The document provides information on the design of electrical systems for residential buildings according to British standards. It includes sections on generation, transmission, distribution and residential modeling. It also covers topics like sockets, lighting circuits, power factor calculations, voltage drop analysis, short circuit calculations and street lighting design parameters. Diagrams and examples are provided to illustrate how to calculate the number of lights and sockets needed, size circuit breakers and cables based on load calculations.

1. According to
British Standard
1

2.  Prof. Dr. Eng. Almoataz Abdelaziz
 Dr. Eng. Mahmoud Abdallah
2

3.  Ahmed Abdelshafy Mohamed
 Ahmed Khaled Afifi
 Aya Mohamed Adel
 Aya Mostafa Mohamed
 Kareem Hamdy Sakr
 Ziad Adel Gad
3

4. Generation Transmission Distribution
4

5.  Residential Model
 Pillar and Transformer
 Voltage Drop
 Short Circuit
 Earthing System
 Street Lighting
 Cost Estimation
 Bill of Quantity
5

6. Sockets
Ahmed Khaled
6

7. 1. Normal Sockets
7
Room Purpose Minimum Number of
sockets (Twin)
Living room 4
Dining room 3
Kitchen 5
Bedroom 3
Corridor 1
Terrace 1
Garage 1

8. 8

9.  for domestic loads: 100% of the largest point + 40%
of the remainder.
For example
Circuit(LS1): Reception (area1) & Terrace (area10)
I(LS1)=1+0.4(1*9))=4.6 A
I(C.B)=
4.6∗1.25
0.89∗0.79
= 8.17 𝐴 ≈ 10A .:. C.S.A = 2mm² C.S.A = 4mm²
9
Grouping FactorTemp. Factor

10. Circuit
No.
Current C.B C.S.A
LS1 4.6 10A 4mm²
LS2 4.6 10A 4mm²
LS3 4.6 10A 4mm²
LS4 4.6 10A 4mm²
LS5 4.6 10A 4mm²
10
Summary

11. 2. Power Sockets
For example
Circuit (LP1): Bedroom(area7)
I(LP1)=
4∗746
230∗0.85∗0.8
=19.08 A P.F=0.85 , Eff=0.8 , V=230
I(C.B)=
19.08∗1.25
0.89∗0.79
= 33.92 A ≈ 40A .:. C.S.A = 10mm²
Summary
11
LP1 19.08 40A 10mm
²
LP2 10.23 26A 6mm²

12. 12

13. Lighting
Aya Mohamed Adel
13

14. Area Illumination Level
[lux]
Salon 150
Bedroom 100
Bathroom 200
Kitchen 300
Corridor 100
Terrace 100
14

15. 15
Lighting calculation approach
1. Manual
 N=
𝐸∗𝐴
𝐿𝐿𝐹∗𝑈𝐹∗𝑓
 I(LL)=
𝑁𝑥 𝑝
𝑉 𝐶𝑜𝑠𝜑

16. 16
 K=
𝑎∗𝑏
ℎ(𝑎+𝑏)
𝜌 𝑐𝑒𝑖𝑙𝑖𝑛𝑔
𝜌 𝑤𝑎𝑙𝑙
𝜌 𝑓𝑙𝑜𝑜𝑟
K
0.7
0.5
0.2
0.7
0.3
0.2
0.75 0.38 0.32
1.00 0.44 0.38

17. 17

18. For example
 Circuit (LL1): Reception(area1) +Terrace(area10)
+Corridor(area9)+Kitchen(area3)+Bedroom(area4)
 Reception (Area1)
N =
150 𝑥 32
0.4 𝑥 0.6 𝑥 6000
= 3.3 =4 Luminaires
18

19.  Corridor (Area9)
K =
1.2𝑥6.9
1.2+6.9 (3)
=0.34 ≈0.75.:. UF = 0.38
N =
100 𝑥 6
0.38 𝑥 0.6 𝑥 1800
= 1.4 = 2
19

20. 2. Dialux
20

21. 21

22. 22

23. 23

24. 24

25. 25

26.  I(LL1)=
2𝑥600 + 1 𝑥 22 + 2𝑥22 + 3 𝑥 28 +(1𝑥300)
230 𝑥 0.85
= 8.45A
I(C.B)=
8.45 𝑥 1.25
0.89 𝑥 0.79
= 15 ≈ 16A .:. C.S.A =2.5 mm²
C.S.A = 3mm²
26
Circuit
No.
Current C.B C.S.A
LL1 8.45 16A 3mm²
LL2 8.33 16A 3mm²
LL3 8.225 16A 3mm²

