Per Unit System Lecture Notes.pptx

PER UNIT SYSTEM

Per unit system  A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer impedances to the different sides of the transformers

This problem is avoided by a normalization of all variables. This normalization is known as per unit analysis In power system analysis, it is common practice to use per-unit quantities for analyzing and communicating voltages, current, power, and impedance values.

Per unit system  These per-unit quantities are normalized or scaled on a selected base, allowing engineers to simplify power system calculations with multiple voltage transformations Therefore; actual quantity quantity in per unit = base value of quantity

Per Unit Conversion Procedure Pick a 1f VA base for the entire system, SB Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral. Calculate the impedance base, ZB= (VB)2/SB Calculate the current base, IB = VB/ZB Convert actual values to per unit

Per unit system Note:  Per unit conversion affects magnitudes, Angles are not affected If the nominal value is chosen as the base voltage, a “normal” voltage value will be close to 1.0 p.u  Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u. volts) except under certain conditions

Advantages of per unit system ordinary parameters vary considerably with variation of physical size,

terminal voltage and power rating etc. while per unit parameters are independent of these quantities over a wide range of the same type of apparatus. In other words, the per unit impedance values for the apparatus of like ratings lie within a narrow range

It provide more meaningful information. The chance of confusion between line and phase values in a three-phase

balanced system is reduced. A per unit phase quantity has the same numerical value as the corresponding per unit line quantity regardless of the three-phase connection whether star or delta.

Advantages of per unit system cont’d Impedances of machines are specified by the manufacturer in terms of

per unit values. The per unit impedance referred to either side of a single-phase transformer is the same.  The per unit impedance referred to either side of a three -phase transformer is the same regardless of the connection whether they are ∆-∆, Y-Y or ∆-Y. The computation effort in power system is very much reduced with the use of per unit quantities.  Usually, the per unit quantities being of the order of unity or less can easily be handled with a digital computer. Manual calculation are also simplified. Per unit quantities simply theoretical deduction and give them more generalizes forms.

example •Example: nominal voltage at bus bar A is 132 kV and actual voltage at A is 127 kV. The per unit voltage at A is:

Rules for Choosing Base Values  Choose ONE base power for the entire system Arbitrary choice Commensurate with the system e.g. 400V use 100kVA or 1MVA, 11kV use 10MVA,  132KV use 100MVA, 400kV use 1000MVA  Choice also depends on rating of generators/transformers  e.g. commercial/industrial building = mainly 400V use 100kVA  distribution utility = mainly 11kV use 10MVA

Rules for Choosing Base Values cont’d Choose ONE base voltage for voltage level “of most interest” Choose the nominal voltage, Each voltage level must have one base voltage, between two base voltages the connecting transformer turns ratio must be maintained

Other base values chosen to get the same relations between per unit quantities as between actual quantities – electrical laws (ohms law,….) will not be broken under per unit system With the base power and base Voltage determined, the base current and base impedance can be calculated

Rules for Choosing Base Values cont’d • That is;

• For a three phase system,

Per unit example • Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV, respectively.

Per unit example cont’d Z BLeft

82 (kV) 2 = = 0.64W 100MVA

Z BMiddle Z BRight

802 (kV) 2 = = 64W 100MVA

162 (kV) 2 = = 2.56W 100MVA

Same circuit, with values expressed in per unit

Per unit example cont’d 1.0а 0 I = = 0.22�- 30.8�p.u. (not amps) 3.91 + j 2.327 VL =а1.0 �-0��0.22 2.327�30.8 90� = 0.859�- 30.8�p.u. 2

VL SL = = = 0.189 p.u. Z SG =а�а 1.0 =0.22 0 0.22 а 30.8 VL I L*

30.8 p.u.

Per unit example cont’d To convert back to actual values just multiply the per unit values by their per unit base V LActual = 0.859�- 30.8� �16 kV = 13.7�- 30.8�kV S LActual =а� 0.189 =а 0 SGActual =а� 0.22=а30.8

100 MVA 18.9 0 MVA 100 MVA

22.0 30.8 MVA

I BMiddle

100 MVA = = 1250 Amps 80 kV

Actual I Middle

= 0.22�- 30.8� �1250 = 275�- 30.8�A

Calculation in 3 phase “Homework” Solve for the current, load voltage and load power in the previous circuit, assuming:

–a 3f power base of 300 MVA,  –and line to line voltage bases of 13.8 kV, 138 kV  and 27.6 kV  –the generator is Y­connected so its line to line voltage is 13.8 kV.    

Change of MVA Base in per unit Parameters

for equipment are often given using power rating of equipment as the MVA base To analyze a system all per unit data must be on a common power base OriginalBase Z pu

Hence

NewBase � Z actual � Z pu

OriginalBase Z pu

2 Vbase

2 Vbase

� OriginalBase / NewBase = S Base S Base

NewBase S Base

NewBase ZOriginalBase � = Z pu pu OriginalBase S Base

NewBase Z pu

example 

Example: A 350 MVA, 230/20 kV transformer has leakage reactance of 10%. What is p.u. value on 100 MVA base? What is value in ohms (230 kV)? 100 X e = 0.10 � = 0.0286 p.u. 350 230 2 0.0286 � = 15.1 W 100

• In general for change of MVA base,

• In cases where the new base Voltage equates to the old base voltage, then;

Calculation example questions • A 3phase two winding transformer is rated at 75MVA, 11.8/145kV, 50Hz and the equivalent leakage impedance referred to the 11.8kV • Using the transformer ratings as base values, determine the per-unit leakage impedances referred to the 11.8kV winding and the 145kV winding. • winding is Zeq = 0.22Ω

Example question 2 For the power network shown below calculate using per-unit quantities the voltage in kV, the current in kA and the power in kW received by the load • • • • •

G1 = synchronous generator = 75MVA, 11.8kV, sync reactance Xd = j1.83 p.u. T1 = generator transformer = 75MVA, 11.8kV/145kV X = j0.125 p.u. L1 = 132kV transmission line: Z = 0.18 + j0.40 Ω/km, length = 20km. T2 = step-down transformer = 45MVA, 132kV/11kV, X = j0.125 p.u. ZL = 11kV load = 10MVA, cosɸ = 0.8 (lagging).