A 260 kVA transformer has a primary winding resistance of 0.35 Q and a secondary winding resistance of 0.002 Q. The iron loss is 7.5 kW and the primary and secondary voltages are 2 kV and 290 V respectively. If the power factor of the load is 0.6,then determine the efficiency of the transformer on 25 % load. 11 12 Pc Efficiency

A 260 kVA transformer has a primary winding resistance of 0.35 Q and a secondary winding resistance of 0.002 Q. The iron loss is 7.5 kW and the primary and secondary voltages are 2 kV and 290 V respectively. If the power factor of the load is 0.6,then determine the efficiency of the transformer on 25 % load. 11 12 Pc Efficiency

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.9P

See similar textbooks

Related questions

Q: Determine Vg, given Vin=25V, V1= - 8V, R;=3k0, R2=5k0 For Ge VK= 0.3 V For Si 0.7 V Ge R1 Si V:. Vo…

A: Find the voltage Vo All diode forward bias voltage given.

Q: 12. Find the resistance of a piece of 12 AWG aluminum wire that is 550 ft long. Assume a temperature…

A: Given data: 12 AWG aluminum wire Length l=550 ft Temperature 20° C

Q: Power Triangle

Q: How did they o from SNRDAE go e CSNR)aR=10 log, CONR) waltker 1.8 (SNR)w266Eio To where is 63 from?…

Q: . Explain the environmental impacts of Hydropower plant?

Q: 14.) True or False: A load with low power factor draws less current than a load with a high power…

A: A load with low power factor draws high current than a load with high power factor for the same…

Q: Calculate the value of V1 and V2?

Q: b. three The allowable ampacity of a 4 AWG THHN conductor is 95 amperes. If we have four 4 AWG THHN…

Q: (d) For the circuit shown in Figure Q1(d), the inductors have the following value: L1 = 1 mH, L2 =…

A: In this question we need to find a equivalent inductance between terminal ab and bc

Q: VDD M1 M2 IIN IOUT

Concept explainers

KVL and KCL

KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.

Sign Convention

Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.

Question

A 260 kVA transformer has a primary
winding resistance of 0.35 Q and a
secondary winding resistance of 0.002 Q.
The iron loss is 7.5 kW and the primary and
secondary voltages are 2 kV and 290 V
respectively. If the power factor of the load
is 0.6,then determine the efficiency of the
transformer on 25 % load.
11
12
Pc
Efficiency

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions

For a short-circuit test on a 2-winding transformer, with one winding shorted, can you apply the rated voltage on the other winding? (a) Yes (b) No
An ideal transformer has no real or reactive power loss. (a) True (b) False
A 23/230-kV step-up transformer feeds a three-phase transmission line, which in turn supplies a 150-MVA,0.8 lagging power factor load through a step-down 230/23-kV transformer. The impedance of the line and transformers at 230kVis18+j60. Determine the tap setting for each transformer to maintain the voltage at the load at 23 kV.
A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.
For an ideal transformer, the efficiency is (a) 0 (b) 100 (c) 50
The symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.
Consider an ideal transformer with N1=3000andN2=1000 turns. Let winding 1 be connected to a source whose voltage is e1(t)=100(1| t |)volts for 1t1ande1(t)=0 for | t |1 second. A2- farad capacitor is connected across winding 2. Sketch e1(t),e2(t),i1(t),andi2(t) versus time t.
The ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.
A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
For an ideal 2-winding transformer, an impedance Z2 connected across winding 2 (secondary) is referred to winding 1 (primary) by multiplying Z2 by (a) The turns ratio (N1/N2) (b) The square of the turns ratio (N1/N2)2 (c) The cubed turns ratio (N1/N2)3
Rework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at (a) unity power factor, (b) 0.8 power factor leading. Compare the results with those of Problem 3.14. -
For the transformer in Problem 3.10. The open-circuit test with 11.5 kV applied results in a power input of 65 kW and a current of 30 A. Compute the values for GcandBm in siemens referred to the high-voltage winding. Compute the efficiency of the transformer for a load of 10 MW at 0.8 p.f. lagging at rated voltage.