মঙ্গলবার, ২৯ মে, ২০১২
International system of Power Developments: Phenomenon of Corona
International system of Power Developments: Phenomenon of Corona: Phenomenon of Corona :-Electrical Transmission overhead line provides one of the most important property, which is a specific line voltag...
সোমবার, ২৮ মে, ২০১২
Phenomenon of Corona
Phenomenon of Corona:-Electrical Transmission overhead line provides one of the most
important property, which is a specific line voltage reveres occur. Transmission
efficiency and therefore provides a degree of power in the Loss decreased.
Moreover, because the line to get his prey
life span can be much reduced. And bodies created by the induced current harmonics line charging current increases and the
nearest line of the unwanted noise is telecommunication . Also If the
line is dirty or rough weather - rainy when the bodies of thousands of adverse
effects. in line due to the need to design the appropriate safety system, which provides line protection from
the harmful effects can be.
What is theCorona
the makes a bodies due to bad weight gases (ozone), theLos Angeles
They are less than the amount of voltage increase. The applied voltage can be increased if the
What is the
the makes a bodies due to bad weight gases (ozone), the
They are less than the amount of voltage increase. The applied voltage can be increased if the
.
break-down an upgrade, then air
insulation, the break in the hot blazing Conductors occurred over the Flash.
Effects ofCorona
i. Conductor all side Purple - Glow (violet glow) is observed.
ii. This hissing sound (hissing noise) generated by.
iii. Provides a certain amount of power is wasted. Which can be measured using meters. The transmission efficiency decreases.
iv. Provides for the weight of the gas line to the wire with the chemical reaction. The loss is received.
v. Max is rough and dirty conductors glow.
Factors affecting corona: -
i. Atmosphere.
ii. Conductor size.
iii. Spacing between conductors.
iv. Line Voltage.
Advantages and Disadvantages of corona: -
Benefit
. The Flash - over the possibility of reduced and system performance is improved.
. This and other reasons caused lightning reduce transient effect. It provides safety for the surgeon as the bulb .
Difficulty
. Provides the energy dissipation, or power loss, which affects the Transmission skills.
. Provides for the weight of gas (O3) is created. The loss of the metal parts.
. Bus bar particularly damaging for the bodies. The general decrease in the discharge because switchgear bus bar area.
Effects of
i. Conductor all side Purple - Glow (violet glow) is observed.
ii. This hissing sound (hissing noise) generated by.
iii. Provides a certain amount of power is wasted. Which can be measured using meters. The transmission efficiency decreases.
iv. Provides for the weight of the gas line to the wire with the chemical reaction. The loss is received.
v. Max is rough and dirty conductors glow.
Factors affecting corona: -
i. Atmosphere.
ii. Conductor size.
iii. Spacing between conductors.
iv. Line Voltage.
Advantages and Disadvantages of corona: -
Benefit
. The Flash - over the possibility of reduced and system performance is improved.
. This and other reasons caused lightning reduce transient effect. It provides safety for the surgeon as the bulb .
Difficulty
. Provides the energy dissipation, or power loss, which affects the Transmission skills.
. Provides for the weight of gas (O3) is created. The loss of the metal parts.
. Bus bar particularly damaging for the bodies. The general decrease in the discharge because switchgear bus bar area.
. Disruptive of the relation for Disruptive critical voltage, Visual critical Voltage and Power loss due to corona: -
. Disruptive critical voltage: - Minimum phase - to – neutral voltage bodies occur in the critical voltage Vc is the DISRUPTIVE.
Vc = mo go δ r loge D ÷ r kv / phase.
mo = irregularity factor
= 1 for polished conductors
= 0.98 to 0.92 for dirty conductors.
= 0.87 to 0.8 for standard conductors.
D = conductor spacing
r = conductor
radius
go = wind dies - electric strength
= 30 kv / cm (max) = 21.21 kv / cm (r.m.s.)
δ = Air density factor = 3.92 b ÷ 273 + t
= 1 standard condition
Corona
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5} kw / km / ph
Where
go = wind dies - electric strength
= 30 kv / cm (max) = 21.21 kv / cm (r.m.s.)
δ = Air density factor = 3.92 b ÷ 273 + t
= 1 standard condition
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5} kw / km / ph
Where
p = bodies found in Los Angeles
f = supply frequency in (Hz)
V = face to neutral voltage (r.m.s.)
DISRUPTIVE Vc = voltage / phase (r.m.s.)
