WLCOM TO TH LCTUR ON TRNFORMR
Single Phase Transformer Three Phase Transformer
Transformer transformer is a stationary electric machine which transfers electrical energy (power) from one voltage level to another voltage level. Unlike in rotating machines, there is no electrical to mechanical energy conversion. transformer is a static device and all currents and voltages are C. The transfer of energy takes place through the magnetic field.
Transformer Principles t has electric circuits called primary and secondary. magnetic circuit provides the link between primary and secondary. When an C voltage is applied to the primary winding (Vp)of the transformer, an C current will result (p). p sets up a time-varying magnetic flux φ in the core. voltage is induced to the secondary circuit (Vs) according to the Faraday s law of lectromagnetic nduction.(emd/dt)
Transformer Transformer is a device that: Transfers electric power from one circuit to another. t does so without a change of frequency t accomplishes this by electromagnetic induction and Where the two circuits are in mutual inductive influence of each other.
Transformer Core Types The magnetic (iron) core is made of thin laminated steel. The reason of using laminated steel is to minimizing the eddy current loss by reducing thickness (t): kh tf P e ( ) max common cross section of core is square or rectangular for small transformers and circular (stepped ) for the large and 3 phase transformers.
Transformer Construction Type of Transformers: ) Core (U/) Type: is constructed from a stack of U- and -shaped laminations. n a core-type transformer, the primary and secondary windings are wound on two different legs of the core. ) Shell (/) Type: shell-type transformer is constructed from a stack of - and -shaped laminations. n a shell-type transformer, the primary and secondary windings are wound on the same leg of the core, as concentric windings, one on top of the other one.
lementary Theory of an deal Transformer For a transformer to be an ideal one, the various assumptions are as follows: ) Winding resistances are negligible, i.e. no losses. ) There is no magnetic leakage and hence which has no R and core losses. 3) ll the flux set up by the primary links the secondary windings, i.e. all the flux is confined to the magnetic core. 4) The core has constant permeability, i.e. the magnetization curve for the core is linear.
lementary Theory of an deal Transformer (cont.) Secondary is open and primary is connected to sinusoidal alternation voltage, V Since secondary is open, the primary draws the magnetizing current µ only. µ is small in magnitude and lags V by 0 90. The alternating current produces an alternating flux Φ which is proportional to the current and is in phase with it. This flux produces self-induced e.m.f. in the primary. The self-induced e.m.f. equal to and in opposition to V is also known as counter e.m.f. or back e.m.f. µ V N Φ V N
lementary Theory of an deal Transformer (cont.) Similarly there is produced in the secondary an induced e.m.f. which is known as mutually-induced e.m.f. This e.m.f. is anti phase with V and its magnitude is proportional to the rate of change of flux and the number of secondary turns. The vectorial representation of the effective values of the above quantities is shown in the right figure. V 0 90 0 90 µ Φ
.M.F. quation of a Transformer N No. of turns in primary N No. of turns in secondary Φ m Maximum flux in core in webers f frequency of a.c. input in Hz The figure shows flux increases from its zero value to maximum value Φ m in one quarter of the cycle i.e. ¼ second. Φm verage rate of change of flux / 4 f Now, rate of change of flux per turn means induced e.m.f. in volts verage e.m.f./turn 4 fφmvolt /4f cycle T/f Time r. m. s. value Form factor. averageval ue r.m.s. value of e.m.f./turn. 4 fφ 4. 44 m fφ m
.M.F. quation of a Transformer (cont.) r.m.s. value of the induced e.m.f. in the whole of primary winding(induced e.m.f./turn) No.of primary turns 4.44 fφ m N 4.44 fm N...( i) r.m.s. value of the induced e.m.f. in the whole of secondary winding(induced e.m.f./turn) No.of secondary turns 4.44 fφ m N 4.44 fm N...( ii) From (i) and (ii) we get, N 4. 44 N fφ m n deal transformer on no-load, V andv
