PER UNIT CALCULATION

PER UNIT CALCULATION

a per-unit system is the expression of system quantities as fractions of a defined base unit quantity. Calculations are simplified because quantities expressed as per-unit do not change when they are referred from one side of a transformer to the other. This can be a pronounced advantage in power system analysis where large numbers of transformers may be encountered. Moreover, similar types of apparatus will have the impedance lying within a narrow numerical range when expressed as a per-unit fraction of the equipment rating, even if the unit size varies widely. Conversion of per-unit quantities to volts, ohms, or amperes requires a knowledge of the base that the per-unit quantities were referenced to. The per-unit system is used in power flow, short circuit evaluation, motor starting studies etc.

The main idea of a per unit system is to absorb large differences in absolute values into base relationships. Thus, representations of elements in the system with per unit values become more uniform.

There are several reasons for using a per-unit system:

• Similar apparatus (generators, transformers, lines) will have similar per-unit impedance and losses expressed on their own rating, regardless of their absolute size. Because of this, per-unit data can be checked rapidly for gross errors. A per unit value out of normal range is worth looking into for potential errors.
• Manufacturers usually specify the impedance of apparatus in per unit values.
• Use of the constant ${\displaystyle \textstyle {\sqrt {3}}}$ is reduced in three-phase calculations.
• Per-unit quantities are the same on either side of a transformer, independent of voltage level
• By normalizing quantities to a common base, both hand and automatic calculations are simplified.
• It improves numerical stability of automatic calculation methods.
• Per unit data representation yields important information about relative magnitudes.

The PU system deals primarily with values of power, voltage, current, and impedance. Any two variables are selected to be base values. The selection of two base values then fixes the other values. A per unit quantity then becomes the ratio of a selected parameter to a selected base value. Basically,

Per Unit = Present Value/ Base Value

Relationship between units

Per Unit System Example of Three Phase System

Example 1

Consider a three-phase transformer with a rating of 700 MVA and a secondary voltage of 145 kV. Determine I_base, Z_base, and Y_base.There is no obligation to choose the transformer’s ratings as the base values, but it is an option. Using the transformer’s rated values of apparent power and secondary voltage,

S_base = 700 MVA

V_base = 145 kV

The corresponding values of I_base, Z_base, and Y_base are readily determined.

$$I_{base}=\frac{S_{bas\mathrm{e\; /}}}{{\mathrm{(V}}_{base}\times \sqrt{\mathrm{3)}}}=\frac{700MVA}{\mathrm{/\; (145}kV\times \sqrt{\mathrm{3)}}}=2.78kA$$

$$Z_{base}=\frac{V_{base}}{{\mathrm{/(I}}_{base}\times \sqrt{3}})\; =\frac{145k\mathrm{V/(}}{2.78kA\times \sqrt{\mathrm{3)}}}=30.1\Omega$$

$$Y_{base}=\frac{1}{30.1}=0.0332S$$

$$\mathrm{If,\; say,\; a\; secondary\; voltage\; of\; 130\; kV\; is\; under\; consideration,\; then\; the\; per\; unit\; value\; of\; that\; voltage\; is}$$

Example 2

Figure shows a 33kV load fed from a supply through a double circuit line.

(a) Using a common base of 132kV, 100MVA at the transmission line, convert all impedance to per unit.

Solution

(a) Using the common base as 100 MVA with voltage 132 kV on transmission line,

Generator

New base voltage at generator = 13.5 kV

xg1 = 0.18 × sq(13.8/13.5)  x (100/160) = 0.115 pu

xg2 = 0.15 × sq(13.8/13.5)  x (100/160) = 0.096 pu

xg0 = 0.05 × sq(13.8/13.5)  x (100/160) = 0.032 pu

Voltage may be assumed as 1 pu in fault studies since no information given (or 13.8/13.5 = 1.022 pu)

Transformer T1,

xT11 = 0.10× (100/150) = 0.067 pu

Each line

xL1 = x×(MVAbase/sq(kVbase ) ) = 30×(100/sq132 ) = 0.172 pu

xL0 = x×(MVAbase/sq(kVbase )) = 100×(100/sq1322 ) = 0.574 pu

Transformer T2,

xT21 = 0.08× (100/125) = 0.064 pu

Balanced Load New base voltage at generator = 33 kV

Zpu = 1.0 pu

p.f. = 0.8 lag → Z = 0.8 + j 0.6 pu

PER-UNIT AND BASE IMPEDANCE CALCULATOR  CLICK HERE

𝅴‫