Good weather is a 132 kv line corona loss 0.6 kw / km / ph. But the weather - rain, when it increased to 5/6 times it is. The 400 kv double circuit line, almost double the loss of 10/12 times.
Methods for minimizing corona: - 66 kv or big voltage provides the intense effect. Of the hands for protection from sub-stations of 66 kv high voltage equipment and bus-bar rating is determined. Ways provides the following effects can be reduced: -
1. By increasing conductor Size: - overhead line ACSR of using a relatively coarse.
2. By increasing conductor spacing: - If the conductor spacing provides increased voltage increases. It provides reduced effect. However, the conductor spacing should be increased to not more so. So that the support structure, such as - Cross - arm, support, and cost increases.
Solved problems on power loss due to corona:
1. 3-φ line 132 kv, 50Hz transmission line conductors 3 meter interval between is 2.5km, but standard conductors per km of line in determining the tax bodies in corona Loss.
Answer: - 3-φ line 132 kv standard conductors
r = 2.5 cm, D = 3 cm, t = 60 º, b = 90cm
corona loss P =?
We know,
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5} kw / km / ph
Here
δ = Air density factor = 3.92 b ÷ 273 + t
= 3.92 × 90 ÷ 273 + 60
= 1.06
go = wind dies - and the electric strength irregularity factor, respectively
go = 21.21 kv / cm (r.m.s.)
mo = 0.85 standard conductors
DISRUPTIVE critical voltage / phase, Vc = mo go δ r loge D ÷ r kv
= 0.85 × 21.2 × 1.06 × 2.5 loge (300 ÷ 2.5)
Supply voltage / phase V = 132 ÷ √ 3 = 76.21 kv
Power loss
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5}
P = 242.2 (50 +25 ÷ 1.06) √ 2.5 ÷ √ 300 (76.21-228.61) ² × 10 ¯ 5
= 363.33 kw / km / phase
Per km line total 3-φ corona loss
= 3 × 363.33
= 1090 kw (Ans.)
f = supply frequency in (Hz)
V = face to neutral voltage (r.m.s.)
DISRUPTIVE Vc = voltage / phase (r.m.s.)
Good weather is a 132 kv line corona loss 0.6 kw / km / ph. But the weather - rain, when it increased to 5/6 times it is. The 400 kv double circuit line, almost double the loss of 10/12 times.
Methods for minimizing corona: - 66 kv or big voltage provides the intense effect. Of the hands for protection from sub-stations of 66 kv high voltage equipment and bus-bar rating is determined. Ways provides the following effects can be reduced: -
1. By increasing conductor Size: - overhead line ACSR of using a relatively coarse.
2. By increasing conductor spacing: - If the conductor spacing provides increased voltage increases. It provides reduced effect. However, the conductor spacing should be increased to not more so. So that the support structure, such as - Cross - arm, support, and cost increases.
Solved problems on power loss due to corona:
1. 3-φ line 132 kv, 50Hz transmission line conductors 3 meter interval between is 2.5km, but standard conductors per km of line in determining the tax bodies in corona Loss.
Answer: - 3-φ line 132 kv standard conductors
r = 2.5 cm, D = 3 cm, t = 60 º, b = 90cm
corona loss P =?
We know,
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5} kw / km / ph
Here
δ = Air density factor = 3.92 b ÷ 273 + t
= 3.92 × 90 ÷ 273 + 60
= 1.06
go = wind dies - and the electric strength irregularity factor, respectively
go = 21.21 kv / cm (r.m.s.)
mo = 0.85 standard conductors
DISRUPTIVE critical voltage / phase, Vc = mo go δ r loge D ÷ r kv
= 0.85 × 21.2 × 1.06 × 2.5 loge (300 ÷ 2.5)
Supply voltage / phase V = 132 ÷ √ 3 = 76.21 kv
Power loss
P = {242.2 (f +25 ÷ δ) √ r ÷ √ D (V-Vc) ² × 10 ¯ 5}
P = 242.2 (50 +25 ÷ 1.06) √ 2.5 ÷ √ 300 (76.21-228.61) ² × 10 ¯ 5
= 363.33 kw / km / phase
Per km line total 3-φ corona loss
= 3 × 363.33
= 1090 kw (Ans.)
.
শুক্রবার, ২৫ মে, ২০১২
Voltage Distribution of Suspension Insulator
Voltage
Distribution of Suspension Insulator.Overhead line to the mechanical design is considered that the important issue of the suspension and the Voltage Insulator Division of the disk.
