Electrical

Depreciation Factor with Formulas and examples

Depreciation Factor

During the operation of any work, illumination of any light source keeps on reducing below the calculated illumination. During the application, lamps tend to blacken gradually, or a dust sheath forms on lamp globes, reflectors, or walls, due to which the quantity of illumination is affected. As dust has a tendency to absorb some portion of the refracted, dispersed, or emitted light, therefore, another factor, which is called depreciation factor, is also kept into consideration at the time of planning any lighting scheme.

Depreciation factor (p) = Illumination under actual condition/ Illumination when everything is perfectly clean

Its average value tends to be 0.8. Sometimes, a depreciation factor may also be described as follows;

The ratio between the illumination when everything happens to be clean and illumination during a typical cleaning, is called depreciation factor. In this case, the value of depreciation factor tends to be higher than unity or one, and it tends to be between 1.3 to 1.4.

The depreciation factor is also known as a maintenance factor. For example, if in a newly constructed room, which is adequately plastered, a perfectly new bulb is installed, the bulb emits 3000 illumines. After a usage of about six months, a sheath of dirt and dust builds up on bulb and walls, now the bulb output reduces to 2000 illumines. Under such a situation, the depreciation factor of the bulb will be as follows;

Depreciation factor = 2000/3000 = 0.66

In excellent atmospheric conditions, this factor tends to be 0.8, and under unpleasant dusty, smoky, and cloudy conditions, this factor reduces to 0.4.




Space-Height-Ratio

The ratio of a horizontal distance between two lamps, and vertical mounting height of lamps, is known as height ratio, i.e.,

Space-height-ratio= Horizontal distance between two lamps/ Vertical mounting height of lamps. The value of this ratio actually depends on the type of reflectors being used. In order to achieve same illumination on a working plane, selection of an accurate value of this ratio, tends to be important. The reflectors, which are used for indoor lighting, this ratio keeps changing generally to be between 1, and 2. Whenever, any lighting scheme is devised, it is very essential to take into account this factor.

Examples:

Example 1: A small assembly shop 15m long, 9m wide, and 3m up to trusses, is to be illuminated to a level of 200 lux. The coefficient of utilization is 0.75, and maintenance factor is 0.8. Calculate the number of lamps required to illuminate the whole area if the lumen output of the lamp selected is 3000 lumens.

Solution;

Working area A = 15m x 9m

Required illumination E = 200lux

Lumen output of one lamp O=3000 lumens

Coefficient of utilization CU =0.75

Maintenance factor MF= 0.8

Number of lamps required N= E x A/O x CU x MF

= 200 x 15 x 9/3000 x0.75 x 0.8 =15



Example 2: An office 30m x 15m is illuminated by 40W fluorescent lamps of lumen output 2700 lumens. The average illumination required at the work place is 200 lux. Calculate the number of lamps required to be fitted in the office. Assume coefficient of utilization to be 0.6 and depreciation factor 1.25.

Solution:

Working area = 30 m x 15 m

Required illumination E = 200 lux

Lumen output of one lamp O= 2700 lumens

Coefficient of utilization CU= 0.6

Depreciation factor DF = 1.25

Maintenance factor MF = 1/1.25 = 0.8

Number of lamps required N= E x A/ O x CU x MF

=200 x 30 x 15/2700 x 0.6 x 0.8

Example 3: The conference hall measuring 30m x 10m is to be provided with illumination scheme at 250 lux, and to be provided with double tube set of 40 watt each. If the utilization factor is 0.5, and maintenance factor is 0.9, determine the number of tube sets of two 40 watts fluorescent tubes required. The coefficient of each tube set is 60 lumens per watt.

Solution:

Total area to be illuminated A= 30 m x 10 m = 300m2

Illumination required E = 250 lux

Total lumens required = A x E = 300 x 250 = 75,000 lumens

Utilization factor = 0.5

Maintenance factor = 0.9

Gross lumens required = A x E / UF x MF = 75000/0.5 x 0.9 = 166666 lumens

Output of each set = 60 x (40 x2) = 60 x 80 = 4800 lumens

Total number of sets required = Total lumens required / Lumen output of each set

Or N =E x A /O x CU x MF = 75000/ 4800 x 0.5 x 0.6 = 1,66,666 lumens/4800 = 34 sets



Example 4; The illumination of a drawing office 30m x 10m is to have a value of 250 lux, and is to be provided by a number of 300W filament lamps. If the utilization factor is 0.4, and the depreciation factor is 0.9, determine the number of lamps required. The efficiency of each lamp is 14 lumens per watt.