Voltage Transformation Ratio (K) N N K This constant K is known as voltage transformation ratio. (i) f N Ni. e. transformer. K (ii) f N Ni. e. transformer. K,then transformer is called step-up,then transformer is called step-down
Transformer with Losses but no Magnetic Leakage Two cases: (i) Transformer on no-load: When the transformer in on no-load, the primary input current is not wholly reactive. The primary input current under no-load has to supply, (i) iron loss in the core and (ii) a very small amount of copper loss in primary (there being no Cu loss in secondary as it is open). Hence, the no-load primary input current is not at behind but lags it by an angle φ 90 0 No-load input power cosφ 0 0 W V 0 0 cosφ0 0 0 90 Where is primary power factor under no-load conditions. V
Transformer with Losses but no Magnetic Leakage (cont.) No-load condition of an actual transformer is shown vectorially in the figure. The primary current has two components: V i. One in phase with. This is known as active or working or iron loss component w. t mainly supplies the iron loss plus small quantity of primary Cu loss. V φ w 0 µ i 0 Φ ii. w 0 sinφ 0 The other component is in quadrature with V and is known as magnetizing component µ. ts function is sustain the alternating flux in the core. t is wattless. µ 0 cosφ 0 0 µ w
Transformer with Losses but no Magnetic Leakage (cont.) The following points should be carefully noted: 0 The no-load primary current is very small as compared to the full-load primary current. t is about percent of the full-load current. Owing to the fact that the permeability of the core varies with the instantaneous value of the exciting current, the wave of the exciting or magnetizing current is not truly sinusoidal. s such it should not be represented by a vector because only sinusoidally varying quantities are represented by rotating vectors.ut, in practice, it makes no appreciable difference. s 0 is very small, the no-load primary Cu loss is negligibly small which means that no-load primary input is practically equal to the iron loss in the transformer. s it is principally the core-loss which is responsible for shift in the current vector, angle is known as hysteresis angle of advance. φ 0
Transformer with Losses but no Magnetic Leakage (cont.) (ii) Transformer on load: φ 0 φ φ V φ ' φ φ φ
Transformer with Losses but no Magnetic Leakage (cont.) When the secondary is loaded, the secondary current is set up. The magnitude and phase of, with respect to V is determined by the characteristics of the load. Current is in phase with V if load is resistive, it lags if load is inductive and it leads if load is capacitive. The secondary current sets up its own m.m.f. and hence its own flux Φ which is opposite to the main primary flux Φ. The secondary amp-turns N are known as demagnetizing amp-turns. The opposing flux Φ weakens. Φ V gains the upper hand over and hence causes more current to flow in primary. ' The additional current is known as load component of primary current. ' The additional primary m.m.f. sets up its own flux which is opposition to Φ and is equal to it in magnitude. Hence, the two cancel each other. Whatever the load conditions, the net flux passing through the core is approximately the same as at no-load. Φ
Transformer with Losses but no Magnetic Leakage (cont.) Φ ' Φ N K ' ' N N N Hence, when the transformer is on-load, the primary winding has two currents in it. The total primary current is the vector sum of and ' 0
Transformer with Losses but no Magnetic Leakage (cont.) Vector diagrams: V V V ' ' Φ ' 0 Φ 0 0 Φ 0 Φ Φ 0 0 Φ 0 0 Negligible Φ Φ Φ K K K
Transformer with Losses but no Magnetic Leakage (cont.) Under the assumption: 0 is negligible, Φ Φ N ' N ' N N K
