CALCULATION OF POWER WASTE IN ELECTRICAL NETWORKS IN THE TERRITORY OF OUR REPUBLIC.

Аннотация

Качество электроэнергии на территории нашей республики в основном зависит от тока и наргуза с учетом взаимодействия магнитных токов, используемых при контролеиуправлениитокамиэлектросетей,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 ХУДУДИДAГИЕЛЕКТР ТAРМОҚЛAРИДAҚУВВAТ ИСРОФЛAРИНИ ҲИСОБЛAШ. НаримановБаходир 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 юкламагабоғлиқ. Калит сўзлар: электр, токлар, қувват сарфи, бошқарув, кучланиш, магнит оқим, элемент, Роговский тасмаси - новда, қамиш калити, иш ҳолатининг эҳтимоллиги, модел, ишончлиликкўрсаткичлари,ишлаш. https://ijmri.de/index.php/jmsi volume 4,issue4,2025 499 DР =3I2r =3(I2 + I2 )r (1) а р DQ =3I2x=3(I2 +I2)x (2) а р Here r and x are active and inductive resistances of the line; I and I – active and reactive a R components ofthefull loadcurrent I. As isknown, Р= 3UIcosj; Q= 3UIsinj. (3) Full current through itsactive and reactivecomponents (4) weexpres sc:os = , sin = I and I in(3): a P Р = 3I U, Q = 3I U . (5) a p From nowon P Q I = ; I = Substituting the expressions (1) and (1) we get the following important a p 3U 3U expressions: P2 Q2 P2 +Q2 S2 DР=3I2r =3( + )r = r = r. (6) 3U2 3U2 U2 U2 P2 Q2 P2 +Q2 S2 DQ=3I2x=3( + )x= x= x. (7) 3U2 3U2 U2 U2 Here is Sfullpower. Based on theexpressions obtained above, wemake thefollowing conclusions: Thewaste ofactive andreactive power depends onPand Q. DP =DP + DP + DP + + DP , z 1 2 3 n DQ =DQ + DQ + DQ + +DQ . z 1 2 3 n Here,DP ,DP ,…andDQ ,DQ ,… are determined byexpressions(6) and (7), respectively. 1 2 1 2 theloadon alineper unitlength byi 0, https://ijmri.de/index.php/jmsi volume 4,issue4,2025 500 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 а) L l i i that is i 0 =I/L , A / km .Provider of the linea4t.t1h-e d р l bасeмgi.nninбg)l of a name oflength q dl within the distancedownload idl to is equal to.[3] The power dissipation resulting from the flow of current through the resistance r dl of the line 0 length dl is: d(DP)=3(il)2r dl 0 To determine the total power dissipationDP in the entire line of length L , we add up all the very smalldissipation values d(DP )in theinterval 0–L ,that is: L L L3 L P2 +Q2 DР=Ѓз3(i l)2r dl =3i 2r Ѓlз2dl =3i 2r = I2r = r. (8) 0 0 0 0 0 0 3 U2 0 0 0 In theabove order P2 +Q2 DQ=I2x= x. (9) U2 Thus, when the load is uniformly distributed along the line, the power loss is three times less than when the load isat theend oftheline. S For three-phase networksS = 3UI , I = . (10) 3 3 3U Forsingle-phase networks S S =UI , I = j. (11) 1 1 U Powerdissipationfor athree-phase network DP = 3I2r , DQ = 3I x . (12) 3 3 3 3 3 Powerdissipationfor asingle-phase network DP = 2I2r , DQ = 2I2x . (13) 1 1 1 1 1 1 Substituting(10) and(11) into(12) and (13), respectively, weget: https://ijmri.de/index.php/jmsi volume 4,issue4,2025 501 powerdissipation fora three-phase network S S2 DР = 2 r , DQ = x ; (14) 3 U 3 3 U2 3 2 powerdissipation fora single-phasenetwork 2S2 2S2 DР = r , DQ = x . (15) 1 U2 1 1 U2 1 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 =1.5 r . Substituting this value into the 3 1 expressionforDR ,we obtain thefollowing: 3 DР =(1,5S2/U2)r 3 1 Therefore, power loss in single-phase networks is 2/1.5=1.33 times more than in three-phase networks. [3] Here D R is the active power dissipated in the steel of the transformer (that is, in the core, money which is usuallymade ofsteel. [4] I %S DQ =DQ = с н =U2b (17) пул с т 100 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): Р2 +Q2 DР = r (18) т U2 т н In the same way, the loss of reactive power caused by the spread of the magnetic flux can be determined as in theformula(7): P2 +Q2 DQ = x (19) т U2 т н S2 DP =3I2r = r . т т U2 т н therelationshipDR /DR we form thefollowing expression: t k DР =DР (S / S )2. (20) т к Н