Balanced Sinusoidal Supply Configuration Biology Essay

Published:

Figure 1 illustrates a balanced sinusoidal supply configuration consisting of three line-neutral voltages having equal magnitudes and a phase difference of 120ï‚°. Figure 2 however illustrates an unbalanced configuration wherein there may exist a difference in either the magnitude of the three voltages or in the phase difference between the voltages[1].

There are several reasons for this unbalance to occur. These are

• Incomplete transposition of transmission lines

• Open delta transformer connections

• Single-phase loads

• Blown fuses on capacitor banks

• Railway traction loads

If a balanced three-phase load is connected to an unbalanced supply (voltage) configuration, it leads to an unbalance in the currents drawn by the load. Practically, it is unfeasible for the supply configurations to be perfectly balanced and hence it is very important to try to minimize the level of unbalance as much as possible so as to reduce its effects on consumer loads[2].

Lady using a tablet
Lady using a tablet

Professional

Essay Writers

Lady Using Tablet

Get your grade
or your money back

using our Essay Writing Service!

Essay Writing Service

1.2 Theory of Symmetrical Components

The Theory of Symmetrical Components mathematically represents a set of balanced vectors in terms of a set of unbalanced vectors or vice-versa. These balanced vectors are classified as:

Positive Sequence Components

Negative Sequence Components

Zero Sequence Components

In a perfectly balanced system both the negative and zero sequence systems do not exist. Figure 3 below shows the different symmetrical components of an unbalanced system of voltages[5].

The physical significance of these components is as follows. When a three-phase induction motor is supplied with a positive sequence set of voltages, it rotates in the counter-clockwise direction. However, the direction of rotation becomes clockwise if the three-phase induction motor is supplied with negative sequence voltages instead. Since the zero sequence components are co-phasal and have no phase difference between them, they are unable to produce a rotating magnetic field to run the motor. Hence, the supply of zero sequence voltages to the motor results in no rotation at all[6].

1.2.1 Analysis

Consider the following equations:

Now Let,

Where,  is the phase rotation operator defined to rotate a phasor vector forward by 120 degrees or radians, i.e.

From the above set of equations, we get

where,

Hence, we can represent an unbalanced set of vectors (VP) in terms of a balanced set of vectors (VS) by using the forward transformation:

Alternately, we can also represent the set of balanced vectors in terms of the unbalanced set of vectors by using the reverse transformation:

where,

Similarly,

Hence,

where,

1.3 Induction Motor

1.3.1 Operating principle

Operation of 3-phase induction motors is based upon the application of Faraday's Law and the Lorentz Force on a conductor.

Let us consider a series of conductors of length L represented by bars A and B in figure 4 whose extremities are shorted. A permanent magnet with its north pole facing the conductors travels at a speed v above the conductors resulting in a magnetic field across the conductors.

Figure 4 Operating principle of an induction motor

The following sequence of events takes place:

(a) According to Faraday's Law

This voltage (E) is induced in each conductor as a result of it being cut by the flux.

(b) Circulating currents are produced due to the induced voltage which travel in a loop through the bars and around the conductors.

(c) According to Lorentz force, the current-carrying conductors experience a mechanical force since they lie in a magnetic field.

(d) The mechanical force acts in the direction of the magnetic field and drags the conductor along with the field[9].

1.3.2 Components

A 3-phase induction motor has two main parts:

(a) A stator whose windings are housed in slots on the stator's internal circumference. The stator is made up of a steel frame that supports a void, cylindrical core of stacked laminations.

(b)A rotor, with rotor slots for the rotor winding, also composed of punched laminations.

There are two-types of rotor windings:

(a)Squirrel-cage windings, which are the most common and produce a squirrel-cage induction motor. The squirrel cage rotor consists of copper bars, somewhat longer than the rotor, which are hard-pressed into the slots. The extremities are welded to copper end rings, so that all the bars are short-circuited.

Lady using a tablet
Lady using a tablet

Comprehensive

Writing Services

Lady Using Tablet

Plagiarism-free
Always on Time

Marked to Standard

Order Now

In small motors, the bars and end-rings are die- cast in aluminium to form an integral block.

(b)Conventional 3-phase windings prepared of insulated wire for special applications, which produce a wound-rotor induction motor.

A wound rotor has a 3-phase winding, similar to the stator winding.

The rotor winding terminals are linked to three slip rings which rotate with the rotor. The slip rings/brushes allow peripheral resistors to be attached in series with the winding.

The external resistors are primarily used through start-up - under standard running circumstances the windings are short- circuited superficially. Figure 5 below shows the various components of an induction motor[9].

Figure 5 Components of an induction motor

1.3.3 Operation

(a)Locked rotor: When the rotor is at a standstill, the field rotates at a rate of recurrence (comparative to the rotor) equivalent to the supply rate of recurrence. This induces a huge voltage - hence huge currents surge inside the rotor, producing a strapping torque.

(b)Acceleration: When unrestricted, the rotor accelerates quickly. As speed increases, the relative rate of recurrence of the magnetic field decreases. As a result, the induced voltages and currents decrease quickly as the motor accelerates.

(c)Synchronous speed: The relative rate of recurrence of the rotating field is nil, so the induced currents and voltages are also nil. Therefore, the torque is nil too. It concludes, that induction motors are incapable of reaching synchronous velocity due to losses such as friction.

(d)Motor under load: The motor speed decreases until the relative rate of recurrence is big enough to produce adequate torque to stabilize the load torque[7].

1.3.4 Slip

The difference between the synchronous speed and rotor speed can be expressed as a percentage of synchronous speed, known as the slip.

where,

s = slip,

Ns = synchronous speed (rpm),

N = rotor speed (rpm)

• At no-load, the slip is nearly zero (<0.1%).

