Partial preview of the text
Download Reports and production and more Cheat Sheet Reporting and Production in PDF only on Docsity!
188 Chapter 10 Problem Solutions 10.1 A 60-Hz alternating voltage having a rms value of 100 V is applied to a series RL circuit by closing a switch. The resistance is 15 9 and the inductance is 0.12 H. “ae unumrtec eve nageanae circ (a) Find the value of the de component of current upon closing the switch if the instantaneous value of the voltage is 50 V at that time. (b) What is the instantaneous value of the voltage which will produce the maximum de component of current upon closing the switch? i (c} What is the instantaneous value of the voltage which will result in the absence of any de component of current upon closing the switch? (d) If the switch is closed when the instantaneous voltage is zero, find the instantaneous current 0.5, 1.5 and 5.5 cycles later. i Solution: : (a) q v = Vasin(wt + a) Fort=0 50 = V2x 100sine a = 20.70° or 159.30° Z = 13452m x 60x 012 = 47.667 71.66 . 100 x V2 ° ALt=0 ig ~ “Frag sin (20.7° ~ 71.667) = 2.305 A or ige — xe sin (159.3° — 71.66°) = —2.965A (max.) {b) Maximum dc component occurs when sin(a — 6) = +41 or when (a — 6) = +90° when @ = 161.66° or —18.34°. v = 100V2sin161.66° = 100V/2sin—18.34° = 444.5 V {¢} No de component will occur when a — 6 = 0, or 180°, i.e. when a = 71.66° or 251.66°. v = 100V2sin71.66° = 100V3sin 251.66 = 4134.24 V (d) For v = 0 when t = 0 and a = 0. 0.5 cycles later wt = = rad. r i = — = 0.0083 t = 575 = 0.008335 100V3 - i= 200V2 fain 180° — 11.66") — «78 t.cote x sin (~71.66°)| 100V2 Free (b+ 7141) sin (-71.66°) = 3.810 4 10.2 10.3 189 Similarly, 1.5 cycles later: wt = 37 t = 0025s i = 2940A and 5.5 cycles later: ut = Llz ~ = 0091675 ti = 2817A Note that the de component has essentially disappeared after 5.5 cycles. (5 time constants = 0.04 s). A generator connected through a 5-cycle circuit breaker to a transformer is rated 100 MVA, 18 kV, with reactances of Xf = 19%, Xj = 26% and Xz = 130%. It is operating at no load and rated voltage when a three-phase short circuit occurs between the breaker and the transformer. Find (a) the sustained short-circuit current in the breaker, (b) the initial symmetrical rms current in the breaker and (c) the maximum possible de component of the short-circuit current in the breaker. Solution: Base current = 100,000 3207.5 A v3 x 18 (a) aL x 30075 = 2,467 A fi3 1 (b) Foye * 2207.5 = 16,882 A (ce) V3 x 16,882 = 23,874 A The three-phase transformer connected to the generator described in Prob. 10.2 is rated 100 MVA, 240Y/18A kV, X = 10%. If a three-phase short circuit -occurs on the high-voltage side of the transformer at rated voltage and no load, find (a) the initial symmetrical rms current in the transformer windings on the high-voltage side and (b) the initial symmetrical rms current in the line on the low-voltage side. Solution: 10 I" = = 33. . F(19 + 0.10) $0.10) 73.448 per unit Base Iny = 22000 woe A V3 x 240 100, 000 Base ly = Vixis = 3207.5 A 191 I = Lit = 90:36 = —1.03 — 73.34 per unit ™ 30.35 Tp = I+ TY = —0.55 — 76.58 per unit (8) ees By replacing J} 7 by a current source and then applying the principle of superposition, ” , 70.35 ar 4 . A = OO. z =—— (-0.55 — j6.58) = 0.48 — 73. yy 0.8 + 0.6 + eh 0.55 — 76.58) 0.48 — 33.24 per unit I = 08-706 90.25 —0.55 — 76.58) = -1.03 ~ 73.34 per unit mo F H0.6 + 60 . 56.58) = -11 53.34 per uni 10.6 Two synchronous motors having subtransient reactances of 0.80 and 0.25 per i unit, respectively, on a base of 480 V, 2000 kVA are connected to a bus. This motor is connected by a line having a reactance of 0.023 2 to a bus of a power system. At the power-system bus the short-circuit megavoltamperes of the power system are 9.6 MVA for a nominal voltage of 480 V. When the voltage t at the motor bus is 440 V, neglect load current and find the initial symmetrical i rms current in a three-phase fault at the motor bus. Solution: Power Sys Bus 8 N- } Xu P O aan x _ MOTORS i GEN. Onn 70.25 P: fault point 0.48? BaseZ = > = 0.1152 2 0.023 Xp, = =o = 0: i L Oaiie2 0.20 per unit 2 Xsc = 7 0.208 per unit 1 Xp = 4 t <— = 0.130 per unit vat om + wa0e 2 192 440/480 . y= 130 = 7.05 per unit 2000 or 7.05 x ————. = _ 17, 000 A V3 x 0.48 10.7 The bus impedance matrix of a four-bus network with values in per unit is 50.15 j0.08 70.04 70.07 70.08 70.15 70.06 70.09 i 50.04 70.06 70.13 70.05 i 90.07 70.09 70.05 70.12 Zou = Generators connected to buses () and @ have their subtransient reactances included in Zpus. If prefault current is neglected, find the subtransient current in per unit in the fault for a three-phase fault on bus @. Assume the voltage at the fault is 1.02.0° per unit before the fault occurs. Find also the per-unit current from generator 2 whose subtransient reactance is 0.2 per unit. Note to Instructor: This short problem is easily varied by assuming the fault to occur on other buses. Solution: At bus @, I o — 1 __, , e iF = jo .j8.33 per unit 0.09 WY = 10-—> = 0: i cy 10 023 0.25 per unit From generator 2, 4 Wy = 1025 it = —3.75 per unit j0.2 eer 194 10.9 For the network shown in Fig. 10.17 determine Ypu; and its triangular factors. Use the triangular factors to generate the elements of Zpus needed to solve Prob. 10.8. Solution: ~jl2 5 2 Yous = gS -j7.5 92.5 | per unit g2 725 -j85 -jl2 : : 1 -0.4167 —0.1667 = gS 75.4167 : . 