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Department of Technical Education Andhra Pradesh

NameDesignationBranchInstituteYear/SemesterSubjectSubject CodeTopicDurationSub TopicTeaching Aids

: P. Balanarsimlu: Lecturer: Electrical Engineering: Govt. Polytechnic Nizamabad: VI Semester: Electrical Utilisation and Automation: EE605A: Electric Lighting : 100 Mins: Problems on Laws of Illumination : PPT, Diagrams, Animations

Revised By : K. Chandra Sekhar, L/EEE, GPT, HYD

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Recap

In the last class you have learnt about

• Types of Lighting Schemes• Advantages of Lamp Fittings• Laws of Illumination

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Objectives

On completion of this topic you would be able to know

• Calculation of Illumination using Laws of Illumination

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Problem 1: The candle power of a lamp placed normal to a working plane is 30cp. Find the distance, if the illumination is 15 lux.

Given data Candle power, I = 30 CP Illumination E = 15 luxSolution :

we know, E =

d = = = 1.414 m Ans.

Id2

IE

30 15

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Problem 2: When a 250V lamp takes a current of 0.8 ampere, it produces a total flux of 3,260 lumens. Calculate (i) MSCP (ii) efficiency of lamp

Given data

Voltage , V = 250VCurrent, I = 0.8 AmpFlux, = 3260 lumensFind MSCP Efficiency of lamp ?

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(i) MSCP = = = 259.42 Ans.

(ii) Lam efficiency, η =

= = =16.3 Lm/Watt, Ans

lumen 4π

3260 4π

Total flux output

Electrical input

lumen VI cosФ

3260 250x0.8x1

Solution

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Problems on Indoor Lighting

• The luminous intensity of a lamp is 400 candela is

placed in the middle of a 10mx5mx4m room. Calculate

the illumination, a) in each corner of the room, b) the

middle of 5m wall

Problem 3

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Solutionθ d

4m

10m

5m

L

CB

AD

O

Fig

Given data

Luminous Intensity, I = 400 Candela

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From figure AO = BO

(5.59)2 + 42 =

= 2

√

5.59m

d

102 + 52 √

= 6.874 m

cos θ = 4

6.874 0.5819 =

Contd…Solution

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illumination at each corner

=

= 400

6.8742 0.5819

EA EB EC ED = = =

Id2 = x cos θ

x

= 4.926 flux

Contd…solution

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θ d1

4m

10m

5m

L

CB

AD

O5m

Fig:2B

Contd…

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√ d1 = 6.403 m

cos θ = 4

6.403 0.6247 =

(5)2 + 42 =

= 400

6.4032 0.6247

Id1

2 = cos θ

x

= 6.095 flux

illumination in the middle of 5m wall is

x

Contd…

solution

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Outdoor Lighting

Fig.3

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Problems on Outdoor Lighting

• In street lighting scheme, lamps having luminous

intensity of 600 candela are hung at a height of 6m. The

distance between two lamp posts is 8m. Find the

illumination under the lamp and at center in between the

lamp posts.

Problem 4

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Given Data Intensity,I = 600 cpHeight, h =6m Distance between two posts, D = 8m

Required Data Illumination under the lamp ? Illumination at center?

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C BA4m

8m

6 mL2L1

d1

d2

Ø1

Ø2

Fig.4

Contd…

Solution

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62 + 42 = √ d1 = 7.211 m

cos θ1 =

62 + 82 = √ d2 = 10 m

h d1

= 6

7.211 = 0.832

cos θ2 = h d2

= = 0.6 6 10

Contd…

Solution

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Illumination at ‘C’ due to L1,

E1 = I d1

2 x cos θ1 = 600

(7.211)2 x 0.832

= 9.6 lux

Illumination at ‘C’ due to L2, E2 will be same as E1

Illumination at ‘C’ due to L1&L2, EC =E1+E2

= 9.6+9.6=19.2 lux

Contd… Solution

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Illumination at ‘B’ due to L1,

= 600(10)2

x 0.6 3.6 lux =

Illumination at ‘B’ due to L2,

= 600(6)2

16.667 lux =

Illumination at ‘B’ due to L1&L2,

= 3.6+16.667 20.267 lux =

Contd… Solution

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• Two amps having luminous intensity of 150 candela

200 candela are hung at height 10m and 15m

respectively. The distance between two lamp posts is

30m. Find the illumination at center in between the lamp

posts.

Problem 5

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Given Data

Intensities = 150 cp & 200 cp

Heights =10m & 15m

Distance between two posts, D = 30m

Required Data

Illumination at center?

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C BA15m

30m

10 m200 cp

150 cp

d1

d 2

Ø1

Ø2

15 m

L2

L1

Fig.5

Contd…

Solution

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102 + 152 = √ d1 = 18.028 m

cos θ1 =

152 + 152 = √ d2 = 21.213m

h1

d1

= 10

18.028 = 0.5547

cos θ2 = h2

d2

= = 0.7071 15

21.213

Contd…

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Illumination at ‘P’ due to L1,

E1 = I

d12 x cos θ1 = 150

(18.028)2 x 0.5547

= 0.256 lux Illumination at ‘P’ due to L2,

E2 = I

d22

x cos θ2 = 200(21.213)2

x 0.7071

= 0.3143 lux

Total illumination E = E1+E2

=0.256+0.3143=0.5703 lux

Contd…

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Problem 6

The luminous intensity of a lamp is 400candela in all directions below the lamp. The lamp mounting height is 4m. Find the illumination (a) just below the lamp (b) 5m horizontally away from the lamp on the ground. (c) total luminous flux in a area of 1m dia around the lamp on the ground.

Given data:Luminous intensity, I = 400 candela height, h = 4m

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Illumination just below the lamp

EA = I

H2

= 400

(4)2

= 25 Lux

d = h 2+ 52 = 42+52

= 6.403

Cos = 5/6.403 = 0.7809

5 mB

d

A

lamp

Solution

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Illumination at 5m horizontally from lamp

E B = 2

Id

Cos

= 2

400 0.7809 7.6186.403

lux

From fig

Surface area = 2

4D

21 0.78544

m2

4 m

1 m

lamp

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Total flux = EA X Area

= 25x 0.7854 = 19.635 lumens

Contd…

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Summary

In this class we have solved the problems on

• Laws of Illumination

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QUIZ

1. The illumination at any point on a surface is proportional to

(a) Cos (b) Sin (c) Tan (d) None of the above

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QUIZ

2. The illumination of a surface is inverse proportional to the square of the distance between surface and source

TRUEFLASE

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Frequently Asked Questions1. The Luminous intensity of a lamp is 250 Candela and is

maintained at a height of 5m from the center of a circular area 4m dia. Find the (a) maximum, (b) minimum and

(c) average illumination Ans. 10 lux, 8 lux, 8.94 lux

2. In street lighting scheme, lamps having luminous intensity of 100 candela are hung at a height of 6m. The distance between two lamp posts is 16m. Find the illumination under the lamp and at center in between the lamp posts.

Ans. 2.9 lux, 1.2 lux

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THANK YOU

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