Transformer with winding resistance but no Magnetic Leakage n ideal transformer has no resistance, but an actual transformer has winding resistances. Due to this resistance, there is some voltage drop in the two windings. The result is that: V The secondary terminal voltage is vectorially less than the secondary induced e.m.f. by an amount R where is the resistance of the R secondary winding. Hence, V is equal to the vector difference of and resistive voltage drop R. V R V Similarly, primary induced e.m.f. is equal to the vector difference of and R where R is the resistance of the primary winding. V R
Transformer with winding resistance but no Magnetic Leakage (cont.) R V R V V R ' ' K 0 Φ 0 0 Φ 0 Φ Φ 0 0 Φ ' Φ 0 0 Φ V V Φ V Φ R R Non-inductive nductive Capacitive
Resistance of a Transformer transformer whose primary and secondary windings have resistances of R and R respectively is shown in the figure. R R
Magnetic Leakage Leakage Flux: Not all of the flux produced by the primary current links the winding, but there is leakage of some flux into air surrounding the primary. Similarly, not all of the flux produced by the secondary current (load current) links the secondary, rather there is loss of flux due to leakage. These effects are modeled as leakage reactance in the equivalent circuit representation. transformer with magnetic leakage is equivalent to an ideal transformer with inductive coils connected in both primary and secondary circuits as shown in the next slide. The internal e.m.f. in each inductive coil is equal to that due to the corresponding leakage flux in the actual transformer. X e L and X e L
Transformer with Magnetic Leakage mportant points: The leakage flux links one or other winding but not both, hence it in no way contributes to the transfer of energy from the primary to the secondary winding. The primary voltage V will have to supply reactive drop X in addition to R. Similarly, will have to supply and X. R n an actual transformer, the primary and secondary windings are not placed on separate legs or limbs as shown below because due to their being widely separated, large primary and secondary leakage fluxes would result.
Transformer with Resistance and Leakage Reactance The primary impedance: The secondary impedance: The Voltage drops: V ( R jx) ( R ) X ( R ) X ( R jx ) V V
Vector diagram of ctual Transformer
Transfer of resistances & reactances to any side
Transfer of resistances & reactances to any side (cont.)
Transfer of resistances & reactances to any side (cont.)
Transfer of resistances & reactances to any side (cont.) 0 R0 X 0 mpedance referred to primary mpedance referred to secondary
quivalent Circuit of a Transformer
quivalent circuit of a Transformer referred to primary/xact quivalent circuit Where, and impedance of the exciting circuit ( ) ' ' L m ( ) ( ) ' ' ' ' L m L m ' ' ' jx R m ( ) ( ) ' ' ' ' L m L m V ' ' ',, K K X X K R R K V V K ' ',
pproximate quivalent Circuit Transfer exciting circuit as shown in figure:
pproximate quivalent circuit with secondary impedances transferred to primary
quivalent Circuit (cont.) Neglecting the exciting circuit: 0 ' R0 X 0 V ' L
Why Transformer Rating in kv? Cu loss of a transformer depends on current. ron loss of a transformer depends on voltage. Hence, total transformer loss depends on voltampere (V) and not on phase angle between voltage and current i.e.total transformer loss is independent of load power factor. That s why transformer rating is in kv and not in kw.
Percentage Resistance, Reactance and mpedance Usually measured by the voltage drop at full-load current. i. Percentage resistance at full-load: R R V % 0 % R % R 00 00 00 %R%Cu Loss at full-load %R%Cu Loss v r R V 0 R V 0
Percentage Resistance, Reactance and mpedance (cont.) ii. Percentage reactance at full-load: % X X V % X 0 X V 00 0 00 iii. Percentage impedance at full-load: iv. So, % V % V 0 V 00 0 00 ( % R % ) % X v x
Percentage Resistance, Reactance and mpedance (cont.) Now, Now, % R R 0 R V % X X 0 00 % R V 00 0 X V 0 % CuLoss V 00 00 % X V 00 vx V 00 Similarly, Similarly, % R R 0 % X X R V 0 00 % R V 00 0 X V 0 % CuLoss V 00 00 % X V 00 v x V 00 Percentage resistance, reactance and impedance have the same value whether referred to primary or secondary.
