Question

(POWER SYSTEM ANALYSIS) Load + Load Figure 1. One line diagram of a three bus power transmission system. In Figure 1, the power flows in the transmission lines are to be found. In this system, 100 MVA and 154 kV are the base values. The generator bus, which is shown as Bus 1, is considered to be an infinite bus and the voltage magnitude and the angle are 1 and 0°, respectively. The voltages of Bus 2 and Bus 3 are not known, but they will be determined after the load flow. The per unit values of transmission lines are given in Table 1. Table 1. Nominal equivalent circuit parameters of transmission lines Transmission Line Between bus 1 and 2 0.04+j 0.14 0.002 Between bus 1 and 3 0.02+ 0.1 0.001 Between bus 2 and 3 0.01+j 0.25 % 0.003 The loads on Bus 2 and Bus 3 can be modelled according to the load type: Constant impedance, constant curent, or constant power. The power flows of the transmission lines can be determined based on the model used. The model often used in power systems is the constant power representation of the load. To see the effect of these load models, the following questions have been prepared. Please, use the Nominal- A equivalent circuit model of the transmission lines when you are answering the questions. Question) If the load impedance of Bus 2 is 0.9+j 0.3 pu and the load impedance of Bus 3 is 0.6+j 0.4 pu, find the power flow of transmission lines and the voltages of busses in MVA and kV, respectively. Check if the generator power is equal to total of load powers.

Answer 1

Page ③ L® v3 =2 60° 30.02 +30.1 E 0.04 tj 0.14 © V3 mm 0.01 +90.25 Note: IN A 7t-model Aclmittances between the buses are equally shared to each bus. may between I and 2 is 90.002 o Hence Yat bus I = jo.001 0 (0o) Siemens Yat busz = 00001 0 B) Siemens Similary! toy between Daund ② is jo.001 Y at ☺ is j0.0005(S) Y at ☺ is jo. 000565) Y between bus 2 and ③ is ;0.003 Yat is 10.0015 Y at ③ is jo.0015. ground in a î-model Y at respective buses are shunted to Bus h Zrotal komentan Total 5 Hotel 2 *-model.

Page Hence the Network now gett modified as follows llous X= 1400 v0-001 900050 90-0013 290.00050 → LOAD LOAD 1 mm jo.00155 io.00150 mm tig.- Y shared at each Bus equally as posen model Let's say bus @ has Voltage Vz hati say bus @ has Noltage V3 bus @ hos Voltage of 1 Loº as menteoned above. Question 1 L mm Load impedance of Bus@ = 0.9 +30.3.2 in pu. Load impedance of Bus = 0.6+jo.4. inpu. s We need to construct Voltage dependent equation using Kirchoff's current law at each bocks. Bus and solue for V2, V3 =) Once we obtain V₂, V3 we can thus calculate Pflow in - to loads and currents flowing in to lines (1-2), (2-3), (1-3) => Please refce diogeam in Page ☺ and their respective Bus Equation (Nodal equations). like can treat each bus as a each node in the Poure transmission system.

Page . VA - 1 Loºm j 0.0010 HH Lj0.0005 v 0.04+0:14 | : 0.02 +0.1 jo.001 jo.0005 V 0.04 +0.25 1 0.6 tjo.lt ů 0.00155 - $0.9+jo-3pu Nodal Equation Armbus 1 - Som of leaving corrente Sumof entering. ) 160 (90.0005) + 3 (0.004) 1 Lo + * 120-V2 + 160 - V3 = Ia 0-04 +90.14 0.02 tj.!_n (GR) sum of currenter leaving at a node is algeboracally.co. I added Nodal equatuoin at bus @ :- V2 Ci 0.0015) + V2 (0.001) + V2-0 + V2-960 Y2_V3 0.9 +90-3 0.04 +jo-14 0.01 +30.25 Nodal Equation at bus : L . Vg (G90005) + Vg (j0.0015) + V3-0 + Yg - 160 + Mz - V2 = 0.6tjo-4 0.02tjol boltjo.25 L→ equation ©

Page one needs to solve Equations and Ⓡ) to obtoun V, V3 Values and from V, Vg information we con calculate sat bus© and s at buss Sc complexe powder. powe polas form converlidl. Vellices are. Bus (1-2) = %= 0.04+90.14 = 0.1456 | 74.054° Bus (1-3) = Z = 0.02 + 90.1 = 0.1019 | 78.69° Bus (2-3) = 0.01 + $0.25 = 0.2501 187.709 Flood at bus@ = 0.9 +90.3 = 0.9486 | 18.4349° Zload at bus ®) = 0,6+40.4 = 0.72111 | 33.69º.

