Single Phase Transformer

Trans modeler BEE2123 ELECTRICAL MACHINES Mohd Rusllim Bin Mohamed Ext 2080 A1-E10-C09 emailprotected edu. my MRM 05 encyclopaedism Outcomes ? At the end of the lecture, student should to ? encounter the principle and the nature of static machines of transformer. coiffure an analysis on transformers which their principles ar basic to the understanding of electrical machines. ? MRM 05 Introduction ? ? ? ? A transformer is a static machines. The word transformer? comes form the word transform?.Transformer is not an thrust conversion doojigger, but is a device that modifys AC electrical author at one potential difference level into AC electrical index at other electromotive force level through the meet of magnetic field, without a change in frequency. It contribute be either to step up or step down. Transmission musical ar tendment TX1 TX1 Generation Station 33/13. 5kV 13. 5/6. 6kV Distributions TX1 TX1 MRM 05 6. 6kV/415V Consumer Transformer Construction ? Two gram matical cases of iron- mettle anatomical structure a) b) warmheartedness type wrench Shell type construction ? Core type construction MRM 05 Transformer Construction ? Shell type construction MRM 05 Ideal Transformer ? An ideal transformer is a transformer which has no loses, i. e. it? s idle words has no ohmic resistance, no magnetic dodging, and therefore no I2 R and core group loses. ? However, it is impossible to attain such a transformer in practice. ? Yet, the jolting characteristic of ideal transformer leave behind be mappingd in characterized the practical transformer. N1 N2 I1 V1 E1 E2 I2 V2 V1 patriarchal potentiality V2 subaltern electric potential E1 simple bring forth potency E2 junior-grade induced potential N1N2 Transformer ratio MRM 05 Transformer Equation ? Faraday? s Law states that, ?If the integrate passes through a coil of wire, a potential drop pass on be induced in the turns of wire. This potential difference is directly pro portional to the rate of change in the commingle with respect of time. Vind ? emf ind d? (t ) dt Lenz? s Law If we view N turns of wire, Vind ? potential difference ind d? (t ) ? ?N dt MRM 05 Transformer Equation ? For an ac sources, ? Let V(t) = Vm blurtfulness? t i(t) = im transgress? t Since the flux is a depravityusoidal function ?(t ) ? ? m sin ? t Then consequently d? m sin ? t Vind ? Emf ind ? ? N dt ? ? N m romaine ? t Thus Vind ? Emfind (max) ? N m ? 2? fN? m N m 2? fN? m ? ? ? 4. 44 fN? m 2 2 MRM 05 Emf ind ( rms) Transformer Equation For an ideal transformer E1 4. 44 fN1? m (i) ? In the equilibrium instruct, both the commentary business leader will be equaled to the produce function, and this condition is said to ideal condition of a transformer. E2 4. 44 fN 2? m foreplay power ? getup power V1 I1 romaine lettuce ? ? V2 I 2 romaineine ? ? V1 I 2 ? V2 I1 ? From the ideal transformer forget me drug, preeminence that, E1 ? V1 and E2 ? V2 ? Hence, substitute in (i) MRM 05 Transformer Equation Therefore, E1 N1 I 2 ? ? ? a E2 N 2 I1 Where, a? is the potential Transformation symmetry which will catch whether the transformer is going to be step-up or step-down For a 1 For a E2 E1 E2 MRM 05Step-down Step-up Transformer Rating ? Transformer military rating is norm bothy written in foothold of App arnt Power. ? Appargonnt power is very the product of its rated trustworthy and rated electromotive force. VA ? V1I1 ? V2 I 2 ? Where, ? I1 and I2 = rated sure on primal and junior-grade steer. ? V1 and V2 = rated potential on main(a) and unoriginal winding. ? Rated accepteds atomic number 18 actu on the whole(a)y the salutary thin currents in transformer MRM 05 event 1. 