HANOI UNIVERSITY OF SCIENCE AND TECHNOLOGY

SCHOOL OF ELECTRIC AND ELECTRICAL ENGINEERING

REPORT POWER DELIVERY SYSTEM

Instructor: Mr. Lê Việt Tiến

Student: Kiều Việt Anh

ID student: 20212395

Ha Noi, 2025
I. Load characteristics
STT Shop name Installed demand Load
(kW) criticality

1 Cotton ginning shop 1400 Vital

2 Textile shop 2500 Vitual
3 Dyeing and prinfting shop 1200 Vital
4 Cleaning and packing shop 600 Vitual
5 Repair shop 200 Normal
6 Wood work shop 150 Normal
7 Pump store 100 Normal
8 Administration building 150 Normal
9 Ware house 50 Normal
Table1. Load characteristics of the factory

II. Load Calculation

* Estimating shop’s loads

Demand Shop name Installed demand Pshop (kW)

factor (kW)
0.8 Cotton ginning shop 1400 1120
0.8 Textile shop 2500 2000
0.8 Dyeing and prinfting shop 1200 960
0.8 Cleaning and packing shop 600 480
0.6 Repair shop 200 120
0.6 Wood work shop 150 90
0.6 Pump store 100 60
0.6 Administration building 150 90
0.6 Ware house 50 30
* Estimating total load of the plant
=> Choose diversity factor 𝐹𝐷 = 0.85

𝑃𝑝𝑙𝑎𝑛𝑡 = 𝐹𝐷 . ∑ 𝑃𝑠ℎ𝑜𝑝 𝑖 = 4207.5 𝑘𝑊

𝑖=1

* Estimating load of lighting(Choose 𝜎 = 20 W/𝑚2 )

- Example for calculating in research and development SHOP
− 𝑃𝑐 = 50𝑘𝑊 and 𝑃𝑠ℎ𝑜𝑝 = 30𝑘𝑊
 − 𝐷𝑒𝑚𝑎𝑛𝑑 𝐴𝑟𝑒𝑎: 𝑆 = 2 200𝑚2
− 𝑃𝑙𝑖𝑔ℎ𝑡 = 𝜎. 𝑆 = 15. 2200 = 33(𝑘𝑊)
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑜𝑤𝑒𝑟: 𝑃𝑝 = 𝑃𝑐 + 𝑃𝑙𝑖𝑔ℎ𝑡 = 64 + 33 = 97(𝑘𝑊)
=> The same calculation with other workshops
=> We have table

DF Shop name Installed demand Pshop 𝑃𝑙𝑖𝑔ℎ𝑡 S

(kW) (kW) (kW) (𝑚2 )

0.8 Cotton ginning shop 1400 1120 33.57 1678.5

0.8 Textile shop 2500 2000 32.5 1625
0.8 Dyeing and prinfting 1200 960 34.375 1718.75
shop
0.8 Cleaning and packing 600 480 10.3125 515.625
shop
0.6 Repair shop 200 120 6.5 325
0.6 Wood work shop 150 90 15 750
0.6 Pump store 100 60 8.75 437.5
0.6 Administration building 150 90 15.75 787.5
0.6 Ware house 50 30 16.5 825

* Load Center

Dynamic load

Lighting load angle

0
DF Shop name Installed Pshop 𝑃𝑙𝑖𝑔ℎ𝑡 Ptt(KW) S 𝛼𝑐𝑠
demand (kW) (kW) (𝑚2 )
(kW)
0.8 Cotton ginning 1400 1120 33.57 1153.57 1678.5 10.8
shop
0.8 Textile shop 2500 2000 32.5 2032.4 1625 5.85
0.8 Dyeing and 1200 960 34.375 994.375 1718.75 12.9
prinfting shop
0.8 Cleaning and 600 480 10.3125 490.3125 515.625 7.73
packing shop
0.6 Repair shop 200 120 6.5 126.5 325 19.5
0.6 Wood work shop 150 90 15 105 750 60
0.6 Pump store 100 60 8.75 68.75 437.5 52.5
0.6 Administration 150 90 15.75 105.75 787.5 63
building
0.6 Ware house 50 30 16.5 46.5 825 198

