pump the same volume always. Thegreater the depth of the water, the lesser is the flow from the pump. Also, when it pumps againstincreasing pressure, the less it will pump. For these reasons it is important to select a centrifu-gal pump that is designed to do a particular job.

Since the pump is a dynamic device, it is convenient to consider the pressure in terms ofhead i.e. meters of liquid column. The pump generates the same head of liquid whatever thedensity of the liquid being pumped. The actual contours of the hydraulic passages of theimpeller and the casing are extremely important, in order to attain the highest efficiency possi-ble. The standard convention for centrifugal pump is to draw the pump performance curvesshowing Flow on the horizontal axis and Head generated on the vertical axis. Efficiency, Power& NPSH Required (described later), are conventionally shown on the vertical axis, plottedagainst Flow, as illustrated in Figure 6.2.

6. Pumps and Pumping System


Figure 6.2 Pump Performance Curve

Given the significant amount of electricity attributed to pumping systems, even smallimprovements in pumping efficiency could yield very significant savings of electricity. Thepump is among the most inefficient of the components that comprise a pumping system, includ-ing the motor, transmission drive, piping and valves.

Hydraulic power, pump shaft power and electrical input power

Hydraulic power Ph = Q (m3/s) x Total head, hd - hs (m) x ρ (kg/m3) x g (m/s2) / 1000

Where hd – discharge head, hs – suction head, ρ – density of the fluid, g – acceleration due to gravity

Pump shaft power Ps = Hydraulic power, Ph / pump efficiency, ηPump

Electrical input power = Pump shaft power Ps

ηMotor

sents a large portion of the total head, caution should be used in deciding whether to use VFDs.Operators should review the performance of VFDs in similar applications and consult VFDmanufacturers to avoid the damage that can result when a pump operates too slowly againsthigh static head.

For many systems, VFDs offer a means to improve pump operating efficiency despitechanges in operating conditions. The effect of slowing pump speed on pump operation is illus-trated by the three curves in Figure 6.22. When a VFD slows a pump, its head/flow and brakehorsepower (BHP) curves drop down and to the left and its efficiency curve shifts to the left.This efficiency response provides an essential cost advantage; by keeping the operating effi-ciency as high as possible across variations in the system's flow demand, the energy and main-tenance costs of the pump can be significantly reduced.

VFDs may offer operating cost reductions by allowing higher pump operating efficiency,but the principal savings derive from the reduction in frictional or bypass flow losses. Using asystem perspective to identify areas in which fluid energy is dissipated in non-useful work oftenreveals opportunities for operating cost reductions.

For example, in many systems, increasing flow through bypass lines does not noticeablyimpact the backpressure on a pump. Consequently, in these applications pump efficiency doesnot necessarily decline during periods of low flow demand. By analyzing the entire system,however, the energy lost in pushing fluid through bypass lines and across throttle valves can beidentified.

Another system benefit of VFDs is a soft start capability. During startup, most motors expe-rience in-rush currents that are 5 – 6 times higher than normal operating currents. This high cur-rent fades when the motor spins up to normal speed. VFDs allow the motor to be started with alower startup current (usually only about 1.5 times the normal operating current). This reduceswear on the motor and its controller.

6.7 Energy Conservation Opportunities in Pumping Systems

■ Ensure adequate NPSH at site of installation■ Ensure availability of basic instruments at pumps like pressure gauges, flow meters.■ Operate pumps near best efficiency point. ■ Modify pumping system and pumps losses to minimize throttling. ■ Adapt to wide load variation with variable speed drives or sequenced control of multiple

units. ■ Stop running multiple pumps - add an auto-start for an on-line spare or add a booster pump

in the problem area. ■ Use booster pumps for small loads requiring higher pressures. ■ Increase fluid temperature differentials to reduce pumping rates in case of heat

exchangers. ■ Repair seals and packing to minimize water loss by dripping. ■ Balance the system to minimize flows and reduce pump power requirements. ■ Avoid pumping head with a free-fall return (gravity); Use siphon effect to advantage: ■ Conduct water balance to minimise water consumption ■ Avoid cooling water re-circulation in DG sets, air compressors, refrigeration systems,

cooling towers feed water pumps, condenser pumps and process pumps.



■ In multiple pump operations, carefully combine the operation of pumps to avoid throttling■ Provide booster pump for few areas of higher head ■ Replace old pumps by energy efficient pumps ■ In the case of over designed pump, provide variable speed drive, or downsize / replace

impeller or replace with correct sized pump for efficient operation.■ Optimise number of stages in multi-stage pump in case of head margins ■ Reduce system resistance by pressure drop assessment and pipe size optimisation



8. LIGHTING SYSTEM


SyllabusLighting System: Light source, Choice of lighting, Luminance requirements, and Energyconservation avenues

8.1 Introduction

Lighting is an essential service in all the industries. The power consumption by the industriallighting varies between 2 to 10% of the total power depending on the type of industry.Innovation and continuous improvement in the field of lighting, has given rise to tremendousenergy saving opportunities in this area.

Lighting is an area, which provides a major scope to achieve energy efficiency at the designstage, by incorporation of modern energy efficient lamps, luminaires and gears, apart from goodoperational practices.

8.2 Basic Terms in Lighting System and Features

Lamps

Lamp is equipment, which produces light. The most commonly used lamps are describedbriefly as follows:

• Incandescent lamps:

Incandescent lamps produce light by means of a filament heated to incandescence bythe flow of electric current through it. The principal parts of an incandescent lamp, alsoknown as GLS (General Lighting Service) lamp include the filament, the bulb, the fill gasand the cap.

• Reflector lamps:

Reflector lamps are basically incandescent, provided with a high quality internal mirror, whichfollows exactly the parabolic shape of the lamp. The reflector is resistant to corrosion, thusmaking the lamp maintenance free and output efficient.

• Gas discharge lamps:

The light from a gas discharge lamp is produced by the excitation of gas contained in either atubular or elliptical outer bulb.

The most commonly used discharge lamps are as follows:

• Fluorescent tube lamps (FTL) • Compact Fluorescent Lamps (CFL) • Mercury Vapour Lamps • Sodium Vapour Lamps • Metal Halide Lamps

Luminaire

Luminaire is a device that distributes, filters or transforms the light emitted from one ormore lamps. The luminaire includes, all the parts necessary for fixing and protecting thelamps, except the lamps themselves. In some cases, luminaires also include the necessarycircuit auxiliaries, together with the means for connecting them to the electric supply. Thebasic physical principles used in optical luminaire are reflection, absorption, transmissionand refraction.

Control Gear

The gears used in the lighting equipment are as follows:

• Ballast:

A current limiting device, to counter negative resistance characteristics of any discharge lamps.In case of fluorescent lamps, it aids the initial voltage build-up, required for starting.

• Ignitors:

These are used for starting high intensity Metal Halide and Sodium vapour lamps.

Illuminance

This is the quotient of the illuminous flux incident on an element of the surface at a point ofsurface containing the point, by the area of that element.

The lighting level produced by a lighting installation is usually qualified by theilluminance produced on a specified plane. In most cases, this plane is the major planeof the tasks in the interior and is commonly called the working plane. The illuminanceprovided by an installation affects both the performance of the tasks and the appearanceof the space.

Lux (lx)

This is the illuminance produced by a luminous flux of one lumen, uniformly distributed overa surface area of one square metre. One lux is equal to one lumen per square meter.

Luminous Efficacy (lm/W)

This is the ratio of luminous flux emitted by a lamp to the power consumed by the lamp. It is areflection of efficiency of energy conversion from electricity to light form.

Colour Rendering Index (RI)

Is a measure of the degree to which the colours of surfaces illuminated by a given light sourceconfirm to those of the same surfaces under a reference illuminent; suitable allowance havingbeen made for the state of Chromatic adaptation.

8.3 Lamp Types and their Features

The Table 8.1 shows the various types of lamp available along with their features.

8. Lighting System


8. Lighting System


8.4 Recommended Illuminance Levels for VariousTasks / Activities / Locations

Recommendations on Illuminance

Scale of Illuminance: The minimum illuminance for all non-working interiors, has beenmentioned as 20 Lux (as per IS 3646). A factor of approximately 1.5represents the smallest significant difference in subjective effect ofilluminance. Therefore, the following scale of illuminances isrecommended.

20–30–50–75–100–150–200–300–500–750–1000–1500–2000, … Lux

Illuminance ranges: Because circumstances may be significantly different for differentinteriors used for the same application or for different conditions forthe same kind of activity, a range of illuminances is recommendedfor each type of interior or activity intended of a single value ofilluminance. Each range consists of three successive steps of therecommended scale of illuminances. For working interiors the

Lumens / Watt Color Typical Type of Lamp Range Avg. Rendering Typical Application Life

Index (hours)

Incandescent 8–18 14 Excellent Homes, restaurants, 1000general lighting,emergency lighting

Fluorescent Lamps 46–60 50 Good w.r.t. Offices, shops, 5000coating hospitals, homes

Compact fluorescent 40–70 60 Very good Hotels, shops, 8000–10000lamps (CFL) homes, offices

High pressure 44–57 50 Fair General lighting in 5000mercury (HPMV) factories, garages,

car parking, floodlighting

Halogen lamps 18–24 20 Excellent Display, flood 2000–4000lighting, stadiumexhibition grounds,construction areas

High pressure sodium 67–121 90 Fair General lighting 6000–12000(HPSV) SON in factories, ware

houses, streetlighting

Low pressure sodium 101–175 150 Poor Roadways, tunnels, 6000–12000(LPSV) SOX canals, street lighting

TABLE 8.1 LUMINOUS PERFORMANCE CHARACTERISTICS OF COMMONLY USED

LUMINARIES

middle value (R) of each range represents the recommended serviceilluminance that would be used unless one or more of the factorsmentioned below apply.

The higher value (H) of the range should be used at exceptional cases where lowreflectances or contrasts are present in the task, errors are costly to rectify, visual work is criti-cal, accuracy or higher productivity is of great importance and the visual capacity of the work-er makes it necessary.

Similarly, lower value (L) of the range may be used when reflectances or contrasts areunusually high, speed & accuracy is not important and the task is executed only occasionally.

Recommended Illumination

The following Table gives the recommended illuminance range for different tasks and activitiesfor chemical sector. The values are related to the visual requirements of the task, to user's sat-isfaction, to practical experience and to the need for cost effective use of energy.(Source IS3646 (Part I) : 1992).

For recommended illumination in other sectors, reader may refer Illuminating EngineersSociety Recommendations Handbook/

Chemicals

Petroleum, Chemical and Petrochemical worksExterior walkways, platforms, stairs and ladders 30–50–100Exterior pump and valve areas 50–100–150Pump and compressor houses 100–150–200Process plant with remote control 30–50–100Process plant requiring occasional manual intervention 50–100–150Permanently occupied work stations in process plant 150–200–300Control rooms for process plant 200–300–500

Pharmaceuticals Manufacturer and Fine chemicalsmanufacturerPharmaceutical manufacturerGrinding, granulating, mixing, drying, tableting, s 300–500–750terilising, washing, preparation of solutions, filling,capping, wrapping, hardening

Fine chemical manufacturersExterior walkways, platforms, stairs and ladders 30–50–100Process plant 50–100–150Fine chemical finishing 300–500–750Inspection 300–500–750Soap manufacture General area 200–300–500Automatic processes 100–200–300Control panels 200–300–500Machines 200–300–500

8. Lighting System


Paint works General 200–300–500Automatic processes 150–200–300Control panels 200–300–500Special batch mixing 500–750–1000Colour matching 750–100–1500

8.5 Methodology of Lighting System Energy Efficiency Study

A step-by-step approach for assessing energy efficiency of lighting system is given below:Step–1: Inventorise the Lighting System elements, & transformers in the facility as per

following typical format (Table – 8.2 and 8.3).

8. Lighting System


S. No. Plant Lighting Rating in Watts Population No. of hoursLocation Device & Lamp & Ballast Numbers / Day

Ballast Type

TABLE 8.2 DEVICE RATING, POPULATION AND USE PROFILE

S. No. Plant Lighting Numbers Meter Provisions AvailableLocation Transformer Installed Volts / Amps / kW / Energy

Rating (kVA)

TABLE 8.3 LIGHTING TRANSFORMER / RATING AND POPULATION

PROFILE:

In case of distribution boards (instead of transformers) being available, fuse ratings may beinventorised along the above pattern in place of transformer kVA.

Step–2: With the aid of a lux meter, measure and document the lux levels at various plantlocations at working level, as daytime lux and night time lux values alongside the number oflamps "ON" during measurement.

Step–3: With the aid of portable load analyzer, measure and document the voltage, current,power factor and power consumption at various input points, namely the distribution boards orthe lighting voltage transformers at the same as that of the lighting level audit.

Step–4: Compare the measured lux values with standard values as reference and identifylocations as under lit and over lit areas.

Step–5: Collect and Analyse the failure rates of lamps, ballasts and the actual life expectan-cy levels from the past data.

Step–6: Based on careful assessment and evaluation, bring out improvement options, whichcould include :

i) Maximise sunlight use through use of transparent roof sheets, north light roof, etc.

ii) Examine scope for replacements of lamps by more energy efficient lamps, with dueconsideration to luminiare, color rendering index, lux level as well as expected lifecomparison.

iii) Replace conventional magnetic ballasts by more energy efficient ballasts, with dueconsideration to life and power factor apart from watt loss.

iv) Select interior colours for light reflection.

v) Modify layout for optimum lighting.

vi) Providing individual / group controls for lighting for energy efficiency such as:a. On / off type voltage regulation type (for illuminance control)b. Group control switches / unitsc. Occupancy sensorsd. Photocell controls e. Timer operated controlsf. Pager operated controlsg. Computerized lighting control programs

vii) Install input voltage regulators / controllers for energy efficiency as well as longer lifeexpectancy for lamps where higher voltages, fluctuations are expected.

viii) Replace energy efficient displays like LED's in place of lamp type displays in controlpanels / instrumentation areas, etc.

8.6 Case Examples

Energy Efficient Replacement Options

The lamp efficacy is the ratio of light output in lumens to power input to lamps in watts.Over the years development in lamp technology has led to improvements in efficacyof lamps. However, the low efficacy lamps, such as incandescent bulbs, still constitutea major share of the lighting load. High efficacy gas discharge lamps suitable for differ-ent types of applications offer appreciable scope for energy conservation. Typical energyefficient replacement options, along with the per cent energy saving, are given in Table-8.4.

8. Lighting System


Energy Saving Potential in Street Lighting

The energy saving potential, in typical cases of replacement of inefficient lamps with efficientlamps in street lighting is given in the Table 8.5

8. Lighting System


Lamp type Power savingSector

Existing Proposed Watts %

Domestic/Commercial GLS 100 W *CFL 25 W 75 75

Industry GLS 13 W *CFL 9 W 4 31GLS 200 W Blended 160 W 40 20TL 40 W TLD 36 W 4 10

Industry/Commercial HPMV 250 W HPSV 150 W 100 37HPMV 400 W HPSV 250 W 150 35

* Wattages of CFL includes energy consumption in ballasts.

TABLE 8.4 SAVINGS BY USE OF HIGH EFFICACY LAMPS

Existing lamp Replaced units Saving

Type W Life Type W Life W %

GLS 200 1000 ML 160 5000 40 7

GLS 300 1000 ML 250 5000 50 17

TL 2 X 40 5000 TL 2 X 36 5000 8 6

HPMV 125 5000 HPSV 70 12000 25 44

HPMV 250 5000 HPSV 150 12000 100 40

HPMV 400 5000 HPSV 250 12000 150 38

TABLE 8.5 SAVING POTENTIAL BY USE OF HIGH

EFFICACY LAMPS FOR STREET LIGHTING

8.7 Some Good Practices in Lighting

Installation of energy efficient fluorescent lamps in place of "Conventional" fluorescentlamps.

Energy efficient lamps are based on the highly sophisticated tri-phosphor fluorescent powdertechnology. They offer excellent colour rendering properties in addition to the very high lumi-nous efficacy.

Installation of Compact Fluorescent Lamps (CFL's) in place of incandescent lamps.

