Download Example 1 - UniMAP Portal

Electrical Technology (EET 103/ERT105)
Tutorial Module
TUTORIAL 5 – TRANSFORMER
Question 1
The secondary winding of a transformer has a terminal voltage of vs (t) = 282.8
sin 377t V. The turns ratio of the transformer is 100:200. If the secondary current
of the transformer is (t) =7.07 sin (377t - 36.87o) A, the impedances of this
transformer referred to the primary side are:
Req = 0.20Ω
Xeq = 0.750Ω
Rc = 300Ω
XM = 80Ω
Determine:
i.
ii.
the primary current of this transformer.
the voltage regulation and efficiency.
(Ans: I p  11.1  41o A , VR = 6.2%, η = 93.7%)
Question 2
A 20-kVA 8000/480-V distribution transformer has the following resistances and
reactances:
Rp = 32Ω
Rs = 0.05Ω
Xp = 45Ω
Xs = 0.06Ω
RC = 250kΩ XM = 30kΩ
The excitation branch impedances are given referred to the primary side of the
transformer.
a) Find the equivalent circuit of this transformer referred to the primary side.
b) Assume that this transformer is supplying rated load at 480V and 0.8 PF
lagging. What is this transformer’s input voltage, Vp? What is its voltage
regulation, VR?
c) What is the transformer’s efficiency, η under the conditions of part (c)?
(Ans: a) Rs’ = 13.9Ω, Xs’ = 16.7Ω b) Rp = 0.01, Xp = j0.0141, Rs’ = 0.0043, Xs’ =
j0.0052, RC = 78.125, XC = j9.375 c) V p  81850.38o V , VR = 2.31% d) η =
96.6%)
UNIVERSITI MALAYSIA PERLIS
1
Electrical Technology (EET 103/ERT105)
Tutorial Module
Question 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
VOC = 230V
IOC = 0.45A
POC = 30W
Short circuit test
VSC = 19.1V
ISC = 8.7A
PSC = 42.3W
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.
(Ans: a) Req,s =0.140Ω, Xeq,s = j0.532Ω, RC,s =441Ω, XM,s = 134Ω b) VR = 3.3%,
1.1%, -1.5%, η = 94.9%)
UNIVERSITI MALAYSIA PERLIS
2
Electrical Technology (EET 103/ERT105)
Tutorial Module
Question 4
The results of the Open circuit and Short circuit tests are shown in Table 2 for a
20 kVA, 1900/240 V transformer. This test is to determine its excitation branch
components, its series impedances and its voltage regulation.
Open circuit
VOC = 1900V
IOC = 0.45A
POC = 60W
Table 2
Short circuit
VSC = 50V
ISC = 9A
PSC = 250W
a) Calculate the approximate equivalent circuit of this transformer referred to
its primary voltage side.
b) Calculate the approximate equivalent circuit of this transformer referred to
its secondary voltage side.
c) Calculate the full load voltage regulation at Power Factor 0.8 lagging.
d) Calculate the efficiency of the transformer at full load with a Power Factor
0.8 lagging.
UNIVERSITI MALAYSIA PERLIS
3
Electrical Technology (EET 103/ERT105)
Tutorial Module
Question 5
A 15-kVA 8000/230V distribution transformer has impedance referred to the
primary of 80 + j300 Ω. The components of the excitation branch referred to the
primary side are RC = 350kΩ and XM = 70kΩ.
a) If the primary voltage is 7967 V and the load impedance is ZL = 3.0 +
j1.5Ω, what is the secondary voltage of the transformer? What is the
voltage regulation of the transformer?
b) If the load is disconnected and a capacitor of –j4.0Ω is connected in its
place, what is the secondary voltage of the transformer? What is its
voltage regulation under these conditions?
(Ans: (a) Vs  218.8  3.1o V , VR = 4.7% (b) Vs  246.5  1.2 o V , VR = -7.07%)
UNIVERSITI MALAYSIA PERLIS
4
Electrical Technology (EET 103/ERT105)
Tutorial Module
Question 6
UNIVERSITI MALAYSIA PERLIS
5