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
Chapter25: Television, Telephone, And Low-voltage Signal Systems
Section25.2: Telephone System
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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?
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53cd4567-e9c0-49bc-8f29-7fe4db37c68a%2F4371c539-18e2-4fae-b1a6-891b6cb9ab4b%2F67n2mp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Ivan Coronado
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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
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53cd4567-e9c0-49bc-8f29-7fe4db37c68a%2F4371c539-18e2-4fae-b1a6-891b6cb9ab4b%2Fe81xjzf_processed.jpeg&w=3840&q=75)
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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