ELG3311: Tutorial 2 R C = 75 Ω X M = 20 Ω R T = Ω X T = Ω. Solution: The following equivalent circuit is referred to the primary.

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1 ELG3311: Tutrial rblem -1: The secndary winding f a transfrmer has a terminal vltage f v s (t) 8.8 sin 377t. The turns rati f the transfrmer is 50:00 (a 0.5). f the secndary current f the transfrmer is i (t) 7.07 sin (377t ) A, what is the primary current f this transfrmer? What is the vltage regulatin and efficiency? The impedance f this transfrmer referred t the primary side are: R T Ω X T 0.5 Ω R C 75 Ω X M 0 Ω lutin: The fllwing equivalent circuit is referred t the primary. R T X T e R C jx a The secndary vltage and current are A A The secndary vltage referred t the primary side is a The secndary current referred t the primary side is a A The primary circuit vltage is given by 50 0 ( RT jxt ) ( )( 0.05 j0.5) The excitatin current f this transfrmer is 1

2 e C m j The ttal primary current is e The vltage regulatin f the transfrmer a R 100% 100% 7.% a 50 The input pwer is ( ) 857 W in csθ cs The utput pwer is ut csθ 00 5 cs W The efficiency η ut % 100% 93.4% 857 in A

3 rblem -: A 0-kA 8000/77- distributin transfrmer has the fllwing resistances and reactances: R 3 Ω X 45 Ω R C 50 kω R 0.05 Ω X 0.06 Ω X m 30 kω The excitatin branch impedances are given referred t the high-vltage side f the transfrmer a. Find the equivalent circuit f this transfrmer referred t the high-vltage side. b. Find the per-unit equivalent circuit f this transfrmer. c. Assume that this transfrmer is supplying rated lad at 77 and 0.8 F lagging. What is the transfrmer s input vltage? What is its vltage regulatin? d. What is the transfrmer s efficiency under the cnditins f part (c)? lutin: The turns rati a 8000/ The secndary impedances referred t the primary side are R X a a R X ( 8.89) ( 0.05) 41.7 Ω ( 8.89) (0.06) 50.1Ω 3Ω j45 Ω 41.7 Ω j50.1ω /a e 50 kω j30 Ω a (b) The rated KA f the transfrmer is 0 ka, and the rated vltage n the primary side is 8000, therefre, the rated current in the primary side is 0 ka/ A. Accrdingly, the base impedance n the primary side is Z base base base Ω.5 ince Z pu Z actual /Z base, the resulting per unit equivalent circuit is 0.01 j J /a e J9.375 a 3

4 ( c ) The equivalent circuit referred t the primary side is shwn in the fllwing diagram: 3Ω j45 Ω 41.7 Ω j50.1ω /a e 50 kω j30 Ω a 0kA The primary vltage is ( R jx ) T The vltage regulatin is T A ( 73.7 j95.1)( ) R 100% 3.63% 8000 (d) The lsses are cpper ( ) cre RC ( ) R (.5) ( 73.7) The efficiency is η T , W 461 W ut 0kA % 100% 95.6% ut cpper cre 0kA

5 rblem -3: A 000-A, 30/115- transfrmer has been tested t determine it equivalent circuit. The results f the test are shwn belw: OC OC 30 OC 0.45 A OC 30 W C C 13. C 6 A C 0.1 W All data given were taken frm the primary side f the transfrmer. a. Find the equivalent circuit f the transfrmer referred t lw-vltage side. b. Find R at rated cnditins and (1) 0.8 F lagging, () 1.0 F, (3) 0.8 F lagging. lutin: Frm OC Test, we get RC 1763 Ω X 534 Ω m Frm C, we get RT Ω X j.18 Ω T The equivalent circuit referred t the secndary side is shwn in the fllwing diagram a Ω j0.53 Ω e /a 441 Ω 134 Ω (b) The reated secndary current is A 115 (1) When 0.8 F lagging Z T R 100% 3.3% 115 () When 1.0 F Z T ( j0.53)( ) ( j0.53)( ) 5

6 R 100% 1.1% 115 (3) When 0.8 F leading R Z T % 1.5% 115 ( j0.53)( ) The cpper and cre lsses f the transfrmer are: 10.6 W and 3 W. The efficiency is η 94.9%

7 rblem -4: A single-phase pwer system is shwn in Figure -1. The pwer surce feeds a 100-kA 14/.4-k transfrmer thrugh a feeder impedance f 38. j140 Ω. The transfrmer s equivalent series impedance referred t its lw-vltage side is 0.1 j0.5 Ω. The lad n the transfrmer is 90 kw at 0.85 F lagging and 300. G Z lad urce Feeder Transfrmer Lad a. What is the vltage at the pwer surce f the system? b. What is the vltage regulatin f the transfrmer? c. Hw efficient is the verall pwer system? lutin: We will refer this circuit t the secndary. The feeder s impedance referred t the secndary side is Z line ( 38. j 140) 1.1 j4. Ω The secndary current is given by A (a) The pwer at the pwer surce f this system (referred t the secndary lab) is surce Z line ( ) ( 1.1 j 4.11) ( )( 0.1 j0.5) Nw find the vltage at the pwer surce surce Z 14 ( ) k.4 T (b) T find the R, we must find the vltage at the primary side referred t the secndary under full lad cnditin Z T ( )( 0.1 j0.5)

8 Therefre R 100% 0.6% 300 (c) The pwer supplied by the lad is (d) in surce csθ cs kw The efficiency is η u t in 100% % 97.4% 8

9 rblem -14: A 1.4-k single-phase generatr supplies pwer t the lad thrugh a transmissin line. The lad s impedance is Z lad Ω, and the transmissin line s impedance is Z line Ω. Z line G Z lad (a) 1:10 10:1 Z line G Z lad a. f the generatr is directly cnnected t the lad (Figure -3a), what is the rati f the lad vltage t the generatr vltage? What are the transmissin lsses f the system? b. f a 1:10 step-up transfrmer is placed at the utput f the generatr and a 10:1 transfrmer is placed at the lad end f the transmissin line, what is the new rati f the lad vltage t the generated vltage? What are the transmissin lsses f the system nw? lutin: n case f the directly-cnnected lad, the line current is (b) line lad A The lad vltage is lad lad Zlad ( )( ) k The rati f the lad vltage t the generated vltage is 11.16/ The transmissin lsses are (.3) ( 30) 14.9 kw lss linezlad (b) Tw transfrmers are used. The impedance f the transmissin line becmes 9

10 ( ) lad 1 1 Zline Zline Ω lad line lad A The lad vltage is lad lad Zlad ( )( ) k The rati f lad vltage t generated vltage is 1.386/ The current in the transmissin line is 1 1 line lad ( 4.77).477 A The lsses in the transmissin line are (.477) ( 30) 184 W lss liner line Transmissin lsses have decreased by a factr f mre than