Solution;

Area to be illuminated A = 30 x 10 = 300m2

Illumination required E = 250 lux

Total lumens required = A x E = 300 x 250 = 75000

Utilization factor = 0.4

Gross lumens required = A x E/ U.F x D.F = 75000/0.4 x 0.9 = 2,08,333

Total wattage required = Gross lumens required / Lamp efficiency = 2,08,333/14 = 14,880

No. of lamps required = Total wattage required / Wattage of each lamp = 14,880/300 = 50 Ans.

Example 5; A factory hall 36m x 20m is to be illuminated so as to get an illumination of 20 candles/m2 on a working plane. Assume a suitable space-height ratio, mounting height utilization factor, and depreciation factor, calculate the number of lamps, wattage of each lamp, and their disposition.

Solution;

Assume the luminous efficiency = 15lm /W

Mounting height = 3m. space-height ratio = 1.3

Utilization factor (UF)= 0.5

Depreciation factor (DF) = 75%

Illumination required = Area x Illumination / UF x DF = (36 x 20) 20 / 0.5 x 0.75 = 65066l m

Total wattage = 65066/ 15 = 4388W

Horizontal distance between lamps (see figure below)

Space ratio = Spacing of lamps / Mounting height = 1.3 x 3 = 3.9, say 4m

No. of lamps along the length = 36/4 = 9 (distance at the two ends 2 m)

No. of lamps along the width = 20/ 4 = 5 (distance at the two ends = 2 m)

Total number of lamps = 9 x 5 = 45

Wattage of each lamps = 4338/ 45 = 98 W say, 100 W

Figure

depreciation factor



Example 6; A drawing hall 30m x 15m with a ceiling height of 5m is to be provided with an illumination of 120 lux. Taking the coefficient of utilization as 0.5, and the depreciation factor as 1.4, determine the number of fluorescent tubes required, their mounting height, spacing, and disposition.

Solution;

Total lumens required = Area x Illumination x D.F/ U.F

= (30 x 15) 120 x 1.4 / 0.5 = 151200lm

Total wattage with 80-watt tube (40lm / W)

=151200/40 = 3780 W

No. of lamps = 3780 /80 = 47.250, or say 48

48 tubes can be arranged in 24 sets as shown in figure below.

Assuming that the mounting height = 3 m

Space-height ratio= 3.34/3 = 1.11

This is within the limits

Figure

depreciation factor

Example 7; An illumination on the working plane of 75 lux is required in a room 72m x 15m in size. The lamps are required to be hunh 4m above the working bench. Assuming a suitable space-height ratio, a utilization factor of 0.5, a lamp efficiency of 14 lumens per watt and a candle power depreciation of 20%, estimate the number, rating, and disposition of lamps.

Solution;

Area to be illuminated A = 72 x 15 = 1080m2

Illumination required E= 75 lux

Total lumens required = A x E= 1080 x 75 = 81000

Depreciation factor = 1 – candle power depreciation

= 1- 0.2 = 0.8

Gross lumens required = A x E/ U.F x D.F = 81000/ 0.5 x 0.8 = 2,02,500

Total wattage required =  / Lamp efficiency

= 2,02,500/14 = 14,464

80 lamps in 4 rows, each having 20 lamps can be used giving spacings of 3.6 meters in length, and 3.75 meters in width, and space-height ratio of 0.9, and 0.9375 respectively.

Wattage of each lamp = 14464/ 80 = 180.08 = 200 W (say) Ans.

Disposition of lamps is shown in figure below.

Figure

depreciation factor




Example 8; A hall of 30 m long and 12 m wide is to be illuminated and illumination required is 50 m-candle. Five types of lamps having lumens outputs as given below are available;

Watts 100 200 300 500 1000
Lumens 1615 3650 4700 9950 21500

Taking a depreciation factor of 1.3, and utilization coefficient of 0.5, calculate the number of lamps needed in each case to produce required illumination. Out of above five types of lamps, select most suitable type and design a suitable scheme, and make a sketch showing location of lamps. Assume a suitable mounting height and calculate space-height ratio of lamps.

Solution;

Area to be illuminated A = 30 x 12 = 360 m2

Illumination required = 50 m – candles

Depreciation factor = 1.3

Utilization coefficient = 0.5

Gross lumens required = A x E x DF/ UF

=360 x 50 x 1.3 / 0.5 = 46, 800

If 100-watt lamps are used, the number of lamps required = 46800/ 1615 = 29 Ans.