• At full load, the slip for large motors rarely exceeds 0.5%. For small

motors at full load, it rarely exceeds 5%.

• The slip is 100% for locked rotor.

The rate of recurrence induced in the rotor depends on the slip[9]:

where,

fR = rate of recurrence of voltage and current in the rotor

f = rate of recurrence of the supply and stator field

s = slip

1.3.5 Power Flow Equations

Efficiency by definition, is the ratio of output / input power:

Rotor copper losses:

Mechanical power:

Motor torque[7]:

Figure 6 Power flow in an induction motor

1.3.6 Effects of Voltage Unbalance

The greatest result of voltage unbalance is on three-phase induction motors. Three phase induction motors are one of the most widespread loads on the system and are observed in large numbers particularly in industrial environments. When a three-phase induction motor is supplied by an unbalanced system the resultant line currents show an extent of unbalance that is numerous times the voltage unbalance. This can be explained with mention to the two contra-rotating fields established when the motor is subjected to voltage unbalance.

In relation to the positive sequence set of voltages if the motor slip is:

where,

Ns - synchronous speed

Nr - the rotor speed

Then the slip corresponding to the negative sequence set of voltages would be

Slip s2 can be expressed in terms of slip s1 and hence,

As the positive sequence slip s1 is usually very small (close to zero) the negative sequence slip s2 would be very large (close to 2). From the basic theory of induction motors the impedance of an induction motor is very dependent on the slip where at high slip (eg. at start or under locked rotor conditions) it is small and conversely at low slip it is very large. Hence it can be approximately stated that the ratio of the positive sequence impedance to negative sequence impedance is given by:

As the positive sequence current is given by

and the negative sequence current is given by

it can be quickly shown that:

As an example, a motor with a locked rotor current that is 6 times the running current would give rise to a very significant 30% unbalance in the motor line current if the voltage unbalance is 5%[10].

Lady using a tablet
Lady using a tablet

This Essay is

a Student's Work

Lady Using Tablet

This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

Examples of our work

chapter 2 Problem Description

To Investigate the Theory of Symmetrical Components applied to a three-phase induction motor working under unbalanced condition.

2.1 LAB EXPERIMENT

The performance of a 3 phase induction motor was observed under both balanced and unbalanced operating conditions. The following steps were undertaken to achieve the desired result:

No load test under balanced condition

No load test under unbalanced condition

Blocked rotor test under balanced condition

Measurement of stator resistance

Load test under balanced condition

Load test under unbalanced condition

Calculation of positive & negative sequence currents

Determination of machine parameters

Calculation of positive & negative sequence torque

2.2 observations

(a) No Load Test - Balanced Operation

Table 1 No load test (balanced) observations

Voltage

V0 = 415 V

Current

I0 = 2.41 A

Power

P0 = 0.27 kW

(b) No Load Test - Unbalanced Operation

Table 2 No load test (unbalanced) observations

Voltage

V1 = 415 V

Current

I1 = 3.66 A

Power

P1 = 0.32 kW

(c) Blocked Rotor Test

Table 3 Blocked rotor test observations

Voltage

VBR = 80 V

Current

IBR = 2.8 A

Power

PBR= 0.146 kW

(d) Winding Resistance Measurement

Table 4 Winding resistance measurement observations

V (V)

2

4

6

8

I (A)

0.19

0.34

0.49

0.64

Average Value of R1 = 11.93ï-

R1 (eq.) = 7.95ï-

(e) Load Test - Balanced Operation

Table 5 Load test (balanced) observations

Voltage

V2= 415 V

Current

I2 = 3.36 A

Power

P2= 1.51 kW

Speed

n = 1464 rpm

(f) Load Test - Unbalanced Operation

Table 6 Load test (unbalanced) observations

Voltage

V3= 415 V

Current

I3 = 5.45A

Power

P3= 1.66 kW

Speed

n1 = 1444 rpm

(g) Machine Nameplate Ratings

Table 7 Machine nameplate ratings

3 Phase Induction Motor

DC Generator

P = 2.2 kW

P = 2 kW

V = 415 V

V = 220 V

ns = 1500 rpm

n = 1500 rpm

I = 4.8 A

I = 9A

2.3 calculations

2.3.1 machine parameters[4]

2.3.2 current

(a) Positive Sequence Current

(b) Negative Sequence Current

2.3.3 torque

Positive Sequence Torque[3]

Figure 7 Positive sequence equivalent circuit

(b) Negative Sequence Torque[3]Figure 8 Negative sequence equivalent circuit

CHAPTER 3 results

Percentage reduction in torque due to negative sequence component

Percentage reduction in speed due to negative sequence component

CHAPTER 4 Conclusion

When one of the phases of an induction motor is suddenly blown (single-phasing), we are faced with what is basically a current unbalance problem.

As a result of this current unbalance, a negative sequence current is developed. This current produced a negative sequence torque, which acts as a load torque in the direction opposite to the normal positive sequence electro magnetic torque.

As a result, the net torque decreases, resulting in a drop in speed, i.e. the negative sequence torque acts as a virtual load. Hence, load increases and speed must drop even if the machine is running on free shaft.

The effects of negative sequence component on performance of induction motor are:

• The motor takes longer time to run up.

• It increases the thermal stress in the motor which leads to loss in life.

• The net torque is reduced and if full load is still demanded, then the motor is forced to operate at a higher slip, thus increasing the rotor losses and heat dissipation.

• The reduction in the peak torque reduces the ability of motor to ride through dips and sags, thus affecting the stability of the entire system.

Premature failure can only be prevented by derating of the machine to

allow it to operate within the thermal limitations[8].