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Lighting & power design citam

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Basic Lighting and power distribution calculations.

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Lighting & power design citam

  1. 1. Edson Engineers Page 1 EDSON ENGINEERS. DESIGN CALCULATIONS. LIGHTING AND POWER DESIGN CALCULATIONS. BY FRED BUTETE WAMBASI. PROJECT: CITAM KAREN BOYS AND GIRLS TOILETS JANUARY 2015 Power Design Calculations Lighting
  2. 2. Edson Engineers Page 2 1.0 Lighting Design. 1.1 Formulae and Fundamental Considerations. Lighting design aims at determining the number of lighting fixtures (luminaries) required in order to achieve the recommended illumination for a given task. Key considerations prior to design are;  The dimensions of the room; Length, Width and mounting height of the luminaire from working plane.  The nature of ceiling, walls, and floors in terms of coluor and material.  The functionally/use of the design area to be illuminated.  Non analytical factors affecting the choice of luminaires, aesthetics and natural lighting. In determining the number of lighting fixtures (luminaries) required, key among other factors; the following has to be established:  Recommended illuminance (Lux) in Lumens/m²  Utilization factor  Maintenance factor of the Luminaire  Nominal lamp Luminous flux/output in Lumens. Quantity of light here is specified by illuminance which is measured in lux (lm/m2 of illuminated surface). The Lumen method formula used is as below (as per CIBSE Lighting guide and a textbook on building services engineering by David V. Chadderton: Building Services Engineering Fifth edition (2007) - Taylor and Francis Group - New York ).
  3. 3. Edson Engineers Page 3 = Øn xn xMf xUf Where: E = Average recommended illuminance (Lux) in Lumens/m² A = Area of the working plane (m²) Øn= Nominal Lamp Luminous Flux Mf = Maintenance factor Uf = Utilization factor N= Number of Luminaires n= No. of luminaire’s lamps 1.2 Definitions: a. Utilization factor (Uf) This represents the proportion of luminous flux of the lamp that reaches the working plane and is dependent on the following:  Luminaire efficiency  Lighting fitting distribution  Reflectances of the room surfaces i.e. ( Ceiling, walls and floor)  Room index. The room index represents the geometrical ratio of the area of the horizontal surfaces to that of the Vertical surfaces measured from the working plane in the room and is expressed as: Ri = (L x W) ( + ) Where: L = Length of the room
  4. 4. Edson Engineers Page 4 W = Width of the room Hm = Height of the Luminaire above the working plane. The room reflectances depend on the room surface finishes. In my design the ceiling is painted white, floor is terrazzo/pvc tiles in some areas and the walls are painted white.The design reflectance ratio used is C: W: F = 0.7: 0.5: 0.2. Hm of 2.05m is used herein in design calculations. Standard photometric tables for a combination of various values of room indices and surface reflectances exist (given by manufacturers) from which the value of the utilization factor (Uf) was directly obtained or extrapolated for values of Room indices that were not integers as follows: = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) where: (U= upper Value& L= Lower Value) b. Maintenance factor (Mf) This gives the proportion of illuminance provided by a luminaire in normal working conditions (dirty conditions) of both the luminaire and the room surfaces to the illuminance of the same luminaire in clean conditions. For this design, I used a maintenance factor of 0.8 on the assumption that the room surfaces will be maintained very clean most of the times given this is children’s toilet where high standards of cleanliness are desired. c. Nominal Lamp Luminous Flux (Øn) The value was obtained from lamp photometric data usually provided by manufacturers in the catalogues. d. Illumination level (E) These were read from a chart guide used for obtaining recommended illuminance (Lux) prepared by the Chartered Institute of Building Services
