Solve this math :A 210-km, 192-KV, 60 Hz three-phase line has a positive-sequence series impedance z= 0.06+j0.4 Ω/km and a positive-sequence shunt admittance y= j4.33 X 10-6 S/km. At full load, the line delivers 200 MW at 0.92 p.f. lagging and at 185 KV. Using the nominal π circuit, calculate:(a) the ABCD parameters, (b) the sending end voltage and current, and(c) the percent of voltage regulation. Also explain, in which case the voltage regulation would be negative and why. Note: In the below i have added similer type of math and its ans.In the above math, there is just changes of values. Now solve the math according to the given way solve the math.A 200-km, 230-KV, 60 Hz three-phase line has a positive-sequence series impedance z= 0.08+j0.48 Ω/km and a positive-sequence shunt admittance y= j3.33 X 10-6 S/km. At full load, the line delivers 250 MW at 0.99 p.f. lagging and at 220 KV. Using the nominal π circuit, calculate:(a) the ABCD parameters, (b) the sending-end voltage and current, and (c) the percent of voltage regulation. Also explain, in which case the voltage regulation would be negative and why.--- This math solves are given in the picture. Tal ABCD partameters :Sending end Voltage (Vs) & Cunnent (Is):Given,series impe den ee, 2 = 0.08+ jo.48 /km.shund admittance,we know:Vg= AVR + BTRIg = CVR + DIR220 10Criven,Full toadV =220KV .y = j3.33 X 10-6S/Km.and, the lengthL= 200 Km. mediumlength LineLine to neutral ô VRwe Knocw,and,Z = z.L= (6.02+jo.48)x(200)= 127.017 L0° KV,L-eos"0.99)N3 x 220(0.99)= 0.656+0.0934i= 0.6629 L-8.10分KA.IR= PR L-cos?Cef)V3 VR (Pf)Y = Y.L%3Dニ=(j a.33 x0 6x2=36.66x10298Firom the generealized equation of medium lengthLine,weKnow oSo,Vs =AVR+B IR= (0.968L0.315°)x (127L0')+67.32 180. 54)xA = D= ZY +1"Vs=AVe +B1Rand B=2; e=Y(1+ ZY) eA= D=(0.08+jo.48)xQ 00) × ( j 3.3 9x 16 x 200 L98).and B= 2;Is=CVR + DIRit%3D(0.6627-8:105)S0,+1= 142 14+j62.16 = 155 •4 L23.58° KV2-N2= (0.0324 L170.5") + 1> A=D = 0.968+ jo. 00539 Amm:Line-to-Line, Vg =455.4 X J3)= 269.2 KVAn)ウA=D= 0,968 + jo.o0533NOw,s0, B =Z = (0.084j0.48)x200) .Is = CVR + DIR16.96j = 163780.54° = 9. 32L80.540Ame)= (6.553x10 290.15523)+ (0.968L-0.315°) ×%3DC=Y(1+Z)(6.66 x10-4296) ×{1 +0.08+jo48)xo0 x 6.66 X161 2g07(6.66 x16"L96)x (1+ 0.0162/ 170-5')= 6-553 x10-4/90. 155°s(Am)(o*66 23L-8-11)= 0.625356 -j2.6714文102= 0.6353 L-0.3310° KA As:)10 Perleent of VHage Regulation 8we know:VRNL -VRFLVRFL% VR =X100Hereyy VRNL =VsAgivery VRAL = 220 KV.269.2%3D0.968= 278.1 KVSO,% VR =278.1 - 220X 100%220= (6. 264 X 1 o %= 26.4 % Am)If He by any chanee the full load voltagegets highen than the no-load voltage,the voltage regulation cwill be negative.In thatthe machine worekscasereë vert sely. Itsee the machine igenerca ting pousen reathan than consuming,than comspowen,S0, only aboue this eause can makevoltage reegulation negative.