Compact fluorescent lamps are generally considered best for replacement of lower wattageincandescent lamps. These lamps have efficacy ranging from 55 to 65 lumens/Watt. The aver-age rated lamp life is 10,000 hours, which is 10 times longer than that of a normal incandescent

lamps. CFL's are highly suitable for places such as Living rooms, Hotel lounges, Bars,Restaurants, Pathways, Building entrances, Corridors, etc.

Installation of metal halide lamps in place of mercury / sodium vapour lamps.

Metal halide lamps provide high color rendering index when compared with mercury &sodium vapour lamps. These lamps offer efficient white light. Hence, metal halide is thechoice for colour critical applications where, higher illumination levels are required. Theselamps are highly suitable for applications such as assembly line, inspection areas, paintingshops, etc. It is recommended to install metal halide lamps where colour rendering is morecritical.

Installation of High Pressure Sodium Vapour (HPSV) lamps for applications where colourrendering is not critical.

High pressure sodium vapour (HPSV) lamps offer more efficacy. But the colour rendering prop-erty of HPSV is very low. Hence, it is recommended to install HPSV lamps for applicationssuch street lighting, yard lighting, etc.

Installation of LED panel indicator lamps in place of filament lamps.

Panel indicator lamps are used widely in industries for monitoring, fault indication, signaling,etc. Conventionally filament lamps are used for the purpose, which has got the following dis-advantages:

• High energy consumption (15 W/lamp) • Failure of lamps is high (Operating life less than 1,000 hours) • Very sensitive to the voltage fluctuations Recently, the conventional filament lamps are

being replaced with Light Emitting Diodes (LEDs).

The LEDs have the following merits over the filament lamps.

• Lesser power consumption (Less than 1 W/lamp)• Withstand high voltage fluctuation in the power supply. • Longer operating life (more than 1,00,000 hours)

It is recommended to install LEDs for panel indicator lamps at the design stage.

Light distribution

Energy efficiency cannot be obtained by mere selection of more efficient lamps alone. Efficientluminaires along with the lamp of high efficacy achieve the optimum efficiency. Mirror-opticluminaires with a high output ratio and bat-wing light distribution can save energy.

For achieving better efficiency, luminaires that are having light distribution characteristicsappropriate for the task interior should be selected. The luminaires fitted with a lamp shouldensure that discomfort glare and veiling reflections are minimised. Installation of suitable lumi-naires, depends upon the height - Low, Medium & High Bay. Luminaires for high intensity dis-charge lamp are classified as follows:

• Low bay, for heights less than 5 metres. • Medium bay, for heights between 5 – 7 metres. • High bay, for heights greater than 7 metres.

8. Lighting System


System layout and fixing of the luminaires play a major role in achieving energy efficien-cy. This also varies from application to application. Hence, fixing the luminaires at optimumheight and usage of mirror optic luminaries leads to energy efficiency.

Light Control

The simplest and the most widely used form of controlling a lighting installation is "On-Off"switch. The initial investment for this set up is extremely low, but the resulting operational costsmay be high. This does not provide the flexibility to control the lighting, where it is notrequired.

Hence, a flexible lighting system has to be provided, which will offer switch-off or reduc-tion in lighting level, when not needed. The following light control systems can be adopted atdesign stage:

• Grouping of lighting system, to provide greater flexibility in lighting control

Grouping of lighting system, which can be controlled manually or by timer control.

• Installation of microprocessor based controllers

Another modern method is usage of microprocessor / infrared controlled dimming or switchingcircuits. The lighting control can be obtained by using logic units located in the ceiling, whichcan take pre-programme commands and activate specified lighting circuits. Advanced lightingcontrol system uses movement detectors or lighting sensors, to feed signals to the controllers.

• Optimum usage of daylighting

Whenever the orientation of a building permits, day lighting can be used in combination withelectric lighting. This should not introduce glare or a severe imbalance of brightness in visualenvironment. Usage of day lighting (in offices/air conditioned halls) will have to be very limit-ed, because the air conditioning load will increase on account of the increased solar heat dissi-pation into the area. In many cases, a switching method, to enable reduction of electric light inthe window zones during certain hours, has to be designed.

• Installation of "exclusive" transformer for lighting

In most of the industries, lighting load varies between 2 to 10%. Most of the problems faced bythe lighting equipment and the "gears" is due to the "voltage" fluctuations. Hence, the lightingequipment has to be isolated from the power feeders. This provides a better voltage regulationfor the lighting. This will reduce the voltage related problems, which in turn increases the effi-ciency of the lighting system.

• Installation of servo stabilizer for lighting feeder

Wherever, installation of exclusive transformer for lighting is not economically attractive, servostabilizer can be installed for the lighting feeders. This will provide stabilized voltage for thelighting equipment. The performance of "gears" such as chokes, ballasts, will also improveddue to the stabilized voltage.

This set up also provides, the option to optimise the voltage level fed to the lighting feeder.In many plants, during the non-peaking hours, the voltage levels are on the higher side. Duringthis period, voltage can be optimised, without any significant drop in the illumination level.

8. Lighting System


• Installation of high frequency (HF) electronic ballasts in place of conventional ballasts

New high frequency (28–32 kHz) electronic ballasts have the following advantages over thetraditional magnetic ballasts:

Energy savings up to 35% Less heat dissipation, which reduces the air conditioning load

• Lights instantly • Improved power factor • Operates in low voltage load • Less in weight • Increases the life of lamp

The advantage of HF electronic ballasts, out weigh the initial investment (higher costs whencompared with conventional ballast). In the past the failure rate of electronic ballast in IndianIndustries was high. Recently, many manufacturers have improved the design of the ballastleading to drastic improvement in their reliability. The life of the electronic ballast is high espe-cially when, used in a lighting circuit fitted with a automatic voltage stabiliser.

The Table 8.6 gives the type of luminaire, gear and controls used in different areas of industry.

8. Lighting System


Location Source Luminaire Gear Controls

Plant HID/FTL Industrial rail reflector: Conventional/low Manual/electronicHigh bay loss electronic

Medium bay ballastLow bay

Office FTL/CFL FTL/CFL Electronic/low Manual/autoloss

Yard HID Flood light Suitable Manual

Road HID/PL Street light luminaire Suitable Manualperipheral

TABLE 8.6 TYPES OF LUMINAIRE WITH THEIR GEAR AND CONTROLS

USED IN DIFFERENT INDUSTRIAL LOCATIONS

8. Lighting System


QUESTIONS

1. What are the types of commonly used lamps?

2. What do the following terms mean?– Illuminance– Luminous efficacy– Luminaire– Control gear– Colour rendering index

3. What is the function of ballast in a lighting system?

4. Rate the following with respect to their luminous efficacy– GLS lamp– FTL– CFL– HPSV– LPSV

5. Rate the following with respect to colour rendering index– GLS lamp– HPSV lamp– Metal halide lamps– LPSV lamp

6. Briefly describe the methodology of lighting energy audit in an industrial facility?

7. List the energy savings opportunities in industrial lighting systems.

8. Explain how electronic ballast saves energy?

9. A CFL can replace a) FTL b) GLS c) HPMV d) HPSV

10. Explain briefly about various lighting controls available?

REFERENCES1. NPC Experiences

9. DG SET SYSTEM


SyllabusDiesel Generating system: Factors affecting selection, Energy performance assessment ofdiesel conservation avenues

9.1 Introduction

Diesel engine is the prime mover, which drives an alternator to produce electrical energy. Inthe diesel engine, air is drawn into the cylinder and is compressed to a high ratio (14:1 to25:1). During this compression, the air is heated to a temperature of 700–900°C. A meteredquantity of diesel fuel is then injected into the cylinder, which ignites spontaneously becauseof the high temperature. Hence, the diesel engine is also known as compression ignition (CI)engine.

DG set can be classified according to cycle type as: two stroke and four stroke. However,the bulk of IC engines use the four stroke cycle. Let us look at the principle of operation of thefour-stroke diesel engine.

Four Stroke - Diesel Engine

The 4 stroke operations in a diesel engine are: induction stroke, compression stroke, ignitionand power stroke and exhaust stroke.

1st : Induction stroke - while the inlet valve is open, the descending piston draws infresh air.

2nd : Compression stroke - while the valves are closed, the air is compressed to a pressure ofup to 25 bar.

3rd : Ignition and power stroke - fuel is injected, while the valves are closed (fuel injectionactually starts at the end of the previous stroke), the fuel ignites spontaneously andthe piston is forced downwards by the combustion gases.

4th : Exhaust stroke - the exhaust valve is open and the rising piston discharges the spentgases from the cylinder.

Figure 9.1 Schematic Diagram of Four-Stroke Diesel Engine

9. DG Set System


Since power is developed during only one stroke, the single cylinder four-stroke engine hasa low degree of uniformity. Smoother running is obtained with multi cylinder engines becausethe cranks are staggered in relation to one another on the crankshaft. There are many variationsof engine configuration, for example. 4 or 6 cylinder, in-line, horizontally opposed, vee or radi-al configurations.

DG Set as a System

A diesel generating set should be considered as a system since its successful operation dependson the well-matched performance of the components, namely:a) The diesel engine and its accessories.b) The AC Generator.c) The control systems and switchgear.d) The foundation and power house civil works.e) The connected load with its own components like heating, motor drives, lighting etc.

It is necessary to select the components with highest efficiency and operate them at theiroptimum efficiency levels to conserve energy in this system.

Fig 9.2 DG Set System

Selection Considerations

To make a decision on the type of engine, which is most suitable for a specific application,several factors need to be considered. The two most important factors are: power and speedof the engine.

The power requirement is determined by the maximum load. The engine power ratingshould be 10 – 20 % more than the power demand by the end use. This prevents overload-ing the machine by absorbing extra load during starting of motors or switching of sometypes of lighting systems or when wear and tear on the equipment pushes up its powerconsumption.

Speed is measured at the output shaft and given in revolutions per minute (RPM). Anengine will operate over a range of speeds, with diesel engines typically running at lower

9. DG Set System


speeds (1300 – 3000 RPM). There will be an optimum speed at which fuel efficiency willbe greatest. Engines should be run as closely as possible to their rated speed to avoid poorefficiency and to prevent build up of engine deposits due to incomplete combustion - whichwill lead to higher maintenance and running costs. To determine the speed requirement ofan engine, one has to again look at the requirement of the load.

For some applications, the speed of the engine is not critical, but for other applicationssuch as a generator, it is important to get a good speed match. If a good match can beobtained, direct coupling of engine and generator is possible; if not, then some form of gear-ing will be necessary - a gearbox or belt system, which will add to the cost and reduce theefficiency.

There are various other factors that have to be considered, when choosing an engine fora given application. These include the following: cooling system, abnormal environmentalconditions (dust, dirt, etc.), fuel quality, speed governing (fixed or variable speed), poormaintenance, control system, starting equipment, drive type, ambient temperature, altitude,humidity, etc.

Suppliers or manufacturers literature will specify the required information when purchasingan engine. The efficiency of an engine depends on various factors, for example, load factor (per-centage of full load), engine size, and engine type.

Diesel Generator Captive Power Plants

Diesel engine power plants are most frequently used in small power (captive non-utility) sys-tems. The main reason for their extensive use is the higher efficiency of the diesel engines com-pared with gas turbines and small steam turbines in the output range considered. In applicationsrequiring low captive power, without much requirement of process steam, the ideal method ofpower generation would be by installing diesel generator plants. The fuels burnt in dieselengines range from light distillates to residual fuel oils. Most frequently used diesel engine sizesare between the range 4 to 15 MW. For continuous operation, low speed diesel engine is morecost-effective than high speed diesel engine.

Advantages of adopting Diesel Power Plants are:

■ Low installation cost■ Short delivery periods and installation period■ Higher efficiency (as high as 43 – 45 %)■ More efficient plant performance under part loads■ Suitable for different type of fuels such as low sulphur heavy stock and heavy fuel oil in

case of large capacities.■ Minimum cooling water requirements, ■ Adopted with air cooled heat exchanger in areas where water is not available■ Short start up time

A brief comparison of different types of captive power plants (combined gas turbine andsteam turbine, conventional steam plant and diesel engine power plant) is given in Table 9.1.It can be seen from the Table that captive diesel plant wins over the other two in terms ofthermal efficiency, capital cost, space requirements, auxiliary power consumption, plantload factor etc.

9. DG Set System


TABLE 9.1 COMPARISON OF DIFFERENT TYPES OF CAPTIVE POWER PLANT

Description Units Combined Conventional Diesel EngineGT & ST Steam Plant Power Plants

Thermal Efficiency % 40 – 46 33 – 36 43 – 45

Initial Investment of Rs./kW 8,500 – 10,000 15,000 – 18,000 7,500 – 9,000Installed Capacity

Space requirement 125 % (Approx.) 250 % (Approx.) 100 % (Approx.)

Construction time Months 24 – 30 42 – 48 12 – 15

Project period Months 30 – 36 52 – 60 12

Auxiliary Power % 2 – 4 8 – 10 1.3 - 2.1Consumption

Plant Load Factor kWh/kW 6000 – 7000 5000 – 6000 7200 – 7500

Start up time from cold Minutes About 10 120 – 180 15 – 20

Diesel Engine Power Plant Developments

The diesel engine developments have been steady andimpressive. The specific fuel consumption has comedown from a value of 220 g/kWh in the 1970s to a valuearound 160 g/kWh in present times.

Slow speed diesel engine, with its flat fuel consump-tion curve over a wide load range (50%–100%), comparesvery favourably over other prime movers such as mediumspeed diesel engine, steam turbines and gas turbines.

With the arrival of modern, high efficiency tur-bochargers, it is possible to use an exhaust gas driventurbine generator to further increase the engine rated out-put. The net result – lower fuel consumption per kWhand further increase in overall thermal efficiency.

The diesel engine is able to burn the poorest qualityfuel oils, unlike gas turbine, which is able to do so withonly costly fuel treatment equipment.

Slow speed dual fuel engines are now available usinghigh-pressure gas injection, which gives the same thermal efficiency and power output as a reg-ular fuel oil engine.

9.2 Selection and Installation Factors

Sizing of a Genset:

a) If the DG set is required for 100% standby, then the entire connected load in HP / kVAshould be added. After finding out the diversity factor, the correct capacity of a DG setcan be found out.

Figure 9.3 Turbocharger

9. DG Set System


Example :Connected Load = 650 kWDiversity Factor = 0.54(Demand / connected load)Max. Demand = 650 x 0.54 = 350 kW% Loading = 70Set rating = 350/0.7 = 500 kWAt 0.8 PF, rating = 625 kVA

b) For an existing installation, record the current, voltage and power factors (kWh / kVAh)reading at the main bus-bar of the system at every half-an-hour interval for a period of2–3 days and during this period the factory should be having its normal operations. Thenon-essential loads should be switched off to find the realistic current taken for runningessential equipment. This will give a fair idea about the current taken from which therating of the set can be calculated.

For existing installation:

kVA = √3 V IkVA Rating = kVA / Load Factorwhere Load factor = Average kVA / Maximum kVA

c) For a new installation, an approximate method of estimating the capacity of a DG set isto add full load currents of all the proposed loads to be run in DG set. Then, applying adiversity factor depending on the industry, process involved and guidelines obtainedfrom other similar units, correct capacity can be arrived at.

High Speed Engine or Slow/Medium Speed Engine

The normal accepted definition of high speed engine is 1500 rpm. The high speed sets have beendeveloped in India, whereas the slow speed engines of higher capacities are often imported. Theother features and comparison between high and medium / slow speed engines are mentioned below:

Factor Slow speed engine High speed engine

Break mean effective pressure - therefore Low Highwear and tear and consumption of spares

Weight to power ratio- therefore sturdiness More Lessand life

Space High Less

Type of use Continuous use Intermittent use

Period between overhauls* 8000 hours 3200

Direct operating cost (includes lubricating Less Highoils, filters etc.* Typical recommendations from manufacturers

Keeping the above factors and available capacities of DG set in mind, the cost of econom-ics for both the engines should be worked out before arriving at a decision.

9. DG Set System


Capacity Combinations

From the point of view of space, operation, maintenance and initial capital investment,it is certainly economical to go in for one large DG set than two or more DG sets in parallel.

Two or more DG sets running in parallel can be a advantage as only the short-fall inpower–depending upon the extent of power cut prevailing - needs to filled up. Also, flexibilityof operation is increased since one DG set can be stopped, while the other DG set is generatingat least 50% of the power requirement. Another advantage is that one DG set can become 100%standby during lean and low power-cut periods.

