Energy is the capability to produce motion ,force, work ,change in shape change in form etc….

Conventional Energy Sources:-

Coal , fire wood, water at high level ,Radioactive substances & Petroleum

Classification (Energy conversion)

1.Hydro Electric power station2.Steam power station3.Diesel power station4.Nuclear power station.

HYDRO ELECTRIC POWER STATIONPOTENTIAL ENERGY—ELECTRICAL ENERGY.SITE SELECTION FACTORS1. Adequate water available at good head.2.continuous water during summer ,therefore convenient

accommodation for a erection of dam must be available.3.The reservoir must have a large catchment area ,so that level of

water in the reservoir may not fall below the minimum requirement.4.The land should be cheap in cost and rocky in order to stand the

weight of large building and heavy machines.5.Sufficient transportation facility must be available.6.There should be possibility of steam diversion during period of

construction

Hydro plant general block diagram

Turbine: P.E -----M.EAlternator: M.E----E.E

DamTurbin

ealternat

orPower house

Steam power station Heat energy -----Electrical energy.Site selection factors.Nearness of the load centre.Supply of waterAvailability of coalLand requirementTransportation FacilityAvailability of fuel.

Nuclear power station Heat energy -----steam energy-----mechanical

energy-----electrical energy. Site selection factorsAvailability of water.Distance from populated areaNearness to load centre.Availability of space for waste proposal

Reactor---fuel rod(uranium),Moderator (Graphite rods) , Control rods (Cadmium) & coolant (sodium metal)

Heat exchanger---Coolant gives up the heat to heat exchanger which is utilize in raising the steam. After giving up heat the coolant is again fed to the reactor.

Nuclear reactor

Heat Exchange

r

Turbine

alternator

Advantages:-The amount of fuel required is very small.Less space is requiredCan be located near the load centre.Most economical.

Dis Advantage.Capital cost is very high.The fuel is expensive and difficult to recover.The disposal of radio active waste is a big

problem.

Non Conventional Sources:-1.Geo-thermal energy2.Tidal Energy3.Wind Energy4.Magneto Hydrodynamic generation.

5.Solar energy.

Advantages of utilizing renewable energy sources

1.These resources are locally available and everlasting.

2.The conversion does not involve in emission of green houses

3.The power generation is at local level and does not involve in T&D losses.

4.The systems are modular in nature.

Geo Thermal energy:-Heat energy of interior earth -----Electrical energy.Line Diagram:-Earth surface/pipe line output steam

separator steam Turbine Alternator inverter D.C

Advantages:-It is versatile in its use.It is cheaperThe plants have highest annual load factor (85-

90%)Pollution less.

Tidal Energy:-Tide is periodic rise and fall of the water level

of sea which are carried by the action of sun and moon on the water of the earth.

Tide can be used to produce electrical power which is known as tidal power.

The water is above the mean sea level is called Flood tide.

The water level is below the mean sea level is called EBB tide.

The average tide range is 8.5m and maximum tide is 13.5m

LINE DIAGRAM:-

Differential head ----Hydraulic turbine----ac generator----ac/dc supply

TYPES:-1.Single-basin arrangement.2.Double basin arrangement

Single basin arrangement:-There is only one basin.The basin and sea are separated by a dam and

(water) flow between them is through slice-ways located conveniently along the dam.

HIGH TIDE:Sluice gate opened------sea water enter to

basin(till maximum reservoir level)--------sluice gate are closed/turbine opened----conversion take place-----energy

The possible tide projects:-Gulf of Kutch and campay and sunderbans

region of the bay of Bengals 

Magneto Hydro Dynamic Generator:-(MHD)Thermal energy -----electrical energy.Principle:-When an electric conductor moves across a

magnetic field ,a voltage is induced in it which produce an electric current.

(Here conductor may be solid /fluid/ionized gas/liquid metal)(helium /argon/potassium)

Line diagram:-Hot ionized gas----electrodes----dc-----inverter-----

ac 

Advantages:-Power generation is simple.Large power and temp handling capacity.No moving partsHigh reliability.Time taken for full cycle---45 sec 

Example:-An experimental power plant of 5MW

(thermal I/p) has been commissioned at Trichy

Wind EnergyIt is indirect solar electric technology.It is energy from the WINDSite selection consideration:-1. High annual average wind speed.2. Availability of anemometer data.3. Availability of wind curve at the proposed

site.4. Local Ecology.5. Distance to road or railway.6. Nature of ground.7. Favorable land cost

AdvantagesIt is renewable sources of energyNon polluting systemAvoid fuel provision and transport.On large scale costs can be competitive with

conventional electricity and lower costs could be achieved by mass production.

