3.7 Maximum Efficiency It may be shown that the efficiency of a transformer is a maximum when then the variable copper loss (i.e. I P2 × R P + I S2 × RS ) is equal to the constant iron losses. Example(3.2): A (5 KVA) single-phase transformer has a turns ratio of (10:1) and is fed from a (2.5 KV) supply. Neglecting losses, determine (a)The full-load secondary current (b) The primary current at full-load. Solution: (a) N P 10 = , NS 1 Secondary voltage since N P VP = N S VS ⎛N VS = VP ⎜⎜ S ⎝ NP ⎞ ⎟⎟ ⎠ ⎛1⎞ VS = 2500⎜ ⎟ = 250 volt ⎝ 10 ⎠ The transformer rating in volt-amperes(secondary) = VS I S (at full-load), i.e. 5000 = 250 × I S , IS = hence full-load secondary 5000 = 20 A 250 (b) NP IS = NS IP From which primary current ⎛N I P = I S ⎜⎜ S ⎝ NP ⎞ ⎟⎟ ⎠ ⎛1⎞ I P = 20⎜ ⎟ = 2 A ⎝ 10 ⎠ ********************************************************** 85 Example (3.3): A (2.2 KVA), (230/115 V) transformer has been tested to determine its equivalent circuit. The results of the tests are shown below. Open-circuit test Short- circuit test VOC=230 V VSC=19.1 V IOC=0.45 A ISC=8.7 A POC=30 W PSC=42.3 W All data given were taken from the primary side of the transformer. (a) Find the approximate equivalent circuit of this transformer referred to the high-voltage side of transformer. Solution: Open circuit test: P = VI cos(φ ) 30 ⎛ ⎞ ⎟ , ⎝ 230 × 0.45 ⎠ φ = 73.15 o φ = cos −1 ⎜ I C = I o cos(φ ) I m = I o sin(φ ) RC = I C = 0.45 cos(73.15) = 0.13 A , , I m = 0.45 sin(73.15) = 0.43 A 230 = 1769Ω , 0.13 Xm = 230 = 534.88Ω 0.43 Short circuit test: P = I 2 × Req Req = 42.3 (8.7 )2 = 0.558Ω Z eq = VS .C 19.1 = = 2.195Ω I S .C 8.7 X eq = (2.195)2 − (0.558)2 = 2.12Ω ********************************************************** 86 3.8 Three-Phase Transformers Almost all the major power generation and distribution systems in the word today are three-phase A.C systems. Since three- phase systems play such an important role in modern life, it is necessary to understand how transformers are used in them. Transformers for three-phase circuits can constructed in one of two ways. One approach is simply to take three single-phase transformers and connect them in a three-phase bank. An alternative approach is to make a three-phase transformer consisting of three sets of windings wrapped on a common core. The construction of a single three-phase transformer is the preferred practice today. Since it is lighter, smaller, cheaper, and slightly more efficient. The older construction approach was to use three separate transformers. That approach had the advantage that each unit in the bank could be replaced individually in the event of trouble, but that does not outweigh the advantages of a combined threephase unit for most applications. However, there are still a great many installations consisting of three single-phase units service. 3.9 Three-Phase Transformer Connections There are various methods available for transforming 3-phase voltage to higher or lower 3-phase voltage (i.e.) for handling a considerable amount of power. The most common connections are. (1) Υ - Υ (2) Δ - Δ (3) Δ - Υ (4) Υ -Δ (5) open-delta or V-V. Terminal Markings The primary winding bears capitals letters A,B and C for threephase. The secondary winding bears small letters a, b and c. 87 3.9.1 WYE-WYE Connection: The Υ - Υ connection of three-phase transformer voltage on each phase of the transformer is given by ( VφP = V LP 3 ). The primary-phase voltage is related to the secondary-phase voltage by the turns ratio of the transformer. The phase voltage on the secondary is them related to the line voltage on the secondary by ( VLS = 3 × VφS ). Therefore, overall the voltage ratio on the transformers A1 a1 A2 3 × VφP VLP =a = VLS 3 × VφS B1 B2 a2 b1 b2 C1 c1 C2 c2 The Υ - Υ connection is seldom used because of difficulties with exciting current phenomena. And if loads on the transformer circuit are unbalanced, then the voltages on the phases of the transformer can become severely unbalanced. And third-harmonic voltage can be large. Fig.