For the system below: ¥381 G₁: T₁: T₂: M: Line: 60 MVA 50 MVA 50 MVA 43.2 MVA Line 20 kV 20/200 kV 200/20 kV 18 KV 200 kV T₂ X = 9% X = 10% X= 10% X = 8% Z = 120+j200 f M a) Draw an impedance diagram in per unit on 100 MVA, 20kV bases b) If the motor is drawing 45MVA, 0.8 lagging PF at a line to line terminal voltage of 18kV, determine the terminal voltage and the internal emf of the generator in per unit and in kv.

For the system below: ¥381 G₁: T₁: T₂: M: Line: 60 MVA 50 MVA 50 MVA 43.2 MVA Line 20 kV 20/200 kV 200/20 kV 18 KV 200 kV T₂ X = 9% X = 10% X= 10% X = 8% Z = 120+j200 f M a) Draw an impedance diagram in per unit on 100 MVA, 20kV bases b) If the motor is drawing 45MVA, 0.8 lagging PF at a line to line terminal voltage of 18kV, determine the terminal voltage and the internal emf of the generator in per unit and in kv.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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1 PROBLEM 20020° V rms, Vbn 1. A balanced three-phase Y-Y system has the phase voltages of Van 200/120° V rms, and Ven = 2002(-120°) Vrms. The load impedance per phase is Z₁ = 50 - j100 n. a Van Vcn n Vbn Ia Ib Ic B Zu (b) Determine the line currents Īa, Ip and Īc. (c) Determine the total average power delivered to the three loads. N C = ZL (a) The balanced three-phase circuit has a negative phase sequence. Circle your choice. (1) True (ii) False
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The steady-state voltage drop between the load and the sending end of the line seen in (Figure 1) is excessive Suppose that V-4950/0° V (rms). A capacitor is placed in parallel with the 192 kVA load and is adjusted until the steady-state voltage at the sending end of the line has the same magnitude as the voltage at the load end, that is, 4950 V (rms). The 192 kVA load is operating at a power factor of 0.8 lag. Part A Calculate the size of the capacitor in microfarads if the circuit is operating at 60 Hz. In selecting the capacitance, use the value that results in the lowest possible power loss in the line. Express your answer in microfarads to three significant figures. ▸ View Available Hint(s) Figure 1 of 1 202 1100 192 kVA 0.8 lag ΜΕ ΑΣΦ. 11 vec C= 22.8 Submit Previous Answers Incorrect; Try Again; 5 attempts remaining Provide Feedback ? F
For the system as shown in the following figure, convert the components to same base voltage and apparent power, by choosing 100 kVA as the base kVA and 2000 V as the base voltage for the generators. Redraw the single line diagram with the new per unit values. 15 KVA 2000 V Z=j0.3 pu 25 KVA 2000 V Z=j0.15 pu G1 G2 35 KVA 2000/8000 V Z=j0.2 pu Z=30+j100 Z=60+j120 pu O pu leee 35 KVA 8000/2000 V Z=j0.3 pu 35 kVA 2000/8000 V Z=j0.25 pu G3 10 KVA 2000V Z=j0.35 pu M 30 KVA 2000 V
7
Q: Draw the per unit impedance diagram for the system shown in figure below (Assume SB = 200 MVA). G₁ T₁ T.L T3 Ô • †• © · †• — • †•· Ø • † • Ö 어 (3+j9) Ω 110 MVA 120 MVA 75 MVA 60 MVA 6.5 kV j 0.25 10.2 kV j0.2 (11/33) KV j0.1 90 MVA (11/33) KV j0.08 80 MVA 10.8 kV j0.18 0 T₂ T4 Load (33/6.6) kV j0.08 40 MVA (33/11) KV j0.1 34 MVA 9.9KV 0.87 lag
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Q1. For the system shown in Figure Q1, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 KVA 2400 V G₁ Z=j0.2pu 20 KVA 2400 V Z=j0.2pu G₂ T₁ Transmission Line Z= (50+j200) 40 KVA 2400/9600 V Z=j0.1pu Fig Q1 T₂ 80 KVA 10/5 kV Z=j0.1pu M 25 KVA 4 kV
A system consists of four idenital 400 MVA generating units feed- ing a total load of 1016 MW. The intertia constant of each unit is 5 on 400 MVA base. The load changes by 1.5% for a 1% change in frequency. When there is sudden drop in load by 16 MW, obtain the power system time constant (T) with constants H and D expressed on 1600 MVA base.
What is the max load demand of the following feeders 4152 MW, 2.891 MW, 815.20 KW, 905.12 KWand 1.972 MW A) 10,735.32 KW B) 12,815.32 KW C) 10,571 MW D) 12.052 MW