It is clean and plentiful source of energy.It produces hundreds of mega watts.It’s also used for water pumping,

telecommunication power supply and irrigation.

Line diagram:-Wind----turbine-----alternator----electrical energy.

Types:-1. Vertical axis machines(The axis of rotation is vertical, the blades also

be vertical)2. Horizontal axis machines(the axis of rotation is horizontal and in the

aeroturbine plane is vertical facing the wind.

Wind direction

To Grid Wind

ws speed csignal temp

o/p

Aero turbin

e

Gearing

couplingElectric Generat

or

Controller

Solar Energy:-It is energy from the sun.The energy in the form of heat and light.It is defined as the energy produced by the

sun and collected on earth.Solar –Components:-1.solar collector2.storage unit.Line diagramSolar----solar rays----solar collector----

electricity/heat.

Types:-

1.Solar photovoltaic system power plants.2.Solar concentration thermionic power

plants

Solar photovoltaic system:-It is a direct conversion of light to electricity.Solar radiation-----PV cell---dc current.(Each PV cell generate the electricity

approximately 6-12volts)Basic photovoltaic system for power

generation.1.Solar array.(rays---DC power)2.Blocking diode.3.Battery storage.4.inverter/converter5.Appropriate switches and circuit breaker

Basic Photovoltaic system integrated with power grid

Blocking Diode

Solar cell

array

Inverter/Converter

From Utility Feeder

Battery Storage

Local Load

Solar cell or photovoltaic cell is a large area electronic device that converts solar energy into electric energy by the photovoltaic effect.

Photovoltaic's are the only method of converting sunlight directly into electrical energy.

The efficiency of a photovoltaic system is measured as the ratio of electrical power produced to the energy of the incident solar radiation.

It strongly depends on the quality of the semi conducting materials used for the fabrication of solar cells.

SOLAR CELL :-

Solar concentration thermic power plants:-Principle:-Concentrating the sunrays using mirrors into

small area (focal point) where the produced high heat is used to generated steam and electricity.

Thermal solar energy can be used for water heating,cooling,drying ,water distillation,refrigeration and space heating and cooling

Advantages:;It is free pollution.The plant requires little maintenance or help

after setupIt is EconomicalWhen it is connected to the grid, solar energy

can overtake the highest cost electricity at peak demand and can also reduce grid loading,apart from getting rid of the need for local battery power in darkness.

Combined operation of power station.:-

In an interconnected power system which has number of power stations of different types operating in parallel it is necessary to co-ordinate the different station for the best possible economic operation.

Advantages of INTER CONNECTED OPERATION:-

1.Increased reliability of supply In the event of power failure at one station the system can be fed from other stations.

2.Reduction in total installed capacity,,The amount of reductions depends on the characteristics of inter connected system and desired degree of service reliability.The reseve capacity in any system is usually equal to the largest size of the plants in the system.

3.Economic operation4.Spinning reserve is reduced.

Distributed Generated System:-Distributed generation generates electricity

from many small energy sources.It is defined as the integrated use of small

generation units directly connected to a distribution system or inside the facilities of a customer.

It has also been called On site generationDispersed generationEmbedded generationDecentralized generation

DG is a another approach ,It reduces the amount of energy lost in transmitting electricity because the electricity is generated very near where it is used ,perhaps even in the same building..This also reduces the size and number of power lines that must be constructed.

Example—DG1.Micro turbine,Small Wind turbine2.Photovolatic cell ,solar water heating.3.Small gas turbine.Benefits-DG1.Enhance the reliability of the power

supply.2.Reduce the need for T & D investments

to serve the growing demand.3.Encourage efficient investments in

electricity reliability by offering a cost effective alternative in many situations.

UNIT IIEconomics of generation:-  The method of finding the per unit cost of

production of electrical energy is known as economics of power generation.

Connected load: It is a sum of continuous rating of all the

equipments connected to supply system. Maximum demand 

It is the greatest demand of load on the power station during a given Period

Demand factor:-D.F= (Maximum demand) / Connected loadThe value of demand factor is usually less than one

Average Load:-The average of load occurring on the power station in a given period is known as average load.

Load factor:-L.F=(Average load ) / (Maximum Demand) During

a given period.It is always less than one.If the load factor is higher the cost per unit of

generation is lesser.

