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Sub- DC Machine & Transformer Enroll No. 130610109052 130610109053 130610109054 130610109055 130610109056 140613109011 Experiment Setup: In the open circuit test, transformer rated voltage is applied to the primary voltage side of the transformer with the secondary side left open. Measurements of power, current, and voltage are made on the primary side. Since the secondary side is open, the input current IOC is equal to the excitation current through the shunt excitation branch. Because this current is very small, about 5% of rated value, the voltage drop across the secondary winding and the winding copper losses are neglected. Parameter determination of the transformer • Open circuit test – Provides magnetizing reactance and core resistance – Obtain components are connected in parallel loss Admittance I oc Yoc Voc Open circuit Power Factor PF cos Poc Voc I oc Open circuit Power Factor Angle cos 1 Poc Voc I oc Angle of current always lags angle of voltage by Yoc I oc 1 1 GC jBM j Voc RC XM Short circuit test – Provides combined leakage reactance and winding resistance – Obtain components are connected in series Experiment Setup: In the short circuit test, the secondary side is short circuited and the primary side is connected to a variable, low voltage source. Measurements of power, current, and voltage are made on the primary side. The applied voltage is adjusted until rated short circuit currents flows in the windings. This voltage is generally much smaller than the rated voltage. Impedances referred to the primary side Z sc Vsc I sc Power Factor of the current Psc PF cos Vsc I sc Angle Power Factor cos 1 Psc Vsc I sc Therefore Z sc Vsc 0 0 I sc 0 Vsc 0 I sc Z sc Req jX eq R p a 2 Rs j X p a 2 X s Exercise The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the primary side of the transformer and the following data were taken: Open- circuit test (on primary) Voc = 8000 V Short- circuit test (on primary) Vsc = 489 V Ioc = 0.214 A Isc = 2.5 A Poc = 400 W Psc = 240 W Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit Per unit System The per unit value of any quantity is defined as Actual Quantity Per Unit, pu Base value of quantity Quantity – may be power, voltage, current or impedance Two major advantages in using a per unit system: 1. It eliminates the need for conversion of the voltages, currents, and impedances across every transformer in the circuit; thus, there is less chance of computational errors. 2. The need to transform from three phase to single phase equivalents circuits, and vise versa, is avoided with the per unit quantities; hence, there is less confusion in handling and manipulating the various parameters in three phase system. Per Unit (pu) in Single Phase System Pbase ,Qbase , Sbase VbaseI base Vbase Z base I base Ybase I base Vbase ( Vbase )2 Z base Sbase Voltage Regulation (VR) The voltage regulation of a transformer is defined as the change in the magnitude of the secondary voltage as the current changes from full load to no load with the primary held fixed. VR VS ,nl VS , fl VS , fl Req X 100% At no load, VS VP a Vp VR a VS , fl VS , fl Xeq + + Vp/a Vs - X 100% Is - Phasor Diagram Vp/a jIsXeq Vs IsReq Is Lagging power factor Vp/a jIsXeq Is Vs IsReq Unity power factor Vp/a jIsXeq Is IsReq Vs Leading power factor Efficiency The efficiency of a transformer is defined as the ratio of the power output (Pout) to the power input (Pin). Pout X 100% Pin Pout X 100% Pout P losses Vs I s cos X 100% Vs I s cos Pcu Pcore Pcore = Peddy current + Physteresis And Pcu=Pcopper losses Copper losses are resistive losses in the primary and secondary winding of the transformer core. They are modeled by placing a resistor Rp in the primary circuit of the transformer and resistor Rs in the secondary circuit. Core loss is resistive loss in the primary winding of the transformer core. It can be modeled by placing a resistor Rc in the primary circuit of the transformer. Assignment 5 A 15-kVA, 2400/240-V transformer is to be tested to determine its excitation branch components, its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer: Open- circuit test Short- circuit test Voc = 2400 V Vsc = 48 V Ioc = 0.25 A Isc = 6.0 A Poc = 50 W Psc = 200 W Assignment 5 The data have been taken by using the connections of open-circuit test and short-circuit test: a) Find the equivalent circuit of this transformer referred to the high-voltage side. b) Find the equivalent circuit of this transformer referred to the low-voltage side c) Calculate the full-load voltage regulation at 0.8 lagging power factor and 0.8 leading power factor d) What is the efficiency of the transformer at full load with a power factor of 0.8 lagging?