Page Equation A - 110° (30.0005) +9 (0.001) 4L0 + 1-12 + 1-V3 0.04 +0.14 0.02 +j6-1 => 0.0015198 + ( 6868) |-74.054) C1-ya) +.8135 |-78.69) ( 1–va).-IG 0.0015196 + 6.868) | -74.054 -(6-86814-74.054) V2 + 9.8135 |-78.690 – (9-8135 -78.69) V3 = IG. = 16.666 _-76-7803º - V2 ( 6-868) |-74.054) -13 (9-8135|-78. 69) = IG Simplified Eq 0. Equation :- V2 (0.0025190) + V2 0.9486 1843490 +(12-1) + 04456 | 74.054 /2-1/3 =o. 0.2501187.709 >> V2 [0.0025 L90 +4-05411-18.4349° + 6-8681 |_74.054 +3.9984 1-87-709] + VB I-D (3.9984 1-87-709] = 6.8681 | -74.054 => V2 [11-3465 -74.4240 ] + v3 [3.9984 192.241] = 6-8681)-74.054° => v2 (1.3465 (-74-4240) + v3 (3-9984 | 22:41°) = 6-3 689 | -740549 L &quction ® simplified.

Equation :-3 Page ① Valio.002) + + n. _V3 0.7211 |33.69 V3-V2 V3-1 0.2501 187.709 6.1019478.69 => v3 (0.002|40) + V3 (1-38671-33.69) + (3.9984 1-87.709) (v3-V2) + (13-1) (9.8135 |-78-690 ) = 0. =) V8 [1407455 |-77.2805] + V2 [3.9984 192-291° ] = 9.8135 |-78.69 Simplified version of Eq@. Now equation @ and need to be solved to to get V2, V3 Voltages. These can be solved by using coamers method of matrices Eq@ if amplified Version) V2 (11-34651_74.4240) + V3 ( 3.9984192.291) = 6-868) | -74.0549 V2 [3.9984 192.291°) + V3 [1407455 |_77.2805] 1:1 = 9.8135 |-78.69 MATRIX FORM OF E92 and 893';.; -3465|-747.47240 3.9984 192.2017 168681|-74.054) b 3.9984 192-291914.7455 |-77.2807 Vgl. 19.8135|-78.699

A: (ad-bc) , Page - (11-3465 | -74-4248) (1407455 |-77.2805) - (3.4984 192.241°) ( 3.9984 192.291) (167.9092 |- 1517845) – 15.9872 | +184582 = 152.8077. L-149.2930° SA- a = 6-8681 |-74.054 b= 3.9984 192.291 Ce 908135 178.699 d=14-7455 |-77.2805 i A1 = 56-8681) |_74.054"] [14.7455 | -77.2805] L - [63.9984) (92.241] [9.8135 -78.690] dis (104.273 | -151-3345) – (39.2382 (+ 13.6019) sie 139.5358 | -155.525° IMI V2 = Aliz 139.535€ |–155.625 152.8077 | -149.2938 V2 = 0:91317–6*232° pu. L . .- --...

