SOLUTIONS Question 1 A 1000-VA 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 ππππππ = 230 V πππ π π π = 19.1 V πΌπΌππππ = 0.45 A πΌπΌπ π π π = 8.7 A ππππππ = 30 W πππ π π π = 42.3 W All data given were taken from the primary side of the transformer. (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer. (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging. Solution a) Open circuit test βπππΈπΈ β πππΈπΈ = π π ππ = = πΌπΌππππ β πΊπΊππ − πππ΅π΅ππ β = ππππππ ∠ − cos−1 ( 1 1 πΊπΊππ = ππππππ ππππππ πΌπΌππππ πΌπΌππππ ππππππ 0.45π΄π΄ 230ππ = 0.001957 ) = 0.001957∠ −73.15° = 0.000567- j0.001873 = 1763 πΊπΊ 0.000567 = and ππππ = 1 0.001873 = 534 πΊπΊ Short circuit test β ππππππ β = β π π ππππ + ππππππππ β= Ρ³ = cos−1 πππ π π π πππ π π π πΌπΌπ π π π = cos −1 19.1ππ 8.7π΄π΄ 42.3 = 2.2 πΊπΊ (19.1ππ )(8.7π΄π΄) = 75.3° ππππππ = π π ππππ + ππππππππ = 2.2∠75.3° = 0.558+j2.128 π π ππππ = 0.558 πΊπΊ ππππππ = 2.128 πΊπΊ 1 To convert the equivalent circuit to the secondary side, divide each impedance by the square of the turns ratio (a = 230/115 = 2). We get the following circuit π π ππππππ = 0.140 πΊπΊ ππππππππ = j0.532πΊπΊ ππππ,ππ = 134 πΊπΊ π π ππ,ππ = 441 πΊπΊ b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is πΌπΌππ = 1000ππππ 115ππ = 8.70 A We now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor. (1) 0.8 PF Lagging: ππππ ππ = ππππ + ( ππππππ,π π * πΌπΌππ )= 115∠0 V + ( 0.140+j0.532) ( 8.70∠-36 87° A) VR = ππππ ππ − ππππ ππππ x 100% = ππ 118.8−115 115 = 3.3 % (2) 1.0 PF: ππππ = 118.8∠1.4° V × 100% = ππππ + ( ππππππ,π π * πΌπΌππ ) = 115∠0 V + (0.140+j0.532) ( 8.70∠0° A) VR = 116.3−115 115 = 1.1% × 100% = 116.3∠2.28° V 2 (3) 0.8 PF Leading: ππππ ππ = ππππ + ( ππππππ,π π * πΌπΌππ ) = 115∠0 V + (0.140+j0.532)( 8.70∠36.87° A) = VR = 113.3−115 115 = - 1.5% × 100% 113.3∠2.24° V c) At rated conditions and 0.8 PF lagging, the output power of this transformer is ππππππππ = ππππ πΌπΌππ cos Ρ³ = 115x 8.70 x cos Ρ³ = 800 W The copper and core losses of this transformer are ππππππ = (πΌπΌπ π )2 π π ππππ,π π = 10.6 W ππππππππππ = Ζ= ( ππππ 2 ) ππ π π ππ,π π ππππππππ = ππππππππ +ππππππππππ +ππππππ Question 2 (118.8ππ)2 441 πΊπΊ 800 = 32.0 W = 800+32.0+10.6 x 100 % = 94.9% A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances: π π ππ = 32 πΊπΊ π π π π = 0.05 πΊπΊ ππππ = 45 πΊπΊ πππ π = 0.06 πΊπΊ π π ππ = 250 k πΊπΊ ππππ = 30 k πΊπΊ The excitation branch impedances are given referred to the high-voltage side of the transformer. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Assume that this transformer is supplying rated load at 480 V and 0.8 PF lagging. What is this transformer’s input voltage? What is its voltage regulation? (c) What is the transformer’s efficiency under the conditions of part (b) Solution a) The turns ratio of this transformer is a = 8000/480 = 16.67. Therefore, the secondary impedances referred to the primary side are π π ππ ′ = ππ2 π π ππ = (16.67)2 (0.05 πΊπΊ) = 13.9 πΊπΊ 3 ππππ ′ = ππ2 ππππ =(16.67)2 (0.06 πΊπΊ) = 16.7 πΊπΊ The resulting equivalent circuit is shown below b) To simplify the calculations, we use the simplified equivalent circuit referred to the primary side of the transformer and the excitation branch is moved to the left : The secondary current in this transformer is given by : πΌπΌπ π = 20ππππππ 480ππ ∠ −36 87° A = 41.67∠ −36.87 A The secondary current referred to the primary side is πΌπΌππ ′ = πΌπΌππ ππ = 2.50∠ −36.87 A Therefore, the primary voltage on the transformer is ππππ = ππππ ′ + (π π ππππ,ππ + j ππππππ,ππ ) πΌπΌππ ′ ππππ = 8000∠0° V + (45.9+j61.7)( 2.50∠ −36 87° A)= 8185∠ 0.38 ° V The voltage regulation of the transformer under these conditions is VR = c) 8185−8000 8000 = 2.31% ππππππππ = Scos Ρ³ = (20kVA)(0.8) = 16 kW ππππππ = (πΌπΌππ ′ )2 π π ππππ,ππ = (2.5)2 x 45.9 =287W ππππππππππ = Ζ= (ππππ )2 π π ππ ππππππππ = ππππππππ +ππππππππππ +ππππππ (8185ππ)2 250000 πΊπΊ 16000 = 268 W = 16000+287+268 x 100 % = 96.6% 4 Question 3 A 10-kVA, single-phase transformer has its primary connected to a 2000-V supply. It has 60 turns on the secondary winding and voltage across it is found to be 240 V. Assuming the transformer to be ideal, calculate (a) the number of turns on its primary winding; (b) the fullload primary and secondary currents. Solution a) ππππ πππ π ππππ = ππππ = ππππ πππ π x πππ π = b) πΌπΌππ = πΌπΌπ π = ππ πππ π πππ π 2000 240 x 60 = 500 turns ππ ππππ = = 10000ππππ 2000ππ = 5A 10000ππππ 240ππ = 41.67 A 5