Diversity factor:-D.F=(Sum of individual maximum demands)

-----------------------------------------------(Maximum demand on power station)

It is always greater than one.If the Diversity factor is higher the cost per unit of generation is lesser

Capacity factor:-C.F= Average demand

---------------------- Plant capacity

Reserve Capacity = Plant Capacity – Maximum demand.Plant use factor:-PUF= (Station output in kWh) / (Plant capacity x Hours of use)

Problem:1.A generating station has a connected load of 40MW and Maximum Demand of 20 MW,,The units generated being 60 x 106 Find

a.Demand factor. b) Load factor. solution D.F=0.5AD = unit generated per year / hours in a

year = 60 x 106 / 8760 =6849.31 KWL.F=0.3424 OR 34.24%

2. A generating station supplies the following loads :15000Kw,12000Kw,8500Kw,6000Kw and 450Kw.The station has a maximum demand of 22000kw.The annual load factor of the station is 48%

Calculate. a.the number of units supplied

annually. b.diversity factor. c.the demand factor.

Solution:-Demand factor = MD/CL=22000/41950 = 52.44%Diversity factor=Sum of individualMD / MD=1.906Number of units supplied per year = AL X Hours in a

year. = MD X LF X 8760

= 925 X 105 kWh3.A generating station is to supply four regions of

load whose peak loads are 10MW,5MW,8MW and 7MW.The diversity factor at the station is 1.5 and the average annual load factor is 60%. Calculate

1.MD on the station. 2.Annual energy supplied by the station,

suggest the installed Capacity and no.of units?

Solution:- MD=20MWUnits generation/yr = AL X Hours in a year. = MD X LF X 365 X24=105.12 X

106 kWhInstalled capacity = 1.2 x maximum

demand= 24 MWSuitable size of units = 6MW/plant Number of plants = 24 /6 =4 UNIVERSITY PROBLEM4.The maximum demand of power plant is

40MW .The capacity factor is 0.5 and the utilisation factor is 0.8

find 1.Load factor2.Plant Capacity 3.Reserve Capacity.

(Average load =24 MW)

Solution:-Capacity factor = units generated daily / plant

capacity x hours in a dayLoad factor = Average Load / Maximum

demand.=0.6Units generated daily = average load x 24=576 MWReserve capacity = Plant capacity –Maximum

demand.=8MW

Time 0-6 6-10 10-12 12-16 16-20 20-24

load 20 25 30 25 35 20

5.A generating station has the following daily load cycle, Draw a load curve and calculate a.MD b. units generated/day c. average load d. load factor.Ans35 MW,600000KWH ,25000KWH,&0.714

Load curveIt gives/show the load demand of a consumer against time in hours of the day is known as a load curve.

Load duration Curve:-When the load elements of a load curve are arranged in the Order of Descending magnitudes

Base Load:-The unvarying load which occurs almost the whole

day on the station is known as the base load.Peak Load:-The various peak demands of load over and above

the base load of the station is known as peak load

Cost of Electrical Energy:-The cost/unit of electrical energy generated

depends on the Investment in the plants Distribution systemsOperation &Maintenance cost.

Types:-1.Fixed cost2.Running cost (or) cost of energy.

Tariff:-Different methods of charging

consumers are known as tariff (or)

Rates of payment for consumption of electricity.

Objectives of tariff:-1.cost of operations, supplies , maintenance

and losses must be recovered.2.cost of capital investment in generation ,t&d

equipment must be recovered.3.tariff should be uniform.4.it should provide incentive for using power

during the off peak hours.5.it should have provision of penalty for low

power factors.

Desirable characteristics of tariff1.proper return. This will help the EB to ensure continuous and

reliable service to the consumer.2.simplicity.

It should be simple so that ordinary consumer can easily understand it.

3.fairnessA big consumer should be charged at a lower rate than a small consumer.

4.Attractive.The traiff should be attractive ,,so that large no of consumers are encouraged to use electrical energy,

Types of tariffs:- 1.simple tariffWhen there is a fixed rate per unit of energy

consumed.it does not vary with increase or decrease in no.of.units consumed.

2.flat rate tariffWhen different types of consumers are

charged different uniform per unit rates.3.block rate tariffWhen a given block of energy is charged at

a specified rate and succeeding block of energy are chaged at reduced at different rates.

4.two part tariffTariff is charged on the basis of

maximum demand on the consumer and the units consumed.