1 —0.6154 52 73.3333 36.1154 : 1 L U ; Yous’ = U-'L7! where j 1 0.4167 0.4231 70.0833 iy Ul = 1 0.6154 L-? = | 70.0769 70.1846 3 . . 1 50.0692 70.1006 70.1635 Hence, Zou = You? = U“IL“1 = Be 0.2465 70.1006 j0.1447 70.1195 70.0692 per unit 70.0692 70.1006 70.1635 10.10 If a three-phase fault occurs at bus @ of the network of Fig. 10.5 when there is no load (all bus voltages equal 1.02.0° per unit), find the subtransient current in the fault, the voltages at buses @, @ and @, and the current from the generator connected to bus @. Use equivalent circuits based on Zpus of Example 10.3 and similar to those of Fig. 10.7 to illustrate your calculations. i Solution: i 1.020° 1.020° . . . Es [bs = = 54, : 7 ae 502436 4.105 per unit f : - © | 195 During the fault, v% Zn Ys nogor| 1 | -1y! Zs Y, Za 1 1 1 1 70.1938 0.24447 0° 1.0202 | 1 | ~ (—74.105) | 70.1544 | = | 0.36622 0° | per unit 1 70.1456 0.4023 70° Current from generator at bus @ is calculated to be (EX -Va) _ (1.0 - 0.4023) _ 70.30 = 39 = —71.992 per unit . jo. 0.30 ti cs 1 = 1 10.11 The network of Fig. 10.8 has the bus impedance matrix given in Example 10.4. If a short-circuit fault occurs at bus @ of the network when there is no load (all bus voltages equal 1.0/ 0° per unit), find the subtransient current in the fault, the voltages at buses (@ and @), and the current from the generator connected to bus @). Use equivalent circuits based on Zpus and similar to those of Fig. 10.7 to illustrate your calculations. Solution: 10202 1.0208 = = = —j7.47: i t Bao 0.1338 j7 474 per unit ! o i ain —_] | } Zs Hl “° . ° bn 1 — ) i Zs2 ® i H | 197 10.13 Figure 9.2 shows the one-line diagram of a single power network which has the line data given in Table 9.2. Each generator connected to buses @ and @ has a subtransient reactance of 0.25 per unit. Making the usual fault-study assumptions, summarized in Sec. 10.6, determine for the network (a) Ybus, (b) Zyus, (c) the subtransient current in per unit in @ three-phase fault on bus @) and (d) the contributions to the fault current from line Q-@) and from line @-@. Solution: Reactance diagram: Admittance diagram: : ie) ° | J0.0584 30.0372 rey © 718.88 728.88 { : _ -34.0 . Hl 78-28 30.0372 30.0636 70-25 ; 526.88 318.72 740 i + a mis + | le) fo) i (a) 6 ® 8 @® @ [750.72 719.84 726.88 = 0 i yu. = © | 71984 346.72 50 526.88 | bus = | 726.88 jO 42.60 715.72 ® gO j26.88 += 15.72 —746.60 (b) @ @ @ ® 0.1357 90.1234 70.1278 70.1143 90.1284 70.1466 40.1246 70.1266 GO.1278 70.1246 70.1492 70.1222 JO.1143 70.1266 70.1222 70.1357 1.0702 _ 1.020% Zs3. «70.1492 = —76.702 per unit (d} During the fault, a of 3 un} Zia ta = 1020 [3] -7 Zz: 0.1278 6.14352 0° _ ofl], _ = rocoe| | | ( 6.702) | 353209 | = Beira per unit Current flow in line Q-@ is calculated to be (Vi Vs} _ (0.1435 - 0) is 7 00s = ~73.857 per unit 198 Current flow in line @-@ is calculated to be (Vs— V3) _ (0.1810-0) . fia 0083 = ~Jj2.846 per unit where the sum of these currents is —j6.703(& 7). : 10.14 A 625-kV generator with XJ = 0.20 per unit is connected to a bus through a circuit breaker as shown in Fig. 10.18. Connected through circuit breakers to the same bus are three synchronous motors rated 250 hp, 2.4 kV, 1.0 power factor, 90% efficiency, with X7 = 0.20 per unit. The motors are operating at full load, unity power factor and rated voltage, with the load equally divided ve between the machines. (a) Draw the impedance diagram with the impedances marked in per unit on a base of 625 kVA, 2.4 kV. (b) Find the symmetrical short-circuit current in amperes which must be in- terrupted by breakers A and B for a three-phase fault at point P. Simplify the calculations by neglecting the prefault current. (c) Repeat part (8) for a three-phase fault at point Q. (d) Repeat part (5) for a three-phase fault at point R. He Solution: {a} 40.905 C 5 j020 j0.905 ; On nC) MOTORS : GEN. 2 50.905 P © .746 } Motor input = BOx OTS = 207.2kVA e “625 i XP = 02x TS = 0.603 per unit : : £ For interrupting current use iv 1.5Xm_ = 1.5% 0.603 = 0.905 per unit 625 i Base] = Wixaa = 150.4 per unit 70.905 /3)j0.2 Zap = E905/3)902 59 1903 per unit 70.905/3 + 70.2 1 50.1203 = —78.315 per unit 
