9EE605A.7to81 Department of Technical Education Andhra Pradesh Name Designation Branch Institute Year/Semester Subject Subject Code Topic Duration Sub Topic Teaching 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

9EE605A.7to82 Recap In the last class you have learnt about Types of Lighting Schemes Advantages of Lamp Fittings Laws of Illumination

9EE605A.7to83 Objectives On completion of this topic you would be able to know Calculation of Illumination using Laws of Illumination

9EE605A.7to84 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 lux Solution : we know, E = d = = = m Ans. I d 2 I E 30 15

9EE605A.7to85 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 = 250V Current, I = 0.8 Amp Flux, = 3260 lumens Find MSCP Efficiency of lamp ?

9EE605A.7to86 (i) MSCP = = = Ans. (ii) Lam efficiency, η = = = =16.3 Lm/Watt, Ans lumen 4π π Total flux output Electrical input lumen VI cosФ x0.8x1 Solution

9EE605A.7to87 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

9EE605A.7to88 Solution θ d 4m 10m 5m L C B A D O Fig Given data Luminous Intensity, I = 400 Candela

9EE605A.7to89 From figure AO = BO (5.59) = = 2 √ 5.59m d √ = m cos θ = = Contd… Solution

9EE605A.7to810 illumination at each corner = = E A E B E C E D = = = I d2d2 = x cos θ x = flux Contd… solution

9EE605A.7to811 θ d1d1 4m 10m 5m L C B A D O Fig:2B Contd…

9EE605A.7to812 √ d1 d1 = m cos θ = = (5) = = I d12d12 = cos θ x = flux illumination in the middle of 5m wall is x Contd… solution

9EE605A.7to813 Outdoor Lighting Fig.3

9EE605A.7to814 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

9EE605A.7to815 Given Data Intensity,I = 600 cp Height, h =6m Distance between two posts, D = 8m Required Data Illumination under the lamp ? Illumination at center?

9EE605A.7to816 C BA 4m 8m 6m L2L2 L1L1 d1d1 d2d2 Ø1Ø1 Ø2Ø2 Fig.4 Contd… Solution

9EE605A.7to = √ d1 d1 = m cos θ 1 = = √ d2 d2 = 10 m h d1 d1 = = cos θ 2 = h d2 d2 = = Contd… Solution

9EE605A.7to818 Illumination at ‘C’ due to L 1, E1 E1 = I d12 d12 x cos θ 1 = 600 (7.211) 2 x = 9.6 lux Illumination at ‘C’ due to L 2, E 2 will be same as E 1 Illumination at ‘C’ due to L 1 &L 2, E C =E 1 +E 2 = =19.2 lux Contd… Solution

9EE605A.7to819 Illumination at ‘B’ due to L 1, = 600 (10) 2 x lux = Illumination at ‘B’ due to L 2, = 600 (6) lux = Illumination at ‘B’ due to L 1 &L 2, = lux = Contd… Solution

9EE605A.7to820 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

9EE605A.7to821 Given Data Intensities = 150 cp & 200 cp Heights =10m & 15m Distance between two posts, D = 30m Required Data Illumination at center?

9EE605A.7to822 C BA 15m 30m 10m 200 cp 150 cp d1d1 d2d2 Ø1Ø1 Ø2Ø2 15m L2L2 L1L1 Fig.5 Contd… Solution

9EE605A.7to = √ d1 d1 = m cos θ 1 = = √ d2 d2 = m h1 h1 d1 d1 = = cos θ 2 = h2 h2 d2 d2 = = Contd…

9EE605A.7to824 Illumination at ‘P’ due to L 1, E1 E1 = I d12 d12 x cos θ 1 = 150 (18.028) 2 x = lux Illumination at ‘P’ due to L 2, E2 E2 = I d22 d22 x cos θ 2 = 200 (21.213) 2 x = lux Total illumination E = E 1 +E 2 = = lux Contd…

9EE605A.7to825 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

9EE605A.7to826 Illumination just below the lamp E A = I H2H2 = 400 (4) 2 = 25 Lux d = h = = Cos  = 5/6.403 = m B d  A lamp Solution

9EE605A.7to827 Illumination at 5m horizontally from lamp E B = Cos  = lux From fig Surface area = m2m2 4 m 1 m lamp

9EE605A.7to828 Total flux = E A X Area = 25x = lumens Contd…

9EE605A.7to829 Summary In this class we have solved the problems on Laws of Illumination

9EE605A.7to830 QUIZ 1. The illumination at any point on a surface is proportional to (a)Cos  (b)Sin  (c)Tan  (d)None of the above

9EE605A.7to831 QUIZ 2. The illumination of a surface is inverse proportional to the square of the distance between surface and source TRUE FLASE

9EE605A.7to832 Frequently Asked Questions 1. 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

9EE605A.7to833 THANK YOU 9EE605A.07TO 08