BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 6: Single Phase Transformer (Continued)

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Presentation transcript:

BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 6: Single Phase Transformer (Continued)

Open circuit test –Provides magnetizing reactance and core loss resistance –Obtain components are connected in parallel Parameter determination of the transformer

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 I OC 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.

Admittance Open circuit Power Factor Open circuit Power Factor Angle Angle of current always lags angle of voltage by 

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 Power Factor of the current Angle Power Factor Therefore

Exercise The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60- Hz 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: Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit Open- circuit test (on primary) Short- circuit test (on primary) Voc = 8000 VVsc = 489 V Ioc = AIsc = 2.5 A Poc = 400 WPsc = 240 W

Per unit System The per unit value of any quantity is defined as Quantity – may be power, voltage, current or impedance

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. Two major advantages in using a per unit system:

Per Unit (pu) in Single Phase System

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. At no load,

Phasor Diagram Lagging power factor Unity power factor

Leading power factor

Efficiency The efficiency of a transformer is defined as the ratio of the power output (P out ) to the power input (P in ). P core = P eddy current + P hysteresis And P cu =P copper losses Copper losses are resistive losses in the primary and secondary winding of the transformer core. They are modeled by placing a resistor R p in the primary circuit of the transformer and resistor R s 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 R c 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 testShort- circuit test Voc = 2400 VVsc = 48 V Ioc = 0.25 AIsc = 6.0 A Poc = 50 WPsc = 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?