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.
Cable A
Cable A is a coaxial cable of constant cross section. The metal
regions are shaded in grey and are made of copper. The solid central
wire has radius a = 5mm, the outer tube inner radius b = 20mm and
thickness t = 5mm. The dielectric spacer is Teflon, of relative
permittivity &r = 2.1 and breakdown strength 350kV/cm. A potential
difference of 1kV is applied across the conductors, with centre
conductor positive and outer conductor earthed.
Before undertaking any COMSOL simulations we'll first perform some theoretical analysis
of Cable A based on the EN2076 lectures, to make sense of the simulations. Calculate the
radial electric field of cable A at radial positions r b. Also calculate the
maximum operating voltage of cable A, assuming a safety margin of ×2, and indicate where
on the cable's cross section dielectric breakdown is most likely to occur.
: For the gravity concrete dam shown in the figure, the following data are available:
The factor of safety against sliding (F.S sliding)=1.2
Unit weight of concrete (Yconc)=24 KN/m³
- Neglect( Wave pressure, silt pressure, ice force and earth quake force)
μ=0.65, (Ywater) = 9.81 KN/m³
Find factor of safety against overturning (F.S overturning)
6m3
80m
Sm
I need help checking if its correct
-E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a)
R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b)
R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c)
(R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d)
Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be:
E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a)
-R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b)
-R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c)
R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d)
These two equations will be implemented to the matrix formula I = inv(A) * b
R11 R12
(R1 + R3 + R4)
-R4
-R4
R4 + R2
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