a) A six-pole, lap-wound, 220 V, shunt-excited d.c. machine takes an armature currentof 2.5 A when unloaded at 950 r/min. When loaded, it takes an armature current of54 A from the supply and runs at 950 r/min. The resistance of the armature circuit is0.18 and there are 1044 armature conductors. For the loaded condition, calculate:b) Use the information provided in Part (a) above to determine:i.The constant loss, CThe output power when the machine is loaded.ii.iii.The output torque when the machine is loaded.Hint: Draw the power flow diagram.

In this diagram:The Voltage Source represents the 220V DC power supply.The Armature Circuit…

Answered: a) A six-pole, lap-wound, 220 V,…

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

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5. In relation to a compound d.c machine, choose the right statement from the following. (A) The resistance of the shunt field coil will be many times higher than the series field coil. (B) The resistance of the shunt field coil and the series field coil are of the same order. (C) The resistance of the series field coil will be many times higher than the shunt field coil. (D) Either of (A), (B) and (C) is true statement.
a) A six-pole, lap-wound, 220 V, shunt-excited d.c. machine takes an armature current of 2.5 A when unloaded at 950 r/min. When loaded, it takes an armature current of 54 A from the supply and runs at 950 r/min. The resistance of the armature circuit is 0.18 S2 and there are 1044 armature conductors. For the loaded condition, calculate: the induced e.m.f. CS Scanned with CamScanner i. ii. the useful flux per pole. iii. the torque developed by the machine in Nm.
A thicker wire is used in the DC series field winding than DC shunt field winding in DC machines. * O because it carries the load current. O to prevent mechanical vibrations. O to provide strength. O to produce large flux. The speed of a DC motor can be changed by varying - armature resistance circuit O flux per pole O voltage source O all of above
I:01)
I need the complete solution.
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.
Two shunt generations each having an armature resistance of 0.05 ohm and field resistance 40 ohms respectively are connected in parallel. The e.m.f include in one machine is 250 V and that induced in the other machine is 260 V. they supply together a load current of 2000 A. Determine:(i) Terminal voltage; and(ii) Output of each machine.
The armature circuit of a free excited direct current motor with a nominal voltage of 600 V is fed from an adjusted direct voltage source. The excitation circuit is also powered from a regulated voltage source. It is used in an application that requires motor speed adjustment. The speed of the motor at the nominal armature voltage was measured as 1600 rpm. All losses of the motor and voltage source and the effect of armature reaction are neglected. It is accepted that the engine is not saturated.Answer the questions below.a) Draw the armature and excitation circuit of the motor and write the formulas for the equivalent circuit.b) If the load moment of the motor at the nominal armature voltage is 420 Nm, then calculate the armature current and the input power of the motor.c) If the armature voltage of the motor is kept constant at 600 V and the excitation current is decreased until the speed reaches 4000 rpm, then calculate the shaft moment. Show this working area on the M = f (W) axis.
q8
dc machines
THE ARMATURE OF A 4-POLE, 128 SLOTS SHUNT GEN. IS LAP WOUND WITH FOUR CONDUCTORS PER SLOT. THE FLUX PER POLE IS 50 mWb. THE ARMATURE AND SHUNT FIELD RESISTANCES ARE 0.04 OHM AND 50 OHMS RESPECTIVELY. DETERMINE THE SPEED OF THE MACHINE WHEN SUPPLYING 400 A AT A TERMINAL VOLTAGE OF 250V.
The excitation circuit of a free-excited DC motor is a single-phase full- wave uncontrolled rectifier and The armature circuit is fed with a single-phase full-wave controlled rectifier. both rectifiers for AC voltage voltage 220 V(rms) ,is 50 Hz. Excitation O resistance of motor Rf = 100, armature force resistance Ra = 2 0 and motor voltage constant Kv = 0.25 V/(A - r/s). The motor is a fixed 10Nm buy under load a) Draw the driver (converter-motor) power circuit. To rotate with 1200 d/d of the motor, b) Calculate the thyristor assembly. c) When the engine is heavily loaded, with a suitable probability value for supply voltage, armature voltage, armature flow wavelengths draw below. d) Calculate the motor power (total input power). (Note: Va = IaRa + ea, Vf = IfRf, ea = Kv Ifw and Te = Kv Ifla)

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