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SCHOOL OF ENGINEERING AND DESIGN UG AND PG TAUGHT PROGRAMMES Examination MODEL SOLUTIONS and MARKS Scheme Module code ME5004 Module title Electrical Service and Lighting Design Examiner(s) Section A: Dr. Y Ge ; Section B: Dr. M Kolokotroni Stationery No Special Requirements Special requirements None Please start each question on a new sheet. Solutions may be handwritten. Question No. Marks 1 MARKS Question A1 (a) To be protected against fault current, all conductors of the cable need to have: k 2 S 2 I 2t 2 where I t is the energy let-through to the particular conductor by the circuit protective device under fault conditions, k is a constant for the type of conductor, and S is the conductor crosssectional area. [2] Live conductors need to be protected against fault currents up to the value of the prospective short circuit current. [2] Protective conductors need to be protected against the earth fault current resulting from the worst-case earth fault loop impedance. [2] (b) For a three-phase fault the impedance to the fault current is simply the phase impedance. transforme r impedance 3.8 j17.3 m [2] At the Intake Switchboard: Total phase impedance = transformer phase impedance + feeder cable impedance The feeder cable comprises two cables in parallel, therefore the lowest impedance (when the conductors are cold) will be half that of a single cable. Thus: 1 feeder cable impedance 0.075 j0.075 20 m 2 0.75 j0.75 m [2] Total phase impedance = 4.55 + j18.05 m and prospective short circuit current = 433 / 3 4.552 18.052 13.4 kA [2] At the Load: Total phase impedance = impedance at intake switchboard + sub-main cable impedance The sub-main cable lowest impedance (when the conductors are cold) will be: sub - main cable impedance 0.866 j0.084 50 m 43.3 j4.2 m [2] Total phase impedance = 4.55 + j18.05 + 43.3 + j4.2 = 47.85 + j22.25 m and prospective short circuit current = (c) We need S I 2t k2 433 / 3 47.852 22.252 3 10 10.7 mm 2 2 51 4.7 kA [2] 5 Question A2 2 [4] (a) The total monthly cost is given by: cost unit charge kWh used MD charge kVA MD unit charge kWh used MD charge kW MD pf [2] where pf is the power factor. cost kW MD equation 1 unit charge MD charge kWh kWh used pf kWh used But, lf , where T is the metering period - in this case one month. kW MD T kW MD 1 kWh used lf T [2] Substituting into equation 1 gives: cost MD charge unit charge kWh pf lf T equation 2 [2] (b)(i) T = 30 days 24 hours/day = 720 hours, and lf = 0.4 which, substituted into equation 2, gives: cost 252 1 5 5 kWh pf 0.35 720 pf cost 6.67 p Before improvement, pf = 0.6 giving kWh Improved cost will be 95% of this, i.e. 6.33 p, which, substituted into equation 2, gives: 1 6.33 5 pf 1 pf 0.75 1.33 [1] [2] [2] (b)(ii) At 0.6 pf: apparent power = 300 kVA real power = 300 0.6 = 180 kW (which will not change as pf is improved) reactive power 300 2 180 2 240 kVAr At 0.75 pf: 180 240 kVA 0.75 real power = 180 kW (still) apparent power = 3 [2] reactive power 240 2 180 2 158.7 kVAr reactive power from the capacitor = 240 – 158.7 = 81.3 kVAr [3] The capacitors in star are connected to the line voltage, 400 V, and therefore the capacitive reactance per phase will be given by: V2 400 2 Xc 1.97 VAr 81255 (b)(iii) Harmonic distortion results in the flow of higher frequency currents. Capacitive reactance is inversely proportional to frequency, so the capacitors present a low impedance [2] path to the high frequency components. Larger harmonic currents may then flow and overload the capacitors. (2) [2] 4 Question B1 (a) MARKS Hm=4.15-0.8= 3.35m Room Index = LxW 25 x 25 = = 625/167.5=3.73 Hm( L W ) 3.35 x(25 25) RI should be taken as 3.5 as we always round down to consider worse case scenario. From Table B1.3 with Reflectance 0.5, 0.5 and 0.2 Utilisation Factor = 0.53 From the Lumen formula for required Illuminance = 350 lux E= F n N UF MF A EA 350 x(25 x 25) = = 218750/2342.6=93.4 2600 x 2 x0.53 x0.85 F n UF MF luminnaires So N = For required Illuminance = 380 lux [5] N=93.4x380/350=101.4 luminnaires (b) 100 Luminnaires should provide the required illuminance within the specified range. As the floor plan is square they should be arranged 10 x10 A floor plan diagram should be shown with spacing 25/10 = 2.5m and 1.25m [4] at the edges of the room. (c) Average illuminance at the top of desks would be 350 x 100 / 93.4= 374.7 Lux which is within the specified range. [1] [2] (d) SHR = 2.5 /3.35 = 0.746. This is less than the SHR max stated (1.62) so uniformity of illuminance should be satisfactory. 