Air Cooling Vs. Water Cooling

The general feeling has been that a water cooled DG set is better than an air cooled set, as mostusers are worried about the overheating of engines during summer months. This is to someextent is true and precautions have to be taken to ensure that the cooling water temperature doesnot exceed the prescribed limits. However, from performance and maintenance point of view,water and air cooled sets are equally good except that proper care should be taken to ensurecross ventilation so that as much cool air as possible is circulated through the radiator to keepits cooling water temperature within limits.

While, it may be possible to have air cooled engines in the lower capacities, it will be nec-essary to go in for water cooled engines in larger capacities to ensure that the engine does notget over-heated during summer months.

Safety Features

It is advisable to have short circuit, over load and earth fault protection on all the DG sets.However, in case of smaller capacity DG sets, this may become uneconomical. Hence, it isstrongly recommended to install a circuit protection. Other safety equipment like high tem-perature, low lube oil pressure cut-outs should be provided, so that in the event of any ofthese abnormalities, the engine would stop and prevent damage. It is also essential to pro-vide reverse power relay when DG sets are to run in parallel to avoid back feeding from onealternator to another.

Parallel Operation with Grid

Running the DG set in parallel with the mains from the supply undertakings can be done in con-sultation with concerned electricity authorities. However, some supply undertakings ask theconsumer to give an undertaking that the DG set will not be run in parallel with their supply.The reasons stated are that the grid is an infinite bus and paralleling a small capacity DG setwould involve operational risks despite normal protections like reverse power relay, voltage andfrequency relays.

Maximum Single Load on DG Set

The starting current of squirrel cage induction motors is as much as six times the rated currentfor a few seconds with direct-on-line starters. In practice, it has been found that the starting cur-rent value should not exceed 200 % of the full load capacity of the alternator. The voltage andfrequency throughout the motor starting interval recovers and reaches rated values usuallymuch before the motor has picked up full speed.

9. DG Set System


In general, the HP of the largest motor that can be started with direct on line starting isabout 50 % of the kVA rating of the generating set. On the other hand, the capacity of theinduction motor can be increased, if the type of starting is changed over to star delta or toauto transformer starter, and with this starting the HP of the largest motor can be upto 75 %of the kVA of Genset.

Unbalanced Load Effects

It is always recommended to have the load as much balanced as possible, since unbalancedloads can cause heating of the alternator, which may result in unbalanced output voltages. Themaximum unbalanced load between phases should not exceed 10 % of the capacity of the gen-erating sets.

Neutral Earthing

The electricity rules clearly specify that two independent earths to the body and neutralshould be provided to give adequate protection to the equipment in case of an earth fault,and also to drain away any leakage of potential from the equipment to the earth for safeworking.

Site Condition Effects on Performance Derating

Site condition with respect to altitude, intake temperature and cooling water temperature der-ate diesel engine output as shown in following Tables: 9.2 and 9.3.

TABLE 9.2 ALTITUDE AND INTAKE TEMPERATURE CORRECTIONS

Correction Factors For Engine Output

Altitude Correction Temperature Correction

Altitude Meters Non Super Super Charged Intake °C Correction Factorover MSL Charged

610 0.980 0.980 32 1.000

915 0.935 0.950 35 0.986

1220 0.895 0.915 38 0.974

1525 0.855 0.882 41 0.962

1830 0.820 0.850 43 0.950

2130 0.780 0.820 46 0.937

2450 0.745 0.790 49 0.925

2750 0.712 0.765 52 0.913

3050 0.680 0.740 54 0.900

3660 0.612 0.685

4270 0.550 0.630

4880 0.494 0.580

9. DG Set System


9.3 Operational Factors

Load Pattern & DG Set Capacity

The average load can be easily assessed by logging the current drawn at the main switchboard onan average day. The 'over load' has a different meaning when referred to the D.G. set. Overloads,which appear insignificant and harmless on electricity board supply, may become detrimental to aD.G.set, and hence overload on D.G.set should be carefully analysed. Diesel engines are designedfor 10% overload for 1 hour in every 12 hours of operation. The A.C. generators are designed tomeet 50% overload for 15 seconds as specified by standards. The D.G.set/s selection should be suchthat the overloads are within the above specified limits. It would be ideal to connect steady loadson DG set to ensure good performance. Alongside alternator loading, the engine loading in termsof kW or BHP, needs to be maintained above 50%. Ideally, the engine and alternator loading con-ditions are both to be achieved towards high efficiency.

Engine manufacturers offer curves indicating % Engine Loading vs fuel Consumption ingrams/BHP. Optimal engine loading corresponding to best operating point is desirable for ener-gy efficiency.

Alternators are sized for kVA rating with highest efficiency attainable at a loading of around70% and more. Manufacturers curves can be referred to for best efficiency point and corre-sponding kW and kVA loading values.

Sequencing of Loads

The captive diesel generating set has certain limits in handling the transient loads. Thisapplies to both kW (as reflected on the engine) and kVA (as reflected on the generator). Inthis context, the base load that exists before the application of transient load brings downthe transient load handling capability, and in case of A.C. generators, it increases the tran-sient voltage dip. Hence, great care is required in sequencing the load on D.G.set/s. It isadvisable to start the load with highest transient kVA first followed by other loads in thedescending order of the starting kVA. This will lead to optimum sizing and better utilisationof transient load handling capacity of D.G.set.

Load Pattern

In many cases, the load will not be constant throughout the day. If there is substantial variationin load, then consideration should be given for parallel operation of D.G.sets. In such a situa-tion, additional D.G. set(s) are to be switched on when load increases. The typical case may be

TABLE 9.3 DERATING DUE TO AIR INTER COOLER

WATER INLET TEMPERATURE

Water Temperature °C Flow % Derating %

25 100 0

30 125 3

35 166 5

40 166 8

9. DG Set System


an establishment demanding substantially different powers in first, second and third shifts. Byparallel operation, D.G. sets can be run at optimum operating points or near about, for optimumfuel consumption and additionally, flexibility is built into the system. This scheme can be alsobe applied where loads can be segregated as critical and non-critical loads to provide standbypower to critical load in the captive power system.

Load Characteristics

Some of the load characteristics influence efficient use of D.G.set. These characteristics areentirely load dependent and cannot be controlled by the D.G.set. The extent of detrimental influ-ence of these characteristics can be reduced in several cases.

– Power Factor:

The load power factor is entirely dependent on the load. The A.C. generator is designed forthe power factor of 0.8 lag as specified by standards. Lower power factor demands higherexcitation currents and results in increased losses. Over sizing A.C. generators for operationat lower power factors results in lower operating efficiency and higher costs. The econom-ical alternative is to provide power factor improvement capacitors.

– Unbalanced Load:

Unbalanced loads on A.C. generator leads to unbalanced set of voltages and additional heat-ing in A.C. generator. When other connected loads like motor loads are fed with unbalancedset of voltages additional losses occur in the motors as well. Hence, the load on the A.C.generators should be balanced as far as possible. Where single phase loads are predominant,consideration should be given for procuring single phase A.C. generator.

– Transient Loading:

On many occasions to contain transient voltage dip arising due to transient load application,a specially designed generator may have to be selected. Many times an unstandardcombination of engine and A.C. generator may have to be procured. Such a combinationensures that the prime mover is not unnecessarily over sized which adds to capital cost andrunning cost.

– Special Loads:

Special loads like rectifier / thyristor loads, welding loads, furnace loads need an applica-tion check. The manufacturer of diesel engine and AC generator should be consulted forproper recommendation so that desired utilisation of DG set is achieved without any prob-lem. In certain cases of loads, which are sensitive to voltage, frequency regulation, voltagewave form, consideration should be given to segregate the loads, and feed it by a dedicatedpower supply which usually assumes the form of DG motor driven generator set. Such analternative ensures that special design of AC generator is restricted to that portion of theload which requires high purity rather than increasing the price of the D.G.set by speciallydesigned AC generator for complete load.

Waste Heat Recovery in DG Sets

A typical energy balance in a DG set indicates following break-up:

9. DG Set System


Input : 100% Thermal EnergyOutputs : 35% Electrical Output

4% Alternator Losses33% Stack Loss through Flue Gases24% Coolant Losses4% Radiation Losses

Among these, stack losses through flue gases or the exhaust flue gas losses on account ofexisting flue gas temperature of 350°C to 550°C, constitute the major area of concern towardsoperational economy. It would be realistic to assess the Waste Heat Recovery (WHR) potentialin relation to quantity, temperature margin, in kcals/Hour as:

Potential WHR = (kWh Output/Hour) x (8 kg Gases / kWh Output)x 0.25 kcal/kg°C x (tg – 180°C)

Where, tg is the gas temperature after Turbocharger, (the criteria being that limiting exit gastemperature cannot be less than 180°C, to avoid acid dew point corrosion), 0.25 being the spe-cific heat of flue gases and kWh output being the actual average unit generation from the setper hour. For a 1100 KVA set, at 800 KW loading, and with 480°C exhaust gas temperature, thewaste heat potential works out to:

800 kWh x 8 kg gas generation / kWh output x 0.25 kCal/kg°Cx (480 – 180), i.e., 4,80,000 kCal/hr

While the above method yields only the potential for heat recovery, the actual realisablepotential depends upon various factors and if applied judiciously, a well configured waste heatrecovery system can tremendously boost the economics of captive DG power generation.

The factors affecting Waste Heat Recovery from flue Gases are:

a) DG Set loading, temperature of exhaust gasesb) Hours of operation andc) Back pressure on the DG set

* Consistent DG set loading (to over 60% of rating) would ensure a reasonable exit fluegas quantity and temperature. Fluctuations and gross under loading of DG set results inerratic flue gas quantity and temperature profile at entry to heat recovery unit, therebyleading to possible cold end corrosion and other problems.

TABLE 9.4 TYPICAL FLUE GAS TEMPERATURE AND FLOW PATTERN IN A 5-MW DG SET

AT VARIOUS LOADS

100% Load 11.84 kgs/Sec 370°C

90% Load 10.80 kgs/Sec 350°C

70% Load 9.08 kgs/Sec 330°C

60% Load 7.50 kgs/Sec 325°C

If the normal load is 60%, the flue gas parameters for waste heat recovery unit would be 320°C inlet tempera-ture, 180°C outlet temperature and 27180 kgs/Hour gas flow.

At 90% loading, however, values would be 355°C and 32,400 kgs/Hour, respectively

9. DG Set System


* Number of hours of operation of the DG Set has an influence on the thermal perfor-mance of waste heat Recovery unit. With continuous DG Set operations, cost benefitsare favourable.

* Back pressure in the gas path caused by additional pressure drop in waste heat recoveryunit is another key factor. Generally, the maximum back pressure allowed is around250–300 mmWC and the heat recovery unit should have a pressure drop lower than that.Choice of convective waste heat recovery systems with adequate heat transfer area areknown to provide reliable service.

The configuration of heat recovery system and the choice of steam parameters can be judi-ciously selected with reference to the specific industry (site) requirements. Much good work hastaken place in Indian Industry regarding waste heat recovery and one interesting configuration,deployed is installation of waste heat boiler in flue gas path along with a vapour absorptionchiller, to produce 8°C chilled water working on steam from waste heat.

The favourable incentives offered by Government of India for energy efficient equipmentand technologies (100% depreciation at the end of first year), make the waste heat recoveryoption. Payback period is only about 2 years

9.4 Energy Performance Assessment of DG Sets

Routine energy efficiency assessment of DG sets on shopfloor involves following typical steps:1) Ensure reliability of all instruments used for trial.2) Collect technical literature, characteristics, and specifications of the plant.3) Conduct a 2 hour trial on the DG set, ensuring a steady load, wherein the following mea-

surements are logged at 15 minutes intervals.a) Fuel consumption (by dip level or by flow meter)b) Amps, volts, PF, kW, kWhc) Intake air temperature, Relative Humidity (RH)d) Intake cooling water temperaturee) Cylinder-wise exhaust temperature (as an indication of engine loading)f) Turbocharger RPM (as an indication of loading on engine)g) Charge air pressure (as an indication of engine loading)h) Cooling water temperature before and after charge air cooler (as an indication of cool-

er performance)i) Stack gas temperature before and after turbocharger (as an indication of turbocharger

performance)4) The fuel oil/diesel analysis is referred to from an oil company data.5) Analysis: The trial data is to be analysed with respect to:

a) Average alternator loading.b) Average engine loading.c) Percentage loading on alternator.d) Percentage loading on engine.e) Specific power generation kWh/liter.f) Comments on Turbocharger performance based on RPM and gas temperature differ-

ence.g) Comments on charge air cooler performance.

9. DG Set System


h) Comments on load distribution among various cylinders (based on exhaust tempera-ture, the temperature to be ± 5% of mean and high/low values indicate disturbedcondition).

i) Comments on housekeeping issues like drip leakages, insulation, vibrations, etc.

A format as shown in the Table 9.5 is useful for monitoring the performance

DG Electricity Derated Type of Average Specific SpecificSet Generating Electricity Fuel Load as % Fuel Cons. Lube OilNo. Capacity Generating used of Derated Lit/kWh Cons.

(Site), kW Capacity, kW Capacity Lit/kWh

1. 480 300 LDO 89 0.335 0.007

2. 480 300 LDO 110 0.334 0.024

3. 292 230 LDO 84 0.356 0.006

4. 200 160 HSD 89 0.325 0.003

5. 200 160 HSD 106 0.338 0.003

6. 200 160 HSD

7. 292 230 LDO 79 0.339 0.006

8. 292 230 LDO 81 0.362 0.005

9. 292 230 LDO 94 0.342 0.003

10. 292 230 LDO 88 0.335 0.006

11. 292 230 LDO 76 0.335 0.005

12. 292 230 LDO 69 0.353 0.006

13 400 320 HSD 75 0.334 0.004

14. 400 320 HSD 65 0.349 0.004

15. 880 750 LDO 85 0.318 0.007

16. 400 320 HSD 70 0.335 0.004

17. 400 320 HSD 80 0.337 0.004

18. 880 750 LDO 78 0.345 0.007

19. 800 640 HSD 74 0.324 0.002

20. 800 640 HSD 91 0.290 0.002

21. 880 750 LDO 96 0.307 0.002

22. 920 800 LDO 77 0.297 0.002

TABLE 9.5TYPICAL FORMAT FOR DG SET MONITORING

9. DG Set System


9.5 Energy Saving Measures for DG Sets

a) Ensure steady load conditions on the DG set, and provide cold, dust free air at intake (useof air washers for large sets, in case of dry, hot weather, can be considered).

b) Improve air filtration.

c) Ensure fuel oil storage, handling and preparation as per manufacturers' guidelines/oil com-pany data.

d) Consider fuel oil additives in case they benefit fuel oil properties for DG set usage.

e) Calibrate fuel injection pumps frequently.

f) Ensure compliance with maintenance checklist.

g) Ensure steady load conditions, avoiding fluctuations, imbalance in phases, harmonic loads.

h) In case of a base load operation, consider waste heat recovery system adoption for steamgeneration or refrigeration chiller unit incorporation. Even the Jacket Cooling Water isamenable for heat recovery, vapour absorption system adoption.

i) In terms of fuel cost economy, consider partial use of biomass gas for generation. Ensuretar removal from the gas for improving availability of the engine in the long run.

j) Consider parallel operation among the DG sets for improved loading and fuel economythereof.

k) Carryout regular field trials to monitor DG set performance, and maintenance planning asper requirements.

9. DG Set System


QUESTIONS

1. Explain the principle of a four stroke diesel engine.

2. The efficiency of a Genset ranges between:a) 20 – 25% (b) 0 – 20% (c) 40 – 45% (d) 60 – 70%

3. What are the components of a DG Set System?

4. List briefly the salient developments in DG Plants.

5. Connected load of a plant is 1200 kW and Diversity factor is 1.8. What is the desir-able set rating with respect to 0.8 PF and the set load factor of 75%?