𝑆𝑡𝑜𝑡𝑎𝑙 = 5708

3 4
1718.75 515.625

1 2
1678.5 1625

5
325

9
825
6
750

8
787.5
7
437.5
II. FACTORY HIGH VOLTAGE NETWORK DESIGN
1. Select the power supply voltage level for the factory's high-voltage network

U= 4.43√𝑙 + 0.016𝑃 (kV)

With :
P: calculated power of the plant [kW]
L : distance from the intermediate substation to the factory [km]

U = 4.43√10 + 0.016 × 4207.5 = 38.95

The intermediate substation has voltage levels of (6,10,22,35) kV
Selecting the power supply voltage level for the plant : 35 kV.

2. Proposing the power supply diagram of the factory's high-voltage network

2.1. Select the power supply scheme from the factory power supply
- Determine load center:
∑𝑛
𝑖=1 𝑠𝑖 𝑥𝑖 ∑𝑛
𝑖=1 𝑠𝑖 𝑦𝑖 ∑𝑛
𝑖=1 𝑠𝑖 𝑧𝑖
𝑥0 = ∑𝑛
; 𝑦0 = ∑𝑛
; 𝑧0 = ∑𝑛
𝑖=1 𝑠𝑖 𝑖=1 𝑠𝑖 𝑖=1 𝑠𝑖

+ 𝑥0, 𝑦0 , 𝑧0 : load center coordinates

+ 𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 : load center coordinates i

+ 𝑠𝑖 : load capacity i

No. Shop name x y s

1 Cotton ginning shop 37.5 87.5 1274.6
2 Textile shop 70 87.5 2251.2
3 Dyeing and printing shop 102.5 105 1097.6
4 Cleaning and packing shop 122.5 105 542
5 Repair shop 150 72.5 139.2
6 Wood work shop 142.8 45 113.7
7 Pump store 137.5 25 74.6
8 Administration bulding 50 32.5 114.4
9 warehouse 90 57.5 48.7
- 𝑥0 = 489.3

- 𝑦0 = 817.6
= > M(489.3; 817.6)

2.2. Choose the workshop substation option

Method of using intermediate TBA:
- Placed at the intermediate substation 2 transformers with selected capacity as follows:
𝑆𝑡𝑜𝑡𝑎𝑙
𝑆𝑑𝑚 = 2
= 2854

- Check the capacity of the machine when an overload occurs: when a fault occurs in an ap transformer,
we can temporarily stop supplying power to all class III loads in the plant. Therefore, it is easy to see that
the selected transformer satisfies the condition when a fault occurs.
- So at the temporary intermediate transformer will place 2 transformers Sdm = 4000kV - 35/10.5 kV.
Method of using central distribution station.
- Power from the system is supplied to the workshop substations through the central distribution station.
As a result, the management and operation of the plant's high-voltage power network is more convenient,
the investment capital is reduced, the reliability of power supply is increased, but the investment capital
for the network is also large.

Table 2.2. Distance from stations to workshops

 Distances(m)
TBATT-B1 42.5
TBATT-B2 10
TBATT-B3 53
TBATT-B4 73
TBATT-B5 117.5
B1-PX8 70.5
B2-PX9 53
B4-PX5 63
B5-PX7 15.8
 STT Shop name Installed Pshop 𝑃𝑙𝑖𝑔ℎ𝑡 Ptt(KW) Symbol Sttpx
demand (kW) (kW)
(kW)
1 Cotton ginning 1400 1120 33.57 1153.57 B1 1263.32
shop
8 Administration 150 90 15.75 105.75
building
2 Textile shop 2500 2000 32.5 2032.4 B2 2078.9
9 Ware house 50 30 16.5 46.5
3 Dyeing and 1200 960 994.375 B3 994.375
prinfting shop 34.375
4 Cleaning and 600 480 10.3125 490.3125 B4 616.8125
packing shop
5 Repair shop 200 120 6.5 126.5
6 Wood work shop 150 90 15 105 B5 173.75
7 Pump store 100 60 8.75 68.75