Most of the disk and the metal pin comprising insulators Porcelain. Insulator strings, so all across the capacitor and the pin - to - pin and the pin - to - Earth Capacitance influence. Voltage Insulator strings are equally active in the Division as well as a string of calculated predominates Capacitance. The pin - to - Earth (shunt)Capacitance because the string can not be everywhere at alls voltage equally. But what kind of steps required to improve the efficiency of the string it all needs to know.
Efficient string: -
N-number of units with a unit spark over voltage (s.o.v.) and over-voltage N times the spark Efficiency Ratio is a string. The N - number of units s.o.v. means that the maximum lateral line units.
Efficient string η = N - number of units s.o.v.
N × one units s.o.v.
= String of S.O.V. / N × line units s.o.v.(maximum voltage)
= (V / NVN × 100)
Efficiency of the entire unit if the strings above the string Equals the spark over voltage (S.O.V.) is out. Efficiency is an important factor for the overhead line as a string of design work.
Efficient string upgrade:-
Insulator metal parts comprising units of the string, as well as mutual Capacitance Cross - arm, etc., exist in the shunt Capacitance. The voltage per unit Capacitance effected imposed equally on the whole string can not exist. It can be seen, most reactions are line voltage unit. The decrease Cross - most are low voltage arm near the top of the unit. This significantly reduces the efficiency of the string Efficiency.
Systematic way, therefore each unit is equal to the voltage is raised through a string of Efficiency. String Efficiency is the development of methods, namely: -
1. By using longer cross-arms.
2. By grading the insulator units.
3. By using guard ring.
- By conducting glaze method.
5. By conducting glaze method.
Method of voltage grading in suspension insulators:
Different dimension of this process are different units chain (string) in the sorting or grading (grading) is, so that the maximum is Capacitance of minimum unit. Since the amount of voltage to current flow Capacitance vary. (V = i / ωC), line voltage will reduce the unit. On the other hand, other units of voltage increase. Thus, the grading of the Division and the voltage per unit to upgrade the skills of the string.
Now,
C = top of the unit mutual Capacitance
C1 = per unit Earth Capacitance. If the ratio m of the C1 and C,
C1 = m C
C2, C3, C4 , or 3rd and 4th units of mutual Capacitance.
V = voltage across each unit.
nowa and b point to karsoff’s Low the current formula,
I2 = I1 + Ia
ωC2 V = ω CV + ω m C V
C2 = C (1 + m) ............................. (i)
Or
I3 = I2 + Ib
ω C3 V = ω C2 V + 2ω m C V
C3 = C2 + 2 m C = C (1 + m) + 2m C = 3mC
C3 = C (1 + 3m) ........................... (ii)
C4 = C (1 + 6m) ............................ (iii)
Thus, a four-unit Insulator string of top-line unit to the unit replaced Capacitance ratio 1: (1 + m): (1 + 3m): (1 + 6m) will be equal to the per unit voltage.
Solved problems on string efficiency and voltage distribution: -
1. An overhead line suspension Insulator consists of 4 units . Transport and land are the differences between the 66 kV splendor. If the land in each insulators mutual Capacitance and Capacitance ratio 5: 1, the voltage across the string to determine the efficiency and the per unit tax.
Answer: - providedN = 4, V = 66 kv
m = C1 / C
= 1/5 = 0.2
String efficiency of η =?
V1, V2, V3 & V4 =?
V2 = V1 (1 + m)
= V1 (1 + 0.2)
= 1.2 V1
V3 = V1 (m² + 3m + 1)
= V1 [(0.2) ² + 3 × 0.2 + 1]
= 1.64 V1
V4 = V1 (m³ + 5m ² + 6m + 1)
= [(0.2) ³ + 5 × (0.2) ² + 6 × 0.2 + 1]
= 2.408 V1
= V1 + V2 + V3 + V4 = V1 (1 + 1.2 + 1.64 + 2.408) = V
Or
6.248 V1 = 66
V1 = 66 / 6.248
= 10.563kV
V2 = 1.2 × 10.563 = 12.676 kV
V3 = 1.64 × 10.563 = 17.324 kV
V4 = 2.408 × 10.563 = 25.4367 kV
Efficient string η = N - number of units s.o.v.
N × one units s.o.v.
% η = V / 4 × V4 × 100
= 66/4 × 25.4367 × 100
= 64.87% Ans:
2. Insulator with a 4 - units set and the 2nd or 1st Units, 10 kv and 12 kv, respectively, when the supply voltage and string to determine efficiency. Ans: - N = 4
V1 = 10 kv V2 = 12 kv
VL =? η =?
