Solved Example of DC Shunt Motor
A 240 V DC shunt motor having an armature resistance of 0.3 Ω carries an armature current of 40 Amps and runs at 600 RPM. If the flux is decrease by 15 % by the field rheostat. Find the speed of the motor expecting the load torque remains the same.
Case 1:
N1 = 600 rpm, Ia1 = 40 Amp
therefore, Eb1 = V – Ia1Ra
= 240 – (40 x 0.4)
= 224 Volts
Case 2:
12% reduced in the flux
(Phi2)Φ2 = 0.85 2
(Phi1) Φ1 / (Phi2)Φ2 = 0.85
since the load torque is given to be constant,
(Phi1) Φ1 Ia1 = (Phi2) Φ 2Ia2
Ia2 = (Phi1) Φ 1 / (Phi2) Φ 2 x Ia1
=1 / 0.85 x 40
= 47.05 Amps
therefore, Eb2 = V-Ia2Ra
=240- (47.05 x 0.4)
=221.18 Volts
therefore, N2 / N1 = Eb2 / Eb1 = (Phi1) Φ 1/ (Phi2)Φ 2
N2 = 221.18 / 228 x 1 / 0.85 x 600
N2 = 696.99 Rpm
DC Shunt Motor
In the world of electric motors, the DC shunt engine remains a demonstration of dependability, flexibility, and perseverance. Whether you are in a modern setting, driving a conveyor belt, controlling the speed of a rolling mill, or changing the volume on an old stereo system, odds are good that you’ve experienced a DC shunt motor at work. In electrical motors, series circuits, and equal circuits are ordinarily known as series and shunt. Hence, in DC motors, the associations of the field windings, as well as the armature, should be possibly equal, which is known as DC shunt motors. The principal contrast between DC series motors as well as DC shunt motors primarily incorporates the construction, operation, and speed characteristics. This motor gives highlights like simple turning around control and speed guidelines, and it is low to begin force. Hence, this motor can be utilized for belt-driven applications inside autos as well as modern applications.
Table of Content
- DC Shunt Motor
- Construction and Working Principles
- Back EMF
- Speed Control
- Brake Test
- Characteristics
- Advantages
- Disadvantages
- Applications
- Example Problem