Transformer Equivalent Circuit & Auto Transformer PDF

Ministry of Higher Education and Scientific Research Al-Mustaqbal University College of Engineering & Technology Medical Instrumentation Techniques Engineering Department Electrical Technology Third Class Weeks 5 & 6 Equivalent Circuit of a Transf...

Ministry of Higher Education and Scientific Research Al-Mustaqbal University College of Engineering & Technology Medical Instrumentation Techniques Engineering Department Electrical Technology Third Class Weeks 5 & 6 Equivalent Circuit of a Transformer By Osamah Jaber Ghayyib 1. Equivalent Circuit of a Single-Phase Transformer The analysis of a transformer can be carried out by using an equivalent circuit. Equivalent circuit is derived considering the following:- The primary and secondary windings have finite resistances considered as lumped parameters. The leakage fluxes are modelled as leakage reactance in the equivalent circuit. The core-loss component of current is modelled using a shunt resistance. The magnetization of the core is modelled using a magnetizing reactance as a shunt branch. Z1 Z2 R1 X1 I'2 R2 X2 I1 I0 I2 I1 I2 Iw Iµ V1 V2 ZL V1 E1 E2 V2 R0 X0 E1 E2 Ideal Transformer (a) (b) Fig.1 equivalent circuit diagram for the transformer with load The transformer shown diagrammatically in Fig.1 (a) can be resolved into an equivalent circuit in which the resistance and leakage reactance of the transformer are imagined to be external to the winding whose only function then is to transform the voltage. Generally there are two modes of operations in the transformers which are: 1- No-load operation. 2- On load operation. 1  The voltages, currents and elements of the circuit can be defined as in Table 1. SYMBOL DEFINITION R1, R2 Primary and secondary resistances X1, X2 primary and secondary leakage reactances R0 Exciting resistance X0 Exciting reactance I1 Primary side current I0 No-load/Excitation current component of primary current Iw (or Ic1) Core-loss component of no-load current Iµ (or Im1) Magnetizing component of no-load current I'2 Load component of primary current I2 Secondary side current (load current) V1 Applied primary side voltage V2 Secondary side terminal voltage E1 Primary side induced emf E2 Secondary side induced emf Table 1 Quantities of equivalent circuit. 1.1 Transformer at No-load A transformer is said to be on no-load when its secondary winding is kept open and no- load is connected across it. As such, no current flows through the secondary i.e., I2 = 0. Hence, the secondary winding is not causing any effect on the magnetic flux set-up in the core or on the current drawn by the primary. But the losses cannot be ignored. At no-load, a transformer draws a small current I0 (usually 2 to 10% of the rated value). This current has to supply the iron losses (hysteresis and eddy current losses) in the core and a very small amount of copper loss in the primary (the primary copper losses are so small as compared to core losses that they are generally neglected moreover secondary copper losses are zero as I2 is zero). Therefore, current I0 lags behind the voltage vector V1 by an angle I0 (called hysteresis angle of advance) which is less than 90‫؛‬, as shown in Fig. 2. The angle of lag depends upon the losses in the transformer. 