Given data: z=0.06+j0.4 Ω/kmy=j4.33×10-6 S/kmL=210 kmP=200 MWpf=0.92 lagVR=185 kV

Answered: Solve this math : A 210-km, 192-KV, 60…

Introductory Circuit Analysis (13th Edition)

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Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Solve this math :

A 210-km, 192-KV, 60 Hz three-phase line has a positive-sequence series impedance z= 0.06+j0.4 Ω/km and a positive-sequence shunt admittance y= j4.33 X 10^-6 S/km. At full load, the line delivers 200 MW at 0.92 p.f. lagging and at 185 KV. Using the nominal π circuit, calculate:

(a) the ABCD parameters, (b) the sending end voltage and current, and
(c) the percent of voltage regulation. Also explain, in which case the voltage regulation would be negative and why.

Note: In the below i have added similer type of math and its ans.In the above math, there is just changes of values. Now solve the math according to the given way solve the math.

A 200-km, 230-KV, 60 Hz three-phase line has a positive-sequence series impedance z= 0.08+j0.48 Ω/km and a positive-sequence shunt admittance y= j3.33 X 10^-6 S/km. At full load, the line delivers 250 MW at 0.99 p.f. lagging and at 220 KV. Using the nominal π circuit, calculate:

(a) the ABCD parameters, (b) the sending-end voltage and current, and
(c) the percent of voltage regulation. Also explain, in which case the voltage regulation would be negative and why.--- This math solves are given in the picture.

Tal ABCD partameters :
Sending end Voltage (Vs) & Cunnent (Is):
Given,
series impe den ee, 2 = 0.08+ jo.48 /km.
shund admittance,
we know:
Vg= AVR + BTR
Ig = CVR + DIR
220 10
Criven,
Full toad
V =220KV .
y = j3.33 X 10-6
S/Km.
and, the length
L= 200 Km. medium
length Line
Line to neutral ô VR
we Knocw,
and,
Z = z.L
= (6.02+jo.48)x(200)
= 127.017 L0° KV,
L-eos"0.99)
N3 x 220(0.99)
= 0.656+0.0934i
= 0.6629 L-8.10分KA.
IR= PR L-cos?Cef)
V3 VR (Pf)
Y = Y.L
%3D
ニ
=(j a.33 x0 6x2
=36.66x10298
Firom the generealized equation of medium length
Line,
we
Know o
So,
Vs =
AVR+B IR
= (0.968L0.315°)x (127L0')+67.32 180. 54)x
A = D= ZY +1
"Vs=AVe +B1R
and B=2; e=Y(1+ ZY) e
A= D=(0.08+jo.48)xQ 00) × ( j 3.3 9x 16 x 200 L98).
and B= 2;
Is=CVR + DIR
it
%3D
(0.6627-8:105)
S0,
+1
= 142 14+j62.16 = 155 •4 L23.58° KV2-N
2
= (0.0324 L170.5") + 1
> A=D = 0.968+ jo. 00539 Amm:
Line-to-Line, Vg =455.4 X J3)= 269.2 KVAn)
ウA=D
= 0,968 + jo.o0533
NOw,
s0, B =Z = (0.084j0.48)x200) .
Is = CVR + DIR
16.96j = 163780.54° = 9. 32L80.540
Ame)
= (6.553x10 290.15523)+ (0.968L-0.315°) ×
%3D
C=Y(1+Z)
(6.66 x10-4296) ×{1 +0.08+jo48)xo0 x 6.66 X161 2g07
(6.66 x16"L96)x (1+ 0.0162/ 170-5')
= 6-553 x10-4/90. 155°s(Am)
(o*66 23L-8-11)
= 0.625356 -j2.6714文102
= 0.6353 L-0.3310° KA As:)
10

Perleent of V
Hage Regulation 8
we know:
VRNL -VRFL
VRFL
% VR =
X100
Herey
y VRNL =
Vs
A
givery VRAL = 220 KV.
269.2
%3D
0.968
= 278.1 KV
SO,
% VR =
278.1 - 220
X 100%
220
= (6. 264 X 1 o %
= 26.4 % Am)
If He by any chanee the full load voltage
gets highen than the no-load voltage,
the voltage regulation cwill be negative.
In that
the machine woreks
case
reë vert sely. It
see the machine i
generca ting pousen reathan than consuming,
than coms
powen,
S0, only aboue this eause can make
voltage reegulation negative.

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