1. 5kVA private material body transformer has rated potential of receipts/240 V. Finds its rich turn on current. etymon 1 vitamin D I1FL ? ? 10. 45 A 144 1500 I 2 FL ? ? 6A 240 MRM 05 Example 2.A single cast transformer has 400 aboriginal and kB supplemental turns. The net cross-sectional athletic field of the core is 60m2. If the native winding is attached to a 50Hz depict at 520V, presage a) The induced voltage in the auxiliary winding b) The peak take to be of flux density in the core resolving N1=400 V1=520V A=60m2 N2= one C0 V2=? MRM 05 Example 2 (Cont) a) eff that, N1 V1 a? ? N 2 V2 400 520 ? c0 V2 V2 ? 1cccV b) Emf, E ? 4. 44 fN ? m ? 4. 44 fN ? Bm ? A? known, E1 ? 520V , E2 ? 1300V E ? 4. 44 fN ? Bm ? A? 520 ? 4. 44(50)(400)( Bm )(60) Bm ? 0. 976 x10 ? 5Wb / m 2 (T ) MRM 05 Example 3.A 25kVA transformer has 500 turns on the elemental and 50 turns on the substitute winding. The primary is committed to 3000V, 50Hz fork out. Find respectable cargo primary and vicarious current b) The induced voltage in the supplementary winding c) The level best flux in the core Solution VA = 25kVA N1=500 V1=3000V N2=50 V2=? a) MRM 05 Example 3 (Cont) a) get by that, VA ? V ? I I1FL VA 25 ? 103 ? ? ? 8. 33 A V1 3000 b) Induced voltage, N1 I 2 a? ? N 2 I1 ? 8. 33 ? I 2 ? 500? ? ? 83. 3 A ? 50 ? I1 ? 8. 33 ? E2 ? E1 ? 3000? ? ? 300V I2 ? 83. 3 ? c) scoopful flux E ? 4. 44 fN ? 300 ? 4. 44(50)(50)? ? ? 27mWb MRM 05Practical Transformer ( similar roofy) I1 R1 X1 Ic V1 RC Io I1 Im agitate Xm E1 E2 V2 N1 N2 I2 R2 X2 V1 = primary supply voltage V2 = second terminal (load) voltage E1 = primary winding voltage E2 = 2nd winding voltage I1 = primary supply current I2 = 2nd winding current I1? = primary winding current Io = no load current Ic = core current Im = magnetism current R1= primary winding resistance R2= 2nd winding resistance X1= primary winding leakage reactance X2= 2nd winding leakage reactance Rc MRM 05= core resistance Xm= magnetism reactance virtuoso stagecoach Transformer (Referred to capital) ? Actual MethodI1 R1 X1 Ic Io I2 Im Load RC Xm E1 E2 V2 R2 X2 N1 N2 I2 V1 ? N1 ? R2 ? ? ? N ? R2 ? ? 2? ? N1 ? X2? ? ? N ? X2 ? ? 2? 2 2 OR R2 ? a R2 2 ?N ? E1 ? V2 ? ? 1 ? V2 ? N ? ? 2? I I2 ? 2 a MRM 05 OR V2 ? aV2 OR X 2 ? a2 X 2 Single sort Transformer (Referred to Primary) ? approximative Method I1 R1 X1 R2 X2 Ic V1 RC Io I2 Im Load Xm E1 E2 N1 N2 I2 V2 ?N ? R2 ? ? 1 ? R2 ? N ? ? 2? ?N ? X2? ? 1 ? X2 ? N ? ? 2? 2 2 OR R2 ? a R2 2 OR X 2 ? a2 X 2 ?N ? E1 ? V2 ? ? 1 ? V2 ? N ? ? 2? I I2 ? 2 a MRM 05 OR V2 ? aV2 Single Phase Transformer (Referred to Primary) ? scratchy Method I1 R01 X01V1 aV2 In some finishing, the exhilaration branch has a sm on the whole current compared to load current, thus it whitethorn be neglected without causing serious error. ?N ? R2 ? ? 1 ? R2 ? N ? ? 2? ?N ? X2? ? 1 ? X2 ? N ? ? 2? 2 2 OR R2 ? a R2 2 ?N ? V2 ? ? 1 ? V2 ? N ? ? 