2  V1 The basic equations in this mode can viewed as follows:- Iw I0 Working component, 𝐼𝑤 = 𝐼0 cos 𝜙0 𝜙𝑚 Magnetizing component, 𝐼𝜇 = 𝐼0 sin 𝜙0 Iµ 𝐼0 = √𝐼𝜇 2 + 𝐼𝑤 2 E1 No-load current, E2 𝑰𝟎 = 𝐼𝑤 + 𝑗𝐼𝜇 Fig.2 No-load current Primary p.f. at no-load, No-load power input, 𝑉1 Exciting resistance, 𝑅0 = 𝐼𝑤 𝑉1 Exciting reactance 𝑋0 = 𝐼𝜇 Table 2. No-load parameters in transformer. Examples 1: A 230/110 V single-phase transformer has a core loss of 100 W. If the input under no-load condition is 400 VA, find core loss current, magnetizing current and no-load power factor angle. Solution 400 No- load current 𝑉1 𝐼0 = 400 VA, 𝐼0 = = 𝟏. 𝟕𝟑𝟗 𝐀 230 𝑃𝑖 100 Core loss current 𝐼𝑤 = = = 𝟎. 𝟒𝟑𝟒𝟖 𝐀 𝑉1 230 Magnetizing current 𝐼𝜇 = √𝐼0 2 − 𝐼𝑤 2 = √(1.739)2 − (0.4348)2 = 𝟏. 𝟔𝟖𝟒 𝐀 𝐼𝑤 0.4348 No-load power factor, cos 𝜙0 = = = 𝟎. 𝟐𝟓 lag 𝐼0 1.739 3  No-load power factor angle 𝜙0 = cos−1 0.25 = 𝟕𝟓. 𝟓𝟐𝝄 Example 2: At open circuit, transformer of 10 kVA, 500/250 V, 50 Hz draws a power of 167 watt at 0.745 A, 500 V. Determine the magnetizing current, Ro current, no-load power factor, hysteresis angle of advance, equivalent resistance and reactance of exciting circuit referred to primary side. Solution 𝑉1 = 500 𝑉 , 𝐼0 = 0.745 𝐴, 𝑃0 = 167 𝑊 𝑃0 167 𝐼𝑤 = = = 𝟎. 𝟑𝟑𝟒 𝐀 𝑉1 500 𝐼𝜇 = √𝐼0 2 − 𝐼𝑤 2 = √(0.745)2 − (0.334)2 = 𝟎. 𝟔𝟔𝟔 𝐀 𝐼𝑤 0.334 cos 𝜙0 = = = 𝟎. 𝟒𝟒𝟖 𝐥𝐚𝐠 𝐼0 0.745 𝜙0 = cos −1 0.448 = 𝟔𝟑. 𝟑𝟔° 𝐥𝐚𝐠 𝑉1 500 𝑅0 = = = 𝟏𝟒𝟗𝟕 𝛀 𝐼𝑤 0.334 𝑉1 500 𝑋0 = = = 𝟕𝟓𝟎 𝛀 𝐼𝜇 0.666 1.2 Transformer on load When the secondary is loaded, the secondary current I2 is set up. The magnitude and phase of I2 with respect to V2 is determined by the characteristics of the load. Current I2 is in phase with V2 if load is non-inductive, it lags if load is inductive and it leads if load is capacitive. A set of equation can be drawn using basic theory of electric circuit analysis. Applying KVL in the primary side of equivalent circuit:- 4  𝑉1 = 𝐸1 + 𝐼1 (𝑅1 + 𝑗𝑋1 ) Applying KVL in the secondary side of equivalent circuit:- 𝐸2 = 𝑉2 + 𝐼2 (𝑅2 + 𝑗𝑋2 ) 𝐸1 𝐼2 1 Since = =𝑎= using 𝐸1 = 𝑎𝐸2 we get: 𝐸2 𝐼2′ 𝑘 𝑉1 = 𝑎𝐸2 + 𝐼1 (𝑅1 + 𝑗𝑋1 ) Then we can rewrite the equation of 𝑉1 by substituting the equation of 𝐸2. 𝑉1 = 𝑎[𝑉2 + 𝐼2 (𝑅2 + 𝑗𝑋2 )] + 𝐼1 (𝑅1 + 𝑗𝑋1 ) Since 𝐼2 = 𝑎𝐼2′ ⟹ 𝑉1 = [𝑎𝑉2 + 𝐼2′ 𝑎2 (𝑅2 + 𝑗𝑋2 )] + 𝐼1 (𝑅1 + 𝑗𝑋1 ) Let 𝑅2′ = 𝑎2 𝑅2 and 𝑋2′ = 𝑎2 𝑋2 Where 𝑅2′ is called secondary resistance referred to the primary side and 𝑋2′ is called secondary leakage reactance referred to the primary side. Another modification to 𝑉1 applying the same rules above then: 𝑉1 = 𝑉2′ + 𝐼2′ (𝑅2′ + 𝑗𝑋2′ ) + 𝐼1 (𝑅1 + 𝑗𝑋1 ) Where 𝐼2 𝑉2′ = 𝑎𝑉2 is called secondary voltage referred to the primary side, 𝐼2′ = is called 𝑎 secondary current referred to the primary side. In Table 3 shown the quantities of the secondary side with their referred quantities. 