SMJE 2103 Electrical Power System 3- Ph Power Apparatus

Power Apparatus Transmission Lines Transformers Circuit Breaker Switchgear Rotating Machine (Generator – Motor)

Introduction to power TX Connection Types and symbol Functionality Application Structure Material Equivalent Circuit Testing 3-Phase Tx Per-Unit System Transformer

1.Step up 2.Step down 3.Distribution Tx 4.Special purpose (PT, CT)

Transformer

Transformer -Connection Types and Symbol

Transformer -Operation (Ideal x – Turn Ratio)-

Transformer -Operation (Ideal Tx – Power)- P in = V p I p cos θ p P out = V s I s cos θ s θ p - θ s = θ P out = V s I s cos θ Also, V s = V p /a and I s = a I p P out = V p I p cos θ = P in Also same for power Q and power S

Transformer -Operation (Ideal x – Impedance)-

Z L ’ = a 2 Z L

Transformer -Operation (Ideal Tx – Analysis)- (a)If the power system is exactly as described above in Figure (a), what will the voltage at the load be? What will the transmission line losses be? (b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step-down transformer is placed at the load end of the line (Figure (b)). What will the load voltage be now? What will the transmission line losses be now?

Transformer -Operation (Real Tx - flux)- -Flux generated -The flux from each coil -Average flux level

Transformer -Operation (Real Tx – Voltage Across)- Primary Side: Hence, Or rewritten, Pri or sec voltage due to mutual flux Relationship, Vp and Vs

Transformer -Operation (Real Tx – Magnetization)- Magnetization current, i M – current required to produce flux in the core. Core-loss current, i h+e – current required to compensate hysteresis and eddy current losses.

Example

Transformer -Application-

Transformer -Structure-

Transformer -Structure (Shell Form)-

Transformer -Structure (Core Form)-

Transformer -Structure-

Transformer -Material- Copper conductors Silicon iron/laminated piece of steel – eddy current Oil Insulation materials

Transformer -Equivalent Circuit-

Transformer -Equivalent Circuit (simplified)-

Example/Tutorial The secondary winding of a transformer has a terminal voltage of v s (t) = sin 377t V. The turns ratio of the transformer is 100:200 (a = 0.5). If the secondary current of the transformer is i s (t) =7.07 sin (377t o ) A, what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are

Transformer -Testing- To know the value of the inductance and resistance

Transformer -Short circuit test- Power factor, PF = cos θ = P SC / V SC I SC (lagging) The series impedance Z SE = R eq + jX eq = (R P + a 2 R S ) + j(X P + a 2 X S )

Transformer -Open circuit test- Conductance of core loss Susceptance of magnetizing Total admittance Admittance referred to primary Angle of admittance

Transformer -Equivalent Circuit (Example)- Open circuit test (primary)Short circuit test V OC = 8000 VV SC = 489 V I OC = AI SC = 2.5 A P OC = 400 WP SC = 240 W The equivalent circuit impedances of a 20kVA, 8000/240V, 50Hz 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 impedance of the approximate equivalent circuit referred to the primary side, and sketch the circuit.

Example/Tutorial A 1000 VA 230/115 V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below All data given taken from the primary side of the transformer. (a)Find the equivalent circuit of this transformer referred to 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.

Transformer -Impulse voltage test-

Transformer - Voltage Regulation-

Transformer - Efficiency- Types of losses incurred in a transformer: Copper I 2 R losses Hysteresis losses Eddy current losses

3-Phase Transformer Transformers for 3-phase circuits can be constructed in two ways: connect 3 single phase transformers Three sets of windings wrapped around a common core The primaries and secondaries of any three-phase transformer can be independently connected in either a wye (Y) or a delta (∆) The important point to note when analyzing any 3-phase transformer is to look at a single transformer. The impedance, voltage regulation, efficiency, and similar calculations for three phase transformers are done on a per- phase basis.

Circuit Configuration Three-phase two winding Y-Y transformer bank

Circuit Configuration Three-phase two winding Y- ∆ transformer bank

Example

Transformer (Per-Unit System) To express the power system quantities in percent of specified base values. Reason – the quantities do not change when the are referred from one side to the others. Obvious solving when involving hundreds of transformers going to be encountted.

Transformer (Per-Unit System)

Example

Three zones of a single-phase circuit are identified in the such figure. The zones are connected by transformer T1 and T2, whose ratings are also shown. Using base value of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and determine the per-unit impedance and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected.

Solution

Example A balanced –Y- connected voltage source with E ab = 480  0 o volts is applied to a balanced -∆ load with Z ∆ = 30  40 o Ω. The line impedance between the source and load is Z L = 1  85 o Ω for each phase. Calculate the per-unit and actual current in phase a of the line using S base  = 10 kVA and V baseLL = 480 V.

Example

For the circuit shown, determine V out (t)

Example