6. What is the effect of altitude and intake air temperature on DG set output?

7. What is the function of turbo charger in DG set?

8. Draw a typical energy balance of a DG Set.

9. How do you assess waste heat recovery potential in a DG set?

10. What are the factors affecting waste heat recovery from DG sets?

11. What is the role of an energy manager/auditor for energy efficiency in DG plants ofan industrial unit?

12. List the energy savings opportunities in an industrial DG set plant.

REFERENCES1. Proceedings of National Workshop on Efficient Captive Power Generation with

Industrial DG Sets2. NPC Case Studies3. Wartsila-NSD Literature

10. ENERGY EFFICIENT TECHNOLOGIES INELECTRICAL SYSTEMS


Syllabus

Energy Efficient Technologies in Electrical Systems: Maximum demand controllers,Automatic power factor controllers, Energy efficient motors, Soft starters with energysaver, Variable speed drives, Energy efficient transformers, Electronic ballast,Occupancy sensors, Energy efficient lighting controls, Energy saving potential of eachtechnology.

10.1. Maximum Demand Controllers

High-tension (HT) consumers have to pay a maximum demand charge in addition to the usualcharge for the number of units consumed. This charge is usually based on the highest amountof power used during some period (say 30 minutes) during the metering month. The maximumdemand charge often represents a large proportion of the total bill and may be based on onlyone isolated 30 minute episode of high power use.

Considerable savings can be realised by monitoring power use and turning off or reduc-ing non-essential loads during such periods of high power use.

Maximum DemandController (See Figure10.1)is a device designed to meet

the need of industries con-scious of the value of loadmanagement. Alarm issounded when demandapproaches a preset value. Ifcorrective action is nottaken, the controller switch-es off non-essential loads ina logical sequence. Thissequence is predeterminedby the user and is pro-grammed jointly by the userand the supplier of thedevice. The plant equip-ments selected for the loadmanagement are stoppedand restarted as per thedesired load profile. Demand control scheme is implemented by using suitable control contac-tors. Audio and visual annunciations could also be used.

Figure 10.1 Maximum Demand Controller

10.2 Automatic Power Factor Controllers

Various types of automatic power factor controls are available with relay / microprocessorlogic. Two of the most common controls are: Voltage Control and kVAr Control

Voltage Control

Voltage alone can be used as a source of intelligence when the switched capacitors areapplied at point where the circuit voltage decreases as circuit load increases. Generally, wherethey are applied the voltage should decrease as circuit load increases and the drop in voltageshould be around 4 – 5 % with increasing load.

Voltage is the most common type of intelligence used in substation applications, whenmaintaining a particular voltage is of prime importance. This type of control is independent ofload cycle. During light load time and low source voltage, this may give leading PF at the sub-station, which is to be taken note of.

KILOVAR Control

Kilovar sensitive controls (seeFigure 10.2) are used at loca-tions where the voltage level isclosely regulated and not avail-able as a control variable. Thecapacitors can be switched torespond to a decreasing powerfactor as a result of change insystem loading. This type ofcontrol can also be used to avoidpenalty on low power factor byadding capacitors in steps as thesystem power factor begins tolag behind the desired value.Kilovar control requires twoinputs - current and voltage fromthe incoming feeder, which arefed to the PF correction mecha-nism, either the microprocessoror the relay.

Automatic Power Factor Control Relay

It controls the power factor of the installation by giving signals to switch on or off power fac-tor correction capacitors. Relay is the brain of control circuit and needs contactors of appropri-ate rating for switching on/off the capacitors.

There is a built-in power factor transducer, which measures the power factor of the installation and converts it to a DC voltage of appropriate polarity. This is compared witha reference voltage, which can be set by means of a knob calibrated in terms of power fac-tor.

10. Energy Efficient Technologies in Electrical System


Figure 10.2

When the power factor falls below setting, the capacitors are switched on in sequence. Therelays are provided with First in First out (FIFO) and First in Last Out (FILO) sequence. Thecapacitors controlled by the relay must be of the same rating and they are switched on/off in lin-ear sequence. To prevent over correction hunting, a dead band is provided. This setting deter-mines the range of phase angle over which the relay does not respond; only when the PF goesbeyond this range, the relay acts. When the load is low, the effect of the capacitors is more pro-nounced and may lead to hunting. Under current blocking (low current cut out) shuts off therelay, switching off all capacitors one by one in sequence, when load current is below setting.Special timing sequences ensure that capacitors are fully discharged before they are switchedin. This avoids dangerous over voltage transient. The solid state indicating lamps (LEDS) dis-play various functions that the operator should know and also and indicate each capacitorswitching stage.

Intelligent Power Factor Controller (IPFC)

This controller determines the rating of capacitance connected in each step during the first hourof its operation and stores them in memory. Based on this measurement, the IPFC switches onthe most appropriate steps, thus eliminating the hunting problems normally associated withcapacitor switching.

10.3 Energy Efficient Motors

Minimising Watts Loss in Motors

Improvements in motor efficiency can beachieved without compromising motor per-formance - at higher cost - within the limitsof existing design and manufacturing tech-nology.

From the Table 10.1, it can be seen thatany improvement in motor efficiency mustresult from reducing the Watts losses. Interms of the existing state of electric motortechnology, a reduction in watts losses can beachieved in various ways.

All of these changes to reduce motorlosses are possible with existing motordesign and manufacturing technology.They would, however, require addi-tional materials and/or the use of higherquality materials and improved manufacturing processes resulting in increased motor cost.

Simply Stated: REDUCED LOSSES = IMPROVED EFFICIENCY



Figure 10.3 Energy Efficient Motor

Thus energy-efficient electric motorsreduce energy losses through improveddesign, better materials, and improved manu-facturing techniques. Replacing a motor maybe justifiable solely on the electricity costsavings derived from an energy-efficientreplacement. This is true if the motor runscontinuously, power rates are high, the motoris oversized for the application, or its nomi-nal efficiency has been reduced by damage orprevious rewinds. Efficiency comparison forstandard and high efficiency motors is shownin Figure 10.4

Technical aspects of Energy EfficientMotors

Energy-efficient motors last longer, andmay require less maintenance. At lower temperatures, bearing grease lasts longer; required time between re-greasing increases. Lower temperatures translate to long lasting insulation. Generally, motor life doubles for each 10°C reduction in operatingtemperature.

Select energy-efficient motors with a 1.15 service factor, and design for operation at 85% ofthe rated motor load.

Electrical power problems, especially poor incoming power quality can affect the operationof energy-efficient motors.



TABLE 10.1 WATT LOSS AREA AND EFFICIENCY IMPROVEMENT

Watts Loss Area Efficiency Improvement

1. Iron Use of thinner gauge, lower loss core steel reduces eddy current losses. Longercore adds more steel to the design, which reduces losses due to lower operatingflux densities.

2. Stator I2 R Use of more copper and larger conductors increases cross sectional area of stator windings. This lowers resistance (R) of the windings and reduces losses due to current flow (I).

3. Rotor I2 R Use of larger rotor conductor bars increases size of cross section, lowering conductor resistance (R) and losses due to current flow (I).

4. Friction & Windage Use of low loss fan design reduces losses due to air movement.

5. Stray Load Loss Use of optimised design and strict quality control procedures minimizes stray load losses.

Figure 10.4 Efficiency Range for Standard andHigh Efficiency Motors

Speed control is crucial in some applications. In polyphase induction motors, slip is a measureof motor winding losses. The lower the slip, the higher the efficiency. Less slippage in energyefficient motors results in speeds about 1% faster than in standard counterparts.

Starting torque for efficient motors may be lower than for standard motors. Facility managersshould be careful when applying efficient motors to high torque applications.

10.4 Soft Starter

When starting, AC Induction motor develops more torque than isrequired at full speed. This stress is transferred to the mechanical trans-mission system resulting in excessive wear and premature failure ofchains, belts, gears, mechanical seals, etc. Additionally, rapid accelera-tion also has a massive impact on electricity supply charges with highinrush currents drawing +600% of the normal run current.

The use of Star Delta only provides a partial solution to theproblem. Should the motor slow down during the transition period,the high peaks can be repeated and can even exceed direct on linecurrent.

Soft starter (see Figure 10.5) provides a reliable and economical solution to these problemsby delivering a controlled release of power to the motor, thereby providing smooth, steplessacceleration and deceleration. Motor life will be extended as damage to windings and bearingsis reduced.

Soft Start & Soft Stop is built into 3 phase units, providing controlled starting and stoppingwith a selection of ramp times and current limit settings to suit all applications (see Figure 10.6).



Figure 10.5 Soft Starter

Figure 10.6 Soft Starter: Starting current, Stress profile during starting

Advantages of Soft Start

– Less mechanical stress– Improved power factor. – Lower maximum demand. – Less mechanical maintenance

10.5 Variable Speed Drives

Speed Control of Induction Motors

Induction motor is the workhorse of the industry. It is cheap rugged and provides high power toweight ratio. On account of high cost-implications and limitations of D.C. System, inductionmotors are preferred for variable speed application, the speed of which can be varied by chang-ing the supply frequency. The speed can also be varied through a number of other means,including, varying the input voltage, varying the resistance of the rotor circuit, using multispeed windings, using Scherbius or Kramer drives, using mechanical means such as gears andpulleys and eddy-current or fluid coupling, or by using rotary or static voltage and frequencyconverters.

Variable Frequency Drive

The VFD operates on a simple principle. The rotational speed of an AC induction motordepends on the number of poles in that stator and the frequency of the applied AC power.Although the number of poles in an induction motor cannot be altered easily, variable speed canbe achieved through a variation in frequency. The VFD rectifies standard 50 cycle AC linepower to DC, then synthesizes the DC to a variable frequency AC output.

Motors connected to VFD provide variable speed mechanical output with high efficiency.These devices are capable of up to a 9:1 speed reduction ratio (11 percent of full speed), and a3:1 speed increase (300 percent of full speed).

In recent years, the technology of AC variable frequency drives (VFD) has evolved intohighly sophisticated digital microprocessor control, along with high switching frequency IGBTs(Insulated Gate Bi Polar Transistors) power devices. This has led to significantly advancedcapabilities from the ease of programmability to expanded diagnostics. The two most signifi-cant benefits from the evolution in technology have been that of cost and reliability, in additionto the significant reduction in physical size.

Variable Torque Vs. Constant Torque

Variable speed drives, and the loads that are applied to, can generally be divided into twogroups: constant torque and variable torque. The energy savings potential of variable torqueapplications is much greater than that of constant torque applications. Constant torque loadsinclude vibrating conveyors, punch presses, rock crushers, machine tools, and other applica-tions where the drive follows a constant V/Hz ratio. Variable torque loads include centrifugalpumps and fans, which make up the majority of HVAC applications.

Why Variable Torque Loads Offer Greatest Energy Savings

In variable torque applications, the torque required varies with the square of the speed, and thehorsepower required varies with the cube of the speed, resulting in a large reduction of horse-power for even a small reduction in speed. The motor will consume only 25% as much energyat 50% speed than it will at 100% speed. This is referred to as the Affinity Laws, which definethe relationships between speed, flow, torque, and horsepower. The following laws illustratesthese relationships:

❖ Flow is proportional to speed❖ Head is proportional to (speed)2



❖ Torque is proportional to (speed)2

❖ Power is proportional to (speed)3

Tighter process control with variable speed drives

No other AC motor control method compares to variable speed drives when it comes to accu-rate process control. Full-voltage (across the line) starters can only run the motor at full speed,and soft starts and reduced voltage soft starters can only gradually ramp the motor up to fullspeed, and back down to shutdown. Variable speed drives, on the other hand, can be pro-grammed to run the motor at a precise speed, to stop at a precise position, or to apply a specif-ic amount of torque.

In fact, modern AC variable speed drives are very close to the DC drive in terms of fasttorque response and speed accuracy. However, AC motors are much more reliable and afford-able than DC motors, making them far more prevalent.

Most drives used in the field utilize Volts/Hertz type control, which means they provideopen-loop operation. These drives are unable to retrieve feedback from the process, but aresufficient for the majority of variable speed drive applications. Many open-loop variablespeed drives do offer slip compensation though, which enables the drive to measure its out-put current and estimate the difference in actual speed and the set point (the programmedinput value). The drive will then automatically adjust itself towards the set point based on thisestimation.

Most variable torque drives have Proportional Integral Differential (PID) capability for fanand pump applications, which allows the drive to hold the set point based on actual feedbackfrom the process, rather than relying on estimation. A transducer or transmitter is used to detectprocess variables such as pressure levels, liquid flow rate, air flow rate, or liquid level. Then thesignal is sent to a PLC (Programmable Logic Controllers), which communicates the feedbackfrom the process to the drive. The variable speed drive uses this continual feedback to adjustitself to hold the set point.

High levels of accuracy for other applications can also be achieved through drives thatoffer closed-loop operation. Closed-loop operation can be accomplished with either a field-oriented vector drive, or a sensor less vector drive. The field-oriented vector driveobtains process feedback from an encoder, which measures and transmits to the drive thespeed and/or rate of the process, such as a conveyor, machine tool, or extruder. The drivethen adjusts itself accordingly to sustain the programmed speed, rate, torque, and/or position.

Extended equipment life and reduced maintenance

Single-speed starting methods start motors abruptly, subjecting the motor to a high startingtorque and to current surges that are up to 10 times the full-load current. Variable speed drives,on the other hand, gradually ramp the motor up to operating speed to lessen mechanical andelectrical stress, reducing maintenance and repair costs, and extending the life of the motor andthe driven equipment.

Soft starts, or reduced-voltage soft starters (RVSS), are also able to step a motor up grad-ually, but drives can be programmed to ramp up the motor much more gradually and smooth-ly, and can operate the motor at less than full speed to decrease wear and tear. Variable speeddrives can also run a motor in specialized patterns to further minimise mechanical and electri-



cal stress. For example, an S-curve pattern can be applied to a conveyor application forsmoother control, which reduces the backlash that can occur when a conveyor is acceleratingor decelerating.

Typical full-load efficiencies are 95% and higher. High power units are still more efficient.The efficiency of VSDs generally decreases with speed but since the torque requirement alsodecreases with speed for many VSD applications, the absolute loss is often not very significant.The power factor of a VSD drops drastically with speed, but at low power requirement theabsolute kVAr requirement is low, so the loss is also generally not significant. In a suitable oper-ating environment, frequency controllers are relatively reliable and need little maintenance. Adisadvantage of static converters is the generation of harmonics in the supply, which reducesmotor efficiency and reduces motor output - in some cases it may necessitate using a motor witha higher rating.

Eddy Current Drives

This method employs an eddy-current clutch to varythe output speed. The clutch consists of a primarymember coupled to the shaft of the motor and a freelyrevolving secondary member coupled to the loadshaft. The secondary member is separately excitedusing a DC field winding. The motor starts with theload at rest and a DC excitation is provided to the sec-ondary member, which induces eddy-currents in theprimary member. The interaction of the fluxes pro-duced by the two currents gives rise to a torque at theload shaft. By varying the DC excitation the outputspeed can be varied to match the load requirements. The major disadvantage of this system isrelatively poor efficiency particularly at low speeds. (see Figure 10.7)

Slip Power Recovery Systems

Slip power recovery is a more efficient alternative speed control mechanism for use with slip-ring motors. In essence, a slip power recovery system varies the rotor voltage to control speed,but instead of dissipating power through resistors, the excess power is collected from the sliprings and returned as mechanical power to the shaft or as electrical power back to the supplyline. Because of the relatively sophisticated equipment needed, slip power recovery tends to beeconomical only in relatively high power applications and where the motor speed range is 1:5or less.

Fluid Coupling

Fluid coupling is one way of applying varying speeds to the driven equipment, without chang-ing the speed of the motor.

Construction

Fluid couplings (see Figure 10.8) work on the hydrodynamic principle. Inside every fluid coupling are two basic elements – the impeller and the runner and together they con-



Figure 10.7 Eddy Current Drive

stitute the working circuit. One can imagine theimpeller as a centrifugal pump and the runner as aturbine. The impeller and the rotor are bowlshaped and have large number of radial vanes.They are suitably enclosed in a casing, facing eachother with an air gap. The impeller is connected tothe prime mover while the rotor has a shaft boltedto it. This shaft is further connected to the drivenequipment through a suitable arrangement.

Thin mineral oil of low viscosity and good-lubricating qualities is filled in the fluid couplingfrom the filling plug provided on its body. A fusibleplug is provided on the fluid coupling which blowsoff and drains out oil from the coupling in case ofsustained overloading.