  5. 5. Edson Engineers Page 5 Engineers (CIBSE) and the Philips Lighting design Manual. These values entirely depend on the type of use the room is put into. The illuminance are as summarized in the tables below: The following are the recommended Illumination Levels; a) Hospital Item Description Recommended Lux (E) in Lumens/M2 1.0 Corridors: Night Daytime/ Evening 50 200 2.0 Wards: Circulation at night Observation at night General Lighting Simple Examination/Reading 30 5 150 300 3.0 Examination Rooms: General Lighting Local Examination Lighting 500 1000 4.0 Intensive therapy: Bedhead Observation 50 750 5.0 Nurses Stations 300 6.0 Operating Theaters: Pre-Op room General theater lighting - 500 1000 - 7.0 Laboratories & Pharmacies: General Lighting Local 750 1000 8.0 Consulting Rooms: General Lighting Local 500 750 9.0 Autopsy Rooms: General Lighting Local 750 5000
  6. 6. Edson Engineers Page 6 b) Offices Item Description Recommended Lux E in Lumens/M2 1.0 General offices with typing, computers e.t.c 500 2.0 Conference rooms 300 3.0 Archives 200 c) General Building areas Item Description Recommended Lux E in Lumens/M2 1.0 Circulation areas, corridors 100 2.0 Cloak rooms, toilets 100 3.0 Stores, Stock rooms 100 4.0 Stairs, escalators 150 d) Kitchen Block Item Description Recommended Lux E in Lumens/M2 1.0 Servery 300 2.0 Kitchen 500 3.0 Food stores 150 4.0 Food preparation area 500 5.0 Cold store 300 6.0 Office 500 7.0 Kitchen yard 30
  7. 7. Edson Engineers Page 7 e) Laundry Block Item Description Recommended Lux E in Lumens/M2 1.0 Pressing 500 2.0 Sewing/Mending 750 3.0 Gents/Ladies 100 f) Workshop Block Item Description Recommended Lux E in Lumens/M2 1.0 Welding 300 2.0 Machine work, coil winding 500 3.0 Fine bench & Machine work 750 4.0 Testing/ adjusting Electrical components 1000 2.0 Lighting Design Task. 2.1 Boys Toilets. 2.1.1 Area Outside toilets having whbs within toilet block. The room dimensions are: Length of the room, L = 8.9m Width of the room, W = 5.8m
  8. 8. Edson Engineers Page 8 For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 1200mm 1x36W ELECTRONIC BALLAST single batten fluorescent fitting with acrylic diffuser as pierlite. For this fitting, the nominal luminous flux Øn = 5400 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 8.9 x 5.8 = 1.40 2.05(8.9+5.8) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) = 59 + (1.40 − 1.25) . . =62.09% = 0.6209 The number of luminaires is given as, = 100 51.62 3450 x1 x0.8 x0.6209 = 3.0122
  9. 9. Edson Engineers Page 9 Thus we use 3 No.Luminaires,1200mm 1x36W Electronic Ballast single batten fluorescent fitting with acrylic diffuser as pierlite to provide lighting for the area. This gives an illuminance of : = Øn xn xMf xUfxN A = 3450 x1 x0.8 x0.6209x3 51.62 = 99.59 Thislevel of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required. 2.1.2 Area inside toilets having wcs but within toilet block. The room dimensions are: Length of the room, L = 4.5m Width of the room, W = 2.0m For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m
  10. 10. Edson Engineers Page 10 Therefore the room index Ri = L x W Hm (L + W) = 4.5 x 2.0 = 0.555351682 2.05(4.5+2.0) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) = 48 + (0.56 − 0.75) . . =44.10% = 0.441 The number of luminaires is given as, = 100 9.08 1250 x1 x0.8 x0.441 = 2.058 Thus we use 2 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area. . This gives an illuminance of: = Øn xn xMf xUfxN A
  11. 11. Edson Engineers Page 11 = 1250 x1 x0.8 x0.441x2 9.08 = 97.14 Thislevel of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required. 2.1.3 Paraplegic toilets. The room dimensions are: Length of the room, L = 2.38m Width of the room, W = 1.5m For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 2.38 x 1.5 = 0.448830 2.05(2.38+1.5) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( )
  12. 12. Edson Engineers Page 12 = 48 + (0.449 − 0.75) . . =41.98% = 0.42 The number of luminaires is given as, = 100 3.57 1250 x1 x0.8 x0.42 = 0.85 Thus we use 1 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area. . This gives an illuminance of: = Øn xn xMf xUfxN A = 1250 x1 x0.8 x0.42x1 3.57 = 117.64 This level of illumination is acceptable (within + 30%) as the luminaires will therefore provide adequate lighting for the area required. 2.1.4 Showers. The room dimensions are: Length of the room, L = 2.0m Width of the room, W = 1.5m
