Illumination Engineering

LessonAbout 1.5 h

Illumination Engineering is the branch of engineering that deals with the design, measurement, application, and installation of electric lighting systems to produce suitable visual conditions.

1. Point-by-Point Method (2-Dimension)

Inverse Square Law: illumination varies inversely as distance squared
Cosine Law: illumination varies directly as cosine of angle
Normal Line: perpendicular line to plane of illumination
Incident Ray: direct line from light source to point

Point-by-Point Inverse Square and Cosine Law

\[ E_p = \dfrac{I}{d^2} \cos\theta \]

\[ I = \dfrac{\Phi}{4\pi} \]

Where: \(E_p\) = intensity of illumination (\(\text{lux}\); \(\text{fc}\) in FPS), \(I\) = luminous intensity (\(\text{cd}\)), \(d\) = distance from lamp to point (\(\text{m}\); \(\text{ft}\) in FPS), \(\theta\) = angle of incidence between incident ray and normal line (\(\text{deg}\)), \(\Phi\) = total luminous flux (\(\text{lm}\))

EE Board April 1992

Example. A piece of paper lies on a table \(2\ \text{m}\) away from a point directly below a bulb of \(100\ \text{cd}\) and is \(4\ \text{m}\) above the table. Calculate the illumination on the center of the paper in lux.

Solution.

\[ E_p = \left(\dfrac{I}{h^2 + s^2}\right) \cos\left(\tan^{-1}\left(\dfrac{s}{h}\right)\right) = \left(\dfrac{100}{4^2 + 2^2}\right) \cos\left(\tan^{-1}\left(\dfrac{2}{4}\right)\right) = 4.47\ \text{lux} \approx 4.5\ \text{lux} \]

Answer: \(4.5\ \text{lux}\)

EE Board April 1989

Example. A medium unshaded lamp hangs \(8\ \text{m}\) directly above a table. To what distance should it be lowered to increase the illumination to \(4.45\) times its former value?

Solution.

\[ d_2 = \dfrac{d_1}{\sqrt{E_{p2}/E_{p1}}} = \dfrac{8}{\sqrt{4.45}} = 3.79\ \text{m} \]

Answer: \(3.79\ \text{m}\)

2. Point-by-Point Method (3-Dimension)

Three-Dimensional Coordinates: lamp positioned at coordinates \((x, y, z)\)
Slant Distance: \(d = \sqrt{x^2 + y^2 + z^2}\)
Cosine Angle Ratio: \(\cos\theta = z / d\)

Three-Dimensional Point-by-Point Illumination

\[ E_p = \dfrac{I z}{(x^2 + y^2 + z^2)^{3/2}} \]

\[ d = \sqrt{x^2 + y^2 + z^2} \qquad\qquad \cos\theta = \dfrac{z}{d} \]

Where: \(E_p\) = illumination at point \(P\) (\(\text{lux}\); \(\text{fc}\)), \(I\) = luminous intensity (\(\text{cd}\)), \(x, y\) = horizontal distances along perpendicular axes (\(\text{m}\); \(\text{ft}\)), \(z\) = elevation height of lamp above plane (\(\text{m}\); \(\text{ft}\)), \(d\) = 3D slant distance (\(\text{m}\); \(\text{ft}\))

EE Board October 1991

Example. A lamp of \(500\ \text{cd}\) is placed at the center of a room \(20\ \text{m} \times 10\ \text{m} \times 5\ \text{m}\). Calculate the illumination in each corner of the floor.

Solution.

\[ E_p = \dfrac{I z}{(x^2 + y^2 + z^2)^{3/2}} = \dfrac{500(2.5)}{(10^2 + 5^2 + 2.5^2)^{3/2}} = 0.83\ \text{lux} \]

Answer: \(0.83\ \text{lux}\)

3. Illumination from More Than One Light Source

Superposition Principle: total illumination equals arithmetic sum of individual illuminations
Multiple Luminaires: multiple sources contributing light to single point
Equi-Illumination Point: location where light levels from opposing sources balance

Superposition of Multiple Light Sources

\[ E_{p,\text{total}} = E_{p1} + E_{p2} + E_{p3} + \dots + E_{pn} \]

Where: \(E_{p,\text{total}}\) = total combined illumination at point \(P\) (\(\text{lux}\); \(\text{fc}\)), \(E_{p1}, E_{p2}, E_{pn}\) = illuminations from individual light sources \(1, 2, \dots, n\) (\(\text{lux}\); \(\text{fc}\))

Example. Two lamps A and B having intensities of \(300\ \text{cp}\) and \(500\ \text{cp}\) respectively are situated \(12\ \text{ft}\) apart. A screen is placed between them such that illuminations on both sides are equal. Calculate this illumination.

