A shunt-wound DC motor with the field coils and rotor connected in parallel (see

Question

A shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) (Figure 1) operates from a 140 V DC power line. The resistance of the field windings, Rf, is 216 . The resistance of the rotor, Rr, is 4.30 . When the motor is running, the rotor develops an emf E. The motor draws a current of 4.46 A from the line. Friction losses amount to 46.0 W .

A- Compute the field current If.

B- Compute the rotor current Ir.

C- Compute the emf E.

D- Compute the rate of development of thermal energy in the field windings.

E- Compute the rate Pth,rotor of development of thermal energy in the rotor.

F- Compute the power input to the motor Pin.

G- Compute the efficiency of the motor.

Explanation / Answer

The field current is: V/R = 140/216 = .648A

The only other current flowing is the rotor current: Ir = Itot - Ifld
Ir = 4.46 - .648 = 3.812

The emf: The rotor current multiplied by the rotor resistance is the difference between the line voltage and the counter emf:

emf = Vline - Irotor*Rrotor = 140 - 3.812*4.30 =123.6048V

"rate of development of thermal energy" is assumed to be the dissipative losses, i.e. the ohmic power. For the field, it's: Ifld^2*Rfld = .648^2*216 = 90.7W

For the rotor electrical losses: Irot^2*Rrot = 3.812^2*4.30 = 62.48W

The power output is the power input minus the losses:
140*4.46 - 90.7 - 62.48 - 46.0 = 383.22W

The efficiency is Pout/Pin = 317.4/(105*4.6) = 65.7%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.