Total charge = Rs( b x Kw + C X KWh)5.maximum demand tariff.6.power factor tariff.7.three part tariff.Total charge = Rs( a + b x Kw + C X

kWh)

Problems1.The monthly reading of a consumer`s meter are

given asMaximum demand = 16kVAEnergy consumed =24000kWhReactive energy=15600 KvarIf the tariff is Rs 250 per kVA of maximum demand

plus Rs3.50 Per unit ..1% power factor penalty charge will be charged from the customer for every 0.01 drop power factor from the recommended value power factor 0.9.Find out over all cost.(monthly basis)

Solution:-Tanφ= kVAR / active powerAverage reactive power =reactive energy / total no of

hrs in a month

Average reactive power =15600 / 24 x 30 = 21.67 KvarActive power or active load =

Energy consumed /No.of hrs in a month.=24000/24 x 30 =33.33kW

Tanφ =21.67 /33.33 = 0.65Phase angle φ = 33.02Cosφ = 0.84Power factor diff from recommended value = 0.9 –

0.84 =0.06 = 6%Normal cost (with out penalty) = 16 x 250

+24000x3.50=88000Over all cost with penalty =88000 + .06

x88000=93280

CONSERVATIONPrinciple:-

Save money using less energy.

OR

Using energy more efficiently.

Need for EE ConservationEnergy demand in india –9 to 10 % per

annum.

Generating capacity --- 5 t0 6% per annum

Energy saved is energy generated(2 times)

Potential way for bridging the energy gap.

Effect of energy conservationOptimal utilization energy.

Prolong the usage of energy available in the earth.

Reduce the energy cost per unit of production.

Reduce the green house emission

Minimize the global warming.

Advantages of EE motorReduced power consumption for the

same mechanical load.Lower temperature rise and

increased service life.Efficiency curve flat between 50-

100% of load factor.A marginal increase in operating

speed.

Replacement policy for EE motorIdeal for new installation.

Replace with over sized motor.

Replace with 3 to 4 times rewound motor.

Consider up to 37kw motor.

Major parts affecting overall performance of the motor

Stator frameCoreWindingsRotorInsulation.Friction & wind age losses.

COST EFFECTIVINESS

OF ENERGY EFFICIENCY MOTOR

FACTORS STD MOTOR(IS8789) EE MOTOR(IS 12165)

RATING 15 KW 15KW

POLES 4 4

SPEED 1460 1480

EFFICIENCY 86 89

I/P POWER 13.39 12.64

MOTOR COST 22,000 30,000

ENERGY COST 428480 404480

PROBLEMS OF OVER SIZED MOTOR

1.HIGHER INVESTMENT COST.2.HIGHER RUNNING COST.3.HIGHER MAXIMUM DEMAND.4.HIGHER SWITCH GEAR COST.5.HIGHER SPACE REQUIREMENT.6.HIGHER INSTALLATION COST7.HIGHER REWINDING COST

NO

DESIGN IMPROVEMENTS

REDUCTION IN

1 More copper & larger rotor conductor

Copper loss

2 Thinner steel, longer rotor & stator core length, precise air gap

Iron loss

3 Improved lubricating system & efficient fans

Friction & windage loss

4. Optimum slot geometry & minimum overhang of stator

Stray load losses.

Economics of motor selection for an application. 

EEmotor is 20 t0 30% costlier than conventional induction motor.

The capacital cost of the motor=Initial cost+((1-ήat running load )x Load

factorxAnnuityfactorxECxoperating hours/Annum x motor rating)

WhereAnnuity factor = (1+i)n-1 / i(1+i)n 

I=rate of interest,, N= NO.of years of operation.EC=Energy cost(rs/kwh).

Energy Management:-The strategy of optimizing energy by using

approach and procedures to reduce energy requirement per unit of output,without compromising quality and quantity of the product.

Energy audit:-A technical analysis for identifying and

quantifying Energy utilisation pattern An attempt to balance total energy inputs

with its usesAn assessing mechanism to determine the

current performance the plant.

In generalEnergy audit translates the ENCON ideas in

to realities by blending technically feasible solutions with economic and other organizational considerations with in specified time frame.

 ENCON ----Energy conservation.

Why Energy Audit:-To Identify areas of Energy wastage and

means for rectifying the same. Acquire positive orientation towards

Energy cost reduction. Decide on appropriate energy mix. To chalk out financially sound

alternatives for reduced SEC Create Energy Conservation awareness

among workers.