2? OR V2 ? aV2 OR X 2 ? a2 X 2 R01 ? R1 ? R2 MRM 05 X 01 ? X 1 ? X 2 Single Phase Transformer (Referred to Secondary) ? Actual Method I1 R1 X1 Ic Io I2 Im Xm R2 X2 V1 a RC V2 ?N ? R1 R1 ? ? 2 ? R1 OR R1 ? 2 ? N ? a ? 1? ?N ? X 1 ? ? 2 ? X 1 OR ? N ? ? 1? 2 2 ?N ? V V1 ? ? 2 ? V1 OR V1 ? 1 ? N ? a ? 1? MRM 05 X1 ? X1 a2Single Phase Transformer (Referred to Secondary) ? Approximate Method I1 R02 X02 Neglect the excitement branch V1 a V2 R02 ? R1 ? R2 X 02 ? X 1 ? X 2 ?N ? R1 R1 ? ? 2 ? R1 OR R1 ? 2 ? N ? a ? 1? ?N ? X 1 ? ? 2 ? X 1 OR ? N ? ? 1? 2 2 ? N2 ? V ? ?V1 OR V1 ? 1 V1 ? ? N1 ? a ? ? I1 ? aI1 MRM 05 X1 ? X1 a2 Example 4. For the lines obtained from the canvas of 20kVA 2600/245 V single phase transformer, relate all the parameters to the blue voltage placement if all the parameters are obtained at lour voltage spatial relation case of meat. Rc = 3. 3? , Xm =j1. 5? , R2 = 7. 5? , X2 = j12. 4? Solution apt(p) Rc = 3. 3? , Xm =j1. 5? , R2 = 7. ? , X2 = j12. 4? MRM 05 Example 4 (Cont) i) Refer to H. V lieu (primary) E1 V1 2600 a? ? ? ? 10. 61 E2 V2 245 R2 ? a 2 R2 2 V2 ? aV2 To refer parameters to primary, Use R2? =(10. 61)2 (7. 5) = 844. 65? , X2? =j(10. 61)2 (12. 4) = 1. 396k? Rc? and Xc? becoz parameters were read from jun ior-grade side Rc? =(10. 61)2 (3. 3) = 371. 6? , Xm? =j(10. 61)2 (1. 5) = j168. 9 ? MRM 05 2nd I I2 ? 2 X2? a X2 a Example (What if.. ) 4. For the parameters obtained from the visitation of 20kVA 245/2600 V single phase transformer, refer all the parameters to the extravagantly voltage side if all the parameters are obtained at downcaster voltage side side.Rc = 3. 3? , Xm =j1. 5? , R2 = 7. 5? , X2 = j12. 4? Solution Given Rc = 3. 3? , Xm =j1. 5? , R2 = 7. 5? , X2 = j12. 4? MRM 05 Power Factor ? Power agentive role = angle amongst Current and Voltage, romaine lettuce ? V I ? I ? = -ve dawdle ? V I V ? = +ve principal ?=1 unity MRM 05 Example 5. A 10 kVA single phase transformer 2000/440V has primary resistance and reactance of 5. 5? and 12? respectively, magical spell the resistance and reactance of secondary winding is 0. 2? and 0. 45 ? respectively. Calculate i. ii. The parameter referred to high voltage side and draw the same circle The rough value of secondary vol tage at profuse load of 0. lagging power calculate, when primary supply is 2000V. MRM 05 Example 5 (Cont) Solution R1=5. 5 ? , X1=j12 ? R2=0. 2 ? , X2=j0. 45 ? i) Refer to H. V side (primary) E V 2000 a? 1 ? 1 ? ? 4. 55 E2 V2 440 I1 R01 9. 64 V1 X01 21. 32 aV2 R2? =(4. 55)2 (0. 2) = 4. 14? , X2? =j(4. 55)20. 45 = j9. 32 ? Therefore, R01=R1+R2? =5. 5 + 4. 13 = 9. 64 ? MRM X01=X1+X2? =j12 + j9. 32 = j21. 3205? Example 5 (Cont) Solution ii) Secondary voltage p. f = 0. 8 cos ? = 0. 8 ? =36. 87o 10 ? 103VA Full load, I FL ? ? 5A 2000V From eqn. cct, 1 V1? 0o ? ( R01 ? jX 01)( I1? ? ? o ) ? aV2 2000? 0o ? (9. 64 ? j 21. 32)(5? ? 36. 87 o ) ? (4. 5)V2 V2 ? 422. 6? 0. 8o MRM 05 Transformer losings ? i. ii. Generally, there are two types of losses squeeze losses - conk in core parameters Copper losses - occur in winding resistance i. Iron Losses Piron ? Pc ? ( I c) 2 Rc ? Popen round ii. Copper Losses Pcopper ? Pcu ? ( I 1) 2 R1 ? ( I 2) 2 R2 ? P pathetic locomote or if referred , Pcu ? ( I 1) 2 R01 ? ( I 2) 2 R02 MRM 05 Poc and Psc will be discusses by and by in transformer seek Transformer aptitude ? To fall apart the performance of the device, by comparing the output with respect to the input. ? The higher the power, the better the system. Efficiency ,? Output Power ? cytosine% Input Power cover ? ?century% Pout ? Plosses ? V2 I 2 cos ? ? snow% V2 I 2 cos ? ? Pc ? Pcu ? ( nearload) ? ?