L#5 Transformer Per Unit

L#5 Transformer Per Unit - Transformers Per Unit Masoud...

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Transformers Per Unit Masoud Fathizadeh, PhD, PE Department of Engineering Technology Purdue Calumet Hammond, Indiana 46323
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The per-unit system actual value Quantity per unit base value of quantity   2 ,, base base base base base base base base base base P Q or S V I V V Z IS  Another approach to solve circuits containing transformers is the per-unit system . Impedance and voltage-level conversions are avoided. Also, machine and transformer impedances fall within fairly narrow ranges for each type and construction of device while the per-unit system is employed. The voltages, currents, powers, impedances, and other electrical quantities are measured as fractions of some base level instead of conventional units. (1) Usually, two base quantities are selected to define a given per-unit system. Often, such quantities are voltage and power (or apparent power). In a 1-phase system: (2) (3)
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The per-unit system base base base I Y V (3a) Ones the base values of P (or S ) and V are selected, all other base values can be computed form the above equations. In a power system, a base apparent power and voltage are selected at the specific point in the system . Note that a transformer has no effect on the apparent power of the system, since the apparent power into a transformer equals the apparent power out of a transformer. As a result, the base apparent power remains constant everywhere in the power system. On the other hand, voltage (and, therefore, a base voltage) changes when it goes through a transformer according to its turn ratio. Therefore, the process of referring quantities to a common voltage level is done automatically in the per-unit system.
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The per-unit system: Example Example 1 : A simple power system is given by the circuit: The generator is rated at 480 V and 10 kVA. a) Find the base voltage, current, impedance, and apparent power at every points in the power system; b) Convert the system to its per-unit equivalent circuit; c) Find the power supplied to the load in this system; e) Find the power lost in the transmission line (Region 2).
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The per-unit system: Example 1 1 1 1 1 1 10 000 20.83 480 480 23.04 20.83 base base base base base base S IA V V Z I 1 2 1 480 4800 0.1 base base V VV a a. In the generator region: V base 1 = 480 V and S base = 10 kVA The turns ratio of the transformer T 1 is a 1 = 0.1 ; therefore, the voltage in the transmission line region is The other base quantities are
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The per-unit system: Example 2 2 2 10 10 000 2.083 4800 4800 2304 2.083 base base base S kVA IA Z 2 4800 240 20 base base V VV a The turns ratio of the transformer T 2 is a 2 = 20 ; therefore, the voltage in the load region is The other base quantities are
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The per-unit system: Example 10 10 000 41.67 240 240 5.76 41.67 base base base S kVA IA Z , 480 0 1.0 0 480 G pu V pu    , 20 60 0.0087 0.026 2304 line pu j Z j pu b. To convert a power system to a per-unit system, each component must be divided by its base value in its region. The generator’s per-unit voltage is The transmission line’s per-unit impedance is
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L#5 Transformer Per Unit - Transformers Per Unit Masoud...

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