Steps in Energy Auditing:-

1.Planning2.Basic Data collection3.Data Analysis4.Energy Conservation study5.Identification of Low / No cost Measures.6.Estimation of technical feasibility and Economic Viability of Energy saving proposals involving capital investment.7.Action plan / implementation Schedule.8.Energy Audit Report.

 

Economics of power factor improvement:-Cosφ = Real power inkw / Real + Reactive power in kVA

= Real power / Apparent power = W/VADisadvantage of low p.fLPF ,the current drawn from the source will

be higher and this leads to many problrmesLloss –directly proportional to i2

loss

Leads to excess voltage drop and needs extra voltage

regulating devices to maintain terminal voltage to the required level

How to Improve power factorInstalling static power capacitorsOperating Synchronous motor in over excited

condition.Installing static VAR components.Advantages of power factor improvement:-Reduce Maximum Demand(MD)Reduce T&D lossesReduce ohmic drop in T&D Systems. 

Method of Power Factor Compensation:-1.Centralized /Group compensation.

All the capacitors are connected in a single place at main distribution board or substation and controlled depending upon the reactive power requirement.

2.Distributed /Individual compensation.Capacitors are connected in parallel with the all

the inductive equipments.The reactive current drawn by the inductive

equipment will be compensated at the terminal of the equipment.

Since the reactive current requirement is compensated on the spot.the line loss is reduced and voltage drop in the distribution system is less.

3.Mixed Compensation.Part of the capacitor used for compensation is

connected across the inductive equipment terminal and remain connected at the main distribution board or sub station.

POWER QUALITY

The supply of power with in permitted variation of voltage & frequency and with out any distortion of sinusoidal wave form in balanced condition.

Depends on

Voltage

Frequency

Harmonics.

Harmonics effect

Reduced equipment life timeReduced equipment energy efficiencyInterference with other electronics

devices.Interference with other electrical

measurement apparatusOver heating of motors, transformer,

Switchgear etc

UNIT IIIILLUMINATION

LIGHT:-The radiant energy from a hot body which

produces the visual sensation upon the human eye.

It is expressed in lumen-hours and is analogous to Watt-hour.Luminous flux:-

The total quantity of light energy emitted per second from a luminous body.

It is represented by symbol F or φ and is measured in lumens.

F or φ = Q / t

Luminous Intensity:-(I)It is the luminous flux emitted by the source

per unit solid angle measured in the required direction.

It is denoted by I and measured in candela.I = F/ωLamp Efficiency:-(Efficacy)It is defined as the ratio of luminous flux to

the power input.It is expressed in Lumens / watt.

Candle power:-It is the light rating capacity of a source in a

given direction.It is defined as the no.of lumens given out by

the source in a unit solid angle. CP = Lumens / ω

Specific consumption :-It is defined as the ratio of power input to

the average candle power.It is Expressed in Watt /Candle.

Space-Height ratio:-It is defined as the ratio of horizontal distance

between adjacent lamps and height of their mountings.

SH RATIO=Horizontal distance between adjacent lamps -----------------------------------------------------------

Mounting height of the lamps above working plane.

GLARE:-It is defined as the brightness with in the field of

vision of such a character as to cause annoyance,discomfort ,intferference with vision or eye – fatigue.

MEAN HORIZONTAL CANDLE POWER (MHCP)

It is defined as the mean of candle powers in all directions and on all planes from the light source.

MEAN HEMI SPHERICAL CANDLE POWER(MHCP)

It is defined as the mean of candle powers in all direction above or below the horizontal plane passing through the light source.

PLANE ANGLE:-When two straightlines lying in the same plane

meet at a point ,there will be an angle between these converging lines at the meeting point

The angle is termed as plane angle.It is represented in radians .Θ = ARC / RADIUS.SOLID ANGLE:-The angle subtended at a point in space by an

area is termed as solidangle.It is expressed in steradiansSolid angle ω= area / (radius)2

UTILIZATION FACTOR:-U.F=Total lumens on the working plane/Total

lumens emitted by the lamp.U.F various from 0.4 t0 0.6REFLECTION FACTOR :-R.F = Illumination when everything is

clean /Illumination under normal working condition.

Relation between Θ and ωω = 2π (1- cos Θ/2 )LAWS OF ILLUMINATION :-Inverse Square Law.Lamberts Cosine Law. 

Inverse Square Law;-The Illumination of a received surface is

inversely proportional to the square of the distance of the surface from the source of light provided.