If 200-watt lamps are used, the number of lamps required = 46800/ 3650 = 13 Ans.

If 300-watt lamps are used, the number of lamps required = 46800 / 4700 = 10 Ans.

If 500-watt lamps are used, the number of lamps required = 46800/ 9950 = 5 Ans.

If 1000-watt lamps are used, the number of lamps required= 46800/ 21500 = 2 Ans.

Let the mounting height be of 5 meters.

Most suitable type of lamps will be 300-watt lamps requiring 10 lamps in two rows, each row having 5 lamps giving spacing of 6 meters in length as well as in width and space-height ratio of 6/5, i.e., 1.2

Lamps of lower wattage will require a greater number of fittings and therefore, will increase the cost whereas of higher wattages will be few in requirement giving space-height ratio much more than desirable. Arrangement is shown I figure below.

Figure

depreciation factor



Example 9; An illustration on the working plane of 75 lux is required in a room of 72 m x 15 m in size. The lamps are required to be hung 4m above the working bench. Take utilization factor 0.5 and depreciation factor 0.8. The lamp efficiency is 14 lumens/watts. Calculate;

(a). Total wattage required

(b). No. of lamps used

Solution;

Illumination required E = 75 lux

Area of room (A) = 72 x 15 m2

Height of lamps (H) = 4 m

Utilization factor (UF) = 0.5

Depreciation factor (DF) = 0.8

Lamp efficiency (  = 14 lumens/ watt

Flux required = A x E/ U.F x D.F = 72 x 15 x 75 / 0.5 x 0.8 = 202500 lumens

(a). Total wattage required = Total flux /  = 202500 /14 = 14464.285 watt

(b). No. of 100 W lamps = 14464.285 / 100 = 144. 64 No.

Or,

No. of 200 W lamps = 14464.285 /200 = 72.32 No.

Or,

No. of 500 W lamps = 14464.285 / 500 = 28.92 (say) 29 No.

We will use 72 lamps of 200 W for a suitable arrangement, in 4 rows and 18 lamps in each row.

Length wise spacing = Total length / No. of lamps = 72 / 18 = 4

Width wise spacing = Total width / No. of rows = 15 / 4 = 3.75 m

Space – height ratio width wise = 3.75 / 4 = 0.93

The scheme is suitable because space – height ratio is within limit.

Note; Space – height ratio depends upon many factors such as type of light source, Flux intensity on the surface, and type of reflector. For suitable arrangement and uniform illumination, it can vary in between 0.66 to 1.5 limit.

Example 10; A room of 12 m x 12 m x 3 m size is to have direct lighting, giving illumination of 80 lux. Coefficient of utilization is 0.5, and depreciation factor is 0.8. if the efficiency of lamps available is 14.75 lumens per watt, find the No. of lamps and their ratings. Give the sketch by having suitable space–height ratio.

Solution;

A = 12 x 12 = 144 m2, E = 80 lux

Efficiency = 14.75 lumens / watt

U.F = 0.5, D.F = 0.8

Flux required = E.A / U.F x D.F = 80 x 144 / 0.5 x 0.8 = 28800 lumens

Total wattage required = Flux / Efficiency = 28800 / 14.75 = 1952 .54 watt

No. of 100 W lamps = Total wattage / Power of one lamp = 1952. 54 / 100

= 19.5254 say 20 No. Ans.

We will use 4 lamps / row, and 5 such rows length wise

Length wise spacing = 12/5 = 2.4,

Width wise spacing = 12 / 4 = 3

Length – wise space – height ratio = Length wise spacing / Mounting height = 2.4 / 3 = 0.8

Width – wise space – height ratio = width wise spacing / Mounting height = 3/3 = 1

The scheme is suitable because space – height ratio is between 0.8 and 1, which is normal. The spacing arrangement can be as shown in figure below;

Figure

depreciation factor



Example 11; It is desired to illuminate a drawing hall with an average illumination of 200 lux.  The hall is 30 m x 20 m2 in size. The lamps are to be fitted 4 m from ground floor. Find the number of lamps and wattage/lamp for the lighting scheme. Given efficiency of the lamps available as 25 lumens/watt, depreciation factor 0.8, and coefficient of utilization of 0.75, space height ratio between 0.8, and 1.2. give satisfactory spacing arrangement.