Losses in a Transformer Two types of losses i. Core or ron Loss: t includes both hysteresis and eddy current loss. ecause the core flux in a transformer remains practically constant for all loads. The core is practically same at all loads. Hysteresis Loss W h ddy current Loss η.6 max W e P fv max watt; watt. These losses are minimized by using steel of high silicon content for the core and by using very thin laminations. ron or Core loss is found from the Open Circuit test. f The input of the transformer when on no-load measures the core loss. t
Losses in a Transformer (cont.) ii. Copper Loss: This loss is due to the ohmic resistance of the transformer windings. Total Cu loss R R R0 R0 So, Cu loss is proportional to ( current ) or ( kv) Cu loss is found from the short-circuit test.
fficiency of a Transformer η η Output Output Losses Output Output CuLoss ronloss nput Losses η nput Losses nput fficiency is based on power output in watts and not in voltamperes, although losses are proportional to V. Thus, efficiency depends on power factor and maximum at a power factor of unity.
Condition for Maximum fficiency Cu Loss R 0 ron Loss hysteresis loss eddy current loss Primary nput V Cosφ Wi V η V η Cosφ Losses V Cos φ η V Cosφ R V Cosφ R0 Wi η V Cosφ V Cosφ 0 R0 Wi Cosφ V Cosφ W i
Condition for Maximum fficiency (cont.) dη d d d dη d R W 0 i V Cos V Cos φ φ» R0 Wi 0 V Cosφ V Cosφ» or» Therefore,» The output current corresponding to maximum efficiency is R0 0 0 V Cosφ V R0 V Cosφ R R 0 0 W i W i V Wi Cosφ Wi Cos φ R 0 Wi Cu loss ron Loss φ ( W ) i R 0
ll-day fficiency The distribution transformer have their primaries energized all the twenty-four hours, although their secondaries supply little or no-load much of the time during the day except during the house lighting period. The core loss occurs throughout the day, the Cu loss occurs only when the transformer is loaded. η all day Output in kwh nput in kwh (for 4 hours)
xamples x-: (a),00/00 V transformer draws a no-load primary current of 0.6 and absorbs 400 watts. Find the magnetizing and iron loss currents. (b) 00/50 V transformer takes 0.5 at p.f. of 0.3 on open circuit. Find magnetizing and working components of no-load primary current.
xamples x-: transformer has a primary winding of 800 turns and a secondary winding of 00 turns. When the load current on the secondary is 80 at 0.8 power factor lagging, the primary current is 5 at 0.707 power factor lagging. Determine the no-load current of the transformer and its phase.
xamples x-3: 30 kv, 400/0-V, 50Hz transformer has a high voltage winding resistance of 0.Ω and a leakage reactance of 0.Ω. The low voltage winding resistance is 0.035Ω and the leakage reactance is 0.0Ω. Find the equivalent winding resistance, reactance and impedance referred to the (i) high voltage side and (ii) the low-voltage side.
xamples x-4: n a 5 kv, 000/00-V, single phase transformer, the iron and full-load Cu losses are 350W and 400W, respectively. Calculate the efficiency at unity power factor on (i) full-load, (ii) half full-load.
Parallel Operation of a Single Phase Transformer Conditions:. Primary winding of a transformer should be suitable for the supply system voltage and frequency.. The transformer should be properly connected with regard to polarity. 3. The voltage ratings of both primaries and secondaries should be identical.n other words, the transformers should have the same turn ratio i.e. transformation ratio. 4. The percentage impedance should be equal in magnitude and have the same X/R ratio in order to avoid the circulating currents and operation at different power factors. 5. With transformers having different kv ratings, the equivalent impedances should be inversely proportional to the individual kv rating if circulating currents are to be avoided.
