Single Phase Transformer

Transformer BEE2123 ELECTRICAL MACHINES Mohd Rusllim stack away Mohamed Ext 2080 A1-E10-C09 email&160protected edu. my MRM 05 Learning Outcomes ? At the extirpate of the lecture, student should to ? Understand the principle and the nature of unchanging machines of substituteer. Perform an analysis on transformers which their principles argon basic to the understanding of electric machines. ? MRM 05 Introduction ? ? ? ? A transformer is a static machines. The word transformer? comes form the word transform?.Transformer is not an energy conversion device, plainly is a device that changes AC electrical occasion at one electric potential cipher into AC electrical antecedent at another potentiality level through the action of magnetised field, without a change in frequency. It can be either to gradation-up or step down. Transmission System TX1 TX1 Gene balancen seat 33/13. 5kV 13. 5/6. 6kV Distributions TX1 TX1 MRM 05 6. 6kV/415V Consumer Transformer Construction ? twa in types of iron- heart and soul device a) b) lens nucleus type construction mash type construction ? Core type construction MRM 05 Transformer Construction ? Shell type construction MRM 05 Ideal Transformer ? An saint transformer is a transformer which has no loses, i. e. it? s tip has no ohmic immunity, no magnetic leakage, and in that locationfore no I2 R and cell nucleus loses. ? However, it is impossible to realize such a transformer in practice. ? Yet, the approximate characteristic of perfect transformer provide be used in characterized the practical transformer. N1 N2 I1 V1 E1 E2 I2 V2 V1 Primary electromotive force V2 substitute potential difference E1 Primary generate potency E2 substitute generate electric potential N1N2 Transformer ratio MRM 05 Transformer Equation ? Faraday? s Law states that, ?If the blend in passes through a coil of wire, a potential drop will be induce in the turns of wire. This potential difference is directly propo rtionate to the rate of change in the mix with enjoy of time. Vind ? potentiality ind d? (t ) ?? dt Lenz? s Law If we have N turns of wire, Vind ? Emf ind d? (t ) ? ?N dt MRM 05 Transformer Equation ? For an ac sources, ? Let V(t) = Vm depravity? t i(t) = im depravity? t Since the magnetic field is a sinningusoidal function ?(t ) ? ? m sin ? t Then Therefore d? m sin ? t Vind ? Emf ind ? ? N dt ? ? N?? m romaine lettuce ? 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 warning transformer E1 ?? 4. 44 fN1? m (i) ? In the proportionality condition, both the input power will be equaled to the output power, and this condition is tell to ideal condition of a transformer. E2 ?? 4. 44 fN 2? m Input power ? output power V1 I1 romaine ? ? V2 I 2 romaine ? ? V1 I 2 ? V2 I1 ? From the ideal transformer tour, note 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 potentiality Transformation Ratio which will observe 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 rating is usu every last(predicate)y written in terms of obvious reason. ? App bent power is actu onlyy the harvest-time of its rated period and rated potential difference. VA ? V1I1 ? V2 I 2 ? Where, ? I1 and I2 = rated underway on essential and secondary crook. ? V1 and V2 = rated electromotive force on old and secondary weave. ? Rated circulating(prenominal)s atomic number 18 in truth the luxuriant lade up-to-the-minutes in transformer MRM 05 mannikin 1. 1. 5kVA iodine stagecoach transformer has rated potential difference of 144/240 V. Finds its broad lodge current. lead 1500 I1FL ? ? 10. 45 A 144 1500 I 2 FL ? ? 6A 240 MRM 05 usage 2.A bingle phase transformer has four hundred chief(a) and li ght speed0 secondary turns. The net cross-sectional ara of the kernel is 60m2. If the primary quill flatus is connected to a 50Hz supply at 520V, calculate a) The induced potential drop in the secondary braid b) The peak honour of flux absorption in the kernel rootage N1=400 V1=520V A=60m2 N2= degree Celsius0 V2=? MRM 05 typeface 2 (Cont) a) Know that, N1 V1 a? ? N 2 V2 400 520 ? one hundred0 V2 V2 ? 1300V 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 physical exertion 3.A 25kVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000V, 50Hz supply. Find Full gist primary and secondary current b) The induced emf in the secondary winding c) The maximum flux in the core Solution VA = 25kVA N1=500 V1=3000V N2=50 V2=? a) MRM 05 ensample 3 (Cont) a) Know that, VA ? V ? I I1FL VA 25 ? 103 ? ? ? 8. 33 A V1 3000 b) bring forth potential drop, 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) Max flux E ? 4. 44 fN ? 300 ? 4. 44(50)(50)? ? ? 27mWb MRM 05Practical Transformer (Equivalent dress circle) I1 R1 X1 Ic V1 RC Io I1 Im Load Xm E1 E2 V2 N1 N2 I2 R2 X2 V1 = primary supply potential difference V2 = second terminal ( adulterate) electric potential E1 = primary winding emf E2 = 2nd winding voltage I1 = primary supply current I2 = 2nd winding current I1? = primary winding current Io = no agitate current Ic = core current Im = magnetism current R1= primary winding defense R2= 2nd winding underground X1= primary winding leakage reactance X2= 2nd winding leakage reactance Rc MRM 05= core subway Xm= magnetism reactance Single build Transformer ( describered to Primary) ? real(a) systemI1 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 Phase Transformer (Referred to Primary) ? approximative manner 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) ? Approximate Method I1 R01 X01V1 aV2 In some application, the excitation carve up has a small current compared to shoot down current, thus it may be neglect 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 excitation discriminate 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 spokesperson 4. For the line of reasonings obtained from the examen of 20kVA 2600/245 V single phase transformer, stir all the logical arguments to the luxuriously voltage attitude if all the parameters are obtained at lower voltage stead nerve. Rc = 3. 