.1. Power System Representation Single Line Diagram: Almost all modern power systems are three phase systems with the phases of equal magnitude and equal phase difference (i.e., 10 o ). These three phase balanced systems are always solved as a single phase circuit of one of the three lines and neutral return and this is sufficient to give a complete analysis. Such a simplified diagram of an electric system is called a one-line diagram or single-line diagram. Combined with a standard set of symbols for electric components, such one-line diagrams provide a compact way to represent information. The purpose of the one-line diagram is to supply in concise form, the significant information about the system, the importance of different features of a system varies with the problem under consideration, and the amount of information included on the diagram depends on the purpose for which the diagram is intended. Figure 1 shows the symbols for representing the components of a three phase power systems. Generator or Motor (Rotating Machine) Circuit breaker (oil or liquid) Air Circuit breaker Two winding power transformer Three-winding power transformer Current transformer Fuse or 3-phase delta connection
3-phase star connection (neutral ungrounded) 3-phase star connection (neutral grounded) Disconnecting Switch Reactor Lighting arrestor Auto transformer Fig 1: Symbols for representing single line diagram 1 3 T A T B Load A Load B Fig : Single-line diagram of an electrical power system Above figure shows the single line diagram of an electrical power system. Two generators grounded through reactors are connected to a bus and through a step up transformer to a transmission line. Another generator, grounded through reactor, is connected to a bus and through a transformer to the opposite end of the transmission line. A load is connected at each bus. Impedance and Reactance Diagrams: The impedance diagram on single phase basis for use under balanced operating conditions can be easily drawn from the online diagram. For the system of figure, the impedance diagram is shown in figure 3.
E 1 E E 3 Generators 1 and Load A Transformer T A Transmission Line Transformer T B Load B Generator 3. Fig 3: The per-phase impedance diagram. No currents flows in the ground under balanced conditions and the neutral of the generators are at the potential of the neutral of the system, so the impedance diagram does not include the current limiting impedances shown in the one line diagram between the neutral of the generator and ground. Since the shunt current of a transformer is usually insignificant compared with the full load current, the shunt admittance is usually omitted in the equivalent circuit of the transformer. The inductive reactance of a system is much larger than its resistance. So, the resistance is neglected in fault calculations. Synchronous motor loads are always included in making fault calculations, since their generated emfs contribute to the short-circuit current. Induction motors are represented by a generated emf in series with an inductive reactance if the diagram is to be used to determine the current immediately after the occurrence of a fault. Induction motors are ignored in computing the current a few cycles after the fault occurrence because the current contributed by an induction motor dies out very quickly after the induction motor is short circuited. Per phase reactance diagram after neglecting all static loads, resistances, shunt admittances of each transformer and the capacitance of the transmission line is shown in figure 4. The per phase impedance and reactance diagrams are sometimes called the per phase positive sequence diagram.
X T A X L X T B X 1 X X 3 E 1 E E 3 Fig 4: Simplified reactance diagram.. Per Unit System The quantities in a power systems (i.e., voltage, current, voltampers and impedance) are often expressed as a percent or per unit of a or reference value specified for each. The per unit value of any quantity is defined as the ratio of the actual quantity to its quantity. the actual value in any units Per Unit Value = the or reference value in the same units The ratio in percent is 100 times the value in per unit. Both the percent and per-unit methods of calculation are similar and more informative than the use of actual quantities. The per-unit method has an advantages over percent calculation method because the product of two quantities expressed in per unit is expressed in per unit itself, but the product of two quantities expressed in percent must be divided by 100 to obtain the result in percent. There are several reasons for using a per-unit system 1. Similar apparatus (generators, transformers, lines) will have similar per-unit impedances and losses expressed on their own ratings, regardless of their absolute size.. Use of the constant 3 is reduced in three phase calculations. 3. Per-unit quantities are the same on either side of a transformer, independent of voltage level. 4. By normalizing quantities to a common, the calculations are simplified.
A per unit system provides units for power, voltage, current and impedance. Only two of these are independent, usually power and voltage. Generally values of power and voltage are chosen. Once the power and the voltage are chosen, the current and the impedances are determined by the natural laws of electrical circuits. The relationship between quantities in a per-unit system depends on whether the system is single phase or three phase. Single Phase: Assuming that the independent values are power and voltage. Base voltampers = S = 1pu Base voltage = V = 1pu Base Active power = P = S.cosφ Base reactive power = Q = S.sinφ ( VA) S Base current = I = = = 1pu V V V Base Impedance = = I V Base Admittance = Y = i= V = S If the actual impedance is (ohms), its per unit value is given by ( ohms) ( VA) ( pu) = = V Base ( ohms) S = V Three-Phase: Power and voltage are specified in the same way as single-phase system. But the difference is that the power is specified as total power (not per phase), and voltage is line to line voltage.