Page Az =) a = 11.3465 |-74.4240 be 6.8681 L-74.054°: r ca 3.9984 192-2919 do 9.8135 _-78.69º. A2= (1.3465)(9.8135) -74.4240 –78.699 (6.8681) (3.9984) |–74.054°+92: 291 Az = 1.3488.-153414° ~ 27.4614 18-234° = 138.559 1-154-821° Ug = = (1963:50+) 1-149.2930° +154.821 V2 = 0.9067 +5.5280 pu. There come dopo Questiona - Ve .0.9131 | -6•232° Pu. V3 = 0.9067 50528° PU "Salt, bucz : $ = 1 - (0.91313 NOTE :- 0.9486 418.4349 22=0.9486 | +18.4349 77* = 0.9486 [18.4349. :: Son = 0.8789 | +18.4349° pu. Andet (pot +18.43

Page ① Sg = 1 Vgl. (0.9067) zt (0.92111 |-33.69°) S2 = 1.8400 1 +33.690 pic - 1o 1400 ) m At buls 2!- S2 = 0.8789 | +18.43.49* P4 P = 0.8789 cos (18.4349) PU Q = 0.8789 Sin (18-4349). pu eu : i , Säcteuer - Spou x Sbase = 0.8769/18.4349) (100 MVA) Bus@ 5) 87-89 118-4349 MVA Szactical = Ss pu x Shave: (1.1400 | 33.690) (100 MVA) = 114.0 | 33.69 MVA..j At bus : Bus .P2 = 87.89 Cos (18.4349) MW = 83-3797 MW Q2 = 87.89 sin (18.4349). = 27.7931 MVAR.. At Bus 3 Pa = 114 cos (33-69) = 94.8538 MWI Gan 114 sin (33.69) = 63.2357 MVAR.

IG = 16.666 |_76.7803° V2 (6-8681L-74-054) : Page Ciri -(9.8135 L-78-69) V2 = 0.9131 |– 6:232° py ..13 = 0.9067 | +5.528° pu. Ja = 16.666 _767803° - (0.91312-6-232) (6.8681 7. 4-054) -6.9067 L5-528) (9.8135 |-78.69). Is = 16.666 L-76-7803 - 6.27126 1-80.286 – 8.8979 | -73.162 Ta = 1.5367 (-8304344, s at bail = V IGA EV= 1.109 IG"= 1.5367 | -83.4344. S. 1.5367 | +83.4344. Pu. BUSAI SACTUAL - (Spul x(Sbave) ...:) (1.5367) COOMYA) | +83.4344° 153.67.183.4344° MVA: (1.53 Pactual BUSI Q Biasi = 153.67 CAS (83.4344) = 19,57 mW 153.67 sin (83.4344) = 152.66 MVAR.

IG = 1.5367 | +88043 pue flow of s' 'm nes TRANSMiss mm Transmission lines ! Son - 105567 1-83.43. I 21 7 8 V₂ -VA Ziz 0.91311–67232-1 19 Igy - 0.1456 174.056. 0.8789 1-1 M400 4418-43 Note: The direction of 'S' flow specifies direction of Current flow. mm 1- 0.9067 | 5.528 0.019 | 78.69. 1.82 = 09067L5-528 – 0.9131 1–6.232 022501 | 87.709 121 = 0.9299 _-207.006. . I1 = 1.2845 L-120.54 . 132. = 0.7457 | 3.894, Sin te dine 1-2 = Izel z = 0.1259 |-74-054 Son Tx line 2-3. - Inaella zone = 0.1390 1-87.709. Sin Txline 13 = (113). Zona = 0.1681 |-78.69 STOTAL Tx line .= 0.431 L-80.23.

Verification of power s by Admittance: At Busis = 1V1Y* = 0.0015 02 L-90° Bos 2 S - lv, Y* = (6.9131°(0-00 29-10 Bus 3 S = 1 4312 Y* = (0.967) 2 (0.002) L-90° Stotal of Admittance = 0.0052 L-90 SGenerală = 1.53 -83.43. S Busza 0.8789 | +18:43 Sg Bulo 1.1400 | 33.69 SGenerator + S Admittance + Styline = 1.965 1-82-72°- Sbus 2 ts bus 3 = 1.99 +27-0909°, Almost. Sgenerated by Sload +Slosses in Tx line. TX lines in actual values. Sbase = 100 MVA Si pu: Su-2 actuced sbase Sv-2 actual = 12.59 -74.054 MVA similas S23 actual = 13.90 1- 87.709 MVA Sing actual = @ 16,81 -7869 MVA:

V, = I PU LO° 9x154 KV = 154 KV L0° V2 = 0.91311-6:232° 140 Kv 1–6.232º, V2 - 0.9067/5-528° = 139•63 kv 1 +5,528