ECE320, Spring 2010, Exam 4a 12thApril, 2010

Name
PID Number
Problem / MaxPoints / Points
1 / 40
2 / 30
3 / 30
Total / 100

Instructions:

  • Closed book, but you are allowed to have one-page single-sided handwritten notes.
  • Write your solutions and steps neatly and orderly. No credits for random side writings.
  • Box your final answers/values when needed.

Problem 1.

a)Draw the equivalent circuit of an IM, label each parameter and its meaning. Indicate which elements (or parameters) represent the copper losses and mechanical output, respectively. (20 pts)

SOLUTION:

Rs = Stator winding resistance

Xls= Stator leakage inductance

Xlr= Rotor leakage inductance

Xm = Magnetizing inductance

Rr = Rotor winding resistance

Ir2*Rr((1-s)/s) = mechanical output power

Is2*Rs = Stator winding loss

Ir2*Rr = Rotor winding loss

Is2*Rs + Ir2*Rr = Copper Loss.

Ir2*Rr((1-s)/s) = mechanical output power.

b)Fill out the following blanksto complete the sentences. (20 pts)

For an IM, the stator current consists of two components: magnetizing current and torque current. The magnetizing current is determined by the stator Voltage and

stator frequency .

The magnetizing current produces a Rotating Magnetic Field . The torque current component is related to the Rotor current.

When the rotor is rotating at its synchronous speed, the rotor current becomes zero , the torque produced becomes zero , and the stator currentbecomes the

magnetizing current component only. When the rotor is rotating slightly slower than the synchronous speed with a slip, the rotor current starts to flow and torque is produced.

The torque produced at a speed near the synchronous speed is proportional to the slip speed (or slip frequency) .

An IM’s parameters can be determined by Locked Rotor and No Load_ tests,which are equivalent to transformer’s Short Circuit and__Open Circuit tests, respectively.

Problem 2a)A 3-phase IM is fed by a 60 Hz power and operating at a slip of s = 0.05, the rotor current (referred to the stator side), I2 is 12 A, and rotor resistance is 2 , determine

i)the mechanical output power (4 pts), rotor speed (4 pts), and shaft torque (2 pts), given that the IM is a 2-pole machine, and

ii)the mechanical output power (2 pts), rotor speed (2 pts), and shaft torque (2 pts), given that the IM is a 4-polemachine.

SOLUTION:

i)

ii)

b)A 3-phase 2-pole 60-Hz IM is rated at 240 V, 30A and 10 kW and its stator connection is delta and its rate rotor speed is 3,300 rpm. What are the rated voltage, rated current, rated speed, and rated power after reconfigured to Y connection? (14 pts)

SOLUTION:

Rated voltage :

Rated current :

Rate speed: 3,300rpm (same)

Rated power : 10 KW (same)
Problem 3. (30 points)SOLUTION:

a) The circuit below shows a single-phase inverter circuit using four switches to convert a dc voltage into ac to power an inductive load. Draw a control sequence for each of the four switches by using logic “0” (for off) and “1” for on, respectively, to complete the waveform figure. (20 pts)

b) In the given circuit below, assume the switch is used to dim the light by turning on and off at 1 kHz. When the switch is turned on for 100% of each switching period (i.e., turn on the switch all the time), the light bulb consumes 60 W. What is the duty cycle of the switch when the light is dimmed to 20 W? (The duty cycle is defined as Ton/Tsw, the ratio of turn-on interval, Ton versus the switching period, Tsw) (10 pts)

SOLUTION:

When S is on, PL = 60 W, and when off, PL = 0 W.

Therefore, the average power, PL = 20 W = D·(60 W) and D = 20/60 = 1/3.