A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. » Single Phase TransFarmer -> S=ASOkva -> TUIA Ratio n= -> %3D 60 4[60 -> Primary winding VP = 33000v, > Fregvancy f = 50 Hz a) we Know Vs ニ So Vs = 4Vp = 4x33000 = 132000V ã 132KV gining secondary vatage (Vs) = 132 kp b) Primary siole エpニア CUsrent 4 SO10-3 ニ VP = 13.6363A 33, secoundary Side cullent. 4S016-3 132 10-3 エs Is = V = 3.409A
A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. » Single Phase TransFarmer -> S=ASOkva -> TUIA Ratio n= -> %3D 60 4[60 -> Primary winding VP = 33000v, > Fregvancy f = 50 Hz a) we Know Vs ニ So Vs = 4Vp = 4x33000 = 132000V ã 132KV gining secondary vatage (Vs) = 132 kp b) Primary siole エpニア CUsrent 4 SO10-3 ニ VP = 13.6363A 33, secoundary Side cullent. 4S016-3 132 10-3 エs Is = V = 3.409A
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 3.9P
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Are my Calculations on these two questions correct? I feel like I'm off somewhere?
![A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to
33000V, 50HZ. supply. Calculate:
a. the secondary voltage on no load
b. the approximate values of the primary and secondary currents on full load.
-» Single Phase Transfaramer
-> S= ASOkva
-> Tun Ratio n=
N2
->
60
-> Primary winding VP = 33000V,
-> Freguancy
{ = 50 Hz
%3D
Vp
a) we Know
Vs
So Vs = 4 Vp = 4x33000 = 132000V % 132 Kv
%3D
giving secondary vortage (Vs) = 132 kv
b) Primary sSile
Ipニ。
CUsrent
4SO103
= 13.6363A
VP
33,03
secondeery Side cullet.
45016-3
132 10-3
Is =
= 3.409A](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e004b1d-cdd8-4674-8e71-c95051c592ae%2F10d18ed5-f6e4-46b0-8c7e-2239b1f3f1f0%2Flny6vwf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to
33000V, 50HZ. supply. Calculate:
a. the secondary voltage on no load
b. the approximate values of the primary and secondary currents on full load.
-» Single Phase Transfaramer
-> S= ASOkva
-> Tun Ratio n=
N2
->
60
-> Primary winding VP = 33000V,
-> Freguancy
{ = 50 Hz
%3D
Vp
a) we Know
Vs
So Vs = 4 Vp = 4x33000 = 132000V % 132 Kv
%3D
giving secondary vortage (Vs) = 132 kv
b) Primary sSile
Ipニ。
CUsrent
4SO103
= 13.6363A
VP
33,03
secondeery Side cullet.
45016-3
132 10-3
Is =
= 3.409A
![A 15KVA single-phase transformer, for 10000V/440V at no load, has resistances and leakage reactance
as follows. Primary winding: resistance 42, reactance 162. Secondary winding: resistance 0.152,
reactance 0.45Q.
Determine the approximate value of the secondary voltage at full load, 0.85 power factor (lagging),
when the primary supply voltage is 10000V.
Ignore core resistance and magnetization reactance.
-> Singie Phase 1runs Folme
-> S='15kva
→ 10'000 /44OV
Is
R,
Jx,
0.15 JO.45
Vs
2
(440
R, =
100001
x 4 = 7.744,
°440
=(42)x 16= 0030976 r
l000.
440
V =
X 10000 = 440V
10000
If Power Factes Of Load =
085(1aging)
Ø = Cos Co85) = 31.7883°
ISo00
at Full Load curent |Isd =SI
=34.09A
440
So Is = 34.09<-31.788°A
giving,
Vs = Vp'- (R,+o.15+j(x,+o.45) Iş
* 440-(>,74,340.15+jco.03097640.45) 34.094 - 36.56°
to.-
giving Secondaty Side vartage =425.974 <-1.3}'](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e004b1d-cdd8-4674-8e71-c95051c592ae%2F10d18ed5-f6e4-46b0-8c7e-2239b1f3f1f0%2F4f6gy28_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 15KVA single-phase transformer, for 10000V/440V at no load, has resistances and leakage reactance
as follows. Primary winding: resistance 42, reactance 162. Secondary winding: resistance 0.152,
reactance 0.45Q.
Determine the approximate value of the secondary voltage at full load, 0.85 power factor (lagging),
when the primary supply voltage is 10000V.
Ignore core resistance and magnetization reactance.
-> Singie Phase 1runs Folme
-> S='15kva
→ 10'000 /44OV
Is
R,
Jx,
0.15 JO.45
Vs
2
(440
R, =
100001
x 4 = 7.744,
°440
=(42)x 16= 0030976 r
l000.
440
V =
X 10000 = 440V
10000
If Power Factes Of Load =
085(1aging)
Ø = Cos Co85) = 31.7883°
ISo00
at Full Load curent |Isd =SI
=34.09A
440
So Is = 34.09<-31.788°A
giving,
Vs = Vp'- (R,+o.15+j(x,+o.45) Iş
* 440-(>,74,340.15+jco.03097640.45) 34.094 - 36.56°
to.-
giving Secondaty Side vartage =425.974 <-1.3}'
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