5  Referred to the primary Referred to the secondary 𝐸2 𝐸1 = 𝐸2′ = 𝐸1′ = 𝐸2 = 𝑘𝐸1 𝑘 𝑉2 𝑉2′ = 𝑉1′ = 𝑘𝑉1 𝑘 𝐼1 𝐼2′ = 𝑘𝐼2 𝐼1′ = 𝑘 𝑋2 𝑋2′ = 𝑋1′ = 𝑘 2 𝑋1 𝑘2 𝑅2 𝑅2′ = 2 𝑅1′ = 𝑘 2 𝑅1 𝑘 Table 3. Referred quantities. I0 Iw Iµ R0 X0 Fig 3. Modified equivalent circuit of a single-phase transformer. 1.3 Exact Equivalent Circuit The transformer circuit can be moved to the right or left by referring all quantities to the primary or secondary side, respectively. This is almost invariably done. The equivalent circuit moved to primary is shown in Fig. 3. If we shift all the impedances from one winding to the other, the transformer core is eliminated and we get an equivalent electrical circuit. Various voltages and currents can be readily obtained by solving this electrical circuit. 6 1.4 Approximate Equivalent Circuit The equivalent circuit can be simplified by assuming small voltage drop across the primary impedance and V1 = E1. If the applied voltage and the induced emf are the same then the shunt branch can be moved across the source voltage and the approximate equivalent circuit is drawn as shown in Figure 4. Z 01 R1 R2 X1 X I1 2 I 2 I0 R 01 X 01 Iw Iµ V1 V 2 R0 X0 Fig 4. Approximate equivalent circuit From the equivalent circuit the following relations are obtained 𝑅01 = 𝑅1 + 𝑅2′ , 𝑋01 = 𝑋1 + 𝑋2′ 𝒁𝟎𝟏 = 𝑅01 + 𝑗𝑋01 Example 3: A 30 kVA, 2400/120-V, 50-Hz transformer has a high voltage winding resistance of 0.1 Ω and a leakage reactance of 0.22 Ω. The low voltage winding resistance is 0.035 Ω and the leakage reactance is 0.012 Ω. Find the equivalent winding resistance, reactance and impedance (only magnitude) referred to : 1- High Voltage Side. 2- Low-Voltage Side. Solution 120 𝑘= = 1/20, 𝑅1 = 0.1 Ω , 𝑋1 = 0.22 Ω 2400 7  𝑅2 = 0.035 Ω and 𝑋2 = 0.012 Ω 1- For high voltage side 𝑅2 0.035 𝑅01 = 𝑅1 + 𝑅2′ = 𝑅1 + = 0.1 + = 𝟏𝟒. 𝟏 𝛀 𝑘2 1 2 ( ) 20 𝑋2 0.012 𝑋01 = 𝑋1 + 𝑋2′ = 𝑋1 + 2 = 0.22 + = 𝟓. 𝟎𝟐 𝛀 𝑘 1 2 ( ) 20 2 2 𝑍01 = √𝑅01 + 𝑋01 = √14.12 + 5.022 = 𝟏𝟓 𝛀 2- For low voltage side 1 2 𝑅02 = 𝑅2 + 𝑅1′ 2 = 𝑅2 + 𝑘 𝑅1 = 0.035 + ( ) × 0.1 = 𝟎. 𝟎𝟑𝟓𝟐𝟓 𝛀 20 1 2 𝑋02 = 𝑋2 + 𝑋1′ 2 = 𝑋2 + 𝑘 𝑋1 = 0.012 + ( ) × 0.22 = 𝟎. 𝟎𝟏𝟐𝟓 𝛀 20 2 2 𝑍02 = √𝑅02 + 𝑋02 = √0.03252 + 0.012552 = 𝟎. 𝟎𝟑𝟕𝟒 𝛀 Example 4: The parameters of a 2300/230 V, 50-Hz transformer are given below: 𝑅1 = 0.286 𝛺, 𝑅2′ = 0.319 𝛺 , 𝑅0 = 250 𝛺 𝑋1 = 0.73 𝛺 , 𝑋2′ = 0.73 𝛺 , 𝑋0 = 1250 𝛺 The secondary load impedance 𝑍𝐿 = 0.387 + 𝑗 0.29 Ω. Using the exact equivalent circuit Calculate the following. 1- Primary power factor. 4- Cu loss. 2- Power input. 5- Efficiency. 