4 Economic analys

*Cross-section of medium voltage cable

Choose Tmax = 5200
Using armored copper wire XLPE, Jkt = 2.7
𝑍 = (𝑎𝑣ℎ + 𝑎𝑡𝑐 ). 𝐾 + ∆𝐴. 𝑐 > MIN
+ Choose
𝑎𝑣ℎ = 0.1
+ Choose
𝑎𝑡𝑐 = 0.1
c: money price for 1kWH drop ∆𝐴: Voltage drop
* Drop voltage:

𝑇max=5200ℎ => 𝜏 = 3633ℎ

- Choose cable from TBATT to B2:
𝑆𝑡𝑡𝑝𝑥𝐵1 2078.9
Ilvmax = = 2√3 ∗35 = 17.15 (A)
2√3 ∗𝑈

Ilvmax
Fkt = = 6.35 (mm2)
Ikt
We choose XLPE cable with a cross section of 16mm2 and ICp=105A
Check the selected cable cross-section according to the heating condition :
KdIcp≥Isc
When
Isc is the current that occurs when a fault breaks a cable
Kd is the correction factor according to actual installation conditions, Kd = K1K2
K1 is the correction factor according to temperature, we take K1=1, K2 is the correction factor for
the number of cables placed in the same cable trench, in low-voltage networks, two cables are
placed in the trenches and the distance between the wires is 300 mm. K2 = 0.93
-Test:
0.93*Icp=97.65(A)>2*Imax=34.30 (satisfy)
Since the distance from TBATT to B2 is small, there is no need to check condition ∆Ucp .
- Choose cable from TBATT to B5:
𝑆𝑡𝑡𝑝𝑥𝐵1 173.75
Ilvmax = = 2√3 ∗35 = 1.43 (A)
2√3 ∗𝑈

Ilvmax
Fkt = = 0.53 (mm2)
Ikt

We choose XLPE cable with a cross section of 16mm2 and ICp=105A

Check the selected cable cross-section according to the heating condition :
KdIcp≥Isc
When
Isc is the current that occurs when a fault breaks a cable
Kd is the correction factor according to actual installation conditions, Kd = K1K2
K1 is the correction factor according to temperature, we take K1=1, K2 is the correction factor for
the number of cables placed in the same cable trench, in low-voltage networks, two cables are
placed in the trenches and the distance between the wires is 300 mm. K2 = 0.93
-Test:
0.93*Icp=97.65(A)>2*Imax=1.06 (satisfy)
-Check condition ∆Ucp
+Choose allowable voltage loss of 4%
4∗35000
 ∆Ucp = = 1400(V)
100
-XLPE cable with a cross section of 16mm2 has x0 = 0.128 Ω/km , r0 = 1.47 Ω /km

-The resistance and reactance of the line connecting the TBATT to B5 are:
R = r0 *l = 1.47 * 0.1175 = 0.173 Ω
X = x0* l = 0.128 * 0.1175 = 0.015 Ω

-Thus, the voltage loss on the line is:

𝑃∗𝑅+𝑄∗𝑋 173.75∗0.173
∆U = = = 0.86 (V) < 1400 (V) (satisfy)
𝑈𝑑𝑚 35

It is the same with other Subtations

We have table
Branch Udm(kV) STBA(kVA) I(A) Ikt(A/mm2) Fkt(mm2) Choose Icp(A)
F(mm2)

TBATT- 35 1263.32 10.42 2.7 3.86 16 105

B1
TBATT- 35 2078.9 17.15 2.7 6.35 16 105
B2
TBATT- 35 994.375 8.20 2.7 3.04 16 105
B3
TBATT- 35 616.8125 5.09 2.7 1.89 16 105
B4
TBATT- 35 173.75 1.43 2.7 0.53 16 105
B5