Operating Principle

There is no mechanical inter-connection betweenthe impeller and the rotor and the power istransmitted by virtue of the fluid filled in the coupling. When the impeller is rotated by theprime mover, the fluid flows out radially and then axially under the action of centrifugalforce. It then crosses the air gap to the runner and is directed towards the bowl axis and backto the impeller. To enable the fluid to flow from impeller to rotor it is essential that there is difference in head between the two and thus it is essential that there is difference inRPM known as slip between the two. Slip is an important and inherent characteristic of afluid coupling resulting in several desired advantages. As the slip increases, more and morefluid can be transferred. However when the rotor is at a stand still, maximum fluid istransmitted from impeller to rotor and maximum torque is transmitted from the coupling.This maximum torque is the limiting torque. The fluid coupling also acts as a torque limiter.

Characteristics

Fluid coupling has a centrifugal characteristic during starting thus enabling no-load start up ofprime mover, which is of great importance. The slipping characteristic of fluid coupling pro-vides a wide range of choice of power transmission characteristics. By varying the quantity ofoil filled in the fluid coupling, the normal torque transmitting capacity can be varied. The max-imum torque or limiting torque of the fluid coupling can also be set to a pre-determined safevalue by adjusting the oil filling. The fluid coupling has the same characteristics in both direc-tions of rotation.

10.6 Energy Efficient Transformers

Most energy loss in dry-type transformers occurs through heat or vibration from the core. Thenew high-efficiency transformers minimise these losses. The conventional transformer is madeup of a silicon alloyed iron (grain oriented) core. The iron loss of any transformer depends on



Figure 10.8 Fluid Coupling

the type of core used in the transformer.However the latest technology is to useamorphous material - a metallic glass alloyfor the core (see Figure 10.9). The expectedreduction in energy loss over conventional(Si Fe core) transformers is roughly around70%, which is quite significant. By usingan amorphous core- with unique physicaland magnetic properties- these new type oftransformers have increased efficiencieseven at low loads – 98.5% efficiency at35% load.

Electrical distribution transformersmade with amorphous metal cores provideexcellent opportunity to conserve energyright from the installation. Though thesetransformers are a little costlier thanconventional iron core transformers, theoverall benefit towards energy savingswill compensate for the higher initialinvestment. At present amorphous metalcore transformers are available up to 1600 kVA.

10.7 Electronic Ballast

Role of Ballast

In an electric circuit the ballast acts as a stabilizer. Fluorescent lamp is an electric dischargelamp. The two electrodes are separated inside a tube with no apparent connection betweenthem. When sufficient voltage is impressed on these electrodes, electrons are driven from oneelectrode and attracted to the other. The current flow takes place through an atmosphere of low-pressure mercury vapour.

Since the fluorescent lamps cannot produce light by direct connection to the power source,they need an ancillary circuit and device to get started and remain illuminated. The auxillarycircuit housed in a casing is known as ballast.

Conventional Vs Electronic Ballasts

The conventional ballasts make use of the kick caused by sudden physical disruption of current in an inductive circuit to produce the high voltage required for starting the lamp and then rely on reactive voltage drop in the ballast to reduce the voltage appliedacross the lamp. On account of the mechanical switch (starter) and low resistance offilament when cold the uncontrolled filament current, generally tend to go beyond the limitsspecified by Indian standard specifications. With high values of current and flux densities the operational losses and temperature rise are on the higher side in conventionalchoke.



Figure 10.9 1600 kVA Amorphous Core Transformer

The high frequency electronic ballast overcomes the above drawbacks. The basic functionsof electronic ballast are:

1. To ignite the lamp 2. To stabilize the gas discharge3. To supply the power to the lamp

The electronic ballasts (see Figure 10.10)make use of modern power semi-conductordevices for their operation. The circuit compo-nents form a tuned circuit to deliver power to thelamp at a high resonant frequency (in the vicinityof 25 kHz) and voltage is regulated through an in-built feedback mechanism. It is now well estab-lished that the fluorescent lamp efficiency in thekHz range is higher than those attainable at lowfrequencies. At lower frequencies (50 or 60 Hz)the electron density in the lamp is proportional tothe instantaneous value of the current because theionisation state in the tube is able to follow theinstantaneous variations in the current. At higherfrequencies (kHz range), the ionisation state cannot follow the instantaneous variations of thecurrent and hence the ionisation density is approximately a constant, proportional to the RMS(Root Mean Square) value of the current. Another significant benefit resulting from this phe-nomenon is the absence of stroboscopic effect, thereby significantly improving the quality oflight output.

One of largest advantages of an electronic ballast is the enormous energy savings it pro-vides. This is achieved in two ways. The first is its amazingly low internal core loss, quiteunlike old fashioned magnetic ballasts. And second is increased light output due to the excita-tion of the lamp phosphors with high frequency. If the period of frequency of excitation issmaller than the light retention time constant for the gas in the lamp, the gas will stay ionizedand, therefore, produce light continuously. This phenomenon along with continued persistenceof the phosphors at high frequency will improve light output from 8–12 percent. This is possi-ble only with high frequency electronic ballast.

10.8 Energy Efficient Lighting Controls

Occupancy Sensors

Occupancy-linked control can be achieved using infra-red, acoustic, ultrasonic ormicrowave sensors, which detect either movement or noise in room spaces. These sensorsswitch lighting on when occupancy is detected, and off again after a set time period, whenno occupancy movement detected. They are designed to override manual switches and toprevent a situation where lighting is left on in unoccupied spaces. With this type of systemit is important to incorporate a built-in time delay, since occupants often remain still or quietfor short periods and do not appreciate being plunged into darkness if not constantly mov-ing around.



Figure 10.10 Electronic Ballast

Timed Based Control

Timed-turnoff switches are the least expensive type ofautomatic lighting control. In some cases, their lowcost and ease of installation makes it desirable to usethem where more efficient controls would be tooexpensive (see Figure 10.11).

Types and Features

The oldest and most common type of timed-turnoffswitch is the "dial timer," a spring-wound mechanicaltimer that is set by twisting the knob to the desired time.Typical units of this type are vulnerable to damagebecause the shaft is weak and the knob is not securelyattached to the shaft. Some spring-wound units make anannoying ticking sound as they operate. Newer types oftimed-turnoff switches are completely electronic andsilent. Electronic switches can be made much morerugged than the spring-wound dial timer. These unitstypically have a spring-loaded toggle switch that turns on the circuit for a preset time interval.Some electronic models provide a choice of time intervals, which you select by adjusting a knoblocated behind the faceplate. Most models allow occupants to turn off the lights manually. Somemodels allow occupants to keep the lights on, overriding the timer. Timed-turnoff switches areavailable with a wide range of time spans. The choice of time span is a compromise. Shorter timespans waste less energy but increase the probability that the lights will turn off while someone isin the space. Dial timers allow the occupant to set the time span, but this is not likely to be donewith a view toward optimising efficiency. For most applications, the best choice is an electronicunit that allows the engineering staff to set a fixed time interval behind the cover plate.

Daylight Linked Control

Photoelectric cells can be used either simply to switch lighting on and off, or for dimming. Theymay be mounted either externally or internally. It is however important to incorporate time delaysinto the control system to avoid repeated rapid switching caused, for example, by fast movingclouds. By using an internally mounted photoelectric dimming control system, it is possible toensure that the sum of daylight and electric lighting always reaches the design level by sensingthe total light in the controlled area and adjusting the output of the electric lighting accordingly. Ifdaylight alone is able to meet the design requirements, then the electric lighting can be turned off.The energy saving potential of dimming control is greater than a simple photoelectric switchingsystem. Dimming control is also more likely to be acceptable to room occupants.

Localized Switching

Localized switching should be used in applications which contain large spaces. Local switchesgive individual occupants control over their visual environment and also facilitate energy sav-ings. By using localized switching it is possible to turn off artificial lighting in specific areas,while still operating it in other areas where it is required, a situation which is impossible if thelighting for an entire space is controlled from a single switch.



Figure 10.11 Timed Turnoff Switch



QUESTIONS

1. Explain how maximum demand control works.

2. Explain the principle of automatic power factor controller .

3. What are the advantages of energy efficient motors?

4. What are the precautions to be taken in the case of energy efficient motor application ?

5. Explain the working of a soft starter and its advantage over other conventionalstarters.

6. Explain why centrifugal machines offers the greatest savings when used withVariable Speed Drives.

7. Hydrodynamic principle for speed control is used ina) DC drives b) Fluid coupling c) Pulse width modulation d) Eddy Current Drive

8. Typical loss in conventional magnetic chokes for a 40 W FTL is of the order ofa) 8 Watts b) 14 Watts c) 20 Watts d) 6 Watts

9. Which method uses infrared, acoustic, ultrasonic or microwave sensors for lightingcontrol?a) Time-based control b) Daylight-linked control c) Occupancy-linked controld) Localized switching

10. Slip Power Recovery system is used ina) All kinds of motors b) Synchronous motors c) Slip - Ring Induction motord) None of the above

REFERENCES 1. Energy Management Supply and Conservation, Butterworth Heinemann, 2002 – Dr. Clive

Beggs.2. Handbook of Energy Engineering, The Fairmont Press, INC. – Albert Thumann & Paul

Mehta.

5. ENERGY PERFORMANCE ASSESSMENT OF MOTORSAND VARIABLE SPEED DRIVES


5.1 Introduction

The two parameters of importance in a motor are effi-ciency and power factor. The efficiencies of inductionmotors remain almost constant between 50% to 100%loading (Refer figure 5.1). With motors designed toperform this function efficiently; the opportunity forsavings with motors rests primarily in their selectionand use. When a motor has a higher rating than thatrequired by the equipment, motor operates at part load.In this state, the efficiency of the motor is reduced.Replacement of under loaded motors with smallermotors will allow a fully loaded smaller motor to oper-ate at a higher efficiency. This arrangement is general-ly most economical for larger motors, and only whenthey are operating at less than one-third to one-halfcapacity, depending on their size.

5.2 Performance Terms and Definitions

Efficiency :

The efficiency of the motor is given by

Where Pout – Output power of the motorPin – Input power of the motorPLoss – Losses occurring in motor

Motor Loading :

Motor Loading % =

Pout Ploss

η = �� = 1 � ��Pin Pin

Actual operating load of the motorRated capacity of the motor

Figure 5.1 Efficiency vs. Loading

x 100

5. Energy Performance Assessment of Motors and Variable Speed Drives


5.3 Efficiency Testing

While input power measurements are fairly simple, measurement of output or losses need a labo-rious exercise with extensive testing facilities. The following are the testing standards widely used.

Europe: IEC 60034-2, and the new IEC 61972US: IEEE 112 - Method BJapan: JEC 37

Even between these standards the difference in efficiency value is up to 3%.For simplicity nameplate efficiency rating may be used for calculations if the motor load is inthe range of 50 -100 %.

Field Tests for Determining Efficiency

(Note: The following section is a repeat of material provided in the chapter-2 on ElectricalMotors in Book-3.)

No Load Test :

The motor is run at rated voltage and frequency without any shaft load. Input power, current,frequency and voltage are noted. The no load P.F. is quite low and hence low PF watt metersare required. From the input power, stator I2R losses under no load are subtracted to give thesum of Friction and Windage (F&W) and core losses. To separate core and F & W losses, testis repeated at variable voltages. It is worthwhile plotting no-load input kW versus Voltage; theintercept is F & W kW loss component.

F&W and core losses = No load power (watts) – (No load current)2 x Stator resistance

Stator and Rotor I2R Losses :

The stator winding resistance is directly measured by a bridge or volt amp method. The resis-tance must be corrected to the operating temperature. For modern motors, the operating tem-perature is likely to be in the range of 100°C to 120°C and necessary correction should be made.Correction to 75°C may be inaccurate. The correction factor is given as follows :

R2 235 + t2—– = ———– , where, t1 = ambient temperature, °C & t2 = operating temperature, °C.R1 235 + t1

The rotor resistance can be determined from locked rotor test at reduced frequency, but rotorI2R losses are measured from measurement of rotor slip.

Rotor I2R losses = Slip x (Stator Input - Stator I2R Losses - Core Loss)

Accurate measurement of slip is possible by stroboscope or non-contact type tachometer. Slipalso must be corrected to operating temperature.

Stray Load Losses :

These losses are difficult to measure with any accuracy. IEEE Standard 112 gives a complicatedmethod, which is rarely used on shop floor. IS and IEC standards take a fixed value as 0.5 % of



output. It must be remarked that actual value of stray losses is likely to be more. IEEE - 112 spec-ifies values from 0.9 % to 1.8 %.

Motor Rating Stray Losses

1 – 125 HP 1.8 %125 – 500 HP 1.5 %501 – 2499 HP 1.2 %2500 and above 0.9 %

Points for Users :

It must be clear that accurate determination of efficiency is very difficult. The same motor test-ed by different methods and by same methods by different manufacturers can give a differenceof 2 %.

Estimation of efficiency in the field can be summarized as follows:a) Measure stator resistance and correct to operating temperature. From rated current

value, I2R losses are calculated.b) From rated speed and output, rotor I2R losses are calculatedc) From no load test, core and F & W losses are determined for stray loss

The method is illustrated by the following example :

Example :

Motor Specifications

Rated power = 34 kW/45 HPVoltage = 415 VoltCurrent = 57 AmpsSpeed = 1475 rpmInsulation class = FFrame = LD 200 LConnection = Delta

No load test Data

Voltage, V = 415 VoltsCurrent, I = 16.1 AmpsFrequency, F = 50 HzStator phase resistance at 30°C = 0.264 OhmsNo load power, Pnl = 1063.74 Watts

a) Calculate iron plus friction and windage losses

b) Calculate stator resistance at 120°C

235 + t2

R2 = R1 x ————235 + t1



c) Calculate stator copper losses at operating temperature of resistance at 120°C

d) Calculate full load slip(s) and rotor input assuming rotor losses are slip times rotor input.

e) Determine the motor input assuming that stray losses are 0.5 % of the motor rated power

f) Calculate motor full load efficiency and full load power factor

Solution

a) Let Iron plus friction and windage loss, Pi + fwNo load power, Pnl = 1063.74 WattsStator Copper loss, P st-30°C (Pst.cu)= 3 x (16.1 / √3)2 x 0.264= 68.43 WattsPi + fw = Pnl - Pst.cu= 1063.74 – 68.43= 995.3 W

b) Stator Resistance at 120°C,

120 + 235R120°C = 0.264 x —————

30 + 235

= 0.354 ohms per phase

c) Stator copper losses at full load, Pst.cu 120°C= 3 x (57 / √3)2 x 0.354= 1150.1 Watts

d) Full load slipS = (1500 – 1475) / 1500

= 0.0167

Rotor input, Pr = Poutput/ (1-S)= 34000 / (1-0.0167)= 34577.4 Watts

e) Motor full load input power, P input = Pr + Pst.cu 120°C + (Pi + fw) + Pstray

= 34577.4 + 1150.1 + 995.3 + (0.005* x 34000)= 36892.8 Watts*where, stray losses = 0.5% of rated output (assumed)

f) Motor efficiency at full loadPoutput

Efficiency = ——– x 100Pinput

34000= ——–

36892.8

= 92.2%



= PinputFull Load PF = —————–

= √3 x V x Ifl

= 36892.8= ——————–= √3 x 415 x 57

= 0.90

Comments :

a) The measurement of stray load losses is very difficult and not practical even on test beds.b) The actual value of stray loss of motors up to 200 HP is likely to be 1 % to 3 % compared

to 0.5 % assumed by standards.c) The value of full load slip taken from the nameplate data is not accurate. Actual measure-

ment under full load conditions will give better results.d) The friction and windage losses really are part of the shaft output; however, in the above

calculation, it is not added to the rated shaft output, before calculating the rotor inputpower. The error however is minor.

e) When a motor is rewound, there is a fair chance that the resistance per phase wouldincrease due to winding material quality and the losses would be higher. It would be inter-esting to assess the effect of a nominal 10 % increase in resistance per phase.