27. 27

28. Power Factor
Aya Mohamed Adel
28

29.  𝑃𝑆𝑜𝑐𝑘𝑒𝑡𝑠 = ⅀𝐼 𝑥 𝑉 𝑥 0.8 , 𝑄 𝑆𝑜𝑐𝑘𝑒𝑡𝑠 = 𝑃𝑆𝑜𝑐𝑘𝑒𝑡𝑠 x tan 36.86
 𝑃ℎ𝑒𝑎𝑡𝑒𝑟 = 2000 watt , 𝑄ℎ𝑒𝑎𝑡𝑒𝑟= 0
 𝑃𝐴𝐶 = 𝐻𝑜𝑢𝑟𝑠𝑒 𝑝𝑜𝑤𝑒𝑟x 746 , 𝑄 𝐴𝐶 = 𝑃𝐴𝐶 x tan 36.86
 𝑃𝑖𝑛𝑐 = ⅀ 𝑃 , 𝑄𝑖𝑛𝑐 = 0
 𝑃𝑓𝑙 = ⅀ 𝑃 , 𝑄 𝑓𝑙= ⅀ 𝑃𝑥 tan 36.86
Then we get 𝑃𝑡𝑜𝑡𝑎𝑙 , 𝑄𝑡𝑜𝑡𝑎𝑙 , tan∅ =
𝑄 𝑡𝑜𝑡𝑎𝑙
𝑃 𝑡𝑜𝑡𝑎𝑙
, then we get ∅
and finally
 𝐜𝐨𝐬 ∅ ≈ 0.85
29

30. Main Current Calculations
Aya Mostafa Eldeeb
30

31. Phase Lines Total Current
A
LP1
LL3
27.3A
B
LS1
LS2
LS3
LP2
24.03A
C
LL1
LL2
LS4
LS5
25.98A
31

32.  Panel Balance
Average loading =
27.3+24.03+25.98
3
= 25.77
Deviation PHA = 27.3 – 25.77 = 1.53
Deviation PHB = 24.03 – 25.77 = -1.74
Deviation PHC = 25.98 – 25.77 = 0.21
%unbalance=
𝑚𝑎𝑥.𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝ℎ𝑎𝑠𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔
≤ 10%
=
1.74
25.77
x 100% = 6.75%
 I(CB)= 27.307*1.25 = 34.1 ≈ 40A
∴ CSA= (4x16mm²)+16 mm² (TNCS)
S = 3*V*I = 3*230*27.3 = 19 KVA
32

33. Energy Meter
33

34.  High efficiency
34
Pre paid meter

35. Aya Mostafa Eldeeb
35

36.  I(feeder)= current of flat x no. of floor flats x no. of
floors x no. of buildings x diversity
 S (apparent power)= 3 x V x I
36

37.  I(feeder)= 32.5 x 12 x 1x 0.41 = 159.9A
S = 3 x V x I= 3 x 230 x 99.7 = 110 KVA
 I(feeder)= 32.5 x12 x 2 x 0.35 = 273 A
S = 3 x V x I= 3 x 230 x 273= 188.4 KVA
I(cable)=310 A
Cable rating between Pillar & Building
= 3*300mm²+150mm²
37

38.  All pillars rating is 200 KVA
 All transformers rating is 1 MVA
38

39. 39

40. 40

41. Point of view Egyptian code British code
Diversity Factor For lighting circuit=0.5
For normal sockets=0.4
For heater<3000watt=1
For air conditioner=0.5
I feeder=current of
flat x no. of floor
flats x no. of floors x
no. of buildings x
diversity
41

42. Ahmed Abdelshafy
42

43.  Voltage drop =
𝑚𝑉
𝐴𝑚
(𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑) 𝑥 𝐼(𝐴) 𝑥 𝑙𝑒𝑛𝑔𝑡ℎ(𝑚)
1000
, if CSA ≤
16 𝑚𝑚2
 Voltage drop =
𝐼 𝐴 𝑥 𝑉𝑟
2
+𝑉 𝑋
2
𝑥 𝑙𝑒𝑛𝑔ℎ𝑡(𝑚)
1000
, if CSA >
16 𝑚𝑚2
43

44.  Socket Circuit
CSA= 3x4 𝑚𝑚2, L= 20 m, mV/A/M=9.5 𝐼𝑙𝑜𝑎𝑑 =4.5 A
VD =
4.5 𝑥 20 𝑥 9.5
1000
= 0.87
VD% =
0.87 𝑥 100
230
= 0.37% ≤ 2%
44

45.  Summary
45
Cable Length(m) Voltage
Drop(v)
Voltage
Drop%
I(LS1) (3*4) 20m 0.87 0.37%
I(LS2) (3*4) 14m 0.61 0.26%
I(LS3) (3*4) 15m 0.65 0.28%
I(LS4) (3*4) 25m 1.09 0.47%
I(LS5) (3*4) 17m 0.74 0.32%
I(LL1) (3*3) 20m 2.52 1.09%
I(LL2) (3*3) 20m 2.49 1.08%
I(LL3) (3*3) 25m 3.07 1.33%
I(LP1) (3*10) 13m 0.94 0.40%
I(LP2) (3*6) 8m 0.52 0.42%

46. Ahmed Abdelshafy
46

47.  Short circuit level at transformer = 500MVA
 𝐼𝑠𝑐 =
𝑈 𝑠
𝑍 𝑇
= 1.05
𝑈 𝑁
𝑍 𝑇
 Determination of impedances:
1. Transformers impedance:
From tables:
Transformer of rating 1 MVA:
 Rtr = 1.94milli-ohm
 Xtr = 7.76 milli-ohm
47

