[DC machine with armature reaction] A 300kW, 240V, 900rpm separately excited, non- compensated DC generator is shown in Figure 3(a). The generator has parameters: Ra = 0.002342, Rmerpole = 0.00082, R, = 18.12. The shunt field has 1020 turns/pole, separately excited by DC supply of 120V through a rheostat, Ry. The equivalent demagnetizing ampere-turns caused by armature reaction, when supplying rated load at rated voltage is assumed equal to 12.1% of the shunt field mmf. The magnetization curve is shown in figure 3(b). R, = 18.10 ER=0.002340 V, = 120V Load Ruempole = 0.00082 Figure 3(a) Separately excited DC generator Determine:- a) No load voltage b) Voltage regulation c) Rheostat resistance, R, to obtain rated conditions.

[DC machine with armature reaction] A 300kW, 240V, 900rpm separately excited, non- compensated DC generator is shown in Figure 3(a). The generator has parameters: Ra = 0.002342, Rmerpole = 0.00082, R, = 18.12. The shunt field has 1020 turns/pole, separately excited by DC supply of 120V through a rheostat, Ry. The equivalent demagnetizing ampere-turns caused by armature reaction, when supplying rated load at rated voltage is assumed equal to 12.1% of the shunt field mmf. The magnetization curve is shown in figure 3(b). R, = 18.10 ER=0.002340 V, = 120V Load Ruempole = 0.00082 Figure 3(a) Separately excited DC generator Determine:- a) No load voltage b) Voltage regulation c) Rheostat resistance, R, to obtain rated conditions.

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

Chapter12: Power System Controls

Section: Chapter Questions

Problem 12.4P

See similar textbooks

Related questions

Q: I need the complete solution.

A: Step 1: Solution(7) : **(i) Lap Fashion (Winding)**1. Determine the following formula for DC machine…

Q: A 4 pole d.c. The motor has a wave-wound armature winding. The value of flux per pole is 60 mWb. The…

A: We have an DC motor with wave-wound armature. We have been asked to find The speed of this motor and…

Q: Two identical 2,000 kW alternator operates in parallel. The governor of the first machine is such…

A: It is given that: Rating of alternator=P=2000 KWFrequency drop of first machine 1=∆f1=50 Hz-48 Hz=2…

Q: In a DC generator, the effect of armature reaction on the main field is to * reverse it O distort it…

A: - When current flows through the armature conductors, the flux is set up in the air gap. - The flux…

Q: Suppose we have an ac current flowing through a coil wound on an iron core. Name two mechanisms by…

A: The two mechanisms in which energy is converted to heat in the core material are: 1) Core loss or…

Q: Derive the EMF and Torque equations of DC machines from the first principles.

A: The E.M.F equation of the DC machine is to be derived using the operating principle. Let us consider…

Concept explainers

DC Generator

An electromechanical machine which converts input mechanical energy to respective output electric energy is called a generator.

Question

100%

Magnetization curve 2
280
240
200
160
120
80
40
1
2
3
5
Magnetomotive Force, mmf (A-t/pole x1000)
Figure 3(b) Magnetizing curve of non-compensated DC machine
Generated EMF(V)

[DC machine with armature reaction]
A 300kW, 240V, 900rpm separately excited, non- compensated DC generator is shown in Figure
3(a). The generator has parameters: Ra = 0.002342, Rmerpole = 0.00082, R, = 18.12. The
shunt field has 1020 turns/pole, separately excited by DC supply of 120V through a rheostat, Ry.
The equivalent demagnetizing ampere-turns caused by armature reaction, when supplying rated
load at rated voltage is assumed equal to 12.1% of the shunt field mmf. The magnetization curve
is shown in figure 3(b).
R, = 18.10-
R=0.002340
V, = 120V
Load
Roempole =
0.00082
Figure 3(a) Separately excited DC generator
Determine:-
a) No load voltage
b) Voltage regulation
c) Rheostat resistance, R, to obtain rated conditions.

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 by step

Solved in 2 steps with 1 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

The resistance of the field winding(copper wire) of a DC machine is 0.25 ohm at 25°C. When operating at full-load, the temperature of the winding is 75 °C. The temperature coefficient of resistance of copper is 0.00427 per °C at 0°C. Find the resistance of the field winding at full-load.
A lap-winding DC machine has 24 coils, 18 turns per coil, and 4 poles. The average flux density per pole is 1T. The effective length is 40 cm and the armature radius is 10 cm. The armature velocity is 140.2 rad/sec and the armature current is 24 A. The average flux per pole in Wb is: 6.283 0.06283 O 0.6283 0.006283
(a) Explain briefly what is fringing effect and leakage flux. (b) A ferromagnetic core is shown in Figure 1. The depth of the core is 10 cm. The other dimensions of the core are as shown in the figure. By assuming the magnetic leakage to be negligible and relative permeability of the core is 800. Calculate the total reluctance and find the value of the current that will produce a flux of 0.01 Wb. With this current, calculate the flux density at the top of the core 20 cm 40 cm 10 cm, 30 cm 500 turns 30 cm 30 cm Figure 1
A 220 V , 60 Hz , supply is connected to 200 – turn armature winding of reluctance 1.7x105 At / Wb . Find : the magnetizing reactance the rms values of the ( magneto – motive force , exciting and armature currents , magnetic fluxflu the inductance Draw the phasor diagram of the circuit
A magnetic circuit with a cross-sectional area of 20 cm? is to be operated at 50 Hz from a 120 V rms supply. The number of turns required to achieve a peak magnetic flux density of 1.8 T in the core is
2. A six-pole lap wound armature rotating at 350 rpm is required to generate 260 V. The effective flux per pole is about 0.05 Wb. If the armature has 120 slots, determine the suitable number of conductors per slot and hence determine the actual value of flux required to generate the same voltage.
In DC machines, are made from laminated steel or low carbon steel and used to reduce the armature reaction.
For operation at 60 Hz and a given peak flux density, the core loss of a given core is 1W due to hysteresis and 0.5W due to eddy currents. Estimate the core loss for 400-Hz operation with the same peak flux density.
DC MACHINE TORQUE The armature of a dc motor has 648 conductors, 65 % of which are directly under thepole where the flux density is 48,000 lines per in2. If the core diameter is 7 inchesand its length is 4 inches and the current in each conductor is 20 A. What is thetorque exerted in the armature in N-m?
3Q) Analyze the relationship between flux produced by the moving coil and torque in a PMMC instrument.
I need the complete solution.
Dc machine