Electrical Wiring: Residental - With Plans (Paperback) Package
18th Edition
ISBN: 9781305416376
Author: MULLIN
Publisher: Cengage
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Textbook Question
Chapter 28, Problem 10R
Using the method shown in this unit, what is the approximate short-circuit current available at the terminals of a 50-kVA single-phase transformer rated 120/240 volts? The transformer impedance is 1%.
- a. Line-to-line
- b. Line-to-neutral
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(a) A 5 MVA 230/11.5-kV single-phase power transformer has a per-unit resistance of 1.5%
and a per-unit reactance of 6% (data taken from the transformer's nameplate). The
transformer approximate equivalent circuit is given in the figure below. The open-circuit
test performed on the low-voltage side of the transformer yielded the following data:
Voc = 120 kV
loc = 12 A
Poc = 40 kW
Refer to the low-voltage side of the transformer and calculate:
(i)
The value of Req and Xeq.
(ii)
The value of Rc and Xm.
(ii)
The voltage regulation when secondary voltage Vs = 11 kV and the secondary
current Is = 350 A.
kIp
Reg
jXcq
1,
Vp
Re
V.
(a) A 5 MVA 230/11.5-kV single-phase power transformer has a per-unit resistance of 1.5%
and a per-unit reactance of 6% (data taken from the transformer's nameplate). The
transformer approximate equivalent circuit is given in the figure below. The open-circuit
test performed on the low-voltage side of the transformer yielded the following data:
Voc = 120 kV
loc = 12 A
Poc = 40 kW
Refer to the low-voltage side of the transformer and calculate:
(i)
The value of Req and Xeq.
(ii)
The value of Rc and Xm.
(ii)
The voltage regulation when secondary voltage Vs = 11 kV and the secondary
current Is = 350 A.
kIp
Roa
jXe
Re
Belonging to a phase measured from a 220/48 V phase transformer, no-load (no load)
operation and short-circuit test.
values are given in the table below. Low voltage by applying rated voltage to the high
voltage (primary) winding.
voltage (secondary) winding ends were left unloaded and the results of the no-load
operation test were obtained. low voltage
By applying nominal current to the windings, the high voltage windings are short-circuited
and short-circuit test results are obtained.
has been done. In addition, the transformer with 5 V direct voltage applied to the high
voltage winding is 2.2 A direct current.
attracts. Calculate all equivalent circuit elements moved to the high voltage side.
Boşta çalışma testi
Gerilim
220 V
Güç
80 W
Akım
0,44 A
Kısa devre testi
Gerilim
12 V
Güç
Akım
65 W
16,1 A
Chapter 28 Solutions
Electrical Wiring: Residental - With Plans (Paperback) Package
Ch. 28 - What is the minimum size service for a one-family...Ch. 28 - Prob. 2RCh. 28 - Prob. 3RCh. 28 - Prob. 4RCh. 28 - Prob. 5RCh. 28 - Prob. 6RCh. 28 - Prob. 7RCh. 28 - List the standard sizes of fuses and circuit...Ch. 28 - Prob. 9RCh. 28 - Using the method shown in this unit, what is the...
Ch. 28 - State four possible combinations of service...Ch. 28 - Which Code section states that all overcurrent...Ch. 28 - All electrical components have some sort of...Ch. 28 - Arc-fault damage is closely related to the value...Ch. 28 - The utility company has provided a letter to the...Ch. 28 - While working on the main panelboard that is...Ch. 28 - Repeat problem 16 for a main service panel that...Ch. 28 - NEC ______________________________ generally...Ch. 28 - Prob. 19RCh. 28 - Prob. 20RCh. 28 - Overcurrent devices must be accessible. (True)...
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- 2. What type of potential transformer would you use for a 362-kV grounded-neutral metering application on an interconnection system? A. Bar-type B. Cascade-type C. Window-type D. Phase-shift-typearrow_forwardFind the elements of the equivalent circuit relative to the secondary of a single-phase transformer. The no-load voltages in the primary and secondarypement are 200 and 400 volts, respectively, with a trend of 50 Hz. The no-load current at 2, 50. Voltage 200 volts is equal to 0.7 amperes, and the cores are equal to 70 watts, 1.25 ohms, respectively, the resistance and reactance of the secondary 0.6, 7... Ohms, respectively. Draw the assigned circle Explaining the values of its elements.arrow_forward3.14 A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a step-down transformer at the load end of a 2400-volt feeder whose series impedance is (1.0 + j2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the high- voltage (primary) side. The transformer is delivering rated load at a 0.8 power factor lagging and at a rated secondary voltage. Neglecting the transformer exciting current, determine (a) the voltage at the transformer primary terminals, (b) the voltage at the sending end of the feeder, and (c) the real and reactive power delivered to the sending end of the feeder.arrow_forward
- 17.a. For commercial tests on transformers, direct current may be used for theA. induced-voltage test.B. ratio test.C. temperature test.D. polarity kick back test.17.b. A core-form three-phase delta-delta transformer can provide some service with one phase damaged ifthe damaged phase isA. grounded.B. disconnected and short-circuited.C. open-circuited.D. short-circuited.arrow_forward(a) A 5 MVA 230/10.5-kV single-phase power transformer has a per-unit resistance of 1.5% and a per-unit reactance of 6% (data taken from the transformer's nameplate). The transformer approximate equivalent circuit is given in Figure Q1 (a). The open-circuit test performed on the low-voltage side of the transformer yielded the following data: Voc = 120 kV loc = 10 A Poc = 60 kW %3D Refer to the low-voltage side of the transformer and calculate: (i) The value of Req and Xeq. (i) The value of Rc and Xm. (ii) The voltage regulation when secondary voltage Vs = 11 kV and the secondary %3D current Is = 300 A. %3D 1, Rea jXcu Re Figure Q1 (a)arrow_forwardProblem # 2 A 320-kVA, 240/4800-V, 60 Hz transformer yield the following data from short-circuit test and open-circuit- test measurements. Voltage (V) Power (W) 1200 Current (A) Open-Circuit Test 240 39.5 Short-Circuit Test 195 66.67 3925 Determine the transformer equivalent circuit parameters. Note: Short-circuit test will determine the Reg and Xeq and open-circuit test will determine the Re and Xm.arrow_forward
- answerarrow_forwardPlease solution speedarrow_forwardThree single-phase, 50 kVA, 2300/230 V, 60 Hz transformers are connected to form a three-phase, 4000/230 V transformer bank. The equivalent impedance of each transformer referred to low voltage is 0.04+j0.08 ohms. The three-phase transformer supplies a three-phase, 130 kVA, 230 V, 0.63 lagging-power-factor load. (a) Determine the transformer winding currents. (b) Determine the primary voltage (line-to-line) required. (c) Determine the voltage regulation. (a) Current in primary winding: A (b) Current in secondary winding: A (c) Primary Voltage (line to line): V (d) Voltage Regulation: %arrow_forward
- What are the objectives of conducting an open circuit test on single phase transformer?arrow_forwardCALCULATE REACTANCE( NEED ONLY HANDWRITTEN SOLUTION PLEASE OTHERWISE DOWNVOTE).arrow_forwardThe actual equivalent circuit parameters of a single-phase transformer operating at 220/110 volts are given as follows. R1=0.2 ohm, X1=0.3ohm, RFe=45 kohm, Xm=5 kohm, R2=0.025ohm, X2=0.075 ohm Using the L-equivalent circuit, a) reduce the secondary circuit parameters to the primary b) Since a reduced load of 44.72+j32 ohms is connected on the secondary side, calculate the primary current, the load current, and the total copper losses.arrow_forward
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