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1)
= 1.0 [(0.2) ² + 3 × 0.2 + 1]
= 10 × 1.64
= 16.4 kv
V4 = V1 (m³ + 5m ² + 6m + 1)
= 10 [(0.2) ³ + 5 × (0.2) ² + 6 × 0.2 + 1]
= 10 × 2.408
= 24.08 kv
V = V1 + V2 + V3 + V4
10 + 12 + 16.4 + 24.08
= 62.48 kv
Line Voltage VL = √ 3 × V
= √ 3 × 62.48
108.22 kv
Efficient string η = N - number of units s.o.v.
N × one units s.o.v.
% η = 62.48 / 24.08 × 100
= 64.86% (Ans).
3. A 3-φ overhead line in 3 Insulator a set of (string) is. If the voltage at the top of the 1st unit and 2nd units across Voltage with in 10 kv, 12 kv respectively is but the tax :- 1) m 2) Line Voltage 3) Efficient string.
VL =? η =?
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1)
= 1.0 [(0.2) ² + 3 × 0.2 + 1]
= 10 × 1.64
= 16.4 kv
V4 = V1 (m³ + 5m ² + 6m + 1)
= 10 [(0.2) ³ + 5 × (0.2) ² + 6 × 0.2 + 1]
= 10 × 2.408
= 24.08 kv
V = V1 + V2 + V3 + V4
10 + 12 + 16.4 + 24.08
= 62.48 kv
Line Voltage VL = √ 3 × V
= √ 3 × 62.48
108.22 kv
Efficient string η = N - number of units s.o.v.
N × one units s.o.v.
% η = 62.48 / 24.08 × 100
= 64.86% (Ans).
3. A 3-φ overhead line in 3 Insulator a set of (string) is. If the voltage at the top of the 1st unit and 2nd units across Voltage with in 10 kv, 12 kv respectively is but the tax :- 1) m 2) Line Voltage 3) Efficient string.
Ans :- 3-φ Line,
N = 3
V1 = 10 kv,
V2 = 12kv, m = ? VL = ? η = ?
We know that
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
earth and mutual Capacitance ratio
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1) = 1.64 V1
= 1.64 × 10 = 16.4 kv
V = V1 + V2 + V3
= 10 + 12 + 16.4 = 38.4 kv
Line Voltage VL = √ 3 × V = 66.5 kv
String Efficiency = V / NV3 × 100
= 38.4 / 3 × 16.4 × 100
= 78.05% (Ans).
N = 3
V1 = 10 kv,
V2 = 12kv, m = ? VL = ? η = ?
We know that
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
earth and mutual Capacitance ratio
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1) = 1.64 V1
= 1.64 × 10 = 16.4 kv
V = V1 + V2 + V3
= 10 + 12 + 16.4 = 38.4 kv
Line Voltage VL = √ 3 × V = 66.5 kv
String Efficiency = V / NV3 × 100
= 38.4 / 3 × 16.4 × 100
= 78.05% (Ans).
শনিবার, ১২ মে, ২০১২
Effect of Sag Part:- 02
It was learned that
the key effect sag? We know that mutual spacing of the conductive and Span Length: -
Interval must be sufficient so that the overhead lines of mutual conductive All side of air and air ayounized insulation can not break down.
The distance between two conductive to the two sources are used.
. Mutual interval = (0.01Vkv + 1.24D / W √ d) feet ............... (i)
Where,
Vkv = Voltage (kv)
D = conductor diameter (inch)
W = weight per unit length of conductor (ft)
d = sag (ft)
. Mutual interval = (0.075 √ d + Vkv ² / 20000) meters .................. (ii)
. Where d = sag (meters)
Overhead line conductor of the different types of mutual minimum distance of the voltage. A list is given.
Interval must be sufficient so that the overhead lines of mutual conductive All side of air and air ayounized insulation can not break down.
The distance between two conductive to the two sources are used.