3- Power output. Solution k = 230/2300 = 0.1 , 𝒁𝑳 = 0.387 + 𝑗 0.29 Ω 8  𝒁𝑳⁄ 𝒁′𝑳 = 𝑘 2 =100 (0.387 + j 0.29) = 38.7 + j 29 = 48.4 ∠36.8° Ω 𝒁′𝟐 + 𝒁′𝑳 =(38.7 + 0.319) + j(29 + 0.73) = 39.02 + j29.73 = 49∠ 37.3° Ω 𝟏⁄ 1 1 𝒁𝒎 = ⁄𝑅0 + ⁄𝑗𝑋0 ⟹ 𝒁𝒎 =240 + j48 = 245 ∠11.3° Ω 𝒁𝒎 //(𝒁′𝟐 + 𝒁′𝑳 ) = 245∠11.3° // 49 ∠ 37.3° = 41.4∠33° Ω 𝒁𝒕𝟏 = 𝒁𝟏 + (𝒁𝒎 //(𝒁′𝟐 + 𝒁′𝑳 ))=(0.286+j 0.73)+ 41.4 ∠33° = 42∠33.7° Ω 𝑽𝟏⁄ 𝑰𝟏 = 𝒁𝒕𝟏 =2300∠0 / 42∠33.7° = 54.8∠−33.7° A 𝒁𝒎 + 𝒁′𝟐 + 𝒁′𝑳 =245 ∠11.3° + 49 ∠ 37.3° = 290∠15.6° Ω 𝒁𝒎 𝑰′𝟐 = 𝑰𝟏 × ( ⁄𝒁′ + 𝒁′ + 𝒁 ) = 54.8∠−33.7 × 0.845∠−4.3 = 46.2 ∠ − 38° A 𝟐 𝑳 𝒎 𝒁′𝟐 + 𝒁′𝑳 49 ∠ 37.3 𝑰𝟎 = 𝑰𝟏 × ( ⁄𝒁′ + 𝒁′ + 𝒁 ) = 54.8∠−33.7 × = 9.26∠−12° A 𝟐 𝑳 𝒎 290∠15.6 1- Primary power factor = cos(33.7) = 0.832 lag 2- Power input = 𝑉1 𝐼1 cos 𝜙1 = 2300 × 54.8 ×0.832 = 105 kW 3- Power output = 𝐼2′2 𝑅𝐿′ = 46.22×38.7 = 82.7 kW 4- Primary Cu loss = 54.82×0.286 = 860 W Secondary Cu loss = 46.22×0.319 = 680 W Core Cu loss = 9.262×240 = 20.6 kW 82.7 5- Efficiency = ×100 = 78.8 % 105 Example 5: The equivalent circuit parameters of a single-phase 240/2400, 50 Hz, transformer are R0 =600 Ω, X0 =300 Ω, R01 =0.25 Ω, X01 =0.75 Ω. The transformer is supplying a load of 400+ j 200 Ω. Keeping the primary voltage of 240V, calculate the 1- The secondary terminal voltage 2- Current in the primary winding 9  3- Power factor of the primary side 5- Power Input 4- Power output Solution Z01 0.25 0.75 I1 I2 I0 Iw Iµ 4 240V V2 600 300 2 Since the equivalent circuit is referred to the low voltage (primary side), the load impedance is also transformed to the low voltage side. k = 2400/240 = 10 𝒁𝑳⁄ 𝒁′𝑳 = 2 𝑘 2 =(400 + j200) (0.1) = 4 + j2 Ω 𝒁′𝑳 + 𝒁𝟎𝟏 =0.25 + j0.75 + 4 + j2 = 4.25 + j2.75 = 5.062∠32.9 Ω 𝑽𝟏 240∠0 𝑰′𝟐 = ⁄(𝒁′ + 𝒁 ) = = 47.412∠ − 32.9 = 𝟑𝟗. 𝟖 − 𝐣𝟐𝟓. 𝟕𝟓𝟑 𝐀 𝑳 𝟎𝟏 5.062∠32.9 1- Secondary terminal voltage (without phase) 𝑉2′ = 𝐼2′ 𝑍𝐿 =5.062 × √42 + 22 = 212.03V 2- Primary current: 𝑉1 240 The core loss component of current 𝐼𝑤 = = = 0.4 A 𝑅0 600 𝑉1 240 The magnetizing component of current 𝐼𝜇 = = = 0.8 A 𝑋0 300 The no-load current 𝐼0 = 𝐼𝑤 + 𝑗𝐼𝜇 = 0.4 − j0.8 A 10  The primary current 𝐼1 = 𝐼0 + 𝐼2′ =39.8 − j25.753 + 0.4 − j0.8 = 40.2 − j26.553 = 48.178∠−33.44 A 3- Power factor of the primary current pf = cos(33.44) = 0.834 lagging 4- Power output = 𝐼2′2 𝑅𝐿′ = 47.4122×4 = 8.99 kW 5- Power Input = 𝑉1 𝐼1 cos 𝜙1 = 240 × 48.178 × cos (33.44) = 9.65 kW 2. Auto-transformer It is a transformer with one winding only, part of this being common to both primary and secondary. Obviously, in this transformer the primary and secondary are not electrically isolated from each other as is the case with a 2-winding transformer. But its theory and operation are similar to those of a two-winding transformer. Because of one winding, it uses less copper and hence is cheaper. It is used where transformation ratio differs little from unity. Fig. 5 shows both step down and step-up auto-transformers. Fig 5. Auto-transformer 11  As shown in Fig. 5, AB, is primary winding having N1 turns and BC is secondary winding having N2 turns. Neglecting iron losses and no-load current. 