Problem 108-1    Source:   Problem prepared by the site authors.

Suppose a single-phase, 4-pole induction motor with a nominal power of 1 HP and slip of 3%. It is known that it has rotational losses of 120 W and is connected to a 60 Hz power supply. Find:

a) the synchronous speed of the motor in rpm;

b) the shaft speed of the motor in rpm;

c) the mechanical power of the motor;

d) the induced torque of the motor.

Problem 108-2    Source:   Problem prepared by the site authors.

A 220 V, 60 Hz network supplies 1,100 W to a 4-pole single-phase induction motor. The motor's nameplate states that FP = 0.85 and η = 0.9. Assuming that rotational losses are on the order of 140 W, find:

a) the electric current in the motor;

b) the nominal power of the motor;

c) the mechanical power of the motor.

Problem 108-3    Source:   Problem prepared by the site authors.

Considering the motor data in the previous problem and assuming a slip of s = 0.045, find:

a) the air gap power;

b) the induced torque in the motor;

d) the load torque.

Problem 108-4    Source:   Problem prepared by the site authors.

In a single-phase 6-pole motor with slip, s = 0.05, connected to a 60 Hz, 220 V network, it has an air gap power, P_gap = 800 W. It is also known that R_P = 27.4 Ω and R_R = 2.4 Ω. Determine:

a) the synchronous angular speed of the motor;

b) the electric current I₁ that flows through the motor;

c) the power in the air gap due to the forward field, P_Pgap, and the backward field, P_Rgap;

d) the torque induced in the motor rotor.

Problem 108-5    Source:   Problem prepared by the website authors.

Based on the results of the previous problem:

a) determine the total copper losses of the motor;

b) determine the mechanical power, P_mec, of the motor;

c) assuming FP = 0.8, what is the motor's input power?

d) assuming the motor efficiency is 75%, what is the motor's nominal power?

e) determine the rotational losses of the machine.

Problem 108-6    Source:   Adapted from Example 9-1 - page 595 - CHAPMAN, Stephen J. -    Book: Fundamentals of Electrical Machines - 5th edition - Ed. McGrawHill - 2013.

A split-phase, 1/3 HP, 110 V, 60 Hz motor with 6 poles has the following impedances:

R₁ = 1.52 Ω   X₁ = 2.10 Ω   R₂ = 3.13 Ω   X₂ = 1.56 Ω   and   X_m = 58.2 Ω

The core losses of this motor are 35 W, while the mechanical losses due to friction and ventilation and supplementary losses total 16 W. The motor is operating at nominal voltage and frequency with its starting winding open, and the motor slip is 5%. Find the following quantities under these conditions:

a) the speed in rpm;

b) the stator current;

c) the stator power factor;

d) the machine input power;

e) the machine air gap power;

f) the machine mechanical power.

Problem 108-7    Source:   Adapted from Example 9-1 - page 595 - CHAPMAN, Stephen J. -   Book: Fundamentals of Electrical Machines - 5th edition - Ed. McGrawHill - 2013.

Based on the data and results found in the previous problem "problesol108-6", find the following quantities:

a) the torque induced in the motor;

b) the nominal power of the motor;

c) the torque on the load;

d) the efficiency of the machine;

e) calculate the losses in the copper of the motor and verify if the law of conservation of energy is preserved.

Problem 108-8    Source:   Problem prepared by the site authors.

A split-phase, 127 V, 60 Hz motor with 4 poles has the following impedances:

R₁ = 0.55 Ω   X₁ = 0.83 Ω   R₂ = 1.70 Ω   X₂ = 0.72 Ω   and   X_m = 41.5 Ω

The core losses of this motor are 57 W, while the mechanical losses due to friction and ventilation and supplementary losses total 17.7 W. The motor is operating at nominal voltage and frequency with its starting winding open, and the motor slip is 6.5%. Find the following quantities under these conditions:

a) the stator current;

b) the stator power factor;

c) the machine input power;

d) the machine air gap power;

e) the machine rated power;

f) the machine efficiency.

Problem 108-9    Source:   Example 9.2 - Electrical Machines - Umans, Stephen & Fitzgerald - 7th edition - McGraw Hill Education - 2014

A 1/4 HP, 110 V, 60 Hz, 4-pole capacitor-start motor has the following equivalent circuit parameters:

R₁ = 2.02 Ω  R₂ = 4.12 Ω   X₁ = 2.79 Ω   X₂ = 2.12 Ω   and   X_m = 66.8 Ω  

The motor operates at nominal voltage and frequency with the auxiliary winding open. Knowing that core losses are 24 W, rotational losses are 13 W, and slip s = 0.05, determine:

a) the stator current;

b) the power factor, PF, of the motor;

c) the nominal power;

d) the synchronous angular velocity;

e) the torque on the motor shaft;

f) the efficiency of the machine.

Problem 108-10    Source:   Problem prepared by the website authors.

The design of a split-phase, 127 V motor, with a nominal power of 1 HP, must have an efficiency of 68% and FP = 0.80. Furthermore, it must have the following equivalent circuit parameters:

Z_P = 8.0 + j 5.5 Ω  Z_R = 0.8 + j 0.7 Ω

The motor must operate at nominal voltage and frequency with the auxiliary winding open. It is known that core losses must be 40 W, rotational and supplementary losses 12.4 W. Determine:

a) the stator current;

b) the resistance value R₁ of the main winding;

c) the reactance value X₁ of the main winding.