  13. 13. Edson Engineers Page 13 For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 2.0 x 1.5 = 0.418 2.05(2.0+1.5) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) = 48 + (0.418 − 0.75) . . =41.36% = 0.41 The number of luminaires is given as, = 100 3.0 1250 x1 x0.8 x0.41 = 0.73
  14. 14. Edson Engineers Page 14 Thus we use 1 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area. This gives an illuminance of: = Øn xn xMf xUfxN A = 1250 x1 x0.8 x0.41x1 3.57 = 114.85 This level of illumination is acceptable (within + 30%) as the luminaires will therefore provide adequate lighting for the area required. 2.1.5 Area in front of cleaners’ room (corridor space). The room dimensions are: Length of the room, L = 2.23m Width of the room, W = 2.17m For this toilets, the design illuminance E = 150 Lux as the lighting requirements for this space. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 2.23 x 2.17 = 0.536 2.05(2.23+2.17)
  15. 15. Edson Engineers Page 15 From the photometric tables, and by extrapolation, the utilization factor is obtained as Uf = 48 + (0.536 − 0.75) . . =43.72% = 0.44 The number of luminaires is given as, = 150 3.0 1250 x1 x0.8 x0.44 = 1.023 Thus we use 1 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area. This gives an illuminance of: = Øn xn xMf xUfxN A = 1250 x1 x0.8 x0.44x1 3.0 = 146.67 This level of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required.Being a room that is not used all day illumination levels are however not as significant as in other areas of the block.
  16. 16. Edson Engineers Page 16 2.2 Girls Toilets. 2.2.1 Area Outside toilets having whbs within toilet block. The room dimensions are: Length of the room, L = 11.9m Width of the room, W = 4.02 For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 1200mm 1x36W ELECTRONIC BALLAST single batten fluorescent fitting with acrylic diffuser as pierlite. For this fitting, the nominal luminous flux Øn = 5400 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 11.9 x 4.02 = 1.47 2.05(11.9+4.02) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( )
  17. 17. Edson Engineers Page 17 = 59 + (1.47 − 1.25) . . =63.4% = 0.634 The number of luminaires is given as, = 100 47.84 3450 x1 x0.8 x0.634 = 2.73 Thus we use 3 No.Luminaires, 1200mm 1x36W Electronic Ballast single batten fluorescent fitting with acrylic diffuser as pierlite. to provide lighting for the area. This gives an illuminance of : = Øn xn xMf xUfxN A = 3450 x1 x0.8 x0.634x3 47.84 = 109.73 Thislevel of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required. 2.2.2 Area inside toilets having wcs but within toilet block. The room dimensions are: Length of the room, L = 10.215m Width of the room, W = 2.0m
  18. 18. Edson Engineers Page 18 For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 10.215 x 2.0 = 0.816 2.05(10.215+2.0) From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) = 48 + (0.816 − 0.75) . . =49.32% = 0.4932 The number of luminaires is given as, = 100 20.43 1250 x1 x0.8 x0.4932 = 4.142 Thus we use 4 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area.
  19. 19. Edson Engineers Page 19 This gives an illuminance of: = Øn xn xMf xUfxN A = 1250 x1 x0.8 x0.49x4 20.43 = 95.94 This level of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required. 2.2.3 Paraplegic toilets. The room dimensions are: Length of the room, L = 2.0m Width of the room, W = 1.5m For this toilets, the design illuminance E = 100 Lux as the lighting requirements are for cloak rooms and toilets. The luminaire chosen was 16W Polo Opal C/W lamp downlighter with white lining. For this fitting, the nominal luminous flux Øn = 1250 lumens per lamp. Height of the luminaire above the working plane Hm = 2.05m Therefore the room index Ri = L x W Hm (L + W) = 2.0 x 1.5 = 0.585 2.05(2.0+1.5)
  20. 20. Edson Engineers Page 20 From the photometric tables, and by extrapolation, the utilization factor is obtained as = ( ) + ( − ( ) ( ) − ( ) ( ) − ( ) = 48 + (0.585 − 0.75) . . =44.7%=0.447 The number of luminaires is given as, = 100 3.0 1250 x1 x0.8 x0.447 = 0.67 Thus we use 1 No.Luminaires, as 16W Polo Opal C/W lamp downlighter with White lining to provide lighting for the area. . This gives an illuminance of: = Øn xn xMf xUfxN A = 1250 x1 x0.8 x0.447x1 3.0 = 149 This level of illumination is acceptable as the luminaires will therefore provide adequate lighting for the area required.