3? , Xm =j1. 5? , R2 = 7. 5? , X2 = j12. 4? Solution Given Rc = 3. 3? , Xm =j1. 5? , R2 = 7. ? , X2 = j12. 4? MRM 05 exercising 4 (Cont) i) Refer to H. V billet (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 pick up from secondary slope 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 psychometric evidenceple (What if.. ) 4. For the parameters obtained from the leaven of 20kVA 245/2600 V single phase transformer, refer all the parameters to the mettlesome voltage place if all the parameters are obtained at lower voltage status facial expression.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 precedent Factor ? Power doer = slant between Current and Voltage, romaine lettuceine ? V I ? I ? = -ve Lagging ? V I V ? = +ve take ?=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, while 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 get in the resembling hitch The approximate value of secondary voltage at full point of 0. fall behind power factor, 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 beseech losses - occur in core parameters Copper losses - occur in winding resistance i. Iron Losses Piron ? Pc ? ( I c) 2 Rc ? Popen rotary ii. Coppe r Losses Pcopper ? Pcu ? ( I 1) 2 R1 ? ( I 2) 2 R2 ? Pshort electric circle or if referred , Pcu ? ( I 1) 2 R01 ? ( I 2) 2 R02 MRM 05 Poc and Psc will be discusses later in transformer examine Transformer Efficiency ? To gibe the murder of the device, by comparing the output with respect to the input. ? The higher the efficiency, the better the system. Efficiency ,? payoff Power ? c% Input Power Pout ? ? c% Pout ? Plosses ? V2 I 2 romaineine ? ? coulomb% V2 I 2 romaineine lettuce ? ? Pc ? Pcu ? ( fullload) ? ?(load n ) ? VA cos ? ? one C% VA cos ? ? Pc ? Pcu nVA cos ? ?100% 2 nVA cos ? ? Pc ? n Pcu Where, if ? load, hence 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 ordinance ? The measure of how well a power transformer maintains constant secondary voltage over a range of load currents is called the transformers voltage statute ?The purpose of voltage regulation is basically to larn the part of voltage drop between no load and full load. MRM 05 Voltage mandate ? For calculation of Voltage regulating, terminologies may be quite confusing, hence you wish constantly calculate in current, I (A) read/write head of view Full-load means the point at which the transformer ? is operating at maximum tolerable secondary current ? When connected to load, current being drawn, hence Voltage drop) ? ? No Load means at Rated At no load, current almost zero, so takes Voltage at rated MRM 05 value think like an open hitch) Voltage Regulation Voltage Regulation can be determine based on 3 systems a) b) c) staple fibre Definition Short circuit tally Equivalent rophy MRM 05 Voltage Regulation (Basic Defination) ? In this regularity, all parameter are being referred to primary or secondary side. ? Can be equal in either ? vote out voltage Regulation mark that VNL ? VFL V . R ? ?100% VNL (at Rated Value) VNL ? Up Voltage Regulation VNL ? VFL V . R ? ?100% VFL MRM 05 Voltage Regulation (Short circuit tribulation) ? In this method, direct order can be used. V . R ? V . R ? Vsc cos ?? sc ? ? p. f ? V1 ?100% If s/c sieve on primary side Vsc cos ?? c ? ? p. f ? V2 ?100% If s/c test on primary side melody that ? is for Lagging power factor +? is for leash power factor Must check that Isc must equal to IFL (I at Rated), differently MRM 05 can? t use this formula Voltage Regulation (Equivalent travel ) ? 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 pipeline that +? is for Lagging power factor ? is for Leading power factor MRM 05 assume j terms 0Comment on VR ? Purely repellent Load ? > 3 % is considered poor VR commonly poor than Resistive Load ? inductive Load ? ? Example of application sought after Poor VR ? ? Discharge lighting AC arc welders MRM 05 Example 6. In archetype 5, determine the Voltage regulation 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 Know 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 Circuit I1=5A R01=9. 64? X01 = 21. 32? V1=2000V, 0. 8 lag 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. Determine the voltage regulation at 0. 8 lagging power factor if Vsc =18V Isc =100 Psc=240W Solution Check 7. I FL2 I FL2 VA 10000 ? ? ? 100 A V 100 ? I sc , Hence, we can use short-circuit 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 following entropy were obtained in test on 20kVA 2400/240V, 60Hz transformer. Vsc =72V Isc =8. 33A Psc=268W Poc=one hundred seventyW The cadence 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 efficiency and one-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. Short 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) ? one hundred seventy ? (1) (268) (0. 5)(20000)(0. 8) ? ?100% ? 97. 12% 2 (0. 5)(20000)(0. 8) ? 170 ? (0. 5) (268) ?( half load) MRM 05 Measurement on Transformer ? i. ii.There ar e two test conducted on transformer. bluff Circuit Test Short Circuit test ? ? ? The test is conducted to determine the parameter 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 given 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 sc ? 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 appropriate 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 vo ltage side ? all reading 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 E xample 9 (Cont) Efficiency,? ? ? VA cos ? ? FL ? ? ? ?100% ? VA cos ? ? Psc ? Poc ? ? ? 500 ? 103 )(0. 866) ?? ? ? 100% (500 ? 103 )(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 Final 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 a t 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 Any Questions ??? Test 1 coming soon posit sure you prepared for that MRM 05