【正文】 變電所低壓側通過的總功率為 S =+,試求:當變壓器的 變 比 為KT 1 = KT 2 = 35 / 11KV時 , 每 臺 變 壓 器 的 總 功 率 為 多 少 ? 當KT1 = , KT 2 = 35 /11KV 時,每臺變壓器通過的功率是多少?解2 3 3 2R = PKU N= 2510 (35 10 )( )ST1 2N= ?(8 106 )2XT1= U K (%)U N2100SN (35 10 3 )2=100 8 106= (?)2 3 3 2R = PKU N= 24 10 (35 10 )( )ST 2 2N= ?(2 106 )2XT 2= U K (%)U N2100SN (35 10 3 )2=100 2 106= (?)ZT1 = + (?), ZT 2= + (?)在 1 點處把環(huán)網(wǎng)解開,則:.S Z. *I: ST1 = T 2Z *∑= ( + )( ? )( ? ) + ( ? )= + (MVA).S Z. *ST 2 = T 1Z *∑= ( + )( ? )( ? ) + ( ? )= + (MVA)II: 如右圖,設U 2 = 35KV則U = 35 11 = (KV )1 35 11?U = U 2 ? U1 = (KV )S?C?U *U= NZ *∑= 35( ? ) + ( ? )= ? (MVA)S?T1 = ( + ) + ( ? ) = + (MVA)S?T 2= ( + ) ? ( ? ) = + (MVA)?什么是最大負荷使用時間 Tmax ?什么是最大負荷損耗時間τ max ?2答:電能損耗= I 2 Rt = P+ Q 2Rt ,為某個時段 t 內(nèi)的電能損耗.U 2Tmax :指一年中負荷消費的電能除以一年中的最大負荷 Pmax ,即T = WPmaxmaxτ max :指全年電能損耗除以最大負荷時的功率損耗,即τ max= ?W?Pmax?什么叫經(jīng)濟功率分布?答:自然功率分布是指環(huán)網(wǎng)中功率分布未加任何調(diào)節(jié)手段,完全取決于自然情況進行的分布， 也即按阻抗共軛值成反比分布的。經(jīng)濟功率分布是使有功功率損耗最小的分布，是按線段的電阻分布。318. 某 一 由 發(fā) 電 廠 A 供 電 的 220KV 環(huán) 式 網(wǎng) 絡 , 其 運 算 網(wǎng) 絡 如 圖 所 示 , 圖 中Z1 = 16 +j120? , Z 2 = 33 + j89?, Z 3 = 48 + j120?, Z 4~= 60 + j152?, S1 =~ ~170+j40MVA, S 2 = 50 + j30MVA, S 3 = 40 + j15MVA. 計算:網(wǎng)絡的自然功率分布。網(wǎng)絡的經(jīng)濟功率分布。實現(xiàn)經(jīng)濟功率分布后,每年節(jié)省的電能? 解:在 A 點處把環(huán)網(wǎng)解開,如圖:. .1. S (Z * + Z * + Z * ) + S.(Z * + Z * ) + S Z *I: S A = 2 3 42 3 4 3 4Z * + Z * + Z * + Z *1 2 3 4= + (MVA). .1. S (Z * + Z * + Z * ) + S.(Z * + Z * ) + S Z *S A = 2 3 42 3 4 3 4Z * + Z * + Z * + Z *1 2 3 4= + (MVA). . . . .AS A + S 39。= +( MVA), S 1 + S 2 + S 3 = 260 +j85(MVA). . .S12 = S A ? S1 = ( + ) ? (170 + j40) = ? + (MVA). . .S 23 = S 2 ? S12 = (50 + j30) ? (? + ) = + (MVA). . .SA39。 = S 23 + S 3 = ( + ) + (40 + j15) = + (MVA)II:.S AO.= S1(R2+ R3+ R4.) + S2 (R3+ R4.) + S3 R4R1 + R2+ R3 + R4= (170 + j40)(33 + 48 + 60) + (50 + j30)(48 + 60) + (40 + j15) 60 16 + 33 + 48 + 60= + (MVA). 39。S AO.= S 3(R1+ R2+ R3.) + S2 (R1+ R2.) + S1 R1R1 + R2+ R3 + R4= (40 + j15)(33 + 48 + 16) + (50 + j30)(16 + 33) + (170 + j40) 1616 + 33 + 48 + 60= + (MVA). .AOS AO + S 39。= 260 + (MVA)AO 23?O 3S?