Transformer. Transformer comprises two or more windings coupled by a common magnetic circuit (M.C.).

Transcription

1 . Transformers

2 Transformer Transformer comprises two or more windings coupled by a common magnetic circuit (M.C.). f the primary side is connected to an AC voltage source v (t), an AC flux (t) will be produced in the M.C. rimary side Secondary side

3 deal transformer voltage relationship e e t) ( t) ( d( t) dt d( t) dt e e 3

4 deal transformer symbol v v Another representation n 4

5 There are two common types of transformer construction: (a) core-type (b) shell-type 5

6 . Analysis of the transformer with no-load Secondary side open-circuited: (no sec. current) 6

7 Modelling of Magnetic Core r c : core loss resistance (hysteresis & eddy-current loss) x m : magnetizing reactance e : exciting current g c : core loss conductance g c r c jb m jx m hasor diagram g c -jb m : magnetizing admittance b m : magnetizing susceptance m lags E by 90º e c m c in phase with E 7

8 Modelling of Leakage Flux Let us express the voltage drop due to leakage flux in the primary winding below d dt v d di e die dt L die dt where v L die dt L : leakage inductance of primary winding primary leakage reactance: x ω L ω f Modelling of Copper Loss r : resistance of primary winding 8

9 Equivalent circuit model of primary side primary leakage impedance exciting branch impedance Exciting current is only a few percent of rated primary current of the transformer 9

10 . deal Transformer Operation o leakage fluxes egligible winding internal resistances B-H characteristic of the magnetic material is singlevalued, and linear o hysteresis loss Magnetic core has a very high r, i.e. Core reluctance is negligible. o copper, no core losses (Efficiency = 00%) nterwinding capacitatances are negligle at power frequencies (50Hz, 60Hz) 0

11 Basic Relations. From Faraday s Law: e = d /dt, and e = d /dt So, e e. Since winding resistances & leakage fluxes are negligible: v v

12 3. F = F i i 4. deal transformer symbol

13 4. o power loss Conservation of power: v i vi 5. Under Load 3

14 4 deal transformer under load v v i v i v Z v i i i Z n Z Z Eqv. crt. referred to primary Secondary impedance referred to primary side:

15 5 Terminology secondary quantities referred to primary side :,, primary quantities referred to secondary side :,, secondary quantities actual :,, primary quantities actual :,, Z Z Z Z,, Z Z,, Z Z

16 3. Equivalent circuit representation of a practical transformer Transformer under load 6

17 Equivalent Circuit representation primary leakage impedance secondary leakage impedance exciting branch ideal transformer r : rimary winding internal resistance () x : rimary winding leakage reactance () r : Secondary winding internal resistance () x : Secondary winding leakage reactance () r c : Core-loss resistance () x m : Magnetizing reactance () 7

18 Equivalent circuit referred to primary side r : Secondary winding internal resistance referred to primary side x : Secondary winding leakage reactance referred to primary side : Secondary winding current referred to primary side : Secondary winding voltage referred to primary side Z L : Load impedance referred to primary side n r n x r x n Z n L n Z L E n E ne 8

19 Equivalent circuit referred to secondary side r : rimary winding internal resistance referred to secondary side x : rimary winding leakage reactance referred to secondary side : rimary winding current referred to secondary side : rimary winding voltage referred to secondary side r c : Core-loss resistance referred to sec. side x m : Magnetizing reactance referred to sec. side r r n x x n r x c m r n x n c m n n n E E E n 9

20 Equivalent circuit referred to secondary side g c : Core-loss conductance referred to sec. side b m : Magnetizing admittance referred to sec. side n g c r c jb m jx m g c n g c b m n b m 0

21 Simplification Th, Z Th Let us apply Thévenin equivalent circuit to the primary side r c jx m g c jb m r jx where r jx r jx c m Th c m r jx rc jxm Th r jx r jx Z r jx Th c m Z Th r jx

22 Simplification E Z Th Th where Th Z Th r jx E Error made in approximations is at most % % compared to actual values.

23 Approximate Equivalent Circuit referred to primary side Further simplification gives us the figure below r eq r r x eq x x 3

24 Approximate Equivalent Circuit referred to secondary side Further simplification gives us the figure below r eq r r x eq x x 4

25 Approximate Equivalent Circuits for Large Transformers (referred to primary side) (of a few 00 kas) r c jxm is very large (in the MA range) x eq r eq (4-0 times) 5