48. 2. Aluminum cable from transformer to pillar:
Type of cable 2(4*300 mm²) and length equals (24.7 m)
 R=1.3585 (m.ohm)
 X= 1.729 (m.ohm)
48

49. 3. Aluminum cable from pillar to coffree:
Type of cable (4*300 mm²) and length equals (46 m)
 R= 5.06 (m.ohm)
 X= 3.22 (m.ohm)
49

50. 4. Aluminum cable from coffree to distribution
board:
Type of cable (4*16 mm²) and length equals (13 m)
 R= 26.8 (m.ohm)
 X is Neglected as C.S.A is smaller than 25 mm²
50

51. Short circuit current calculation:
1. Calculation of s.c current just after transformer.
 Ztransformer=8(m.ohm)
 Itransf=
230 𝑥 1.05
8
=30.1875 ≈31 KA
51

52. 2. Calculation of s.c current from transformer to pillar.
(Impedance of transformer + impedance of first Al
cable) =A
 Za= 1.3585 + 1.84 2 + (7.76 + 1.729)²=10.05(m.ohm)
 IscA=
230 𝑥 1.05
10.05
=24 KA
52

53. 3. Calculation of s.c current from pillar to coffree.
(Impedance of transformer + impedance of first Al cable +
impedance of second Al cable) =B
 Zb= 5.06 + 1.94 + 1.3585 2 + (7.76 + 1.729 + 3.22)² =
15.22(m.ohm)
 IscB=
230 𝑥 1.05
15.22
=15.87 ≈16 KA
53

54. 4. Calculation of s.c current from coffree to
riser.(Impedance of transformer + impedance of first
Al cable + impedance of second Al cable+ impedance
of riser) =C
 Zc= 5.06 + 1.3585 + 1.94 + 21 2 + (7.76 + 1.729 + 3.22)² =
31.6(m.ohm)
 IscC=
230 𝑥 1.05
31.6
=7.6≈ 10KA
54

55. Ahmed Abdelshafy
55

56. 56
 We are using TNC-S earthing system which gathers
safety and cost

57. 57

58. Kareem Hamdy
58

59.  Purpose
 Types of lamp and poles
59

60. SODIUM LAMP
Advantages:
High efficacy
Low power consumption
Low cost consumption
60

61. Normal pole
6m : Side streets, public gardens
8m&10m : Traffic routes
12m&15m : High speed dual carriageways
High mast
18M or more : Airports, stadiums and large industrial areas
61

62. Factors affecting street lighting design
A. Distance between poles
B. Height(1)
C. Overhang(2)
D. Boom angle(3)
E. Setback(4)
62

63. Design speed
(Km/h)
Setback (m) Height (m)
50 0.8 5 or 6
80 1 8 or 10
100 1.5 12 or 15
63
According to British standard code:
How to calculate setback, arm length ,
overhang?

64. Arm length =
1
4
height of pole
Overhang = Arm length-setback
64

65.  Cont. Factors affecting street lighting design
65

66. Street Lighting Feeder Pillar
 Responsible for feeding street lighting circuits
 Contains three phase breakers connected on 4 core
cable to feed poles
 Contains a Photocell/Timer
66

67. Design using DIALux
67

68. File ⊲ Wizards ⊲ Quick Street Planning
68

69. 69

70. Dialux select the class of street lighting according to:
 How high is the typical speed of the main user of the street?
 Weather type ?
 How many vehicles are there per day ?
 Does a conflict zone exist ?
 How is the street connected to other streets ?
70

71. 71

72. Street Lighting requirements
A. Average luminance (L)
B. Overall uniformity (U0)
C. Longitudinal uniformity (U1)
D. Threshold increment (TI)
E. Surrounding ratio (SR)
These requirements depends on Road Type
72

73. Select luminaire
73

74. 74

75. 75

76. 76

77. Street Lighting Summery
77
CSA(mm²)I(feeder
)
S(KVA
)
P(kw)No of
column
s
PillarTransform
er
1625.4617.647.530P5T5
1626.318.215.562P1T8
1617.812.310.542P2T8
1641.128.524.297P1T10
1637.3525.82290P1T11
1624.1916.7614.2
5
57P3T12
1628.4319.716.7
5
67P4T14
Total number of poles used = 445 poles

78. 78

79.  Example(1):
Indoor Lighting Circuit
79

80.  Example(2):
Outdoor Cables
80

81. Summary
81

82. 82

83. 83

84. 84

85. 85

86. ‫البريطانى‬ ‫للكود‬ ‫السعر‬ ‫اعمال‬ ‫نموذج‬
16,462574 Type A
6,178850 Type B
11,572256 Type C
8,600368 Type D
24,999214 Layout
67,813263 ‫االجمالى‬ ‫السعر‬‫للمشروع‬
86

87. 87