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Electrical Estimation & Energy Auditing

Question Code – 612

October – 2019

Max. Marks: 75

[N.B: (1) Q.No.8 in PART – A and Q.No.16 in PART – B are compulsory. Answer any FOUR questions from the

remaining in each PART – A and PART – B.

(2) Answer division (a) or division (b) of each questions in PART – C.

(3) Each question carries 2 marks in PART – A, 3 marks in PART – B and 10 marks in PART – C.

(4) Cost of material may be assumed suitably

(5) Electrical estimation tables are permitted.)

PART - A

1. Draw the symbols of (i) UG cable (ii) Earth.

UG cable: Earth:

2. State the warning signs of safety.

Warning signs give safety information. These are triangular yellow signs with a black border. They give

warning of a hazard or danger.

3. Convert 10 HP into watts.

7460 watts

4. Define energy audit.

It is defined as "the verification, monitoring and analysis of use of energy including submission of

technical report containing recommendations for improving energy efficiency with cost benefit analysis and an

action plan to reduce energy consumption”.

5. Define colour rendering index.

It is the measure of effect of light on the colour of the object in a more natural form. Qualification of the

ability of the light source to render colour of surface accurately is colour rendering index.

6. Define luminous efficacy.

It is the ratio of luminous flux emitted by a lamp to the power consumed by the lamp.

Unit: lumens per watt (lm/W)

7. What is power factor controller?

It controls the power factor of the installation by giving signals to switch on or off power factor correction

capacitors. Relay is the brain of control circuit and needs contactors of appropriate rating for switching on/off the

capacitors.

8. What is soft starter?

Soft starter provides a reliable and economical solution to these problems by delivering a controlled release

of power to the motor, thereby providing smooth, stepless acceleration and deceleration. Motor life will be extended as damage to windings and bearings is reduced.

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PART – B

9. Write the IE rule with reference to voltage.

Except with the written consent of the consumer or with the previous sanction of the State Government

a supplier shall not permit the voltage at the point of commencement of supply as defined under rule 58 to vary

from the declared voltage.

LOW- Where the voltage does not exceed 250 volts under normal conditions.

MEDIUM- Where the voltage does not exceed 650 volts under normal conditions.

HIGH- Where the voltage does not exceed 22000 volts under normal conditions EXTRA HIGH-When the

voltage exceeds 22000 volts.

10. Briefly describe the necessity of earthing.

Necessity of Earthing:

1) To save human life from danger or shock or death by blowing of fuse of any apparatus which becomes

leaky.

2) To protect all machines fed from overhead lines from lightning arrestors.

3) To protect large building from atmospheric lightning.

4) To maintain the line voltage constant.

11. What is diversity factor for sub circuits?

In a building, the whole load of Electrical Wiring installation doesn’t use at the same time. In other

words, we do not use all the loads at once at home. Therefore, keep in mind this (diversity factor) factor while

selecting the proper and most suitable size of cable for electrical wiring installation.

Diversity factor may be defined as:

Diversity Factor = Total Connected Load / Actual Maximum Load

12. State the types of energy audit.

i) Preliminary Audit

ii) Detailed Audit

13. State the different losses in transformer.

i) Core loss or Constant loss

(a) Eddy current loss

(b) Hysteresis loss

ii) Copper loss or Variable loss

14. Write short notes on rewinding.

It is common practice in industry to rewind burnt-out motors. The population of rewound motors in

some industries exceeds 50% of the total population. Careful rewinding can sometimes maintain motor

efficiency at previous levels, but in most cases, losses in efficiency result.

For example, a common problem occurs when heat is applied to strip old windings are the insulation

between laminations can be damaged, thereby increasing eddy current losses. A change in the air gap may affect

power factor and output torque.

15. Write a note on lumen.

The lumen (symbol: lm) is the SI derived unit of luminous flux, a measure of the total quantity of visible

light emitted by a source per unit of time. Luminous flux differs from power (radiant flux) in that radiant flux

includes all electromagnetic waves emitted, while luminous flux is weighted according to a model (a "luminosity

function") of the human eye's sensitivity to various wavelengths. Lumens are related to lux in that one lux is one

lumen per square meter.

16. Describe the energy efficient motor. Improvements in motor efficiency can be achieved without compromising motor performance. Any

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improvement in motor efficiency must result from reducing the Watts losses. A reduction in watts losses can be

achieved in various ways. All of these changes to reduce motor losses are possible with existing motor design

and manufacturing technology. They would, however, require additional materials and/or the use of higher

quality materials and improved manufacturing processes resulting in increased motor cost.

