VIETNAM GENERAL CONFEDERATION OF LABOR
TON DUC THANG UNIVERSITY
FACULTY OF ELECTRICAL ENGINEERING
E-LEARNING (30%)
COURSE ON ELECTRICAL SAFETY
ASSIGNMENT: DESIGN GROUNDING AND LIGHTING PROTECTION SYSTEMS
SUPERVISOR: DR. NGUYEN CONG TRANG
GROUP: 8
STUDENTS: 1. PHAN PHUOC SON, STUDENT ID:42300361
2. LE KHANH BINH, STUDENT ID:42300264
3. NGUYEN HUU THUAN, STUDENT ID:42300373
HO CHI MINH CITY, YEAR 2024
1
Table Report
No.
Full Name
Student ID
1
Phan Phuoc Son
42300361
2
Le Khanh Binh
42300264
3
Nguyen Huu Thuan
42300373
2
Task
Design lighting
protection
system using
ESE air terminal,
Design lighting
protection
system using
multiple
Franklin air
terminals, draw
AutoCad, type
Word
Grounding
system
design,type
Word, translate
Design lighting
protection
system using
Franklin air
terminal
Completion
percentage
100%
100%
100%
TABLE OF CONTENTS
Table Report ................................................................................................................................................................... 2
Part A:DESIGN OF GROUNDING SYSTEM ....................................................................................................................4
I: SELECTION AND DRAWING OF GROUNDING SYSTE CONFIGURATION ...............................................................4
II: MANUAL CALCULATION WITHOUT GEM CHEMICALS ........................................................................................5
III. SIMULATION USING SOFTWARE ........................................................................................................................6
IV. COMMENTS AND EVALUATION .........................................................................................................................7
PART B: DESIGN OF LIGHTNING PROTECTION SYSTEM ..............................................................................................8
Project 1 ..................................................................................................................................................................8
Question 1 ..........................................................................................................................................................8
Question 2 ..........................................................................................................................................................9
Question 3 ........................................................................................................................................................15
Project 2 ................................................................................................................................................................17
Question 1 ........................................................................................................................................................17
Question 2: .......................................................................................................................................................19
Question 3 ........................................................................................................................................................21
Project 3 ................................................................................................................................................................23
Question 1 ........................................................................................................................................................24
Question 2 ........................................................................................................................................................25
Question 3 ........................................................................................................................................................26
3
Part A:DESIGN OF GROUNDING SYSTEM
I: SELECTION AND DRAWING OF GROUNDING SYSTE CONFIGURATION
+ Draw the configuration:
4
II: MANUAL CALCULATION WITHOUT GEM CHEMICALS
We have condition resistance: Ryc  4()
Ptt  K m * pd  1.5*1440  2160(m)
+ We choose the grounding method in a loop circuit: because the transformer is over
1000kVA, the ground current is high, so it is suitable for distribution transformers with
capacity. We design it to include 100 grounding rods.
+ Resistance of 1 rod:
rc 
Ptt
4* Lc
2* h  Lc
2160
4*2.4
2*0.5  2.4
*[ln(
)]*

[ln(
)*
]  674()
3
2 * Lc
1.36* d
4* h  Lc 2 * 2.4
1.36*16*10
4*0.5  2.4
With n=100, we have:
a
6

 2.5  3
L 2.4
Check to the table => c  0.62,th  0.33
+ Resistance of each row: rc ' 
rc
674

 10.87()
n *c 100*0.62
+ Total length of the wire: Lt : Lt  9*6*10*2  1080(m)
d
4* S


4*70*106

 9.4*106 ( m)
Ptt
4* Lt
2160
4*1080
*[ln(
)  1] 
*[ln(
)  1]  6.4()
 * Lt
 *1080
h*d
0.5*9.4*10 3
r
6.4
Rth  th 
 19.4()
th 0.33
rth 
Rc 
rc
674

( )
nc *c 100 *0.62
RndHT 
Rth * Rc 10.9*19.4

 7  4()
Rth  Rc 10.9  19.4
Does not meet the grounding requirements for large capacity transformers, so it is
necessary to use GEM chemicals.
5
III. SIMULATION USING SOFTWARE
1. USING RODS
Rc =10.636(  )
2. USING WIRE
Rth =5.297(  )
RndHT 
Rth * Rc 5.297 *10.636