E = I / r2

Lamberts Cosine law:-If a surface is inclined at an angle (90- Θ) to

the direction of luminous flux ,then the illumination of the surface is reduced from, that given according to inverse square law in the ratio Cos Θ/1

TUNGSTEN FILAMENT LAMP:-

Its provides good lighting interms of colour correctness.

Low initial cost.Easy to replace.Inefficient in energy use and have a

short life than other types.5% light 95% heat.

TUNGSTEN FILAMENT LAMP:-Principle:When a body is heated ,it emits radiant energy in the form of

light

The amount of light emitted is proportional to the absolute temp.

Filament –salient features

High melting point.

Sufficient mechanical strength.

Thickness is about 0.01 millimetres

It is covered with inert gas (85%Argon & 15% Nitrogen)

Life time of the filament(lamp) is 1000 hours.

FLUORESCENT LAMP:-

Provides good lighting interms of colour correctness.

Medium priced lampsMore energy efficientHave a longer life than incandescent

lampElongated shape.

Principle:-When a low pressure mercury gas start

conducting will radiate ultra violate rays,this uv rays is converted in to visible light with the help of phosphor coating coated on the glass tube.

The type of phosphor material used decides the conversion efficiency of ultra violate in to visible light.

Starting methodsApplying high voltage with the help of starter and

electromagnetic ballastApplying high frequency with the help of

electronics ballast.

COMPACT FLUORESCENT LAMP.

CFLs are smaller versions of standard fluorescent lamps

Power consumption between 5 to 40 watts

Brightness and color rendition comparable to incandescent lights.

CFLS can directly replace standard incandescent bulbs.

Lighting Design

More light does not mean better

light !!!Quality of light is equally important to

quantity of light.

Lighting design

What is purpose of Light?How much light I really need?What are long term energy cost? Identify

critical and non –critical areas.Use combination of narrow beam and wide

beam luminaries.Locate the fixture where it is required.Use color contrast to attract the attention

instead of higher light levels.

STEPS-INVOLVED IN DESIGN OF LIGHTING SYSTEM:-

Calculate area to be illuminated.Decide the level of illuminationTotal illumination = Area x illumination level.

Select Utilization factor and depreciation factor

Divide total illumination by utilization factor and depreciation Factor.

If D.F is greater than one ,then to find gross lumen ,it is multiplied With total lumen instead of division.

Select lamp and luminaries Calculate no. of lamps Decide arrangements of lamps for uniform distribution

Considering space height ratio

A hall measuring 20mx50m is to be illuminated by suitable lamps to give an average illumination of 45 lux .Mounting height from the working plane = 3 m,UF=0.65, DF=1.3

The lamps are to be choosen from the following groups

Watts 75 100 150 200Lumens 800 1200 2000 2800 Calculate the number of lamps of each type.&

arrangement position.

Step1: Calculate area to be illuminated.Area = 20 x 50 = 1000 (meter )2Step 2:Decide the level of illumination = 45 lux

(given)Step 3:Total illumination = Area x illumination

level = 1000 x 45 = 45000 lumensStep 4: Divide total illumination by utilization

factor and depreciation Factor.Gross lumens required =45000 x 1.3 / 0.65=

90,000(If D.F is greater than one ,then to find gross

lumen ,it is multiplied With total lumen instead of division.)

Lamp1 75 watts & 800 lumensRequired lamps=90,000/800 = 112.5 =112 0r 113 lampsArrangementsAssume 8 r0ws & 14 columns.

Lamp1 100 watts & 1200 lumensRequired lamps=90,000/1200 = 75 Assume 5 r0ws & 15 columnsLllly next two case answer is 45 & 32 lamps.

Building illumination levelLocation Illumination level

Entrance 100

Dining room 150

Kitchen 200

Bathroom 100

Stair 100

Study room 300

Bed room 300

A hall 30m long and 12 m wide is to be illuminated and illumination required 50 lumen / m2.calculate the number ,the wattage of each unit and location and mounting height of the units. taking a D.F of 1.3 and U.F of 0.5. given that the o/p of different types of lamp are given below.

Watts 100 200 300

Lumens 1615 3650 4700

Problems:-1.A point source of light is found to produce

on illumination 100 lux at a distance of x meters from the source and 1 lux at a distance of x+36 m from the source .Determine the two distance s and evaluate the luminous intensity of the source if the source has same intensity in all directions.