(3.11) 88 3.9.2 WYE-DELTA Connection: The Υ -Δ connection of three-phase transformers is shown in fig.(3.12). In this connection, the primary line voltage is related to the primary phase voltage by ( VLP = 3 × VφP ), while the secondary line voltage is equal to the secondary phase voltage ( V LS = VφS ). The voltage ratio of each phase is ( VφP VφS = a ), so the overall relationship between the line voltage on the primary side of the bank and the line voltage on the secondary side of the bank is. V LP = V LS 3 × VφP a1 A1 Vφ S A2 B1 VLP = 3×a VLS C1 a2 b1 b1 c1 B2 C2 c2 The secondary voltage is shifted (300) relative to the primary voltage of the transformer. The Υ -Δ connection is commonly used in stepping down from a high voltage to a medium or low voltage. Fig.(3.12) 89 3.9.3 DELTA -WYE Connection: A Δ - Υ connection of three-phase Δ - Υ connection, the primary transformers is shown in fig.(3.13). In a line voltage is equal to the primary-phase voltage ( V LP = VφP ), which the secondary voltage are related by ( V LS = 3 × Vφ S ). Therefore, the line-toline voltage ratio of this transformer connection is VφP VL P = VLS 3 × Vφ .S VLP a = VLS 3 A1 a1 A2 B1 a2 b1 B2 C1 b2 c1 C2 c2 This connection has the same phase shift as Υ -Δ transformer. The Δ - Υ connection is commonly used for stepping up to a high voltage. The neutral of the secondary is grounded for providing 3-phase 4-wire service. Fig.(3.13) 90 3.9.4 DELTA - DELTA Connection: The Δ - Δ connection is shown in fig.(3.14). In Δ - Δ connection, ( V LP = VφP ) and ( VLS = Vφ S ), so the relationship between primary and secondary line voltage is VLP VφP = =a VLS Vφ .S A1 a1 A2 B1 a2 b1 B2 C1 b2 c1 C2 c2 This transformer has no phase shift associated with it and no problem with unbalance loads or harmonics. The Δ - Δ connection has the advantage that one transformer can be removed for repair or maintenance while the remaining two continue to function as a threephase bank with the rating reduced to 58 percent of that of the original bank, this is known as the open-delta, or V, connection. Fig.(3.14) 91 Example(3.4): A three-phase transformer bank is to handle (600 KVA) and have a (34.5/13.8 KV) voltage ratio. Find the rating of each individual transformer in bank (high voltage, low voltage, turns ratio, and (a) Υ - Υ apparent power) if the transformer bank is connected to (b) Δ - Δ (c) Δ - Υ (d) Υ -Δ? Solution: For Υ - Υ connection Primary voltage = Turns ratio = 34.5 3 = 19.9 KV ,Secondary voltage = 13.8 3 = 7.97 KV 19.9 = 2.5 7.97 For Δ - Δ connection Primary voltage =34.5 KV Turns ratio = , Secondary voltage =13.8 KV 34.5 = 2 .5 13.8 For Δ - Υ connection Primary voltage =34.5 KV Turns ratio = , Secondary voltage = 13.8 3 = 7.97 KV 34.5 = 4.33 7.97 For Υ -Δ connection Primary voltage = Turns ratio = 34.5 3 = 19.9 KV , Secondary voltage =13.8 KV 19.9 = 1.44 13.8 Apparent power = (600 ) = 200 KVA 1 3 ********************************************************* 92 Example(3.5): The following are the light-load test readings on a 3phase, 100 KVA, (400/6600)V, star/delta transformer. Open circuit: supply to low-voltage side 400 V, 1250 W. Short-circuit supply to high-voltage side 314 V,1600W, full-load current Calculate the efficiencies at full-load, 0.8 power factor and at half-full load, u.p.f. Calculate also the maximum efficiency, u.p.f. Solution: η= output nKVA cos(φ ) × 100 = × 100 input nKVA cos(φ ) + PCore + n 2 ( PCu ) The O.C. test gives the normal PCore loss since rated voltage is applied and the S.C. test gives the PCu loss at load since rated current is flowing. For full-load, 0.8 P.f. η= (1 × 100 × 10 3 × 0.8) × 100 = 96.56 0 0 (1 × 100 × 10 3 × 0.8) + 1250 + (1) 2 (1600) For half- full load, u.p.f η= (0.5 × 100 × 10 3 × 1) × 100 = 96.8 0 3 2 0 (0.5 × 100 × 10 × 1) + (1250) + (0.5) × (1600) Maximum efficiency when PCore = n 2 PCu n= 1250 = 0.88 1600 and power factor is unity. (0.884 × 100 × 10 3 × 1) × 100 = 97.25 0 Max. η = 0 (0.884 × 100 × 10 3 × 1) + (1250) + (0.884) 2 (1600) 93