5.4 Determining Motor Loading

1. By Input Power Measurements

• First measure input power Pi with a hand held or in-line power meterPi = Three-phase power in kW

• Note the rated kW and efficiency from the motor name plate

• The figures of kW mentioned in the name plate is for output conditions.So corresponding input power at full-rated load

Nameplate full rated kWPir = ————————————————

ηfl

ηfl = Efficiency at full-rated loadPir = Input power at full-rated load in kW

• The percentage loading can now be calculated as follows

PiLoad = — x 100%

Pir



Example

The nameplate details of a motor are given as power = 15 kW, efficiency η = 0.9. Using a powermeter the actual three phase power drawn is found to be 8 kW. Find out the loading of the motor.

Input power at full-rated power in kW, Pir = 15 /0.9= 16.7 kW

Percentage loading = 8/16.7= 48 %

2. By Line Current Measurements

The line current load estimation method is used when input power cannot be measured and onlyamperage measurements are possible. The amperage draw of a motor varies approximately lin-early with respect to load, down to about 75% of full load. Below the 75% load point, powerfactor degrades and the amperage curve becomes increasingly non-linear. In the low loadregion, current measurements are not a useful indicator of load. However, this method maybe used only as a preliminary method just for the purpose of identification of oversized motors.

Input load current% Load = ———————— *100 (Valid up to 75% loading)

Input rated current

3. Slip Method

In the absence of a power meter, the slip method can be used which requires a tachometer. Thismethod also does not give the exact loading on the motors.

SlipLoad = —— *100%

Ss–Sr

Where:Load = Output power as a % of rated powerSlip = Synchronous speed - Measured speed in rpmSs = Synchronous speed in rpm at the operating frequencySr = Nameplate full-load speed

Example: Slip Load Calculation

Given: Synchronous speed in rpm = 1500 at 50 HZ operating frequency.(Synchronous speed = 120f/P) f: frequency, P: Number of polesNameplate full load speed = 1450Measured speed in rpm = 1480Nameplate rated power = 7.5 kW

Determine actual output power.

1500 – 1480Load = ————— *100% = 40%

1500 – 1450

From the above equation, actual output power would be 40% x 7.5 kW = 3 kW



The speed/slip method of determining motor part-load is often favored due to its simplicityand safety advantages. Most motors are constructed such that the shaft is accessible to atachometer or a strobe light.

The accuracy of the slip method, however, is limited. The largest uncertainty relates tothe accuracy with which manufacturers report the nameplate full-load speed. Manufacturersgenerally round their reported full-load speed values to some multiple of 5 rpm. While 5 rpm isbut a small percent of the full-load speed and may be considered as insignificant, the slipmethod relies on the difference between full-load nameplate and synchronous speeds. Given a40 rpm "correct" slip, a seemingly minor 5 rpm disparity causes a 12% change in calculatedload.

Slip also varies inversely with respect to the motor terminal voltage squared. A voltage cor-rection factor can, also, be inserted into the slip load equation. The voltage compensated loadcan be calculated as shown

SlipLoad = ———————– x 100%

(Ss – Sr) x (Vr/V)2

Where:

Load = Output power as a % of rated power

Slip = Synchronous speed - Measured speed in rpm

Ss = Synchronous speed in rpm

Sr = Nameplate full-load speed

V = RMS voltage, mean line to line of 3 phases

Vr = Nameplate rated voltage

5.5 Performance Evaluation of Rewound Motors

Ideally, a comparison should be made of the efficiency before and after a rewinding. Arelatively simple procedure for evaluating rewind quality is to keep a log of no-load inputcurrent for each motor in the population. This figure increases with poor quality rewinds. Areview of the rewind shop's procedure should also provide some indication of the quality ofwork. When rewinding a motor, if smaller diameter wire is used, the resistance and the I2Rlosses will increase.



5.6 Format for Data Collection

The motor loading survey can be performed using the format given below:

Motor Field Measurement Format

Company_________________________ Location_______________________Date ________ Process________________________

Department_____________________

General Data

Driven Equipment__________________ Motor Operating Profile:

Motor Name Plate Data No of hours of operationManufacturer ______________________ I Shift _____________ Model ___________________________ II Shift _____________ Serial Number _____________________ III Shift _____________Type :Squirrel cage/Slp ring__________Size (hp/kW)______________________ Annual Operating Time ______ hours/yearSynchronous Speed (RPM) ___________Full-Load Speed (RPM) _____________ Type of load Voltage Rating _____________________ 1.Load is quite steady, motor "On" during shiftFull-Load Amperage ________________ 2.Load starts, stops, but is constant when "On"Full-Load Power Factor (%) __________ 3.Load starts, stops, and fluctuates when "On"Full-Load Efficiency (%) ____________Temperature Rise __________________Insulation Class ____________________

From Test Certificate

Load 100% 75% 25% No Load

Current

PF

Efficiency

Stator resistance per phase =

Measured DataSupply VoltageBy VoltmeterVRY ________VYB ________ V avg ______VBR ________Input AmpsBy AmmeterA a __________A b __________ A avg ______A c __________Power Factor (PF) _____________________Input Power (kW) ______________________

Motor Operating Speed ____________RPMAt frequency of __________Driven Equipment Operating Speed__________RPMType of Transmission (Direct/Gear/Fluid coupling)

Rewound � Yes ,if yes How manytimes rewound ?---

� NoMotor Loading %_________________



The monitoring format for rewound motor is given below:

5.7 Application of Variable Speed Drives (VSD)

Although there are many methods of varying the speeds of the driven equipment such ashydraulic coupling, gear box, variable pulley etc., the most possible method is one ofvarying the motor speed itself by varying the frequency and voltage by a variablefrequency drive.

5.7.1 Concept of Variable Frequency Drive

The speed of an induction motor is proportional to the frequency of the AC voltage applied toit, as well as the number of poles in the motor stator. This is expressed by the equation:

RPM = (f x 120) / p

Where f is the frequency in Hz, and p is the number of poles in any multiple of 2.

Therefore, if the frequency applied to the motor is changed, the motor speed changes indirect proportion to the frequency change. The control of frequency applied to the motor is thejob given to the VSD.

The VSD's basic principle of operation is to convert the electrical system frequency and volt-age to the frequency and voltage required to drive a motor at a speed other than its rated speed.The two most basic functions of a VSD are to provide power conversion from one frequency toanother, and to enable control of the output frequency.

VSD Power Conversion

As illustrated by Figure 5.1,there are two basic components,a rectifier and an inverter, toaccomplish power conversion.

The rectifier receives the50-Hz AC voltage and con-verts it to direct current (DC)voltage. A DC bus inside theVSD functions as a "parkinglot" for the DC voltage. The Figure 5.1 Components of a Variable Speed Drive



DC bus energizes the inverter, which converts it back to AC voltage again. The inverter canbe controlled to produce an output frequency of the proper value for the desired motor shaftspeed.

5.7.2 Factors for Successful Implementation of Variable Speed Drives

a) Load Type for Variable Frequency Drives

The main consideration is whether the variable frequency drive application require a variabletorque or constant torque drive. If the equipment being driven is centrifugal, such as a fan orpump, then a variable torque drive will be more appropriate. Energy savings are usually the pri-mary motivation for installing variable torque drives for centrifugal applications. For example,a fan needs less torque when running at 50% speed than it does when running at full speed.Variable torque operation allows the motor to apply only the torque needed, which results inreduced energy consumption.

Conveyors, positive displacement pumps, punch presses, extruders, and other similar typeapplications require constant level of torque at all speeds. In which case, constant torque vari-able frequency drives would be more appropriate for the job. A constant torque drive shouldhave an overload current capacity of 150% or more for one minute. Variable torque variablefrequency drives need only an overload current capacity of 120% for one minute since cen-trifugal applications rarely exceed the rated current.

If tight process control is needed, then you may need to utilize a sensor less vector, or fluxvector variable frequency drive, which allow a high level of accuracy in controlling speed,torque, and positioning.

b) Motor Information

The following motor information will be needed to select the proper variable frequency drive:

Full Load Amperage Rating. Using a motor's horsepower is an inaccurate way to size vari-able frequency drives.

Speed Range. Generally, a motor should not be run at any speed less than 20% of its specifiedmaximum speed allowed. If it is run at a speed less than this without auxiliary motor cooling,the motor will overheat. Auxiliary motor cooling should be used if the motor must be operatedat very slow speeds.

Multiple Motors. To size a variable frequency drive that will control more than one motor, addtogether the full-load amp ratings of each of the motors. All motors controlled by a single drivemust have an equal voltage rating.

c) Efficiency and Power Factor

The variable frequency drive should have an efficiency rating of 95% or better at full load.Variable frequency drives should also offer a true system power factor of 0.95 or better

across the operational speed range, to save on demand charges, and to protect the equipment(especially motors).

d) Protection and Power Quality

Motor overload Protection for instantaneous trip and motor over current.



Additional Protection: Over and under voltage, over temperature, ground fault, control ormicroprocessor fault. These protective circuits should provide an orderly shutdown of the VFD,provide indication of the fault condition, and require a manual reset (except under voltage)before restart. Under voltage from a power loss shall be set to automatically restart after returnto normal. The history of the previous three faults shall remain in memory for future review.

If a built-up system is required, there should also be externally-operated short circuit protec-tion, door-interlocked fused disconnect and circuit breaker or motor circuit protector (MCP)

To determine if the equipment under consideration is the right choice for a variable speeddrive:

The load patterns should be thoroughly studied before exercising the option of VSD. In effectthe load should be of a varying nature to demand a VSD ( refer figure 5.3 & 5.4).

Figure 5.3 Example of an excellent variablespeed drive candidate

Figure 5.4 Example of a poor variable speeddrive candidate

The first step is to identify the number of operating hours of the equipment at various loadconditions. This can be done by using a Power analyzer with continuous data storage or by asimple energy meter with periodic reading being taken.

5.7.3 Information needed to Evaluate Energy Savings for Variable Speed Application

1. Method of flow control to which adjustable speed is compared: o output throttling (pump) or dampers (fan) o recirculation (pump) or unrestrained flow (fan) o adjustable-speed coupling (eddy current coupling) o inlet guide vanes or inlet dampers (fan only) o two-speed motor.

2. Pump or fan data: o head v's flow curve for every different type of liquid (pump) or gas (fan) that is

handled o Pump efficiency curves.



3. Process information: o specific gravity (for pumps) or specific density of products (for fans) o system resistance head/flow curve o equipment duty cycle, i.e. flow levels and time duration.

4. Efficiency information on all relevant electrical system apparatus: o motors, constant and variable speed o variable speed drives o gears o transformers.

If we do not have precise information for all of the above, we can make reasonable assump-tions for points 2 and 4.



QUESTIONS

1) Define motor efficiency.

2) Why it is difficult to measure motor efficiency at site?

3) Describe the various methods by which you calculate motor loading.

4) If no instrument other than tachometer is available, what method you would suggestfor measuring the motor load?

5) A 20 kW rated motor is drawing actual measured power of 14 kW. If the rated effi-ciency is 92%, determine the motor loading?

6) What are the limitations of slip method in determining motor loading?

7) A 4 pole motor is operating at a frequency of 50 Hz. Find the RPM of the motor?

8) What are the two factors influencing the speed of induction motor?

9) A fan's operating hours and loading are given below:

15 hours at 100% load

8 hours at 95% load

1 hour at 40% load

Is the application suitable candidate for application of VSD?

11) The losses in a variable speed drive is a) 12% b) 8% c) <5% d) no losses at all

REFERENCES1. Motor challenge: Office of Industrial Technologies, Department of Energy, USA2. Energy audit Reports of National Productivity Council

10. ENERGY PERFORMANCE ASSESSMENT OF LIGHTINGSYSTEMS


10.1 Introduction

Lighting is provided in industries, commercial buildings, indoor and outdoor forproviding comfortable working environment. The primary objective is to provide therequired lighting effect for the lowest installed load i.e highest lighting at lowest powerconsumption.

10.2 Purpose of the Performance Test

Most interior lighting requirements are for meeting average illuminance on a horizontal plane,either throughout the interior, or in specific areas within the interior combined with generallighting of lower value.

The purpose of performance test is to calculate the installed efficacy in terms oflux/watt/m² (existing or design) for general lighting installation. The calculated value canbe compared with the norms for specific types of interior installations for assessingimprovement options.

The installed load efficacy of an existing (or design) lighting installation can be assessedby carrying out a survey as indicated in the following pages.

10.3 Performance Terms and Definitions

Lumen is a unit of light flow or luminous flux. The lumenrating of a lamp is a measure of the total light output of thelamp. The most common measurement of light output (orluminous flux) is the lumen. Light sources are labeled withan output rating in lumens.

Lux is the metric unit of measure for illuminance of a sur-face. One lux is equal to one lumen per square meter.

Circuit Watts is the total power drawn by lamps and ballasts in a lighting circuit underassessment.

Installed Load Efficacy is the average maintained illuminance provided on a horizontal work-ing plane per circuit watt with general lighting of an interior. Unit: lux per watt per squaremetre (lux/W/m²)

Lamp Circuit Efficacy is the amount of light (lumens) emitted by a lamp for each watt ofpower consumed by the lamp circuit, i.e. including control gear losses. This is a moremeaningful measure for those lamps that require control gear. Unit: lumens per circuitwatt (lm/W)

Installed Power Density. The installed power density per 100 lux is the power needed persquare metre of floor area to achieve 100 lux of average maintained illuminance on a horizon-

10. Energy Performance Assessment of Lighting Systems


tal working plane with general lighting of an interior. Unit: watts per square metre per 100 lux(W/m²/100 lux)

100Installed power density (W/m²/100 lux) = —————————————–

Installed load efficacy (lux/W/m²)

Installed Load Efficacy Ratio (ILER)= Actual Lux/W/m² Target W/m²/100lux

——————— or ————————Target Lux/W/m² Actual W/m²/100lux

Average maintained illuminance is the average of lux levels measured at various points in adefined area.

Color Rendering Index (CRI) is a measure of the effect of light on the perceived color of objects.To determine the CRI of a lamp, the color appearances of a set of standard color chips are measuredwith special equipment under a reference light source with the same correlated color temperature asthe lamp being evaluated. If the lamp renders the color of the chips identical to the reference lightsource, its CRI is 100. If the color rendering differs from the reference light source, the CRI is lessthan 100. A low CRI indicates that some colors may appear unnatural when illuminated by the lamp.

10.4 Preparation (before Measurements)

Before starting the measurements, the following care should be taken:

• All lamps should be operating and no luminaires should be dirty or stained.

• There should be no significant obstructions to the flow of light throughout the interior,especially at the measuring points.

• Accuracies of readings should be ensured by– Using accurate illuminance meters for measurements– Sufficient number and arrangement of measurement points within the interior– Proper positioning of illuminance meter– Ensuring that no obstructions /reflections from surfaces affect measurement.

• Other precautions– If the illuminance meter is relatively old and has not been checked recently, it

should be compared with one that has been checked over a range of illuminances,e.g. 100 to 600 lux, to establish if a correction factor should be applied.

– that the number and arrangement of measurement points are sufficient andsuitable to obtain a reasonably accurate assessment of the average illuminancethroughout an interior. The procedure recommended in the CIBSE Code forsuch site measurements is as follows:

The interior is divided into a number of equal areas, which should be as square as possible.The illuminance at the centre of each area is measured and the mean value calculated. Thisgives an estimate of the average illuminance on the horizontal working plane.



10.5 Procedure for Assessment of Lighting Systems

10.5.1 To Determine the Minimum Number and Positions of Measurement Points

Calculate the Room Index: RI = L x W————–Hm(L + W)

Where L = length of interior; W = width of interior; Hm = the mounting height, whichis the height of the lighting fittings above the horizontal working plane. The working planeis usually assumed to be 0.75m above the floor in offices and at 0.85m above floor level inmanufacturing areas.

It does not matter whether these dimensions are in metres, yards or feet as long as thesame unit is used throughout. Ascertain the minimum number of measurement points fromTable10.1.

TABLE 10.1 DETERMINATION OF

MEASUREMENT POINTS

Room Index Minimum number ofmeasurement points

Below 1 9

1 and below 2 16

2 and below 3 25

3 and above 36

To obtain an approximately "square array", i.e. the spacing between the points oneach axis to be approximately the same, it may be necessary to increase the number ofpoints.