Solution :- Source---------------------A--------------------B----------- X---------------------------------------------------------X +36 -------

As per illumination first law E=I / d2 at point A E1 = I1 / X2

100 = I1 / X2----------(1)

At point B E2= I2 / (X +36)2

1 = I2 / (X +36)2---------(2)

Given I1 = I2

Compare (1) and (2)X = 4 mX+36 = 40 mDistance b/w source and point A = 4 mDistance b/w source and point B = 40m 

Solution: B

I O II

Two lamps one 200cp and another 500cp are hung at a height of 10metres and 25metres respectively. The horizontal distance between the poles is 80metres. Determine the illumination at the mid point between the poles on the ground.

A

From the figureTriangle OAI

HEATING:-When current passes through a medium (solid,

liquid & gas) heat is produced. H = I2Rt joules. H = I2Rt / 4186 KILO CALORIES 1 kcal = 4186 joules Domestic application:- Heater , Electric Iron ,Oven ,Electric drier

etc.Industrial application:- Electric welding, Melting of metals

&Enamelling of copper conductor etc.

WELDINGIt is the process of joining metal of similar

composition by heating to suitable temperature with or with out application of pressure and addition of filer material

PRESSURE WELDING:-The metal parts to be joined are heated to a

plastic state and then joined by applying external force is known as plastic or pressure welding.

FUSION WELDING:-The metal parts to be joined are heated to a

molten state and then allowed to solidify joining a localized homogeneous union of the two is known as fusion welding.

RESISTANCE WELDING

The process(pressure welding) make use of the heat generated by a heavy electric current passing through the material being joined.

The electrical resistance between the surface in contact being mainly responsible for the rise in temperature.

The heat generated bring these surfaces to the plastic state,at this point mechanical pressure is applied to squeeze the material together thus forming the joint.

TYPES:-SPOT WELDINGSEAM WELDINGPROJECTION WELDINGBUTT WELDINGFLASH WELDING

SPOT WELDING:-

The electric current is passed through the metal ,and the resistance to the flow of current produce heat in the metal up to welding temperature.

It is used to weld together two or more overlapping pieces.

Required elements:-Step down transformer It is used to reduce a high voltage low

current supply to low voltage high current.Two electrodes. Made of cadmium-copper or chromium

copper.—can carry high currents and have sufficient strength to work under high pressure.

Metals to be welded.

Operation:- When two pieces are held together ,pressing

two work pieces ,a large electric current flows from one electrode through the metals to be welded together to second electrodes.

The spot welding controlled by the current pressure of the electrodes and the duration of flow of current.

APPLICATIONS:- Automatic welding process Joining automatic body sections.,cabinets

and other sheet metal assemblies.

ELECTRIC ARC WELDING:- Electric energy is converted at the arc in

to heat energy.

PRINCIPLE:- The electric produced by bringing two

conductors connected to a suitable source of electric current ,momentarily in contact and then separating by a small distance ..the current continuous flow across a small gap and gives intense heat.

The heat developed is utilized by the metal part of the work piece and the filer metal and thus form the joint.

SETUP One terminal is connected to the

electrode(work piece1 or filer metal) and other to the work piece and the circuit is completed through air gap.

Distance between the electrode and work piece is 3mm to 6mm

Due to the interruption by the air gap or gas heat is produced and the temperature attained varies from 3700 to 4000

Current rate 30 to 600A Voltage rate 60 to 100v Temp range 3000 to 4000

APPLICATIONS:- Used for joining steel sheets. Used for repairing steel castings. Used for welding stainless steel Used for welding non-ferrous metals.

UNIT IVThe electric traction means a locomotion in

which the driving force is obtained from electric motors.

REQUIREMENTS:- 1.Equipment should be capable of over

loads for short periods. 2.The wear caused on the track should

be minimum. 3.Braking should be possible with out

excessive wear on brake shoes. 4.If possible ,the braking energy should

be generated and returned to the supply.

ADVANTAGES:-

It is the cheapest method than all other methods of traction.

It has smooth and rapid acceleration and braking.

DIS-ADVANTAGES:- Capital cost is high. Failure of supply is a problem to be

faced in electric traction. For achieving electric braking and

control ,additional equipment is required .In case of dc series motors ,the regenerative braking can not be easily achieved.

BASIC DEFINITIONS USED IN TRACTION.CREST SPEED(VC)Maximum speed attained by the train during run.