′ = S? + S? = ( + ) + (40 + j15) = + (MVA)SA 2S12 2S23 2S′A 2UIII: ?Pi1 = ( )NR1 + ( )UNR2 + (UN) R3 + ( ) R4UN= + + + = (KW )SA?O2S12? O) 2S23?O) 2R + (3S A′ ?O) 2( )U NR1 + (U NR2 + (U NU N?Pi2 = R 4= + + + = (KW )?W = ?Pi1t ? ?Pi 2t= ( ? ) 8760= 8760 = (J) 20KM,35KV 的雙回平行電力線路,其負荷為 10MW, cos? = .電力線路導線 型號為 LGJ70, ,三相導線的幾何平均距離為 ,在用戶變電所內(nèi)裝 設 有 兩 臺 SFZ7500/35 型 變 壓 器 并 聯(lián) 運 行 ,SN = 7500KVA, UN 為35 / 11KV , Pk= 75KW , UK (%) = , P0= , I 0 (%) = .兩臺變壓器全年投入運行,:①電力網(wǎng)中的有功功率損耗。②電力網(wǎng)的網(wǎng)損率。③電力網(wǎng) 一年中的電能損耗。④當用戶分別接功率因數(shù)為 和 運行時(用戶的有功功率不變)重 復前三項計算.ρ解: r1 = s= = 0,45(?/ KM ) 70r = = (mm) 2Dm = = 3500(mm)Dm x1 = r+ = 3500 + = (? / KM ) b1 = D 10? 6= 10?63500= 10?6(s / km)lg mrlgR = 1 r l = 1 20 = (?)L 2 1 2X = 1 x l = 1 20 = (?)L 2 1 2B = 2b l = 1 10?6 20 = 10? 4 (S)L 1 22 3 3 2R = PKU N= 75 10 (35 10 )= (?)SNT 2 ( 106 )2U (%) U 2X = K N75 (35 103 )2== (?)NT 100S100 106G = PO = 103= 10? 6 (S)UNT 2 (35 103 )2NB = I 0 (%)S N T 100U 2 10 6=100 (35 103 )2= 10?5 (S)變壓器在額定負荷下的損耗為:?PKT= ?P0= = (MW )?QKT= I 0 (%)SN100= 100= (MVar)?PZT= ?PK= (MW )?QKT= U K (%)SN100= 100= (MVar)等效電路如下:I: cos? = a. S =10cos?∠arccos? = 10 ∠ = ∠? = 10 + (MvA)?S?T= ?S?KT + ?S?ZT = 2(+ ) + 2[( )2 2 + j( )2 2 ]= + + + = + (MvA)S?A = S?+ ?S?T= 10 + + + = + (MVA)?S?P 2 + Q 2= A A (R+ jX ) =+  ( + ) = + (MVA)U1 2 1 1N352? ? ?S1 = SA + ?S1 = + + + = + (MVA)?S?b.= ?S?T + ?S?1 = + + + = + (MVA)S = 10 ∠ = ∠? = 5 + (MVA) ?ST= ?S?KT + ?S?ZT = 2(+ ) + 2[( )2 2 + j( )2 2 ]= + + + = + (MVA)S?A = S?+ ?S?T= 5 + + + = + (MVA)?S?P 2 + Q 2= A A (R+ jX) =+  ( + ) = + (MVA)U1 2 1 1N352? ? ?S1 = SA + ?S1 = + + + = + (MVA)?S?= ?S?T + ?S?1 = + + + = + (MVA)C.S? = 10 ∠arccos? = 10 ∠ = ∠?= + (MvA)cos??S?T= ?S?KT + ?S?ZT = 2(+ ) + 2[( )2 2 + j( )2 2 ]= + + + = + (MVA)S?A = S?+ ?S?T= + + + = + (MVA)2 2 2 2?S?= PA + QA (R+ jX) = +  ( + ) = + (MVA)U1 2 1 1N352? ? ?S1 = SA + ?S1 = + + + = + (MVA)?S?= ?S?T + ?S?1 = + + + = + (MVA)有功功率損耗:?P1 = (MVA),?P2 = (MVA),?P3= (MVA)網(wǎng)損率:?P1 100 =P1 100 = ?P2 100 = 100 = P2 ?P3 100 =P3100 = ?W = ?P1t1 + ?P2t2 + ?P3 t3 = 2000+ 3000+ 3760= + + = 2262640(KWh)II. cos? = a. S? = 10 ∠ = ∠?= 10 + (MVA)?S?T= ?S?KT + ?S?ZT= 2( + ) + 2[( )2 2