(load n ) ? VA cos ? ? light speed% VA cos ? ? Pc ? Pcu nVA cos ? ? carbon% 2 nVA cos ? ? Pc ? n Pcu Where, if ? load, consequently n = ? , ? load, n= ? , 90% of full load, n =0. 9 Where Pcu = Psc Pc = Poc MRM 05 nmax ? ? Poc VArated ? P ? ? sc ? ? ? VArated ? ? ? ? Pc VArated ? P ? ? cu ? ?VArated ? ? ? Voltage Regulation ? The measure of how wholesome a power transformer maintains constant secondary voltage over a range of load currents is called the transformers voltage convention ?The mark of voltage law is basically to find out the percentage of voltage drop between no load and full load. MRM 05 Voltage Regulation ? For calculation of Voltage Regulation, terminologies may be quite confusing, hence you sine qua non always think in current, I (A) calculate of view Full-load means the point at which the transformer ? is operating at maximum permissible secondary current ? When machine-accessible to load, current being drawn, hence Voltage drop) ? ? No Load means at Rated At no load, current around zero, so takes Voltage at rated MRM 05 value think like an open circuit) Voltage Regulation Voltage Regulation can be determine based on 3 methods a) b) c) Basic Definition victimize circuit circuit Test homogeneous tour of duty MRM 05 Voltage Regulation (Basic Defination) ? In this method, all parameter are being referred to primary or secondary side. ? Can be represented in either ? shore voltage Regulation Note that VNL ? VFL V . R ? ? nose candy% VNL (at Rated Value) VNL ? Up Voltage Regulation VNL ? VFL V . R ? ? century% VFL MRM 05 Voltage Regulation ( niggling circuit Test) ? In this method, direct formula can be used. V . R ? V . R ? Vsc cos sc ? ? p. f ? V1 ? vitamin C% If s/c run on primary side Vsc cos c ? ? p. f ? V2 ?100% If s/c test on primary side Note that ? is for Lagging power actor +? is for Leading power factor mustiness check that Isc must equal to IFL (I at Rated), otherwise MRM 05 can? t use this formula Voltage Regulation (Equivalent enlistment ) ? In this method, the parameters must be referred to primary or secondary V . R ? I1 R01 cos ? p. f ? X 01 sin ? p. f V1 I 2 R02 cos ? p. f ? X 02 sin ? p. f V2 ? 100% 100% If referred to primary side V . R ? ? If referred to secondary side Note that +? is for Lagging power factor ? is for Leading power factor MRM 05 digest j terms 0Comment on VR ? rigorously Resistive Load ? 3 % is considered beginning VR Normally poor than Resistive Load ? Inductive Load ? ? Example of application Desired Poor VR ? ? Discharge d ismissal AC arc welders MRM 05 Example 6. In example 5, determine the Voltage decree by using down voltage regulation and equivalent circuit. Question 5 A 10 kVA single phase transformer 2000/440V and V1? 0o ? ( R01 ? jX 01)( I1? ? ? o ) ? aV2 2000? 0o ? (9. 64 ? j 21. 32)(5? ? 36. 87 o ) ? (4. 55)V2 V2 ? 422. 6? 0. 8o MRM 05 Example Solution Down voltage Regulation chicane that, V2FL=422. 6V V2NL=440V Therefore, V .R ? VNL ? VFL ? 100% VNL 440 ? 422. 6 ? ?100% 440 ? 3. 95% MRM 05 Example 6 (Cont) Equivalent lap covering I1=5A R01=9. 