PART – C

17.(a) Explain joint box, looping back and tree system of wiring.

i) Looping back system:

Figure shows a simple looping back method which is commonly employed.

The phase or live conductors are looped in switchboard and neutrals are looped either in

switchboard or from light or fan. It will be seen that the live terminal of the loads (lamps and fan) are

connected through switch and neutral for next load is looped in previous lamp/fan point. In this system no

joints are made in the run of wire or cables.

Advantages:

1. Junction boxes are not required.

2. Fault location is easy.

Drawback:

1. Length of cable required is more.

Joint box system:

The arrangement of the joint box system is shown in figure.

In this system the phase or line conductors are looped in switchboard and neutrals are looped

either in junction box provided at the center of all light points. This system is only economical in small

houses having few lighting points because looping is only done in the joint box.

Tree system:

In this system smaller branches are taken from the main branch. From the main switch rising

mains are run to every floor. Rising mains are of copper or aluminium strips which go up from the main

switch to all floors.

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The rising mains are usually provided in the centre of the load for the purpose of economy. In

this system the supply for the switches or appliances can be taken by tapping connection from any point

of running circuit wires (rising mains).

Drawback: • Voltage across all the lamps does not remains same.

• Involves more number of joints.

• Fuses are scattered.

(Or)

(b). With neat sketch, explain the construction and working of ELCB.

Construction and working of ELCB :

Earth Leakage Circuit Breaker:

The differential current is immediately sensed by the current operated ELCB and protecting the

people from dangerous electrical shocks.

Construction:

It consists of a toroid type core with two primary windings. One is connected in series with the

phase and the other is connected in series with the neutral. Observe that the supply coil, the neutral coil

and the search coil all wound on a common transformer core. Specification for ELCB should contain

normal rated current, leakage current and the time duration within which the ELCB should trip.

Working:

During normal condition:

As far as there is no leakage current, the phase or line current is equal to the neutral current and

the magnetic flux produced by the two primary windings oppose and cancel each other. Thus the

secondary winding which is connected to trip (operating) coil does not induce any voltage.

During fault:

However, when there is any leakage in the circuit, the line current differs from neutral current,

thus inducing a voltage in the secondary and the trip coil opens the circuit. Working of the ELCB could

be checked by test button at intervals. Since the trip current is driven by the residual magnetic flux

between the phase and the neutral coils, it is also called as residual current devise.

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19.(a) Discuss in detail about (i) Electrical load management

(ii) Maximum demand control

Electrical load management:

In a macro perspective, the growth in the electricity use led to shortfalls in capacity to meet

demand. As capacity addition is costly, better load management at user end helps to minimize peak

demands on the utility infrastructure.

The utilities (distribution companies) use power tariff structure to influence end user in better

load management through measures like time of use tariffs, penalties on exceeding allowed maximum

demand, night tariff concessions etc., Load management is a powerful means of efficiency improvement

both for end user as well as utility.

As the demand charges constitute a considerable portion of the electricity bill, there is a need for

integrated load management to effectively control the maximum demand.

Maximum demand control:

i) Load curve generation:

Presenting the load demand of a consumer against time of the day is known as a ‘load curve’. If

it is plotted for the 24 hours of a single day, it is known as an ‘hourly load curve’ and if daily demands

plotted over a month, it is called ‘daily load curve’.

ii) Rescheduling of loads:

Rescheduling of large electric loads and equipment operations, in different shifts can be planned

and implemented to minimize the simultaneous maximum demand. For this purpose, it is advisable to

prepare an operation flow chart and a process chart.

iii) Storage of products/in process material/process utilities like refrigeration:

It is possible to reduce the maximum demand by building up storage capacity of products/

materials, water, chilled water/ hot water, using electricity during off peak periods. Off peak hour

operations also help to save energy due to favourable conditions such as lower ambient temperature etc.

Example: Ice bank system is used in milk & dairy industry. Ice is made in lean period and used in

peak load period and thus maximum demand is reduced.

iv) Shedding of Non-Essential Loads:

When the maximum demand tends to reach preset limit, shedding some of non-essential loads

temporarily can help to reduce it. It is possible to install direct demand monitoring and control systems

which will switch off non-essential loads when a preset demand is reached.

v) Operation of captive generation and diesel generation sets:

When diesel generation sets are used to supplement the power supplied by the electric utilities,

it is advisable to connect the D.G sets for durations when demand reaches the peak value. This would

reduce the load demand to a considerable extent and minimize the demand charges.

vi) Reactive power compensation:

The maximum demand can also be reduced at the plant level by using capacitor banks and

maintaining the optimum power factor. Capacitor banks are available with microprocessor based control

systems. These systems switch on and off the capacitor banks to maintain the desired power factor of system.