AVERAGE SPEED(VA)Maximum speed maintained by train during the

runAverage speed = Distance between the stops in

km / Actual time of run in hour.Va=D / T

SCHEDULE SPEED (VS)

VS= Distance b/w stop in km / (actual time run in hour + stop time in hour)

SPEED-TIME CURVE:-The curve between speed(km/hr-yaxis) and

time (sec-xaxis) is called speed time curve.ADVANTAGES:-Complete information about the motion of

train.Slope of curve at any point gives the speed at

the instant.The area under the curve gives total distance

travelled by the train.It consists of Acceleration period Free run period Coasting period Braking period (or) Retardation period.

Acceleration period:-Notching up(OA)Starting resistance is gradually cut out ,so that

motor current is limited to certain value and voltage across the motor gradually increased

Acceleration remains constant during this period.

Speed – Curve running.(AB)The current starts decreasing with increasing

speed according to the characteristics of series motor and finally the current taken by the motor becomes constant.

Acceleration decreases with increasing speed and becomes zero at point B

Free running period:-(BC)During this period (on level track )power

output from the driving axle balances the rate at which energy is expented against the resistance to motion.

At end of speed curve running train reaches maximum speed.

Coasting Period(CD)At the end of free running period power supply

to the motor is cut off and train is allowed to run under is own momentum.

Due to train resistance speed of the train gradually decreases

 

Retardation Period(DE)At the end of coasting period,brakes are

applied to bring the train to stop. 

BRAKING:-The electrical device is disconnected from

the supply and slows down and come to rest.TYPES OF BRAKING1.Mechanical braking. 2.Electrical braking.

Advantage of Electrical braking:1.Little maintenance is required.2.No dust is produced.3.Replacement & adjustment of brake shoe is

eliminated.4.Smooth braking can be obtained.5.Some energy feedback to supply.6.Heat is dissipated in a convenient place.

Dis Advantage of electrical braking.1.For some cases the rating of braking

equipment required is higher than the rating required for motor alone.

2.In some cases additional pieces of equipment is required.

3.Economical consideration.

Types-Electrical braking.1.Reverse current braking.(plugging)2.Rheostatic braking (dynamic braking)3.Regenerative braking.Plugging:-Motor now develops torque in opposite

direction(by interchanging the polarity of armature and field winding) to the movement of rotor .

System speed will decrease will zero speed is reached.

InefficientK.E of the moving masses is wasted.It requires additional energy from supply for

the development of counter torque.In EffectiveIn case of failure of supply ,this method of

braking becomes ineffective.Operation:-We have to reverse either magnetic field (or)

armature field.

UNIT- VDRIVES AND THEIR INDUSTRIAL

APPLICATIONS Introduction motor selection and related factors loads – types characteristics steady state and transient characteristics load equalization industrial applications modern methods of speed control of industrial

drives.

Types of Electric drivesIndividual Drives.  An single electric motor is used to drive one

individual machine.

Group Drives. It is a drive in which a single electric motor

drives a line of shaft by means of which an entire group of machine may be operated.

 

Advantages of Electrical Drives 1.They are clean in operation ,providing better

working conditions2.Low cost & high operating efficiency.3.Electric drives are more economical as they can be

designed as per requirements. There is less wastage of power and consequently less power consumption.

4.Suitable for group drives as well as individual drives .Even high speed machines can be directly coupled to the motor ,with out needing gear arrangements.

5.Electric motors are compact units ,strong in construction and required comparatively less maintenance and repairs.

 

SL.NO GROUP DRIVE INDIVIDUAL DRIVE

1 Initial cost is low Initial cost is high

2 Power consumption is low

Comparatively higher consumption of power

3 Less space is required

More space is required

4. It has low power factor

It has high power factor.

5 Constant speed is not possible

Constant speed is possible

6 It is not reliable It is more reliable7 Efficiency is low Efficiency is high.

Factors affecting choice of motors The various factors that are to be considered

in the selection of electric drive for a particular application are

Electrical CharacteristicsMechanical Characteristics.

a.Types of Enclosure.b.Types of transmission for Drives.

LOAD TYPES:-Active Load

Active loads are those which are due to the force of gravity, tension

It is independent of loads.Example;-Paper mill drive, fans

&compressorsPassive Load

Passive loads are those which are due to friction, cutting and deformation of in elastic bodies

Example:-High speed hoists,traction,centrifugal pumps

Load Characteristics1. Static characteristics2. Dynamic Characteristics.

Electrical characteristics:-1. Running Characteristics2. Starting characteristics3. Speed control4. Braking.

Mechanical Characteristics:-1.Types of enclosures2.Types of Transmission for Drive.

Load Equalization:;Certain load fluctuates very widely with in

short interval of time ,(like electric hammers, reciprocating pumps