  21. 21. Edson Engineers Page 21 3.0 Power Design. 3.1 Formulae and Fundamental Considerations. The following formulae as provided in the IEE Wiring regulations 17th Edition (BS 7671:2008) and its guide: Design and Verification of Electrical Installations by Brian Scaddan( IEng, MIET) – Newnes Publishers UK (2008) were used in calculations pertaining to power. 3.1.1 Design current I b This is defined as ‘the magnitude of the current to be carried by a circuit in normal service ’ , and is either determined directly from manufacturers’ details or calculated using the following formulae as provided by 17th Edition of IEE wiring regulations : 3.1.1 (a) Single phase loads. = or = 3.1.1 (b) Three phase load = √ or = √ % In both cases where: P _ power in watts V _ line to neutral voltage in volts V L _ line to line voltage in volts Eff% _ efficiency PF _ power factor.
  22. 22. Edson Engineers Page 22 3.1.2 Diversity The application of diversity to an installation permits, by assuming that not all loads will be energized at the same time, thus a reduction in main or distribution circuit cable sizes. The IEE Regulations Guidance Notes or On-Site Guide tabulate diversity in the form of percentages of full load for various circuits in a range of installations. However, it is for the designer to make a careful judgement as to the exact level of diversity to be applied by consideration of the actual design intended, most a times designer judgement is employeds . 3.1.3 Nominal rating or setting of protection In. In is that it should be greater than or equal to I b. We can select for this condition from IEE Regulations Tables 41.2, 41.3 or 41.4. For types and sizes outside the scope of these tables, details from the manufacturer will need to be sought. 3.1.4 Selection of suitable conductor size . Based on the quantity of current anticipated. This is obtainable from calculations and reference to standard cable rating tables. 3.1.5 Sizing the protective device According to IEE regulations: Ib≤ II ≤ Ic Where: Ib= Design current of the circuit II = Nominal current or current setting of the protective device Ic = Current carrying capacity of the conductor in the particular installation conditions 3.1.6 Voltage drop In many instances this may well be the most onerous condition to affect cable sizes. The Regulations require that the voltage at the terminals of fixed equipment should be greater than the lower limit permitted by the British Standard for that equipment,
  23. 23. Edson Engineers Page 23 or in the absence of a British Standard, that the safe functioning of the equipment should not be impaired. These requirements are fulfilled if the voltage drop between the origin of the installation and any load point does not exceed the following values (IEE Regulations, Appendix 12) ( Table 3.1 Below ). Standard cable current ratings are tabulated against voltage drop in milli-volts (mV) dropped for every ampere of design current (A), for every metre of conductor length(m), case of Tables 9D1, 9D2, 9D3, 9E1 etc in BS Code i.e. Volt drop = mV/A/m (IEE 17th Edition) Table 3.1 Lighting Power 3% 5% 240V Single phase 7.2V 12V 415V three phase 12.5V 20.8V or fully translated with Ib for A and L (length in metres) as below formulae: Formulae. = volts = Vd = I x ∂ x L Volts (IEE 17th Edition) 3.2 Power Calculations for Both toilet Blocks. 3.2.1 Fixed power loads on distribution boards calculations, power supply cable and protective switch gear sizing. The ratings of the cable size and protective switch gear at the distribution board for individual equipment /appliance is established from its load current during starting operation voltage drops are pertinent in proper cable sizing and safety of operation.