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(Or)

(b) Explain in detail about performance assessment of power factor capacitor.

Performance Assessment of Power Factor Capacitors:

i) Voltage effects:

Ideally capacitor voltage rating is to match the supply voltage. If the supply voltage is lower, the 2 2

reactive KVAr produced will be the ration V1 /V2 where V1 is the actual supply voltage, V2 is the rated voltage. One the other hand, if the supply voltage exceeds rated voltage, the life of the capacitor is

adversely affected.

ii) Material of capacitors:

Power factor capacitors are available in various types by dielectric material used as;

paper/polypropylene etc., The watt loss per KVAr as well as life vary with respect to the choice of the

dielectric material and hence is a factor to be considered while selection.

iii) Connections:

Shunt capacitor connections are adopted for almost all industry/ end user applications, while

series capacitors are adopted for voltage boosting in distribution networks.

iv) Operational performance of capacitors:

This can be made by monitoring capacitors charging current. Portable analysers can be used for

measuring KVAr delivered as well as charging current. Capacitors consume 0.2 to 6.0 watt per KVAr,

which is negligible in comparison to benefits.

Some checks that need to be adopted in use of capacitors are:

• Nameplates can be misleading with respect to ratings. It is good to check by charging currents.

• Capacitor boxes may contain only insulated compound and insulated terminals with no capacitors

elements inside.

• Capacitors for single phase motor starting and those used for lighting circuits for voltage boost

are not power factor capacitor units and these cannot withstand power system conditions.

20.(a)

Write the various losses in induction motor.

Losses:

The losses can be divided into two parts.

• Constant losses or core losses.

• Variable losses or copper losses.

Core Losses:

▪ Core losses = Hysteresis losses + Eddy current losses

▪ The flux density in the magnetic structure influences the magnetic core losses.

▪ Core losses can be reduced by:

o Increasing the length of the magnetic structure and therefore, the flux density gets reduced. This increases the weight of the motor.

o Using steel containing silicon up to 4% in the lamination. The better magnetic property reduces the losses by 10 to 25%. But permeability will be low and this will increase the magnetizing current and reduces the power factor.

Eddy current losses:

▪ Eddy current losses can be reduced by:

o Using thinner laminations

o Ensuring adequate insulation between the laminations to reduce the eddy current flow.

Friction and Windage Losses:

Friction Losses: ▪ This loss is due to friction in bearing of the motor.

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▪ The friction loss in the bearing depends on the bearing size, type, speed of the motor, load on the

motor and lubrication used. ▪ The friction loss is small and fixed for a particular design of motor.

Windage Losses:

▪ This loss takes place due to power consumed by the ventilation fan and other rotating elements of

the motor.

▪ It depends on design of ventilation fan and the amount of ventilation required for removing the

heat produced due to various other losses.

Copper Losses in Stator and Rotor:

Copper losses constitute almost 50% of total losses in a motor. Stator copper losses make up to

66% of total copper losses in the motor. The reduction in copper losses results in reduction in overall

losses and increase in efficiency. The quality of copper and cross section area of copper wire is more in

an energy efficient motor so that the resistance of stator winding gets reduced. The stator slots in an

energy efficient motor can accommodate approximately 20% more copper in comparison of a standard

motor.

Rotor current depends on the slip and sometimes rotor losses are also called slip losses. If slip is

reduced, rotor losses will also get reduced. This is done by increasing the mass of the rotor conductor i.e.

rotor bars and end rings.

Copper losses can be reduced by:

▪ Using Copper of higher conductivity

▪ Using die cast aluminium bars resulting in reduction in slip.

▪ Replacement of a standard efficiency motor with an energy efficient motor.

Stray Load Losses:

▪ Stray load losses are the losses due to leakage fluxes induced by load currents and increases with

load. ▪ These losses are difficult to determine by direct measurement or calculation.

▪ Factors which influence the stray load losses are:

❖ Stator winding design

❖ Ratio of gap length to rotor slot opening

❖ Ratio of number of rotor slots to stator slots

❖ Air gap flux density

▪ Stray load losses can be reduced by careful design of the motor.

(Or)

(b). Discuss about the various factors affecting the performance of motor.