For example, the dimensions of an interior are:

Length = 9m, Width = 5m, Height of luminaires above working plane (Hm) = 2m

Calculate RI = 9 x 5 = 1.6072(9 + 5)

From Table 10.1 the minimum number of measurement points is 16

As it is not possible to approximate a "square array" of 16 points within such a rectan-gle it is necessary to increase the number of points to say 18, i.e. 6 x 3. These should bespaced as shown below:



Therefore in this example the spacing between points along rows along the length of theinterior = 9 ÷ 6 = 1.5m and the distance of the 'end' points from the wall = 1.5 ÷ 2 = 0.75m.

Similarly the distance between points across the width of the interior = 5 ÷ 3 = 1.67m withhalf this value, 0.83m, between the 'end' points and the walls.

If the grid of the measurement points coincides with that of the lighting fittings, large errorsare possible and the number of measurement points should be increased to avoid such anoccurrence.

STEP 1 Measure the floor area of the interior: Area = -------------------- m²

STEP 2 Calculate the Room Index RI = -----------------------

STEP 3 Determine the total circuit watts of the installation by a power Total circuit watts = --------meter if a separate feeder for lighting is available. If the actualvalue is not known a reasonable approximation can be obtained bytotaling up the lamp wattages including the ballasts:

STEP 4 Calculate Watts per square metre, Value of step 3 ÷ value of step 1 W/m² = ----------------------

STEP 5 Ascertain the average maintained illuminance by using lux meter, Eav. Maintained Eav.maint. = ----------------

STEP 6 Divide 5 by 4 to calculate lux per watt per square Metre Lux/W/m² = ---------------

STEP 7 Obtain target Lux/W/m² lux for type of the type ofinterior/application and RI (2): Target Lux/W/m² =

STEP 8 Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ). ILER =

10.5.2 Calculation of the Installed Load Efficacy and Installed Load Efficacy Ratio of aGeneral Lighting Installation in an Interior



The principal difference between the targets for Commercial and Industrial Ra: 40-85(Cols.2 & 3) of Table 10.2 is the provision for a slightly lower maintenance factor for the lat-ter. The targets for very clean industrial applications, with Ra: of 40 -85, are as column 2.

10.5.3 ILER Assessment

Compare the calculated ILER with the information in Table 10.3.

Ra : Colour rendering index

TABLE 10.3 INDICATORS OF

PERFORMANCE

ILER Assessment

0.75 or over Satisfactory to Good

0.51 – 0.74 Review suggested

0.5 or less Urgent action required

TABLE 10.2 Target lux/W/m² (W/m²/100lux) values for

maintained illuminance on horizontal

plane for all room indices and applica-

tions:

ILER Ratios of 0.75 or more may be considered to be satisfactory. Existing installationswith ratios of 0.51 - 0.74 certainly merit investigation to see if improvements are possible. Ofcourse there can be good reasons for a low ratio, such as having to use lower efficacy lamps orless efficient luminaires in order to achieve the required lighting result -but it is essential tocheck whether there is a scope for a more efficient alternative. Existing installations with anILER of 0.5 or less certainly justify close inspection to identify options for converting theinstallation to use more efficient lighting equipment.



Having derived the ILER for an existing lighting installation, then the difference betweenthe actual ILER and the best possible (1.0) can be used to estimate the energy wastage. For agiven installation:

Annual energy wastage (in kWh)= (1.0 - ILER) x Total load (kW) x annual operating hours (h)

This process of comparing the installed load efficacy (ILE) with the target value for theRoom Index and type of application can also be used to assess the efficiency of designs for newor replacement general lighting installations. If, when doing so, the calculated ILE (lux/W/m²)is less than the target value then it is advisable to ascertain the reasons. It may be that therequirements dictate a type of luminaire that is not as efficient as the best, or the surfacereflectances are less than the normal maxima, or the environment is dirty, etc., Whatever thereasons, they should be checked to see if a more efficient solution is possible.

10.6 Example of ILER Calculation (for the room as mentioned in paragraph 10.5.1)

STEP 1 Measure the floor area of the interior: Area = 45 m²

STEP 2 Calculate the Room Index RI = 1.93

STEP 3 Determine the total circuit watts of the installation by a powermeter if a separate feeder for lighting is available. If the actualvalue is not known a reasonable approximation can be obtained bytotaling up the lamp wattages including the ballasts:

STEP 4 Calculate Watts per square metre, 3 ÷1 : W/m² = 22

STEP 5 Ascertain the average maintained illuminance, Eav. Maintained(average lux levels measured at 18 points) Eav.maint. = 700

STEP 6 Divide 5 by 4 to calculate the actual lux per watt per square Metre Lux/W/m² = 31.8

STEP 7 Obtain target Lux/W/m² lux for type of the type of interior/application and RI (2):(Refer Table 10.2) Target Lux/W/m² = 46

STEP 8 Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ). ILER = 0.7

Total circuit watts = 990 W

Referring to table 3, ILER of 0.7 means that there is scope for review of the lighting system. Annual energy wastage = (1 - ILER) x watts x no. of operating hours

= (1 - 0.7) x 990 x 8 hrs/day x 300 days= 712 kWh/annum

10.7 Areas for Improvement

• Look for natural lighting opportunities through windows and other openings

• In the case of industrial lighting, explore the scope for introducing translucent sheets

• Assess scope for more energy efficient lamps and luminaries

• Assess the scope for rearrangement of lighting fixtures



10.8 Other Useful Information

10.8.1 IES - Recommendations

The Illuminating Engineering Society (IES) has published illuminance recommendations forvarious activities. These tables cover both generic tasks (reading, writing etc), and 100's ofvery specific tasks and activities (such as drafting, parking, milking cows, blowing glass andbaking bread).

All tasks fall into 1 of 9 illuminance categories, covering from 20 to 20,000 lux, (2 to 2000foot candles). The categories are known as A - I, and each provide a range of 3 iluminance val-ues (low, mid and high). See Table 10.4.

ACTIVITY CATEGORY LUX FOOTCANDLES

Public spaces with dark surroundings A 20-30-50 2-3-5

Simple orientation for short temporary visits B 50-75-100 5-7.5-10

Working spaces where visual tasks are only C 100-150-200 10-15-20occasionally performed

Performance of visual tasks of high contrast D 200-300-500 20-30-50or large size

Performance of visual tasks of medium E 500-750-1000 50-75-100contrast or small size

Performance of visual tasks of low contrast F 1000-1500-2000 100-150-200or very small size

Performance of visual tasks of low contrast G 2000-3000-5000 200-300-500or very small size over a prolonged period

Performance of very prolonged and exacting H 5000-7500-10000 500-750-1000visual tasks

Performance of very special visual tasks of I 10000-15000-20000 1000-1500-2000extremely low contrast

TABLE 10.4 IES ILLUMINANCE CATEGORIES AND VALUES - FOR GENERIC INDOOR

ACTIVITIES

A-C for illuminances over a large area (i.e. lobby space)

D-F for localized tasks

G-I for extremely difficult visual tasks

10.8.2 Example Using IES Recommendations

Let us determine the appropriate light level for a card file area in a library.

Step 1: The visual task is reading card files in a library. A number of tasks are accomplished inthe room. In such a cases, a category is chosen based on the generic descriptions in the IESIlluminance Category and Illuminance table discussed in step 3. For example, offices will usu-ally require Category E: 500-750-1000 lux.



Step 2: More detailed task descriptions are given in the recommended illuminance level tablesin the IES Handbook. (For an intensive lighting survey) Under the task category "Libraries,"subheading "Card files," the illuminance category is E.

Step 3: From the IES Illuminance Category and Ranges table, find category E and choose 500-750-1000 lux for the range of illuminance recommended. The first column in the table is illuminancevalues in units of lux, the metric version of footcandle. Notice that categories A through C are forgeneral illumination throughout the area, but D through I are for illuminance on the task. CategoriesG through I would require a combination of general lighting and task lighting.

Step 4: Use the weighting factors to decide which of the values in the illuminance range to use.Since libraries are public facilities, there may be many individuals over 55 years of age so selectthe category 'Over 55' for a weighting factor of +1.

Next, decide whether the demand for speed and accuracy is not important, important or crit-ical. Filing of cards correctly is not a critical activity, so the weighting factor of zero (0) isselected. An example of critical might be drafting work. The task background reflectance forblack type on a white page is 85%. So choose "greater than 70 percent" for a weighting factorof -1. The total weighting factor is 0. So use the middle recommended illuminance, or 750 lux.

For more detailed information on this the IES handbook may be referred.

10.8.3 Characteristics of Different Types of Lamps

Type Lamp Lumens Lamp Efficiency Choke Life of Capacitor Colorof Wattage (Lumens/Watt) Rating Lamp Rating Rendering

Lamp (Watts) (Watts) (Hours) Required Index(Microfarads)

HPSV 70 5600 80 13 15000 - 0.2 - 0.39 1220000

HPSV 150 14000 93 20 15000 - 0.2 - 0.39 2020000

HPSV 250 25000 100 20 15000 - 0.2 - 0.39 3220000

HPSV 400 47000 118 40 15000 - 0.2 - 0.39 4520000

HPSV 70 --- --- --- --- --- ---Super

HPSV 100 9500 95 18 15000 - 0.2 - 0.39 ---Super 20000

HPSV 150 15500 103 20 15000 - 0.2 - 0.39 ---Super 20000

HPSV 250 30000 120 25 15000 - 0.2 - 0.39 ---Super 20000

HPSV 400 54000 129 40 15000 - 0.2 - 0.39 ---



Super 20000

HPSV 600 --- --- --- --- --- ---Super

HPMV 80 3400 43 9 4000 - 0.6 - 0.69 85000

HPMV 125 6300 50 12 4000 - 0.6 - 0.69 105000

HPMV 250 13000 52 16 4000 - 0.6 - 0.69 185000

HPMV 400 22000 55 25 4000 - 0.6 - 0.69 185000

Metal 70 4200 84 26 10000 0.9 - 0.93 ---Halide

Metal 150 10500 70 20 10000 0.9 - 0.93 ---Halide

Metal 250 19000 76 25 10000 0.9 - 0.93 ---Halide

Metal 400 31000 76 60 10000 0.9 - 0.93 ---Halide

Metal 1000 80000 80 65 10000 0.9 - 0.93 ---Halide

FTL 40 2400 60 15 4400 0.8 - 0.89 3.2 - 3.8

FTL 36 3250 90 5 14000 0.8 - 0.89 3.2 - 3.8Super



QUESTIONS

1) What is circuit watts?

2) Define ILER and its significance.

3) Distinguish between lux and lumens.

4) What do you understand by the term colour rendering index?

5) Define room index?

6) For a room of length 10 m and width 20 m, calculate room index?

7) For a room of 9 x 6 m, determine the appropriate number of measuring points for luxlevels?

8) What possible improvement measures you would look for in a general lighting sys-tem?

9) Which of the following lamps has the maximum lamp efficiency?(lumes/Watt) a) Metal Hallide b) Fluorescent c) Incandescent d) HPSV

REFERENCES 1. Illumination engineering for energy efficient luminous environments by Ronald N.

Helms, Prentice-Hall, Inc. 2. The 'LIGHTSWITCH' programme, Energy Saving Trust, UK

11. PERFORMING FINANCIAL ANALYSIS


11.1 Introduction

When planning an energy efficiency or energy management project, the costs involvedshould always be considered. Therefore, as with any other type of investment, energy man-agement proposals should show the likely return on any capital that is invested. Consider thecase of an energy auditor who advises the senior management of an organisation that capi-tal should be invested in new boiler plant. Inevitably, the management of the organisationwould ask:

• How much will the proposal cost?• How much money will be saved by the proposal?

These are, of course, not unreasonable questions, since within any organisation there aremany worthy causes, each of which requires funding and it is the job of senior management toinvest in capital where it is going to obtain the greatest return. In order to make a decision aboutany course of action, management needs to be able to appraise all the costs involved in a pro-ject and determine the potential returns.

This however, is not quite as simple as it might first appear. The capital value of plantor equipment usually decreases with time and it often requires more maintenance as it getsolder. If money is borrowed from a bank to finance a project, then interest will have to bepaid on the loan. Inflation too will influence the value of any future energy savings thatmight be achieved. It is therefore important that the cost appraisal process allows for allthese factors, with the aim of determining which investments should be undertaken, and ofoptimising the benefits achieved. To this end a number of accounting and financial appraisaltechniques have been developed which help energy managers and auditors make correct andobjective decisions.

The financial issues associated with capital investment in energy saving projects are inves-tigated in this chapter. In particular, the discounted cash flow techniques of net present valueand internal rate of return are discussed in detail.

11.2 Fixed and Variable Costs

When appraising the potential costs involved in a project it is important to understand the dif-ference between fixed and variable costs. Variable costs are those which vary directly with theoutput of a particular plant or production process, such as fuel costs. Fixed costs are those costs,which are not dependent on plant or process output, such as site-rent and insurance. The totalcost of any project is therefore the sum of the fixed and variable costs. Example 1 illustrateshow both fixed and variable costs combine to make the total operating cost.

Example 1

The capital cost of the DG set is Rs.9,00,000, the annual output is 219 MWh, and the mainte-nance cost is Rs.30,000 per annum. The cost of producing each unit of electricity is 3.50Rs./kWh. The total cost of a diesel generator operating over a 5-year period, taking into con-sideration both fixed and variable cost is:

From Example 1, it can be seen that the fixed costs represent 21.5% of the total cost. In fact,the annual electricity output of 219 MWh assumes that the plant is operating with an averageoutput of 50 kW. If this output were increased to an average of 70 kW, then the fuel cost wouldbecome Rs. 53,65,500, with the result that the fixed costs would drop to 16.37% of the total.Thus the average unit cost of production decreases as output increases.

The concept of fixed and variable costs can be used to determine the break-even pointfor a proposed project. The break-even point can be determined by using the followingequation.

11. Performing Financial Analysis


Item Type of cost Calculation Cost

Capital cost of generator Fixed - 9,00,000

Annual maintenance Fixed 30,000 x 5 (years) 1,50,000

Fuel cost Variable 219,000 x 3.50 x 5 3,83,2500

Total cost 4,88,2500

Example 2

If the electricity bought from a utility company costs an average of Rs.4.5/kWh, the break-even point for the generator described in Example 1, when the average output is 50 kW isgiven by:

4.5 x 50 x n = (9,00,000 + 150000) + (3.5 x 50 x n)

n = 21000 hours

If the average output is 70 kW, the break-even point is given by:

4.5 x 70 x n = (9,00,000 + 150000) + (3.50 x 70 x n)

n = 15000 hours

Thus, increasing the average output of the generator significantly reduces the break-eventime for the project. This is because the capital investment (i.e. the generator) is being betterutilised.



11.3 Interest Charges

In order to finance projects, organizations often borrow money from banks or other leadingorganizations. Projects financed in this way cost more than similar projects financed fromorganisation's own funds, because interest charges must be paid on the loan. It is thereforeimportant to understand how interest charges are calculated. Interest charges can be calculatedby lending organization in two different ways: simple interest and compound interest.

(i) Simple interest: If simple interest is applied, then charges are calculated as a fixed per-centage of the capital that is borrowed. A fixed interest percentage is applied to each year ofthe loan and repayments are calculated using the equation.

(ii) Compound interest: Compound interest is usually calculated annually (although this isnot necessarily the case). The interest charged is calculated as a percentage of the outstandingloan at the end of each time period. It is termed 'compound' because the outstanding loan isthe sum of the unpaid capital and the interest charges up to that point. The value of the totalrepayment can be calculated using the equation.

Example 3

A company borrows Rs.3,00,00,00 to finance a new boiler installation. If the interest rate is10% per annum and the repayment period is 5 years, let us calculate the value of the totalrepayment and the monthly repayment value, assuming (i) simple interest and (ii) compoundinterest.

(i) Assuming simple interest:

Total repayment = 30,00,000 + (10/100 x 30,00,000 x 5) = Rs.45,00,000Monthly repayment = 45,00,000 / (5 x 12) = Rs.75,000

(ii) Assuming compound interest

Repayment at end of year 1 = 30,00,000 + (10/100 x 30,00,000) = Rs.33,00,000

Repayment at end of year 2 = 33,00,000 + (10/100 x 33,00,000) = Rs.36,30,000

Similarly, the repayments at the end of years 3, 4 and 5 can be calculated:

Repayment at end of year 3 = Rs. 39,93,000

Repayment at end of year 4 = Rs. 43,92,300

Repayment at end of year 5 = Rs. 48,31530

Alternatively, the following equation can be used to determine the compound interest repay-ment value.