  24. 24. Edson Engineers Page 24 3.2.1( a ) Calculations. The loads are single phase and thus single phase distribution supply adopted for economic justification. Formulae for calculation is provided above (also indicated below) as per IEE Wiring standards . = ( ) or = ( ) 3.2.1(b) Hand drier spur point Cable Sizing and Circuit breaker rating. Consideration during design is made for each spur point.This is performed assuming the hand drier is in full operation. Ideal Load due to hand drier is used, Assuming diversity of 1. = = 3000 240 0.8 = 15.63 From IEE Wiring Regulations tables, a cable of 2.5mm2 SC PVC insulated copper cables is sufficient for the operation, however considering the overshoot during start operation 4.0 mm2 SC Insulated by PVC copper cables can be used to cater for the in currents rush as the case of a childrens’ toilet and also considering that the distance from the DB to the spur point is very short, thus negligible economic implications. (BS 6004 , BS 6346) Thus a 20 A Double pole switch is chosen for the hand drier. A protective circuit breaker of 20 A TP MCB is chosen for hand drier circuit.
  25. 25. Edson Engineers Page 25 3.2.2 Current to CU’s, Cable Sizing for Armoured Cable and Choice of MCCB. 3.2.2(a) CU – CA , Girls toilets. Total load on CU – CA in Kw is given in the table below. ITEM NO RATING Total Wattage(KW) Total Wattage x 1.8 Total Wattage for Fluorescent (Watts) Fittings Diversity Applying a Diversity (KW) Flourescent Lights 3 0.036 0.108 0.194 0.9 0.175 Security Lights 4 0.060 0.240 - 0.5 0.120 Down Lights 5 0.016 0.080 - 0.9 0.072 Double Poles 1 3.0 3.000 - 0.7 2.100 TOTAL KW 2.467 ASSUMING PF = 0.8 KVA 1.974 Table3.2 Total load supplied to CU – CA was 2.467 Kw. Allowing for 10% Extra Load for future additional load. 2.467Kw + 10/100*2.467Kw = 2.714 Kw Design Load Current = . = 14.14 A The maximum allowed voltage drop is 5% of 240 V0lts = 12V = volts = Vd = I x ∂ x L Volts Since the CU’s are approximately 45 m from Maintenance workshop, Maximum value of ∂ = Vd /( Ib x L) Volts = 12 / (14.14 x 45) = 18.86 Mv/A/m ,thus a
  26. 26. Edson Engineers Page 26 Armorued cable of 4mm2 with a voltage drop of 12 Mv/A/m is chosen. Giving a total voltage drop of: Vd = 14.14 x 12 x 45 x 10-3 Volts = 7.64 Volts. The MCCB chosen is 20 A. Since ; Ib≤ II ≤ Ic Where: Ib= Design current of the circuit II = Nominal current or current setting of the protective device Ic = Current carrying capacity of the conductor in the particular installation conditions Ib = 14.14 A, II = 20 A Ic = 37 A. 3.2.2(b) CU – CB , Boys toilets. Total load on CU – CA in Kw is given in the table below. ITEM NO RATING Total Wattage(KW) Total Wattage x 1.8 Total Wattage for Fluorescent (Watts) Fittings Diversity Applying a Diversity (KW) Flourescent Lights 3 0.036 0.108 0.194 0.9 0.175 Security Lights 5 0.060 0.300 - 0.5 0.150 Down Lights 6 0.016 0.096 - 0.9 0.086 Double Poles 1 3.0 3.000 - 0.7 2.100 TOTAL KW 2.511 ASSUMING PF = 0.8 KVA 2.009 Table3.3 Total load supplied to CU – Cb was 2.511 Kw. Allowing for 30% Extra Load for future additional load.
  27. 27. Edson Engineers Page 27 2.511Kw + 10/100*2.511Kw = 2.762 Kw Design Load Current = . = 14.39 A The maximum allowed voltage drop is 5% of 240 V0lts = 12V = volts = Vd = I x ∂ x L Volts Since the CU’s are approximately 45 m from Maintenance workshop, Maximum value of ∂ = Vd /( Ib x L) Volts = 12 / (14.39 x 45) = 18.53 Mv/A/m ,thus a Armorued cable of 4mm2 with a voltage drop of 12 Mv/A/m is chosen. Giving a total voltage drop of: Vd = 14.39 x 12 x 45 x 10-3 Volts = 7.77 Volts. The MCCB chosen is 20 A. Since ; Ib≤ II ≤ Ic Where: Ib= Design current of the circuit II = Nominal current or current setting of the protective device Ic = Current carrying capacity of the conductor in the particular installation conditions Ib = 14.39 A, II = 20 A Ic = 37 A. ……………………………………………..end……………………………………………

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