Power supply Quality: Motor performance is affected considerably by the quality of input power that is

the actual volts and frequency available at motor terminals. Motors in India must comply with standards set by the Bureau of Indian Standards (BIS) for tolerance to variations in input power quality. The BIS standards specify that a motor should be capable of delivering its rated output with a voltage variation of +/- 6 % and frequency variation of =/- 3%. Fluctuations much larger than these are quite common in utility supplied electricity in India. Voltage fluctuations can have detrimental impacts on motor performance.

Voltage Unbalance: Voltage unbalance, the condition where the voltages in the three phases are not

equal, can be still more detrimental to motor performance and motor life. Unbalance typically occurs as a result of supplying single-phase loads disproportionately from one of the phases. It can also result from the use of different sizes of cables in the distribution system. An example of

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the effect of voltage unbalance on motor performance is shown in table.

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Example of the Effect of Voltage Unbalance on Motor Performance

Parameter % unbalance in voltage

0.30 2.30 5.40

Unbalance in current (%) 0.4 17.7 40.0

Increased temperature rise(°C) 0.18 10.6 58

The NEMA (National Electrical Manufactures Association of USA) standard definition of voltage unbalance is given by the following equation:

Common Causes of Voltage Unbalance:

o Unbalanced incoming utility supply

o Unequal transformer tap settings

o Large single phase distribution transformer on the system

o Open phase on the primary of a 3 phase transformer on the distribution system

o Faults or grounds in the power transformer

o Open delta connected transformer banks

o A blown fuse on a 3 phase bank of power factor improvement capacitors

o Unbalanced distribution of single phase loads such as lighting

o Heavy reactive single phase loads such as welders

Effect of voltage unbalance:

▪ It is recommended that the voltage unbalance at the motor terminals not exceed 1%, anything above this will lead to derating of the motor.

▪ Voltage unbalance is probable the leading power factor problem that results in motor overheating and premature motor failure.

▪ Voltage unbalance causes extremely high current imbalance. The magnitude of current imbalance may be 6 to 10 times as large as the voltage imbalance.

▪ A motor will run hotter when operating on a power supply with voltage unbalance.

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Motor Loading:

% 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 =

or

𝐴𝑐𝑡𝑢𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟

𝑅𝑎𝑡𝑒𝑑 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟

𝑥 100

% 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 =

Or

𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 𝑑𝑟𝑎𝑤𝑛 𝑏𝑦 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟 (𝐾𝑊)𝑎𝑡 𝑒𝑥𝑖𝑠𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 𝑋 100

√3 𝑥 𝐾𝑉 𝑋 𝐼 𝑥 𝐶𝑜𝑠 ∅

𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 𝑑𝑟𝑎𝑤𝑛 𝑏𝑦 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟 (𝐾𝑊)𝑎𝑡 𝑒𝑥𝑖𝑠𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 𝑋 100 % 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 =

𝑁𝑎𝑚𝑒 𝑝𝑙𝑎𝑡𝑒 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝐾𝑊 𝑟𝑎𝑡𝑖𝑛𝑔/𝑛𝑎𝑚𝑒 𝑝𝑙𝑎𝑡𝑒 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑚𝑜𝑡𝑜𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

▪ Never assume power factor ▪ Loading should not be estimated as the ratio of current

Reducing Under-loading: Under-loading results in lower efficiency and power factor. Under-loading of the

motor may also occur from under utilisation of the equipment. Another common reason for under loading is selection of a larger motor to enable the output to be maintained at the desired level even when input voltages are abnormally low. Finally, under-loading also results from selecting a large motor for an application requiring high starting torque where a special motor, designed for high torque, would have been suitable.

For motors, which consistently operate at loads below 40 % of rated capacity, an inexpensive and effective measure might be to operate in star mode. Operating in the star mode leads to a voltage reduction by a factor of '√3'. Motor is electrically downsized by star mode operation, but performance characteristics as a function of load remain unchanged. Thus, full-load operation in star mode gives higher efficiency and power factor than partial load operation in the delta mode. However, motor operation in the star mode is possible only for applications where the torque-to-speed requirement is lower at reduced load.

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 the highest anticipated load. Rather than selecting a motor of high rating, a motor would be selected with a rating slightly lower than the peak anticipated load and would operate at overload for a short period of time. Since operating within the thermal capacity of the motor insulation is of greatest concern in a motor operating at higher than its rated load.