Total repayment value = 30,00,000 x (1 + 10 / 100)5 = Rs.48,31,530

4831530Monthly repayment = = Rs.80,525

5 x 12

It can be seen that by using compound interest, the lender recoups an additional Rs.33,1530.It is not surprisingly lenders usually charge compound interest on loans.

11.4 Simple Payback Period

This is the simplest technique that can be used to appraise a proposal. The simple payback peri-od can be defined as 'the length of time required for the running total of net savings beforedepreciation to equal the capital cost of the project'. In theory, once the payback period hasended, all the project capital costs will have been recouped and any additional cost savingsachieved can be seen as clear 'profit'. Obviously, the shorter the payback period, the moreattractive the project becomes. The length of the maximum permissible payback period gener-ally varies with the business culture concerned. In some companies, payback periods in excessof 3 years are considered acceptable.

The payback period can be calculated using the equation



The annual net cost saving (AS) is the least savings achieved after all the operational costs havebeen met. Simple payback period is illustrated in Example 4.



Example 4

A new small cogeneration plant installation is expected to reduce a company's annual energybill by Rs.4,86,000. If the capital cost of the new boiler installation is Rs.22,20,000 and theannual maintenance and operating costs are Rs. 42,000, the expected payback period for theproject can be worked out as.

Solution

PB = 22,20,000 / (4,86,000 – 42,000) = 5.0 years

11. 5 Discounted Cash Flow Methods

The payback method is a simple technique, which can easily be used to provide a quick evalu-ation of a proposal. However, it has a number of major weaknesses:

• The payback method does not consider savings that are accrued after the payback periodhas finished.

• The payback method does not consider the fact that money, which is invested, shouldaccrue interest as time passes. In simple terms there is a 'time value' component to cashflows. Thus Rs.1000 today is more valuable than Rs.1000 in 10 years' time.

In order to overcome these weaknesses a number of discounted cash flow techniques havebeen developed, which are based on the fact that money invested in a bank will accrue annualinterest. The two most commonly used techniques are the 'net present value' and the 'internalrate of return' methods.

Net Present Value Method

The net present value method considers the fact that a cash saving (often referred to as a'cash flow') of Rs.1000 in year 10 of a project will be worth less than a cash flow of Rs.1000in year 2. The net present value method achieves this by quantifying the impact of time onany particular future cash flow. This is done by equating each future cash flow to its currentvalue today, in other words determining the present value of any future cash flow. The pre-sent value (PV) is determined by using an assumed interest rate, usually referred to as a dis-count rate. Discounting is the opposite process to compounding. Compounding determinesthe future value of present cash flows, where" discounting determines the present value offuture cash flows.

In order to understand the concept of present vale, consider the case described in Example 3.If instead of installing a new cogeneration system, the company invested Rs.22,20,000 in a

bank at an annual interest rate of 8%, then:

The value of the sum at the end of year 1 = 22,20,000 + (0.08 x 22,20,000) = Rs.23,97,600

The value of the sum at the end of year 2 = 23,97,600 + (0.08 x 23,97,600) = Rs.25,89,408

The value of the investment would grow as compound interest is added, until after n yearsthe value of the sum would be:

Example :

The future value of the investment made at present, after 5 years will be:

FV = 22,20,000 x (1 + 8/100)5 = Rs.32,61,908.4

So in 5 years the initial investment of 22,20,000 will accrue Rs.10,41,908.4 in interest and willbe worth Rs.32,61,908.4. Alternatively, it could equally be said that Rs.32,61908.4 in 5 yearstime is worth Rs.22,20,000 now (assuming an annual interest rate of 8%). In other words thepresent value of Rs.32,61,908.40 in 5 years time is Rs.22,00,000 now.

The present value of an amount of money at any specified time in the future can be deter-mined by the following equation.



The net present value method calculates the present value of all the yearly cash flows (i.e.capital costs and net savings) incurred or accrued throughout the life of a project, and summatesthem. Costs are represented as a negative value and savings as a positive value. The sum of allthe present values is known as the net present value (NPV). The higher the net present value,the more attractive the proposed project.

The present value of a future cash flow can be determined using the equation above.However, it is common practice to use a discount factor (DF) when calculating present value.The discount factor is based on an assumed discount rate (i.e. interest rate) and can be deter-mined by using equation.

DF = (1 + IR/100)–n

The product of a particular cash flow and the discount factor is the present value.

PV = S x DF

The values of various discount factors computed for a range of discount rates (i.e. interest rates)are shown in Table 11.1. The Example 5 illustrates the process involved in a net present valueanalysis.

Example 5

Using the net present value analysis technique, let us evaluate the financial merits of the proposedprojects shown in the Table below. Assume an annual discount rate of 8% for each project.



TABLE 11.1COMPUTED DISCOUNT FACTORS

Discount rate % (or interest rate %)

Year 2 4 6 8 10 12 14 16

0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

1 0.980 0.962 0.943 0.926 0.909 0.893 0.877 0.862

2 0.961 0.825 0.890 0.857 0.826 0.797 0.769 0.743

3 0.942 0.889 0.840 0.794 0.751 0.712 0.675 0.641

4 0.924 0.855 0.792 0.735 0.683 0.636 0.592 0.552

5 0.906 0.822 0.747 0.681 0.621 0.567 0.519 0.476

6 0.888 0.790 0.705 0.630 0.564 0.507 0.456 0.410

7 0.871 0.760 0.665 0.583 0.513 0.452 0.400 0.354

8 0.853 0.731 0.627 0.540 0.467 0.404 0.351 0.305

9 0.837 0.703 0.592 0.500 0.424 0.361 0.308 0.263

10 0.820 0.676 0.558 0.463 0.386 0.322 0.270 0.227

11 0.804 0.650 0.527 0.429 0.350 0.287 0.237 0.195

12 0.788 0.625 0.497 0.397 0.319 0.257 0.208 0.168

13 0.773 0.601 0.469 0.368 0.290 0.229 0.182 0.145

14 0.758 0.577 0.442 0.340 0.263 0.205 0.160 0.125

15 0.743 0.555 0.417 0.315 0.239 0.183 0.140 0.108

16 0.728 0.534 0.394 0.292 0.218 0.163 0.123 0.093

17 0.714 0.513 0.371 0.270 0.198 0.146 0.108 0.080

18 0.700 0.494 0.350 0.250 0.180 0.130 0.095 0.069

19 0.686 0.475 0.331 0.232 0.164 0.116 0.083 0.060

20 0.673 0.456 0.312 0.215 0.149 0.104 0.073 0.051

Project – 1 Project – 2

Capital cost (Rs.) 30 000.00 30 000.00

Year Net annual saving (Rs.) Net annual saving (Rs.)

1 +6 000.00 +6 600.00

2 +6 000.00 +6 600.00

3 +6 000.00 +6 300.00

Solution

The annual cash flows should be multiplied by the annual discount factors for a rate of 8% todetermine the annual present values, as shown in the Table below:



4 +6 000.00 +6 300.00

5 +6 000.00 +6 000.00

6 +6 000.00 +6 000.00

7 +6 000.00 +5 700.00

8 +6 000.00 +5 700.00

9 +6 000.00 +5 400.00

10 +6 000.00 +5 400.00

Total net saving at +60 000.00 + 60 000.00end of year 10

Year Discount Project 1 Project 2Factor for Net Present Net Present

8% savings value (Rs.) savings value (Rs.)(a) (Rs.) (a x b) (Rs.) (a x c)

(b) (c)

0 1.000 –30 000.00 –30 000.00 –30 000.00 –30 000.00

1 0.926 +6 000.00 +5 556.00 +6 600.00 +6 111.60

2 0.857 +6 000.00 +5 142.00 +6 600.00 +5 656.20

3 0.794 +6 000.00 +4 764.00 +6 300.00 +5 002.20

4 0.735 +6 000.00 +4 410.00 +6 300.00 +4 630.50

5 0.681 +6 000.00 +4 086.00 +6 000.00 +4 086.00

6 0.630 +6 000.00 +3 780.00 +6 000.00 +3 780.00

7 0.583 +6 000.00 +3 498.00 +5 700.00 +3323.10

8 0.540 +6 000.00 +3 240.00 +5 700.00 +3 078.00

9 0.500 +6 000.00 +3 000.00 +5 400.00 +2 700.00

10 0.463 +6 000.00 +2 778.00 +5 400.00 +2 500.20

NPV = +10 254.00 NPV = +10 867.80

It can be seen that over a 10 year life-span the net present value for Project 1 isRs.10,254.00, while for Project 2 it is Rs.10,867.80. Therefore Project 2 is the preferentialproposal.

The whole credibility of the net present value method depends on a realistic prediction offuture interest rates, which can often be unpredictable. It is prudent therefore to set the discountrate slightly above the interest rate at which the capital for the project is borrowed. This willensure that the overall analysis is slightly pessimistic, thus acting against the inherent uncertainties in predicting future savings.

Internal rate of return method

It can be seen from Example 5 that both projects returned a positive net present value over 10years, at a discount rate of 8%. However, if the discount rate were reduced there would come apoint when the net present value would become zero. It is clear that the discount rate whichmust be applied, in order to achieve a net present value of zero, will be higher for Project 2 thanfor Project 1. This means that the average rate of return for Project 2 is higher than for Project1, with the result that Project 2 is the better proposition.



Example 6 illustrates how an internal rate of return analysis is performed.

Example 6

A proposed project requires an initial capital investment of Rs.20 000. The cash flows generat-ed by the project are shown in the table below:

Year Cash flow (Rs.)

0 –20,000.00

1 +6000.00

2 +5500.00

3 +5000.00

4 +4500.00

5 +4000.00

6 +4000.00

Given the above cash flow data, let us find out the internal rate of return for the project.

It can clearly be seen that the discount rate which results in the net present value being zerolies somewhere between 12% and 16%.

For12% discount rate, NPV is positive; for 16% discount rate, NPV is negative. Thusfor some discount rate between 12 and 16 percent, present value benefits are equated topresent value costs. To find the value exactly, one can interpolate between the two ratesas follows:

459.5Internal rate of return = 0.12 + (0.16 – 0.12) x

(459.5 – (–1508.5))

459.5Internal rate of return = 0.12 + (0.16 – 0.12) x = 12.93%

(459.5 + 1508.5)

Thus the internal rate of return for the project is 12.93 %. At first sight both the net presentvalue and internal rate of return methods look very similar, and in some respects are. Yet thereis an important difference between the two. The net present value method is essentially a com-parison tool, which enables a number of projects to be compared, while the internal rate ofreturn method is designed to assess whether or not a single project will achieve a target rate ofreturn.

Profitability index

Another technique, which can be used to evaluate the financial viability of projects, is the prof-itability index. The profitability index can be defined as:



Cash 8% discount rate 12% discount rate 16% discount rateflow Discount Present Discount Present Discount Present(Rs.) factor value factor value factor value(Rs.) (Rs.) (Rs.)

0 –20000 1.000 –20000 1.000 –20000 1.000 –20000

1 6000 0.926 5556 0.893 5358 0.862 5172

2 5500 0.857 4713.5 0.797 4383.5 0.743 4086.5

3 5000 0.794 3970 0.712 3560 0.641 3205

4 4500 0.735 3307.5 0.636 3862 0.552 2484

5 4000 0.681 2724 0.567 2268 0.476 1904

6 4000 0.630 2520 0.507 2028 0.410 1640

NPV = 2791 NPV = 459.5 NPV = –1508.5

Solution

x 100

x 100

The application of profitability index is illustrated in Example 7.

Example 7

Determine the profitability index for the projects outlined in Example 5

10254For Project 1: Profitability index = = 0.342

30,000

10867For Project 2: Profitability index = = 0.362

30,000

Project 2 is therefore a better proposal than Project 1.

11.6 Factors Affecting Analysis

Although the Examples 5 and 6 illustrate the basic principles associated with the financialanalysis of projects, they do not allow for the following important considerations:

• The capital value of plant and equipment generally depreciates over time• General inflation reduces the value of savings as time progresses. For example, Rs.1000

saved in 1 year's time will be worth more than Rs.1000 saved in 10 years time.

The capital depreciation of an item of equipment can be considered in terms of its salvagevalue at the end of the analysis period. The Example 8 illustrates the point.

Example 8

It is proposed to install a heat recovery equipment in a factory. The capital cost of installing theequipment is Rs.20,000 and after 5 years its salvage value is Rs.1500. If the savings accrued bythe heat recovery device are as shown below, we have to find out the net present value after5 years. Discount rate is assumed to be 8%.



Data

Year 1 2 3 4 5

7000 6000 6000 5000 5000

Solution



Year Discount Capital Net PresentFactor for Investment Savings Value

8% (Rs.) (Rs.) (Rs.)(a) (b) (c) (a) x (b + c)

0 1,000 –20,000.00 –20,000.00

1 0.926 +7000.00 +6482.00

2 0.857 +6000.00 +5142.00

3 0.794 +6000.00 +4764.00

4 0.735 +6000.00 +3675.00

5 0.681 +1,500.00 +5000.00 +4426.50

NPV = +4489.50

It is evident that over a 5-year life span the net present value of the project is Rs.4489.50.Had the salvage value of the equipment not been considered, the net present value of the pro-ject would have been only Rs.3468.00.

Real value

Inflation can be defined as the rate of increase in the average price of goods and services. Insome countries, inflation is expressed in terms of the retail price index (RPI), which is deter-mined centrally and reflects average inflation over a range of commodities. Because of infla-tion, the real value of cash flow decreases with time. The real value of sum of money (S)realised in n years time can be determined using the equation.

RV = S x (1 + R/100)–n

Where RV is the real value of S realized in n years time. S is the value of cash flow in nyears time and R is the inflation rate (%).

As with the discount factor it is common practice to use an inflation factor when assessingthe impact of inflation on a project. The inflation factor can be determined using the equation.

IF = (1 + R/100)–n

The product of a particular cash flow and inflation factor is the real value of the cash flow.

RV = S x IF

The application of inflation factors is considered in Example 9.

Example 9

Recalculate the net present value of the energy recovery scheme in Example 8, assuming thediscount rate remains at 8% and that the rate of inflation is 5%.

Solution

Because of inflation; Real interest rate = Discount rate – Rate of inflationTherefore Real interest rate = 8 – 5 = 3%



Year Capital Net real Inflation Net real Real PresentInvestment Savings Factor Savings Discount Value

(Rs.) (Rs.) For 5% (Rs.) Factor (Rs.)For 3%

0 –20,000.00 1.000 –20,000.00 1.000 –20,000.00

1 +7000.00 0.952 +6664.00 0.971 +6470.74

2 +6000.00 0.907 +5442.00 0.943 +5131.81

3 +6000.00 0.864 +5184.00 0.915 +4743.36

4 +5000.00 0.823 +4145.00 0.888 +3654.12

5 +1500.00 +5000.00 0.784 +5096.00 0.863 +4397.85

NPV = +4397.88

The Example 9 shows that when inflation is assumed to be 5%, the net present value of theproject reduces from Rs.4489.50 to Rs.4397.88. This is to be expected, because general infla-tion will always erode the value of future 'profits' accrued by a project.



QUESTIONS

1. Why fresh investments are needed for energy conservation in industry ?

2. Cost of an heat exchanger is Rs.1.00 lakhs. Calculate simple pay back period consid-ering annual saving potential of Rs.60,000/- and annual operating cost ofRs.15,000/-.

3. What is the main draw back of simple pay back method?

4. Calculate simple pay back period for a boiler that cost Rs.75.00 lakhs to purchaseand Rs.5 lakhs per year on an average to operate and maintain and is expected tosave Rs.30 lakhs.

5. What are the advantages of simple pay back method?

6. What do you understand by the term " present value of money"?

7. Define ROI.

8. What is the objective of carrying out sensitivity analysis?

9. You are investing Rs.100 in a bank. The bank gives 10% interest per year for twoyears. What is the present value and what is the future value?

REFERENCES 1. Energy Management, Supply and Conservation, Dr. Clive Beggs, .Butterworth

Heinemann

Documents

Energy_Audit (Sir Has Said to Use Dis File for Audit)