In three phase systems the relations P = S.cosφ and Q = S sinφ holds good. The apparent power S equals S = 3 V I S I = = 1 pu 3 V V Phase V = = = 1pu I S 1 S Y = = = 1 pu V If the actual impedance is (ohms), its per unit value is given by ( ohms) S pu = = V If the impedance has to be represented in a new value denoted as new (Referred to S new and V new ) V old S -new pu new = pu old V -new S-old Note: pu S pu 1 V Example: Single line diagram of a power system is shown in the figure. The system contains one generator, 4 transformers and two transmission lines. The system ratings and reactances are indicated in the figure. Draw the equivalent impedance diagram for the given system and per unit equivalent diagram with S = 100MVA and V = kv.
T 1 1 3 Line 1 4 T Gen ~ 90MVA kv X pu=0.18 T 3 0kV 5 Line 1 110kV T 4 T 1:50MVA, /0kV. X pu=0.1, T :40MVA, 0/11kV. X pu=0.0, T 3:40MVA, /110kV. X pu=0.04, T 4:40MVA, 110/11kV. X pu=0.08, Line 1:48.4 Ω, Line : 5.43Ω Reactance Diagram Motor. 5MVA 10.45kV X pu=0.185 3ph load 10.45kV Absorbs 57MVA at 0.pf (lag) 1 3 4 X T1 X L1 X T Xsg 5 X T3 X L X T4 X sm Eg ~ R L ~ ~ E m Base Impedance with given Base Values V = S For Section II For Section I 1 0 10 = = 484Ω 100 10 110 10 For Section III = = 11Ω 3 100 10 11 10 For Section IV = = 1. 1Ω 4 100 10 10 = = 4. 84Ω 100 10
Per unit impedance calculations pu= actua S new V old punew = puold S old V new For generator 1, new per unit reactance 100 Xsg = Xsg,old = 0.pu 90 48.4 For Transmission line 1, Xl pu = = 0.1pu 484 5.43 For Transmission line, Xl pu = = 0.5pu 11 100 For Transformer 1, XT = 0.1 = 0. pu 1 50 For Transformer, X T = 100 = 0.0 40 0.15pu For transformer 3, 100 0 XT3 = 0.04 = 0.1pu 40 0 For transformer 4, 100 0 XT = 0.08 = 0.pu 4 40 0 For motor, 100 10.45 XSm = 0.185 = 0.5pu.5 11 For 3-phase load: Power factor: cos -1 (0.) = 53.13 o
Load S 3φ = 57 53. 13 V rated 10.45 act = = = 1.1495+ j1. 53Ω S * 57 53.13 Per unit impedance of 3- load 1.1495 + j1.53 = 1.1 = 0.95 + j1.7pu X t1=j0. X tl1=j0.1 X t=j 0.15 X sg=j0i C X sm=j0.5 X t3=j0.1 X tl=j0.5 X t4=j0. Eg ~ 0.95 j1. ~ E m Per unit impedance diagram Examples:.01. The current and voltage of a 345kV system are chosen to be 3000A and 300kV, respectively, the impedance of the system is (A) 115Ω (B) 100Ω (C) 10Ω (D) 0.01Ω 300 10 Sol. Base impedance = 3000 3 V = b Ib = 100Ω Choice (B).0. Let a 10KVA, 400/00-V transformer be approximately represented by a 4Ω reactance referred to the low-voltage side, considering the rated values as quantities, what is the transformer reactance as a per unit quantity. (A) 1pu (B) 0.5pu (C) 0.5pu (D) 1.5pu
Sol. Base impedance ( B ) V 00 = = = 4Ω S 10000 B The per unit reactance referred to the low-voltage side is 4 Per unit reactance = = 1pu Choice (A) 4.03. The per unit impedance of an alternator corresponding to values 13.kV and 5 MVA is 0.18 pu. The per unit value of the impedance for values are 13.8kV and 0MVA in pu will be (A) 0.395pu (B) 0.47pu (C) 0.39pu (D) 0.08pu Sol. V new VA old pu.new= pu.old Vold VAnew 0 13. = 0.18 = 0.395pu Choice (A) 5 13.8.04. A single phase system is shown in figure below A B C 500Ω 13.8 / 138 kv 15 MVA x = 10% 13.8 / 9 kv 15 MVA x = 8% With the in A circuit chosen as 13.8kV and 15MVA, the impedance diagram is (A) j0.1 j0.08 (B) j0.1 j0.08 500 0.34
(C) j0.08 (D) j0.1 j0.08 1.57 1.57 Sol. Chosen values 13.8kV and 15MVA Transformer 1 X pu = 0.1pu [old and new values are same] Transformer X pu = 0.08pu [old and new values are same] S Load pu = actual V = 15 10 = 500 1.57pu Choice (D) 3 ( 9 10 )