26 AC ower : angle between voltage and current Real ower : = rms rms cos [W, Watts] Reactive ower : (maginary ower) Q = rms rms sin [AR, olt Ampere Reactive] Complex ower : S = jq Apparent ower : S = rms rms [A, olt Ampere] ower Factor : cos S 6

27 Transformer ower Flow in = cos = = cos out load = in cu core cu load cu : Copper loss core : Core loss cu r r core E g c E g c core g c cu r r 7

28 4. Short-circuit and Open-circuit Tests Measure voltage (), current () and power () in order to determine the equivalent circuit parameters of the transformer: For leakage impedance parameters with secondary short circuited For exciting branch parameters with secondary open circuited 8

29 sc, sc, i. Short-circuit Test sc A reduced voltage sc of % - 0 % of rated voltage is applied to allow rated primary current. where g c jb m r eq jx eq Short circuit equivalent circuit 9

30 Short-circuit Test r eq sc sc z eq sc sc where z eq r eq j x eq x eq z eq r eq r eq r r where r r r r r eq x eq x x where x x x x x eq Form factor at 50 Hz r. r AC. r r AC DC DC Measured DC resistance 30

31 oc, oc, oc ii. Open-circuit Test Rated voltage is applied to the transformer under no-load and exciting current flows, which is a few percent of rated current. where 0 and e Open circuit equivalent circuit 3

32 3 oc oc c g oc oc c Y m c c j b g Y where c c m g Y b Open-circuit Test c c r g where m m x b and

33 5. oltage regulation (R%) The change in secondary terminal voltage (load voltage) from no-load to full-load expressed as a percentage (%) of the rated value ideally R% = 0. R% (rated) 00% or R% (rated) 00% 33

34 Simplification by approximation req j x eq r cos x sin eq eq R% r eq cos x eq sin 00% 34

35 Zero regulation, i.e. R% = 0 For zero regulation, phase angle ( ) of the load is given by R% r eq cos x sin eq 00% r eq cos x eq tan r x sin 0 eq eq ote that < 0, so Load must be capacitive for zero regulation 35

36 OTE For an inductive load, Z L = R L + jx L Always >, i.e. R% > 0 For a capactive load, Z L = R L jx L Usually, i.e. R% 0 ( where x sin r cos ) eq eq 36

37 6. Efficiency (%) The ratio of the output power given to the load and the input power taken from the electrical supply expressed as a percentage (%) out in 00% out out losses 00% 37

38 38 00% losses out out 00% core cu cu out out core cu losses 00% cos cos g c r r r eq cu 00% cos cos eq g c r cu cu cu 00% cos cos eq g c r or

39 Maximum efficiency Let us find the value of which maximizes the efficiency where d 0 d cos cos r eq core 00% d 0 d cos r cos r cos 0 cos eq core eq cos r 0 cos req core eq core req 0 core r eq core cu i.e., gc req req Thus, maximum efficiency is achieved if core loss equals to the copper loss. 39

40 Examples. A ka, 0/440, 50 Hz single phase transformer has the following test data: o-load test: 0, A, 65W (measured at primary side) Short-circuit test:, 5A, 60W (measured at secondary side) a) Calculate the equivalent circuit parameters referred to primary side b) Calculate the primary terminal voltage on full-load at a power factor of: 0.8 pf lagging. 40

41 . Given a 50kA, 460:480, 60 Hz transformer, the following parameters are obtained by tests r = 0.09 and x =.7 r =. 0-3 and x = eglecting core losses, a) Calculate the primary voltage and voltage regulation for rated load at 76% pf lagging b) Repeat a) for a load of 76% pf leading c) Calculate the transformer efficiency for parts a) and b) with a core loss core = 547W. 4

42 3. The parameters of the exact equivalent circuit of a 50kA, 400/40 transformer are r = r = 0. x = x = 0.45 r c = 0 k and x m =.55 k Using both the exact and approximate equivalent circuit of the transformer, determine a) oltage regulation b) Efficiency for rated load at 0.8pf lagging 4

43 4. At 0kA, 8000:30 transformer has a leakage impedance referred to primary of 90+ j400. Exciting branch parameters are r c = 500 k and x m = 60 k a) f primary voltage = 7967 and actual load impedance Z L = , find the secondary voltage of the transformer b) f the load is disconnected, and a capacitor of j6 is connected in its place, what will be the load voltage? 43