𝑉2 𝑁2 𝐼1 = = =𝐾 𝑉1 𝑁1 𝐼2 2.1 Saving of Cu Volume and hence weight of Cu, is proportional to the length and area of the cross-section of the conductors. Now, length of conductors is proportional to the number of turns and cross- section depends on current. Hence, weight is proportional to the product of the current and number of turns. Wt. of Cu in section AC is ∝ (N1 − N2) I1. Wt. of Cu in section BC is ∝ N2 (I2 − I1). ∴ Total Wt. of Cu in auto-transformer ∝ (N1 − N2) I1 + N2 (I2 − I1) If a two-winding transformer were to perform the same duty, then Wt. of Cu on its primary ∝N1I1 ; Wt. of Cu on secondary ∝N2I2 Total Wt. of Cu ∝ N1I1 + N2I2 Wt. of Cu in auto − transformer (𝑁1 − 𝑁2 ) 𝐼1 + 𝑁2 (𝐼2 − 𝐼1 ) = =1−𝐾 Wt. of Cu in ordinary transformer 𝑁2 𝐼1 + 𝑁2 𝐼2 Wt. of Cu in auto-transformer (Wa) = (1 − K) × (Wt. of Cu in ordinary transformer W0) ∴ Saving = W0 −Wa = W0 − (1 − K) W0 = KW0 ∴ Saving percent = K × 100 % Hence, saving will increase as K approaches 12  unity. 2.2 Advantages of autotransformers 1- An autotransformer requires less Cu than a two-winding transformer of similar rating. 2- An autotransformer operates at a higher efficiency than a two-winding transformer of similar rating. 3- An autotransformer has better voltage regulation than a two-winding transformer of the same rating. 4- An autotransformer has smaller size than a two-winding transformer of the same rating. 5- An autotransformer requires smaller exciting current than a two-winding transformer of the same rating. 2.3 Applications of Auto transformer The various applications of an auto-transformer are, 1- For safely starting the machines like induction motors, synchronous motors i.e. as a starter. 2- To give a small boost to a distribution cable to compensate for a voltage drop i.e. as a booster. 3- As a furnace transformer to supply power to the furnaces at the required supply voltage. 4- For interconnecting the systems which are operating roughly at same voltage level. Example 5: In Figure shown an auto transformer used to supply a load of 2 kW at 230 V from a 400 V a.c. supply. Find the currents in parts AC and BC, neglecting losses and no load 13 current. Also find the copper saving percent due to the use of autotransformer instead of using two winding transformer: Assume purely resistive load. As the load is resistive, cos 𝜙𝐿 = 1 𝑃𝑜𝑢𝑡 = 𝑉2 𝐼2 cos 𝜙2 = 230 × 𝐼2 × 1 = 2 × 103 ⟹ 𝐼2 = 𝟖. 𝟔𝟗𝟓𝟔 𝑨 𝑉2 230 =𝐾= = 𝟎. 𝟓𝟕𝟓 𝑉1 400 𝐼1 = 𝐾 ⇒ 𝐼1 = 𝟓 𝑨 which is the current in AC part 𝐼2 Current in BC = 𝐼2 − 𝐼1 = 𝟑. 𝟔𝟗𝟓𝟔 𝑨 14  Copper saving = 𝐾 × 100% = 𝟓𝟕. 𝟓% Example 6: A 10 kVA, 230/110 V transformer is to be used as an autotransformer. What will be the voltage ratio and output rating of an autotransformer. Solution 𝑉1 = 230 𝑉 , 𝑉2 = 110 𝑉 , 𝑘𝑉𝐴 = 10 𝑘𝑉𝐴 𝑉𝐴 10×103 Current through 230 V = = = 43.478 A 230 230 𝑉𝐴 10×103 Current through 110 V = = = 90.909 A 110 110 Now as secondary voltage of two winding transformer is 110 V, let us assume. that autotransformer output voltage required is 110 V. So it can be connected as an autotransformer as shown in the Figure The part AC is primary of two winding while BC is secondary of two winding transformer. 𝑉1 = 230 + 110 = 𝟑𝟒𝟎 𝑽, 𝑉2 = 𝟏𝟏𝟎 𝑽 𝑉2 110 𝐾= = = 𝟎. 𝟑𝟐𝟑𝟓 𝑉1 340 15  𝐼1 43.478 𝐾= ⟹ 𝐼2 = = 𝟏𝟑𝟒. 𝟑𝟖𝟔 𝑨 𝐼2 0.3235 Output rating = 𝑉2 × 𝐼2 = 110 x 134.386 = 14.782 kVA 16