Solution.

\[ \dfrac{300}{x^2} = \dfrac{500}{(12 - x)^2} \implies \dfrac{12 - x}{x} = \sqrt{\dfrac{500}{300}} = 1.291 \implies x = 5.238\ \text{ft} \]

\[ E_p = \dfrac{300}{(5.238)^2} = 10.92\ \text{fc} \]

Answer: \(10.92\ \text{fc}\)

4. Mounting Height to Give Maximum Illumination

Optimum Height Ratio: height \(h = 0.707 s\) gives maximum illumination
Derivative Condition: maximum illumination found by differentiating \(E_p\) with height \(h\)
Fixed Distance: applies for fixed horizontal distance \(s\) from source

Mounting Height for Maximum Illumination

\[ h = \dfrac{\sqrt{2}}{2} s = 0.707 s \]

Where: \(h\) = optimal mounting height above surface (\(\text{m}\); \(\text{ft}\)), \(s\) = horizontal offset distance to point of interest (\(\text{m}\); \(\text{ft}\))

EE Board April 1985

Example. A light bulb is to be placed on a wall in order to obtain maximum brightness of illumination at a point on the floor that is \(3.6\ \text{m}\) from the wall. How high on the wall should the bulb be placed?

Solution.

\[ h = \left(\dfrac{\sqrt{2}}{2}\right) s = \left(\dfrac{\sqrt{2}}{2}\right)(3.6) = 2.54\ \text{m} \]

Answer: \(2.54\ \text{m}\)

5. Zonal Cavity or Average Illumination Method

Lumen Method: calculates average maintained illumination over entire floor area
Coefficient of Utilization: fraction of total lamp lumens reaching work plane
Maintenance Factor: factor accounting for lamp dirt and lumen depreciation
Depreciation Factor: reciprocal of maintenance factor (\(d.f. = 1 / M_f\))

Average Illumination (Lumen Method)

\[ E_{\text{maintained}} = \dfrac{\Phi \times N \times C_u \times M_f}{A} = \dfrac{\Phi \times N \times C_u}{A \times d.f.} \]

\[ E_{\text{initial}} = \dfrac{\Phi \times N \times C_u}{A} \]

Where: \(E_{\text{maintained}}\) = maintained average illumination (\(\text{lux}\); \(\text{fc}\)), \(E_{\text{initial}}\) = initial average illumination (\(\text{lux}\); \(\text{fc}\)), \(\Phi\) = luminous flux output per lamp (\(\text{lm}\)), \(N\) = total number of lamps, \(C_u\) = coefficient of utilization, \(M_f\) = maintenance factor, \(d.f.\) = depreciation factor (\(d.f. \ge 1\)), \(A\) = total floor area illuminated (\(\text{m}^2\); \(\text{ft}^2\))

EE Board October 1989

Example. The floor area of a room is \(10\ \text{m} \times 10\ \text{m}\). The desired illumination level is \(150\ \text{lux}\). Assume a coefficient of utilization of \(0.51\), a maintenance factor of \(0.95\), and fluorescent luminaires providing \(1200\ \text{lumens}\) output each. Determine the number of luminaires needed.

Solution.

\[ N = \dfrac{E \times A}{\Phi \times C_u \times M_f} = \dfrac{150(10 \times 10)}{1200(0.51)(0.95)} = 25.8 \approx 26\ \text{lamps} \]

Answer: \(26\ \text{lamps}\)

EE Board April 1990

Example. A hall \(30\ \text{m}\) long and \(12\ \text{m}\) wide requires an illumination of \(50\ \text{lux}\). Calculate the number of lamps required taking a depreciation factor of \(1.3\) and utilization coefficient of \(0.5\), given that each lamp output is rated \(1615\ \text{lumens}\).

