2/12/2013. Topics. 14-Per Unit Text: Per Unit. Per Unit. Base Selection. Per Unit

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1 //03 Topics 4 Text: Definitions Analysis & Transformers ThreePhase ECEGR 45 Power ystems Per unit () simplifies system calculations eliminates partitions introduced by transformers erroneous values are easily spotted Normalizes values in the system per unit quantity = actual quantity base value of quantity oltage ase Current ase Power ase mpedance ase Note: base values are scalars 3 4 ase election if is selected to be 5,000 volts and = 37,500 at zero degrees, then 37, ,000 Note: units cancel, so is technicay unitless, but it is common to add a per unit at the end Proper selection of base quantities is essential to simplify the problem base must be a real scalar select base quantities to obey Ohm s Law therefore: 5 6

2 //03 We foow the same procedure to define the power base P Q P jq P Q note error in the book For impedance: R X Y R R X jx 7 8 Let = 00k and = 50A, compute and. Let = 00k and = 50A, compute and MA 9 0 ase election There are two equations that govern the relationship between base values There are four base values that can be assigned,,, Convention is to select and and compute and is typicay selected such that the voltage levels are between 0.95 to.05 Let = 77 = 69.5 A Compute and = 77 R = 554

3 //03 Let = 77 = 69.5 A Compute and A = 77 R = 554 = R = 0.5 We can easily solve the circuit j0 We may wish to convert back to current and power: = /0.5 = 0.5A = /69.5 = 38.5A = R = & Transformers One of the most important advantages of using per unit is realized when transformers are involved n side bases side bases a n & Transformers We then get n,a'n',an n,a'n',,an, And can remove the ideal transformer a a 5 6 Transformer nameplate: k Y/3.8 k Y rated = 60 MA (3 phase) jx L = j Ohms Load: R = 5 Ohms Generator: rated = k rated = 60 MA Find the power out of the generator Transformer nameplate: k Y/3.8 k Y rated = 60 MA (3 phase) jx L = j Ohms Load: R = 5 Ohms Generator: rated = k rated = 60 MA Assign bases generator side = k/(.73) =.7 k =0 MA load side =3.8k/(.73) = 7.9 k = 0 MA 7 8 3

4 //03 Assigned bases generator side = k/(.73) =.7 k =0 MA load side =3.8k/(.73) = 7.9 k = 0 MA Computed bases = ( ) / = 8.07 = ( ) / =3.74 = / =574.6 A = 50. A Transformer impedance () X l = /8.07= 0.0 Load (): R = (5/3.74) =.576 Generator (): =.7k/.7k = 9 0 Transformer X l = 0.0 Load: R =.575 Generator: = olving j MA MA.00 j0.0 R = j0.0 R =.575 To find current: A A Notes on the Transformer impedances are usuay given in (expressed as %) based on threephase power and lineline voltage i.e. X l = % = 0.0 A perphase diagram using per unit values is caed an impedance diagram We considered a YY case. For normal systems, phase shifts are ignored. ee text page 60 for other xfmr connection notes 3 4 4

5 //03 ThreePhase o far we have dealt with per unit as defined by single phase expressions nterested in extension to threephase lineline voltages are used for threephase power used for ThreePhase As with three phase circuits, convention is to always give power base as threephase, and voltage base as lineline Extra notation is given in this lecture and is omitted later ( ) ( ) Change of ase mpedance of machines may not be given in the same base as used in the system ase conversion of impedance can save time new new actual new new new new new (single vs threephase) Given: line = j00, load = 75 (per phase) and =3.8 k (lineline) First without using per unit 7.967k j j0.406 MA 3,an j.9 MA,an = k line =j00 perphase,an load = (single vs threephase) Now using per unit with = 3.8/sqrt(3), = MA Load Line j00 j line =j.575,an =.0 perphase,an load =.8 (single vs threephase) Now find the singlephase and threephase power from the generator: pu j j0.406 pu,pu pu j0.406 MA pu j.9 3,pu pu 3 pu 0.9 j.9 MA

6 //03 (single vs threephase) (single vs threephase) We found the singlephase power to be: pu j j0.406 MA and the threephase power to be: 0.9 j.9 3,pu j.9 MA We define the threephase base power as: 3 3 MA 3 3,pu 0.9 j.9 MA j MA f the single phase power is j0.406, the three phase power is also j0.406 A voltage of.0 in the single phase circuit implies that the lineline voltage is.0 When we do a perphase analysis, the results can be applied the threephase quantities; just remember to use the correct base when converting back to the actual values f we dealing with per unit values only, you need not distinguish between per phase and three phase power, nor lineline voltage and lineneutral voltage 3 3 (single vs threephase) For example: you are doing a problem with = k (lineline), and = 00 MA (threephase). You do a perphase analysis and find:,an,pu = 0.90 pu =.5 (per phase power) What is the lineline voltage at? 0.9 k 9.8k What about the three phase power? MA 50 MA (transformer connections) f we use the bases corresponding to the nameplate ratings, the per unit reactances have the same numerical values in the YY, DY, DD and YD connections as in the single phase case ee example 5. for details Analysis Procedure (normal systems). Pick a A base for the entire system (usuay largest generator rating). Arbitrarily select one voltage base. Relate a others by the ratio of the magnitudes of the opencircuit line voltages of each transformer bank 3. Find the impedance bases in the different sections and express a impedances in consistent per unit terms. 4. Draw the impedance diagram for the entire system, and solve for the desired per unit quantities. 5. Convert back to actual quantities, if needed. 35 6