. Mutual interval = (0.01Vkv + 1.24D / W √ d) feet ............... (i)
Where,
Vkv = Voltage (kv)
D = conductor diameter (inch)
W = weight per unit length of conductor (ft)
d = sag (ft)
. Mutual interval = (0.075 √ d + Vkv ² / 20000) meters .................. (ii)
. Where d = sag (meters)
Overhead line conductor of the different types of mutual minimum distance of the voltage. A list is given.
|
Operating
voltage
|
Vertical
spacing
|
Horizontal
spacing
|
Spacing
between conductor & support
|
|
400/300
V
|
46 cm
|
38 cm
|
15 cm
|
|
11 KV
|
76cm
|
115
cm
|
31 cm
|
|
22 KV
|
91cm
|
137
cm
|
46 cm
|
|
33 KV
|
122cm
|
153
cm
|
61 cm
|
|
66 KV
|
200cm
|
325
cm
|
76 cm
|
|
110
KV
|
315cm
|
495
cm
|
110
cm
|
|
132
KV
|
370cm
|
590
cm
|
130
cm
|
|
230
KV
|
645cm
|
1028
cm
|
226
cm
|
If it is possible to create low-cost Transmission lines may be useful. The maximum limit for the line span (optimum) to be. The support and Fixture such as the number of Insulator decreases, the cost will be lower. However, due to the increase in October sag support height will increase, while the conductor spacing, increasing cross - arm's length will increase. Span length of the line to create a way to control costs should be increased, and the other kit is a high cost.
Below the economic optimum span Powered_by_gitorious (most economical span length) and a different type and height of the support line and voltage sag and span of the Table: -
|
Line voltage (Voltage)
|
Type of support(Meters)
|
Hight of supports
(Meters)
|
Approx. Sag
(Meters)
|
Approx. Span
(meters)
|
|
400/300 V
|
Wooden,concrete,tubular
|
8.5-9.0
|
0.5-1.0
|
50-70
|
|
11000 V
|
Tubular, rail
|
9-10
|
1.0-1.5
|
80-120
|
|
33000 V
|
Rail pole
|
10-12
|
1.4-1.8
|
110-130
|
|
66000 V
|
Tower
|
20-25
|
2.0-3.0
|
180-240
|
|
132000 V
|
Do
|
25-30
|
4.5-6.0
|
250-350
|
|
2220000 V
|
Do
|
30-40
|
7.0-9.0
|
300-400
|
|
380000 V
|
Do
|
35-45
|
8.0-10
|
350-430
|
|
|
|
|
|
|
Notes:1. - An overhead line span 200m. Conductor cable cross 3cm ² meters and weight of 0.85kg. 2000 kg per cm ² and a safety factor of 2 when the maximum meaningful tension wire sag calculate tax?
Ans: - here,
m = 220 m, A = 2cm ², W / m = 0.85 kg.
The maximum meaningful tension = 2000 kg / cm ², safety factor = 2
d =?
We know, meaningful tenasana = maximum strength (strain) × cross / Safety Factor
T = 2000kg/cm ² × 3 cm ² / 2
= 3000 kg
Sag d = Wm ² / 8T
Ans. = 0.85 × (200) ² / 8 × 3000 = 1.71 m.
Notes: 2 - a river on a transmission line to the water surface and 30 m height, with the help of two tower has been constructed. Horizontal distance between towers 300 m. If the maximum strain carrying weight of 1590 kg and 0.8 kg All meters is, but how much above the river between the water surface may be conductive?
Ans: - P1 = 30 m, P2 = 40 m
T = 1590 kg, W / m = 0.8 kg
h = P1-P2 = 10 m
m = 300 m, H =?
Now,
X1 = m / 2 - Th / Wm
= 300/2 - 1590 × 10 / 0.8 × 300 = 83.75 m.
X2 = m / 2 + Th / Wm
= 300/2 + 1590 × 10 / 0.8 × 300 = 216.25 m.
X = m / 2-X1
= 150 - 83.75 = 66.25 m
d = W X ² 2/2T = 0.8 × (66.25) ² / 2 × 1590 = 1.1 m.
d2 = W X ² 2/2T = 0.8 × (216.25) ² / 2 × 1590 = 11.76 m
H = P2 - d2 + d = 40 - 11.76 +1.1
Ans = 29.34 m.
Tips: - 1. How many and what size sag?
Ans = There are three different ways, namely: -
1. Sag board method.
2. Dynamo meter method
3. Stop watch or return wave method.
Tips: - 2. There are two common species to carrying out the distance between you?
Ans = Mutual interval = (0.75 √ d + Vkv ² / 20000) meters.
Tips: - 3. Sag span of approximately 132 kv line support and how much?
Ans = 132 kv for -
Sag scale: 4.5 - 6 m.
Scale span: 250 - 350 m.
Tips: - 4. Span is allowed and the 220 kv sag?
Ans = Respectively 8 – 10 m. and 350 - 430 m.
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