 3.5  4()
Rth  Rc 5.297  10.636
Meets the grounding requirements for large capacity transformers.
6
IV. COMMENTS AND EVALUATION
 Comments:
• It is necessary to review the data, calculations, and select a reasonable grounding
system according to the assignment requirements.
• Calculate to find the most optimal method (number of rods, GEM chemicals) to save
costs while still ensuring safety conditions as per regulations.
• It is advisable to use software due to its quick calculations, high accuracy,
optimization of the grounding system, and high complexity, especially regarding strict
technical requirements. However, manual calculations are required with simple
systems that can be flexibly adjusted to specific project requirements to save costs,
when there is no software support.
 Evaluation:
1. After manual calculation and using gem software to simulate the results have many
deviations:
o Manual calculation does not meet the requirements for grounding large
capacity transformers.
o Calculation using GEM software meets the requirements for grounding large
capacity transformers.
2. For transformer neutral grounding systems, if high accuracy and optimization are
required, using GEM software will bring better efficiency. However, if the project
conditions are not complicated and the cost is low, manual calculation can still be a
suitable option.
7
PART B: DESIGN OF LIGHTNING PROTECTION SYSTEM
Project 1
-Block 1(Yellow): D1  R1  C1=80m  50m  10m
-Block 2(Green): D2  R2  C2=70m  50m  5m
-Block 3(Blue): D3  R3  C3=50m  30m  7m
Question 1: Design and calculate for arranging Franklin air terminal to protect the entire
building? Comment?
Location of air terminal
8
 hx  10m
 amax 
(80  70) 2  (50  50)2
 90.14m
2
 r  1.6  h 
 1.6  h 
h  hx
 p  amax
h  hx
h  10
1  90.14
h  10
 h  73.95m Select h=74m
 hkim  h  hx =74-10=64m
In conclusion: The air terminal is too heigh, It’s not feasible in reality
Question 2: Design and calculation of lightning protection with central point and lightning
protection with classical air terminal:
Block 1
a=7  (h-hx)  p=7  2  1=14
Consider 80m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
80
 1 =6.71
14
 Select N=7
9
a=
80
=13.33m<14m
6
 have imaginary air terminal, no need to install more air terminal
h01 = h1 -
13.33
a
= 12 
=10.1m>10m
7 p
7
Consider 50m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
50
 1 =4.57 kim
14
 Select N=5 kim
a=
50
=12.5m<14m
4
 have imaginary air terminal, need to install more air terminal
h02 = h1 -
12.5
a
= 12 
=10.21m>10m
7 p
7
Block 3
h h
12  9
a11' = 1.6  h1  1 3  p = 1.6 12 
=2.74
h1  h3
12  9
 a1'3 =50- a11' =50-2.74=47.23m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
47.23
 1 =4.37
14
10
 Select N=5
a=
47.23
=11.8m<14m
4
h03 = h3 -
11.8
a
=9 
=7.31m>7m
7 p
7
Consider 30m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
30
 1 =3.14
14
 Select N=4
a=
30
=10m<14m
3
 have imaginary air terminal, no need to install more air terminal
h04 = h3 -
10
a
= 9  =7.57m>7m
7 p
7
Block 2
h h
12  7