Solution;

E, required = 200 lux or lumen / m2

Floor area = 30 x 20 = 600 m2

Total flux = E x A = 200 x 600 = 120000 lumens

Desired light output = Total flux / DF x CU = 120000/ 0.8 x 0.75 = 200, 000 lumens

Wattage of lamps =   = 200000 / 25 = 8000 W Ans.

If we take 200 W lamps, then No. of lamps = 8000/ 200 = 40 Ans.

Rating / lamp = 200 W Ans.

Taking 5 rows of eight lamps, we have total distribution of 40 lamps.

Now length – wise spacing comes to be = 30 / 8 = 3.75 m

Width – wise spacing comes to be = 20 / 5 = 4 m

The lamps near the walls are kept at half the calculated spacing to keep uniformity of light. If for example, this distance is kept equal, then light between two adjacent lamps will be more as compared to illumination on the wall.

Figure

depreciation factor

Now length wise space – height ratio = 3.75 / height = 3.75 / 4 = 0.94 approx.

Width wise space – height ratio = 4 /4 = [Space / Height] = 1

Note; If in the given example, we choose 80 lamps of 100 W. Let us make 8 rows of 10 bulbs.

Width – wise space = 20/ 8 = 2.5

Space – height ratio = Space / Height = 2.5 / 4 = 0.625

Length – wise space comes to = 30 / 10 = 3 m

Space – height ratio = ¾ = 0.75, which does not satisfy the conditions.

If on the other hand, we choose say, 500-watt lamps, then

The number of bulbs comes to = 8000 / 500 = 16

Let us make 4 rows of 4 lamps each.

The length – wise spacing comes = 30 / 4 = 7.5 m

Space to height ratio = 7.5 / 4 = 1.87

Width – wise space = 20 / 4 = 5 m

Space to height ratio = 5 /4 = 1.25

So, it also fails to satisfy the given conditions of space height ratio, lying between 0.8 to 1.2. Thus, the scheme (five rows of eight lamps) given in the problem is correct.



Example 12; It is required to provide an illumination of 100 lumens / m2, in a workshop hall 40 m x 10 m. Assume depreciation factor is 0.8 and coefficient of utilization is 0.4, and efficiency of lamps is 14 lumens/watt. Calculate number of lamps and their positions when seven trusses are provided at a mutual distance of 5 m.

Solution;

E desired = 100 lumens /m2, Area = 40 x 10 = 400 m2

Lumens required = E x A = 100 x 400 = 40000 lumens.

Lamps output necessary = / C.U x D.F = 40000/ 0.4 x 0.8 = 125, 000 lumens

Efficiency watt = 14 lumens

Wattage = 125000 / 14 = 8929 watts

If we take 500 W lamps, the number of lamps = 8929 / 500 = 18 lamps.

Since breadth – wise trusses are provided, we have to use these trusses for fixing these lamps.

The scheme of fixing these lamps has been shown in figure below.

Figure

depreciation factor

Example 13; A reading room 50 m x 6 m, requires illumination of 40 meter – candle on the reading table. Assuming a space – height ratio of 1.2, calculate (i) the number of lamps required (ii) C.P of each. Assuming the utilization factor as 0.4, depreciation factor as 0.75, efficiency of each lamp as 0.75, watts per candle power and the height of lamps above reading table as 4 m. Draw a sketch of arrangement of lamps. 

Solution;

Area of the room = 50 x 15 = 750 m2

E desired = 40 m- candle = 40 lux = 40 lumens / m2

Flux reaching the surface = E x A = 40 x 750 = 30000 lumens

Now C.U = 0.4, and D.F = 0.75

Flux necessary to be produce = Flux required / C.U x D.F = 30000 / 0.4 x 0.75 = 100000 lumens.

C.P desired =  = 100000 / 4

= 0.75 watts / C.P

Total wattage required = C.P x  = 100000 / 4  = 0.75 = 5965 watts

If we use 200 watts lamps

Number of lamps = 5965 / 200 = 30 Ans.

(i). C.P of each lamp = Wattage /  (when  = W / C.P)

= 200 / 0.75 = 266. 67 C.P. Ans.

(ii). Height above the working surface = 4 m

Space height ratio = 1.2

Thus, space between adjacent lamp = 1.2 x 4 = 4.8 m (say, 5 m)




Example 14; An illumination of 300 lux is to be provided in a classroom 200 m x 10 m with 40 W fluorescent lamps. Determine the number and layout of lamps in the lighting installation. Assume data not given.