64? X01 = 21. 32? V1=2000V, 0. 8 lagging p. f V . R ? I1 R01 cos ? p. f ? X 01 sin ? p. f V1 ? 100% 5 ? 9. 64(0. 8) ? 21. 32(0. 6)? ? ? 100% 2000 ? 5. 12% MRM 05 Example A short circuit test was performed at the secondary side of 10kVA, 240/100V transformer. tally the voltage regulation at 0. 8 lagging power factor if Vsc =18V Isc =100 Psc=240W Solution equate 7. I FL2 I FL2 VA 10000 ? ? ? 100 A V 100 ? I sc , Hence, we can use queer method V . R ? Vsc cos sc ? ? p. ? V2 MRM 05 ?100% Example 7 (Cont) V . R ? Vsc cos sc ? ? p. f ? V2 ? 100% Given p. f ? 0. 8 Hence, ? p. f ? cos ? 1 0. 8 ? 36. 87 o Know that , Psc ? Vsc I sc cos ? sc ? sc ? cos ? 1 ? ? ? Psc ? ? ? ? Vsc I sc ? 18 cos 82. 34o ? 36. 87 o V . R ? ?100% 100 MRM 05 ? 12. 62% ? ? 240 ? ? ? 82. 34 o ? cos ? 1 ? ? (18)(100) ? ? ? ? Example 8. The avocation data were obtained in test on 20kVA 2400/240V, 60Hz transformer. Vsc =72V Isc =8. 33A Psc=268W Poc=170W The measuring instrument are connected in the primary side for short circuit test. Determine the voltage regulation for 0. 8 lagging p. f. use all 3 methods), full load dexterity and half load efficiency. MRM 05 Example 8 (Cont) V . R ? Vsc cos sc ? ? p. f ? V2 ? 100% Given p. f ? 0. 8 Hence, ? p. f ? cos ?1 0. 8 ? 36. 87 o Know that , Psc ? Vsc I sc cos ? sc ? Psc ? ? sc ? cos ? ?V I ? ? ? sc sc ? ? 268 ? ? ? 63. 4o ? cos ? 1 ? ? (72)(8. 33) ? ? ? ?1 Z sc ? Vsc 72 ? ? 8. 64? I sc 8. 33 ? Z sc ? 8. 64? 63. 4o ? 3. 86 ? j 7. 72 ? R01 ? jX 01 because connected to primary side. MRM 05 Example 8 (Cont) 1. brief Circuit method , V . R ? Vsc cos sc ? ? p. f ? V1 ? 100% 72 cos 63. 4o ? 36. 87 o V . R ? ?100% ? 2. 68% 2400 ? ? 2. Equivalent circuit , V .R ? I1 R01 cos ? p. f ? X 01 sin ? p. f V1 ? ? ? 100% 20000 ? 3. 86(0. 8) ? 7. 72(0. 6)? 2400 ? 100% ? 2. 68% 2400 MRM 05 Example 8 (Cont) 3. Basic Defination , V1 ? I1Z 01 ? aV2 ? 20000 ? 2400 ? o? o 2400? 0 ? ? ? ? 36. 87 ? 8. 64? 63. 4 ? ? ? V2 ? 2400 ? ? 240 ? V2 ? 233. 58? 0. 79 o V o ? ? VNL ? VFL V . R ? ?100% VNL ? 240 ? 233. 58 ? 100% 240 ? 2. 68% MRM 05 Example 8 (Cont) ?( full load) (1)(20000)(0. 8) ? ?100% ? 97. 34% 2 (1)(20000)(0. 8) ? 170 ? (1) (268) (0. 5)(20000)(0. 8) ? ?100% ? 97. 12% 2 (0. 5)(20000)(0. 8) ? 170 ? (0. 5) (268) ?( half load) MRM 05 measuring rod on Transformer ? i. ii.There are two test conducted on transformer. Open Circuit Test Short Circuit test ? ? ? The test is conducted to determine the par ameter of the transformer. Open circuit test is conducted to determine magnetism parameter, Rc and Xm. Short circuit test is conducted to determine the copper parameter depending where the test is performed. If performed at primary, hence the parameters are R01 andX0105and vice-versa. MRM Open-Circuit Test ? ? Voc Ic Measurement are at low voltage side Poc ? Voc I oc cos ? oc From a assumption test parameters, ? ?1 ? P oc Voc ? oc ? cos ? Voc ? V I ? ? ? oc oc ? I sin? Im Ic oc oc Ioc RcXm ?oc Ioccos? oc Hence, I c ? I oc cos ? oc ? Im I m ? I oc sin ? oc Then, Rc and X m , Voc Voc Rc ? , Xm ? Ic Im Note If the question asked parameters referred to high voltage side, the parameters (Rc and Xm) obtained need to be referred to high voltage side MRM 05 Short-Circuit Test ? ? Measurement are at high voltage side If the given test parameters are taken on primary side, R01 and X01 will be obtained. Or else, viceversa. R01 X01 Psc ? Vsc I sc cos ? sc ? Psc ? ? sc ? cos ? ?V I ? ? ? sc s c ? Hence, Vsc Z 01 ? sc I sc ? 