a11' = 1.6  h1  1 2  p = 1.6 12 
=5.05m
h1  h2
12  7
 a1'2 =50- a11' =50-5.05=44.95m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
44.95
 1 =4.21
14
11
 Select N=5
a=
44.95
=11.24m<14m
4
 have imaginary air terminal, no need to install more air terminal
h05 = h2 -
11.8
a
=7
=5.31m>5m
7 p
7
h h
97
a33' = 1.6  h3  3 2  p = 1.6  9 
=1,8m
h3  h2
97
 a23' =50- a33' =20-1.8=18.2m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
18.2
 1 =2.3
14
 Select N=3
a=
18.2
=9.1m<14m
2
 have imaginary air terminal, no need to install more air terminal
h06 = h2 -
9.1
a
= 7  =5.61m>5m
7 p
7
Consider 50m>14m
12
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
50
 1 =4.5
14
 Select N=5
a=
50
=12.5m<14m
4
 have imaginary air terminal, no need to install more air terminal
h07 = h2 -
12.5
a
=7
=5.21m>5m
7 p
7
Consider 70>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
70
 1 =6
14
 Select N=6
a=
70
=14m=14m
5
 have imaginary air terminal, no need to install more air terminal
h08 = h2 -
14
a
= 7  =5m
7 p
7
Consider the overall protection of block 1
D1 =
502  13.332 =51.74m> 8  (h1  hx1 )  p =16
 Not satisfied, need to install more air terminal
D1 ’= 12.52  13.332 =18.25
 can accept
Consider the overall protection of block 2
D2 =
502  9.12 =51.82m> 8  (h2  hx 2 )  p =16
 Not satisfied, need to install more air terminal
13
D1 ’= 12.52  9.12 =15.45<16
 satisfied
Consider the overall protection of block 3
D2 =
302  11.82 =32.24m> 8  (h2  hx 2 )  p =16
 Not satisfied, need to install more air terminal
D1 ’= 102  11.82 =15.47<16
 satisfied
12  10
h h
r1 = 1.6  h1  1 x1  p = 1.6 12 
=1.74m
12  10
h1  hx1
97
h h
r3 = 1.6  h2  2 x 2  p = 1.6  9 
=1.8m
97
h2  hx 2
75
h h
r2 = 1.6  h3  3 x 3  p = 1.6  7 
=1.87m
75
h3  hx 3
10.1  10
h h
r01 = 1.6  h01  01 x1  p = 1.6 10.1
=0.08m
10.1  10
h01  hx1
10.21  10
h h
r02 = 1.6  h02  02 x1  p = 1.6 10.21
=0.17m
10.21  10
h02  hx1
7.31  7
h h
r03 = 1.6  h03  03 x 3  p = 1.6  7.31
=0.25m
7.31  7
h03  hx 3
7.57  7
h h
r04 = 1.6  h04  04 x 3  p = 1.6  7.57 
=0.47m
7.57  7
h04  hx 3
5.31  5
h h
r05 = 1.6  h05  05 x 2  p = 1.6  5.31
=0.26m
5.31  5
h05  hx 2
5.61  5
h h
r06 = 1.6  h06  06 x 2  p = 1.6  5.61
=0.52m
5.61  5
h06  hx 2
5.21  5
h h
r07 = 1.6  h07  07 x 2  p = 1.6  5.21
=0.17m
5.21  5
h07  hx 2
55
h h
r08 = 1.6  h08  08 x 2  p = 1.6  5 
=0m
55
h08  hx 2
14
Total 91 air terminals
Draw protection radius
Question 3: Design and calculation of lightning protection using modern ESE air terminal:
D  60m, I  15kA
amax 
1502  1002
 90.14( m)
2
Select hkim  4m, T  60 S  L  60m
R p (5)  5*(2* D  5)  L(L  2 D)  148,54m
4
R p1 (4)  * R p (5)  118,54m  90,14m
5
R p2 (7)  7*(2* D  7)  L(L  2 D)  149, 64m  90,14m
15
R p3 (9)  9*(2* D  9)  L(L  2 D)  150,33m  amax  62, 65m
R pGround (14)  14*(2* D  14)  L(L  2 D)  151,93  90,14m
Drawing Faraday cage
16
Project 2
Question 1: Design and calculate for arranging Franklin air terminal to protect the entire
building? Comment?
Location of air terminal
17
602  7 2
 amax 
 30.2m
2
h  hx
 r  1.6  h 
 p  amax
h  hx
 1.6  h 
h  45 5, 5