Under extreme load changes, e.g. frequent starts / stops, or high inertial loads, this method of calculating 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 control strategy employed can have a significant impact on motor electricity use. More efficient speed control mechanisms include multi-speed motors, eddy-current couplings, fluid couplings, and solid-state electronic variable speed drives.

Power Factor Correction:

Induction motors are characterized by power factors less than unity, leading to lower overall efficiency. Capacitors connected in parallel with the motor are typically used to improve the power factor. The impacts of PF correction include reduced kVA

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demand, reduced I2R

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losses in cables, reduced voltage drop in the cables, and an increase in the overall efficiency of the plant electrical system.

The size of capacitor required for a particular motor depends upon the no-load reactive power (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.

Maintenance: Inadequate maintenance of motors can significantly increase losses and lead to

unreliable operation. For example, improper lubrication can cause increased friction in motor. Resistance losses in the motor, which rise with temperature, would increase. Providing adequate ventilation and keeping motor cooling ducts clean can help dissipate heat to reduce excessive losses.

Good maintenance practices: ▪ 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.

▪ Lubricating appropriately. Manufacturers generally give recommendations for how and when to lubricate their motors. Over lubrication can also create problems, e.g. excess oil or grease.

▪ 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 to both the motor and the driven equipment.

▪ Ensuring that supply wiring and terminal box are properly sized and installed.

21.(a) Discuss about (i) Energy performance assessment of diesel conservation avenues (ii) Electronic

ballast.

Energy Performance Assessment of DG Sets:

Routine energy efficiency assessment of DG sets on shop floor 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

measurements are logged at 15 minutes intervals. a. Fuel consumption (by dip level or by flow meter)

b. Amps, volts, PF, kW, kWh

c. Intake air temperature, Relative Humidity (RH)

d. Intake cooling water temperature

e. 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

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 difference.

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g. Comments on charge air cooler performance

h. Comments on load distribution among various cylinders (based on exhaust temperature,

the temperature to be ± 5% of mean and high/low values indicate disturbed condition).

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

Electronic Ballast:

Role of Ballast:

In an electric circuit the ballast acts as a stabilizer. Fluorescent lamp is an electric discharge

lamp. The two electrodes are separated inside a tube with no apparent connection between them. When

sufficient voltage is impressed on these electrodes, electrons are driven from one electrode 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 auxiliary circuit housed in

a casing is known as ballast.

Electronic Ballasts:

The basic functions of electronic ballast are:

1. To ignite the lamp

2. To stabilize the gas discharge

3. To supply the power to the lamp

The electronic ballasts make use of modern power semi-conductor devices for their operation.

The circuit components form a tuned circuit to deliver power to the lamp at a high resonant frequency (in

the vicinity of 25 kHz) and voltage is regulated through an inbuilt feedback mechanism. It is now well

established that the fluorescent lamp efficiency in the kHz range is higher than those attainable at low

frequencies. At lower frequencies (50 or 60 Hz) the electron density in the lamp is proportional to the

instantaneous value of the current because the ionization state in the tube is able to follow the Figure.

Electronic Ballast instantaneous variations in the current. At higher frequencies (kHz range), the

ionization state cannot follow the instantaneous variations of the current and hence the ionization density

is approximately a constant, proportional to the RMS (Root Mean Square) value of the current. Another

significant benefit resulting from this phenomenon is the absence of stroboscopic effect, thereby

significantly improving the quality of light output.

(Or)

(b) Write in detail about (i) Occupancy sensors (ii) Energy efficient lighting controls.

(i) Occupancy Sensors:

Occupancy-linked control can be achieved using infra-red, acoustic, ultrasonic or microwave

sensors, which detect either movement or noise in room spaces. These sensors switch lighting on when

occupancy is detected, and off again after a set time period, when no occupancy movement detected.

They are designed to override manual switches and to prevent a situation where lighting is left on in

unoccupied spaces.

(ii) Energy efficient lighting controls:

Timed Based Control:

Timed-turnoff switches are the least expensive type of automatic lighting control. In some cases,

their low cost and ease of installation makes it desirable to use them.

Timed-turnoff switches are available with a wide range of time spans. The choice of time span is

a compromise. Shorter time spans waste less energy but increase the probability that the lights will turn

off while someone is in the space. Dial timers allow the occupant to set the time span, but this is not

likely to be done with a view toward optimizing efficiency. For most applications, the best choice is an

electronic unit 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. They

may be mounted either externally or internally. By using an internally mounted photoelectric dimming

control system, it is possible to ensure that the sum of daylight and electric lighting always reaches the design level by sensing the total light in the controlled area and adjusting the output of the electric

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