A single-phase l 0 -kVA , 23 00 /23 0 -volt , 6 0 -Hz two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage = 23 0 V , input current = 0. 45A , input power = 7 0 W . Short-circuit test with the high-voltage terminals excited: Applied voltage = 12 0 , input current = 4 . 5A , input power = 24 0 W . Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.
A single-phase l 0 -kVA , 23 00 /23 0 -volt , 6 0 -Hz two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage = 23 0 V , input current = 0. 45A , input power = 7 0 W . Short-circuit test with the high-voltage terminals excited: Applied voltage = 12 0 , input current = 4 . 5A , input power = 24 0 W . Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.
Solution Summary: The author illustrates the schematic diagram when delivering full load at rated voltage. The primary voltage value remains the same while making an autotransformer of specified rating.
A single-phase
l
0
-kVA
,
23
00
/23
0
-volt
,
6
0
-Hz
two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer:
Open-circuit test with the low-voltage terminals excited:
Applied voltage
=
23
0
V
, input current
=
0.
45A
, input power
=
7
0
W
.
Short-circuit test with the high-voltage terminals excited:
Applied voltage
=
12
0
, input current
=
4
.
5A
, input power
=
24
0
W
.
Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.
The current coil of a wattmeter is connected in the red
line of a three-phase system. The voltage circuit can be
connected between the red line and either the yellow
line or the blue line by means of a two-way switch.
Assuming the load to be balanced, show with the aid
of a phasor diagram that the sum of the wattmeter
indications obtained with the voltage circuit connected
to the yellow and the blue lines respectively gives the
total active power.
A wattmeter has its current coil connected in the yellow
line, and its voltage circuit is connected between the
red and blue lines. The line voltage is 400 V and the
balanced load takes a line current of 30 A at a power
factor of 0.7 lagging. Draw circuit and phasor diagrams
and derive an expression for the reading on the wattmeter
in terms of the line voltage and current and of the phase
difference between the phase voltage and current.
Calculate the value of the wattmeter indication.
ANS:
. Line amperes × line volts × sin φ = 8750 var
4. The circuit shown below shows an infinite impedance (open circuit) in phase B
of the Y-connected load. Find the phasor voltage VOB if the system is 208 V,
sequence ABC.
-j100 Q
100 Ω
B
5. Three identical impedances of Z = 15260°2 are connected in Y to a three-phase,
three-wire, 240 V, ABC system. The lines between the supply and the load have
impedances of 2 +j 1 Q2. Find the line voltage magnitudes at the load. Find the
new values when a set of capacitors with reactance of -j10 Q (Y-connection) is
connected in parallel with the load. Draw the vector diagram for the load current,
the capacitor current and the system line current.
Chapter 3 Solutions
Power System Analysis and Design (MindTap Course List)
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