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.
Practice1
A single-phase step-down transformer of 83 kVA, nominal voltages 24kV/230 V, frequency 60 Hz is available.The following test parameters are available:Pfe = 216 W, Io = 2% Pcc = 1083 W, Vcc = 4%
Determine:a. Parameters Rcc, Xcc and Rfe of the equivalent circuit referring to the secondary.b. Relative voltage drops. εcc, εrcc, εxcc
A single-phase step-down transformer of 83 kVA, nominal voltages 24kV/230 V, frequency 60 Hz is available.The following test parameters are available:Pfe = 216W, Io = 2%, Pcc = 1083W, Vcc = 4%
Determine:
If the transformer is connected to 24 kV, a load Zc, fp = 0.866 in arrears, is installed in the secondary transformer, which consumes the nominal current. Calculate:• Transformer voltage regulation (perform calculations by PU's)• Maximum efficiency.
The magnetic circuit shown in the figure is made of TRAN-COR material, the flow magnetic power on the right arm (BCDE) is 6 x 10 -4 Wb. (disregard marginal effects anddispersion)
Calculate the current in the 200-turn coil
Chapter 3 Solutions
Power System Analysis and Design (MindTap Course List)
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