Download Question B1

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.
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(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.
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[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]
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(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.
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[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]
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