Refer to page 30, the single line diagram shows 1000 kVA. if the transformer capacity rating to be used is 1500 kVA at 480 volts, and the second transformer rated 225 kVA be changed to 250 KVA at 208 volts, calculate the fault at X1 and fault at X2. Assume all other data are the same.

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.9MCQ: Match the following: (i) Hysteresis loss (a) Can be redud by constructing the core with laminated...
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Refer to page 30, the single line diagram shows 1000 kVA. if the transformer capacity rating to be used is 1500 kVA at 480 volts, and the second transformer rated 225 kVA be changed to 250 KVA at 208 volts, calculate the fault at X1 and fault at X2. Assume all other data are the same.

Point-to-point
Available
utility
500,000 SC KVA was
30-500 kamil
4 per phase Cu
in pvc londuit
Fault X
ONE LINE DIAGRAM
mm 1000 KVA x met
4804, 30,3.5%2
IfL = 1203 A
1600 A Switch
1600 A Fuse
400A switch
400A Fuse
20¹2/0
2 per phase Cu.
in PVC conduit wou
mm
225 KVA xmer
208V, 3-4
1-2%2
Fault XL
Step 0:
IFL= 1000 x 1000 = 1203 A
480 X 1.732
100
step@
multipliet:
=
28.57
3.5
step Isc = 1202 x 28.57 = 34,370 A
Ⓒ
step f
= 0.0348
= 1.732 X30 x 34-370
4 x 26 704 x 480
1
Step M=
= 0.09664
It 0.0348
Step Ⓒ Isc sum Rm 5² 34 370 × 0.09664
= 33215A /
Fault X2
Step 4:
1:
(30)
f=
Hep: M=
1.732 x 20 x 33 215.
2 x 11423 X480
1
1+0.1049
= 0.905
Step Ⓒ: Ise Sum Rins
= 33215 x0.905
= 30059 A
•=0.1049
Transcribed Image Text:Point-to-point Available utility 500,000 SC KVA was 30-500 kamil 4 per phase Cu in pvc londuit Fault X ONE LINE DIAGRAM mm 1000 KVA x met 4804, 30,3.5%2 IfL = 1203 A 1600 A Switch 1600 A Fuse 400A switch 400A Fuse 20¹2/0 2 per phase Cu. in PVC conduit wou mm 225 KVA xmer 208V, 3-4 1-2%2 Fault XL Step 0: IFL= 1000 x 1000 = 1203 A 480 X 1.732 100 step@ multipliet: = 28.57 3.5 step Isc = 1202 x 28.57 = 34,370 A Ⓒ step f = 0.0348 = 1.732 X30 x 34-370 4 x 26 704 x 480 1 Step M= = 0.09664 It 0.0348 Step Ⓒ Isc sum Rm 5² 34 370 × 0.09664 = 33215A / Fault X2 Step 4: 1: (30) f= Hep: M= 1.732 x 20 x 33 215. 2 x 11423 X480 1 1+0.1049 = 0.905 Step Ⓒ: Ise Sum Rins = 33215 x0.905 = 30059 A •=0.1049
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