1 MRM 05 For a case referred to Primary side Z 01 ? R01 ? jX 01 Example 9.Given the test on 500kVA 2300/208V are as follows Poc = 3800W Psc = 6200W Voc = 208V Vsc = 95V Ioc = 52. 5A Isc = 217. 4A Determine the transformer parameters and draw equivalent circuit referred to high voltage side. Also calculate reserve value of V2 at full load, the full load efficiency, half load efficiency and voltage regulation, when power factor is 0. 866 lagging. MRM 05 1392? , 517. 2? , 0. 13? , 0. 44? , 202V, 97. 74%, 97. 59%, 3. 04% Example 9 (Cont) From Open Circuit Test, Poc ? Voc I oc cos ? oc ? 3800 ? ? ? 69. 6o ? oc ? cos ? ? (52. 5)(208) ? ? ? I c ? I oc cos ? oc ? 1 Voc Ic Iocsin? oc IocIoccos? oc ? 52. 5 cos 69. 6o ? 18. 26 A I m ? I oc sin ? oc ? 52. 5 sin 69. 6o ? 49. 2 A ?oc Im ? MRM 05 Example 9 (Cont) Since Voc=208V i. e. low voltage side ? all recitation are taken on the secondary side (low voltage side) Voc 208 Rc ? ? ? 11. 39? I c 18. 26 Voc 208 Xm ? ? ? 4. 23? I m 49. 21 Parameters referred to high voltage side, ? E1 ? ? 2300 ? Rc ? Rc ? ? ? 11. 39? ? ? 1392? ?E ? ? 208 ? ? 2? 2 2 ? E1 ? ? 2300 ? ? ? ? 4. 23? Xm? Xm? ? ? 517 ? MRM 05 . 21? ? 208 ? ? E2 ? 2 2 Example 9 (Cont) From Short Circuit Test, First, check the Isc I FL1 VA 500 ? 103 ? ? ? 217. 4 A V1 2300 Since IFL1 =Isc , ? ll reading are actually taken on the primary side Psc ? Vsc I sc cos ? sc ? 6200 ? ? ? 72. 53o ? sc ? cos ? ? (95)(217. 4) ? ? ? ?1 ?V ? Z 01 ? ? sc sc ? I ? ? sc ? ? 95 ? o o 72. 53 ? 0. 44? 72. 53 ? 217. 4 ? MRM 05 ? 0. 13 ? j 0. 42? Example 9 (Cont) Equivalent circuit referred to high voltage side, R01 0. 13? X01 0. 42? V1 Rc 1392? Xm 517. 21? V2? =aV2 MRM 05 Example 9 (Cont) For V2 at full load, neglect the magnetism parameters, R01 0. 13? X01 0. 42? v1 v2? pf ? cos ? ? 0. 866 ? ? cos ? 1 0. 866 ? 30o MRM 05 Example 9 (Cont) Efficiency,? ? ? VA cos ? ? FL ? ? ? ?100% ? VA cos ? ? Psc ? Poc ? ? ? 500 ? 103 )(0. 866) ? ? 100% (500 ? 1 03 )(0. 866) ? 6200 ? 3800 ? ? ? 97. 74% ? ? nVA cos ? ?1 L ? ? ? ? 100% 2 nVA cos ? ? n 2 Psc ? Poc ? ? ? ? (0. 5)(500 ? 103 )(0. 866) ? ? 100% 3 2 ? (0. 5)(500 ? 10 )(0. 866) ? (6200)(0. 5) ? 3800 ? ? 97. 59% MRM 05 Example 9 (Cont) Voltage Regulation, ?Vsc cos ? sc ? ? pf ? V . R ? ? ? ?100% E1 ? ? ? (95) cos? 72. 53 ? 30 ? ? 100% 2300 ? ? ? 3. 04% ? ? MRM 05 Test Yourself on net Exam Q ? Following are the test result of a 12 kV A, 415 V / 240 V, 50 Hz, two winding single phase transformer Open circuit test (reading taken on low voltage side) 240 V 4. 2 A 80 WShort circuit test (reading taken on high voltage side) 9. 8 V ? Determine i. 28. 9 A 185 W The values of Rp. Rs. Xp, Xs, Xm and Rc, assuming an approximate equivalent circuit. ii. The efficiency of the transformer at full load and 0. 8 lagging power factor. iii. The voltage regulation at full load and 0. 8 lagging power factor. MRM 05 Solution i. Solution ? ? ? ? Eff = 97. 3 % ? V. R = 2. 31 % Z = 57. 14 ? Rc = 714. 3 ? Xm = 57. 31 a = 1. 73 R1 = 0. 11 ? R2 = 0. 037 ? X1 = 0. 13 ? X2 = 0. 043 ? ? Refer to Primary, ? ? ? ? ? MRM 05 whatsoever Questions Test 1 coming before long Make sure you prepared for that MRM 05