Chapter 3 Generator and Transformer Models The Per Unit System Part III 3 13 The Per Unit System Let us say that the measure unit of age is 80 years this is the base unit Then a person who is 40 years old is half of the base unit So we say the person s age is 0 5 per unit age in years Age per unit age base in years Thus the age per unit is dimensionless In power systems there are four base quantities required to define a per unit system These are power voltage current and impedance Given two of these all others can be defined using these two In general for any quantity actual quantity Quantity per unit base value of quantity And for power systems we have S V I Z S pu V pu I pu Z pu SB VB IB ZB N B The value of the numerators above may be a complex number however the denominator the base value is a positive real number Normally we select a three phase power base S B or MVAB and a line to line voltage base VB or kVB From these two the other bases can be computed using circuit laws thus S 3 S IB B B 3 VB 3VB and V ZB B 3I B Other forms for Z B follow from the above equations ZB ZB VB 2 SB 2 kVB MVAB Note that VB is line to line voltage base and I B is line current base Also S B is 3 phase power base Thus in dealing with per phase quantities the 3 and into account 3 need to be taken Per unit quantities obey the circuit laws thus S pu V pu I pu and V pu Z pu I pu In terms of phase quantities the three phase complex power to a load is given by S L 3 3VP I P and the phase current is related to the phase voltage by the load impedance thus V IP P ZP Using this value of phase current in the equation above we have solving for impedance 2 2 3 VP VL L ZP SL 3 S L 3 Therefore the phase impedance in per unit is given by Z pu and since Z B Zp ZB VL L 2 S L 3 Z B VB2 see last page the above expression becomes SB Z pu Zp VL L 2 2 V SB L L Z B S L 3 Z B VB S L 3 or Z pu V pu 2 S L pu 3 14 Change of Base Usually if none are specified the pu values given are on nameplate ratings as base For example a generator whose impedance is 0 4 pu and whose ratings are 110 MVA and 24 kV then the base MVA is assumed 110 MVA and base line to line voltage is 24 kV Often the base for the system is different from the base for each particualr generator or transformer hence it is important to be able to express the pu value in terms of different bases This is derived below old and voltage base VBold We Let Z old pu be the per unit impedance on the power base S B new would like to find Z new and voltage base VBnew Let Z be pu on the new power base S B the ohmic value of the impedance Then we have 2 Z old pu Z SBold Z 2 Z Bold VBold and similarly Z new pu Z S Bnew new Z 2 ZB VBnew By comparison of these two expressions it is evident that 2 S Bnew VBold Z Z new S old B VB This last expression is very useful in transforming per unit values from one set of bases to another new pu old pu Example A transformer has a voltage rating of 110 220 V and a power rating of 1000 VA The impedance of the transformer on the low voltage side is 5 j33 Find the perunit impedance viewed from each side assuming the bases are equal to the voltages and power ratings on each side Solution using Matlab From the give quantities VB1 110 VB2 220 SB1 1000 SB2 1000 Z1 5 j 33 Z2 Z1 4 ZB1 VB1 2 SB1 ZB2 VB2 2 SB2 Zpu1 Z1 ZB1 Zpu2 Z2 ZB2 Zpu1 0 4132 2 7273i Zpu2 0 4132 2 7273i Thus with proper selection of S base and Vbase on each side of a transformer the per unit impedance of the transformer becomes the same viewed from either side Thus the transformer may be removed and replaced with its per unit impedance but we have to keep in mind that now there are two different values of voltage base one on each side of the transformer Rules for selection of voltage bases around a transformer Let the turns ratio of the N transformer be 1 and suppose there is a connection gain of k If this is a Y Y N2 or connected transformer then k 1 If it is a Y connection then k 3 If it is Y then k 1 3 Select a voltage base on side 1 say VB1 Then the voltage base on side 2 must be VB 2 k VB1 The power base of course is the same everywhere in the power system 3 Example 3 7 The one line diagram of a power system is shown in fig 3 29 page 92 Select a common base of 100 MVA and 22 kV on the generator side Draw an impedance diagram with all impedances including the load impedance marked in per unit The manufacturer s data for each device is given on page 92 The three phase load on bus 4 absorbs 57 MVA 0 6 power factor lagging at 10 45 kV Line 1 and line 2 have reactances of 48 4 and 65 43 respectively Outline of solution first find the voltage base in all sections of the system In this case there are five sections Note that in this case there is no mention of how the transformers are connected i e Y or delta Hence we consider only the turns ratio otherwise there would be some 3 factors as well The Matlab program follows VLL 10 45 VB1 22 VB2 22 220 22 VB4 220 11 220 VB5 22 110 22 Xg 0 18 100 90 Xt1 0 1 100 50 Xt2 0 06 100 40 Xt3 0 064 100 40 Xt4 0 08 100 40 Xm 0 185 100 66 5 10 45 11 2 ZB2 220 2 100 ZB5 110 2 100 Xline1 48 4 484 Xline2 65 43 121 SL 57 0 6 j sin acos 0 6 ZL VLL 2 conj SL ZB4 11 2 100 ZLpu ZL ZB4 Xg 0 2000 Xt1 0 2000 Xt2 0 1500 Xt3 0 1600 Xt4 0 2000 Xm 0 2511 ZB2 484 ZB5 121 Xline1 0 1000 Xline2 4 0 5407 SL 34 2000 45 6000i ZL 1 1495 1 5327i ZB4 1 2100 ZLpu 0 9500 1 2667i Note these …