Transformer Parallel Connection V V - - - -
Transformer Parallel Operation Case :deal Case Two transformers having the same voltage ratio. mpedance voltage triangles are identical in size and shape. C V V Load 0 ф V
Transformer Parallel Operation (cont.) ; ; V ) ( ); ( ;, ; lso V
Transformer Parallel Operation (cont.) Case :qual Voltage Ratios Two transformers having the same voltage ratio. mpedance voltage triangles are different in size and shape. No-circulating current. V V Load 0 ф V
Transformer Parallel Operation (cont.) ) ( ); ( ; ; Multiplying both sides by V, Where, V 0-3 S-the combined load kv. S S V V S S V V
Transformer Parallel Operation (cont.) Case 3:Unequal Voltage Ratios Two transformers having different voltage ratio. mpedance voltage triangles are different in size and shape. Circulating current. C V V Load 0 ф V
Transformer Parallel Operation (cont.) ( ) [ ] ; ; ) ;...( ) ( ; ) ( ; ) ( ; ; L L L L iii V V V Substituting this value in eqn (iii) ( ) [ ] ; ( ) { } [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) L L L L L L ; ; ; ( ) C L L L V
Transformer Test
Open Circuit Test
Short Circuit Test
Transformer Regulation Loading changes the output voltage of a transformer. Transformer regulation is the measure of such a deviation. Definition of % Regulation Vno load V V load load *00 V no-load RMS voltage across the load terminals without load V load RMS voltage across the load terminals with a specified load
uto Transformer t is a transformer with one winding only, part of it been common to both primary and secondary. Primary and Secondary winding are not electrically isolated from each other. ts theory and operation are similar to that of a two winding transformer. ecause of one winding it is compact, efficient and cheaper. t is used where transformation ratio differs little from unity. t is also used as an adjustable transformer for both stepping up and stepping down the input voltage.
uto Transformer (cont.) Step up uto transformer Step down uto transformer
Saving of Cu uto Transformer (cont.) Volume and weight of Cu length and area of the cross-section of the conductors. Length of conductor N Cross-section Weight N Wt. of Cu in C section (N -N ) ; Wt. of Cu in C section N ( - ); Total Wt. of Cu in uto-transformer (N -N ) N ( - ); f two-winding transformer perform the same duty, then Wt. of Cu on its primary N ; Wt. of Cu on its secondary N ; Total Wt. of Cu N N ;
uto Transformer (cont.) Wt. of Cu in uto-transformer Wt. of Cu in ordinary transformer ( ) ( ) N N N N N K K N N N So, Saving of Cu Saving of Cu ( ) O a W K W N O O O a O KW W K W W W ) ( O KW
xamples x-5: 5 kv, 300/30-V, 50Hz transformer was tested for the iron losses with normal excitation and Cu losses at full-load and these were found to be 40W and W,respectively. Calculate the efficiency of the transformer at 0.8 power factor for.5kv outputs.
xamples x-6: Find the all-day efficiency of 500kV distribution transformer whose copper loss and iron loss at full load are 4.5kW and 3.5kW,respectively. During a day of 4 hours, it is loaded as under: No. of hours Loading in kw Power factor 6 400 0.8 0 300 0.75 4 00 0.8 4 0 -
xamples x-7: 600kV, -ph transformer has an efficiency of 9% both at full-load and halfload at unity power factor. Determine its efficiency at 60% of full-load at 0.8 power factor lag.
xamples x-8: x: Two -ph transformers with equal turns have impedances of (0.5j3) ohm and (0.6j0) ohm with respect to the secondary. f they operate in parallel, determine how they will share a total load of 00 kw at p.f. 0.8 lagging?
xamples x-9: x: Two 00/0-V, transformers are operated in parallel to share a load of 5kV at 0.8 power factor lagging. Transformers are rated as below: :00 kv;0.9% resistance and 0% reactance. :50 kv;.0% resistance and 5.0% reactance. Find the load carried by each transformer.
xamples x-0: x: Two transformers and are operated in parallel to the same load. Determine the current delivered by each transformer having given: open-circuit e.m.f. 6600V for and 6400V for. quivalent leakage impedance in terms of the secondary 0.3j3 for and 0.j for.the load impedance is 8j6.