Solution.

\[ N = \dfrac{E \times A \times d.f.}{\Phi \times C_u} = \dfrac{50(30 \times 12)(1.3)}{1615(0.5)} = 28.98 \approx 29\ \text{lamps} \]

Answer: \(29\ \text{lamps}\)

6. Illumination Units

System	Illumination (\(E\))	Candlepower (\(I\))	Distance (\(d\))
FPS / English	foot-candle (\(\text{fc}\))	candela (\(\text{cd}\))	foot (\(\text{ft}\))
MKS / SI	lux (\(\text{lx}\))	candela (\(\text{cd}\))	meter (\(\text{m}\))
CGS	phot (\(\text{ph}\))	candela (\(\text{cd}\))	centimeter (\(\text{cm}\))

Quantity	Unit Name	Equivalency / Relation
Luminous Intensity	Candela (\(\text{cd}\))	\(1\ \text{cd} = 1\ \text{candlepower}\ (\text{cp})\)
Luminous Flux	Lumen (\(\text{lm}\))	\(\Phi = 4\pi I\ \text{lm}\) (spherical source)
Foot-Candle	\(\text{fc}\) or \(\text{lm/ft}^2\)	\(1\ \text{fc} = 10.76\ \text{lux}\)
Lux	\(\text{lx}\) or \(\text{lm/m}^2\)	\(1\ \text{lx} = 1\ \text{meter-candle}\)
Phot	\(\text{ph}\) or \(\text{lm/cm}^2\)	\(1\ \text{ph} = 10,000\ \text{lux}\)
Luminous Efficiency	\(\text{lm/W}\)	\(\text{Efficiency} = \Phi / P\)

7. Formula Summary

Formula Name	Formula	Variables
Point-by-Point 2D Illumination	\(E_p = \dfrac{I}{d^2} \cos\theta\)	\(E_p\) = illumination (\(\text{lux}\); \(\text{fc}\)) \(I\) = luminous intensity (\(\text{cd}\)) \(d\) = distance (\(\text{m}\); \(\text{ft}\)) \(\theta\) = incidence angle (\(\text{deg}\))
Total Luminous Flux	\(\Phi = 4\pi I\)	\(\Phi\) = total luminous flux (\(\text{lm}\)) \(I\) = luminous intensity (\(\text{cd}\))
Point-by-Point 3D Illumination	\(E_p = \dfrac{I z}{(x^2 + y^2 + z^2)^{3/2}}\)	\(E_p\) = illumination (\(\text{lux}\); \(\text{fc}\)) \(I\) = intensity (\(\text{cd}\)) \(x, y\) = offset distances (\(\text{m}\); \(\text{ft}\)) \(z\) = lamp elevation (\(\text{m}\); \(\text{ft}\))
Three-Dimensional Slant Distance	\(d = \sqrt{x^2 + y^2 + z^2}\)	\(d\) = 3D slant distance (\(\text{m}\); \(\text{ft}\)) \(x, y, z\) = spatial coordinates (\(\text{m}\); \(\text{ft}\))
Superposition Illumination	\(E_{p,\text{total}} = E_{p1} + E_{p2} + \dots + E_{pn}\)	\(E_{p,\text{total}}\) = total illumination (\(\text{lux}\); \(\text{fc}\)) \(E_{p1}, E_{p2}\) = individual illuminations (\(\text{lux}\); \(\text{fc}\))
Mounting Height for Max Illumination	\(h = \dfrac{\sqrt{2}}{2} s\)	\(h\) = optimal height (\(\text{m}\); \(\text{ft}\)) \(s\) = horizontal offset distance (\(\text{m}\); \(\text{ft}\))
Maintained Average Illumination	\(E_{\text{maintained}} = \dfrac{\Phi \times N \times C_u \times M_f}{A}\)	\(E_{\text{maintained}}\) = maintained illumination (\(\text{lux}\); \(\text{fc}\)) \(\Phi\) = flux per lamp (\(\text{lm}\)) \(N\) = lamp count \(C_u\) = utilization coefficient \(M_f\) = maintenance factor \(A\) = floor area (\(\text{m}^2\); \(\text{ft}^2\))
Average Illumination with Depreciation Factor	\(E_{\text{maintained}} = \dfrac{\Phi \times N \times C_u}{A \times d.f.}\)	\(d.f.\) = depreciation factor (\(d.f. \ge 1\)) \(\Phi\) = flux per lamp (\(\text{lm}\)) \(N\) = lamp count \(C_u\) = utilization coefficient \(A\) = floor area (\(\text{m}^2\); \(\text{ft}^2\))
Luminous Efficiency	\(\text{Efficiency} = \dfrac{\Phi}{P}\)	\(\text{Efficiency}\) = luminous efficiency (\(\text{lm/W}\)) \(\Phi\) = total luminous flux (\(\text{lm}\)) \(P\) = power input (\(\text{W}\))