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499
CALCULATION OF POWER WASTE IN ELECTRICAL NETWORKS IN THE
TERRITORY OF OUR REPUBLIC.
Narimanov Bahodir Absalomovich
Assistant of the Department of Energy
Jizzakh Polytechnic Institute
Аbstract:
The quality of electricity in the territory of our republic mainly depends on current
and narguz, taking into account the interaction of magnetic currents used in the control and
management of currents of power supply networks, frequency, voltage, currents.
Key words:
electricity, currents, power dissipation, control, voltage, magnetic flux, element,
Rogovsky belt - stem, reed switch, probability of working state, model, reliability indicators,
working ability.
РАСЧЕТ ЭНЕРГЕТИЧЕСКИХ ПОТЕРЬ В ЭЛЕКТРИЧЕСКИХ СЕТЯХ НА
ТЕРРИТОРИИ НАШЕЙ РЕСПУБЛИКИ.
Нариманов Баходир Абсаломович
Cтаршый преподаватель кафедра энергетики
Джизакский политехнический институт
Аннотация:
Качество электроэнергии на территории нашей республики в основном
зависит от тока и наргуза с учетом взаимодействия магнитных токов, используемых при
контроле и управлении токами электросетей, частоты, напряжения, токов.
Ключевые слова:
электричество, токи, рассеиваемая мощность, управление, напряжение,
магнитный поток, элемент, ремень Роговского - шток, геркон, вероятность рабочего
состояния, модель, показатели надежности, работопособность.
РЕСПУБЛИКAМИЗ ХУДУДИДAГИ ЕЛЕКТР ТAРМОҚЛAРИДA ҚУВВAТ
ИСРОФЛAРИНИ ҲИСОБЛAШ.
Нариманов Баходир Абсаломович
Энергетика кафедраси катта ўқитувчиси
Аннотация:
Республикамиз ҳудудидаги электр энергиясининг сифати электр тармоқлари
токларини,
частотасини,
кучланишини,
токларини
кузатиш
ва
бошқаришда
қўлланиладиган магнит оқимларнинг ўзаро таъсирини ҳисобга олган ҳолда асосан ток ва
юкламага боғлиқ.
Калит сўзлар:
электр, токлар, қувват сарфи, бошқарув, кучланиш, магнит оқим, элемент,
Роговский тасмаси - новда, қамиш калити, иш ҳолатининг эҳтимоллиги, модел,
ишончлилик кўрсаткичлари, ишлаш.
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volume 4, issue 4, 2025
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Here
r
and
x
are active and inductive resistances of the line;
I
a
and
I
R
–
active and reactive
components of the full load current
I.
As is known,
j
j
sin
3
;
cos
3
UI
Q
UI
Р
=
=
.
(3)
Full current through its active and reactive components
� cos � = �
�
, � sin � = �
�
(4)
we express:
I
a
and
I
P
in (3):
U
I
Q
U
I
Р
p
a
3
,
3
=
=
.
(5)
From now on
U
Q
I
U
P
I
p
a
3
;
3
=
=
Substituting the expressions (1) and (1) we get the following important
expressions:
.
r
U
S
r
U
Q
P
r
)
U
Q
U
P
(
r
I
Р
2
2
2
2
2
2
2
2
2
2
3
3
3
3
=
+
=
+
=
=
D
(6)
.
)
3
3
(
3
3
2
2
2
2
2
2
2
2
2
2
x
U
S
x
U
Q
P
x
U
Q
U
P
x
I
Q
=
+
=
+
=
=
D
(7)
Here is
S
full power.
Based on the expressions obtained above, we make the following conclusions:
The waste of active and reactive power depends on
P
and
Q.
.