Solution;

Given room area = 20 x 10 = 200 m2

E required = 300 lux, wattage of tubes = 40 W

Assume coefficient of utilization = 0.5

Depreciation factor = 0.75

Efficiency of tubes = 40 lumens / watt.

Now flu required = E x A = 300 x 200 = 60000 lumens

Flux necessary to be produced taking different factors into account

 =  = 60000 / 0.5 x 0.75 x 40 = 4000 watts

Thus, number of tubes = 4000 / 40 = 40 W / tube given) = 100

Let us mount tubes in pairs. Thus, the No. of pairs = 100 / 2 = 50

Let us take 5 rows, so that pair in each row = 50 / 5 = 10

Figure

depreciation factor

Example 15; A drawing hall 30 m x 15 m with a ceiling height of 5 m is to be provided with general illumination of 120 lux. Taking a coefficient of utilization of 0.7, and maintenance factor of 0.8. Determine the number of fluorescent tubes required, their spacing and total wattage.

The lamps are mounted at a height of 3.0 m from floor, taking luminous efficiency of fluorescent tubes as 60 lumens / watt for 40 – watt tube.

Solution;

Area to be illuminated = 30 x 15 = 450 m2

Number of tube sets required N = E x A / O x CU x MF

N = 120 x 30 x 15 / 2400 x 0.7 x 0.8

= 54000 / 2400 x 0.7 x 0.8

No. of 40 W tubes required = 40

40 tubes can be converted to 20 tube sets of two tubes each. The placing of tubes should be so uniform that no pocket of the room should be without light. So, there will be three rows of 3 tube sets each. If single tube sets are to be used, the spacing may be calculated as under.

Spacing between lamps fitting:

Mounting height (given) = 3 m

Assuming the space – height ratio to be 1.25 or, Space / Mounting height = 1.25

Spacing between lamp fittings = 1.25 x 3 = 3.75 m

Therefore, No. of tube fittings with the width = 15 / 3.75 = 4

No. of lamp fittings along the length = 40 (single tube sets) / 4 = 10

Figure

depreciation factor



Example 16; An illumination of 75 lux is required on the working plane in a big hall of 40 m x 15 m. The lamps are required to be hung 4 m above the working bench. Assuming a suitable space height ratio, a utilization factor of 0.5, a lamp efficiency of 14 lumens per watt, and a candle power depreciation of 20%. Estimate the number, rating, and disposition of lamps.

Solution;

Total area to be illuminated A = 40 x 15 = 600 square meter

Illumination required E = 75 lux

Total lumens required E = A x E = 600 x 75 = 45000

Depreciation factor = 1 – Candle power dep.

= 1 – 0.2 = 0.8

Gross lumens required = A x E / UF x DF

= 45000 / 0.5 x 0.8 = 1,12,500

Total wattage required = Gross lumen / Lamp efficiency = 1,12,500/ 14 = 8036 W

No. of lamps of 200 W each = 8036 / 200 = 40 Approx.

Or, A x E / O x CU x MF = 600 x 75 / 2800 x 0.5 x0.8 = 40 lamps

(Assuming efficiency of lamp to be 14 lumens / watt) 40 lamps in 4 rows each having 10 lamps can be used giving approx. spacing of 3.75 m either side.

Example 17; A drawing hall 30 m x 15 m with a ceiling height of 5 m is to be provided with a general illumination of 120 lux. Taking a coefficient of utilization of 0.5, and maintenance factor 1.4. Determine the number of fluorescent tubes required, their spacing and mounting height, and total wattage, taking luminous efficiency of fluorescent tube as 40 lumens/watt for 80-watt tube.

Solution;

Area = 30 m x 15 m = 450 m2, Height = 5 m

E = 129 lux U.F = 0.5, D.F = 1.4

(a). No. of tubes required =? (b). Spacing?

©, Mounting height (d). Total wattage?

Efficiency of tube = 40 lumens / watt for 80 W

Flux required = A x E x D.F/ U.F = 450 x 120 x 1.4 /0.5 = 151200 lumens

Total wattage required = Total flux / Efficiency = 151200 / 40 = 3780 W

We will use 80-watt tubes then,

No. of 80-watt tubes = Total wattage / Wattage of one tube

= 3780 / 80 47. 25 = 48 No. (Approx.)