5 (e) Utilisation factor for the walls. Transfer and distribution factor are shown below for RI=3 and reflectance 0.5, 0.5, 0.2. These are determined from Tables B1.2 and B1.3. TF (F,W) TF (W,W) TF (C,W) TF DF Product=Utilisation Factor 0.091 1.172 0.19 0.46 0.09 0 0.04186 0.10548 0 [4] Utilisation Factor, UF(W) = 0.04186 + 0.10548= 0.14734 (f) Wall illuminance for the office Ewall = F n N UF (W ) MF 2600 x 2 x100 x0.14734 x0.85 = = (25 25 25 25)x3.35 wall _ area [2] 65124/335 = 194.4~ 195 lux (g) The ratio of wall to desk top illuminance is 195/375 = 0.52 This is within the required range of 0.5-0.6 as CIBSE recommends Therefore, wall lighting would not need to be supplemented. 6 [2] Question B2 (a) MARKS (i) 1.2 m luminaire 2.1 m Bench E A io sin cos 2h io=total source intensity/source length = 600/1.2= 500 cd/m Therefore tan io 500 119 2h 2 x2.1 1.2 0.5716 2.1 Therefore =29.74o Therefore =0.591 radians sin = 0.4961 cos = 0.8682 sin cos = 0.4307 Therefore E=119 x [0.591 + 0.4307] = 113 lux [6] 7 (a) (ii) 0.9 m 0.3 m 1.2 m luminaire 2.1 m E1 Bench E2 B Total illuminance at specified position is E1 + E2 lux 0.9 m source tan 1 0.9 0.4286 2.1 Therefore 1=23.19 Therefore 1=0.4049 radians sin 1 = 0.3939 cos 1 = 0.9181 sin 1 cos 1 = 0.3621 Therefore E1=119 x [0.4049 + 0.3621] = 91.27 lux 0.3 m source tan 2 0 .3 0.1429 2.1 Therefore 2=8.13o Therefore 2=0.1419 radians sin 2 = 0.1414 8 cos 2 = 0.9899 sin 2 cos 2 = 0.1399 Therefore E2=119 x [0.1419 + 0.1399] = 33.48 lux Therefore total illuminance is 91.27+33.38=130.65 Lux [6] (b) (i) Photopic vision: The cones of the retina of the human eye (some 7 million in number) are responsible for daylight vision, in three dimensions and in colour, with three distinct photo pigments giving sensitivity to red, blue and green. The response of the cones is referred to as photopic vision and this is applicable above luminance levels of about 10 cd/m2. This is the kind of vision that is of greatest interest in the context of building lighting. Scotopic vision: The rods of the retina of the human eye (some 130 million of them) are responsible for scotopic visions, through the photosensitivity of a fourth pigment. This night vision is effective over a range of low perceptible luminance, from about 3 x 10-6 cd/m2tpo about 0.01 cd/m2. The lower luminance limit of scotopic vision is below that due to starlight on a moonless night. The rods are not capable of providing colour vision, and scotopic visions is therefore in black, white and shades of grey. Depth perception is also poor in scotopic vision. Adaptation: The eye commonly takes between 10 and 30 minutes to adapt from photopic to scotopic vision. Hence, the perception of brightness depends not only on the luminance of the source, but on the state of adaptation of the eye. The process of adaptation to a different level of illuminance is a complex one. The adjustment in pupil size and the neurological responses occupy only a second or so. However, the photochemical changes in the cones of the retina commonly take 10 minutes and those in the rods as long as 60 mins. The reverse process of adaptation from a low to a high level of illuminance may take only about 2 minutes. Apparent brightness: The apparent brightness of a given object which determines the human perception of an illuminated object is related to the luminance rather than to illuminance. The relationship is not linear, with apparent brightness varying more slowly than luminance. The apparent brightness of a given object also depends on the state of adaptation of the observer’s eye, which in turn is determined by the luminances and positions of other surfaces in, say, the interior of a room. 9 [4] (b) (ii) Luminance: The luminance of the light emitted from a lamp in a given direction is the intensity in that direction divided by the surface area of the source projected in the direction of light emission. Thus we have: L I cd/m2 As where L=luminance, I= intensity and As the projected area of the source. Lighting task: The term lighting task is used to specify a particular visual activity, for example, recognising the presence of a specific object, or identifying a specific alphabetic character. The definition of a task involves not only the characteristic of the visual display itself, but also the characteristics of the background against which is to be seen, and factors such as the age of the observer, and the importance of speed and accuracy in performance of the task. Contrast: Contrast can be defined formally as: C Lt Lb Lb where Lt is the luminance of the task details, foe example, the decimal point on the printed page under a given illuminance, and Lb is the background luminance, for example, that of the paper on which the decimal point is printed. C has a positive value, regardless the sign of Lt – Lb. [4] 10