,
3
2
1
3
2
1
n
z
n
z
Q
Q
Q
Q
Q
P
P
P
P
P
D
+
+
D
+
D
+
D
=
D
D
+
+
D
+
D
+
D
=
D
Here,
D
P
1
,
D
P
2
, … and
D
Q
1
,
D
Q
2
,…
are determined by expressions (6) and (7), respectively.
the load on a line per unit length by i
0 ,
r
)
I
I
(
r
I
Р
р
а
2
2
2
3
3
+
=
=
D
(1)
x
I
I
x
I
Q
р
а
)
(
3
3
2
2
2
+
=
=
D
(2)
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501
that is
i
0
=I/L
, A / km . Provider of the line at the beginning
l
of a name of length q
dl
within the
distance download
id
l
to is equal to . [3]
The power dissipation resulting from the flow of current through the resistance
r
0
dl of
the line
length
dl is:
dl
r
il
P
d
0
2
)
(
3
)
(
=
D
To determine the total power dissipation
D
P in
the entire line of length
L
, we add up all the very
small dissipation values
d
(
D
P ) in the interval 0 – L , that is:
.
3
3
3
)
(
3
2
2
2
2
0
3
0
2
0
0
2
0
2
0
0
2
0
0
Ѓз
Ѓз
r
U
Q
P
r
I
L
r
i
dl
l
r
i
dl
r
l
i
Р
L
L
L
+
=
=
=
=
=
D
(8)
In the above order
.
2
2
2
2
x
U
Q
P
x
I
Q
+
=
=
D
(9)
Thus, when the load is uniformly distributed along the line, the power loss is three times less
than when the load is at the end of the line.
For three-phase networks
.
3
,
3
3
3
U
S
I
UI
S
=
=
(10)
For single-phase networks
.
,
1
1
j
U
S
I
UI
S
=
=
(11)
Power dissipation for a three-phase network
3
3
3
2
3
3
3
,
3
x
I
Q
r
I
P
=
D
=
D
.
(12)
Power dissipation for a single-phase network
1
2
1
1
1
2
1
1
2
,
2
x
I
Q
r
I
P
=
D
=
D
.
(13)
Substituting (10) and (11) into (12) and (13), respectively, we get:
Р
1
+jQ
1
1
Р
2
+jQ
2
2
Р
3
+jQ
3
ΔР
1
+jΔQ
1
ΔР
2
+jΔQ
2
ΔР
3
+jΔQ
3
Р
a
+jQ
a
Р
в
+jQ
в
Р
с
+jQ
с
4.1- расм
.
L
l
i
dl
i
а)
б)
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volume 4, issue 4, 2025
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power dissipation for a three-phase network
3
2
2
3
3
2
2
3
x
U
S
Q
,
r
U
S
Р
=
D
=
D
;
(14)
power dissipation for a single-phase network
1
2
2
1
1
2
2
1
2
2
x
U
S
Q
,
r
U
S
Р
=
D
=
D
.
(15)
However, in a single-phase system there are two conductors, and in a three-phase system there
are three. To reduce the metal loss, the cross-sectional area of the conductors in a three-phase
network should be reduced by 1.5 times compared to that in a single-phase system. In this case,
the resistance increases by 1.5 times, that is,
r
3
=1.5
r
1
. Substituting this value into the
expression for
D
R
3
, we obtain the following:
1
2
2
3
)
/
5
,1
(
r
U
S
Р
=
D
Therefore, power loss in single-phase networks is 2/1.5=1.33 times more than in three-phase
networks. [3]
Here
D R
money
is the active power dissipated in the steel of the transformer (that is, in the core,
which is usually made of steel. [4]
т
2
100
b
U
S
%
I
Q
Q
н
с
с
пул
=
=
D
=
D
(17)
The active power dissipation in a short-circuit condition spent on heating the coils (this
dissipation is called the power dissipation in the copper) can be found as follows, as in formula
(6):
т
2
2
2
т
r
U
Q
Р
Р
н
+
=
D
(18)
In the same way, the loss of reactive power caused by the spread of the magnetic flux can be
determined as in the formula (7):
т
2
2
2
т
x
U
Q
P
Q
н
+
=
D
(19)
т
2
2
т
2
т
3
r
U
S
r
I
P
н
=
=
D
.
the relationship
D
R
t
/
D
R
k
we form the following expression:
2
т
)
S
/
S
(
Р
Р
Н
к
D
=
D
.
(20)
Literature
1. ..Nabijonovich JA Renewable energy sources in Uzbekistan //ACADEMICIA: An
https://ijmri.de/index.php/jmsi
volume 4, issue 4, 2025
503
International Multidisciplinary Research Journal. - 2020. - T. 10. – no. 11. - S. 769-774.