Figure

depreciation factor

The tubes will be hung at the height of 3 m above the floor

No. of rows = 6, No. of tubes in each row = 8

In case of tubes, the length of tubes should be subtracted from the total distance of room.

Length of one tube = 4 ft = 1.2 m

Length of 8 tubes = 1.2 x 8 = 9.6 m

Length wise spacing = Total length – Length of 8 tubes / No. of tubes in a row

= 30 – 9.6 / 8 = 2.55 m

Width wise spacing = Width of room / No. of rows = 15 / 6 = 2.5 m

Space height ratio length wise = Distance between tubes / Mounting height

= 2.55 / 3 = 0.83

The scheme is suitable because space height ratio is near to 1.



Example 18; Ninety double fixture of 40-watt warm white tubes are to be installed in nine rows, each row having 10 double features. How many circuits would you suggest to supply the tubes and, why? Which size of tube and circuit breaker would be feasible? How many wattage of each fixture would be taken for calculation purpose?

Solution;

No. of double fixture tubes = 90

Power of one tube = 40 W

No. of rows = 09

No. of double fixture in each circuit = 7

No. of circuits for supply =?

Size of fuse and circuit breaker =?

Wattage of each fixture =?

80- watt 7 – double fixture can be run on a sub circuit, and each row will be controlled separately.

No. of circuits for supply = No. of double fixture / Double fixture in each circuit

= 90 / 7 = 13 No. Approx.

No. of fluorescent lamp / circuit = Sub – circuit current x Supply voltage / 2 x Wattage of lamp or fixture

Sub – circuit current = No. of double fixture x 2 x Wattage of lamp or double fixture / Supply voltage

= 7 x 2 x 80 / 220 = 5.1 Amp.

10% added for future extension = 0.15 A

Total current = 5.1 +0.51 = 5.61 Amp.

Size of fuse = 5.61 A

Nearest fuse rating of available wire 30 SWG (0.0124-inch diameter) = 8.5 A

Size of circuit breaker = 1.2 x load current

= 1.2 x 5.1 = 6.732 A (refer table 10)

Available size of circuit breaker = 10 A (Fuji)

Wattage of each fixture = 2 x 40 = 80 – watt

Because double fluorescent tube is used in each fixture.

Example 19; Make necessary lighting arrangements of the workshop with 250 lux. The tubes will be of 40 watts each giving 2350 lumens. Draw suitable No. of circuit and connect them to the distribution board in the building. Plan of hall is attached.

Solution;

Lumens required in workshop (E) = 250 lux

Power of tube = 40 – watt

Lumen output of 40 – watt tube = 2350 lumens

Area of workshop = 17 m x 9 m = 153 m2 (Scale 1:100)

No. of circuits =?

Flux = A.E / UF x DF = 153 x 250 / 0.6 x 0.6 = 106250 lumen

No. of tubes = Total lumens / Lumen output of one tube

= 106250 / 2350 = 45.21 tubes

These number of tubes are not suitable for uniform distribution. Hence, 48 tubes will be used, double fixture in eight row each having three fixtures.

Length wise spacing = 17 / 8 = 2.125 m

Width–wise spacing = Total length width wise – Length of three tubes / 3

= 9 – (1.2 x 3) / 3 = 9 – 3.6 / 3 = 5.4 /3 = 1.8 m

Maximum lighting load can be controlled on a subcircuit = 80 watt

We can control two rows on one sub–circuit for suitable arrangement.

No. of circuit = No. of rows / 2 = 8 / 2 = 4 No.



Example 20; Draw the neat diagram showing the position of lamps in a hall of 15 m x 16 m x 4.5 m. The supply voltage 230 – volt single phase and installation 10 – watt /m2.

Solution;

Area = 15 m x 6 m = 90 m2

Illumination level = 10 – watt /m2

No. of lamps 100 – watts =?

Total wattage required in hall = Area x Wattage per m2

= 90 x 10 = 900 – watt

No. of lamps 100 – watt = Total wattage required in hall / Wattage of each lamp

= 900 / 100 = 9 No.

For suitable arrangement, 10 lamps will be hung on 3 – meter height from floor level.

No. of rows = 2

No. of lamps in each row = 5

Length – wise spacing = 15 / 5 = 3 m

Width – wise spacing = 6 / 2 = 3 m

Length wise space – height ratio = Length – wise spacing / Mounting height

Width – wise space – height ratio = 3 / 3 = 1

This scheme is suitable because space – height ratio is unity (1).

Figure

depreciation factor

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