Homework 6
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Transcript of Homework 6
![Page 1: Homework 6](https://reader035.fdocuments.in/reader035/viewer/2022081211/563db7dc550346aa9a8ea7f5/html5/thumbnails/1.jpg)
2-3. A 1000-VA 230/115-V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below.
Open-circuit test Short-circuit test VOC = 230 V VSC = 19.1 V IOC = 0.45 A ISC = 8.7 A POC = 30 W PSC = 42.3 W
All data given were taken from the primary side of the transformer. (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer. (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging.
SOLUTION
(a) OPEN CIRCUIT TEST:
1 1
0.45 0.001957230
30cos cos 73.15(230)(0.45)
0.001975 73.15 0.000567 0.0018731/ 17631/ 543
EX C M
oOC
OC OC
oEX C M
C C
M M
Y G jB S
PV I
Y G jB jR GX B j
θ − −
= − = =
= = =
= − = ∠− = − Ω
= = Ω= = Ω
SHORT CIRCUIT TEST:
1 1
19.1 2.28.7
42.3cos cos 75.3(19.1)(8.7)
2.20 75.3 0.558 2.128
0.558
2.128
EQ EQ EQ
oSC
SC SC
oEQ EQ EQ
EQ
EQ
Z R jX
PV I
Z R jX j
R
X j
θ − −
= − = = Ω
= = =
= + = ∠ = + Ω
= Ω
= Ω
To convert the equivalent circuit to the secondary side, divide each impedance by the square of the turns ratio (a = 230/115 = 2). The resulting equivalent circuit is shown below:
REQ,s = 0.532 Ω X EQ,s = j441 Ω RC,s = 134 Ω XM,s = 134 Ω
(b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is
1000 /115 8.70SI A= = We will now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor. (1) 0.8 PF Lagging:
' 115 0 (0.140 0.532 )(8.7 36.87 )
' 118.8 1.4
o oP S EQ S
oP
V V Z I j
V
= + = ∠ + + Ω ∠−
= ∠
![Page 2: Homework 6](https://reader035.fdocuments.in/reader035/viewer/2022081211/563db7dc550346aa9a8ea7f5/html5/thumbnails/2.jpg)
118.8 115 100% 3.3%115
VR −= × =
(2) 1.0 PF:
' 115 0 (0.140 0.532 )(8.7 0 )
' 116.3 2.28116.3 115 100% 1.1%
115
o oP S EQ S
oP
V V Z I j
V
VR
= + = ∠ + + Ω ∠
= ∠−
= × =
(3) 0.8 PF Leading:
' 115 0 (0.140 0.532 )(8.7 36.87 )
' 113.3 2.24113.3 115 100% 1.5%
115
o oP S EQ S
oP
V V Z I j
V
VR
= + = ∠ + + Ω ∠
= ∠−
= × = −
(c) At rated conditions and 0.8 PF lagging, the output power of this transformer is
cos (115)(8.7)(0.8) 800OUT S SP V I Wθ= = =
The copper and core losses of this transformer are 2 2
,
2 2
(8.7) (0.140) 10.6
( ') (118.8) 32.0441
CU S EQ S
pcore
C
P I R W
VP W
R
= = =
= = =
Therefore the efficiency of this transformer at these conditions is
800100% 100% 94.9%800 10.6 32.0
OUT
OUT CU core
PP P P
η = × = × =+ + + +