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Figure 2-18 shows a one-line diagram of a small 480-V industrial distribution system. The power system supplies a constant line voltage of 480 V, and the impedance of the distribution lines is negligible. Load 1 is a A-connected load with a phase im- pedance of 130° 2, and load 2 is a Y-connected load with a phase impedance of 5L-36.87° N. (a) Find the overall power factor of the distribution system. (b) Find the total line current supplied to the distribution system. Figure 2-18 I The system in Example 2-3. Bus A Load Delta connected 1 Z, = 10230°O Load Wye connected 480 V three-phase 2. Z, = 52-36.87°2

Figure 2-18 shows a one-line diagram of a small 480-V industrial distribution system. The power system supplies a constant line voltage of 480 V, and the impedance of the distribution lines is negligible. Load 1 is a A-connected load with a phase im- pedance of 130° 2, and load 2 is a Y-connected load with a phase impedance of 5L-36.87° N. (a) Find the overall power factor of the distribution system. (b) Find the total line current supplied to the distribution system. Figure 2-18 I The system in Example 2-3. Bus A Load Delta connected 1 Z, = 10230°O Load Wye connected 480 V three-phase 2. Z, = 52-36.87°2

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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For this kind a problem, how do you know how the voltage source is configured ( wye or delta). The problem start getting the phase current for phase 1 and 2 , but do we have to pretend that voltage source to delta load is connected like y-delta to convert the voltage to delta source? Or how does it work?
Ivan Coronado ions Animations Shde Shaw Record Review View Help t can contain virruses Ontes you needto edit, N's'safer to stay in Protected View, Record Pnt Enable Editing Figure 2-18 shows a one-line diagram of a small 480-V industrial distribution systern. The power system supplies a constant line voltage of 480 V, and the impedance of the distribution lines is negligible. Load 1 is a A-connected load with a phase im- pedance of 10430° N, and load 2 is a Y-connected load with a phase impedance of 5L-36.87° N. (a) Find the overall power factor of the distribution system. (b) Find the total line current supplied to the distribution system. Figure 2-18 I The system in Example 2-3. Bus A Load Delta connected 1 Z, = 10230°S2 IL Load Wye connected 480 V three-phase Z, = 52-36.87°2 915 PAA e d0 /29/2022
Transcribed Image Text:Ivan Coronado ions Animations Shde Shaw Record Review View Help t can contain virruses Ontes you needto edit, N's'safer to stay in Protected View, Record Pnt Enable Editing Figure 2-18 shows a one-line diagram of a small 480-V industrial distribution systern. The power system supplies a constant line voltage of 480 V, and the impedance of the distribution lines is negligible. Load 1 is a A-connected load with a phase im- pedance of 10430° N, and load 2 is a Y-connected load with a phase impedance of 5L-36.87° N. (a) Find the overall power factor of the distribution system. (b) Find the total line current supplied to the distribution system. Figure 2-18 I The system in Example 2-3. Bus A Load Delta connected 1 Z, = 10230°S2 IL Load Wye connected 480 V three-phase Z, = 52-36.87°2 915 PAA e d0 /29/2022
ORecord Present in Team enet can contain viruses. Unless you need to edit, it's safer to stay in Protected View, Enable Editing I Solution The lines in this system are assumed impedanceless, so there will be no voltage drops within the system. Since load 1 is A connected, its phase voltage will be 480 V. Since load 2 is Y connected, its phase voltage will be 480/V3 = 277 V. The phase current in load 1 is Z, 480 V = 48 A 10 Ω Z, Therefore, the real and reactive powers of load 1 are P = 3Vl61 Cos e = 3(480 V)(48 A) cos 30° = 59.9 kW Q, = 3Vl61 sin e = 3(480 V)(48 A) sin 30° = 34.6 kvar The phase current in load 2 is 277 V = 55.4 A Therefore, the real and reactive powers of load 2 are P= 3Vele2 cos 0 = 3(277 V)(55.4 A) cos(-36.87°) = 36.8 kW O2 = 3Va2la2 Ssin 0 = 3(277 V)(55.4 A) sin(-36.87°) = -27.6 kvar 915 PM 3/29/2022
Transcribed Image Text:ORecord Present in Team enet can contain viruses. Unless you need to edit, it's safer to stay in Protected View, Enable Editing I Solution The lines in this system are assumed impedanceless, so there will be no voltage drops within the system. Since load 1 is A connected, its phase voltage will be 480 V. Since load 2 is Y connected, its phase voltage will be 480/V3 = 277 V. The phase current in load 1 is Z, 480 V = 48 A 10 Ω Z, Therefore, the real and reactive powers of load 1 are P = 3Vl61 Cos e = 3(480 V)(48 A) cos 30° = 59.9 kW Q, = 3Vl61 sin e = 3(480 V)(48 A) sin 30° = 34.6 kvar The phase current in load 2 is 277 V = 55.4 A Therefore, the real and reactive powers of load 2 are P= 3Vele2 cos 0 = 3(277 V)(55.4 A) cos(-36.87°) = 36.8 kW O2 = 3Va2la2 Ssin 0 = 3(277 V)(55.4 A) sin(-36.87°) = -27.6 kvar 915 PM 3/29/2022
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