 30.2
h  45
h
 h  94.22m Select h=95m
 hkim  h  hx  95  45  50m
In conclusion: The air terminal is too heigh, It’s not feasible in reality
18
Question 2: Design and calculation of lightning protection with central point and lightning
protection with classical air terminal:
a1  7 *(h1  hx )* p  7 *(47  45)*
5.5
 11.23(m)
47
Consider 7m < 11.23m  have imaginary air terminal
h01  h 
a
7
 47 
 45.75m  45m
5.5
7* p
7*
47
Consider 40m>11.23m  need to install more air terminal
N
40
 1  4.56  select N=5 kim
11.23
N 40

 10  11.23
4
4
a
10
h02  h 
 47 
 45.2m  45m
5.5
7* p
7*
47
a
19
h  h 5.5
a11'  1.6* h1 * 1 2 *
 17.36( m)
h1  h2
h1
 a1'2  20  a11'  2.64  a2  7 * ( hx  hx1 ) * p  7 *(26  24) *1  14m
 have imaginary air terminal
h03  h1' 
a
2.64
 26 
 25.62(m) and p=1
7* p
7*1
Consider 7m in block 2 : 7m<14m  have imaginary air terminal
20
h04  h2 
a
7
 26 
 26(m)
7* p
7*1
Consider the protection of the entire building 1
D1  72  102  12.2  8*(hx  hx1 )* p  8*(47  45)*
5.5
 12.84(1)
47
Consider the protection of the entire building 2
D2  72  2.642  7.48  8*(h1  hx1 )* p  8*(26  24)*1  16(2)
From (1),(2)  satisfy
47  45 5.5
h h

 1.31m
r1 = 1.6  h1  1 x1  p = 1.6  47 
47  45
h1  hx1
47
26  24
h h
 1.66m
r2 = 1.6  h2  2 x 2  p = 1.6  26 
26  24
h2  hx 2
45.75  45
5.5
h h

 0.01m
r01 = 1.6  h01  01 x1  p = 1.6  45.75 
45.75  45
h01  hx1
45.75
46.78  45
5.5
h h

 1.17m
r02 = 1.6  h02  02 x1  p = 1.6  46.78 
46.78  45
h02  hx1
46.78
24.38  24
h h
 0.31m
r03 = 1.6  h03  03 x 2  p = 1.6  24.38 
24.38  24
h03  hx 2
25  24
h h
 0.82m
r04 = 1.6  h04  04 x 2  p = 1.6  25 
25  24
h04  hx 2
Total 14 air terminals
Draw protection radius
21
Question 3: Design and calculation of lightning protection using modern ESE needles:
Standard protection level ( I  15kA, D  60m )
amax 
602  7 2
 30.2m
2
hkim  2m, T  40 S  L  40m
R p (5)  5*(2* D  5)  L(L  2 D)  81.46m
2
R p1 (2)  * R p (5)  32.58m  30.2m
5
R p2 (23)  23*(23* D  23)  L(L  2D)  92.9m  30.2m
R p3round (47)  47 *(2* D  47)  L(L  2 D)  99.15m  30.2m
Drawing Faraday cage
22
Project 3
23
Question 1: Design and calculate for arranging Franklin air terminal to protect the entire
building? Comment?
Location of air terminal
482  602
 6 41  38.42m
=> amax 
2
 r  1.6  h 
 1.6  h 
h  hx
 p  amax
h  hx
h  12
1  38.42
h  12
 h  42.7 m Select h=43
 hkim  h  hx  43  12  31m
In conclusion: The air terminal is too heigh, It’s not feasible in reality
24
Question 2: Design and calculation of lightning protection with central point and lightning
protection with classical air terminal:
  tan 1
7
= 25
15
 need to install lightning arrester belt
amax  7  (h  hx )  p  7  2 1  14m
Consider 48m>14m
 no have imaginary air terminal, need to install more air terminal
L
a
N=  1 =
48
 1 =4.43
14
 Select N=5
a=
48
= 12m<14m
4
 have imaginary air terminal, no need to install more air terminal
h01 = h 
a
12
 14   12.3m  12m
7 p
7
r = 1.6  h 
14  12
h  hx
=1.72m
 p = 1.6 14 
14  12
h  hx
25
12.23  12
h h
 0.2m
r01 = 1.6  h01  01 x  p = 1.6 12.23 
12.23  12
h01  hx
Total 10 air terminals
Draw protection radius
Question 3: Design and calculation of lightning protection using modern ESE air terminal:
Standard protection level ( I  15kA, D  60m )
482  302
amax 
 28.3( m)
2
hkim  2m, T  40 S  L  40m
26
R p (5)  5*(2* D  5)  L(L  2 D)  83.52m
2
R p1 (2)  * R p (5)  33.41m  28.3m
5
R p2 (9)  9*(23* D  9)  L(L  2 D)  86,02m  28.3m
R p3round (14)  14*(2* D  14)  L(L  2 D)  88.79m  28.3m
Drawing Faraday cage
27