For the four bus system as shown in Fig. Determine the fault current when the fault takes place at bus no.2 with fault impedance Zr =0 p.u. The Subtransient reactance of the generators and positive sequence reactance of other elements are given. Assume prefault Votage Eth=1Z0° pu Generator X = j0.12p.u; Transmission line X = j0.26p.u; Transformer X = %3D j0.43p.u À Yz Y= Zn in pu Fault current Ir in p.u 625FC6C7-5514JPES 46 (Mc_1850jpg
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- Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance Xeq=0.10perunit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.Three single-phase two-winding transformers, each rated 25MVA,54.2/5.42kV, are connected to form a three-phase Y- bank with a balanced Y-connected resistive load of 0.6 per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance RL in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base.
- The per-unit equivalent circuit of two transformers Ta and Tb connected in parallel, with the same nominal voltage ratio and the same reactan of 0.1 per unit on the same base, is shown in Figure 3.43. Transformer Tb has a voltage-magnitude step-up toward the load of 1.05 times that of Ta (that is, the tap on the secondary winding of Tb is set to 1.05). The load is represented by 0.8+j0.6 per unit at a voltage V2=1.0/0 per unit. Determine the complex power in per unit transmitted to the load through each transformer, comment on how the transformers share the real and reactive powers.Consider a single phase transformer of rating 150 KVA, 1500/150, 50 Hz. Let the series resistance due to windings be 0.2 ohm and 0.002 ohm on the HV and LV sides, respectively. Let the series reactance due to leakage flux be j0.45 ohm and j0.0045 ohm on the HV and LV sides, respectively. While conducting Short Circuit test, the HV side should be excited with Volt Input power factor while conducting Short Circuit test is laggingb) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) = j0.1X p.u. V₁ = 120° p.u. V₂ = 120° p.u. (i) (ii) 0 jX(1) = j0.2 p.u. 1 jx(2) j0.2 p.u. 2 jX1(0) = j0.25 p.u. jXT(1) jXT(2) 종 3 j0.1X p.u. JX3(1) j0.1Y p.u. j0.1X p.u. JX3(2) j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0 = x = 1, jX2(1) j0.2Y p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V3 = 120° p.u. Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: y = 7 = = jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from…
- b) A fault occurs at bus 2 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. JX20 /0.1X p.u. jXa2) 0.1X p.u. JX20 j0.2Y p.u. V,= 120° p.u. V, 120° p.u. V, 120° p.u. jX4-70.2X p.u. jX2 j0.2X p.u. jX o 0.2Y p.u. jXncay J0.25 p.u. jXna J0.25 p.u. 3 jXno0.3 p.u. jXTu) /0.2Y p.u. jXra j0.2Y p.u. - j0.2Y p.u. Xp-10.1X p.u. jXa j0.1X p.u. jXp0)- j0.05 p.u. 0 Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXac1) = j0.22 p.u., jXac2) = j0.22 p.u., and jXaco) = j0.23 p. u. X-2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.The sample large power system network data's are given below, The total number of buses is 5 Three-phase short circuit fault subjected at the bus 5 The initial voltage of the faulted bus is 1.0 p.u The Zbus matrix element Z55 is 0.704 p.u Fault impedance Zf= 0.33 p.u Fault current (If )in p.u ..........No-load operation and short circuit tests were performed on three-phase, 50 Hz 480/220 delta / Y connected transformer. In these testsThe line voltage, line current and three-phase total power have been measured from the high voltage windings and the test results are below.pictures. 100 A DC current when 1.08 V DC voltage from non-transferable software is applied to the low voltage windingshas attracted.Find the single phase equivalent circuit parameters of this transformer.idle test:Voltage: 480V Power:114w current: 0.71Ashort circiut test:V: 10v P: 78w c: 18.3a
- 2) A single-phase power transformer of 150 MVA, 15/200 KV has a resistance in per unit of 1.1% and reactance per unit of 4.5%. The magnetizing impedance is j95 per unit. a) Find the equivalent circuit of the transformer referred to the low side voltage. b) Calculate the voltage regulation of the transformer in nominal conditions and f.p. 0.9 behind.b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…A generator connected through 5 cycle CB to a transformer is rated 8000 kVA with the reactances of X"= 10 %, X₁ = 16% and Xa = 100 %. It is operating at no load and rated voltage when 3 phase short circuit occurs between breaker and transformer. Find: i) sustained short circuit in breaker ii) the initial symmetrical rms current in breaker iii) maximum possible d.c. component of short circuit in breaker iv) the momentary current rating of the breaker v) current to be interrupted by breaker vi) the interrupting kVA